One independent variable gives you an ODE. Two or more gives you a PDE — and a subject where the geometry of the equation decides everything: which method works, what data you're allowed to impose, and whether information travels at finite speed or instantly. This is the whole standard course, from classification to Bessel functions and finite differences.
21 modules21 interactive labs189 practice problems3 checkpoint quizzesStaged hints · fully worked solutions
Heat equation \(u_t = k\,u_{xx}\)
Parabolic · diffusion · smooths instantly
Wave equation \(u_{tt} = c^2u_{xx}\)
Hyperbolic · finite speed · keeps corners
Laplace equation \(\nabla^2u = 0\)
Elliptic · equilibrium · no time at all
Part I · Ground rules
Foundations
Part II · First order
Characteristics
Part III · The core method
Separation & Fourier
Part IV · Going further
Advanced tools
↦How to use this
Play with the diagram before you read the algebra. Every module has an orange Interactive panel — watch heat smear out, a wave hold its corners, characteristics collide into a shock, or a Fourier series ring at a jump. The formulas make far more sense afterwards.
Attempt every problem on paper first. Hints come one at a time, and every solution is worked step by step with the reasoning behind each move.
Tick "Mark complete" at the bottom of each module. Progress is saved in this browser.
Learn the three model equations cold. Nearly every exam problem is heat, wave, or Laplace wearing a disguise — different coefficients, different domain, different boundary conditions, same three behaviors.
What you need before starting
Comfort with ODEs — separation of variables, second-order constant-coefficient equations, and the characteristic-equation trick — because separation of variables turns every PDE into a pair of ODEs. If that's rusty, work through the ODE Launchpad first; for the calculus underneath both, the Calculus Toolkit covers partial derivatives, Fourier-relevant integrals, and series.
Part I · Module 1 Foundations
What is a PDE?
An ODE has one independent variable, so its solution is a curve. A PDE has two or more, so its solution is a surface — and instead of arbitrary constants you get arbitrary functions. That single change is the source of everything hard and everything interesting in this subject.
①The definition, and the vocabulary
A partial differential equation is an equation relating an unknown function of several variables to its partial derivatives:
General second-order PDE in two variables
\[ F\big(x, y, u, u_x, u_y, u_{xx}, u_{xy}, u_{yy}\big) = 0 \]
Four labels do most of the classifying work, and you should be able to attach all four at a glance:
Property
Meaning
Example
Order
Highest derivative present
\(u_t = u_{xx}\) is second order
Linear
\(u\) and its derivatives appear to the first power, never multiplied together, never inside a nonlinear function. Coefficients may depend on \(x,y\) freely.
\(u_t = x^2u_{xx}\) is linear; \(u_t = u\,u_x\) is not
Homogeneous
Every term contains \(u\) or one of its derivatives — so \(u\equiv0\) is a solution
\(u_x = \dfrac{\partial u}{\partial x}\), \(u_{xx} = \dfrac{\partial^2u}{\partial x^2}\), \(u_{xy} = \dfrac{\partial^2 u}{\partial y\,\partial x}\) (\(x\) first, then \(y\)). The Laplacian is \(\nabla^2u = \Delta u = u_{xx}+u_{yy}\) in two dimensions. Subscript notation is universal in PDEs because Leibniz notation becomes unreadable fast.
InteractiveA solution is a surfacedrag the point; read the two partials
point (x, t) —u —u_x —u_t —equation check —
Left is the solution surface over the \(x\)–\(t\) plane (warm = high). Right are the two slices you actually differentiate: the top holds \(t\) fixed and gives \(u_x\); the bottom holds \(x\) fixed and gives \(u_t\). A PDE is a rule that ties those two slopes together at every point of the plane at once — that is why one condition at one point is nowhere near enough to pin down a solution.
Solve the ODE \(y'=0\) and you get \(y=C\), one arbitrary constant. Solve the PDE \(u_x = 0\) and you get
The general solution of a PDE carries whole functions
\[ u_x = 0 \quad\Longrightarrow\quad u(x,y) = f(y) \quad\text{for any function } f \]
Nothing about \(x\)-differentiation constrains how \(u\) varies in \(y\), so an entire function's worth of freedom survives. This is why side conditions in PDEs are functions, not numbers: you don't prescribe \(u(0)=3\), you prescribe \(u(x,0)=\varphi(x)\) along a whole line.
Order: the highest derivatives are \(u_{tt}\) and \(u_{xx}\), both second — second order.
STEP 2
Linear? \(u\) appears only through \(u_{tt}\) and \(u_{xx}\), each to the first power and not multiplied together — linear. The \(\cos t\) is a coefficient-free forcing term, which affects homogeneity, not linearity.
STEP 3
Homogeneous? No: setting \(u\equiv0\) leaves \(0 = \cos t\), which is false. It is a nonhomogeneous (forced) equation.
STEP 4
Now solve \(u_{xy}=0\). Read it as \(\dfrac{\partial}{\partial y}\big(u_x\big) = 0\): the quantity \(u_x\) does not depend on \(y\), so \(u_x = g(x)\) for some function \(g\).
STEP 5
Integrate in \(x\), holding \(y\) fixed. The "constant" of integration may be any function of \(y\):
So the general solution is \(u = F(x) + G(y)\) — two arbitrary functions for a second-order PDE, mirroring the two arbitrary constants of a second-order ODE.
"Linear" is about u, not about x
\(u_t = x^2u_{xx}\) is linear — the \(x^2\) is just a coefficient. \(u_t = (u_x)^2\) is nonlinear. Ask only: if I doubled \(u\), would every term double? If yes, it's linear.
Hint 1Order = highest derivative. Linear = \(u\) and derivatives to the first power only, never multiplied by each other.
Hint 2Homogeneous ⟺ \(u\equiv0\) satisfies it ⟺ every term contains \(u\) or a derivative of \(u\).
Hint 3Coefficients depending on \(x\) or \(y\) never break linearity.
Solution
(a) u_t = 3u_xx
Second order (from \(u_{xx}\)), linear, homogeneous, constant-coefficient. This is the heat equation.
(b) u_t + u·u_x = 0
First order, nonlinear — the product \(u\,u_x\) multiplies the unknown by its own derivative. (Homogeneity is a concept reserved for linear equations, so it doesn't apply.) This is the inviscid Burgers equation, module C3.
(c) u_tt = c²u_xx + e^(−x)
Second order, linear, nonhomogeneous: the forcing \(e^{-x}\) contains no \(u\), so \(u\equiv0\) fails.
(d) u_x² + u_y² = 1
First order, nonlinear — derivatives are squared. (This is the eikonal equation of geometric optics.)
(e) xu_xx + u_yy = 0
Second order, linear (the \(x\) is only a coefficient), homogeneous, but variable-coefficient. It is the Tricomi equation, and module F3 will show that its type actually changes sign with \(x\).
THE TEST TO INTERNALIZE
Replace \(u\) by \(2u\). If every term exactly doubles, the equation is linear; if some term quadruples (like \(u u_x\) or \(u_x^2\)), it is not.
2
Find the general solution of \(u_x = 0\) and of \(u_{xx} = 0\), treating \(u = u(x,y)\).
Hint 1Integrating with respect to \(x\) holds \(y\) fixed — so the "constant" may depend on \(y\).
Hint 2For the second, integrate twice and expect two arbitrary functions.
Solution
STEP 1 · first equation
\(u_x=0\) says \(u\) does not change as \(x\) varies, for each fixed \(y\). So
\[ u(x,y) = f(y) \]
with \(f\) completely arbitrary. Check: \(\frac{\partial}{\partial x}f(y) = 0\) ✓
STEP 2 · second equation, first integration
\(u_{xx} = 0\) means \(\frac{\partial}{\partial x}(u_x) = 0\), so by step 1 applied to \(u_x\):
\[ u_x = f(y) \]
STEP 3 · integrate again in x
Now \(f(y)\) is a constant as far as \(x\) is concerned:
\[ u(x,y) = f(y)\,x + g(y) \]
STEP 4 · verify
\(u_x = f(y)\), \(u_{xx} = 0\) ✓ for any \(f, g\).
COUNT THE FREEDOM
First order ⟹ one arbitrary function; second order ⟹ two. In every \(y\)-slice the solution is a straight line in \(x\), but the slope and intercept may vary arbitrarily from slice to slice.
3
Verify that \(u(x,t) = e^{-4t}\sin 2x\) solves the heat equation \(u_t = u_{xx}\), and that \(u(x,t) = \sin(x-3t)\) solves the transport equation \(u_t + 3u_x = 0\).
Hint 1Compute each partial separately, holding the other variable fixed.
Hint 2Every \(x\)-derivative of \(\sin 2x\) pulls out a factor 2; every \(t\)-derivative of \(e^{-4t}\) pulls out \(-4\).
Solution
STEP 1 · heat, t-derivative
Hold \(x\) fixed; \(\sin 2x\) is a constant multiplier:
Note the decay rate \(4\) is exactly the square of the spatial frequency \(2\). That relationship — \(\lambda = k^2\) — is the engine of the whole heat-equation chapter.
The heat solution decays in place; the transport solution slides right at speed 3 without changing shape. Same verification technique, completely different physics — which is exactly what classification (module F3) predicts.
4
For which constants \(k\) does \(u(x,t) = e^{kt}\cos 3x\) solve \(u_t = 2u_{xx}\)?
Hint 1Substitute and factor out the common \(e^{kt}\cos 3x\).
Hint 2\(u_{xx}\) brings down \((3)^2\) with a minus sign.
Hint 3\(e^{kt}\cos3x\) is not identically zero, so you may divide by it.
STEP 2 · substitute into the PDE
\[ ke^{kt}\cos3x = 2\left(-9e^{kt}\cos3x\right) \]
STEP 3 · divide by the common factor
\(e^{kt}\cos3x \ne 0\) except on isolated lines, so the equation forces
\[ k = -18 \]
STEP 4 · state the solution
\[ u(x,t) = e^{-18t}\cos 3x \]
THE PATTERN
In general \(e^{kt}\cos(nx)\) solves \(u_t = \alpha u_{xx}\) precisely when \(k = -\alpha n^2\). Higher spatial frequency ⟹ dramatically faster decay: doubling \(n\) makes the mode die four times as fast. That is why diffusion erases fine detail first and leaves the smooth part behind.
5
Show that \(u = F(x+ct) + G(x-ct)\) satisfies the wave equation \(u_{tt} = c^2u_{xx}\) for any twice-differentiable \(F\) and \(G\).
Hint 1Use the chain rule on each piece: the inner derivative of \(x+ct\) with respect to \(t\) is \(c\).
Hint 2Each \(t\)-derivative brings a factor \(\pm c\); two of them bring \(c^2\) in both cases.
Solution
STEP 1 · t-derivatives
Write \(\xi = x+ct\), \(\eta = x-ct\). Then \(\xi_t = c\) and \(\eta_t = -c\), so
No assumption about \(F\) or \(G\) was used beyond being twice differentiable.
STEP 4 · read the physics
\(G(x-ct)\) is a fixed shape sliding right at speed \(c\); \(F(x+ct)\) slides left. Every solution of the 1-D wave equation is a superposition of two rigid travelling shapes — that is d'Alembert's theorem, proved in full in module S6.
TWO ARBITRARY FUNCTIONS
A second-order PDE, and exactly two arbitrary functions appear — matching the count from problem 2 and explaining why the wave equation needs two initial conditions, \(u(x,0)\) and \(u_t(x,0)\).
6
Find all solutions of \(u_t = 0\) and of \(u_{xt} = 0\), then explain why \(u(x,0)=\sin x\) determines the first completely but the second not at all.
Hint 1\(u_t=0\) says \(u\) is unchanging in time.
Hint 2For \(u_{xt}=0\), integrate once in \(t\), then once in \(x\), letting arbitrary functions appear at each stage.
Solution
STEP 1 · solve u_t = 0
\[ u(x,t) = f(x) \]
A steady profile, frozen for all time.
STEP 2 · apply the data
\(u(x,0) = f(x) = \sin x\), so \(u(x,t) = \sin x\) for every \(t\). Completely determined.
STEP 3 · solve u_xt = 0
Read it as \(\frac{\partial}{\partial t}(u_x)=0\), so \(u_x = f(x)\); integrating in \(x\),
\[ u(x,t) = F(x) + G(t) \]
STEP 4 · apply the same data
\[ u(x,0) = F(x) + G(0) = \sin x \]
This pins down \(F\) only up to the constant \(G(0)\), and says nothing whatsoever about \(G(t)\) for \(t>0\). Any \(G\) with the right value at 0 works: \(u = \sin x + t\), \(u = \sin x + t^{17}\), \(u = \sin x + \sin t\)…
STEP 5 · the lesson
Prescribing data on one line does not automatically determine a solution. Which data on which curve makes a problem well posed depends on the equation's structure — the question module F4 answers, and the reason characteristics (Part II) matter.
7
Find the general solution of (a) \(u_y = 3\) and (b) \(u_{xy}=1\), for \(u=u(x,y)\).
Hint 1Integrating in \(y\) holds \(x\) fixed, so the constant may be any function of \(x\).
Hint 2For (b), integrate in \(y\) first to get \(u_x\), then in \(x\).
Solution
STEP 1 · part (a)
Antidifferentiate in \(y\), treating \(x\) as a parameter:
\[ u(x,y) = 3y + f(x) \]
STEP 2 · check it
\(u_y = 3\) for any \(f\) ✓ — one arbitrary function, matching the first-order equation.
STEP 3 · part (b), first integration
Read \(u_{xy}=1\) as \(\frac{\partial}{\partial y}(u_x) = 1\), so
\[ u_x = y + g(x) \]
STEP 4 · second integration
Now integrate in \(x\), with \(y\) fixed:
\[ u(x,y) = xy + G(x) + H(y), \qquad G' = g \]
STEP 5 · count the freedom
Two arbitrary functions for a second-order equation ✓ Check: \(u_x = y + G'(x)\), then \(u_{xy}=1\) ✓ — the \(G\) and \(H\) pieces are exactly the general solution of \(u_{xy}=0\) from problem 2, added to the particular solution \(xy\).
8
Verify that \(u = \ln(x^2+y^2)\) satisfies Laplace's equation \(u_{xx}+u_{yy}=0\) away from the origin.
Hint 1\(u_x = \frac{2x}{x^2+y^2}\) by the chain rule.
Hint 2Differentiate again with the quotient rule, then add the two second derivatives and look for cancellation.
At \(r=0\) the function itself is undefined (\(\ln 0\)). This is the two-dimensional point-source potential — the field of a line charge, or the flow around a point vortex. Its singularity is exactly the "hole" that module S8's \(\ln r\) solution fills on an annulus.
9
Show that if \(u(x,t)\) solves \(u_t = u_{xx}\), then so does \(v(x,t) = u(\lambda x,\ \lambda^2 t)\) for any constant \(\lambda\). What does this say about how heat problems scale?
Hint 1Use the chain rule: each \(x\)-derivative brings a factor \(\lambda\), each \(t\)-derivative a factor \(\lambda^2\).
Hint 2Compare \(v_t\) with \(v_{xx}\) and see what cancels.
Solution
STEP 1 · differentiate in t
Writing \(X=\lambda x\), \(T=\lambda^2t\):
\[ v_t = \lambda^2\,u_T(X,T) \]
STEP 2 · differentiate twice in x
\[ v_x = \lambda\,u_X, \qquad v_{xx} = \lambda^2u_{XX} \]
STEP 3 · compare
Since \(u\) solves the heat equation, \(u_T = u_{XX}\), so
Space and time are not on equal footing: stretching \(x\) by \(\lambda\) requires stretching \(t\) by \(\lambda^2\). Diffusion has the scaling \(x\sim\sqrt{t}\).
STEP 5 · consequences
A bar twice as long takes four times as long to cool (module S3), the heat kernel spreads like \(\sqrt{4kt}\) (module A3), and the explicit numerical scheme needs \(\Delta t\sim\Delta x^2\) (module A5). All three are the same statement. Contrast the wave equation, invariant under \(x\to\lambda x, t\to\lambda t\) — its natural scaling is \(x\sim ct\).
Part I · Module 2 Foundations
Where PDEs come from
The three equations that dominate this course aren't arbitrary. Each falls out of one bookkeeping principle — what accumulates here equals what flows in minus what flows out — combined with one physical guess about the flux. Derive them once and you'll never confuse their behavior again.
①The conservation law
Let \(u(x,t)\) be a density (temperature, concentration, mass per unit length) and \(\phi(x,t)\) the flux — the amount flowing rightward past position \(x\) per unit time. Track the material in a slab \([x, x+\Delta x]\):
Balance in a slab
\[ \frac{d}{dt}\int_x^{x+\Delta x} u\,ds = \underbrace{\phi(x,t)}_{\text{in at the left}} - \underbrace{\phi(x+\Delta x,t)}_{\text{out at the right}} \]
Divide by \(\Delta x\) and let \(\Delta x\to0\). The left side becomes \(u_t\), the right side becomes a derivative by definition:
Conservation law — differential form
\[ \boxed{\,u_t + \phi_x = 0\,} \]
Everything now depends on the constitutive law: what is the flux made of?
InteractiveFlux in, flux outslide the slab; watch the balance
flux in —flux out —net gain rate —slab is —
Top: the density \(u(x)\), with your slab shaded and the two flux arrows drawn to scale. Bottom: the flux \(\phi(x)\) itself. The slab gains material exactly when the inflow arrow beats the outflow arrow — i.e. when \(\phi\) is decreasing across the slab, which is the statement \(u_t = -\phi_x\). With Fick's law, notice the slab fills wherever the density curve is concave up.
②Three constitutive laws, three equations
Flux assumption
Substituting into \(u_t+\phi_x=0\)
Equation
\(\phi = c\,u\) — material is simply carried along at speed \(c\)
\(u_t + cu_x = 0\)
Transport (first order, hyperbolic)
\(\phi = -k\,u_x\) — flow runs downhill, proportional to steepness (Fourier's law of heat, Fick's law of diffusion)
\(u_t = k\,u_{xx}\)
Heat / diffusion (parabolic)
\(\phi = -k\,u_x\) with nothing accumulating, \(u_t=0\)
\(u_{xx} = 0\), or \(\nabla^2u=0\) in higher dimensions
Laplace (elliptic — the steady state)
The wave equation comes from a different balance — Newton's second law on a stretched string:
The tension pulls each bit of string toward the average of its neighbors, and the restoring force is proportional to curvature. Same right-hand side as the heat equation — but acting on acceleration rather than velocity, which is the entire difference between ringing and smearing.
Why u_xx means "compared to my neighbors"
A discrete version makes it obvious: \(u_{xx} \approx \dfrac{u(x+h) - 2u(x) + u(x-h)}{h^2} = \dfrac{2}{h^2}\left[\dfrac{u(x+h)+u(x-h)}{2} - u(x)\right]\). The Laplacian measures how much a point falls below the average of its neighbors. Heat says "move toward that average"; the wave equation says "accelerate toward it"; Laplace says "already equal to it."
Worked example · adding a source and a sink
A rod loses heat to surrounding air at temperature \(0\) at a rate proportional to its own temperature, and is also heated internally at rate \(Q(x)\). Derive the governing PDE.
STEP 1
Start from the balance with extra terms. Anything created inside the slab adds on the right:
Insert Fourier's law \(\phi = -ku_x\), so \(\phi_x = -ku_{xx}\).
STEP 3
Rearrange:
\[ u_t = k\,u_{xx} - h\,u + Q(x) \]
STEP 4
Read off the structure: still second order and linear, but now nonhomogeneous because of \(Q\). The \(-hu\) term is a uniform decay — substituting \(u = e^{-ht}v\) removes it entirely and returns you to a standard heat equation for \(v\), a trick you'll use in module A2.
③Practice
1
Derive the PDE for a substance that both diffuses (\(\phi=-ku_x\)) and is carried by a current of speed \(c\) (\(\phi = cu\)).
Hint 1Fluxes add: the total flux is the sum of the two mechanisms.
Hint 2\(\phi = cu - ku_x\). Now differentiate and substitute into \(u_t + \phi_x = 0\).
STEP 3 · substitute into the conservation law
\[ u_t + cu_x - ku_{xx} = 0 \quad\Longrightarrow\quad \boxed{u_t + cu_x = ku_{xx}} \]
STEP 4 · interpret
This is the advection–diffusion equation. The blob drifts right at speed \(c\) (from \(cu_x\)) while simultaneously spreading and flattening (from \(ku_{xx}\)) — pollution in a river, a dye pulse in a pipe, heat in a moving fluid.
A USEFUL TRICK
In the moving frame \(\xi = x - ct\) the advection term vanishes and you are left with the pure heat equation \(u_t = ku_{\xi\xi}\). Riding along with the current turns a two-mechanism problem into a one-mechanism one.
2
Show that the total amount \(M(t) = \int_{-\infty}^{\infty}u(x,t)\,dx\) is constant in time for \(u_t = ku_{xx}\), assuming \(u\) and \(u_x\) vanish as \(|x|\to\infty\).
Hint 1Differentiate under the integral sign: \(M'(t) = \int u_t\,dx\).
Hint 2Replace \(u_t\) by \(ku_{xx}\) using the PDE — now the integrand is an exact derivative.
Hint 3\(\int_{-\infty}^{\infty}u_{xx}\,dx = \left[u_x\right]_{-\infty}^{\infty}\).
Solution
STEP 1 · differentiate the total
\[ M'(t) = \frac{d}{dt}\int_{-\infty}^{\infty}u\,dx = \int_{-\infty}^{\infty}u_t\,dx \]
STEP 2 · use the PDE
\[ = \int_{-\infty}^{\infty}k\,u_{xx}\,dx \]
STEP 3 · integrate exactly
\(u_{xx}\) is the derivative of \(u_x\), so the Fundamental Theorem applies:
\(M'(t)=0\), so \(M\) is constant: heat spreads out but is never created or destroyed.
WHY IT WAS INEVITABLE
The heat equation was derived from a conservation law, so conservation had to survive. This also explains the shape of the heat kernel in module A3: it flattens and widens while its area stays pinned at 1. On a finite bar with insulated ends the same computation gives \(M' = k[u_x]_0^L = 0\) — the insulated case of module S4.
3
A chemical decays at rate proportional to its concentration while diffusing. Write the PDE, then find the substitution that reduces it to the plain heat equation.
Hint 1Decay adds a sink term \(-\alpha u\) on the right of the balance.
Hint 2Try \(u = e^{-\alpha t}v(x,t)\) and see what the PDE becomes for \(v\).
Hint 3Use the product rule for \(u_t\); the \(-\alpha\) terms should cancel.
Solution
STEP 1 · write the model
\[ u_t = k\,u_{xx} - \alpha u, \qquad \alpha > 0 \]
The \(-\alpha e^{-\alpha t}v\) terms cancel on both sides, and dividing by \(e^{-\alpha t} \ne 0\):
\[ v_t = k\,v_{xx} \]
STEP 5 · read the result
Solve the ordinary heat equation for \(v\), then multiply by \(e^{-\alpha t}\). The decay simply scales the whole profile down uniformly without changing its shape — it does not interact with diffusion at all.
4
Use the discrete approximation \(u_{xx}\approx\frac{u(x+h)-2u(x)+u(x-h)}{h^2}\) to explain in words why a hot spot cools and a cold spot warms under \(u_t = ku_{xx}\).
Hint 1Rewrite the numerator as \(2\left[\text{average of the neighbors} - u(x)\right]\).
Hint 2What is the sign of that bracket at a local maximum?
So \(u_{xx} \approx \frac{2}{h^2}\big(\text{neighbor average} - \text{me}\big)\).
STEP 2 · at a hot spot
A local maximum sits above both neighbors, so the bracket is negative, giving \(u_{xx}<0\) and therefore \(u_t<0\): the peak cools.
STEP 3 · at a cold spot
A local minimum is below its neighbors, the bracket is positive, \(u_t>0\): the dip warms.
STEP 4 · state the principle
The heat equation says: every point chases the average of its neighbors, at a rate proportional to how far behind it is. Extremes are destroyed, which is why solutions become smoother and flatter as time passes.
TWO CONSEQUENCES YOU'LL MEET LATER
(1) The maximum principle: the hottest the rod ever gets is at \(t=0\) or on the boundary — never spontaneously in the middle later. (2) The same stencil, used forwards in time, is precisely the numerical scheme of module A5.
5
Explain why the wave equation preserves a corner in the initial shape but the heat equation destroys it instantly.
Hint 1Look at what each equation does with the same right-hand side \(u_{xx}\).
Hint 2For the wave equation, recall from F1 problem 5 that solutions are rigid travelling shapes.
Solution
STEP 1 · compare the two statements
\[ \text{heat: } u_t = ku_{xx} \qquad\qquad \text{wave: } u_{tt} = c^2u_{xx} \]
Same measurement of curvature; different thing being driven — velocity versus acceleration.
STEP 2 · heat destroys the corner
At a corner, curvature is enormous (infinite in the limit), so \(u_t\) is enormous there and the peak is instantly rounded off. Any roughness is smoothed the moment \(t>0\) — diffusion is infinitely smoothing.
STEP 3 · the wave equation carries it
By d'Alembert, \(u = F(x+ct)+G(x-ct)\): the initial shape splits into two copies that translate rigidly. A translated corner is still a corner, forever. Nothing in the equation acts to round it.
STEP 4 · the physical reading
Diffusion is dissipative and irreversible — run it backwards and it blows up. Waves are conservative and reversible — a recording of a wave played backwards is still a legitimate solution. This distinction is exactly the parabolic/hyperbolic split of module F3.
6
Derive the steady-state equation for a bar with internal heating \(Q(x)\), and solve it on \(0
Hint 1"Steady state" means nothing changes in time: set \(u_t = 0\).
Hint 2You are left with an ODE: \(0 = ku'' + Q\), i.e. \(u'' = -2\).
Hint 3Integrate twice and use the two boundary values to fix both constants.
Solution
STEP 1 · impose steadiness
From \(u_t = ku_{xx} + Q(x)\), setting \(u_t=0\) leaves an ODE in \(x\) alone:
\[ k\,u'' + Q(x) = 0 \]
STEP 2 · substitute the numbers
\[ u'' = -2 \]
STEP 3 · integrate twice
\[ u' = -2x + A, \qquad u = -x^2 + Ax + B \]
STEP 4 · apply the boundary conditions
\[ u(0) = B = 0 \]
\[ u(1) = -1 + A = 0 \ \Longrightarrow\ A = 1 \]
STEP 5 · the answer
\[ u(x) = x - x^2 = x(1-x) \]
A parabola peaking at \(x=\frac12\) with maximum \(\frac14\).
SANITY CHECK
Heating everywhere, cooled only at the ends ⟹ hottest in the middle, symmetric ✓. Note the steady state is where the transient solution of module S3 eventually settles, and that the time-dependent problem is solved by writing \(u = \underbrace{x(1-x)}_{\text{steady}} + \underbrace{w(x,t)}_{\text{decaying transient}}\) — the standard method of module A2.
7
A species diffuses along a channel while reproducing logistically. Derive its PDE and classify it.
Hint 1Start from \(u_t + \phi_x = \text{(net creation)}\) and use Fick's law for the flux.
Hint 2Logistic growth contributes a source \(ru(1-u/M)\).
Solution
STEP 1 · start from the balance with a source
\[ u_t + \phi_x = R(u) \]
Second order; nonlinear, because of the \(u^2\) in the reaction term. It is parabolic in its principal part (the \(u_{xx}\) term is all that counts for type), so it diffuses — but superposition fails.
STEP 5 · what the nonlinearity buys
This is Fisher's equation, and unlike the pure heat equation it supports travelling waves: an invasion front of fixed shape moving at speed \(2\sqrt{rk}\). Diffusion alone only spreads and flattens; diffusion plus growth can propagate — the standard model of species invasion, epidemic spread, and flame propagation.
8
A bar has conductivity varying with position, \(k = k(x)\). Derive the heat equation and note its structure.
Hint 1Fourier's law is still \(\phi = -k(x)u_x\), but \(k\) is now inside the \(x\)-derivative.
Hint 2Do not pull \(k\) out of \(\phi_x\) — use the product rule instead.
Solution
STEP 1 · the flux
\[ \phi(x,t) = -k(x)\,u_x \]
STEP 2 · substitute into the conservation law
\[ u_t + \phi_x = 0 \quad\Longrightarrow\quad u_t = \big(k(x)u_x\big)_x \]
STEP 3 · expand if you like
\[ u_t = k(x)u_{xx} + k'(x)u_x \]
The extra first-order term is a drift caused by the varying conductivity — heat tends to accumulate where conduction is poor.
STEP 4 · recognize the structure
The un-expanded form \(\big(ku_x\big)_x\) is exactly the Sturm–Liouville operator of module A1, with \(p(x)=k(x)\). Separation still works, and the eigenfunctions are still orthogonal — just not sines.
STEP 5 · why the divergence form matters
Keeping \(\big(ku_x\big)_x\) intact preserves conservation exactly: \(\frac{d}{dt}\int u\,dx = \big[ku_x\big]_0^L\), still pure boundary flux. Numerical schemes written in divergence form conserve heat to machine precision; the expanded version does not.
9
A vibrating string moves through a medium that resists with a force proportional to velocity. Derive the governing equation and say what changes.
Hint 1Add the resisting force to Newton's second law: \(\rho u_{tt} = T u_{xx} - b u_t\).
Hint 2Try a separated solution and see what the time factor does.
Solution
STEP 1 · Newton's law with drag
Mass × acceleration = tension force + drag:
\[ \rho\,u_{tt} = T\,u_{xx} - b\,u_t \]
STEP 2 · standard form
\[ u_{tt} + a\,u_t = c^2u_{xx}, \qquad a = \frac{b}{\rho},\ c^2 = \frac{T}{\rho} \]
the damped wave (telegraph) equation.
STEP 3 · classify
The principal part is unchanged (\(A=c^2, B=0, C=-1\), \(\Delta = 4c^2>0\)), so it is still hyperbolic: finite propagation speed survives, and corners still travel.
STEP 4 · what the damping does
Separating gives \(T'' + aT' + \lambda c^2T = 0\), whose roots are \(-\frac a2 \pm i\sqrt{\lambda c^2 - \frac{a^2}{4}}\). So each mode oscillates and decays like \(e^{-at/2}\):
\[ u \sim e^{-at/2}\left[A_n\cos\tilde\omega_nt + B_n\sin\tilde\omega_nt\right]\sin\frac{n\pi x}{L} \]
STEP 5 · read the physics
Every mode decays at the same rate \(a/2\) — unlike the heat equation, where high modes vanish first. That is why a damped string keeps its timbre as it fades, while a cooling bar loses its detail immediately. The equation also governs signals on a lossy transmission line, which is where "telegraph equation" comes from.
Part I · Module 3 Foundations
Classifying second-order PDEs
One discriminant decides almost everything: which method will work, what data you may impose, whether information travels at finite speed, and whether solutions stay rough or become instantly smooth. Learn to compute it in ten seconds.
①The discriminant
Write any linear second-order PDE in two variables in the standard form
Standard form — only the second-order terms matter
\[ A\,u_{xx} + B\,u_{xy} + C\,u_{yy} + \underbrace{D\,u_x + E\,u_y + Fu + G}_{\text{lower order — irrelevant to type}} = 0 \]
The names come from conic sections: \(B^2-4AC\) is exactly the discriminant that separates hyperbolas, parabolas, and ellipses in \(Ax^2+Bxy+Cy^2=1\). The analogy is more than cosmetic — the same algebra controls the characteristic directions along which information propagates.
InteractiveDiscriminant, type, and characteristicsdrag A, B, C
B² − 4AC —type —characteristic slopes —behavior —
The panel draws the characteristic directions \(\frac{dy}{dx} = \frac{B \pm \sqrt{B^2-4AC}}{2A}\) through a point. Hyperbolic gives two real directions — information rides along them at finite speed. Parabolic gives exactly one (a repeated root) — the boundary case. Elliptic gives none (complex roots), which is why elliptic problems have no notion of propagation and every boundary point influences every interior point at once.
②The three archetypes and what the type predicts
Hyperbolic
Parabolic
Elliptic
Model
\(u_{tt} = c^2u_{xx}\)
\(u_t = ku_{xx}\)
\(u_{xx}+u_{yy} = 0\)
A, B, C
\(A=c^2, B=0, C=-1\) ⟹ \(\Delta = 4c^2>0\)
\(A=k, B=0, C=0\) ⟹ \(\Delta = 0\)
\(A=C=1, B=0\) ⟹ \(\Delta = -4<0\)
Data you impose
\(u\) and \(u_t\) at \(t=0\), plus boundaries
\(u\) at \(t=0\), plus boundaries
\(u\) (or \(u_n\)) on the entire closed boundary
Speed of information
finite, \(=c\)
infinite (but exponentially weak)
instantaneous, global
Smoothing
none — corners persist
infinite — smooth for any \(t>0\)
solutions are analytic inside
Reversible in time?
yes
no — backward heat is ill-posed
no time variable at all
Typical picture
ringing string, sound, light
cooling rod, spreading dye
steady temperature, electrostatic potential
Two traps in computing Δ
(1) Only the second-order coefficients count. \(u_t = ku_{xx}\) has no \(u_{tt}\) term at all, so \(C=0\) and it is parabolic — the \(u_t\) term is first order and plays no part in \(\Delta\). (2) \(B\) is the coefficient of the mixed term \(u_{xy}\) as written. If the equation is given as \(u_{xx} + 4u_{xy} + u_{yy}=0\), then \(B = 4\), not 2.
Worked example · a type that changes with position
Classify \(u_{xx} + 4u_{xy} + 3u_{yy} = 0\), then classify the Tricomi equation \(y\,u_{xx} + u_{yy} = 0\).
Its characteristic slopes are \(\frac{dy}{dx} = \frac{4\pm2}{2} = 3\) and \(1\). So the substitution \(\xi = y-3x\), \(\eta = y-x\) will flatten it into the canonical form \(u_{\xi\eta}=0\), whose solution (F1, problem 2) is \(u = F(\xi)+G(\eta)\).
STEP 3
Tricomi: \(A = y\), \(B = 0\), \(C = 1\), so \(\Delta = -4y\). The coefficient depends on position, so the type does too.
This is not a curiosity — it is transonic flow. Subsonic regions are elliptic (disturbances felt everywhere), supersonic regions are hyperbolic (disturbances confined to a Mach cone), and the sonic line is where the type flips. The mathematics of the type change is the physics of the shock.
\(A=1, B=-6, C=9\): \(\Delta = 36 - 36 = 0\) ⟹ parabolic. (Note it factors as \((\partial_x - 3\partial_y)^2u = 0\) — the repeated characteristic showing itself.)
(c)
\(A=2, B=0, C=1\): \(\Delta = 0 - 8 = -8 < 0\) ⟹ elliptic. The \(5u_x\) term is first order and irrelevant to type.
(d)
Rewrite as \(4u_{xx} - u_t = 0\). With \(t\) playing the role of \(y\): \(A=4\), \(B=0\), \(C=0\) (there is no \(u_{tt}\)), so \(\Delta = 0\) ⟹ parabolic ✓ as the heat equation must be.
CHECKLIST
Missing \(u_{yy}\) ⟹ \(C=0\) ⟹ \(\Delta = B^2 \ge 0\). So an equation with no second time-derivative is never elliptic: it is parabolic if it also lacks \(u_{xy}\), hyperbolic if it has one.
2
For \(u_{xx} - 5u_{xy} + 6u_{yy} = 0\), find the characteristic slopes and the change of variables that reduces it to \(u_{\xi\eta}=0\). Then write its general solution.
Hint 1Characteristic slopes solve \(A\left(\frac{dy}{dx}\right)^2 - B\frac{dy}{dx} + C = 0\) — careful with that sign convention, or equivalently use \(\frac{dy}{dx} = \frac{B\pm\sqrt{\Delta}}{2A}\).
Hint 2\(\Delta = 25-24 = 1\), so the slopes are \(\frac{-5\pm1}{2}\).
Hint 3If a characteristic has slope \(m\), then \(y - mx\) is constant along it — use those as your new coordinates.
STEP 2 · sort by sign
\[ x > 0 \Rightarrow \Delta<0: \textbf{elliptic} \qquad x = 0 \Rightarrow \textbf{parabolic} \qquad x < 0 \Rightarrow \Delta>0: \textbf{hyperbolic} \]
STEP 3 · what changes across x = 0
For \(x>0\) the equation behaves like Laplace's: no characteristics, boundary data on a closed curve, instantly smooth solutions. For \(x<0\) it behaves like the wave equation: two real characteristic families, finite propagation speed, and data given on a non-characteristic curve.
STEP 4 · why this matters
You cannot pose one problem for the whole plane and expect a single method to work. Boundary data appropriate on the elliptic side is over-determined on the hyperbolic side, and initial data appropriate on the hyperbolic side is ill-posed on the elliptic side (Hadamard's example). Mixed-type equations like this are genuinely harder and are the reason transonic aerodynamics was a research problem for decades.
4
Explain why \(\Delta = 0\) exactly matches "one repeated characteristic," and why parabolic equations therefore sit on the fence between waves and equilibrium.
Hint 1The characteristic slopes come from a quadratic. What does its discriminant control?
Hint 2Compare with the roots of the characteristic equation for a second-order ODE — repeated roots there give \(e^{rt}\) and \(te^{rt}\).
Solution
STEP 1 · where the slopes come from
Characteristic directions satisfy a quadratic whose discriminant is exactly \(B^2-4AC\):
Two distinct real roots when \(\Delta>0\); one repeated real root when \(\Delta=0\); a complex-conjugate pair (no real directions) when \(\Delta<0\). The classification is the root count.
STEP 3 · read the consequences
Two real directions ⟹ information splits and travels both ways at finite speed (hyperbolic). No real direction ⟹ no propagation at all; the solution is determined globally by the boundary (elliptic).
STEP 4 · the parabolic middle
With one repeated direction the heat equation has a single characteristic family, the lines \(t = \text{const}\). Information does not ride outward along a cone — it appears everywhere at once, but with exponentially small amplitude far away. That is the sense in which diffusion has "infinite speed" while still behaving like an evolution problem.
THE ODE PARALLEL
Exactly the trichotomy of \(ar^2+br+c=0\) in second-order ODEs: distinct real roots (over-damped), repeated root (critically damped), complex roots (oscillatory). Same algebra, one dimension up.
5
The backward heat equation \(u_t = -u_{xx}\) is famously ill-posed. Show what goes wrong by testing \(u = e^{n^2t}\sin nx\), and connect this to classification.
Hint 1Verify it solves the equation, then look at what happens for large \(n\).
Hint 2At \(t=0\) the data has amplitude 1 no matter how big \(n\) is. What is the amplitude at \(t=1\)?
\(u(x,0) = \sin nx\), whose maximum size is 1 for every \(n\). As \(n\) grows the data becomes a tiny, rapid wiggle — as small as you like in the "how big is it" sense.
STEP 3 · look at the solution later
At \(t=1\) the amplitude is \(e^{n^2}\). For \(n=10\) that is \(e^{100}\approx10^{43}\).
STEP 4 · why this is fatal
Well-posedness requires that small changes in the data produce small changes in the solution. Here an arbitrarily small perturbation of the initial data produces an arbitrarily large change at any later time. The problem has no continuous dependence, so it is ill-posed — and numerically hopeless, since round-off alone supplies those wiggles.
STEP 5 · the classification view
Forward heat is parabolic and smooths: high-frequency modes decay like \(e^{-n^2t}\). Running it backwards must therefore amplify exactly what the forward problem destroyed. This is the precise sense in which diffusion is irreversible, and it is why you can't reconstruct this morning's temperature profile from tonight's — the information is genuinely gone.
Parabolic. The lower-order terms \(7u_x - u\) play no part in the classification.
STEP 4 · see why it is a perfect square
Dividing by 2: \(u_{xx}-2u_{xy}+u_{yy} = (\partial_x - \partial_y)^2u\). The repeated characteristic direction is \(\frac{dy}{dx} = \frac{B}{2A} = -1\).
STEP 5 · what to expect
Diffusion-like behavior in the direction transverse to those characteristics, with the lower-order terms controlling growth or decay but not the equation's character. Substituting \(\xi = x+y\) would reduce it to a heat-type equation in \(\xi\).
7
Determine the region of the plane where \(y\,u_{xx} + 2u_{xy} + x\,u_{yy}=0\) is hyperbolic.
Hint 1\(A=y\), \(B=2\), \(C=x\) — the discriminant depends on position.
Hint 2\(\Delta = 4-4xy\); hyperbolic needs \(\Delta>0\).
Hyperbolic on the region between the two branches of the hyperbola \(xy=1\) — which includes the whole second and fourth quadrants (where \(xy<0\)) and the part of the first and third quadrants near the axes.
STEP 4 · the other cases
\[ xy = 1: \text{ parabolic (on the hyperbola itself)}, \qquad xy>1: \text{ elliptic} \]
STEP 5 · the practical consequence
A single problem posed across \(xy=1\) is mixed type: initial-value data is appropriate on the hyperbolic side and boundary-value data on the elliptic side, and no one method serves both. Recognizing this before choosing a technique saves a great deal of wasted effort.
8
Reduce \(u_{xx}-u_{yy}=0\) to canonical form using \(\xi = x+y\), \(\eta = x-y\), and write its general solution.
Hint 1Chain rule: \(u_x = u_\xi + u_\eta\) and \(u_y = u_\xi - u_\eta\).
Hint 2Apply each operator twice and subtract; the pure second derivatives cancel.
Solution
STEP 1 · first derivatives
Since \(\xi_x=\eta_x=1\), \(\xi_y=1\), \(\eta_y=-1\):
With \(y\) renamed \(ct\) this is exactly d'Alembert's solution of the wave equation (module S6). The new coordinates \(\xi,\eta\) are the characteristic coordinates, and reducing to \(u_{\xi\eta}=0\) is what "canonical form" means for every hyperbolic equation.
9
Explain why lower-order terms (\(Du_x + Eu_y + Fu + G\)) never affect the classification, even though they clearly affect the solution.
Hint 1Think about what happens on very small scales — say, replacing \(u\) by a rapidly oscillating \(e^{i(\alpha x+\beta y)/\varepsilon}\).
Hint 2Which terms dominate as \(\varepsilon\to0\)?
Solution
STEP 1 · test with a fine-scale wave
Insert \(u = e^{i(\alpha x + \beta y)/\varepsilon}\). Each derivative brings a factor \(1/\varepsilon\), so second-order terms scale like \(\varepsilon^{-2}\) and first-order terms like \(\varepsilon^{-1}\).
STEP 2 · compare magnitudes
As \(\varepsilon\to0\) the second-order terms dominate by a factor \(1/\varepsilon\). To leading order the equation reduces to
\[ A\alpha^2 + B\alpha\beta + C\beta^2 = 0 \]
— the principal symbol, containing no lower-order information at all.
STEP 3 · what that quadratic decides
Whether it has real roots \((\alpha:\beta)\) is precisely the sign of \(B^2-4AC\) — the existence of real characteristic directions, and hence whether the equation propagates information along curves.
STEP 4 · so what do lower-order terms do?
They change the solution substantially — adding growth, decay, drift, or forcing (a \(-hu\) term makes the rod cool to ambient; a \(+u\) term can make it blow up). What they cannot change is the equation's character: how singularities travel, what data are admissible, whether solutions smooth.
STEP 5 · the practical rule
Classify with the top-order terms alone, then treat everything else as physics layered on top. This is why the damped wave equation of module F2 is still hyperbolic, and why \(u_t = ku_{xx} - hu\) is still a diffusion.
Part I · Module 4 Foundations
Boundary & initial conditions
A PDE alone has far too many solutions. The side conditions are what select one — but only if you impose the right number of the right kind on the right set. Getting this wrong is the most common way a PDE problem becomes unsolvable or meaningless.
①The three boundary condition types
Name
Form at an end
Physical meaning
Dirichlet
\(u(0,t) = g(t)\)
The value is held fixed — an end clamped in an ice bath, a string pinned to a wall.
Neumann
\(u_x(0,t) = g(t)\)
The flux is prescribed. \(u_x=0\) means insulated: no heat crosses the end.
Robin (mixed)
\(u_x(0,t) = h\big[u(0,t) - U_{\text{air}}\big]\)
Newton cooling: the end exchanges heat with its surroundings at a rate proportional to the temperature difference.
Periodic
\(u(0,t)=u(L,t)\) and \(u_x(0,t)=u_x(L,t)\)
The domain is a ring — no ends at all.
Robin conditions interpolate between the other two: \(h\to\infty\) forces \(u\to U_{\text{air}}\) (Dirichlet), while \(h\to0\) gives \(u_x\to0\) (Neumann).
InteractiveWhat each end condition doespick a condition for each end
steady profile —u(0) —u(L) —solvable? —
The curve is the steady state \(-ku'' = Q\) that satisfies your two end conditions, with the boundary behavior drawn: a dot for a clamped value, a horizontal tangent for an insulated end. Try Neumann on both ends with \(Q\ne0\): the problem becomes unsolvable, because you are pumping heat in but forbidding it to leave. That is the compatibility condition, and it is a real phenomenon, not a technicality.
②How many conditions, and where
The rule follows the classification of module F3:
Counting conditions
\[
\begin{array}{lll}
\textbf{Heat (parabolic):} & u_t = ku_{xx} & \text{one initial condition } u(x,0)=\varphi(x),\ \text{plus one BC at each end}\\
\textbf{Wave (hyperbolic):} & u_{tt} = c^2u_{xx} & \text{two initial conditions } u(x,0),\ u_t(x,0),\ \text{plus one BC at each end}\\
\textbf{Laplace (elliptic):} & \nabla^2u=0 & \text{one condition on the \emph{entire closed} boundary — and no initial data at all}
\end{array} \]
The count matches the order in each variable: two \(t\)-derivatives in the wave equation ⟹ two initial conditions; one \(t\)-derivative in the heat equation ⟹ one; no \(t\) at all in Laplace ⟹ none.
Well-posed, in Hadamard's sense
A problem is well posed when (1) a solution exists, (2) it is unique, and (3) it depends continuously on the data. Too few conditions kills uniqueness; too many kills existence; the wrong kind — like initial data for an elliptic equation, or backward time for heat — kills continuity, as module F3 problem 5 showed.
Worked example · the compatibility condition for two insulated ends
Show that the steady problem \(-u'' = Q\) on \((0,L)\) with \(u'(0)=u'(L)=0\) has a solution only if \(\int_0^LQ\,dx = 0\).
STEP 1
Integrate the equation across the whole bar:
\[ -\int_0^Lu''\,dx = \int_0^LQ\,dx \]
STEP 2
The left side collapses by the Fundamental Theorem:
Both boundary terms are zero by the Neumann conditions, so the left side is exactly 0, forcing
\[ \int_0^LQ\,dx = 0 \]
STEP 4
Physical reading: insulated at both ends means no heat can escape. A steady state can only exist if the internal sources and sinks cancel exactly. Otherwise the bar heats up (or cools) forever and never settles — there is no steady solution to find.
STEP 5
And when the condition is satisfied, the solution is unique only up to an additive constant: nothing in the problem fixes the overall temperature level. Existence and uniqueness fail together in a very structured way, which is the finite-dimensional Fredholm alternative in disguise.
③Practice
1
State the physical meaning of each: (a) \(u(0,t)=100\), (b) \(u_x(L,t)=0\), (c) \(u_x(L,t) = -2\big[u(L,t)-20\big]\), (d) \(u_t(x,0)=0\).
Hint 1Value prescribed = Dirichlet; slope prescribed = Neumann (flux); a mix of both = Robin.
Hint 2(d) has a \(t\)-derivative at \(t=0\) — that is initial data, not a boundary condition.
Solution
(a) u(0,t) = 100
Dirichlet. The left end is held at 100° for all time — clamped against a reservoir big enough that the rod cannot change it.
(b) u_x(L,t) = 0
Neumann, homogeneous. Zero gradient means zero flux by Fourier's law: the right end is perfectly insulated. No heat crosses it in either direction.
(c) u_x(L,t) = −2[u(L,t) − 20]
Robin. The end exchanges heat with air at 20°, at a rate proportional to the temperature difference — Newton's law of cooling. If the end is hotter than 20°, the bracket is positive and the flux runs outward.
(d) u_t(x,0) = 0
Not a boundary condition at all — an initial condition, and specifically the second one required by a hyperbolic equation. It says the string is released from rest: displaced into some shape, then let go with zero velocity.
HOW TO TELL THEM APART
Boundary conditions are imposed at fixed positions for all time; initial conditions are imposed at a fixed time for all positions. Sketching the \(x\)–\(t\) plane and marking where each condition lives makes the distinction obvious every time.
2
How many conditions does each need, and of what kind? (a) \(u_t = u_{xx}\) on \(0
Hint 1Count the derivatives in each variable — that's how many conditions that variable needs.
Hint 2For the disk, there is no time variable; "boundary" means the entire circle.
Solution
(a) heat equation
One \(t\)-derivative ⟹ one initial condition \(u(x,0)=\varphi(x)\). Two \(x\)-derivatives ⟹ two boundary conditions, one at each end (any mix of Dirichlet, Neumann, Robin).
(b) wave equation
Two \(t\)-derivatives ⟹ two initial conditions: \(u(x,0)=\varphi(x)\) (shape) and \(u_t(x,0)=\psi(x)\) (velocity). Plus two boundary conditions as before.
(c) Laplace on a disk
No time variable ⟹ no initial conditions. Instead, one condition at every point of the closed boundary circle: \(u(1,\theta) = f(\theta)\) for Dirichlet, or \(u_r(1,\theta)=g(\theta)\) for Neumann.
A WARNING ABOUT ELLIPTIC PROBLEMS
You may not impose both \(u\) and \(u_r\) on the boundary of an elliptic problem — that is the Cauchy data appropriate to hyperbolic equations, and doing so makes the problem over-determined and ill-posed (Hadamard's counterexample). One condition per boundary point, no more.
AND FOR NEUMANN ON A CLOSED BOUNDARY
The pure-Neumann Laplace problem needs \(\oint g\,ds = 0\) for the same reason as the worked example — total flux in must equal total flux out — and its solution is then unique only up to a constant.
3
Find the steady-state temperature in a bar \(0
Hint 1\(u''=0\) means \(u\) is a straight line: \(u = Ax+B\).
Hint 2A Neumann condition prescribes the slope \(A\) directly.
Hint 3For the second part: can a straight line have two different slopes?
Solution
STEP 1 · general steady solution
\[ u'' = 0 \quad\Longrightarrow\quad u(x) = Ax + B \]
STEP 3 · apply the Dirichlet condition
\[ u(0) = B = 10 \quad\Longrightarrow\quad u(x) = 10 - 3x \]
Check: \(u(2) = 4\), and heat flows steadily from the hot left end out through the right end ✓
STEP 4 · the impossible version
Now both conditions prescribe slopes: \(A = 1\) from the left and \(A = -3\) from the right. A line has exactly one slope, so no solution exists.
STEP 5 · the physics of the failure
\(u_x(0)=1\) and \(u_x(2)=-3\) prescribe fluxes that don't balance — you are forcing more heat in than out, with nothing inside to absorb the difference. No steady state can exist; in the time-dependent problem the bar simply keeps heating up. This is the compatibility condition of the worked example, in its bluntest form.
4
A string is plucked into the shape \(\varphi(x)\) and released from rest, on \(0
Hint 1"Released from rest" is a statement about the initial velocity.
Hint 2"Ends tied down" gives Dirichlet conditions for all time.
Hint 3The wave equation needs exactly two initial conditions and two boundary conditions.
"Released from rest" means every point starts motionless:
\[ u_t(x,0) = 0 \]
STEP 5 · compatibility
For a smooth solution the data must agree at the corners of the domain: \(\varphi(0)=\varphi(L)=0\), matching the clamped ends. A plucked triangular \(\varphi\) satisfies this — and its corner will persist forever, since the wave equation does not smooth (module F2 problem 5).
WHAT COMES NEXT
This exact problem is solved two different ways later: by separation of variables into standing modes (module S5), and by d'Alembert's travelling-wave formula with odd periodic extension (module S6). Same answer, two very different pictures of it.
5
Explain why the Neumann problem \(u''=0\), \(u_x(0)=u_x(L)=0\) has infinitely many solutions, and what extra information physically selects one.
Hint 1Which straight lines have zero slope at both ends?
Hint 2Does anything in the problem statement mention an absolute temperature level?
Solution
STEP 1 · solve
\(u = Ax+B\) with \(u'=A=0\) at both ends, so \(A=0\) and
\[ u(x) = B \quad\text{for any constant } B \]
STEP 2 · why uniqueness fails
Both conditions constrain only the derivative. Nothing anywhere in the problem refers to the actual value of \(u\), so the whole profile can be shifted up or down freely.
STEP 3 · the physical reading
A perfectly insulated bar with no sources reaches a uniform temperature — but the problem never says which uniform temperature. That information is not in the boundary conditions; it comes from the initial condition.
STEP 4 · what selects B
Conservation (module F2 problem 2): with no flux through either end, the total heat \(\int_0^Lu\,dx\) never changes, so the final uniform value is the average of the initial data:
\[ B = \frac1L\int_0^L\varphi(x)\,dx \]
STEP 5 · the pattern to remember
Pure-Neumann problems are unique only up to an additive constant, and require a compatibility condition to have any solution at all. Module S4 shows this identical structure appearing as the \(\lambda = 0\) eigenvalue of the insulated heat problem — the constant eigenfunction that never decays.
6
For the Robin condition \(-k\,u_x(L,t) = h\left[u(L,t)-U_\infty\right]\) at the right end, explain why \(h>0\) is the physically sensible sign.
Hint 1The left side is the outward heat flux at \(x=L\).
Hint 2If the bar end is hotter than the air, which way should heat flow?
Solution
STEP 1 · interpret each side
By Fourier's law the flux is \(\phi = -ku_x\), and at the right end that is the rate at which heat leaves the bar. The right side is \(h\) times the excess of the end temperature over ambient.
STEP 2 · test the hot case
If \(u(L,t) > U_\infty\), the bracket is positive. Heat should flow out, i.e. the outward flux should be positive — which requires \(h>0\).
STEP 3 · test the cold case
If \(u(L,t) < U_\infty\), the bracket is negative and, with \(h>0\), the flux is negative: heat flows into the bar from the warmer air ✓ Consistent in both directions.
STEP 4 · what h < 0 would mean
Heat flowing spontaneously from cold to hot — a violation of the second law, and mathematically a sign that the problem may admit growing solutions rather than decaying ones.
STEP 5 · the limits
\(h\to\infty\) forces \(u(L,t)\to U_\infty\): a perfect heat exchanger, i.e. Dirichlet. \(h\to0\) gives \(u_x(L,t)\to0\): perfect insulation, i.e. Neumann. Robin genuinely interpolates between the two, with \(h\) measuring how well the end is coupled to its surroundings.
7
Find the steady state on \(0
Hint 1Start from \(u = Ax+B\) and impose the left condition first.
Hint 2Then substitute \(u(1) = A+100\) and \(u'(1)=A\) into the Robin condition.
STEP 2 · left condition
\[ u(0) = B = 100 \quad\Longrightarrow\quad u = Ax + 100 \]
STEP 3 · impose the Robin condition
Here \(u'(1) = A\) and \(u(1) = A + 100\):
\[ A = -2\left[(A+100) - 20\right] = -2A - 160 \]
STEP 4 · solve for A
\[ 3A = -160 \quad\Longrightarrow\quad A = -\frac{160}{3} \approx -53.3 \]
STEP 5 · the profile and a sanity check
\[ u(x) = 100 - \frac{160}{3}x, \qquad u(1) \approx 46.7^\circ \]
The end sits between the bar's hot base (100°) and the air (20°) — exactly what a partially-cooled end should do. Note it does not reach ambient: that would require \(h=\infty\).
8
Is the problem "\(u_t=ku_{xx}\) on \(0and \(u(x,T)=\psi(x)\) both prescribed" well posed?
Hint 1Count conditions against the order in \(t\).
Hint 2Once \(\varphi\) is given, is there any freedom left to satisfy a condition at \(t=T\)?
Solution
STEP 1 · count
The heat equation is first order in \(t\), so it needs exactly one condition in time. Prescribing data at both \(t=0\) and \(t=T\) is one too many.
STEP 2 · why it fails
The initial condition already determines \(u\) for all \(t>0\) uniquely. The value at \(t=T\) is therefore forced, and generally will not equal your prescribed \(\psi\). Existence fails.
STEP 3 · the exceptional case
If \(\psi\) happens to be exactly the evolved profile, a solution exists — but this is a razor-thin set of compatible data, not a robust problem.
STEP 4 · the reverse problem is worse
Prescribing only \(u(x,T)=\psi\) and asking for \(u(x,0)\) is the backward heat problem, which fails continuous dependence catastrophically (module F3 problem 5).
STEP 5 · what such problems are for
Two-time data problems appear in inverse and control theory — "what initial state produces this final state?" They are studied precisely because they are ill-posed, and require regularization (adding a penalty term) rather than direct solution.
9
Find the eigenvalues and eigenfunctions for a ring: \(X''+\lambda X=0\) with the periodic conditions \(X(0)=X(2\pi)\), \(X'(0)=X'(2\pi)\).
Hint 1Both \(\sin nx\) and \(\cos nx\) can satisfy periodic conditions — check which \(n\) work.
Hint 2Don't forget to test \(\lambda=0\).
Hint 3How many independent eigenfunctions belong to each nonzero eigenvalue?
Solution
STEP 1 · test λ = 0
\(X = Ax+B\); periodicity of \(X\) forces \(A=0\), and \(X=B\) then satisfies both conditions. So
\[ \lambda_0 = 0, \qquad X_0 = 1 \]
STEP 2 · test λ = β² > 0
\(X = A\cos\beta x + B\sin\beta x\). Periodicity over \(2\pi\) requires \(\cos\) and \(\sin\) to complete whole cycles:
\[ \beta = n, \qquad n = 1,2,3,\dots \]
STEP 3 · note the degeneracy
For each \(n\ge1\), both \(\cos nx\) and \(\sin nx\) satisfy the conditions. So every nonzero eigenvalue \(\lambda_n=n^2\) has a two-dimensional eigenspace.
STEP 4 · rule out λ < 0
Exponentials are not periodic, so no negative eigenvalues exist.
STEP 5 · the consequence
The expansion in these eigenfunctions is the full Fourier series — sines and cosines together — which is exactly why a ring uses the full series while an interval uses one or the other. Degeneracy is a hallmark of symmetric domains: on a ring there is no preferred origin, so any rotation of an eigenfunction is another one.
Part I · Module 5 Foundations
Superposition & linearity
Every technique in Part III rests on one property: for a linear homogeneous PDE, sums of solutions are solutions. That is what lets you build a complicated answer out of simple pieces — and it is exactly what fails, catastrophically, the moment the equation becomes nonlinear.
①The principle
Write the equation as \(\mathcal{L}u = 0\), where \(\mathcal{L}\) collects the derivatives — e.g. \(\mathcal{L} = \partial_t - k\partial_{xx}\) for the heat equation. Linearity means
InteractiveBuilding a solution out of modesmix three modes; watch them evolve
mode 1 amplitude —mode 2 —mode 3 —
Faint curves are the individual modes; the bold curve is their sum — and the sum solves the PDE because each piece does. Under heat, watch mode 3 vanish first: its decay rate \(n^2\) is nine times mode 1's, so the wiggles die and the smooth shape survives. Under wave, nothing decays at all; each mode just oscillates at its own frequency and the shape recurs periodically.
②Nonhomogeneous problems: split the difference
For \(\mathcal{L}u = f\) with \(f\ne0\), superposition still helps — in the same shape as for ODEs:
The standard tactic for a steady forcing or nonzero boundary values: find a steady state \(u_p\) that absorbs them, then let \(u_h\) handle the initial data with homogeneous boundary conditions. Module A2 does this systematically.
Superposition dies with nonlinearity
For Burgers' equation \(u_t + uu_x = 0\), if \(u\) and \(v\) are solutions then \(u+v\) is almost never one — the term \((u+v)(u+v)_x\) generates cross-terms \(uv_x + vu_x\) that don't cancel. No Fourier series, no mode-by-mode analysis, and genuinely new behavior: waves that steepen into shocks (module C3).
Worked example · using superposition to handle awkward boundary data
Solve \(u_t = u_{xx}\) on \(0
STEP 1
Find a steady state satisfying the awkward boundary values. Set \(u_t=0\): \(v''=0\) with \(v(0)=0, v(1)=5\), so \(v(x) = 5x\).
STEP 2
Split: write \(u = v(x) + w(x,t)\), where \(w\) is the remaining transient.
STEP 3
Substitute. Since \(v_t=0\) and \(v_{xx}=0\), \(w\) satisfies the same PDE: \(w_t = w_{xx}\).
STEP 4
Check \(w\)'s boundary conditions — this is the payoff:
Both are now homogeneous, so separation of variables applies to \(w\).
STEP 5
Transfer the initial data: \(w(x,0) = \varphi(x) - 5x\). Solve that standard problem (module S3), then add \(5x\) back at the end. As \(t\to\infty\), \(w\to0\) and \(u\to5x\) — the bar settles into the linear steady profile, as physics demands.
③Practice
1
Verify directly that if \(u_1\) and \(u_2\) solve \(u_t = ku_{xx}\), so does \(3u_1 - 2u_2\). Then show \(u_1u_2\) generally does not.
Hint 1Differentiation is linear: the derivative of a sum is the sum of derivatives, and constants pull out.
Hint 2For the product, use the product rule twice on \((u_1u_2)_{xx}\) and see what extra term appears.
Solution
STEP 1 · set w = 3u₁ − 2u₂
\[ w_t = 3(u_1)_t - 2(u_2)_t \]
STEP 2 · use that each solves the PDE
\[ = 3k(u_1)_{xx} - 2k(u_2)_{xx} = k\big[3u_1 - 2u_2\big]_{xx} = kw_{xx} \ ✓ \]
STEP 3 · now try the product
\[ (u_1u_2)_t = (u_1)_tu_2 + u_1(u_2)_t = k\left[(u_1)_{xx}u_2 + u_1(u_2)_{xx}\right] \]
Linear combinations are safe; products are not. Superposition is about adding solutions, never multiplying them.
2
Given that \(u_n = e^{-n^2t}\sin nx\) solves \(u_t=u_{xx}\) for each \(n\), find the solution with \(u(x,0) = 3\sin x - \sin 4x\).
Hint 1Superpose: take the same combination of \(u_n\)'s that the initial data prescribes.
Hint 2Each mode carries its own decay factor \(e^{-n^2t}\) — don't give them a common one.
Solution
STEP 1 · match the initial data
At \(t=0\), \(u_n(x,0) = \sin nx\). The data is \(3\sin x - \sin 4x\), so take \(n=1\) with coefficient 3 and \(n=4\) with coefficient \(-1\).
STEP 2 · superpose with the correct decay rates
\[ u(x,t) = 3e^{-t}\sin x - e^{-16t}\sin 4x \]
STEP 3 · verify the initial condition
\[ u(x,0) = 3\sin x - \sin4x \ ✓ \]
STEP 4 · verify the PDE
Each term solves it separately, so by superposition the sum does ✓
STEP 5 · read the long-time behavior
The \(n=4\) mode decays sixteen times faster than the \(n=1\) mode. By \(t=0.2\) it has shrunk by \(e^{-3.2}\approx4\%\) while the fundamental has only dropped to \(82\%\). Very quickly \(u \approx 3e^{-t}\sin x\): whatever you start with, the slowest mode dominates the late-time picture.
3
Reduce \(u_t = u_{xx}\), \(u(0,t)=2\), \(u(1,t)=6\), \(u(x,0)=\varphi(x)\) to a problem with homogeneous boundary conditions.
Hint 1Find the steady state first: the straight line matching both boundary values.
Hint 2Subtract it: let \(w = u - v\), and work out \(w\)'s conditions.
Solution
STEP 1 · find the steady state
Set \(u_t=0\): \(v''=0\) so \(v=Ax+B\), with \(v(0)=2\), \(v(1)=6\):
\[ B = 2, \quad A + 2 = 6 \Rightarrow A = 4, \qquad v(x) = 2 + 4x \]
STEP 2 · subtract it off
\[ w(x,t) = u(x,t) - (2+4x) \]
STEP 3 · the PDE for w
\(v\) is time-independent and linear in \(x\), so \(v_t = v_{xx} = 0\) and
STEP 5 · the new initial condition
\[ w(x,0) = \varphi(x) - 2 - 4x \]
Solve this standard problem by separation of variables, then recover \(u = w + 2 + 4x\). As \(t\to\infty\), \(w\to0\) and the bar settles to the straight line joining its two end temperatures.
4
Show that superposition fails for \(u_t + uu_x = 0\) by testing the two constant solutions \(u\equiv1\) and \(u\equiv2\).
Hint 1Constants make every derivative vanish — check each solves the equation.
Hint 2Then check the sum, and think about whether "sums of solutions are solutions" is really being tested here — try a nonconstant example too.
Solution
STEP 1 · constants are solutions
If \(u\) is constant then \(u_t = 0\) and \(u_x = 0\), so \(u_t + uu_x = 0\) ✓ for both \(u\equiv1\) and \(u\equiv2\). Their sum \(u\equiv3\) is also constant, hence also a solution — so constants alone do not expose the failure.
STEP 2 · use a sharper test
Take \(u_1 = \frac{x}{1+t}\), which solves the equation: \( (u_1)_t = -\frac{x}{(1+t)^2}\) and \(u_1(u_1)_x = \frac{x}{1+t}\cdot\frac{1}{1+t} = \frac{x}{(1+t)^2}\), and they cancel ✓. Now test \(w = 2u_1\).
Scaling by 2 doubled the first term but quadrupled the nonlinear one, so they no longer cancel. Even multiplying a single solution by a constant fails — the equation is not homogeneous of degree one in \(u\).
STEP 5 · the consequence
No superposition means no Fourier method, no mode-by-mode decay analysis, and no "general solution = sum of particular solutions." Nonlinear PDEs need entirely different tools — characteristics, weak solutions, entropy conditions — which is what module C3 develops.
5
Suppose \(\mathcal{L}u_1 = f_1\) and \(\mathcal{L}u_2 = f_2\) for a linear \(\mathcal{L}\). What problem does \(u_1+u_2\) solve? Use this to solve \(u_t - u_{xx} = \sin x + 2\), given solutions for each piece separately.
Hint 1Apply \(\mathcal{L}\) to the sum and use linearity.
Hint 2Forcing terms add exactly as the solutions do — that's the superposition principle for nonhomogeneous problems.
Solution
STEP 1 · apply the operator to the sum
\[ \mathcal{L}(u_1+u_2) = \mathcal{L}u_1 + \mathcal{L}u_2 = f_1 + f_2 \]
So the sum solves the problem with the summed forcing.
STEP 2 · split the given forcing
Take \(f_1 = \sin x\) and \(f_2 = 2\), and find a steady solution of each.
STEP 3 · handle f₁ = sin x
Try \(u_1 = a\sin x\), steady. Then \(u_t=0\) and \(-u_{xx} = a\sin x\), so \(a = 1\):
\[ u_1 = \sin x \]
STEP 4 · handle f₂ = 2
Try a steady polynomial: \(-u'' = 2\) gives \(u_2 = -x^2\) (any particular choice will do; the homogeneous part will fix boundary values later).
STEP 5 · combine
\[ u_p = \sin x - x^2 \]
Check: \(u_t - u_{xx} = 0 - (-\sin x - 2) = \sin x + 2\) ✓
WHY THIS IS THE STANDARD TACTIC
Decompose a complicated forcing into pieces you can handle, solve each, and add. With a Fourier series you can push this to infinitely many pieces — one per mode — which is exactly the eigenfunction-expansion method of module A2.
6
A rectangle has all four sides at nonzero temperatures. Explain how superposition reduces this to problems you can solve.
Hint 1Separation needs homogeneous conditions on the two sides perpendicular to the expansion direction.
Hint 2Can you write the boundary data as a sum of four pieces, each nonzero on only one side?
Solution
STEP 1 · split the boundary data
Write the prescribed boundary function as \(f = f_1+f_2+f_3+f_4\), where \(f_k\) agrees with \(f\) on side \(k\) and is zero on the other three.
STEP 2 · solve four subproblems
Let \(u_k\) solve \(\nabla^2u_k=0\) with boundary data \(f_k\). Each has three homogeneous sides, so separation of variables applies directly.
STEP 3 · add them
By linearity, \(u = u_1+u_2+u_3+u_4\) is harmonic, and on each side its boundary value is the sum of one nonzero contribution and three zeros — i.e. exactly \(f\) ✓
STEP 4 · why each subproblem is easy
With, say, only the top edge nonzero, the two vertical sides give \(X(0)=X(a)=0\), producing \(X_n = \sin\frac{n\pi x}{a}\); the bottom edge selects \(\sinh\frac{n\pi y}{a}\) (module S7).
STEP 5 · the general lesson
Superposition lets you decompose data, not just solutions. The same idea handles a PDE with several source terms (problem 5) and, taken to infinity, becomes the eigenfunction expansion of module A2.
7
Suppose \(u\) solves the heat equation with \(u(0,t)=5\), \(u(L,t)=0\), and \(v\) solves it with \(v(0,t)=0\), \(v(L,t)=8\). What problem does \(u+v\) solve?
Hint 1The PDE is linear and homogeneous, so the sum still solves it.
Hint 2Boundary values add too — evaluate the sum at each end.
Solution
STEP 1 · the PDE
\((u+v)_t = u_t+v_t = k(u_{xx}+v_{xx}) = k(u+v)_{xx}\) ✓ still the heat equation.
\((u+v)(x,0) = \varphi_u(x)+\varphi_v(x)\) — the initial conditions add as well.
STEP 5 · the general principle
For a linear problem, everything superposes: the equation, the boundary data, and the initial data. This is what lets you build the solution of a complicated problem out of a catalogue of simple ones — and it is the entire reason a Fourier series is useful.
8
Solve \(u_t = u_{xx}\) on \(0
Hint 1Each mode carries \(e^{-n^2t}\).
Hint 2Solve \(e^{-25t}=0.01\).
Solution
STEP 1 · read the coefficients
\(b_1=5\), \(b_2=-3\), \(b_5=1\); all others zero — the data is already an eigenfunction combination.
STEP 2 · attach the decay factors
\[ u(x,t) = 5e^{-t}\sin x - 3e^{-4t}\sin2x + e^{-25t}\sin5x \]
STEP 3 · verify
Each term solves the PDE and vanishes at \(x=0,\pi\) ✓, and \(t=0\) reproduces the data ✓
STEP 4 · solve for the 1% time
\[ e^{-25t} = 0.01 \quad\Longrightarrow\quad t = \frac{\ln 100}{25} = \frac{4.605}{25} \approx 0.184 \]
STEP 5 · compare the modes
At that same instant the fundamental has only decayed to \(e^{-0.184}\approx83\%\) of its start. The fine \(n=5\) ripple is gone while the broad hump is barely touched — the \(n^2\) law making the solution smooth.
9
Show that superposition fails for \(u_t = u_{xx} + u^2\).
Hint 1Suppose \(u\) and \(v\) both solve it, and substitute \(w=u+v\).
Hint 2Expand \((u+v)^2\) and see which terms are unaccounted for.
Solution
STEP 1 · assume both solve it
\[ u_t = u_{xx}+u^2, \qquad v_t = v_{xx}+v^2 \]
STEP 2 · add the two equations
\[ (u+v)_t = (u+v)_{xx} + u^2+v^2 \]
STEP 3 · compare with what w = u+v would need
\[ w_t = w_{xx} + w^2 = w_{xx} + u^2 + 2uv + v^2 \]
STEP 4 · identify the discrepancy
The sum falls short by the cross term \(2uv\), which is not generally zero. So \(u+v\) does not solve the equation.
STEP 5 · what changes as a result
No Fourier method, no mode-by-mode decay, no "sum of particular solutions". And genuinely new behavior appears: this equation can blow up in finite time — solutions reaching infinity at a finite \(t\), which no linear heat equation ever does. Nonlinearity is not a technical inconvenience; it changes what can happen.
Part II · Module 1 Characteristics
The transport equation
The simplest PDE with anything to teach: \(u_t + cu_x = 0\). Its entire content is one sentence — the initial profile slides sideways at speed \(c\), unchanged — and understanding why gives you the method of characteristics for free.
①Reading the equation as a directional derivative
Follow an observer moving with velocity \(c\), so their position is \(x(t) = x_0 + ct\). What does the solution look like to them? Differentiate along their path with the chain rule:
The PDE says exactly that this derivative vanishes: \(u\) is constant along every line \(x - ct = \text{const}\). Those lines are the characteristics, and the whole solution follows:
InteractiveProfile and characteristicschange \(c\); press play
u at x = 2 —that value came from x₀ = —travelling —
Top: the profile sliding. Bottom: the \(x\)–\(t\) plane with the characteristic lines \(x - ct = \text{const}\), and your current time marked. Every value of \(u\) simply rides up its own line, unchanged. Notice the square pulse keeps its corners perfectly — no smoothing whatsoever, exactly as a hyperbolic equation should behave.
②Why this is the whole idea
The method generalizes immediately. Any first-order equation of the form
General linear first-order PDE
\[ a(x,t)\,u_x + b(x,t)\,u_t = 0 \]
says that the directional derivative of \(u\) along the vector field \((a,b)\) is zero. Solve the ODE \(\frac{dx}{dt} = \frac{a}{b}\) to find the curves that field traces out, and \(u\) is constant on each one. A PDE becomes an ODE — that is the trade the whole of Part II is built on.
Direction of travel
\(u_t + cu_x = 0\) with \(c>0\) moves the profile right: the solution \(\varphi(x-ct)\) has its features shifted to larger \(x\) as \(t\) grows. A common slip is to read the minus sign backwards. Sanity-check with a bump at the origin: at \(t=1\) it should sit at \(x=c\).
Worked example · transport with a source
Solve \(u_t + 2u_x = u\), \(u(x,0) = e^{-x^2}\).
STEP 1
The left side is still a derivative along characteristics. Parametrize a characteristic by \(x = x_0 + 2t\) and define \(U(t) = u(x_0+2t,\ t)\).
STEP 2
Differentiate:
\[ \frac{dU}{dt} = 2u_x + u_t = u = U \]
The PDE has become an ODE along each characteristic.
Convert back. A point \((x,t)\) sits on the characteristic that started at \(x_0 = x - 2t\), so
\[ u(x,t) = e^{t}\,e^{-(x-2t)^2} \]
STEP 5
Read it: the bump still travels right at speed 2, but now grows exponentially as it goes. The source term changed what happens along the characteristics without changing where they go.
③Practice
1
Solve \(u_t + 3u_x = 0\) with \(u(x,0) = \frac{1}{1+x^2}\), and find \(u(6,2)\).
Hint 1The solution is the initial profile with \(x\) replaced by \(x-ct\).
Hint 2Here \(c=3\), so evaluate \(\varphi\) at \(x-3t\).
Solution
STEP 1 · identify the speed
Comparing with \(u_t + cu_x = 0\) gives \(c = 3\): the profile moves right at 3 units per unit time.
A negative speed means the profile moves left at 4 units per unit time. Sanity check: the zero of \(\sin\) initially at \(x=0\) satisfies \(x+4t=0\), i.e. \(x = -4t\) — moving left ✓
STEP 4 · the characteristic through (1,2)
Characteristics are \(x + 4t = \text{const}\). Through \((1,2)\):
\[ x + 4t = 1 + 8 = 9 \quad\Longrightarrow\quad x = 9 - 4t \]
STEP 5 · use it
Along that line \(u\) is constant, equal to its value at \(t=0\), where \(x=9\): \(u = \sin 9 \approx 0.412\). Check directly: \(u(1,2) = \sin(1+8) = \sin 9\) ✓
3
Solve \(u_t + u_x = -2u\), \(u(x,0) = \cos x\), by reducing to an ODE along characteristics.
Hint 1Set \(U(t) = u(x_0+t,\,t)\) and differentiate with the chain rule.
Hint 2You should get \(U' = -2U\), whose solution is exponential decay.
Hint 3Finally replace \(x_0\) by \(x - t\).
Solution
STEP 1 · ride a characteristic
Characteristics of \(u_t+u_x\) are \(x = x_0 + t\). Define \(U(t) = u(x_0+t,\,t)\).
The wave still travels right at speed 1, while its amplitude decays like \(e^{-2t}\) — transport and damping act independently, because the source term only affects motion along characteristics.
4
A pollutant is released as a square pulse into a river flowing at 2 m/s. Where is it after 30 s, and what does the transport model fail to capture?
Hint 1Pure transport moves the pulse rigidly.
Hint 2What physical mechanism would blur the sharp edges in reality?
Solution
STEP 1 · the model
\[ u_t + 2u_x = 0 \quad\Longrightarrow\quad u(x,t) = \varphi(x - 2t) \]
STEP 2 · locate the pulse
After \(t=30\) s the entire pulse has shifted \(2\times30 = 60\) m downstream, with its width and height exactly unchanged.
STEP 3 · what is missing
Real pollutant also diffuses: molecular and turbulent mixing smear the sharp edges, lowering the peak and widening the pulse as it travels.
STEP 4 · the better model
\[ u_t + 2u_x = k\,u_{xx} \]
the advection–diffusion equation of module F2. The pulse still arrives centered at 60 m, but as a spread-out Gaussian-ish hump rather than a crisp box.
STEP 5 · why the simple model still matters
The arrival time depends only on advection, so \(u_t + cu_x=0\) predicts when the plume reaches a town perfectly well. Diffusion controls the peak concentration — which is what you need for a safety threshold. Different questions need different terms.
5
Solve \(u_t + cu_x = 0\) on \(x>0,\ t>0\) with \(u(x,0)=0\) and boundary data \(u(0,t) = g(t)\), assuming \(c>0\). Why would this problem be ill-posed if \(c<0\)?
Hint 1Characteristics still carry values, but now some of them enter through the boundary \(x=0\) rather than the initial line.
Hint 2A characteristic through \((x,t)\) hits \(x=0\) at time \(t - x/c\).
Hint 3For \(c<0\), which way do characteristics cross the boundary — in or out of the domain?
Solution
STEP 1 · trace the characteristic backwards
Through \((x,t)\) the line is \(x - c\tau\) with \(\tau\) decreasing. It reaches \(x=0\) when \(\tau = t - x/c\).
STEP 2 · split into two cases
If \(t > x/c\), the characteristic entered through the boundary and carries the value \(g(t - x/c)\). If \(t < x/c\), it came from the initial line, where \(u=0\).
STEP 3 · write the solution
\[ u(x,t) = \begin{cases} g\!\left(t - \dfrac{x}{c}\right), & x < ct \\[6pt] 0, & x > ct \end{cases} \]
STEP 4 · interpret
Whatever you inject at the left end travels right at speed \(c\) and arrives at position \(x\) after a delay \(x/c\). Ahead of the front \(x = ct\), nothing has arrived yet — a clean demonstration of finite propagation speed.
STEP 5 · why c < 0 breaks it
With \(c<0\), characteristics travel leftward, so at \(x=0\) they are exiting the domain \(x>0\), not entering it. Values there are already determined by the initial data, and prescribing \(g\) as well over-determines the problem — generally contradicting it. Boundary conditions must be imposed only where characteristics flow inward, a rule that governs inflow/outflow boundaries in every fluid solver.
6
Solve \(u_t+4u_x=0\) with \(u(x,0)=x^2\), and evaluate \(u(3,0.5)\).
Hint 1Replace \(x\) by \(x-ct\) in the initial profile.
The value 1 was sitting at \(x_0=1\) at \(t=0\) and has ridden its characteristic to \(x=3\) ✓
STEP 5 · note the shape
The parabola keeps its shape exactly, sliding right — the vertex moves from \(x=0\) to \(x=4t\). Unbounded data causes no difficulty: transport neither smooths nor amplifies.
7
Solve the two-dimensional transport equation \(u_t + 2u_x + 3u_y = 0\) with \(u(x,y,0)=f(x,y)\).
Hint 1The equation says the derivative along the direction \((2,3)\) in space, moving forward in time, is zero.
Hint 2Characteristics: \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=3\).
Solution
STEP 1 · the characteristic equations
\[ \frac{dx}{dt}=2, \qquad \frac{dy}{dt}=3 \quad\Longrightarrow\quad x = x_0+2t,\ \ y = y_0+3t \]
STEP 2 · u is constant along them
Along a characteristic, \(\frac{du}{dt} = u_t + 2u_x + 3u_y = 0\).
The whole two-dimensional pattern translates rigidly with velocity vector \((2,3)\), at speed \(\sqrt{13}\), without distortion. Everything from module C1 carries over unchanged — the dimension of the space makes no difference to the method.
8
A pulse governed by \(u_t+cu_x=0\) is photographed at \(t=0\) centred at \(x=2\), and again at \(t=1.5\) centred at \(x=8\). Find \(c\), and predict where it is at \(t=4\).
Hint 1The centre rides a characteristic, so it moves at constant speed.
Hint 2Speed = distance ÷ time.
Solution
STEP 1 · use the characteristic through the peak
The peak satisfies \(x = 2 + ct\).
STEP 2 · fit the second observation
\[ 8 = 2 + c(1.5) \quad\Longrightarrow\quad c = \frac{6}{1.5} = 4 \]
STEP 3 · predict
\[ x(4) = 2 + 4(4) = 18 \]
STEP 4 · what else the model asserts
Not just the position: the pulse should have exactly the same height and width at \(t=4\) as at \(t=0\), since transport is shape-preserving.
STEP 5 · how to test the model
If the observed pulse is shorter and wider at \(t=4\), pure transport is the wrong model and a diffusion term is needed: \(u_t+4u_x = ku_{xx}\) (module F2). Measuring the spreading rate then gives you \(k\).
9
Show that every solution of \(u_t+cu_x=0\) has the form \(u = f(x-ct)\), by changing to the variables \(\xi = x-ct\), \(\tau = t\).
Hint 1Apply the chain rule: \(u_t = u_\xi\xi_t + u_\tau\tau_t\) with \(\xi_t = -c\).
STEP 2 · substitute into the PDE
\[ \left(-cu_\xi + u_\tau\right) + c\,u_\xi = u_\tau = 0 \]
STEP 3 · solve the reduced equation
\(u_\tau = 0\) means \(u\) does not depend on \(\tau\) at all:
\[ u = f(\xi) = f(x-ct) \]
for some arbitrary function \(f\).
STEP 4 · note what was proved
Not merely that such functions are solutions (problem 1 did that), but that they are the only ones — the general solution.
STEP 5 · the technique
Choosing coordinates that follow the characteristics turns the PDE into a triviality. It is the same move that reduces hyperbolic equations to \(u_{\xi\eta}=0\) (module F3 problem 8) and underlies d'Alembert's formula.
Part II · Module 2 Characteristics
Method of characteristics
Take the idea from module C1 and let the coefficients vary. The characteristics bend, but the principle survives untouched: find the curves along which the PDE collapses to an ODE, solve that ODE, then thread the answer back through the initial data.
Characteristic ODE. Solve \(\dfrac{dx}{dt} = a(x,t)\) to get the curves, each labelled by where it starts, \(x_0\).
Transport ODE. Along a characteristic, \(\dfrac{du}{dt} = f\). Solve it with initial value \(\varphi(x_0)\).
Invert. Express \(x_0\) in terms of \((x,t)\) and substitute — this is the step where the answer finally becomes a function of \(x\) and \(t\).
InteractiveCharacteristics that bendwatch the profile deform
characteristics —profile —
Bottom: the characteristic curves in the \(x\)–\(t\) plane, with your current time marked. Top: the profile they carry. With \(a=x\) the curves fan apart and the bump is stretched thin; with \(a=-x\) they crowd together and it is compressed and amplified. The values themselves never change — only the spacing of the carriers does, which is what deforms the picture.
Characteristic ODE: \(\dfrac{dx}{dt} = x\), a separable ODE with solution \(x = x_0e^{t}\).
STEP 2
Transport ODE: the right side is 0, so \(\dfrac{du}{dt} = 0\) along each curve, i.e. \(u\) keeps the value it started with, \(\varphi(x_0)\).
STEP 3
Invert: from \(x = x_0e^t\) we get \(x_0 = xe^{-t}\).
STEP 4
Substitute:
\[ u(x,t) = \varphi\!\left(xe^{-t}\right) \]
STEP 5
Check: \(u_t = \varphi'(xe^{-t})\cdot(-xe^{-t})\) and \(xu_x = x\varphi'(xe^{-t})e^{-t}\); they cancel ✓ Interpretation: everything is swept away from the origin exponentially fast, so the profile stretches horizontally without changing its values — a bump of width \(w\) becomes a bump of width \(we^{t}\).
Don't stop before inverting
The most common lost mark: writing "\(u = \varphi(x_0)\), where \(x_0\) is the starting point" and stopping. That is not a solution — it is a description of one. You must express \(x_0\) explicitly in terms of \(x\) and \(t\) and substitute.
②Practice
1
Solve \(u_t + 2t\,u_x = 0\), \(u(x,0)=e^{-x^2}\).
Hint 1Characteristic ODE: \(\frac{dx}{dt} = 2t\). Note the right side depends on \(t\), not \(x\) — so integrate directly.
Parabolas in the \(x\)–\(t\) plane, all congruent, each shifted by its starting point.
STEP 2 · transport along them
The PDE's right side is 0, so \(\frac{du}{dt}=0\): \(u\) is constant on each parabola, equal to \(\varphi(x_0)\).
STEP 3 · invert
\[ x_0 = x - t^2 \]
STEP 4 · substitute
\[ u(x,t) = e^{-(x-t^2)^2} \]
STEP 5 · verify and interpret
\(u_t = -2(x-t^2)(-2t)u\) and \(2tu_x = 2t\cdot(-2)(x-t^2)u\); they sum to zero ✓ The bump keeps its shape exactly but accelerates to the right, since the local speed \(2t\) grows with time.
2
Solve \(u_t - x\,u_x = 0\), \(u(x,0) = \frac{1}{1+x^2}\), and describe what happens to the profile as \(t\to\infty\).
Hint 1\(\frac{dx}{dt} = -x\) gives exponential decay toward the origin.
For any fixed \(x\ne0\), \(x^2e^{2t}\to\infty\) so \(u\to0\). But at \(x=0\), \(u=1\) for all time.
STEP 5 · the picture
The profile is squeezed toward the origin: it becomes an ever-narrower spike of unchanging height 1. Values are never destroyed — the characteristics simply crowd them into a shrinking region. This is the mirror image of the stretching seen in the worked example, and a warning that "the solution tends to 0 pointwise" can hide a spike that never goes away.
3
Solve \(u_t + u_x = x\), \(u(x,0)=0\).
Hint 1Characteristics: \(x = x_0 + t\). Now the transport ODE has a nonzero right side.
Hint 2Along a characteristic, \(\frac{du}{dt} = x = x_0 + t\) — express the right side in terms of \(t\) before integrating.
Hint 3Integrate from 0 to \(t\), using \(u=0\) at \(t=0\).
Solution
STEP 1 · characteristics
\[ \frac{dx}{dt} = 1 \quad\Longrightarrow\quad x = x_0 + t \]
The amplitude is \(e^{-3}\approx0.0498\) everywhere; the maximum occurs where \(\sin(x-1)=1\), i.e.
\[ x - 1 = \frac{\pi}{2} \quad\Longrightarrow\quad x = 1 + \frac{\pi}{2} \approx 2.571 \]
The crest that started at \(\pi/2\) has moved right by 1, while shrinking to 5% of its original height.
5
Explain why the method of characteristics does not work for \(u_t = u_{xx}\), and what that reveals about the difference between first- and second-order equations.
Hint 1The method needs the PDE to be a directional derivative — a combination of first derivatives only.
Hint 2Recall the classification: how many real characteristic directions does a parabolic equation have?
Solution
STEP 1 · what the method requires
Characteristics work because \(a u_x + u_t\) is exactly \(\frac{d}{dt}\) along a curve — a first-order directional derivative. There is no curve along which \(u_{xx}\) becomes an ordinary derivative of \(u\).
STEP 2 · the classification view
From module F3, the heat equation is parabolic with a single repeated characteristic family, \(t = \text{const}\). Those lines carry no information from point to point — they are the time slices themselves, not paths of propagation.
STEP 3 · the physical difference
First-order equations move information along curves at finite speed, so the value at \((x,t)\) depends on exactly one initial point. In the heat equation, \(u(x,t)\) depends on the entire initial profile — the heat kernel of module A3 gives every \(x_0\) a nonzero weight.
STEP 4 · what to use instead
Parabolic and elliptic problems need global methods: separation of variables and Fourier series (Part III), or integral transforms (module A3). Characteristics are for hyperbolic problems — the wave equation's d'Alembert formula (module S6) is precisely the second-order version of this same idea.
THE TAKEAWAY
Match the method to the type. Hyperbolic ⟹ characteristics. Parabolic/elliptic ⟹ eigenfunction expansions or transforms. Choosing the method is the classification question in disguise.
6
Solve \(u_t + e^t\,u_x = 0\) with \(u(x,0)=\varphi(x)\).
Hint 1The speed depends on \(t\) only, so integrate \(\frac{dx}{dt}=e^t\) directly.
Hint 2\(x = x_0 + e^t - 1\), since \(x=x_0\) at \(t=0\).
Solution
STEP 1 · characteristic curves
\[ \frac{dx}{dt} = e^t \quad\Longrightarrow\quad x = e^t + K \]
At \(t=0\), \(x=x_0\), so \(K = x_0-1\) and \(x = x_0 + e^t - 1\).
\(u_t = -e^t\varphi'\), \(u_x = \varphi'\), so \(u_t + e^tu_x = 0\) ✓ The profile keeps its shape but accelerates away exponentially — after \(t=5\) it has moved nearly 150 units. All the characteristics stay parallel, so nothing distorts.
7
Solve \(u_t + x\,u_x = -u\) with \(u(x,0)=\varphi(x)\).
Hint 1Characteristics from \(\frac{dx}{dt}=x\), as in the worked example.
Hint 2Along them, \(\frac{du}{dt}=-u\).
Solution
STEP 1 · characteristics
\[ \frac{dx}{dt} = x \quad\Longrightarrow\quad x = x_0e^t \]
STEP 2 · transport ODE along a characteristic
\[ \frac{dU}{dt} = u_t + xu_x = -u = -U \quad\Longrightarrow\quad U(t) = \varphi(x_0)e^{-t} \]
\(u_t = -e^{-t}\varphi - e^{-t}\varphi'\cdot xe^{-t}\) and \(xu_x = xe^{-t}\varphi'e^{-t}\); adding gives \(-e^{-t}\varphi = -u\) ✓ Two effects at once: the profile is stretched horizontally (from the characteristics fanning out) and damped vertically (from the source term).
8
Find the general solution of the steady equation \(a\,u_x + b\,u_y = 0\) with constants \(a,b\), and describe its level curves.
Hint 1The equation says the directional derivative of \(u\) along the vector \((a,b)\) vanishes.
Hint 2Along which lines is \(u\) therefore constant?
Solution
STEP 1 · read it as a directional derivative
\[ au_x+bu_y = (a,b)\cdot\nabla u = 0 \]
So \(u\) does not change as you move in the direction \((a,b)\).
STEP 2 · find the characteristics
Lines with direction \((a,b)\) satisfy \(\frac{dy}{dx}=\frac ba\), i.e. \(bx - ay = \) constant.
STEP 3 · write the general solution
\[ u(x,y) = f(bx-ay) \]
The level curves are the parallel lines \(bx-ay=\text{const}\), and \(u\) is constant on each. The solution is a "corrugated sheet" — it varies only across the characteristic direction, never along it.
9
For \(u_x + u_y = 0\), explain why prescribing data along the line \(y=x\) is a badly posed problem.
Hint 1Find the characteristics and compare their direction with the data line.
Hint 2What must be true of \(u\) along a characteristic?
Solution
STEP 1 · find the characteristics
Direction \((1,1)\), so the characteristics are the lines \(y = x + \text{const}\) — and the data line \(y=x\) is itself one of them.
STEP 2 · what the PDE forces on that line
\(u\) must be constant along any characteristic. So the only admissible data on \(y=x\) is a constant function.
STEP 3 · non-constant data ⟹ no solution
If you prescribe, say, \(u(x,x)=x\), you are demanding that \(u\) both change and not change along the same line. Existence fails.
STEP 4 · constant data ⟹ no uniqueness
If you prescribe \(u(x,x)=5\), then every function \(f(x-y)\) with \(f(0)=5\) works. Infinitely many solutions. Uniqueness fails.
STEP 5 · the rule
Never give data on a characteristic curve. Data must be prescribed on a non-characteristic curve — one that the characteristics cross transversally, so each carries information away from exactly one data point. Choosing \(y=0\) here works perfectly.
Part II · Module 3 Characteristics
Nonlinear waves & shocks
Let the wave speed depend on the wave's own height and something remarkable happens: tall parts travel faster, overtake the slow parts, and the profile steepens until it goes vertical. A smooth start produces a discontinuity in finite time — the mathematical birth of a shock wave.
①Burgers' equation
The inviscid Burgers equation
\[ u_t + u\,u_x = 0 \]
Compare with \(u_t + cu_x = 0\): the constant speed \(c\) has been replaced by \(u\) itself. Characteristics still carry constant values, but now each one moves at the speed of the value it carries:
Characteristics of Burgers' equation
\[ \frac{dx}{dt} = u \quad\text{and}\quad \frac{du}{dt} = 0 \quad\Longrightarrow\quad x = x_0 + \varphi(x_0)\,t, \qquad u = \varphi(x_0) \]
The characteristics are still straight lines — but with different slopes. Wherever \(\varphi\) is decreasing, faster values sit behind slower ones and the lines must eventually cross.
InteractiveSteepening into a shockrun time forward and watch the crossing
breaking time t* —status —shock speed —
Bottom: characteristics with slopes set by the initial height. Where they converge and cross, the solution tries to become multivalued — physically impossible, so a jump forms instead. Top: the profile steepening. The dashed vertical line marks the shock once it exists; before that, the solution is perfectly smooth. Switch to the increasing ramp to see the opposite case, where characteristics fan out and leave a gap that a rarefaction fills.
②Breaking time and shock speed
Two characteristics starting at \(x_0\) and \(x_0+\delta\) collide when their positions coincide. Working out the first time any pair crosses gives:
After \(t^*\) the classical solution ceases to exist, and you must allow a jump. Conservation determines how fast that jump travels:
Rankine–Hugoniot condition
\[ s = \frac{[\,\text{flux}\,]}{[\,u\,]} = \frac{f(u_L)-f(u_R)}{u_L-u_R}, \qquad f(u)=\tfrac12u^2 \ \Longrightarrow\ s = \frac{u_L+u_R}{2} \]
For Burgers' equation the shock simply travels at the average of the values on either side.
Worked example · when does it break, and where does the jump go?
Take \(u_t + uu_x=0\) with \(\varphi(x)=1-x\) on \(0\le x\le1\), \(\varphi=1\) for \(x<0\), \(\varphi=0\) for \(x>1\).
STEP 1
On the ramp, \(\varphi' = -1\) everywhere, so the minimum slope is \(-1\) and
\[ t^* = \frac{-1}{-1} = 1 \]
STEP 2
Check directly: the characteristic from \(x_0=0\) carries \(u=1\) and moves as \(x = t\); the one from \(x_0=1\) carries \(u=0\) and stays at \(x=1\). They meet when \(t=1\), at \(x=1\) ✓
STEP 3
For \(t<1\) the solution is smooth. Solving \(x = x_0 + (1-x_0)t\) for \(x_0\) gives \(x_0 = \frac{x-t}{1-t}\), so
The slope \(\frac{-1}{1-t}\) blows up as \(t\to1^-\): the profile goes vertical exactly at the breaking time.
STEP 4
For \(t>1\) insert a shock with \(u_L=1\) on the left and \(u_R=0\) on the right. Rankine–Hugoniot gives
\[ s = \frac{1+0}{2} = \frac12 \]
STEP 5
So the jump starts at \(x=1,\ t=1\) and moves right at speed \(\frac12\): \(x_{\text{shock}}(t) = 1 + \frac{t-1}{2}\). Note it travels slower than the fast material behind it and faster than the slow material ahead — characteristics run into it from both sides, which is exactly the entropy condition that makes the shock physically admissible.
Increasing data does the opposite
If \(\varphi\) is increasing, characteristics diverge and never cross — \(t^*=\infty\), no shock. Instead a gap opens where no characteristic reaches, and the physical solution fills it with a smooth rarefaction fan \(u = x/t\). Compression makes shocks; expansion makes fans.
③Practice
1
For \(u_t+uu_x=0\) with \(\varphi(x) = 2 - x\), find the breaking time and the solution for \(t
Hint 1\(t^* = -1/\min\varphi'\), and here \(\varphi'\) is constant.
Hint 2Characteristics: \(x = x_0 + (2-x_0)t\). Solve this for \(x_0\).
Solution
STEP 1 · breaking time
\[ \varphi'(x_0) = -1 \ \text{ for all } x_0 \quad\Longrightarrow\quad t^* = \frac{-1}{-1} = 1 \]
STEP 4 · substitute into u = φ(x₀)
\[ u(x,t) = 2 - \frac{x-2t}{1-t} = \frac{2(1-t) - x + 2t}{1-t} = \frac{2-x}{1-t} \]
STEP 5 · verify and interpret
At \(t=0\), \(u = 2-x\) ✓. The spatial slope is \(\frac{-1}{1-t}\), which \(\to-\infty\) as \(t\to1^-\) — the profile rotates toward vertical and the classical solution dies exactly at \(t^*=1\), as predicted.
2
Find the shock speed for Burgers' equation with \(u_L = 4\) and \(u_R = 1\), and check that characteristics run into the shock from both sides.
Hint 1For \(f(u)=\frac12u^2\), Rankine–Hugoniot simplifies to the average of the two states.
Hint 2Characteristic speed on each side equals the value of \(u\) there. Compare each with \(s\).
STEP 2 · check with the shortcut
\[ s = \frac{u_L+u_R}{2} = \frac{4+1}{2} = 2.5 \ ✓ \]
STEP 3 · characteristic speeds
Left of the shock, characteristics move at speed \(u_L = 4\); right of it, at \(u_R = 1\).
STEP 4 · compare with the shock
\[ u_L = 4 > s = 2.5 > u_R = 1 \]
The left characteristics catch up to the shock and the shock catches up to the right ones: they collide with it from both sides ✓
STEP 5 · why this matters
That inequality \(u_L > s > u_R\) is the Lax entropy condition. It selects the physically meaningful discontinuity — the one where information is destroyed, not created. A jump with \(u_L < u_R\) would satisfy Rankine–Hugoniot too, but it is inadmissible; nature builds a rarefaction fan there instead.
3
Solve the rarefaction problem: \(u_t+uu_x=0\) with \(u=0\) for \(x<0\) and \(u=1\) for \(x>0\).
Hint 1Draw the characteristics: vertical lines on the left, slope-1 lines on the right. What happens in the wedge between them?
Hint 2No characteristic enters the region \(0
Hint 3Look for a solution depending only on the ratio \(x/t\).
Solution
STEP 1 · draw the characteristics
From \(x_0<0\) they carry \(u=0\) and stand still: \(x = x_0\). From \(x_0>0\) they carry \(u=1\) and move as \(x = x_0+t\). The wedge \(0 < x < t\) is reached by neither family.
STEP 2 · look for a self-similar fill
The problem has no length scale, so try \(u = g(x/t)\). Substituting into \(u_t+uu_x=0\):
STEP 3 · take the nonconstant branch
\[ g\!\left(\frac xt\right) = \frac{x}{t} \quad\Longrightarrow\quad u = \frac{x}{t} \ \text{ in the wedge} \]
STEP 4 · assemble the full solution
\[ u(x,t) = \begin{cases} 0, & x \le 0\\[2pt] \dfrac{x}{t}, & 0 < x < t\\[6pt] 1, & x \ge t \end{cases} \]
STEP 5 · check continuity and meaning
At \(x=0\) it gives 0 and at \(x=t\) it gives 1 — continuous ✓ The initial jump immediately spreads into a widening linear ramp, the rarefaction fan. Physically: fast material ahead runs away from slow material behind, so the profile expands rather than compresses. (Traffic-flow reading: a green light, where the queue disperses smoothly.)
4
Find the breaking time for \(\varphi(x) = e^{-x^2}\), and say which part of the bump shocks first.
Hint 1You need the most negative value of \(\varphi'\).
Hint 2\(\varphi' = -2xe^{-x^2}\). Minimize it by setting its derivative to zero.
Hint 3\(\varphi'' = (4x^2-2)e^{-x^2} = 0\) at \(x = 1/\sqrt2\).
Solution
STEP 1 · differentiate the data
\[ \varphi'(x) = -2xe^{-x^2} \]
Negative for \(x>0\) — the back (right-hand, downhill) side of the bump.
STEP 2 · find its minimum
\[ \varphi''(x) = (4x^2-2)e^{-x^2} = 0 \quad\Longrightarrow\quad x = \frac{1}{\sqrt2} \approx 0.707 \]
The shock forms on the trailing (right) side of the bump, at the point of steepest descent. The crest travels fastest (speed 1) and overtakes the low material ahead of it, so the front face steepens into a jump while the rear face stretches into a gentle ramp. Every nonlinear wave with a hump does this: sharpen in front, spread behind — a breaking ocean wave in one line of calculus.
5
In the traffic model \(\rho_t + f(\rho)_x = 0\) with \(f(\rho) = \rho(1-\rho)\), find the wave speed \(f'(\rho)\) and explain why traffic jams travel backwards.
Hint 1For a conservation law \(\rho_t + f(\rho)_x=0\), characteristics move at speed \(f'(\rho)\), not \(\rho\).
Hint 2Differentiate \(f(\rho)=\rho-\rho^2\).
Hint 3When is \(f'(\rho)\) negative?
Solution
STEP 1 · expand the chain rule
\(f(\rho)_x = f'(\rho)\rho_x\), so the equation is \(\rho_t + f'(\rho)\rho_x = 0\) — transport at speed \(f'(\rho)\).
STEP 3 · find where it is negative
\[ 1-2\rho < 0 \iff \rho > \tfrac12 \]
Above half the maximum density, the wave speed is negative.
STEP 4 · interpret
Cars always move forward (their velocity is \(f/\rho = 1-\rho \ge 0\)), but the disturbance — the region of high density — propagates upstream. Each driver brakes slightly before the car ahead did, so the brake-light wave travels backwards through the queue even as every vehicle inches forward.
STEP 5 · shock version
Rankine–Hugoniot for a jam front between \(\rho_L\) and \(\rho_R\) gives \(s = \frac{f(\rho_L)-f(\rho_R)}{\rho_L-\rho_R} = 1 - \rho_L - \rho_R\). With light traffic ahead \((\rho_R = 0.2)\) meeting a jam \((\rho_L=0.9)\), \(s = -0.1\): the back of the jam creeps upstream at a tenth of the free-flow speed — exactly the phantom traffic jam you experience on a motorway.
6
Find the breaking time for \(u_t+uu_x=0\) with \(\varphi(x)=\cos x\) on \([0,\pi]\), and say where the shock first appears.
Hint 1\(t^*=-1/\min\varphi'\), so you need the most negative slope.
Hint 2\(\varphi'=-\sin x\), whose minimum on \([0,\pi]\) is at \(x=\pi/2\).
Solution
STEP 1 · differentiate the data
\[ \varphi'(x) = -\sin x \]
STEP 2 · find its minimum
On \([0,\pi]\), \(\sin x\) is largest at \(x = \frac{\pi}{2}\), so \(\varphi'\) is most negative there:
\[ \min\varphi' = -1 \ \text{ at } x = \tfrac{\pi}{2} \]
The first crossing happens on the characteristic from \(x_0=\frac\pi2\), which carries \(u = \cos\frac\pi2 = 0\) and therefore does not move. So the shock forms at
\[ x = \frac{\pi}{2}, \qquad t = 1 \]
STEP 5 · the picture
The steep descending portion of the cosine steepens until it is vertical exactly at the inflection point. Compare problem 4's Gaussian, where the shock also formed at the steepest descending point — that is the general rule.
7
In the traffic model with \(f(\rho)=\rho(1-\rho)\), a light turns green: \(\rho = 1\) for \(x<0\) (jam) and \(\rho=0\) for \(x>0\) (empty road). Describe the solution.
Hint 1Wave speed is \(f'(\rho)=1-2\rho\). Compute it on each side.
Hint 2Left speed \(-1\), right speed \(+1\) — do the characteristics converge or diverge?
Hint 3A gap opens; fill it with a self-similar fan \(f'(\rho)=x/t\).
The front of the queue accelerates away at speed 1 while the back of the queue starts moving only at speed \(-1\) — the "start-up wave" travelling backwards through the line of cars. This is exactly the everyday experience of waiting several seconds after the light turns green before your own car moves.
8
Show that \(u_t + uu_x=0\) and \(u_t + \left(\tfrac12u^2\right)_x = 0\) are the same equation for smooth solutions, and explain why only the second is correct once a shock forms.
Hint 1Expand the flux derivative with the chain rule.
Hint 2Which form can be integrated across a jump?
Solution
STEP 1 · expand the conservative form
\[ \left(\tfrac12u^2\right)_x = u\,u_x \]
so for differentiable \(u\) the two forms are identical.
STEP 2 · what changes at a jump
At a shock \(u_x\) does not exist, so \(uu_x\) is meaningless. But the conservative form can be integrated over a control volume containing the jump, which requires no derivative of \(u\) at all.
STEP 3 · that integral gives the jump condition
Integrating \(u_t + f(u)_x=0\) across the discontinuity yields
\[ s\left(u_L-u_R\right) = f(u_L)-f(u_R) \]
the Rankine–Hugoniot condition — obtainable only from the flux form.
STEP 4 · why it matters practically
Different-looking but equivalent flux functions give different shock speeds. Writing \(u_t + uu_x=0\) as \(\left(\frac{u^2}{2}\right)_x\) versus, say, multiplying the equation by \(u\) first and conserving \(u^2\) instead, produces genuinely different weak solutions. The physics decides which quantity is conserved.
STEP 5 · the numerical consequence
Shock-capturing schemes are always written in conservative (flux-difference) form. A non-conservative discretization converges to the wrong shock speed — a classic and expensive mistake in computational fluid dynamics.
9
For the ramp problem (\(u_L=1\), \(u_R=0\), shock starting at \(x=1,t=1\), speed \(\tfrac12\)), verify that the shock position conserves total mass.
Hint 1Mass is \(\int u\,dx\), and for this data it should stay constant.
Hint 2Compare the area under the multivalued "would-be" solution with the area under the jump — the equal-area rule.
Solution
STEP 1 · initial mass
With \(u=1\) for \(x<0\) and the ramp \(1-x\) on \([0,1]\), the mass over \([0,\infty)\) is
\[ \int_0^1(1-x)dx = \frac12 \]
STEP 2 · mass at time t > 1
The solution is \(u=1\) up to the shock at \(x_s(t) = 1+\frac{t-1}{2}\), then 0. Relative to the same reference,
\[ \int_0^{x_s}1\,dx - \text{(the part that was already there)} \]
Concretely, at \(t=1\) the shock sits at \(x=1\) and the profile is \(u=1\) on \([0,1]\) — mass 1. But the ramp had mass \(\frac12\).
STEP 3 · resolve the bookkeeping
Mass is conserved on the whole line, not on \([0,\infty)\): material flows in from \(x<0\). Track instead the total to the left of a moving front. The shock speed \(s=\frac{u_L+u_R}{2}\) is derived from \(\frac{d}{dt}\int u = -\left[f(u)\right]\), so conservation holds by construction.
STEP 4 · the equal-area interpretation
Without a shock, characteristics would give a multivalued S-shaped profile. Inserting the jump so that the two "overhang" lobes have equal area removes exactly as much as it adds — and that position is precisely \(x_s = 1+\frac{t-1}{2}\).
STEP 5 · the check that matters
Rankine–Hugoniot is the statement that mass crossing the jump balances. So verifying \(s = \frac{f(u_L)-f(u_R)}{u_L-u_R} = \frac12\) ✓ is the conservation check — the equal-area rule is its geometric picture.
Parts I & II · Checkpoint Quiz
Quiz A — Foundations & characteristics
Ten questions on classification, conditions, and first-order methods. Tap an answer to lock it in and see the reasoning. Aim for 8/10 before starting Part III.
Question 1
Classify \(u_t = 5u_{xx} + \sin x\).
Why
\(u_{xx}\) makes it second order. \(u\) appears only to the first power, so it is linear — \(\sin x\) is a function of the independent variable, which never affects linearity. But \(u\equiv0\) gives \(0=\sin x\), which fails, so it is nonhomogeneous.
Question 2
What is the type of \(4u_{xx} + 4u_{xy} + u_{yy} = 0\)?
Why
\(A=4, B=4, C=1\), so \(\Delta = 16 - 16 = 0\) — parabolic. It is a perfect square, \((2\partial_x+\partial_y)^2u=0\), which is why it has one repeated characteristic direction.
Question 3
The solution of \(u_t + 5u_x = 0\) with \(u(x,0)=\varphi(x)\) is:
Why
The profile translates rigidly to the right at speed 5, so a point at \((x,t)\) carries the value that started at \(x-5t\). Choice B is the same equation with the sign of the speed flipped.
Question 4
How many conditions does \(u_{tt}=c^2u_{xx}\) on \(0
Why
Two \(t\)-derivatives ⟹ two initial conditions (\(u\) and \(u_t\) at \(t=0\)); two \(x\)-derivatives on a finite interval ⟹ one boundary condition at each end. The heat equation, with only one \(t\)-derivative, is choice B.
Question 5
\(u_x(0,t)=0\) physically means:
Why
Flux is \(-ku_x\) by Fourier's law, so zero gradient means zero flux: perfect insulation. Choice A describes the Dirichlet condition \(u(0,t)=0\), which is a completely different physical setup.
Question 6
For \(u_t + xu_x = 0\), the characteristics satisfy:
Why
The characteristic ODE is \(\frac{dx}{dt} = x\), whose solution is exponential growth \(x = x_0e^t\). Choice D would come from \(u_t - xu_x = 0\).
Question 7
Why does the heat equation destroy a corner in the initial data while the wave equation preserves it?
Why
\(u_t = ku_{xx}\) drives velocity by curvature, so the enormous curvature at a corner rounds it off immediately. \(u_{tt}=c^2u_{xx}\) drives acceleration instead, and its solutions \(F(x+ct)+G(x-ct)\) are rigid travelling shapes — a translated corner is still a corner.
Question 8
For \(u_t + uu_x=0\) with \(\varphi(x)=1-x\) on \([0,1]\), the breaking time is:
Why
\(t^* = -1/\min\varphi'\), and \(\varphi' = -1\) throughout, giving \(t^*=1\). Decreasing data always produces a finite breaking time; increasing data (choice C's situation) never does.
Question 9
A shock in Burgers' equation separating \(u_L=6\) from \(u_R=2\) travels at speed:
Why
Rankine–Hugoniot with \(f=\frac12u^2\) gives \(s = \frac{u_L+u_R}{2} = 4\). Note \(6 > 4 > 2\): characteristics run into the shock from both sides, so it satisfies the entropy condition and is physically admissible.
Question 10
The backward heat equation \(u_t = -u_{xx}\) is ill-posed because:
Why
\(u = e^{n^2t}\sin nx\) starts with amplitude 1 and reaches \(e^{n^2}\) at \(t=1\). Continuous dependence on the data fails, which is Hadamard's third requirement — solutions may still exist, but they are useless numerically because round-off supplies exactly those wiggles.
Part III · Module 1 Separation & Fourier
Fourier series
Separation of variables produces solutions built from \(\sin\frac{n\pi x}{L}\) — one for each \(n\). To match an arbitrary initial shape you need to write that shape as a combination of them. Fourier's astonishing claim is that you always can.
Notice \(\sin\frac{n\pi x}{L}\) vanishes at \(x=0\) and \(x=L\) automatically, while \(\cos\frac{n\pi x}{L}\) has zero slope there. The boundary conditions choose the series — not the other way round.
Multiply the series by \(\sin\frac{m\pi x}{L}\) and integrate: every term dies except the \(n=m\) one, which leaves \(b_m\cdot\frac{L}{2}\). Solving gives the formula. It is exactly how you extract one component of a vector by dotting with a basis vector — the same idea, in a space of functions.
InteractiveBuilding a shape out of sinesadd terms; watch Gibbs
bₙ formula —max error —overshoot at jump —
Amber is the target, cyan is the partial sum. For the smooth triangle the fit improves everywhere. For a jump, watch the Gibbs phenomenon: the overshoot near the discontinuity refuses to shrink below about 9% of the jump no matter how many terms you add — it only gets narrower. Also notice which coefficients vanish: the square wave's even harmonics are all zero, by symmetry.
Convergence, and what happens at a jump
For a piecewise-smooth \(f\), the Fourier series converges to \(f(x)\) wherever \(f\) is continuous, and to the midpoint \(\frac{f(x^-)+f(x^+)}{2}\) at a jump. The series can only produce what its terms allow: a sine series always converges to 0 at \(x=0\) and \(x=L\), whatever the original function did there.
Worked example · sine series of a constant
Find the sine series of \(f(x)=1\) on \([0,\pi]\).
Sanity check the endpoints: every term vanishes at \(x=0\) and \(x=\pi\), so the series gives 0 there — not 1. That is not an error: the odd extension of \(f\) jumps from \(-1\) to \(+1\) across the origin, and the series converges to the midpoint, 0. Setting \(x=\pi/2\) recovers Leibniz's formula \(\frac{\pi}{4} = 1-\frac13+\frac15-\cdots\) ✓
③Practice
1
Find the sine series of \(f(x)=x\) on \([0,L]\).
Hint 1\(b_n = \frac2L\int_0^Lx\sin\frac{n\pi x}{L}dx\) — integrate by parts with \(u=x\).
Hint 2The boundary term gives \(-\frac{L}{n\pi}\cos n\pi\), and \(\int\cos\) over a whole number of half-periods vanishes.
Hint 3\(\cos n\pi = (-1)^n\).
Solution
STEP 1 · set up
\[ b_n = \frac{2}{L}\int_0^L x\sin\frac{n\pi x}{L}\,dx \]
STEP 2 · integrate by parts
\(u=x\), \(dv = \sin\frac{n\pi x}{L}dx\), so \(v = -\frac{L}{n\pi}\cos\frac{n\pi x}{L}\):
STEP 5 · write the series
\[ x = \frac{2L}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin\frac{n\pi x}{L}, \qquad 0Note \(b_n \sim 1/n\), the slow decay typical of a function whose odd extension has a jump (at \(x=L\), where \(f\) jumps from \(L\) down to \(-L\)). That slow decay is exactly why this series shows Gibbs ringing near \(x=L\).
2
Find the cosine series of \(f(x)=x\) on \([0,L]\), and explain why its coefficients decay faster than the sine series of the same function.
Hint 1You need \(a_0\) separately: \(a_0 = \frac2L\int_0^Lx\,dx\).
Hint 2For \(n\ge1\), integrate by parts; the boundary term now vanishes and the surviving piece involves \(\cos n\pi - 1\).
Hint 3Think about whether the even extension of \(x\) has any jumps.
Solution
STEP 1 · the average term
\[ a_0 = \frac2L\int_0^Lx\,dx = \frac2L\cdot\frac{L^2}{2} = L \quad\Longrightarrow\quad \frac{a_0}{2} = \frac{L}{2} \]
which is just the mean value of \(x\) on \([0,L]\) ✓
STEP 2 · integrate by parts for n ≥ 1
\[ a_n = \frac2L\int_0^Lx\cos\frac{n\pi x}{L}dx = \frac2L\left(\left[\frac{Lx}{n\pi}\sin\frac{n\pi x}{L}\right]_0^L - \frac{L}{n\pi}\int_0^L\sin\frac{n\pi x}{L}dx\right) \]
STEP 3 · evaluate
The boundary term vanishes (\(\sin n\pi = 0\)), and the integral gives
\[ a_n = \frac{2L}{n^2\pi^2}\big(\cos n\pi - 1\big) = \begin{cases} -\dfrac{4L}{n^2\pi^2}, & n \text{ odd}\\[4pt] 0, & n \text{ even}\end{cases} \]
STEP 4 · the series
\[ x = \frac{L}{2} - \frac{4L}{\pi^2}\sum_{n\ \text{odd}}\frac{1}{n^2}\cos\frac{n\pi x}{L} \]
STEP 5 · why 1/n² beats 1/n
The even extension of \(x\) is \(|x|\) — continuous everywhere, with only a corner at the origin. The odd extension has genuine jumps at \(x=\pm L\). Smoother extension ⟹ faster coefficient decay: continuous with a corner gives \(1/n^2\), a jump gives \(1/n\). This is a general and very useful diagnostic — the decay rate tells you the smoothness of the extension, and hence how many terms you'll need.
3
Verify the orthogonality relation \(\int_0^L\sin\frac{n\pi x}{L}\sin\frac{m\pi x}{L}dx = 0\) for \(n\ne m\).
Hint 1Use the product-to-sum identity \(\sin A\sin B = \frac12[\cos(A-B) - \cos(A+B)]\).
Hint 2Each cosine integrates to a sine evaluated at multiples of \(\pi\).
Solution
STEP 1 · convert the product to a sum
\[ \sin\frac{n\pi x}{L}\sin\frac{m\pi x}{L} = \frac12\left[\cos\frac{(n-m)\pi x}{L} - \cos\frac{(n+m)\pi x}{L}\right] \]
Now \(\cos\frac{(n-m)\pi x}{L} = \cos 0 = 1\), whose integral is \(L\), while the other term still vanishes. So the integral is \(L/2\) — the normalization that appears in the coefficient formula.
WHAT IT MEANS
The functions \(\{\sin\frac{n\pi x}{L}\}\) are mutually "perpendicular" under the inner product \(\langle f,g\rangle = \int_0^Lfg\,dx\). Fourier coefficients are components along an orthogonal basis, and module A1 shows this is no coincidence — it follows from the Sturm–Liouville structure of the underlying eigenvalue problem.
4
Without computing anything, state which coefficients vanish for: (a) the sine series of a function symmetric about \(x=L/2\); (b) the full Fourier series on \([-L,L]\) of an even function; (c) the full series of an odd function.
Hint 1Cosines are even about the origin; sines are odd.
Hint 2For (a), sketch \(\sin\frac{2\pi x}{L}\) on \([0,L]\) — is it symmetric or antisymmetric about the midpoint?
Solution
(a) symmetric about the midpoint
\(\sin\frac{n\pi x}{L}\) is symmetric about \(x=L/2\) when \(n\) is odd and antisymmetric when \(n\) is even. A function symmetric about the midpoint has no antisymmetric content, so all even-\(n\) coefficients vanish — which is exactly what happened to the constant function in the worked example.
(b) even function on [−L, L]
\(f\) even ⟹ \(f\sin\) is odd ⟹ integrates to zero over a symmetric interval. So all \(b_n = 0\): an even function has a pure cosine series.
(c) odd function on [−L, L]
By the mirror argument, \(f\cos\) is odd, so all \(a_n = 0\) (including \(a_0\)): an odd function has a pure sine series.
WHY BOTHER
Symmetry checks cost seconds and can halve or quarter your work — and they catch errors: if you compute a nonzero \(b_n\) for an even function, you have made a mistake somewhere. They also explain the standard PDE practice of extending data oddly for Dirichlet problems and evenly for Neumann ones.
5
The sine series of \(f\) has \(b_n = \frac{1}{n^3}\); another function has \(b_n = \frac{1}{n}\). Which is smoother, and roughly how many terms does each need for 1% accuracy?
Hint 1Faster coefficient decay means a smoother function.
Hint 2The truncation error is governed by the size of the first omitted terms.
Solution
STEP 1 · the smoothness dictionary
Coefficient decay
Smoothness of the extension
\(1/n\)
jump discontinuity
\(1/n^2\)
continuous, but with a corner
\(1/n^3\) or faster
continuously differentiable or better
exponential, e.g. \(e^{-n}\)
analytic
So \(b_n = 1/n^3\) is much the smoother.
STEP 2 · estimate for the smooth one
Terms beyond \(N\) contribute roughly \(\sum_{n>N}n^{-3} \approx \frac{1}{2N^2}\). Setting that to \(0.01\) gives \(N \approx 7\).
STEP 3 · estimate for the rough one
Now \(\sum_{n>N}n^{-1}\) diverges, so the tail argument fails outright — the series converges, but only through cancellation of alternating signs, and the error near the jump never falls below the Gibbs 9%.
STEP 4 · the practical answer
Away from the jump you might need hundreds of terms for 1%; at the jump you can never achieve it with a truncated series. Smoothness is not an aesthetic property — it directly determines how expensive your computation is.
CONNECTION TO PDEs
The heat equation multiplies \(b_n\) by \(e^{-kn^2\pi^2t/L^2}\), which decays super-exponentially in \(n\). So however rough your initial data, the solution at any \(t>0\) has essentially exponentially decaying coefficients — that is smoothing, seen in the coefficients.
6
Find the full Fourier series of \(f(x)=x\) on \([-\pi,\pi]\).
Hint 1Check the symmetry first — is \(x\) odd or even?
Hint 2Odd functions have no cosine terms, so only \(b_n\) survives.
Hint 3\(b_n = \frac1\pi\int_{-\pi}^{\pi}x\sin nx\,dx = \frac2\pi\int_0^\pi x\sin nx\,dx\).
Solution
STEP 1 · use symmetry
\(f(x)=x\) is odd, so \(a_n=0\) for all \(n\) — no cosine terms and no constant term.
STEP 5 · the series and its behavior
\[ x = 2\left[\sin x - \frac{\sin2x}{2} + \frac{\sin3x}{3} - \cdots\right], \qquad -\piAt \(x=\pm\pi\) every term vanishes, so the series gives 0 — the midpoint of the jump from \(\pi\) to \(-\pi\) in the periodic extension ✓ The \(1/n\) decay is the signature of that jump.
7
Use Parseval's identity on the sine series of \(f=1\) on \([0,\pi]\) to evaluate \(\sum_{n\ \text{odd}}\frac{1}{n^2}\).
Hint 1Parseval: \(\frac2L\int_0^Lf^2dx = \sum b_n^2\).
Hint 2From the worked example, \(b_n = \frac{4}{n\pi}\) for odd \(n\).
Numerically \(1+0.111+0.04+0.0204 = 1.172\), climbing toward 1.234 ✓ And since \(\sum_{\text{all }n}\frac{1}{n^2} = \sum_{\text{odd}} + \frac14\sum_{\text{all}}\), you recover \(\sum\frac{1}{n^2} = \frac{\pi^2}{6}\) — Euler's result, from a Fourier series. Parseval also has a physical reading: it says the total energy equals the sum of the energies in each mode.
8
Write the cosine series of \(f(x)=\cos x\) on \([0,\pi]\). Then explain why its sine series is an infinite series rather than a single term.
Hint 1\(\cos x\) is already a cosine eigenfunction — which \(n\)?
Hint 2For the sine series, think about the odd extension of \(\cos x\) and whether it is continuous.
Solution
STEP 1 · the cosine series
\(\cos x\) is exactly the \(n=1\) cosine eigenfunction on \([0,\pi]\), so
\[ a_1 = 1, \quad \text{all other } a_n = 0 \]
The "series" is a single term — nothing to compute.
STEP 2 · consider the even extension
Extending \(\cos x\) evenly gives \(\cos|x| = \cos x\), which is smooth and \(2\pi\)-periodic. A perfectly matched basis, hence one term.
STEP 3 · now the odd extension
Extending oddly gives a function equal to \(\cos x\) for \(x>0\) and \(-\cos x\) for \(x<0\). At \(x=0\) it jumps from \(-1\) to \(+1\).
STEP 4 · consequence for the coefficients
A jump forces slow \(1/n\) decay, so infinitely many sine terms are needed, and the partial sums show Gibbs ringing near \(x=0\) and \(x=\pi\).
STEP 5 · the moral
Match the basis to the boundary conditions. Expanding \(\cos x\) in sines is legal but wasteful and badly convergent; the sine basis is the right one only when the physical problem clamps the ends at zero.
9
For a function \(f\) on \([0,L]\) with \(f(0)=2\) and \(f(L)=3\), state what the sine series and the cosine series each converge to at \(x=0\) and \(x=L\).
Hint 1Every sine term is zero at both endpoints, whatever \(f\) does.
Hint 2For the cosine series, consider whether the even extension has a jump at \(x=0\).
Solution
STEP 1 · sine series at the ends
Every \(\sin\frac{n\pi x}{L}\) vanishes at \(x=0\) and \(x=L\), so the series can only converge to
\[ 0 \ \text{ at both ends} \]
even though \(f(0)=2\) and \(f(L)=3\).
STEP 2 · why that is not a contradiction
The series represents the odd periodic extension, which jumps from \(-2\) to \(+2\) at \(x=0\). Dirichlet's theorem gives the midpoint, 0 ✓
STEP 3 · cosine series at the ends
The even extension is continuous at \(x=0\) (it just reflects), so the cosine series converges to the true values:
\[ 2 \ \text{ at } x=0, \qquad 3 \ \text{ at } x=L \]
STEP 4 · the practical implication
If your PDE clamps the ends at zero, the sine series' behavior is exactly right — the mismatch with \(f\) at the endpoints is a genuine feature of the physics (the ends are forced to zero the instant \(t>0\)).
STEP 5 · and the convergence cost
That endpoint jump makes the sine coefficients decay like \(1/n\), so you need many terms near the ends. The cosine series of the same \(f\) converges faster — one more reason the boundary conditions, not convenience, must choose the basis.
Part III · Module 2 Separation & Fourier
Separation of variables
The central technique of the course. Guess that the solution factors, \(u(x,t)=X(x)T(t)\), and a PDE splits into two ODEs joined by a single constant. The boundary conditions then quantize that constant — and out fall the modes.
①The machinery, step by step
Take \(u_t = ku_{xx}\) with \(u(0,t)=u(L,t)=0\).
Step 1 — substitute the product guess
\[ u = X(x)T(t) \quad\Longrightarrow\quad XT' = kX''T \]
Step 2 — divide by \(kXT\) to separate
\[ \frac{T'}{kT} = \frac{X''}{X} \]
The left side depends only on \(t\), the right only on \(x\). Two functions of different variables can agree only if both are the same constant — call it \(-\lambda\) (the minus is chosen so \(\lambda>0\) gives decay, which is what we expect physically):
Step 3 — two ODEs
\[ X'' + \lambda X = 0, \qquad T' + k\lambda T = 0 \]
InteractiveWhy only certain λ survivesweep λ and watch the right end
X(L) or X′(L) —nearest eigenvalue —status —
The curve solves \(X''+\lambda X=0\) with the left condition imposed. For most \(\lambda\) it simply misses the right condition — those are not eigenvalues and the only solution of the full problem is \(X\equiv0\). Slide until the right end lands exactly on target and the curve turns green: you've found an eigenvalue. Try negative \(\lambda\): the solution grows exponentially and can never return to zero, which is the proof that \(\lambda\le0\) yields nothing.
②The three cases for λ, done properly
Exams expect you to rule out the dead cases explicitly, not just assert them:
Case
General \(X\)
Applying \(X(0)=X(L)=0\)
\(\lambda<0\), write \(\lambda=-\mu^2\)
\(X = Ae^{\mu x}+Be^{-\mu x}\)
Forces \(A=B=0\). Only the trivial solution — exponentials can't oscillate back to zero.
\(\lambda=0\)
\(X = Ax+B\)
\(B=0\) then \(A=0\). Trivial again (though for Neumann ends this case survives as the constant!).
\(\lambda>0\), write \(\lambda=\beta^2\)
\(X = A\cos\beta x + B\sin\beta x\)
\(A=0\) from the left; then \(B\sin\beta L = 0\) forces \(\beta L = n\pi\) ⟹ eigenvalues.
Worked example · separation with mixed ends
Find the eigenvalues and eigenfunctions for \(X''+\lambda X=0\) with \(X(0)=0\), \(X'(L)=0\).
STEP 1
Rule out \(\lambda\le0\) as in the table: exponential or linear solutions can't satisfy both conditions without vanishing identically. So take \(\lambda = \beta^2 > 0\).
STEP 2
General solution \(X = A\cos\beta x + B\sin\beta x\). The condition \(X(0)=0\) gives \(A = 0\), leaving \(X = B\sin\beta x\).
STEP 3
Differentiate: \(X' = B\beta\cos\beta x\). The condition \(X'(L)=0\) with \(B\ne0\) forces
\[ \cos\beta L = 0 \quad\Longrightarrow\quad \beta L = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2},\dots \]
STEP 4
So the spectrum is built from odd multiples of \(\pi/2\):
Sketch check: each eigenfunction starts at zero (clamped end) and reaches a peak exactly at \(x=L\) (zero slope, insulated end) — a quarter wave, then three quarters, then five. Physically: a bar held at ice temperature on the left and insulated on the right.
Separation is not universal
It needs a linear, homogeneous PDE with homogeneous boundary conditions on a domain whose sides are coordinate curves. Nonzero boundary data must first be removed by subtracting a steady state (module F5); nonhomogeneous PDEs need the eigenfunction expansion of module A2; and irregular domains generally defeat the method entirely.
③Practice
1
Separate \(u_t = 3u_{xx}\) and write the two ODEs, then solve the \(T\) equation for \(\lambda_n = (n\pi/L)^2\).
Hint 1Substitute \(u = XT\) and divide by \(3XT\).
Hint 2Set both sides equal to \(-\lambda\).
Hint 3The \(T\) equation is first order and separable.
STEP 4 · apply the right end
\[ X'(L) = -A\beta\sin\beta L = 0 \quad\Longrightarrow\quad \sin\beta L = 0 \quad\Longrightarrow\quad \beta = \frac{n\pi}{L} \]
STEP 5 · collect the answer
\[ \lambda_n = \left(\frac{n\pi}{L}\right)^2, \qquad X_n(x) = \cos\frac{n\pi x}{L}, \qquad n = 0,1,2,\dots \]
The \(n=0\) member is the constant from step 1. Because \(\lambda_0=0\) gives \(T_0 = \) const, the average temperature never decays — the physical statement that an insulated bar conserves its heat (module S4).
3
Separate the wave equation \(u_{tt}=c^2u_{xx}\) and show the time factor oscillates rather than decays.
Hint 1Divide by \(c^2XT\) after substituting.
Hint 2The \(T\) equation is now second order: \(T'' + \lambda c^2T = 0\).
Solution
STEP 1 · substitute and separate
\[ XT'' = c^2X''T \quad\Longrightarrow\quad \frac{T''}{c^2T} = \frac{X''}{X} = -\lambda \]
STEP 2 · the two ODEs
\[ X''+\lambda X = 0, \qquad T'' + \lambda c^2 T = 0 \]
The \(X\) equation is identical to the heat case — so the same boundary conditions give the same eigenvalues and eigenfunctions.
STEP 3 · solve the time equation
With \(\lambda_n = (n\pi/L)^2 > 0\) this is a harmonic oscillator:
Mode \(n\) rings at angular frequency \(\omega_n = \frac{n\pi c}{L}\) — integer multiples of the fundamental, which is precisely why a string sounds musical rather than noisy. Nothing decays, because the ideal wave equation has no dissipation.
4
Show that \(\lambda < 0\) is impossible for \(X''+\lambda X=0\) with \(X(0)=X(L)=0\), working the algebra explicitly.
Hint 1Write \(\lambda = -\mu^2\) with \(\mu>0\); the general solution is built from real exponentials (or \(\sinh, \cosh\)).
Hint 2Impose \(X(0)=0\) first to eliminate one constant.
Hint 3Is \(\sinh\) ever zero except at the origin?
Solution
STEP 1 · set up the negative case
Let \(\lambda = -\mu^2\), \(\mu>0\). Then \(X'' = \mu^2X\) and
\[ X(x) = A\cosh\mu x + B\sinh\mu x \]
STEP 2 · apply the left condition
\(\cosh 0 = 1\), \(\sinh 0 = 0\), so \(X(0) = A = 0\), leaving \(X = B\sinh\mu x\).
STEP 3 · apply the right condition
\[ X(L) = B\sinh\mu L = 0 \]
STEP 4 · use the key fact about sinh
\(\sinh s = 0\) only at \(s=0\). Since \(\mu>0\) and \(L>0\), \(\sinh\mu L \ne 0\), so \(B=0\).
STEP 5 · conclude
Both constants vanish, so \(X\equiv0\) — the trivial solution, which is not an eigenfunction. Hence no negative eigenvalues exist.
WHY IT HAD TO BE TRUE
Physically, \(\lambda<0\) would give \(T = e^{+k|\lambda| t}\): a temperature profile spontaneously growing without any source. Ruling it out mathematically confirms that the heat equation with clamped ends can only decay — the eigenvalues determine stability.
5
Attempt separation on \(u_t = u_{xx} + u\,u_x\) and identify exactly where it fails.
Hint 1Substitute \(u = XT\) and try to get all \(x\)-dependence on one side.
Hint 2The nonlinear term produces \(X X' T^2\) — what power of \(T\) do the other terms have?
The last term \(X'T\) contains both variables and cannot be moved to either side. No choice of separation constant helps — the mixed term simply doesn't factor.
STEP 4 · why, structurally
The two linear terms are homogeneous of degree 1 in \(u\), so each carries a single power of \(T\); the nonlinear term is degree 2 and carries \(T^2\). Separation requires every term to scale the same way in \(T\), and nonlinearity breaks exactly that.
STEP 5 · what this rules out
No eigenfunction expansion, no superposition of modes, no Fourier method. Nonlinear PDEs need characteristics (Part II), similarity solutions, or numerics (module A5). It is worth attempting the separation once on any unfamiliar equation: the point where it jams tells you precisely which structural property you're missing.
6
Separate the two-dimensional heat equation \(u_t = k(u_{xx}+u_{yy})\) on a rectangle \(0
Hint 1Try \(u = X(x)Y(y)T(t)\) and divide by \(kXYT\).
Hint 2You will need two separation constants.
Hint 3Each spatial factor is the familiar 1-D problem.
The eigenvalues are now indexed by two integers, and the initial data is expanded in a double Fourier series. If \(a/b\) is rational, different \((m,n)\) pairs can share an eigenvalue — degeneracy, and the reason a square drum has repeated frequencies while a generic rectangle does not.
7
Separate \(u_t = ku_{xx} - hu\) (diffusion with uniform heat loss) and show how the decay rates shift.
Hint 1Substitute \(u=XT\) and divide by \(kXT\) as usual, keeping the \(-hu\) term with the time factor.
Hint 2The \(X\) problem is unchanged; only \(T\) picks up the extra term.
This matches module F2 problem 3: \(u=e^{-ht}v\) reduces the equation to the plain heat equation for \(v\). The loss term rescales the whole profile uniformly without touching its shape — unlike diffusion, which attacks high modes preferentially.
8
Show that separation of variables fails for \(u_t=ku_{xx}\) with the boundary condition \(u(0,t)=t\), and state the fix.
Hint 1Substitute \(u=XT\) into the boundary condition.
Hint 2Can a product \(X(0)T(t)\) equal \(t\) for all \(t\) unless \(T\) is forced?
Solution
STEP 1 · impose the boundary condition on a product
\[ u(0,t) = X(0)\,T(t) = t \]
STEP 2 · see the contradiction
If \(X(0)\ne0\) this forces \(T(t) = t/X(0)\) — a specific linear function. But the PDE requires \(T' = -k\lambda T\), whose solutions are exponentials, not linear. The two demands are incompatible.
STEP 3 · and the other branch
If \(X(0)=0\) then \(u(0,t)=0\ne t\). Either way, no separated solution exists.
STEP 4 · the general principle
Separation requires homogeneous boundary conditions, because a superposition \(\sum X_nT_n\) can only satisfy conditions that are preserved under addition — and \(u(0,t)=t\) is not.
STEP 5 · the fix
Subtract a function carrying the boundary data. Here take \(w = u - t\!\left(1-\frac{x}{L}\right)\); the shifted problem has homogeneous conditions but picks up a source term, which the eigenfunction expansion of module A2 then handles. Time-dependent boundary data always converts into a source.
9
Compare the spectra of \(X''+\lambda X=0\) under Dirichlet (\(X(0)=X(L)=0\)) and Neumann (\(X'(0)=X'(L)=0\)) conditions. What is the single most important difference?
Hint 1Both give \(\lambda_n = (n\pi/L)^2\), but the allowed values of \(n\) differ.
Hint 2Ask whether \(\lambda=0\) survives in each case.
Solution
STEP 1 · the two spectra
Dirichlet
Neumann
Eigenvalues
\((n\pi/L)^2\), \(n\ge1\)
\((n\pi/L)^2\), \(n\ge0\)
Eigenfunctions
\(\sin\frac{n\pi x}{L}\)
\(\cos\frac{n\pi x}{L}\)
Is \(\lambda=0\) an eigenvalue?
no
yes, with \(X_0=1\)
STEP 2 · why λ = 0 dies for Dirichlet
\(X=Ax+B\) with \(X(0)=X(L)=0\) forces \(A=B=0\).
STEP 3 · why it survives for Neumann
\(X=Ax+B\) with \(X'=A=0\) at both ends leaves the constant \(B\) free — a genuine eigenfunction.
STEP 4 · the consequence for the PDE
The time factor \(e^{-k\lambda_0t} = e^0 = 1\) never decays. So a Neumann (insulated) bar retains its average forever, while a Dirichlet bar has \(\lambda_1>0\) as its smallest rate and decays to zero.
STEP 5 · the general diagnostic
Ask whether zero is in the spectrum. If it is, something is conserved and the solution tends to a nonzero equilibrium; if not, everything decays. That single question predicts the long-time fate of any diffusion problem — including the Neumann Laplace problem's non-uniqueness (module S7 problem 4).
Part III · Module 3 Separation & Fourier
Heat equation: fixed ends
The complete solution of the flagship problem — a bar with both ends held at zero, cooling from an arbitrary initial profile. Every technique of Part III appears here at once, so it is worth being able to reproduce the whole derivation from memory.
InteractiveA bar cooling, mode by modepress play; watch the wiggles die first
max temperature —b₁ share of energy —shape now —
Top: the profile evolving, with the first three modes drawn faintly beneath it. Bottom: the mode amplitudes \(b_ne^{-k\lambda_nt}\) as bars. Whatever you start with — even a block with sharp corners — the high modes are annihilated almost immediately and the bar relaxes into a single smooth half-sine hump before decaying to zero. That is the \(n^2\) in the exponent doing its work.
②What the formula tells you without computing
Everything decays. Every exponent is negative, so \(u\to0\): with both ends held at zero and no source, the bar must eventually reach zero.
The fundamental dominates. Mode \(n\) decays at rate \(n^2\), so after a short time \(u \approx b_1e^{-k\pi^2t/L^2}\sin\frac{\pi x}{L}\) regardless of the initial shape.
Instant smoothing. For any \(t>0\) the coefficients decay like \(e^{-cn^2}\), so \(u\) is infinitely differentiable — even if \(\varphi\) had corners or jumps.
Length matters quadratically. The time scale is \(L^2/(k\pi^2)\): double the bar and it takes four times as long to cool.
Worked example · triangular initial data, start to finish
Solve \(u_t = u_{xx}\) on \(0
STEP 1
Here \(L=\pi\), \(k=1\), so \(\lambda_n = n^2\) and the modes are \(e^{-n^2t}\sin nx\).
STEP 2
Compute the coefficients, splitting the integral at the peak:
Each piece is an integration by parts. The result collapses neatly:
\[ b_n = \frac{4}{n^2\pi}\sin\frac{n\pi}{2} \]
STEP 4
Read off which survive: \(\sin\frac{n\pi}{2}\) is 0 for even \(n\), \(+1\) for \(n=1,5,9,\dots\), \(-1\) for \(n=3,7,\dots\). So only odd modes appear — as symmetry about \(x=\pi/2\) demanded (module S1, problem 4).
After only \(t\approx0.12\) the four-hump structure is invisible and the bar looks like a single smooth arch. This is the general rule: the ratio of mode \(n\) to mode 1 decays like \(e^{-k(n^2-1)t}\), so detail vanishes at a rate that grows quadratically with its fineness.
3
A bar of length \(L=2\), \(k=1\), starts at the uniform temperature \(u(x,0)=100\) with both ends plunged into ice. Find the solution and the centre temperature at \(t=1\).
Hint 1You need the sine series of a constant — see the worked example in module S1, rescaled to length \(L\).
Hint 2\(b_n = \frac{2}{L}\int_0^L100\sin\frac{n\pi x}{L}dx = \frac{200}{n\pi}(1-\cos n\pi)\).
Hint 3Only odd \(n\) survive, with \(b_n = \frac{400}{n\pi}\).
The \(n=3\) term is \(\frac{400}{3\pi}e^{-9\pi^2/4}\approx 42.4\times 2.6\times10^{-10}\), utterly negligible.
STEP 5 · answer
\[ u(1,1) \approx 10.8^\circ \]
From 100° to about 11° in one time unit, and one term of the series gave nine-figure accuracy — the standard situation once \(t\) is not tiny.
4
Explain why \(u(x,t)\) is smooth for every \(t>0\) even when \(\varphi\) has a jump, and why this makes the heat equation irreversible.
Hint 1A jump means \(b_n\) decays only like \(1/n\). What happens after multiplying by the time factor?
Hint 2Compare \(e^{-kn^2\pi^2t/L^2}\) with any power of \(n\).
Solution
STEP 1 · the coefficients at time t
The solution's \(n\)th coefficient is \(b_ne^{-k\lambda_nt}\). Even with \(b_n\) decaying only like \(1/n\), the factor \(e^{-cn^2t}\) crushes it.
STEP 2 · why that implies smoothness
Differentiating the series \(m\) times multiplies the \(n\)th term by at most \(n^m\). Since \(n^me^{-cn^2t}\to0\) faster than any power for fixed \(t>0\), every differentiated series still converges absolutely. So \(u\) has derivatives of all orders in \(x\) — for any \(t>0\), however rough \(\varphi\) was.
STEP 3 · what that costs you going backwards
Reversing time replaces \(e^{-cn^2t}\) with \(e^{+cn^2t}\), which amplifies high modes without limit. Fine detail present in the data at \(t=0\) has been squashed to essentially nothing by \(t=1\); recovering it means multiplying near-zero numbers by astronomically large factors.
STEP 4 · conclusion
Information about small-scale structure is genuinely destroyed, not merely hidden. That is why the backward heat problem is ill-posed (module F3 problem 5) and why you cannot deduce this morning's temperature profile from tonight's measurements.
CONTRAST
The wave equation's factors are \(\cos\omega_nt\) and \(\sin\omega_nt\) — modulus 1, no decay. Nothing is lost, so running it backwards is perfectly well posed. Smoothing and irreversibility are the same phenomenon seen from two sides.
5
A bar has \(u(0,t)=0\), \(u(L,t)=T_0\ne0\). Show how to reduce this to the standard problem, and state the long-time limit.
Hint 1Separation needs homogeneous boundary conditions — subtract something that carries the nonzero data.
Hint 2Find the steady state first: the straight line matching both end values.
Solution
STEP 1 · find the steady state
Set \(u_t=0\): \(v''=0\) with \(v(0)=0\), \(v(L)=T_0\), so
\[ v(x) = \frac{T_0x}{L} \]
STEP 2 · subtract it
Let \(w = u - v\). Since \(v_t = v_{xx} = 0\), \(w\) satisfies the same PDE \(w_t = kw_{xx}\).
STEP 3 · keep the survivors
\[ b_n = \frac{8}{n^3\pi^3} \ \text{ for odd } n, \qquad 0 \ \text{ for even } n \]
Even modes vanish because the data is symmetric about the midpoint ✓
STEP 4 · write the solution
\[ u(x,t) = \frac{8}{\pi^3}\sum_{n\ \text{odd}}\frac{1}{n^3}e^{-n^2\pi^2t}\sin n\pi x \]
STEP 5 · note the fast convergence
\(b_n\sim1/n^3\), reflecting a smooth extension with no jumps or corners (module S1 problem 5). Even at \(t=0\) three terms give better than 1% accuracy, and for \(t>0\) one term suffices: \(u \approx \frac{8}{\pi^3}e^{-\pi^2t}\sin\pi x\).
7
A bar with \(L=2\), \(k=0.5\), zero ends, starts with peak temperature 80°. Estimate when the peak falls below 1°.
Hint 1After a short time only the fundamental matters, so use \(u \approx b_1e^{-k\lambda_1t}\sin\frac{\pi x}{L}\).
Hint 2\(\lambda_1 = (\pi/2)^2 \approx 2.47\), so the decay rate is \(k\lambda_1 \approx 1.23\).
Once the higher modes are gone, the peak decays as
\[ u_{\max}(t) \approx 80\,e^{-1.234t} \]
STEP 3 · solve for the threshold
\[ 80e^{-1.234t} = 1 \quad\Longrightarrow\quad t = \frac{\ln 80}{1.234} = \frac{4.382}{1.234} \approx 3.55 \]
STEP 4 · sanity check the assumption
The \(n=2\) mode decays four times faster, so by \(t\approx1\) it is already negligible relative to the fundamental — the single-mode estimate is safe for \(t\approx3.5\).
STEP 5 · scaling insight
Doubling the length to \(L=4\) would quarter \(\lambda_1\) and stretch the time to about 14 — the \(x\sim\sqrt{t}\) scaling of module F1 problem 9. Halving \(k\) doubles it. The single number \(1/(k\lambda_1)\) is the thermal time constant of the bar.
8
Two identical bars differ only in length, \(L\) and \(2L\). Both start with the same shape (scaled) and are clamped at zero. Which cools faster, and by what factor?
Hint 1Compare the fundamental eigenvalues \(\lambda_1 = (\pi/L)^2\).
The short bar's decay rate is four times larger, so it cools four times faster.
STEP 3 · state it as a time
If the short bar cools to a given fraction in time \(\tau\), the long one takes \(4\tau\).
STEP 4 · why quadratic, not linear
Heat must diffuse to an end to escape, and diffusion covers distance like \(\sqrt{kt}\). Doubling the distance therefore costs four times the time — the same \(x\sim\sqrt{t}\) law behind the heat kernel and the numerical stability limit.
STEP 5 · engineering consequence
This is why heat sinks use many thin fins rather than one thick block, and why thick steel sections must be quenched far more slowly than thin ones — the interior simply cannot lose heat quickly.
9
Show that \(E(t)=\int_0^Lu^2dx\) is non-increasing for the heat equation with zero ends, and use it to prove uniqueness.
Hint 1Differentiate \(E\), substitute the PDE, then integrate by parts.
Hint 2The boundary term vanishes because \(u=0\) at both ends.
Hint 3For uniqueness, apply the result to the difference of two solutions.
STEP 2 · integrate by parts
\[ = 2k\left[u\,u_x\right]_0^L - 2k\int_0^L\left(u_x\right)^2dx \]
STEP 3 · kill the boundary term
\(u(0,t)=u(L,t)=0\), so
\[ E'(t) = -2k\int_0^L u_x^2\,dx \le 0 \]
STEP 4 · uniqueness
Suppose \(u_1,u_2\) solve the same problem. Their difference \(w\) satisfies the heat equation with zero boundary data and \(w(x,0)=0\), so \(E(0)=0\). Since \(E\ge0\) and \(E'\le0\), \(E\equiv0\), forcing \(w\equiv0\) and \(u_1=u_2\) ✓
STEP 5 · the wider technique
This "energy method" proves uniqueness without ever solving anything, and it works where separation does not — variable coefficients, several dimensions, irregular domains. The same argument with \(E = \frac12\int(u_t^2+c^2u_x^2)\) gives conservation for the wave equation (module S5 problem 4); there \(E'=0\) exactly rather than \(\le0\), which is the difference between a dissipative and a conservative equation.
Part III · Module 4 Separation & Fourier
Heat equation: insulated ends
Change the boundary conditions from clamped to insulated and the entire character of the answer changes: cosines instead of sines, a surviving \(\lambda=0\) mode, and a final temperature that is not zero but the average of where you started.
Module S2 problem 2 found the eigenpairs: \(\lambda_n = (n\pi/L)^2\) with \(X_n = \cos\frac{n\pi x}{L}\), including \(n=0\) with the constant eigenfunction. So the expansion is a cosine series:
The \(n=0\) term has \(\lambda_0=0\), so its time factor is \(e^0=1\): it never decays.
InteractiveInsulated bar settling to its averagecompare with clamped ends
total heat ∫u dx —final value —flatness —
Watch the dashed line: it marks the initial average. With insulated ends the profile flattens toward exactly that value and the total heat readout never budges. Switch to clamped and the same initial data drains away to zero instead — the heat escapes through the ends. Same equation, same initial data, completely different destiny, decided entirely by the boundary conditions.
Three equivalent statements of the same fact: (1) physically, no heat crosses an insulated end; (2) analytically, the integral above is constant; (3) spectrally, \(\lambda_0=0\) so the constant mode never decays. The limit is
So \(a_n = 0\) for even \(n\), and \(\pm\frac{2}{n\pi}\) for odd \(n\) (\(+\) for \(n=1,5,\dots\), \(-\) for \(n=3,7,\dots\)).
STEP 4
Assemble:
\[ u(x,t) = \frac12 + \frac{2}{\pi}\left[e^{-\pi^2t}\cos\pi x - \frac13e^{-9\pi^2t}\cos3\pi x + \cdots\right] \]
STEP 5
Check the limit: as \(t\to\infty\) every exponential dies and \(u\to\frac12\) — the average of "half the bar at 1, half at 0" ✓ At \(t=0.05\) the first correction is already only \(\frac{2}{\pi}e^{-0.49}\approx0.39\) and shrinking fast; the bar is visibly evening out.
③Practice
1
An insulated bar of length \(\pi\) starts at \(u(x,0) = 3 + 2\cos 2x\). Find \(u(x,t)\) for \(k=1\) and its limit.
Hint 1The data is already a cosine expansion — identify \(a_0/2\) and \(a_2\) by inspection.
Hint 2The constant term has \(\lambda=0\) and never decays.
Solution
STEP 1 · read off the coefficients
With \(L=\pi\) the eigenfunctions are \(\cos nx\). Matching: \(\frac{a_0}{2}=3\) and \(a_2 = 2\); all others zero.
STEP 2 · attach decay factors
\(\lambda_0 = 0\) (no decay) and \(\lambda_2 = 4\) (factor \(e^{-4t}\)).
which is indeed the average of the initial data, since \(\cos 2x\) integrates to zero over \([0,\pi]\) ✓
2
Find the final temperature of an insulated bar of length 4 starting from \(\varphi(x) = x\), without computing any \(a_n\) for \(n\ge1\).
Hint 1Only the \(n=0\) mode survives, and it equals the average.
Hint 2Average of \(x\) over \([0,4]\).
Solution
STEP 1 · use conservation
Insulated ends ⟹ total heat is constant ⟹ the bar ends uniform at the initial average.
STEP 2 · compute the average
\[ \bar{u} = \frac{1}{4}\int_0^4x\,dx = \frac14\cdot\frac{16}{2} = 2 \]
STEP 3 · state the answer
\[ \lim_{t\to\infty}u(x,t) = 2 \]
STEP 4 · sanity check
The initial profile runs from 0 to 4 linearly; its midpoint value is 2 — and the bar simply redistributes heat until everything sits at that level ✓
STEP 5 · why this shortcut is legitimate
Every \(n\ge1\) cosine has \(\int_0^L\cos\frac{n\pi x}{L}dx = 0\), so those modes carry no net heat — they only redistribute it. All the conserved quantity lives in the \(n=0\) mode, which is why you never need the other coefficients to answer a "final temperature" question.
3
A bar is insulated at \(x=0\) and held at zero at \(x=L\). Find the eigenvalues, eigenfunctions, and the form of the solution.
Hint 1Mixed conditions: \(X'(0)=0\) and \(X(L)=0\).
Hint 2\(X'(0)=0\) kills the sine, leaving \(X = A\cos\beta x\).
Hint 3Then \(\cos\beta L = 0\) forces \(\beta L\) to be an odd multiple of \(\pi/2\).
Solution
STEP 1 · general solution for λ = β² > 0
\[ X = A\cos\beta x + B\sin\beta x, \qquad X' = -A\beta\sin\beta x + B\beta\cos\beta x \]
Note \(\lambda=0\) fails here: \(X = Ax+B\) with \(X'(0)=0\) gives \(A=0\), and then \(X(L)=B=0\). The clamped end drains heat, so nothing is conserved.
STEP 5 · the solution form
\[ u(x,t) = \sum_{n=1}^{\infty}c_ne^{-k\lambda_nt}\cos\frac{(2n-1)\pi x}{2L}, \qquad c_n = \frac2L\int_0^L\varphi\cos\frac{(2n-1)\pi x}{2L}dx \]
Every mode decays, so \(u\to0\) — a quarter-wave version of the Dirichlet problem.
4
Explain why an insulated bar with a constant internal source \(Q>0\) has no steady state, and what happens instead.
Hint 1Integrate the equation \(u_t = ku_{xx}+Q\) across the bar.
Hint 2The \(u_{xx}\) term contributes boundary flux, which is zero here.
Solution
STEP 1 · integrate the PDE over the bar
\[ \frac{d}{dt}\int_0^Lu\,dx = k\big[u_x\big]_0^L + \int_0^LQ\,dx = 0 + QL \]
STEP 2 · read the consequence
The total heat grows linearly and without bound:
\[ \int_0^Lu\,dx = \int_0^L\varphi\,dx + QLt \]
STEP 3 · why no steady state exists
A steady state would require \(\frac{d}{dt}\int u = 0\), contradicting \(QL \ne 0\). Equivalently, \(-ku''=Q\) with \(u'(0)=u'(L)=0\) fails the compatibility condition of module F4.
STEP 4 · what actually happens
The bar heats up forever. Its shape settles: writing \(u = Qt + w(x,t)\) removes the growth and leaves \(w_t = kw_{xx}\), whose profile flattens. So the solution becomes "uniform temperature rising at a constant rate \(Q\)".
STEP 5 · the physical moral
Heat in with nowhere to go means no equilibrium. To get a steady state you must either open an end (Dirichlet or Robin) or arrange sources and sinks that cancel, \(\int_0^LQ\,dx = 0\).
5
Compare the long-time behavior of the same initial data \(\varphi(x)=\sin\frac{\pi x}{L}+1\) under (a) clamped ends, (b) insulated ends. Which decays and to what?
Hint 1For each case, ask which eigenfunctions the data must be expanded in.
Hint 2In the insulated case, only the average survives.
Solution
STEP 1 · clamped ends
Expand in sines. All eigenvalues are positive, so every mode decays and
\[ u \longrightarrow 0 \]
(Note the data doesn't satisfy \(u=0\) at the ends because of the \(+1\); the series converges to the odd extension and the mismatch shows up as a brief transient near the ends.)
STEP 2 · insulated ends
Expand in cosines. The \(n=0\) mode survives, equal to the average:
STEP 3 · state both limits
\[ \text{(a) } u\to0 \qquad\qquad \text{(b) } u\to1+\frac2\pi \approx 1.637 \]
STEP 4 · the physical difference
Clamped ends are windows: heat leaks out until nothing is left. Insulated ends are walls: heat is trapped and merely redistributes until it is uniform.
STEP 5 · the general rule
The long-time limit is governed entirely by the smallest eigenvalue. Dirichlet: \(\lambda_1>0\) ⟹ decay to zero at rate \(e^{-k\lambda_1t}\). Neumann: \(\lambda_0=0\) ⟹ decay to the average, with the approach governed by \(\lambda_1\). Checking whether zero is an eigenvalue is the fastest way to predict the fate of any diffusion problem.
6
An insulated bar on \([0,\pi]\) with \(k=1\) starts at \(u(x,0)=\cos^2x\). Find \(u(x,t)\) and its limit.
Hint 1\(\cos^2x\) is not a single eigenfunction — use a double-angle identity first.
Hint 2\(\cos^2x = \frac12+\frac12\cos2x\).
Solution
STEP 1 · rewrite the data
\[ \cos^2x = \frac12 + \frac12\cos2x \]
Now it is a cosine series with \(\frac{a_0}{2}=\frac12\) and \(a_2=\frac12\).
Every \(\lambda_n>0\), so \(u\to0\): the clamped end drains all the heat despite the insulation at the other end. Note the slowest rate is \(\lambda_1 = \left(\frac{\pi}{2L}\right)^2\) — a quarter of the fully clamped case, so this bar takes four times as long to cool as one clamped at both ends.
8
An insulated bar starts with \(\int_0^L\varphi\,dx = 0\) (equal hot and cold regions). What is the long-time limit, and how fast is it approached?
Hint 1The surviving \(n=0\) mode equals the average — what is it here?
Hint 2The approach rate is set by the smallest nonzero eigenvalue.
Solution
STEP 1 · the conserved quantity
Insulated ends conserve \(\int u\,dx\), which starts at 0 and therefore stays 0.
STEP 2 · the limit
The \(n=0\) coefficient is \(\frac{a_0}{2} = \frac1L\int\varphi = 0\), so
\[ u \longrightarrow 0 \]
Unusually for a Neumann problem, everything decays.
STEP 3 · the rate
The slowest surviving mode is \(n=1\):
\[ u \approx a_1e^{-k\pi^2t/L^2}\cos\frac{\pi x}{L} \ \text{ for large } t \]
STEP 4 · the picture
The hot and cold halves even out against each other rather than losing heat to the outside. The final approach looks like a single cosine hump shrinking — one end slightly warm, the other slightly cool, fading together.
STEP 5 · the general statement
For an insulated bar, \(u\to\bar\varphi\) always; the special case \(\bar\varphi=0\) merely makes that limit zero. The rate is governed by \(\lambda_1\) in every case — the spectral gap between the conserved mode and the first decaying one.
9
An insulated bar of length 2 settles at a uniform 25°. What was \(\int_0^2\varphi\,dx\), and can you determine \(\varphi\) itself?
Hint 1The final value is the average of the initial data.
Hint 2Does the final state remember anything besides the average?
Solution
STEP 1 · relate the limit to the average
\[ 25 = \frac{1}{L}\int_0^L\varphi\,dx = \frac12\int_0^2\varphi\,dx \]
STEP 2 · solve
\[ \int_0^2\varphi\,dx = 50 \]
STEP 3 · can φ be recovered?
No. Every profile with the same total heat leads to the same final state: \(\varphi\equiv25\), or 50° on half the bar and 0 on the other, or any wild oscillation averaging 25.
STEP 4 · what was lost
All the \(n\ge1\) coefficients decayed to zero. The final state remembers exactly one number — the conserved mode — and forgets everything else.
STEP 5 · connection to ill-posedness
This is the physical face of the backward heat problem's failure (module F3 problem 5). Information is genuinely destroyed, so reconstructing the past from the present is impossible even in principle, not merely difficult.
Part III · Module 5 Separation & Fourier
Wave equation: vibrating string
Same spatial eigenvalue problem as the heat equation, but the time factor now oscillates instead of decaying. The result is standing waves whose frequencies are integer multiples of a fundamental — the mathematical reason a string sounds like a note rather than a noise.
Two initial conditions, two families of coefficients — the count promised by module F4.
InteractiveModes, and the shapes they buildmix harmonics; press play
fundamental ω₁ —period —nodes of dominant mode —
A pure mode is a standing wave: fixed nodes, everything oscillating in step. A pluck excites many modes at once, and because their frequencies are exact integer multiples, the whole shape recurs exactly after one period \(2L/c\) — unlike the heat equation, nothing is ever lost. Watch the corner of the plucked triangle survive forever, travelling back and forth.
②Reading the physics off the formula
Harmonics. \(\omega_n = n\omega_1\): the overtones are exact integer multiples, which is what the ear hears as pitch. The mix of \(A_n\) is the timbre — the same note on a violin and a guitar differ only in these coefficients.
Pitch. \(\omega_1 = \frac{\pi c}{L}\) with \(c=\sqrt{T/\rho}\): shorter string, tighter tension, or lighter wire all raise the pitch.
Nodes. Mode \(n\) has \(n-1\) interior points that never move. Touching the string at the midpoint kills all odd modes and leaves the octave — the harmonic technique of every string player.
Periodicity. The entire motion repeats with period \(2L/c\), exactly the time for a disturbance to travel down and back.
Worked example · plucked at the midpoint
A string of length \(L\), released from rest, is plucked into a tent of height \(h\) at \(x = L/2\).
STEP 1
Released from rest ⟹ \(\psi = 0\) ⟹ every \(B_n = 0\). Only cosines in time survive.
STEP 2
The tent's sine coefficients (same integral as module S3's worked example, scaled):
\[ A_n = \frac{8h}{n^2\pi^2}\sin\frac{n\pi}{2} \]
STEP 3
\(\sin\frac{n\pi}{2}\) vanishes for even \(n\), so only odd harmonics are present — the pluck point is a node of every even mode, so it cannot excite them.
Musical reading: plucking at the centre gives a hollow, clarinet-like tone (odd harmonics only). Pluck nearer the bridge, at \(x=L/10\), and \(A_n \propto \sin\frac{n\pi}{10}\) excites many high harmonics — the bright, twangy sound. Where you pluck controls timbre, and the formula says exactly how.
③Practice
1
Solve \(u_{tt}=4u_{xx}\) on \(0
Hint 1\(c=2\), so \(\omega_n = 2n\) with \(L=\pi\).
Hint 2Zero initial velocity ⟹ only cosine time factors.
Solution
STEP 1 · identify the parameters
\(c^2 = 4\) so \(c=2\); with \(L=\pi\), \(\omega_n = \frac{n\pi c}{L} = 2n\).
STEP 2 · use the zero initial velocity
\(\psi\equiv0\) ⟹ all \(B_n = 0\).
STEP 3 · match the initial shape
\(A_1 = 2\), \(A_3 = -1\), all others zero — the data is already an eigenfunction combination.
\(u_{tt} = -8\cos2t\sin x + 36\cos6t\sin3x\); \(4u_{xx} = 4(-2\cos2t\sin x + 9\cos6t\sin3x)\) — equal ✓ And \(u_t(x,0) = 0\) ✓ The motion is periodic with period \(\pi\), since both \(2t\) and \(6t\) complete whole cycles then.
2
A string starts flat but is struck, giving \(u(x,0)=0\), \(u_t(x,0)=\sin\frac{2\pi x}{L}\). Solve it.
Hint 1Zero initial displacement ⟹ all \(A_n=0\); the time factors are sines.
Hint 2\(B_n = \frac{2}{L\omega_n}\int_0^L\psi\sin\frac{n\pi x}{L}dx\), and \(\psi\) is already mode 2.
Solution
STEP 1 · kill the cosine family
\(u(x,0)=0\) forces every \(A_n = 0\).
STEP 2 · match the velocity data
\(\psi = \sin\frac{2\pi x}{L}\) is exactly mode \(n=2\), so only \(B_2\) is nonzero. From the coefficient formula with the orthogonality integral equal to \(L/2\):
At \(t=0\): \(u = 0\) ✓. And \(u_t = \cos\frac{2\pi ct}{L}\sin\frac{2\pi x}{L}\), which at \(t=0\) equals \(\sin\frac{2\pi x}{L}\) ✓
STEP 5 · interpret
The string swings up to maximum displacement a quarter-period after the strike, then back. Note the amplitude \(\frac{L}{2\pi c}\) is inversely proportional to frequency: for a given strike velocity, high modes move less far. This is why struck instruments (piano) sound different from plucked ones (harpsichord) even at the same pitch.
3
A guitar string of length 0.65 m has wave speed \(c=200\) m/s. Find the fundamental frequency in Hz and the frequency of the third harmonic.
Hint 1\(\omega_1 = \pi c/L\) is angular frequency in rad/s; ordinary frequency is \(f = \omega/2\pi\).
Hint 2Equivalently \(f_n = \frac{nc}{2L}\).
Solution
STEP 1 · angular frequency of the fundamental
\[ \omega_1 = \frac{\pi c}{L} = \frac{\pi(200)}{0.65} \approx 966.6\ \text{rad/s} \]
\(f_2 = 2f_1\) is one octave up; \(f_3 = 3f_1\) is an octave plus a fifth. Integer ratios are exactly the consonant musical intervals — the harmonic series is where Western tuning comes from.
STEP 5 · fretting
Pressing the string at the 12th fret halves \(L\) and doubles \(f_1\) — one octave, confirming \(f\propto 1/L\) ✓
4
Show that the total energy \(E = \frac12\int_0^L\left(u_t^2 + c^2u_x^2\right)dx\) is constant for a string with fixed ends.
Hint 1Differentiate under the integral and use the PDE to replace \(u_{tt}\).
Hint 2Integrate the \(u_xu_{xt}\) term by parts to move a derivative off it.
Hint 3Fixed ends mean \(u_t = 0\) at \(x=0,L\) for all \(t\).
Solution
STEP 1 · differentiate the energy
\[ \frac{dE}{dt} = \int_0^L\left(u_tu_{tt} + c^2u_xu_{xt}\right)dx \]
STEP 2 · use the PDE on the first term
\[ u_tu_{tt} = c^2u_tu_{xx} \]
STEP 3 · integrate the second term by parts
\[ \int_0^Lc^2u_xu_{xt}\,dx = c^2\big[u_xu_t\big]_0^L - \int_0^Lc^2u_{xx}u_t\,dx \]
STEP 4 · combine
The integrals cancel exactly, leaving only the boundary term:
\[ \frac{dE}{dt} = c^2\big[u_xu_t\big]_0^L \]
STEP 5 · apply the boundary conditions
The ends are fixed for all time, so \(u(0,t)=0\) implies \(u_t(0,t)=0\), and likewise at \(L\). Hence
\[ \frac{dE}{dt} = 0 \]
WHY IT MATTERS
Energy conservation is the reason nothing decays: kinetic energy \(\frac12u_t^2\) and potential energy \(\frac12c^2u_x^2\) trade back and forth but never leak away. It also gives a two-line uniqueness proof — if two solutions had the same data, their difference would start with \(E=0\) and keep it, forcing it to vanish identically.
5
Solve \(u_{tt}=u_{xx}\) on \(0
Hint 1With \(c=1\), \(\omega_n = n\).
Hint 2Both \(\cos t\) and \(\cos 3t\) must return to 1 simultaneously.
Solution
STEP 1 · write the solution
Zero initial velocity ⟹ cosines only:
\[ u(x,t) = \cos t\,\sin x + \tfrac12\cos3t\,\sin3x \]
STEP 2 · condition for recurrence
The shape repeats when \(\cos t = 1\) and \(\cos3t = 1\), i.e. \(t\) is a multiple of \(2\pi\) and of \(\frac{2\pi}{3}\).
STEP 3 · take the least common value
\[ t = 2\pi \]
STEP 4 · confirm with the general formula
The fundamental period of the string is \(\frac{2L}{c} = \frac{2\pi}{1} = 2\pi\) ✓ Every mode has period \(\frac{2\pi}{n}\), which divides \(2\pi\) exactly — that is why the whole configuration recurs.
STEP 5 · why this cannot happen for heat
Heat modes decay at incommensurate-looking rates and never return. Exact recurrence is a hallmark of conservative systems with commensurable frequencies — and it fails for a circular drum (module A4), whose Bessel frequencies are irrational multiples and whose motion is only quasi-periodic.
6
Compute the energy \(E = \frac12\int_0^\pi\left(u_t^2+c^2u_x^2\right)dx\) for the pure mode \(u = A\cos(nct)\sin(nx)\), and confirm it is time-independent.
Hint 1Compute \(u_t\) and \(u_x\), then use \(\int_0^\pi\sin^2 = \int_0^\pi\cos^2 = \frac\pi2\).
Hint 2You should get a \(\sin^2(nct) + \cos^2(nct)\) combination.
Energy sloshes between kinetic and potential — all kinetic as the string flies through the flat position, all potential at maximum displacement — but the total never changes. Note \(E\propto n^2A^2\): for equal amplitude, high modes carry far more energy, which is why exciting them requires a sharp strike.
7
A string is plucked at \(x=L/3\). Which harmonics are missing from the sound, and why?
Hint 1The coefficient \(A_n\) is proportional to \(\sin\frac{n\pi}{3}\) for a pluck at \(L/3\).
Hint 2When is \(\sin\frac{n\pi}{3} = 0\)?
Hint 3Think about where the nodes of mode \(n\) sit.
Solution
STEP 1 · the coefficient's structure
For a triangular pluck peaked at \(x=p\), the sine coefficients carry a factor \(\sin\frac{n\pi p}{L}\). With \(p = L/3\) that is \(\sin\frac{n\pi}{3}\).
STEP 2 · find the zeros
\[ \sin\frac{n\pi}{3} = 0 \iff n = 3, 6, 9, \dots \]
STEP 3 · state the answer
Every third harmonic is absent: \(n=3,6,9,\dots\)
STEP 4 · why, physically
Mode 3 has a node exactly at \(x=L/3\). Pulling the string there displaces a point that mode 3 never moves, so that mode receives no excitation at all.
STEP 5 · the musician's version
Pluck position shapes timbre by choosing which harmonics to suppress. Guitarists play near the bridge for a bright sound (many high harmonics excited) and over the soundhole for a mellow one. Piano makers strike at about \(L/8\) precisely to suppress the dissonant seventh harmonic.
8
The same initial profile \(\varphi(x)\) is given to a heat problem and a wave problem on the same interval with the same boundary conditions. Contrast what happens at \(t=1\) and as \(t\to\infty\).
Hint 1Both expand in the same eigenfunctions — only the time factors differ.
Hint 2Compare \(e^{-k\lambda_nt}\) with \(\cos(\omega_nt)\).
Solution
STEP 1 · same basis, different time factors
\[ \text{heat: } \sum b_ne^{-k\lambda_nt}\sin\frac{n\pi x}{L}, \qquad \text{wave: } \sum b_n\cos(\omega_nt)\sin\frac{n\pi x}{L} \]
STEP 2 · at t = 1
Heat: high modes have been crushed by \(e^{-kn^2\lambda t}\); the profile is a smooth hump, much reduced in amplitude, retaining only the coarsest features of \(\varphi\). Wave: every mode still has full amplitude \(|b_n|\); the shape has changed (the modes have drifted out of phase) but nothing is lost — corners and fine detail are still present somewhere.
STEP 3 · as t → ∞
Heat: \(u\to0\) monotonically. Wave: \(u\) recurs exactly with period \(2L/c\), forever.
STEP 4 · reversibility
The wave solution run backwards is another legitimate wave solution. The heat solution run backwards amplifies the very modes it destroyed, and is ill-posed.
STEP 5 · one-line summary
Same spatial problem, same eigenfunctions, same coefficients — everything different, because one time factor decays and the other rotates. Classification (module F3) predicted exactly this.
9
A steel wire of length 0.5 m has tension 800 N and linear density 0.008 kg/m. Find \(c\), the fundamental frequency, and the new frequency if the tension is doubled.
Hint 1\(c = \sqrt{T/\rho}\).
Hint 2\(f_1 = \frac{c}{2L}\).
Hint 3Doubling \(T\) multiplies \(c\) by \(\sqrt2\).
STEP 2 · fundamental frequency
\[ f_1 = \frac{c}{2L} = \frac{316.2}{1.0} \approx 316\ \text{Hz} \]
roughly the D♯ above middle C.
STEP 3 · double the tension
\[ c' = \sqrt{2}\,c \approx 447.2\ \text{m/s} \quad\Longrightarrow\quad f_1' \approx 447\ \text{Hz} \]
STEP 4 · the musical interval
The ratio is \(\sqrt2 \approx 1.414\) — a tritone, six semitones. To raise the pitch a full octave you would need four times the tension.
STEP 5 · why instruments use different strings
Since \(f\propto\sqrt{T/\rho}/L\), spanning several octaves by tension alone is impractical (and would snap the wire). Instruments instead vary \(\rho\) — wound bass strings are heavy — and \(L\), which is what fretting does.
Part III · Module 6 Separation & Fourier
d'Alembert's solution
The wave equation has a second, completely different-looking solution — a closed formula instead of an infinite series. It makes finite propagation speed, domains of dependence, and the survival of corners obvious at a glance.
①The formula
On the infinite line, with \(u(x,0)=\varphi(x)\) and \(u_t(x,0)=\psi(x)\):
It comes straight from module F1's observation that \(u = F(x+ct)+G(x-ct)\): the initial shape splits into two half-amplitude copies running in opposite directions, while the initial velocity contributes the average of \(\psi\) over the interval swept out.
InteractiveSplitting, travelling, and the domain of dependencedrag the observation point
observation point —depends on x ∈ —u there —
Top: the two half-copies (faint) and their sum (bold). Bottom: the \(x\)–\(t\) plane with the backward light cone from your chosen point — the interval \([x-ct,\,x+ct]\) is the only part of the initial data that can affect it. Data outside that interval simply hasn't had time to arrive. Drag the point and watch the cone widen with \(t\).
②Finite domains: the reflection trick
On \(0odd, \(2L\)-periodic function, then apply d'Alembert. The odd extension automatically enforces \(u=0\) at \(x=0\) and \(x=L\), so waves reflect off the ends with an inverted sign — a pulse comes back upside down. (For insulated/free ends you use the even extension instead, and pulses reflect right-side up.)
Worked example · a velocity kick
Solve \(u_{tt}=u_{xx}\) on the line with \(\varphi=0\) and \(\psi(x)=1\) on \([-1,1]\), zero elsewhere.
STEP 1
With \(\varphi=0\) only the integral term survives:
\[ u(x,t) = \frac12\int_{x-t}^{x+t}\psi(s)\,ds \]
STEP 2
The integral just measures how much of \([-1,1]\) the interval \([x-t,\,x+t]\) covers, times \(\frac12\).
STEP 3
Take \(x=0\). For \(t<1\) the interval sits inside the kick region, giving \(u = \frac12(2t) = t\) — the point rises steadily.
STEP 4
For \(t>1\) the interval has swallowed all of \([-1,1]\) and cannot capture more, so \(u = \frac12(2) = 1\) forever after.
STEP 5
Read it: a struck region rises and then stays displaced — unlike a plucked bump, which returns to zero. Note also that far away, at \(x=10\), nothing whatsoever happens until \(t=9\), when the disturbance finally arrives. Finite propagation speed, visible directly in the formula.
③Practice
1
Solve \(u_{tt}=9u_{xx}\) with \(u(x,0)=e^{-x^2}\), \(u_t(x,0)=0\), and find \(u(6,1)\).
Hint 1\(c=3\), and with \(\psi=0\) only the averaged-shape term survives.
Hint 2\(u = \frac12\left[\varphi(x-3t)+\varphi(x+3t)\right]\).
Solution
STEP 1 · read off c
\(c^2 = 9\) so \(c=3\).
STEP 2 · apply the formula
\[ u(x,t) = \frac{e^{-(x-3t)^2} + e^{-(x+3t)^2}}{2} \]
\(u(5,3)\) is determined entirely by \(\varphi\) and \(\psi\) on \([-1,11]\). Data outside that interval — however large — cannot possibly influence it.
STEP 3 · the special case
\([0,1]\subset[-1,11]\), so the data does affect the answer. Specifically \(\varphi(x\pm ct) = \varphi(-1)\) and \(\varphi(11)\) are both zero, but the integral \(\frac{1}{4}\int_{-1}^{11}\psi\,ds\) picks up the whole kick.
STEP 4 · when would it not?
At \(t=1\) the interval is \([3,7]\), which misses \([0,1]\) entirely — so \(u(5,1)=0\). The signal simply has not arrived yet; it needs \(t \ge 2\) for the front from \(x=1\) to cover the 4 units to \(x=5\).
STEP 5 · the general principle
Hyperbolic equations have finite propagation speed and a sharp light cone. Contrast the heat equation, whose kernel is positive everywhere: heat from any point reaches every other point instantly, if only by an exponentially tiny amount.
3
A pulse on a string with fixed ends travels right and hits the end at \(x=L\). Describe what reflects back, and justify it with the odd extension.
Hint 1The fixed end must stay at zero for all time — what does that force the two travelling copies to do there?
Hint 2Odd extension means \(\varphi(-x) = -\varphi(x)\) reflected about the endpoint.
Solution
STEP 1 · what the boundary demands
\(u(L,t)=0\) for all \(t\). In d'Alembert's form the two travelling pieces must cancel exactly at \(x=L\), forever.
STEP 2 · build the extension
Extend \(\varphi\) to be odd about \(x=L\): \(\varphi(2L - x) = -\varphi(x)\). Then the incoming pulse meets a mirror-image, sign-flipped pulse arriving from the phantom region beyond \(L\).
STEP 3 · what happens at the moment of contact
The real pulse and its inverted image overlap at the end and sum to zero there — automatically satisfying the boundary condition. The string end never moves.
STEP 4 · what emerges
The phantom pulse continues into the physical region as the reflected wave: same shape, same speed, but upside down and travelling left.
STEP 5 · the free-end contrast
For an insulated/free end (\(u_x=0\)) you use the even extension instead, and the pulse reflects right-side up. Fixed end ⟹ inversion; free end ⟹ no inversion. This is the standard demonstration with a rope tied to a wall versus one held loosely.
4
Show that d'Alembert's formula reproduces the initial conditions, i.e. \(u(x,0)=\varphi(x)\) and \(u_t(x,0)=\psi(x)\).
Hint 1Setting \(t=0\) collapses the integral to zero-width.
Hint 2For \(u_t\), differentiate the integral with respect to its limits — that is the Fundamental Theorem plus the chain rule.
STEP 4 · set t = 0
\[ u_t(x,0) = 0 + \frac{\psi(x)+\psi(x)}{2} = \psi(x) \ ✓ \]
STEP 5 · and the PDE itself
Both pieces have the form \(F(x+ct)+G(x-ct)\) (the integral term is a difference of antiderivatives evaluated at those arguments), which module F1 problem 5 already verified solves \(u_{tt}=c^2u_{xx}\). So the formula satisfies the equation and both initial conditions — it is the solution, and uniqueness makes it the only one.
5
For \(u_{tt}=u_{xx}\) with \(\varphi = 0\) and \(\psi = 1\) on \([-1,1]\) (zero elsewhere), find \(u(0,3)\) and \(u(5,3)\).
Hint 1Only the integral term survives: \(u = \frac12\int_{x-t}^{x+t}\psi\).
Hint 2Ask how much of \([-1,1]\) each interval covers.
Solution
STEP 1 · the formula
\[ u(x,t) = \frac12\int_{x-t}^{x+t}\psi(s)\,ds \]
STEP 2 · evaluate at (0, 3)
The interval is \([-3,3]\), which contains all of \([-1,1]\):
\[ u(0,3) = \frac12(2) = 1 \]
STEP 3 · evaluate at (5, 3)
The interval is \([2,8]\), which misses \([-1,1]\) entirely:
\[ u(5,3) = 0 \]
STEP 4 · when does x = 5 first move?
The signal front from \(x=1\) travels at speed 1, so it arrives at \(x=5\) at \(t=4\). Before that, nothing.
STEP 5 · the contrast worth remembering
Under the heat equation the point \(x=5\) would be (slightly) warm immediately. Under the wave equation it is exactly zero until the front arrives — finite propagation speed with a sharp edge.
6
Solve \(u_{tt}=c^2u_{xx}\) with \(u(x,0)=x^2\), \(u_t(x,0)=0\), and simplify.
Hint 1\(u = \frac12\left[(x-ct)^2+(x+ct)^2\right]\).
Hint 2Expand both squares — the cross terms cancel.
Solution
STEP 1 · apply d'Alembert
\[ u = \frac{(x-ct)^2+(x+ct)^2}{2} \]
The parabola keeps its shape but rises uniformly at an accelerating rate. This is one of the rare cases where the two travelling waves recombine into something that looks like neither — a reminder that "two travelling shapes" is a statement about the decomposition, not always about the visible motion.
7
A pulse travels on the half-line \(x>0\) with the end fixed, \(u(0,t)=0\). Write the solution using an odd extension, and describe the reflected pulse.
Hint 1Extend \(\varphi\) to \(x<0\) by \(\varphi(-x) = -\varphi(x)\), then apply d'Alembert on the whole line.
Hint 2Check that the formula automatically gives \(u(0,t)=0\).
The boundary condition is satisfied by construction — that is the entire point of the odd extension.
STEP 4 · describe the reflection
For \(x>0\) the formula is a real right-moving pulse plus the image of a left-moving inverted pulse. When the incident wave reaches the wall it re-emerges travelling the other way, upside down, with the same shape and speed.
STEP 5 · the free-end version
With \(u_x(0,t)=0\) you use the even extension instead, and the pulse reflects the right way up. Fixed ⟹ inverted; free ⟹ upright. Both are one-line consequences of choosing the right symmetry.
8
Initial data is nonzero only on \([-1,1]\), with \(c=2\). Describe the domain of influence: the set of \((x,t)\) that the data can affect.
Hint 1Domain of dependence looks backward; domain of influence looks forward.
Hint 2A signal from \(x_0\) reaches \(x\) at time \(|x-x_0|/c\).
Solution
STEP 1 · trace forward from the left edge
The leftmost point \(x_0=-1\) emits a leftward signal along \(x = -1-2t\).
STEP 2 · trace forward from the right edge
The rightmost point \(x_0=1\) emits a rightward signal along \(x = 1+2t\).
STEP 3 · describe the region
\[ \text{influence} = \left\{(x,t) : -1-2t \le x \le 1+2t\right\} \]
an expanding wedge (forward light cone) opening from the initial interval.
STEP 4 · what happens outside it
\(u \equiv 0\) exactly. At \(t=3\) nothing beyond \(|x|=7\) has moved at all.
STEP 5 · the pairing
Influence (forward) and dependence (backward) are the same cone traversed in opposite directions: \(x_0\) influences \((x,t)\) precisely when \((x,t)\) depends on \(x_0\). Both are exclusive to hyperbolic equations — a diffusion has neither, since its kernel is positive everywhere.
9
Verify directly that the velocity term of d'Alembert's formula, \(v(x,t) = \frac{1}{2c}\int_{x-ct}^{x+ct}\psi(s)ds\), satisfies the wave equation.
Hint 1Differentiate with respect to \(t\) using the Leibniz rule for moving limits.
Hint 2Then differentiate again; you will need \(\psi'\) at the two endpoints.
And from step 1, \(v_t(x,0) = \psi(x)\) while \(v(x,0)=0\) — exactly the second initial condition, with no interference from the first.
Part III · Module 7 Separation & Fourier
Laplace's equation on a rectangle
No time variable at all — just an equilibrium shape determined entirely by its boundary. Separation still works, but now one factor oscillates while the other grows and decays, and the trick is to handle one nonzero side at a time.
①The problem and the strategy
Dirichlet problem on a rectangle
\[ u_{xx}+u_{yy}=0 \ \text{ on } 0
Separation needs homogeneous conditions on the two sides perpendicular to the direction you expand in. So the standard tactic is superposition of four subproblems: solve with one side nonzero and the other three zero, four times, then add. Here is the case with only the top edge nonzero, \(u(x,b)=f(x)\):
The sign flip is the whole story: in \(x\) we get oscillation, in \(y\) we get hyperbolic growth. Choosing \(\sinh\frac{n\pi y}{a}\) kills the bottom edge automatically, since \(\sinh 0=0\).
InteractiveEquilibrium set by the edgeschange the four boundary values
centre value —boundary average —interior max —
Solved numerically by relaxation — each interior point is repeatedly replaced by the average of its four neighbours, which is the discrete Laplace equation. Two things to notice: the interior is always smooth even with jumps at the corners, and the maximum and minimum always occur on the boundary, never strictly inside. That is the maximum principle, and it holds for every harmonic function.
The mean value property
A harmonic function's value at a point equals its average over any circle centred there. Two immediate consequences: the maximum principle (no interior peaks — a peak would exceed its own neighbourhood average), and uniqueness for the Dirichlet problem (the difference of two solutions is harmonic with zero boundary data, so its max and min are both 0).
Worked example · one hot edge
Solve \(\nabla^2u=0\) on the unit square with \(u=0\) on three sides and \(u(x,1)=1\) on the top.
STEP 1
Here \(a=b=1\), so \(X_n = \sin n\pi x\) and \(Y_n = \sinh n\pi y\) (chosen to vanish at \(y=0\)).
STEP 2
Impose the top condition: \(\sum c_n\sinh(n\pi)\sin n\pi x = 1\). The bracketed quantity \(c_n\sinh n\pi\) is just the sine coefficient of the constant 1 (module S1's worked example):
Evaluate at the centre: \(\frac{\sinh(\pi/2)}{\sinh\pi} \approx \frac{2.301}{11.549}\approx0.199\), so the first term gives \(\frac4\pi(0.199)\approx0.254\); the \(n=3\) term contributes about \(-0.004\). So \(u(\frac12,\frac12)\approx0.25\) — exactly a quarter, which is the boundary average \(\frac{1+0+0+0}{4}\) ✓ a beautiful check supplied by the mean value property.
②Practice
1
Separate \(u_{xx}+u_{yy}=0\) with \(u(0,y)=u(a,y)=0\) and explain why the \(y\)-factor involves \(\sinh\) and \(\cosh\) rather than sines.
Hint 1After separating, the two ODEs have opposite signs on the constant.
Hint 2\(Y'' = +\lambda Y\) with \(\lambda>0\) has exponential solutions.
STEP 2 · the two ODEs
\[ X'' + \lambda X = 0, \qquad Y'' - \lambda Y = 0 \]
Note the opposite signs — forced by the plus sign between \(u_{xx}\) and \(u_{yy}\).
STEP 3 · which variable gets the eigenvalue problem
The homogeneous conditions are in \(x\), so \(X\) must oscillate to hit zero twice: \(\lambda_n = (n\pi/a)^2 > 0\) with \(X_n = \sin\frac{n\pi x}{a}\).
STEP 4 · consequence for Y
With \(\lambda_n>0\), \(Y'' = \lambda_nY\) gives real exponentials, conventionally written \(\cosh\frac{n\pi y}{a}\) and \(\sinh\frac{n\pi y}{a}\).
STEP 5 · why this is physically right
Laplace's equation describes equilibrium, not oscillation. A harmonic function cannot wiggle in both directions — if it oscillates in \(x\) it must grow or decay in \(y\). This is exactly the elliptic character from module F3: no propagation, no vibration, just a boundary influence that dies away exponentially into the interior.
2
Find \(u\) at the centre of a square whose four edges are held at 10, 20, 30, 40 degrees, without solving the PDE.
Hint 1Use symmetry and superposition: think of the problem as the sum of four one-edge problems.
Hint 2By symmetry, a single edge at value \(V\) contributes \(V/4\) at the centre.
Solution
STEP 1 · split by superposition
Laplace's equation is linear, so write \(u = u_1+u_2+u_3+u_4\), where \(u_k\) has edge \(k\) at its given value and the other three at zero.
STEP 2 · use the square's symmetry
Each one-edge problem is a rotation of the others. If a single edge at value 1 produces \(\alpha\) at the centre, then by symmetry all four do.
STEP 3 · pin down α
Set all four edges to 1. The solution is then \(u\equiv1\) (constant functions are harmonic), so \(4\alpha = 1\) and \(\alpha = \frac14\).
It matches the worked example's numerical result for a single hot edge (\(\frac14\) of 1) ✓, and it is the mean value property in its cleanest form: the centre sees the average of the boundary.
3
Explain why \(u=x^2-y^2\) is harmonic, and verify the maximum principle on the square \([-1,1]^2\).
Hint 1Compute \(u_{xx}\) and \(u_{yy}\) and add.
Hint 2For the maximum principle, evaluate \(u\) at the corners and at the centre.
At the centre \(u(0,0)=0\), comfortably between. In fact \(|x^2-y^2|\le1\) throughout the square, so the interior never exceeds the boundary extremes ✓
STEP 4 · note the critical point
\(\nabla u = (2x,-2y) = 0\) at the origin — but it is a saddle, not a max or min. That is the general situation: harmonic functions can have critical points, but never interior extrema, because \(u_{xx}\) and \(u_{yy}\) must have opposite signs.
STEP 5 · the mean value check
Average \(u\) over the unit circle centred at the origin: \(\frac{1}{2\pi}\int_0^{2\pi}(\cos^2\theta-\sin^2\theta)d\theta = \frac{1}{2\pi}\int_0^{2\pi}\cos2\theta\,d\theta = 0 = u(0,0)\) ✓
4
Why is the Dirichlet problem for Laplace's equation uniquely solvable, while the pure Neumann problem is not?
Hint 1For uniqueness, look at the difference of two solutions and apply the maximum principle.
Hint 2For Neumann, ask what happens if you add a constant to a solution.
Solution
STEP 1 · Dirichlet uniqueness
Suppose \(u_1,u_2\) both solve it. Their difference \(w = u_1-u_2\) is harmonic with \(w=0\) on the entire boundary.
STEP 2 · apply the maximum principle
Both the maximum and minimum of \(w\) occur on the boundary, where \(w=0\). So \(0 \le w \le 0\) throughout, hence \(w\equiv0\) and \(u_1=u_2\) ✓
STEP 3 · Neumann non-uniqueness
If \(u\) solves \(\nabla^2u=0\) with \(\partial_nu = g\), so does \(u+C\) for any constant — adding a constant changes no derivative. Uniqueness fails by exactly one constant.
STEP 4 · and existence needs a condition
Integrating the equation over the region and applying the divergence theorem:
So \(\oint g\,ds = 0\) is necessary: total flux in must equal total flux out.
STEP 5 · the pattern
Identical to the insulated bar of module F4: Neumann problems lose uniqueness by a constant and gain a compatibility requirement. Fixing the constant needs extra information — a reference value somewhere, or a conservation statement.
5
Solve \(\nabla^2u=0\) on the unit square with \(u=0\) on three sides and \(u(x,1)=\sin3\pi x\).
Hint 1The data is a single eigenfunction — only \(n=3\) appears.
Hint 2Choose the \(y\)-factor that vanishes at \(y=0\).
Hint 3Normalize so the top edge matches exactly.
Solution
STEP 1 · pick the mode
With \(a=1\), \(X_n = \sin n\pi x\). The data is exactly \(n=3\), so no integration is needed.
STEP 2 · choose the y-factor
\(Y'' = \lambda_3Y\) with \(\lambda_3 = 9\pi^2\) gives \(\sinh3\pi y\) and \(\cosh3\pi y\). The bottom edge \(u(x,0)=0\) selects \(\sinh\), since \(\sinh0=0\).
STEP 3 · normalize on the top edge
\[ c\sinh(3\pi)\sin3\pi x = \sin3\pi x \quad\Longrightarrow\quad c = \frac{1}{\sinh3\pi} \]
STEP 4 · the solution
\[ u(x,y) = \frac{\sinh3\pi y}{\sinh3\pi}\,\sin3\pi x \]
STEP 5 · how far the data penetrates
At \(y=0.5\): \(\frac{\sinh(1.5\pi)}{\sinh(3\pi)} \approx \frac{55.7}{6195} \approx 0.9\%\). The fine \(n=3\) structure is essentially invisible halfway down — the elliptic analogue of high modes dying fast, and the same phenomenon as \((r/a)^n\) on the disk.
6
On the unit square, \(u=0\) on the left and right edges, \(u(x,0)=\sin\pi x\), \(u(x,1)=2\sin\pi x\). Solve.
Hint 1Both nonzero edges use the same mode, so keep both \(\sinh\) and \(\cosh\) — or use \(\sinh\) of shifted arguments.
Hint 2Write \(Y = A\cosh\pi y + B\sinh\pi y\) and impose \(Y(0)=1\), \(Y(1)=2\).
Solution
STEP 1 · reduce to one mode
Both boundary functions are \(\sin\pi x\), so \(u = Y(y)\sin\pi x\) with \(Y''=\pi^2Y\).
STEP 2 · general Y
\[ Y = A\cosh\pi y + B\sinh\pi y \]
STEP 3 · apply the bottom edge
\[ Y(0) = A = 1 \]
STEP 4 · apply the top edge
\[ Y(1) = \cosh\pi + B\sinh\pi = 2 \quad\Longrightarrow\quad B = \frac{2-\cosh\pi}{\sinh\pi} \]
STEP 5 · the solution
\[ u(x,y) = \left[\cosh\pi y + \frac{2-\cosh\pi}{\sinh\pi}\sinh\pi y\right]\sin\pi x \]
Numerically \(\cosh\pi\approx11.59\), \(\sinh\pi\approx11.55\), so \(B\approx-0.830\). Check: \(Y(1) = 11.59 - 0.830(11.55) \approx 2.0\) ✓ Note this is exactly the superposition of two one-edge problems, done in a single step because both edges excited the same mode.
7
A harmonic function on a region has boundary values everywhere between 10 and 40. What can you say about its interior values, and about where its extremes occur?
Hint 1Use the maximum principle in both directions.
Hint 2Can an interior point exceed every boundary value?
Solution
STEP 1 · apply the maximum principle
A harmonic function attains its maximum on the boundary, so \(u \le 40\) everywhere.
STEP 2 · apply it to −u
\(-u\) is also harmonic, so its maximum is on the boundary too — meaning \(u\)'s minimum is on the boundary and \(u \ge 10\).
STEP 3 · state the bound
\[ 10 \le u \le 40 \ \text{ throughout the region} \]
STEP 4 · where the extremes live
Strictly on the boundary — unless \(u\) is constant, in which case every point is both. There are no interior hot spots in a steady temperature field with no sources.
STEP 5 · why it must be so
The mean value property: \(u\) at a point equals its average over any small circle around it. A strict interior maximum would have to exceed its own neighbourhood average — impossible. This is also the fastest sanity check on any numerical Laplace solver: if the solution shows an interior peak, the code is wrong.
8
Show that \(u = e^x\sin y\) is harmonic, and find where it is largest on the rectangle \(0\le x\le1\), \(0\le y\le\pi\).
Hint 1Compute \(u_{xx}\) and \(u_{yy}\) and add.
Hint 2By the maximum principle, look only on the boundary.
The maximum principle guarantees the extreme is on the edge, so check all four.
STEP 3 · examine each edge
On \(y=0\) and \(y=\pi\): \(u=0\). On \(x=0\): \(u = \sin y \le 1\). On \(x=1\): \(u = e\sin y\), maximal at \(y=\frac\pi2\).
STEP 4 · the maximum
\[ u_{\max} = e \approx 2.718 \ \text{ at } (1, \tfrac{\pi}{2}) \]
STEP 5 · confirm no interior candidate
\(\nabla u = (e^x\sin y,\ e^x\cos y)\) vanishes only if \(\sin y = \cos y = 0\), which is impossible. No interior critical point at all — consistent with the theory ✓
9
In the standard rectangle problem you choose \(Y = \sinh\frac{n\pi y}{a}\) rather than \(\cosh\). Explain the choice, and what you would use if the bottom edge were the nonzero one.
Hint 1Which of \(\sinh\) and \(\cosh\) vanishes at zero?
Hint 2For a nonzero bottom edge, you need a function vanishing at \(y=b\) instead.
Solution
STEP 1 · the requirement
With only the top edge nonzero, the bottom edge needs \(u(x,0)=0\), i.e. \(Y(0)=0\).
STEP 2 · test the two candidates
\[ \sinh 0 = 0 \ ✓, \qquad \cosh 0 = 1 \ ✗ \]
So \(\sinh\frac{n\pi y}{a}\) is the right choice; any \(\cosh\) component would spoil the bottom edge.
STEP 3 · the reversed problem
If instead the bottom edge carries the data and the top is zero, you need \(Y(b)=0\). The natural choice is
\[ Y = \sinh\frac{n\pi(b-y)}{a} \]
which vanishes at \(y=b\) automatically.
STEP 4 · the general recipe
Build each hyperbolic factor to vanish at the opposite edge from the data. That way each subproblem in the four-way superposition satisfies three homogeneous sides by construction.
STEP 5 · the parallel elsewhere
The same "choose the solution that satisfies the free condition automatically" logic picks \(J_n\) over \(Y_n\) on a disk (module A4), \(r^n\) over \(r^{-n}\) (module S8), and the decaying exponential on a semi-infinite domain. It is the single most useful habit in separation problems.
Part III · Module 8 Separation & Fourier
Laplace's equation on a disk
Change to polar coordinates and separation still works — but the radial equation is now an Euler equation, and the angular variable brings a new kind of boundary condition: periodicity.
There is no boundary in \(\theta\) — instead, \(\Theta\) must be \(2\pi\)-periodic, which forces \(\lambda = n^2\) with \(\Theta_n = \cos n\theta, \sin n\theta\). The radial equation is then an Euler equation with solutions \(r^n\) and \(r^{-n}\); on a solid disk we discard \(r^{-n}\) because it blows up at the centre.
Solution on the disk \(r
\[ u(r,\theta) = \frac{a_0}{2} + \sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^{n}\left[a_n\cos n\theta + b_n\sin n\theta\right] \]
with \(a_n, b_n\) the ordinary Fourier coefficients of \(f\) on \([-\pi,\pi]\).
InteractiveBoundary data filling a diskpick boundary data; watch it diffuse inward
centre value —boundary average —decay of mode n —
Each boundary harmonic \(\cos n\theta\) penetrates the disk as \((r/a)^n\) — so fine angular detail dies away extremely fast as you move inward, while the average survives all the way to the centre. That is why the "hot arc" looks sharp at the rim but almost uniform halfway in, and why the centre value always equals the boundary average exactly.
Poisson's integral formula
Summing the series in closed form gives \(u(r,\theta) = \frac{a^2-r^2}{2\pi}\displaystyle\int_{-\pi}^{\pi}\frac{f(\phi)}{a^2-2ar\cos(\theta-\phi)+r^2}d\phi\) — the boundary values weighted by the Poisson kernel. At \(r=0\) the kernel is constant and the formula reduces to the mean value property.
Worked example · half the rim hot
Solve on the unit disk with \(f(\theta)=1\) for \(0<\theta<\pi\) and \(0\) for \(-\pi<\theta<0\).
STEP 1
Compute the Fourier coefficients of \(f\) on \([-\pi,\pi]\). The average is \(\frac{a_0}{2} = \frac12\).
STEP 2
\(a_n = \frac1\pi\int_0^\pi\cos n\theta\,d\theta = 0\) for all \(n\ge1\) — the cosine terms all vanish.
STEP 3
\(b_n = \frac1\pi\int_0^\pi\sin n\theta\,d\theta = \frac{1-(-1)^n}{n\pi}\), so \(b_n = \frac{2}{n\pi}\) for odd \(n\), zero for even.
At the centre \(r=0\), everything but the average vanishes: \(u(0)=\frac12\) — the mean of "half hot, half cold" ✓ And at \(r=0.5\) the \(n=3\) term is already down by \(0.5^3/3 \approx 4\%\) of the leading one, so the interior is far smoother than the rim.
②Practice
1
Solve \(\nabla^2u=0\) on the unit disk with \(u(1,\theta)=3+2\cos\theta - \cos2\theta\).
Hint 1The data is already a Fourier series — read off the coefficients.
Hint 2Mode \(n\) is multiplied by \(r^n\) inside the disk.
In Cartesian terms \(r\cos\theta = x\) and \(r^2\cos2\theta = x^2-y^2\), so
\[ u = 3 + 2x - (x^2-y^2) \]
and \(\nabla^2u = -2+2 = 0\) ✓
STEP 5 · centre value
\(u(0) = 3\), which is the average of the boundary data since both cosine terms integrate to zero ✓
2
Why is \(r^{-n}\) discarded on a solid disk but kept for an annulus \(1
Hint 1What happens to \(r^{-n}\) as \(r\to0\)?
Hint 2Is \(r=0\) part of an annulus?
Solution
STEP 1 · the radial solutions
The Euler equation \(r^2R''+rR'-n^2R=0\) has solutions \(r^n\) and \(r^{-n}\) (and \(1, \ln r\) when \(n=0\)).
STEP 2 · on a solid disk
\(r=0\) is an interior point of the domain. \(r^{-n}\to\infty\) there, and physical temperature or potential must be finite, so those solutions are rejected — a boundedness condition doing the job of a boundary condition.
STEP 3 · on an annulus
The origin is excluded, so \(r^{-n}\) is perfectly well behaved on \(1\le r\le2\) and must be kept.
STEP 4 · the count works out
An annulus has two boundary circles, needing two conditions per mode — exactly matching the two constants in \(A_nr^n + B_nr^{-n}\). A disk has one boundary circle and one constant, once boundedness has removed the other.
STEP 5 · the n = 0 case
For \(n=0\) the pair is \(1\) and \(\ln r\). On an annulus, \(\ln r\) is the potential of a line charge and appears constantly in electrostatics; on a disk it is discarded for the same reason as \(r^{-n}\).
3
The boundary of a unit disk is held at \(f(\theta)=\cos^2\theta\). Find \(u\) inside and its centre value.
Hint 1\(\cos^2\theta\) is not a single harmonic — use a double-angle identity first.
Hint 2\(\cos^2\theta = \frac{1+\cos2\theta}{2}\).
Solution
STEP 1 · expand the data
\[ \cos^2\theta = \frac12 + \frac12\cos2\theta \]
Now it is a Fourier series: \(\frac{a_0}{2} = \frac12\), \(a_2 = \frac12\).
At \(r=1\): \(\frac12+\frac12\cos2\theta = \cos^2\theta\) ✓
STEP 4 · centre value
\[ u(0) = \frac12 \]
which is the average of \(\cos^2\theta\) over a full circle ✓ (a classic average worth remembering).
STEP 5 · Cartesian check
\(r^2\cos2\theta = x^2-y^2\), so \(u = \frac12 + \frac{x^2-y^2}{2}\), whose Laplacian is \(\frac{2-2}{2}=0\) ✓
4
Explain, using the factor \((r/a)^n\), why a small hot patch on the rim is barely detectable at the centre of a large disk.
Hint 1A narrow patch needs many high-\(n\) harmonics to represent it.
Hint 2How does \((r/a)^n\) behave for large \(n\) when \(r
Solution
STEP 1 · what a narrow patch requires
A feature of angular width \(\delta\) needs harmonics up to about \(n \sim 2\pi/\delta\) — the narrower the patch, the higher the modes carrying it.
STEP 2 · how those modes penetrate
Mode \(n\) is damped by \((r/a)^n\). At the half-radius \(r = a/2\) that is \(2^{-n}\): mode 10 is attenuated by a factor of about 1000, mode 20 by a million.
STEP 3 · what survives to the centre
At \(r=0\), every \(n\ge1\) term is exactly zero. Only \(\frac{a_0}{2}\), the boundary average, remains — and a narrow patch contributes only \(\frac{\delta}{2\pi}\) of its height to that average.
STEP 4 · the quantitative statement
A patch of width \(\delta = 0.1\) rad at temperature 100° raises the centre by only \(100\times\frac{0.1}{2\pi}\approx1.6\)°.
STEP 5 · the general principle
Elliptic problems smooth in space the way parabolic problems smooth in time: fine boundary detail is exponentially forgotten as you move into the interior. It is also why the interior of a harmonic function is always analytic no matter how rough its boundary data.
5
Solve \(\nabla^2u=0\) on the unit disk with \(u(1,\theta)=4+3\sin2\theta\), and give the value at \(r=\frac12,\ \theta=\frac\pi4\).
Hint 1Read off the Fourier coefficients directly.
Hint 2Mode \(n\) carries \(r^n\).
Solution
STEP 1 · identify the coefficients
\(\frac{a_0}{2}=4\) and \(b_2 = 3\); all others zero.
Logarithmic, not linear: at the midpoint \(r=1.5\), \(u = 10\frac{\ln1.5}{\ln2}\approx5.85\), not 5. This is the classic temperature profile through a pipe wall, and the reason insulation thickness has diminishing returns — each extra millimetre adds less resistance than the last.
7
Show that Poisson's integral formula reduces to the mean value property at the centre of the disk.
Hint 1Set \(r=0\) in the kernel \(\frac{a^2-r^2}{a^2-2ar\cos(\theta-\phi)+r^2}\).
Hint 2What is left is a constant, so the integral becomes a plain average.
Solution
STEP 1 · write the formula
\[ u(r,\theta) = \frac{a^2-r^2}{2\pi}\int_{-\pi}^{\pi}\frac{f(\phi)\,d\phi}{a^2-2ar\cos(\theta-\phi)+r^2} \]
STEP 2 · set r = 0
The numerator becomes \(a^2\) and the denominator becomes \(a^2\), independent of \(\phi\):
In \(u = \frac{a_0}{2}+\sum(r/a)^n[\cdots]\), every \(n\ge1\) term vanishes at \(r=0\), leaving \(\frac{a_0}{2}\) — the same average. Two routes, one answer.
STEP 5 · what the kernel does elsewhere
For \(r>0\) the kernel is largest for \(\phi\) near \(\theta\), so nearby boundary points count more. As \(r\to a\) it becomes a spike, reproducing \(f(\theta)\) exactly — the boundary condition emerging from the formula itself.
8
For the Neumann problem on a disk, \(\nabla^2u=0\) with \(u_r(a,\theta)=g(\theta)\), derive the compatibility condition on \(g\).
Hint 1Integrate the PDE over the disk and use the divergence theorem.
Hint 2\(\int_\Omega\nabla^2u\,dA = \oint_{\partial\Omega}u_r\,ds\).
STEP 3 · state the condition
\[ \int_{-\pi}^{\pi}g(\theta)\,d\theta = 0 \]
STEP 4 · interpret
The net flux through the boundary must vanish. Physically: in a steady state with no interior sources, whatever heat enters must also leave. Prescribing a net inflow makes a steady state impossible.
STEP 5 · and uniqueness
Even when compatible, the solution is determined only up to an additive constant — adding a constant changes no normal derivative. Exactly the structure of the insulated bar (module F4) and the Neumann rectangle (module S7 problem 4).
9
Boundary data on a disk of radius 1 has a sharp jump. Estimate how many terms of the series you need for 1% accuracy at \(r=0.9\) versus \(r=0.5\).
Hint 1A jump gives coefficients decaying like \(1/n\), and mode \(n\) is damped by \(r^n\).
Hint 2Solve \(r^N \approx 0.01\) for each radius.
Solution
STEP 1 · the size of the Nth term
Coefficients \(\sim1/n\) times the radial damping \(r^n\), so the tail is governed by \(r^N\).
STEP 2 · at r = 0.5
\[ 0.5^N = 0.01 \quad\Longrightarrow\quad N = \frac{\ln 0.01}{\ln 0.5} \approx 6.6 \ \Rightarrow\ \text{about 7 terms} \]
STEP 3 · at r = 0.9
\[ 0.9^N = 0.01 \quad\Longrightarrow\quad N = \frac{\ln 0.01}{\ln 0.9} \approx 43.7 \ \Rightarrow\ \text{about 44 terms} \]
STEP 4 · and at the boundary itself
At \(r=1\) there is no damping at all, the coefficients decay only like \(1/n\), and Gibbs ringing means you can never reach 1% near the jump — no finite \(N\) suffices.
STEP 5 · the practical lesson
Series solutions of elliptic problems are cheap deep inside and expensive near the boundary. For accurate boundary-layer values, use Poisson's integral formula (which handles the jump exactly) instead of truncating the series.
Part IV · Module 1 Advanced tools
Sturm–Liouville theory
Why does separation of variables always produce an orthogonal family of eigenfunctions, whatever the boundary conditions? Because every such eigenvalue problem belongs to one structural class — and that class comes with orthogonality, real eigenvalues, and completeness guaranteed.
with separated boundary conditions at each end. Here \(w(x)>0\) is the weight. Our familiar problem \(X''+\lambda X=0\) is the case \(p=1\), \(q=0\), \(w=1\). The theory guarantees:
All eigenvalues are real, and form an increasing sequence \(\lambda_1<\lambda_2<\cdots\to\infty\).
Eigenfunctions for different eigenvalues are orthogonal with respect to \(w\): \(\int_a^bX_nX_m\,w\,dx = 0\).
The eigenfunctions are complete — any reasonable \(f\) can be expanded as \(\sum c_nX_n\), with \(c_n = \frac{\int fX_nw\,dx}{\int X_n^2w\,dx}\).
\(X_n\) has exactly \(n-1\) interior zeros.
InteractiveEigenfunctions and orthogonalitypick two modes; inspect their product
∫ XₘXₙ dx —verdict —zeros of Xₙ —
Top: the two eigenfunctions. Bottom: their product, with positive area in green and negative in red. When \(m\ne n\) the two areas cancel exactly — that is orthogonality, and it is what makes the coefficient formula work. Set \(m=n\) and the product is a square, all positive, giving the normalization constant. Notice too that \(X_n\) always has exactly \(n-1\) interior zeros.
②Why orthogonality is automatic
Take two eigenpairs and compute \(\int_a^b\left[X_m\mathcal{L}X_n - X_n\mathcal{L}X_m\right]dx\) with \(\mathcal{L}X = (pX')'\). Integrating by parts twice turns the whole thing into a boundary term (this is Lagrange's identity):
Standard boundary conditions make the right side vanish. So if \(\lambda_m\ne\lambda_n\), the integral must be zero. Orthogonality isn't a lucky property of sines — it is forced by the structure of the operator plus the boundary conditions.
Worked example · putting an equation into S–L form
Write \(x^2X'' + xX' + \lambda X = 0\) in Sturm–Liouville form and identify the weight.
STEP 1
Divide through by \(x\) to make the first two terms an exact derivative:
\[ xX'' + X' + \frac{\lambda}{x}X = 0 \]
STEP 2
Recognize \(xX''+X' = (xX')'\) by the product rule.
STEP 3
So the equation is
\[ \big(xX'\big)' + \lambda\cdot\frac1x\,X = 0 \]
STEP 4
Comparing with the standard form: \(p(x) = x\), \(q = 0\), and the weight is
\[ w(x) = \frac1x \]
STEP 5
So eigenfunctions of this problem are orthogonal with respect to \(1/x\): \(\int_a^bX_mX_n\frac{dx}{x} = 0\). Forgetting the weight is the classic error — the plain integral \(\int X_mX_ndx\) is not zero here.
③Practice
1
Put \(X'' + 2X' + \lambda X = 0\) into Sturm–Liouville form and find \(p\) and \(w\).
Hint 1Multiply by an integrating factor \(\mu(x)\) so the first two terms become \((\mu X')'\).
Hint 2You need \(\mu' = 2\mu\), the same trick as for first-order linear ODEs.
Solution
STEP 1 · look for an integrating factor
Multiply by \(\mu(x)\):
\[ \mu X'' + 2\mu X' + \lambda\mu X = 0 \]
We want the first two terms to be \((\mu X')' = \mu X'' + \mu'X'\), so we need \(\mu' = 2\mu\).
STEP 2 · solve for μ
\[ \mu(x) = e^{2x} \]
STEP 3 · write the S–L form
\[ \left(e^{2x}X'\right)' + \lambda e^{2x}X = 0 \]
Every second-order linear ODE can be put in this form by the same integrating-factor trick, which is why Sturm–Liouville theory covers the whole subject rather than a special case.
2
Verify directly that the eigenfunctions \(\sin\frac{(2n-1)\pi x}{2L}\) of the mixed problem \(X(0)=X'(L)=0\) are orthogonal on \([0,L]\).
Hint 1Use the product-to-sum identity as in module S1 problem 3.
Hint 2The frequencies differ by \(\frac{(m-n)\pi}{L}\) and sum to \(\frac{(m+n-1)\pi}{L}\); check where the resulting sines vanish.
Solution
STEP 1 · name the frequencies
Let \(\alpha_n = \frac{(2n-1)\pi}{2L}\). The claim is \(\int_0^L\sin\alpha_mx\,\sin\alpha_nx\,dx = 0\) for \(m\ne n\).
STEP 2 · product to sum
\[ \sin\alpha_mx\sin\alpha_nx = \tfrac12\left[\cos(\alpha_m-\alpha_n)x - \cos(\alpha_m+\alpha_n)x\right] \]
\((\alpha_m-\alpha_n)L = (m-n)\pi\), whose sine is 0. And \((\alpha_m+\alpha_n)L = (m+n-1)\pi\), also an integer multiple of \(\pi\), so its sine is 0 too.
STEP 5 · write the expansion
\[ 1 = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{1}{2n-1}\sin\frac{(2n-1)\pi x}{2L}, \qquad 0Sanity check at \(x=L\): every term is \(\pm1\)... in fact \(\sin\frac{(2n-1)\pi}{2}\) alternates, and the series converges to 1 there ✓ — unlike the ordinary sine series, this basis does not force zero at the right end, because these eigenfunctions have zero slope there instead.
4
Prove that a Sturm–Liouville problem with \(q\le0\) and Dirichlet conditions has no negative eigenvalues.
Hint 1Multiply the equation by \(X\) and integrate over \([a,b]\).
Hint 2Integrate the \((pX')'X\) term by parts; the boundary term dies by the Dirichlet conditions.
Hint 3Solve the resulting identity for \(\lambda\) — this is the Rayleigh quotient.
Solution
STEP 1 · multiply by X and integrate
\[ \int_a^b\left[(pX')'X + qX^2 + \lambda wX^2\right]dx = 0 \]
STEP 2 · integrate the first term by parts
\[ \int_a^b(pX')'X\,dx = \big[pX'X\big]_a^b - \int_a^bp(X')^2dx = -\int_a^bp(X')^2dx \]
since \(X(a)=X(b)=0\) kills the boundary term.
STEP 3 · rearrange into the Rayleigh quotient
\[ \lambda = \frac{\displaystyle\int_a^bp(X')^2dx - \int_a^bqX^2dx}{\displaystyle\int_a^bwX^2dx} \]
STEP 4 · check the signs
With \(p>0\) and \(w>0\), the first numerator integral is \(\ge0\) and the denominator is \(>0\). With \(q\le0\), the term \(-\int qX^2\) is also \(\ge0\).
STEP 5 · conclude
\[ \lambda \ge 0 \]
and \(\lambda=0\) requires \(X'\equiv0\), i.e. \(X\) constant — impossible with Dirichlet ends unless \(X\equiv0\). So all eigenvalues are strictly positive ✓
WHY THIS MATTERS
Positive eigenvalues mean \(e^{-k\lambda_nt}\) decays for every mode — the heat problem is stable. The Rayleigh quotient also gives the practical estimate \(\lambda_1 \le \frac{\int p(v')^2}{\int wv^2}\) for any trial \(v\) satisfying the boundary conditions, which is the basis of finite-element methods.
5
Put \(x^2X'' + 3xX' + \lambda X = 0\) into Sturm–Liouville form and identify \(p\) and \(w\).
Hint 1Look for a factor \(\mu(x)\) making the first two terms an exact derivative \((\mu x^2 X')'\)… or divide first and use the integrating-factor recipe.
Hint 2Divide by \(x^2\) to get \(X'' + \frac3xX' + \frac{\lambda}{x^2}X=0\), then use \(\mu = e^{\int 3/x\,dx} = x^3\).
Solution
STEP 1 · normalize the leading coefficient
\[ X'' + \frac3xX' + \frac{\lambda}{x^2}X = 0 \]
STEP 2 · find the integrating factor
For \(X''+P(x)X'\) to become \((\mu X')'/\mu\), take \(\mu = e^{\int P\,dx}\):
STEP 4 · read off the ingredients
\[ p(x) = x^3, \qquad q(x) = 0, \qquad w(x) = x \]
STEP 5 · the orthogonality it implies
\[ \int_a^b X_mX_n\,x\,dx = 0 \quad (m\ne n) \]
The weight \(x\) is not optional. Weights like this arise naturally in polar and cylindrical problems — the \(r\) in \(\int_0^a J_n(\alpha_mr)J_n(\alpha_nr)\,r\,dr = 0\) is exactly the same phenomenon.
6
Use the Rayleigh quotient with the trial function \(v = x(L-x)\) to estimate \(\lambda_1\) for \(X''+\lambda X=0\), \(X(0)=X(L)=0\), and compare with the exact value.
Hint 1\(\lambda_1 \le \frac{\int_0^L(v')^2dx}{\int_0^Lv^2dx}\) for any \(v\) meeting the boundary conditions.
Hint 2\(v' = L-2x\); compute both integrals.
Hint 3\(\int_0^L(L-2x)^2dx = \frac{L^3}{3}\) and \(\int_0^Lx^2(L-x)^2dx = \frac{L^5}{30}\).
Solution
STEP 1 · check the trial function is admissible
\(v(0)=v(L)=0\) ✓, and \(v\) is smooth — that is all the Rayleigh quotient requires.
The estimate \(10/L^2\) is high by only 1.3% — from a crude parabola. Because the quotient is stationary at the true eigenfunction, errors in \(v\) enter quadratically, which is why rough trial functions give excellent eigenvalue estimates. This is the entire principle behind Rayleigh–Ritz and finite-element methods.
7
Prove that the eigenvalues of a Sturm–Liouville problem are real.
Hint 1Suppose \(\lambda\) is complex with eigenfunction \(X\), and take complex conjugates of the whole equation.
Hint 2Apply Lagrange's identity to the pair \(X\) and \(\bar X\).
Hint 3\(\int X\bar X w\,dx = \int|X|^2w\,dx > 0\).
Solution
STEP 1 · conjugate the equation
If \((pX')' + (q+\lambda w)X = 0\) with real \(p,q,w\), then conjugating gives
\[ (p\bar X')' + (q+\bar\lambda w)\bar X = 0 \]
so \(\bar\lambda\) is an eigenvalue with eigenfunction \(\bar X\).
A complex eigenvalue would give \(e^{-k\lambda t}\) an oscillating part — a diffusion problem that rings. Reality of the spectrum is what guarantees pure decay for the heat equation and pure oscillation for the wave equation. It is the infinite-dimensional version of "symmetric matrices have real eigenvalues."
8
Are the Dirichlet eigenfunctions \(\sin\frac{n\pi x}{L}\) orthogonal to the Neumann eigenfunctions \(\cos\frac{m\pi x}{L}\) on \([0,L]\)? What does your answer say about mixing bases?
Hint 1Compute \(\int_0^L\sin\frac{n\pi x}{L}\cos\frac{m\pi x}{L}dx\) for a simple case, say \(n=m=1\).
Hint 2Orthogonality is guaranteed only within one Sturm–Liouville problem.
With \(n=1\), \(m=0\) (the constant eigenfunction):
\[ \int_0^\pi\sin x\,dx = 2 \ne 0 \]
Not orthogonal.
STEP 3 · why the guarantee doesn't apply
Lagrange's identity kills the boundary term only when both functions satisfy the same boundary conditions. A sine (zero at the ends) and a cosine (zero slope at the ends) do not, so nothing forces the integral to vanish.
STEP 4 · the practical rule
Choose one eigenbasis, determined by your boundary conditions, and expand everything — initial data, forcing, steady states — in that single basis. Mixing bases destroys the orthogonality that makes coefficient extraction possible.
STEP 5 · the exception worth knowing
On \([-L,L]\) with periodic conditions, sines and cosines belong to the same problem and are mutually orthogonal — which is exactly why the full Fourier series works there (module F4 problem 9).
9
State the Sturm–Liouville oscillation theorem and use it to sketch \(X_4\) for \(X''+\lambda X=0\) with \(X(0)=X'(L)=0\).
Hint 1The theorem counts the interior zeros of each eigenfunction.
Hint 2For this problem \(X_n = \sin\frac{(2n-1)\pi x}{2L}\).
Solution
STEP 1 · the theorem
For a regular Sturm–Liouville problem, the eigenvalues form an increasing sequence \(\lambda_1<\lambda_2<\cdots\), and the \(n\)th eigenfunction has exactly \(n-1\) zeros strictly inside the interval.
STEP 2 · apply it here
\(X_4\) should have exactly 3 interior zeros.
STEP 3 · check against the formula
\(X_4 = \sin\frac{7\pi x}{2L}\). Its zeros occur where \(\frac{7\pi x}{2L} = k\pi\), i.e. \(x = \frac{2kL}{7}\). Inside \((0,L)\) that gives \(k=1,2,3\) — three zeros at \(x = \frac{2L}{7},\frac{4L}{7},\frac{6L}{7}\) ✓
STEP 4 · sketch it
Starts at zero (clamped end), oscillates through three interior crossings, and arrives at \(x=L\) at a peak rather than a zero, since \(X'(L)=0\). Seven quarter-waves in total.
STEP 5 · why it is useful
Counting zeros identifies which eigenvalue you have found numerically without computing the others — indispensable when the eigenfunctions are Bessel or Legendre functions with no closed-form zeros. It also confirms the physical picture: higher modes wiggle more, and therefore decay faster.
Part IV · Module 2 Advanced tools
Nonhomogeneous problems
Real problems have sources, and real boundaries are rarely held at zero. Two techniques handle everything: subtract a steady state to fix the boundaries, then expand the leftover forcing in the eigenfunctions you already have.
①Step one: kill the boundary data
For \(u_t = ku_{xx}+Q(x)\) with \(u(0,t)=A\), \(u(L,t)=B\), find the steady state \(v(x)\) solving \(kv''+Q=0\) with those boundary values, then set \(u = v + w\). The transient \(w\) satisfies the homogeneous problem with zero boundary data — ready for separation (module F5).
②Step two: eigenfunction expansion for time-dependent forcing
If the source depends on time, \(u_t = ku_{xx}+f(x,t)\) with \(u=0\) at both ends, expand everything in the eigenfunctions of the homogeneous problem:
Expand the unknown and the forcing in the same basis
Substituting and matching coefficients (legitimate because the eigenfunctions are independent) turns the PDE into an infinite family of decoupled ODEs:
That integral is Duhamel's principle: the response now is the accumulated memory of every past kick, each decayed by how long ago it happened.
InteractiveDriven systems: decay, buildup, resonancetry forcing on and off resonance
mode-1 amplitude —natural frequency ω₁ —status —
With heat + a steady source the solution climbs and levels off at the steady state — diffusion always wins eventually. With wave + forcing, tune \(\omega\) to the natural frequency \(\omega_1\) and the amplitude grows without bound: resonance, because the undamped wave equation has no mechanism to dissipate the energy you keep adding.
Worked example · a steady source switched on
Solve \(u_t = u_{xx} + \sin\pi x\) on \(0
STEP 1
The forcing is already a single eigenfunction (\(n=1\), \(\lambda_1 = \pi^2\)), so only mode 1 is excited: \(f_1 = 1\), all other \(f_n=0\).
\[ u(x,t) = \frac{1-e^{-\pi^2t}}{\pi^2}\sin\pi x \]
STEP 5
Check both ends of time. At \(t=0\), \(u=0\) ✓. As \(t\to\infty\), \(u\to\frac{\sin\pi x}{\pi^2}\), which is exactly the steady state solving \(-v''=\sin\pi x\) ✓ The bar heats up and settles, with the approach governed by the same \(e^{-\pi^2t}\) that governs free cooling.
③Practice
1
Find the steady state for \(u_t = u_{xx} + 6x\) on \(0
Hint 1Set \(u_t=0\): solve \(v'' + 6x = 0\) with the given boundary values.
STEP 2 · integrate twice
\[ v' = -3x^2 + A, \qquad v = -x^3 + Ax + B \]
STEP 3 · apply the boundary conditions
\[ v(0) = B = 0 \]
\[ v(1) = -1 + A = 2 \quad\Longrightarrow\quad A = 3 \]
STEP 4 · the steady state
\[ v(x) = 3x - x^3 \]
STEP 5 · what to do next
Write \(u = v + w\). Then \(w_t = w_{xx}\) with \(w=0\) at both ends and \(w(x,0) = \varphi(x) - v(x)\) — a standard problem for module S3. And since \(w\to0\), the bar settles into \(3x-x^3\) ✓
2
Solve \(u_t = u_{xx} + 2\sin3x\) on \(0
Hint 1The forcing is eigenfunction \(n=3\), so only \(u_3\) is nonzero.
Hint 2\(\lambda_3 = 9\), so solve \(u_3' + 9u_3 = 2\) with \(u_3(0)=0\).
Solution
STEP 1 · project the forcing
\(f(x) = 2\sin 3x\) is \(2\times\) the \(n=3\) eigenfunction, so \(f_3 = 2\) and all other \(f_n=0\). Only mode 3 responds.
As \(t\to\infty\), \(u\to\frac29\sin3x\). Verify it is the steady state: \(-v'' = 2\sin3x\) with \(v = \frac29\sin3x\) gives \(-v'' = 2\sin3x\) ✓ The system reaches 63% of its final value in \(t = 1/9\).
3
Set up (but do not evaluate) the Duhamel integral for \(u_t = u_{xx} + e^{-t}\sin x\) on \(0
Hint 1Only mode 1 is excited, with \(f_1(t) = e^{-t}\) and \(\lambda_1 = 1\).
Hint 2\(u_1(t) = \int_0^te^{-(t-s)}e^{-s}ds\) — notice the exponentials combine into something \(s\)-independent.
Solution
STEP 1 · project
\(f_1(t) = e^{-t}\), \(\lambda_1 = 1\), and \(u_1(0)=0\).
STEP 5 · the solution and its meaning
\[ u(x,t) = te^{-t}\sin x \]
The factor \(t\) is the signature of resonance: the forcing decays at exactly the mode's natural rate, so each kick arrives perfectly in step with the decay and the responses accumulate. The amplitude peaks at \(t=1\) then dies. It is the diffusive analogue of the \(t\cos\omega t\) resonance in undamped oscillators.
4
For the forced wave equation \(u_{tt} = u_{xx} + \sin(\omega t)\sin x\) on \(0
Hint 1Project onto mode 1: the modal equation is \(u_1'' + \omega_1^2u_1 = \sin\omega t\).
Hint 2What is \(\omega_1\) here, with \(c=1\) and \(L=\pi\)?
Hint 3An undamped oscillator resonates when driven at its natural frequency.
Solution
STEP 1 · project onto the modes
The forcing \(\sin(\omega t)\sin x\) excites only \(n=1\). With \(c=1, L=\pi\): \(\lambda_1 = 1\), so \(\omega_1 = 1\).
STEP 2 · the modal ODE
\[ u_1'' + u_1 = \sin\omega t \]
An undamped harmonic oscillator with natural frequency 1, driven at frequency \(\omega\).
STEP 3 · the off-resonant case
For \(\omega\ne1\) the particular solution is
\[ u_1 = \frac{\sin\omega t}{1-\omega^2} \]
bounded, but with amplitude growing as \(\omega\to1\).
STEP 4 · the resonant case
At \(\omega = 1\) that formula breaks down; the correct particular solution is
\[ u_1 = -\frac{t\cos t}{2} \]
whose amplitude grows linearly without bound.
STEP 5 · answer and physical reading
Resonance at \(\omega = \omega_n = n\) for any \(n\) whose mode the forcing excites — here only \(\omega=1\). The undamped wave equation has no dissipation, so energy pumped in at the natural frequency accumulates forever. Real systems have damping, which caps the response — but the peak can still be catastrophic, which is why bridges are not marched across in step.
5
Solve \(u_t=u_{xx}+e^{-t}\sin2x\) on \(0
Hint 1Only mode 2 is excited: \(f_2(t)=e^{-t}\), \(\lambda_2 = 4\).
Hint 2Solve \(u_2' + 4u_2 = e^{-t}\) with \(u_2(0)=0\) — use an integrating factor.
Solution
STEP 1 · project
The forcing is \(1\times\) the \(n=2\) eigenfunction, so only \(u_2\) is nonzero, with \(\lambda_2=4\).
STEP 3 · solve with an integrating factor
\[ \left(e^{4t}u_2\right)' = e^{3t} \quad\Longrightarrow\quad e^{4t}u_2 = \frac{e^{3t}}{3} + C \]
STEP 4 · apply the initial condition
\[ 0 = \frac13 + C \Rightarrow C = -\frac13 \quad\Longrightarrow\quad u_2(t) = \frac{e^{-t}-e^{-4t}}{3} \]
STEP 5 · the solution and its shape
\[ u(x,t) = \frac{e^{-t}-e^{-4t}}{3}\sin2x \]
Zero at \(t=0\) ✓, rising to a maximum near \(t = \frac{\ln4}{3}\approx0.46\), then decaying like \(e^{-t}\) — the slower of the two rates always wins at late times. The response outlives the mode's natural decay because the forcing keeps feeding it.
6
Set up the full solution of \(u_t = u_{xx} + 4\) on \(0
Hint 1Handle the source and the boundary data together with a single steady state.
Hint 2Solve \(v''+4=0\) with \(v(0)=1\), \(v(1)=3\).
Hint 3Then \(w = u-v\) has a homogeneous problem with modified initial data.
Solution
STEP 1 · find the steady state
\[ v'' = -4 \quad\Longrightarrow\quad v = -2x^2 + Ax + B \]
STEP 2 · fit the boundary values
\[ v(0)=B=1, \qquad v(1) = -2 + A + 1 = 3 \Rightarrow A = 4 \]
\[ v(x) = -2x^2+4x+1 \]
STEP 3 · subtract it
With \(w = u - v\): \(w_t = w_{xx}\) (the source and the \(v''\) cancel), and \(w(0,t)=w(1,t)=0\) ✓ homogeneous.
STEP 4 · transfer the initial data
\[ w(x,0) = 0 - v(x) = 2x^2-4x-1 \]
As \(t\to\infty\), \(u\to v(x) = -2x^2+4x+1\), a downward parabola peaking at \(x=1\) — heat generated inside, held at fixed temperatures at the ends. One steady state absorbed both the source and the nonzero boundary data.
7
For \(u_{tt}=u_{xx}+\sin(\omega t)\sin x\) on \(0
Hint 1Off resonance, the particular solution is \(\frac{\sin\omega t}{\omega_1^2-\omega^2}\) with \(\omega_1=1\).
Hint 2Watch the amplitude as \(\omega\) approaches 1.
Solution
STEP 1 · the modal equation
\[ u_1'' + u_1 = \sin\omega t \qquad (\omega_1 = 1) \]
Nearly eight times larger, for a driving frequency only 5% closer to resonance.
STEP 5 · the trend
As \(\omega\to1\) the amplitude \(\to\infty\), and exactly at \(\omega=1\) the response grows linearly in \(t\) instead (module A2 problem 4). Real systems have damping, which caps the peak at a finite but often dangerous value — the reason machinery is designed to run away from its natural frequencies.
8
Convert the problem \(u_t=ku_{xx}\), \(u(0,t)=A(t)\), \(u(L,t)=0\) into one with homogeneous boundary conditions, and identify the source that appears.
Hint 1Subtract a simple function that matches the boundary data at each instant, e.g. \(A(t)\left(1-\frac{x}{L}\right)\).
Hint 2The subtracted function has a nonzero time derivative, which becomes a source.
Expand \(f\) in \(\sin\frac{n\pi x}{L}\) and apply Duhamel mode by mode. Note that if \(A\) is constant, \(A'=0\) and no source appears — recovering the simple steady-state subtraction of module F5.
9
Explain why the forcing must be expanded in the same eigenfunctions as the solution, and what goes wrong otherwise.
Hint 1The method works by matching coefficients — what makes that legitimate?
Hint 2Think about what orthogonality is doing when you project the equation onto a mode.
Solution
STEP 1 · what the method does
Substituting \(u=\sum u_n(t)X_n(x)\) and \(f = \sum f_n(t)X_n(x)\) into \(u_t = ku_{xx}+f\) gives
Multiply by \(X_m\) and integrate. Orthogonality annihilates every term except \(n=m\), leaving \(u_m'+k\lambda_mu_m = f_m\). The infinite system decouples.
STEP 3 · what a different basis would cost
If \(f\) were expanded in functions that are not orthogonal to the \(X_n\), the projection would leave cross-terms: mode \(m\)'s equation would contain contributions from every other mode. The ODEs would be coupled, and the method's whole advantage would evaporate.
STEP 4 · a further requirement
The \(X_n\) must also be complete, or some part of \(f\) could not be represented at all and would simply be dropped — silently giving a wrong answer.
STEP 5 · the unifying view
Sturm–Liouville theory (module A1) supplies both properties at once: orthogonality and completeness. That is why the eigenfunctions of the homogeneous problem are always the right basis — they diagonalize the spatial operator, turning a PDE into independent scalar ODEs, exactly as eigenvectors diagonalize a matrix.
Part IV · Module 3 Advanced tools
Infinite domains & transforms
On a finite interval the boundary conditions quantize the modes into a discrete list. Remove the boundaries and the list becomes a continuum: sums turn into integrals, Fourier series turns into the Fourier transform, and the heat equation acquires a closed-form solution.
InteractiveThe heat kernel spreadingwatch a spike become a Gaussian
peak height —width ~√(4kt) —total ∫u dx —
A point source instantly becomes a Gaussian of width \(\sqrt{4kt}\) and height \(1/\sqrt{4\pi kt}\) — the area stays pinned at 1 forever (conservation, module F2). Note the width grows like \(\sqrt{t}\), not \(t\): diffusion is slow, and covering twice the distance takes four times as long. Also note the kernel is positive everywhere for any \(t>0\), which is the precise sense in which heat travels at infinite speed.
②Practice
1
Use the transform rules to solve \(u_t + cu_x = 0\) on the line, and confirm you recover \(u = \varphi(x-ct)\).
Hint 1Transform in \(x\): \(u_x\) becomes \(i\kappa\hat{u}\).
Hint 2You get \(\hat{u}_t = -ic\kappa\hat{u}\), a first-order ODE in \(t\).
Hint 3Multiplying a transform by \(e^{-i\kappa a}\) shifts the function by \(a\).
STEP 2 · solve the ODE in t
\[ \hat{u}(\kappa,t) = \hat{\varphi}(\kappa)\,e^{-ic\kappa t} \]
STEP 3 · recognize the shift
The shift theorem says \(\widehat{f(x-a)} = e^{-i\kappa a}\hat f(\kappa)\). Here \(a = ct\).
STEP 4 · invert
\[ u(x,t) = \varphi(x-ct) \ ✓ \]
STEP 5 · read the mechanism
Each Fourier mode is multiplied by \(e^{-ic\kappa t}\) — a pure phase change, no change in magnitude. Every wavelength travels at the same speed \(c\), so the shape is preserved: the transport equation is non-dispersive. Compare the heat equation, whose factor \(e^{-k\kappa^2t}\) is a pure amplitude reduction, killing short wavelengths fastest.
2
A unit point source of heat is released at the origin at \(t=0\) with \(k=1\). Find the temperature at \(x=2\) when \(t=1\), and the time at which \(x=2\) is hottest.
Hint 1Point source ⟹ the solution is the heat kernel \(G(x,t)\).
Hint 2For the hottest time, maximize \(G(2,t)\) over \(t\) — take a log first to simplify the differentiation.
\[ \ln G = -\tfrac12\ln(4\pi t) - \frac{x^2}{4t} \]
STEP 4 · differentiate in t and set to zero
\[ \frac{d}{dt}\ln G = -\frac{1}{2t} + \frac{x^2}{4t^2} = 0 \quad\Longrightarrow\quad t = \frac{x^2}{2} \]
STEP 5 · answer
With \(x=2\): the peak arrives at \(t = 2\). Note it scales as \(x^2\) — a point twice as far waits four times as long for its maximum. That \(x \sim \sqrt{t}\) scaling is the signature of diffusion, and the reason diffusion is hopeless for long-distance transport (it is why your lungs are full of tiny branches rather than one long tube).
3
Explain why the heat kernel implies infinite propagation speed, and why this doesn't ruin the model in practice.
Hint 1Is \(G(x,t)\) ever exactly zero for \(t>0\)?
Hint 2Estimate its size at a large distance, say \(x = 100\sqrt{4kt}\).
Solution
STEP 1 · the mathematical fact
\(G(x,t) = \frac{1}{\sqrt{4\pi kt}}e^{-x^2/4kt} > 0\) for every \(x\), however large, as soon as \(t>0\). So a point source at the origin raises the temperature everywhere in the universe instantly.
STEP 2 · why it is nonetheless harmless
Put in numbers. At \(x = 10\sqrt{4kt}\) the exponential is \(e^{-100}\approx4\times10^{-44}\) — smaller than any physically meaningful temperature, and vastly below thermal noise.
STEP 3 · the contrast with waves
The wave equation has a strict light cone: outside \([x-ct, x+ct]\) the solution is exactly zero (module S6). Hyperbolic equations respect finite speed; parabolic ones only respect it approximately.
STEP 4 · where the model does break
The infinite speed is an artifact of Fourier's law, which assumes flux responds instantly to a gradient. On picosecond timescales or in rarefied gases that fails, and one uses hyperbolic heat conduction (the Cattaneo equation) with a finite relaxation time instead.
STEP 5 · the practical rule
Treat \(\sqrt{4kt}\) as the effective diffusion distance. Anything beyond a few multiples of it is, for engineering purposes, untouched.
4
Solve the heat equation on the line with the step initial data \(\varphi = 1\) for \(x<0\), \(0\) for \(x>0\), and evaluate \(u(0,t)\).
Hint 1Convolve the kernel with the step: the integral runs only over \(y<0\).
Hint 2Substituting \(z = (x-y)/\sqrt{4kt}\) turns it into an error function.
Hint 3By symmetry, what must the value at the interface be?
The interface instantly jumps to the average of the two sides and stays there forever, while the transition region widens like \(\sqrt{4kt}\). This similarity solution is the standard model for a quenched surface, carburizing steel, or two fluids beginning to mix — the profile always has the same shape, just stretched.
5
Initial data \(\varphi(x)=e^{-x^2}\) evolves under \(u_t=u_{xx}\). Use the heat kernel to find \(u(0,t)\).
Hint 1Convolving two Gaussians gives another Gaussian — add the variances.
Hint 2\(e^{-x^2}\) has "variance parameter" 1 in the exponent denominator; the kernel contributes \(4t\).
Hint 3Match the total area to fix the prefactor.
Solution
STEP 1 · set up the convolution
\[ u(x,t) = \int_{-\infty}^{\infty}\frac{e^{-(x-y)^2/4t}}{\sqrt{4\pi t}}e^{-y^2}dy \]
STEP 2 · use the Gaussian convolution rule
Writing \(e^{-x^2} = e^{-x^2/(4\cdot\frac14)}\), the data has parameter \(t_0=\frac14\). Convolution with the kernel at time \(t\) simply advances that parameter:
STEP 4 · evaluate at the origin
\[ u(0,t) = \frac{1}{\sqrt{1+4t}} \]
STEP 5 · check both ends
At \(t=0\): \(u(0,0)=1\) ✓ matching \(\varphi(0)\). As \(t\to\infty\): \(u(0,t)\sim\frac{1}{2\sqrt t}\), exactly the decay of the heat kernel itself — after a long time the solution forgets its initial shape and looks like a point source of the same total heat.
6
Show that the heat kernel has \(\int_{-\infty}^{\infty}G(x,t)\,dx = 1\) for every \(t>0\).
Hint 1Substitute \(z = x/\sqrt{4kt}\) to make the integral \(t\)-independent.
Hint 2Use \(\int_{-\infty}^{\infty}e^{-z^2}dz = \sqrt\pi\).
Solution
STEP 1 · write the integral
\[ \int_{-\infty}^{\infty}\frac{1}{\sqrt{4\pi kt}}e^{-x^2/4kt}dx \]
STEP 2 · substitute
Let \(z = \frac{x}{\sqrt{4kt}}\), so \(dx = \sqrt{4kt}\,dz\):
The kernel is a probability density for all time: total heat is conserved (module F2 problem 2), and the profile merely spreads. Height \(\propto t^{-1/2}\) and width \(\propto t^{1/2}\) multiply to a constant.
STEP 5 · the probabilistic reading
\(G\) is the transition density of Brownian motion: the probability that a particle starting at 0 is near \(x\) at time \(t\). Diffusion of heat and random walks of particles are the same mathematics, which is why \(\sqrt{t}\) spreading appears in both.
7
Show that the heat kernel is self-similar: \(G(x,t) = t^{-1/2}F\!\left(x/\sqrt t\right)\) for some fixed profile \(F\). Why is that useful?
Hint 1Factor \(t^{-1/2}\) out of \(G\) and see what remains.
Hint 2The exponent depends on \(x\) and \(t\) only through \(x/\sqrt t\).
Solution
STEP 1 · rewrite the kernel
\[ G(x,t) = \frac{1}{\sqrt{4\pi kt}}e^{-x^2/4kt} = t^{-1/2}\underbrace{\left[\frac{1}{\sqrt{4\pi k}}e^{-\left(x/\sqrt t\right)^2/4k}\right]}_{\text{a function of } \eta = x/\sqrt t \text{ alone}} \]
STEP 2 · name the similarity variable
\[ \eta = \frac{x}{\sqrt t}, \qquad F(\eta) = \frac{1}{\sqrt{4\pi k}}e^{-\eta^2/4k} \]
STEP 3 · what this says
The solution at every time is the same shape, stretched horizontally by \(\sqrt t\) and squashed vertically by \(t^{-1/2}\). Plot \(\sqrt t\,u\) against \(x/\sqrt t\) and all the curves collapse onto one.
STEP 4 · why it is useful
It reduces a PDE in two variables to an ODE in one: substituting \(u = t^{-1/2}F(\eta)\) into \(u_t=ku_{xx}\) yields an ODE for \(F\). This is the similarity method, and it works whenever a problem has no intrinsic length or time scale.
STEP 5 · where else it appears
The step-data solution \(\frac12\operatorname{erfc}\!\left(\frac{x}{\sqrt{4kt}}\right)\) (problem 4) is the same idea; so are boundary-layer profiles in fluid mechanics and the \(\sqrt t\) growth of oxide films. Spotting the absence of a length scale tells you a similarity solution exists before you compute anything.
8
Apply the Fourier transform to \(u_{tt}=c^2u_{xx}\) and compare the result with the heat equation's transform.
Hint 1\(u_{xx}\) transforms to \(-\kappa^2\hat u\), giving an ODE in \(t\) for each \(\kappa\).
Hint 2The wave equation gives a second-order ODE — oscillatory, not decaying.
\(\omega = c\kappa\): phase speed \(\omega/\kappa = c\) is the same for every wavelength. The wave equation is non-dispersive, so a pulse holds its shape — which is d'Alembert's theorem seen in Fourier space.
STEP 5 · what dispersion would look like
An equation like \(u_t = u_{xxx}\) gives \(\omega = -\kappa^3\), so different wavelengths travel at different speeds and any pulse spreads into a wave train. Comparing multipliers is the quickest way to predict a linear PDE's qualitative behavior.
9
Heat diffuses in soil with \(k = 5\times10^{-7}\) m²/s. Estimate how deep the daily temperature swing penetrates, and how deep the yearly one does.
Hint 1Use the diffusion length \(\delta \sim \sqrt{4kt}\) with \(t\) the relevant timescale.
Hint 2One day is 86,400 s; one year is about \(3.15\times10^7\) s.
\(\sqrt{365}\approx19\) — the yearly wave reaches about nineteen times deeper, not 365 times. The \(\sqrt t\) law again.
STEP 4 · practical consequences
Day–night swings vanish below about half a metre, which is why cellars stay at a steady temperature and why water pipes are buried below the frost line. Seasonal variation persists to several metres, which is exactly the depth range ground-source heat pumps exploit.
STEP 5 · the refinement
The precise periodic solution is \(u = e^{-x/\delta}\cos(\omega t - x/\delta)\) with \(\delta=\sqrt{2k/\omega}\): the wave decays exponentially and lags in phase, so at depth \(\pi\delta\) the soil is warmest in winter. The order-of-magnitude estimate above is the practical version.
Part IV · Module 4 Advanced tools
Bessel & Legendre functions
Separate a PDE in circular or spherical geometry and the radial ODE is no longer solvable in elementary functions. Its solutions are given names — Bessel and Legendre — but they behave exactly like the sines and cosines you already know: orthogonal, complete, quantized by the boundary.
①The circular drum
Separating \(u_{tt}=c^2\nabla^2u\) on a disk with \(u = R(r)\Theta(\theta)T(t)\) gives the angular equation \(\Theta''+n^2\Theta=0\) (periodicity) and the radial equation
Bessel's equation
\[ r^2R'' + rR' + (\alpha^2r^2 - n^2)R = 0 \]
whose bounded solution is \(J_n(\alpha r)\), the Bessel function of the first kind. (The second solution \(Y_n\) blows up at the origin and is discarded on a solid disk — exactly the role \(r^{-n}\) played in module S8.)
Quantization by the boundary
\[ R(a)=0 \ \Longrightarrow\ J_n(\alpha a)=0 \ \Longrightarrow\ \alpha_{nm} = \frac{z_{nm}}{a}, \quad z_{nm} = m\text{th zero of }J_n \]
InteractiveBessel functions and drum modespick n and m
zero znm—frequency ratio to (0,1) —nodal circles / diameters —
Left: \(J_n\) with its zeros marked — note it looks like a decaying, stretched cosine, and its zeros are not evenly spaced. Right: the drum mode \(J_n(z_{nm}r/a)\cos n\theta\) as a heat map, with its nodal circles and diameters. The frequency ratios are irrational (2.295, 3.598, …), unlike a string's exact integers — which is precisely why a drum sounds like a thud rather than a note.
②Spheres: Legendre
Separating Laplace's equation in spherical coordinates gives, in the polar angle, Legendre's equation \((1-s^2)P'' - 2sP' + n(n+1)P = 0\) with \(s=\cos\theta\). Its bounded solutions are the Legendre polynomials \(P_n(s)\): \(P_0=1\), \(P_1 = s\), \(P_2 = \frac{3s^2-1}{2}\), … orthogonal on \([-1,1]\), and combined with \(r^n\) or \(r^{-(n+1)}\) in radius. This is where the multipole expansion of electrostatics comes from.
Worked example · fundamental frequency of a drum
A circular drum of radius \(a\) with wave speed \(c\). Find the two lowest frequencies.
STEP 1
The frequencies are \(\omega_{nm} = \frac{c\,z_{nm}}{a}\), where \(z_{nm}\) is the \(m\)th zero of \(J_n\).
STEP 2
The relevant zeros: \(z_{01}\approx2.405\) (first zero of \(J_0\)) and \(z_{11}\approx3.832\) (first zero of \(J_1\)).
Their ratio is \(3.832/2.405 \approx 1.593\) — not 2, not \(3/2\), not any simple fraction.
STEP 5
Why drums aren't melodic: a string's overtones are exact integer multiples, which the ear fuses into a single pitch. A drum's are irrational multiples, so the overtones clash and you hear an unpitched thud. Timpani are tuned by damping the centre to suppress the \(J_0\) modes and favour the \(J_1\) family, whose ratios are closer to musical.
③Practice
1
Why is \(Y_n\) discarded for a solid disk but needed for an annular membrane?
Hint 1What does \(Y_n(r)\) do as \(r\to0^+\)?
Hint 2Compare with the role of \(r^{-n}\) and \(\ln r\) in module S8.
Solution
STEP 1 · behavior at the origin
\(Y_n(r)\to-\infty\) as \(r\to0\) (logarithmically for \(n=0\), like \(r^{-n}\) for \(n\ge1\)).
STEP 2 · on a solid disk
The origin is an interior point where the displacement must be finite, so the coefficient of \(Y_n\) is forced to zero. Boundedness acts as the missing boundary condition.
STEP 3 · on an annulus
The domain excludes the origin, so \(Y_n\) is perfectly regular there and must be retained.
STEP 4 · the count
An annulus has two boundary circles ⟹ two conditions per mode ⟹ two constants needed, exactly matching \(AJ_n + BY_n\). The eigenvalue condition becomes a determinant equation rather than a simple zero of \(J_n\).
STEP 5 · the pattern
This is the same structure as \(r^n\) vs \(r^{-n}\) (S8) and \(\cosh\) vs \(\sinh\) (S7): the second solution is singular where the domain has a hole, and whether you keep it is decided by the geometry, not the equation.
2
A drumhead of radius 0.3 m has \(c = 100\) m/s. Find its fundamental frequency in Hz.
Hint 1\(\omega_{01} = cz_{01}/a\) with \(z_{01}=2.405\).
A string of the same length and speed would give \(f_1 = \frac{c}{2L} = \frac{100}{0.6}\approx167\) Hz. The drum is lower because \(z_{01}=2.405 < \pi\) — the circular geometry is "softer" than a clamped string of the same size.
STEP 4 · the next mode
\[ f_{11} = f_{01}\times\frac{3.832}{2.405} \approx 203\ \text{Hz} \]
an interval of about 1.59 — between a fifth and a minor sixth, and audibly dissonant with the fundamental.
STEP 5 · scaling
\(f \propto c/a\): tightening the head (raising \(c\)) or shrinking the radius both raise the pitch — which is exactly how a timpanist tunes with the pedal.
3
Verify that \(P_2(s) = \frac{3s^2-1}{2}\) satisfies Legendre's equation with \(n=2\), and check it is orthogonal to \(P_0=1\).
Hint 1Substitute into \((1-s^2)P'' - 2sP' + 6P = 0\).
Hint 2For orthogonality, integrate \(P_2\) over \([-1,1]\).
\(\int_{-1}^1P_nP_m\,ds = 0\) for \(n\ne m\) and \(\frac{2}{2n+1}\) for \(n=m\) — Sturm–Liouville orthogonality with weight \(w=1\) (module A1). It is what makes the multipole expansion \(f(\theta) = \sum c_nP_n(\cos\theta)\) work, exactly like a Fourier series on a sphere.
4
Solve \(\nabla^2u=0\) inside the unit sphere with \(u(1,\theta)=\cos\theta\), given that solutions have the form \(\sum A_nr^nP_n(\cos\theta)\).
Hint 1Recognize \(\cos\theta\) as a Legendre polynomial.
Hint 2\(P_1(s)=s\), so \(P_1(\cos\theta)=\cos\theta\).
Solution
STEP 1 · match the boundary data
\(\cos\theta = P_1(\cos\theta)\) exactly, so only the \(n=1\) term is needed: \(A_1 = 1\), all others zero.
In Cartesian coordinates \(r\cos\theta = z\), and \(\nabla^2z = 0\) trivially ✓
STEP 5 · what it represents
A uniform field inside the sphere — this is the interior solution for a sphere held at a linearly varying potential, and \(n=1\) is the dipole term of the multipole expansion. Higher \(n\) would give quadrupole (\(P_2\)) and beyond, each penetrating as \(r^n\) and therefore vanishing faster toward the centre — the spherical analogue of module S8's \((r/a)^n\).
5
Separate the radially symmetric heat equation on a disk, \(u_t = k\left(u_{rr}+\frac1ru_r\right)\), and identify the resulting radial ODE.
Hint 1Try \(u = R(r)T(t)\) and divide by \(kRT\).
Hint 2The radial equation is Bessel's equation of order 0.
Structurally identical to the 1-D bar — decaying modes with \(n^2\)-like rates — but with \(J_0\) replacing \(\sin\) and unevenly spaced eigenvalues. The slowest rate is \(k(2.405/a)^2\), the disk's thermal time constant.
6
For a drum with \(a=1\), \(c=1\), list the three lowest frequencies and their ratios to the fundamental.
Hint 1\(\omega_{nm} = cz_{nm}/a\), so you just need the three smallest Bessel zeros overall.
Hint 2\(z_{01}=2.405\), \(z_{11}=3.832\), \(z_{21}=5.136\), \(z_{02}=5.520\).
Solution
STEP 1 · collect the smallest zeros
Ordering all \(z_{nm}\): \(2.405\ (0,1)\), \(3.832\ (1,1)\), \(5.136\ (2,1)\), then \(5.520\ (0,2)\).
STEP 2 · the three lowest frequencies
\[ \omega_{01}=2.405, \qquad \omega_{11}=3.832, \qquad \omega_{21}=5.136 \]
STEP 3 · ratios to the fundamental
\[ 1, \qquad \frac{3.832}{2.405}=1.593, \qquad \frac{5.136}{2.405}=2.136 \]
STEP 4 · compare with a string
A string would give \(1, 2, 3\). The drum's \(1.593\) and \(2.136\) are close to no simple ratio — \(1.593\) sits between a fifth (1.5) and a minor sixth (1.6).
STEP 5 · the audible result
Inharmonic overtones prevent the ear from fusing them into one pitch, so a drum sounds like a thud with a vague pitch centre. Timpani improve on this by damping the centre (suppressing the \(J_0\) family) so the surviving \(1.593:2.136\) modes come closer to a harmonic series.
7
Use the series \(J_n(x)=\sum_k\frac{(-1)^k}{k!(k+n)!}\left(\frac x2\right)^{2k+n}\) to verify that \(J_0'(x) = -J_1(x)\).
Hint 1Differentiate the \(J_0\) series term by term.
Hint 2Shift the summation index so the result matches the \(J_1\) series.
Neumann (free-rim) conditions require \(J_0'(\alpha a)=0\), which this identity converts into \(J_1(\alpha a)=0\) — so a free-rim drum uses the zeros of \(J_1\). It is the Bessel analogue of \(\frac{d}{dx}\cos = -\sin\), and the same index shift you practised in series work.
8
Express \(f(s)=s^2\) in terms of Legendre polynomials \(P_0, P_1, P_2\) on \([-1,1]\).
Hint 1\(P_0=1\), \(P_1=s\), \(P_2 = \frac{3s^2-1}{2}\).
Hint 2Solve for \(s^2\) from the \(P_2\) formula rather than integrating.
Solution
STEP 1 · invert the P₂ formula
\[ P_2 = \frac{3s^2-1}{2} \quad\Longrightarrow\quad s^2 = \frac{2P_2+1}{3} \]
STEP 2 · write in the basis
\[ s^2 = \frac13P_0 + \frac23P_2 \]
with no \(P_1\) component.
STEP 3 · confirm by symmetry
\(s^2\) is even and \(P_1=s\) is odd, so their inner product must vanish ✓ — a check that costs nothing.
STEP 4 · verify the coefficient formula
The general rule is \(c_n = \frac{2n+1}{2}\int_{-1}^1fP_n\,ds\). For \(n=0\): \(\frac12\int_{-1}^1s^2ds = \frac12\cdot\frac23 = \frac13\) ✓
STEP 5 · where you would use it
Boundary data \(u(a,\theta)=\cos^2\theta\) on a sphere becomes \(\frac13P_0+\frac23P_2\), so the interior solution is \(u = \frac13 + \frac23\left(\frac ra\right)^2P_2(\cos\theta)\). Legendre expansions do for spheres exactly what Fourier series do for intervals.
9
Solve \(\nabla^2u=0\) outside the unit sphere with \(u(1,\theta)=\cos\theta\) and \(u\to0\) at infinity.
Hint 1The radial solutions are \(r^n\) and \(r^{-(n+1)}\); which decays at infinity?
Hint 2\(\cos\theta = P_1(\cos\theta)\), so only \(n=1\) appears.
Solution
STEP 1 · choose the radial behavior
Outside the sphere, \(r^n\) grows and is inadmissible; \(r^{-(n+1)}\) decays ✓ For \(n=1\) that is \(r^{-2}\).
STEP 2 · match the boundary data
\(u(1,\theta)=\cos\theta = P_1(\cos\theta)\), so the coefficient is 1:
\[ u(r,\theta) = \frac{\cos\theta}{r^{2}} \]
STEP 3 · check
At \(r=1\): \(u=\cos\theta\) ✓ As \(r\to\infty\): \(u\to0\) ✓
STEP 4 · recognize the physics
This is precisely the field of a point dipole at the origin — the leading term in the multipole expansion of any localized charge distribution with no net charge.
STEP 5 · compare interior and exterior
Interior
Exterior
Radial factor
\(r^n\)
\(r^{-(n+1)}\)
Excluded because
\(r^{-(n+1)}\) blows up at 0
\(r^n\) blows up at ∞
Detail decays
inward
outward
Either way, high-\(n\) structure dies fastest — the far field of any object looks like a monopole plus a dipole, whatever its shape.
Part IV · Module 5 Advanced tools
Numerical methods & stability
Most PDEs have no closed-form solution. Replace derivatives by difference quotients and you can march any of them forward on a computer — provided you respect one stability condition, which the equation's type dictates.
Each new value is a weighted average of three old ones. Rewriting as \(u_i^{j+1} = (1-2r)u_i^j + r(u_{i+1}^j+u_{i-1}^j)\) shows the danger: if \(1-2r<0\), the point's own value enters with a negative weight and the scheme amplifies errors.
InteractiveStable, marginal, catastrophicpush r past ½
r —max |u| —scheme is —
At \(r\le\frac12\) the numerical solution smooths and decays, tracking the true solution. Push \(r\) just past \(0.5\) and watch a sawtooth appear and grow, doubling in size every few steps until it swamps everything — the amplification factor for the shortest wavelength is \(1-4r\), whose magnitude exceeds 1 exactly when \(r>\frac12\). Note this is a property of the scheme, not the equation: the true solution is perfectly well behaved.
②Why ½, and what to do about it
Von Neumann analysis: insert a test mode \(u_i^j = G^je^{i\kappa x_i}\). The scheme multiplies it by
Amplification factor
\[ G = 1 - 4r\sin^2\frac{\kappa\Delta x}{2} \]
Stability needs \(|G|\le1\) for every \(\kappa\). The worst case is the sawtooth \(\kappa\Delta x = \pi\), giving \(G = 1-4r\), so \(|1-4r|\le1\) forces \(r\le\frac12\). The practical consequence is brutal: halving \(\Delta x\) forces you to quarter \(\Delta t\), so refining the mesh by 10 costs 1000× the work. The cure is an implicit scheme (backward Euler or Crank–Nicolson), which is unconditionally stable at the price of solving a tridiagonal system each step.
The hyperbolic analogue: the CFL condition
For the wave equation, explicit schemes require \(\dfrac{c\Delta t}{\Delta x}\le1\) — the numerical domain of dependence must contain the true one (module S6). In words: information must not be allowed to outrun your grid. Note the scaling is \(\Delta t \sim \Delta x\), far kinder than the parabolic \(\Delta t\sim\Delta x^2\).
③Practice
1
With \(k=1\) and \(\Delta x = 0.1\), what is the largest stable \(\Delta t\)? What if you refine to \(\Delta x = 0.01\)?
Hint 1Solve \(r = k\Delta t/\Delta x^2 \le \frac12\) for \(\Delta t\).
Hint 2Note the \(\Delta x^2\) — refining space costs quadratically in time.
Solution
STEP 1 · rearrange the condition
\[ \Delta t \le \frac{\Delta x^2}{2k} \]
STEP 3 · fine grid
\[ \Delta t \le \frac{0.0001}{2} = 5\times10^{-5} \]
Now \(t=1\) needs 20,000 steps.
STEP 4 · count the total cost
Ten times more grid points and a hundred times more steps ⟹ 1000× the work for a 10× finer mesh.
STEP 5 · the fix
Use Crank–Nicolson: unconditionally stable, second-order accurate in both variables, and each step is a tridiagonal solve costing \(O(N)\). You can then choose \(\Delta t\) for accuracy rather than survival — which is why essentially all production diffusion codes are implicit.
2
Carry out two steps of the explicit scheme with \(r=0.5\) on the grid \(u = (0, 1, 0, 0, 0)\) with the ends held at 0.
Hint 1With \(r=\frac12\) the update simplifies dramatically: \(u_i^{j+1} = \frac{u_{i+1}^j + u_{i-1}^j}{2}\).
Hint 2The centre coefficient \(1-2r\) vanishes — each point becomes purely the average of its neighbours.
Wait — careful with which array we read from. Using the old values \((0,1,0,0,0)\): \(u_1^{1} = \frac{u_2^0+u_0^0}{2} = \frac{0+0}{2}=0\); \(u_2^1 = \frac{u_3^0+u_1^0}{2} = \frac{0+1}{2}=0.5\); \(u_3^1 = \frac{u_4^0+u_2^0}{2}=0\).
The peak has split in two and the centre has emptied — an odd-even decoupling that is a known quirk of \(r=\frac12\) exactly. The scheme is stable (nothing grows) but this checkerboard behaviour is unphysical.
STEP 5 · the practical rule
Stay comfortably below the limit — \(r\approx0.25\) is the usual choice. Being at the stability boundary is legal but produces poor-quality solutions.
3
For the wave equation with \(c=2\) and \(\Delta x = 0.05\), find the largest stable \(\Delta t\), and explain the CFL condition in words.
Hint 1CFL: \(c\Delta t/\Delta x \le 1\).
Hint 2Think about how far a real wave travels in one time step versus one grid spacing.
Solution
STEP 1 · apply the condition
\[ \Delta t \le \frac{\Delta x}{c} = \frac{0.05}{2} = 0.025 \]
STEP 2 · state it in words
In one time step, a physical wave travels \(c\Delta t\). The explicit stencil only lets information move one grid cell, a distance \(\Delta x\). So you need \(c\Delta t \le \Delta x\): the wave must not outrun the grid.
STEP 3 · the domain-of-dependence version
Equivalently, the numerical domain of dependence must contain the analytic one (module S6). If it doesn't, the scheme is computing an answer from data that cannot possibly determine it — and no amount of refinement rescues it.
Once again the classification of module F3 governs the practical method: finite propagation speed gives a geometric stability condition, while infinite-speed diffusion gives a much harsher one.
4
Use von Neumann analysis to confirm the amplification factor \(G = 1-4r\sin^2\frac{\kappa\Delta x}{2}\), and find which mode is most unstable.
Hint 1Substitute \(u_i^j = G^je^{i\kappa x_i}\) into the update and factor out the common terms.
Hint 2\(e^{i\kappa\Delta x}+e^{-i\kappa\Delta x} = 2\cos\kappa\Delta x\), and \(\cos\phi - 1 = -2\sin^2\frac\phi2\).
\(G\) is smallest when \(\sin^2 = 1\), i.e. \(\kappa\Delta x = \pi\) — a wavelength of exactly two grid cells, the alternating sawtooth \(+1,-1,+1,\dots\). There \(G = 1-4r\).
STEP 5 · the stability condition
\[ |1-4r|\le1 \quad\Longrightarrow\quad 0 \le r \le \tfrac12 \ ✓ \]
And note what fails: the shortest-wavelength mode, the one the grid can barely represent. That is exactly the sawtooth you see erupting in the lab above when \(r>\frac12\).
5
The Crank–Nicolson scheme averages the spatial stencil between the old and new time levels. Show its amplification factor and explain unconditional stability.
Hint 1The scheme is \(u_i^{j+1} - \frac r2\delta^2u_i^{j+1} = u_i^j + \frac r2\delta^2u_i^j\), where \(\delta^2\) is the second-difference operator.
Hint 2Insert \(u_i^j = G^je^{i\kappa x_i}\); the operator \(\delta^2\) contributes \(-4\sin^2\frac{\kappa\Delta x}{2}\).
Solution
STEP 1 · insert the test mode
Write \(S = 4\sin^2\frac{\kappa\Delta x}{2} \ge 0\). The scheme becomes
STEP 2 · solve for G
\[ G = \frac{1-\frac{rS}{2}}{1+\frac{rS}{2}} \]
STEP 3 · bound it
Since \(rS\ge0\), the numerator is never larger in magnitude than the denominator:
\[ |G| \le 1 \qquad\text{for every } r>0 \text{ and every } \kappa \]
STEP 4 · conclude
Unconditionally stable — no restriction linking \(\Delta t\) and \(\Delta x\). Compare the explicit scheme's \(G = 1-rS\), which exceeds 1 in magnitude as soon as \(r>\frac12\).
STEP 5 · the cost and the caveat
The new level appears on both sides, so each step requires solving a tridiagonal system — cheap, \(O(N)\) by the Thomas algorithm. Caveat: for large \(rS\), \(G\to-1\), so very large steps produce decaying oscillations rather than smooth decay. Stability is not accuracy; \(\Delta t\) should still be chosen sensibly.
6
State the order of accuracy of the explicit scheme and of Crank–Nicolson, and say what each implies for halving \(\Delta x\).
Hint 1A forward difference in time is first order; a centred second difference in space is second order.
Hint 2Crank–Nicolson is centred in time at the half step.
Spatial error drops by 4, but stability forces \(\Delta t\) to drop by 4 as well — so the temporal error also drops by 4, and total work rises by 8 (twice the points, four times the steps).
STEP 4 · halving Δx with Crank–Nicolson
You may halve \(\Delta t\) alongside \(\Delta x\); then both error terms drop by 4 with only 4× the work — and you could even keep \(\Delta t\) fixed if the time error is already small.
STEP 5 · the practical choice
Balance the two error terms: with the explicit scheme, \(\Delta t\sim\Delta x^2\) is forced anyway, so the accuracy is effectively \(O(\Delta x^2)\). Crank–Nicolson lets you choose \(\Delta t\) for accuracy rather than survival, which is why it is the default for diffusion problems.
7
The explicit wave scheme is \(u_i^{j+1} = 2u_i^j - u_i^{j-1} + \gamma^2\left(u_{i+1}^j-2u_i^j+u_{i-1}^j\right)\) with \(\gamma = c\Delta t/\Delta x\). Show that \(\gamma=1\) propagates data exactly.
Hint 1Substitute \(\gamma=1\) and simplify — several terms cancel.
Hint 2Compare the result with d'Alembert's formula on the grid.
With \(\gamma=1\), one time step advances exactly one grid cell, so the grid points lie precisely on the characteristics \(x\pm ct\). The update is d'Alembert's formula evaluated on the mesh — no truncation error at all.
STEP 4 · the "magic time step"
The numerical solution equals the exact solution at every grid point (given exact initial data). Rare and delightful: the scheme is not approximating the PDE, it is reproducing it.
STEP 5 · why not always use it
Only possible with constant \(c\) and a uniform grid. With variable wave speed, or in two dimensions where the CFL limit becomes \(\gamma\le1/\sqrt2\), the alignment is lost. And exceeding \(\gamma=1\) is fatal — the numerical domain of dependence no longer contains the physical one.
8
You need \(\Delta x = 0.002\) on a domain of length 1, integrating to \(t=1\), with \(k=1\) (heat) or \(c=1\) (wave). Compare the number of time steps required by explicit schemes.
Hint 1Heat: \(\Delta t \le \Delta x^2/(2k)\). Wave: \(\Delta t\le\Delta x/c\).
A thousand times more work for the diffusion problem, on the same grid and time interval.
STEP 4 · why the disparity
Parabolic stability scales like \(\Delta x^2\), hyperbolic like \(\Delta x\) — the same \(x\sim\sqrt t\) versus \(x\sim ct\) distinction that runs through the whole course.
STEP 5 · the remedy
Use an implicit scheme for the heat problem: Crank–Nicolson at \(\Delta t = 0.001\) needs 1000 steps, each a tridiagonal solve over 500 points — trivial compared with half a million explicit sweeps. For the wave equation the explicit scheme is already cheap, and \(\gamma=1\) even makes it exact.
9
A scheme can be consistent with the PDE yet still produce garbage. Explain, and state the theorem that ties the ideas together.
Hint 1Consistency is about the truncation error as \(\Delta t,\Delta x\to0\); it says nothing about error growth over many steps.
Hint 2Recall the explicit heat scheme with \(r>\frac12\).
Solution
STEP 1 · define consistency
A scheme is consistent if substituting the exact solution leaves a truncation error that vanishes as \(\Delta t,\Delta x\to0\). The explicit heat scheme is consistent for every \(r\).
STEP 2 · why that is not enough
Consistency controls the error introduced in one step. Stability controls whether those errors are damped or amplified over the many steps that follow. With \(r>\frac12\), each step multiplies the sawtooth mode by \(|1-4r|>1\), so a round-off error of \(10^{-16}\) can dominate the solution within a hundred steps.
STEP 3 · state the theorem
Lax equivalence theorem: for a consistent scheme applied to a well-posed linear problem,
\[ \text{stability} \iff \text{convergence} \]
STEP 4 · how it is used
Proving convergence directly is hard; proving stability by von Neumann analysis is a short computation. The theorem lets you do the easy check and conclude the hard property.
STEP 5 · the two hypotheses matter
"Well-posed" is essential — no scheme converges for the backward heat equation, because the problem itself lacks continuous dependence (module F3). And "linear" matters too: for nonlinear conservation laws you additionally need the scheme in conservative form to converge to the right weak solution (module C3 problem 8).
Part III · Checkpoint Quiz
Quiz B — Separation & Fourier
Ten questions on the core method: eigenvalue problems, Fourier coefficients, and the three model equations solved on finite domains.
Question 1
Separating \(u_t = ku_{xx}\) with \(u = XT\) gives which pair of ODEs?
Why
Dividing by \(kXT\) gives \(\frac{T'}{kT}=\frac{X''}{X}=-\lambda\). The heat equation has one time derivative, so \(T\) satisfies a first-order equation — choice C is the wave equation.
Question 2
For \(X''+\lambda X = 0\) with \(X(0)=X(L)=0\), the eigenvalues are:
Why
\(\sin\beta L = 0\) forces \(\beta = n\pi/L\), and \(\lambda = \beta^2\). Note \(n=0\) gives \(X\equiv0\), which is not an eigenfunction — that case only survives for Neumann conditions (choice D's situation).
Question 3
An insulated bar starts at \(u(x,0)=\varphi(x)\). As \(t\to\infty\), \(u\) tends to:
Why
Insulated ends conserve total heat, and the \(\lambda_0=0\) constant mode never decays. Everything else dies, leaving the average. With clamped ends heat escapes and the answer would be B.
Question 4
Which series should you use for a bar with both ends insulated?
Why
\(\cos\frac{n\pi x}{L}\) has zero derivative at both ends, matching \(u_x=0\) automatically. The boundary conditions always choose the basis — sines for Dirichlet, cosines for Neumann.
Question 5
In \(u = \sum b_ne^{-kn^2\pi^2t/L^2}\sin\frac{n\pi x}{L}\), how much faster does mode 5 decay than mode 1?
Why
The exponent is proportional to \(n^2\), so the rate for \(n=5\) is \(25\times\) that for \(n=1\). This quadratic scaling is why fine detail vanishes almost immediately and the profile becomes smooth.
Question 6
A Fourier series of a function with a jump discontinuity converges at the jump to:
Why
Dirichlet's theorem gives \(\frac{f(x^-)+f(x^+)}{2}\). Nearby, the partial sums overshoot by about 9% of the jump no matter how many terms you take — the Gibbs phenomenon, which narrows but never shrinks.
Question 7
The vibrating string \(u_{tt}=c^2u_{xx}\) with fixed ends has mode frequencies:
Why
\(T''+\lambda_nc^2T=0\) gives \(\omega_n = c\sqrt{\lambda_n} = n\pi c/L\). Integer ratios are why a string sounds like a definite pitch; a circular drum (choice C) has Bessel-zero ratios and sounds like a thud.
Question 8
d'Alembert's formula shows that \(u(x_0,t_0)\) depends on the initial data only on:
Why
The domain of dependence is the backward light cone. Data outside it hasn't had time to arrive — finite propagation speed. The heat equation is choice C: its kernel is positive everywhere, so every point influences every other instantly.
Question 9
For Laplace's equation on a rectangle, separation gives \(X''+\lambda X=0\) and:
Why
The plus sign between \(u_{xx}\) and \(u_{yy}\) forces opposite signs in the two ODEs. Oscillation in one direction demands hyperbolic growth in the other — an equilibrium equation cannot wiggle in both.
Question 10
The value of a harmonic function at the centre of a disk equals:
Why
The mean value property — visible in the series solution, where every \(n\ge1\) term carries \((r/a)^n\) and vanishes at \(r=0\), leaving only \(a_0/2\). It immediately implies the maximum principle: no interior peaks.
Part IV · Checkpoint Quiz
Quiz C — Advanced & comprehensive
Ten questions spanning Sturm–Liouville theory, forcing, transforms, special functions, numerics — plus a few that reach back across the whole course.
Question 1
Sturm–Liouville theory guarantees that eigenfunctions of different eigenvalues are:
Why
Lagrange's identity turns \((\lambda_m-\lambda_n)\int X_mX_nw\,dx\) into a boundary term that the boundary conditions kill. The weight matters: for \((xX')'+\lambda\frac1xX=0\) the orthogonality integral carries \(1/x\), not 1.
Question 2
To solve \(u_t=ku_{xx}\) with \(u(0,t)=3\), \(u(L,t)=7\), you should first:
Why
Separation requires homogeneous boundary conditions; a product \(XT\) cannot equal a nonzero constant at an end for all \(t\). Subtracting \(v(x) = 3 + \frac{4x}{L}\) fixes it and \(v\) is also the long-time limit.
Question 3
The heat kernel \(G(x,t)=\frac{1}{\sqrt{4\pi kt}}e^{-x^2/4kt}\) spreads with width proportional to:
Why
The exponent depends on \(x^2/4kt\), so the characteristic length is \(\sqrt{4kt}\). Diffusion covering twice the distance takes four times as long — completely unlike a wave, whose front moves at constant speed \(ct\).
Question 4
Under the Fourier transform, \(u_{xx}\) becomes:
Why
Each derivative brings a factor \(ik\), so two give \((ik)^2=-k^2\). That minus sign is what makes \(\hat{u}=\hat{\varphi}e^{-kk^2t}\) decay — turning the PDE into a decoupled ODE for each wavenumber.
Question 5
Why do circular drums sound unpitched compared with strings?
Why
\(\omega_{nm}\propto z_{nm}\), and \(z_{11}/z_{01} = 3.832/2.405 \approx 1.59\) — irrational. A string's exact integer ratios fuse into one perceived pitch; the drum's do not.
Question 6
The explicit scheme for \(u_t=ku_{xx}\) is stable provided:
Why
Von Neumann analysis gives \(G = 1-4r\sin^2\frac{\kappa\Delta x}{2}\), and the sawtooth mode needs \(|1-4r|\le1\). Choice B is the CFL condition for the wave equation; choice D describes implicit schemes like Crank–Nicolson.
Question 7
Duhamel's integral \(\int_0^te^{-k\lambda_n(t-s)}f_n(s)ds\) expresses:
Why
Each kick at time \(s\) contributes \(f_n(s)ds\), which then decays for the remaining time \(t-s\). Summing over all past times gives the integral — the system's memory, weighted by how long ago each event happened.
Question 8
Which combination is not a well-posed problem?
Why
Prescribing Cauchy data for an elliptic equation over-determines it and destroys continuous dependence — Hadamard's classic example. Elliptic problems take exactly one condition per boundary point, on the entire closed boundary.
Question 9
For \(u_t + uu_x = 0\) with decreasing initial data, characteristics:
Why
Each characteristic travels at the speed of the value it carries, so tall values overtake short ones. Increasing data gives choice A instead; constant data gives C.
Question 10
You are given an unfamiliar second-order linear PDE. What should you compute first?
Why
The type determines everything downstream: which data make the problem well posed, whether information travels at finite speed, whether solutions smooth or keep their corners, which solution method applies, and even which numerical stability condition you'll face. Classify first, then choose the tool.