Laplace, heat and wave equations: separation of variables

In one sentence

Separation of variables turns a PDE into a Sturm–Liouville problem plus an ODE in time, and the difference between heat and wave is entirely in that time factor — decaying exponential against oscillation.

Why the exam asks it

Because it is the one constructive method that answers a whole family of problems, and because the three equations share a spatial eigenvalue problem while behaving completely differently. Getting the time factor wrong is the standard error.

The idea

Assume $u(x,t) = X(x)T(t)$, substitute, and divide. Each side depends on one variable, so both equal a constant.

Heat equation

$u_t = k u_{xx}$ on $[0,L]$ with $u(0,t)=u(L,t)=0$ gives $X'' + \lambda X = 0$ with $X(0)=X(L)=0$ — precisely the Sturm–Liouville problem from the ODE chapter. So

$$\lambda_n = \frac{n^2\pi^2}{L^2}, \qquad X_n = \sin\frac{n\pi x}{L},$$

and $T' = -k\lambda T$ gives $T_n = e^{-k\lambda_n t}$. Hence

$$u(x,t) = \sum_{n\ge1} b_n \sin\frac{n\pi x}{L}\,e^{-kn^2\pi^2t/L^2},$$

with $b_n$ the Fourier sine coefficients of the initial profile.

Read the behaviour off the exponent: every mode decays, and the $n$-th decays like $e^{-n^2 t}$, so high frequencies vanish fastest. That is the smoothing, quantitatively — after any positive time only the lowest modes survive in any meaningful amount, which is why the solution is analytic for $t>0$ regardless of the initial data. As $t\to\infty$ the solution tends to zero, or to the steady state when the boundary values are non-zero.

Wave equation

$u_{tt} = c^2u_{xx}$ has the same $X_n$ and the same $\lambda_n$ — the spatial problem is identical — but $T'' = -c^2\lambda T$, so

$$T_n = A_n\cos\frac{n\pi ct}{L} + B_n\sin\frac{n\pi ct}{L}.$$

Oscillation, not decay. Energy is conserved, nothing smooths, and the solution is periodic in time with period $2L/c$. The modes are the harmonics of a string, and $n=1$ is the fundamental.

Two initial conditions are needed — $u$ and $u_t$ — because the time equation is second order. Supplying only one is a standard way to make an option incomplete.

Laplace equation

$u_{xx}+u_{yy}=0$ on a rectangle separates into $X''+\lambda X = 0$ and $Y'' - \lambda Y = 0$, so the second factor gives $\sinh$ and $\cosh$ rather than exponential decay or oscillation. On a disc, polar separation gives $r^n(A\cos n\theta + B\sin n\theta)$ and the Poisson integral formula.

Note there is no time: Laplace takes boundary conditions all the way round, and the maximum principle means the solution's extremes are on the boundary.

Which conditions give which eigenfunctions

• Dirichlet $u=0$ at both ends: $\sin\frac{n\pi x}{L}$, $n\ge1$.
• Neumann $u_x=0$ at both ends: $\cos\frac{n\pi x}{L}$, $n\ge0$ — and $\lambda=0$ is an eigenvalue, giving a constant mode that does not decay.
• Mixed: half-integer multiples.
• Periodic: both sines and cosines, so eigenvalues are double.

The Neumann case matters physically: with insulated ends the average temperature is conserved, and that is the $n=0$ mode surviving forever.

Where intuition breaks

"Heat and wave have different eigenfunctions." Identical spatial problem, identical $X_n$ and $\lambda_n$. Only the time factor differs — $e^{-k\lambda t}$ against $\cos$ and $\sin$.

"The heat equation solution oscillates." Every mode decays monotonically. Oscillation in time is the wave equation.

"One initial condition suffices for the wave equation." Two: $u$ and $u_t$. The time ODE is second order.

"With Neumann conditions the solution decays to zero." It decays to the average of the initial profile. The constant mode has $\lambda = 0$ and does not decay — the same $\lambda=0$ that stops a Green's function existing in the ODE chapter.

"Separation works for any PDE." It needs the equation and the domain to separate — a rectangle or a disc, not an arbitrary region, and linear with homogeneous boundary conditions. Inhomogeneous conditions must be subtracted off first via the steady state.

"The series converges to the initial data everywhere." It converges in $L^2$, and pointwise where the data is decent. At a jump it converges to the midpoint — Gibbs overshoot near the discontinuity does not vanish.

The exam's angle

1. Separate, and identify the spatial eigenvalue problem. It is Sturm–Liouville, and its eigenfunctions come from the boundary conditions alone.
2. Write the time factor from the order of the time derivative — first order gives decay, second gives oscillation.
3. Check whether $\lambda = 0$ is an eigenvalue. With Neumann conditions it is, and that mode is the long-run answer.
4. Get the coefficients by orthogonality, exactly as in the Sturm–Liouville chapter.
5. For long-time behaviour, keep only the smallest $\lambda$. Everything else is exponentially negligible, which answers most asymptotic questions in one line.

The night before

• Separation gives a Sturm–Liouville problem in space and an ODE in time.
• Dirichlet on $[0,L]$: $\lambda_n = n^2\pi^2/L^2$, $X_n = \sin(n\pi x/L)$.
• Heat: $T_n = e^{-k\lambda_n t}$, every mode decays, high frequencies fastest. Smoothing.
• Wave: $T_n = A\cos(n\pi ct/L) + B\sin(\cdot)$, oscillation, energy conserved, period $2L/c$, needs two initial conditions.
• Laplace: $\sinh$ and $\cosh$ on a rectangle; $r^n(\cos n\theta, \sin n\theta)$ on a disc.
• Neumann gives cosines including $n=0$; that constant mode never decays, so heat tends to the average.
• Coefficients by orthogonality. Long-run behaviour is the smallest eigenvalue.