Separation of Variables for the Heat Equation

Separation of variables for the heat equation is the entry point into an enormous range of applications. In engineering, it models transient conduction in rods, slabs, and simple 2-D/3-D geometries (with separable domains), giving explicit design formulas for quenching, annealing, and thermal insulation problems. In probability and statistics, the same eigenfunction expansion underlies the transition density of Brownian motion killed at the boundary of an interval, connecting heat conduction to first-passage-time problems and to the spectral theory of Markov diffusions. In quantum mechanics, an almost identical calculation — separating the time-dependent Schrödinger equation into a spatial eigenvalue problem (an infinite square well, formally the same Dirichlet eigenproblem here) and an oscillatory time factor — reproduces the same $\sin(n\pi x/L)$ eigenfunctions with $e^{-iE_nt/\hbar}$ replacing the decaying exponential. In numerical analysis and scientific computing, spectral and pseudo-spectral methods generalize this idea to more complex domains and operators, while finite-difference and finite-element solvers are validated against these closed-form series solutions. In signal processing and image compression, the discrete cousin of this eigenbasis (the discrete sine/cosine transform) is used precisely because it diagonalizes the same kind of second-difference operator that appears here, making it efficient for representing smooth, boundary-constrained signals.

:::quiz[In the Dirichlet problem on $[0,L]$, what happens if we allow $\lambda=0$ in the eigenvalue problem?]

• It gives the eigenfunction $X(x)=1$
• Only the trivial solution $X\equiv0$ satisfies both $X(0)=0$ and $X(L)=0$ ✓
• It gives infinitely many eigenfunctions
• It gives the eigenfunction $X(x)=x$

Ask the tutor: How does this same method change if the boundary conditions are Neumann or mixed (Robin) instead of Dirichlet?

Picture the rod as a one-dimensional strip held at zero temperature at both ends, with some initial heat distribution $f(x)$ in between. Ordinary intuition says the heat should diffuse until everything reaches the boundary temperature, zero, and indeed that is what the theorem below proves. But how it decays is the interesting content: the temperature profile does not shrink uniformly — it decays as a sum of standing spatial waves $\sin(n\pi x/L)$, each shrinking at its own exponential rate $e^{-\alpha^2(n\pi/L)^2 t}$. High-frequency wiggles (large $n$) die out fastest, because the factor $(n\pi/L)^2$ in the exponent grows like $n^2$. This is why an initially jagged temperature profile smooths out almost immediately, while the coarse, overall shape (dominated by $n=1$) persists longest. The eigenfunctions $\sin(n\pi x/L)$ are exactly the shapes for which a solution can be pure exponential decay in time with no change of spatial shape — plug $X_n(x)T_n(t)$ back in and you see the profile just rescales uniformly, never redistributing between different sine shapes. Because these sine functions form a complete orthogonal system on $[0,L]$ (a fact from Fourier analysis foreshadowed here and made rigorous under Sturm-Liouville Theory), any reasonable initial data $f(x)$ can be written as an infinite combination of them, and then each piece evolves independently and exponentially. Separation of variables is therefore really a change of basis: rewrite $f$ in the eigenbasis where the dynamics become diagonal (uncoupled, exponential decay mode by mode).

:::svg[The first three Dirichlet eigenfunctions on $[0,L]$, $L=1$: $\sin(\pi x)$, $\sin(2\pi x)$, $\sin(3\pi x)$]

Example 1 — an asymmetric sawtooth-type initial profile. Solve

i.e. $\alpha^2=3$, $L=2$, and a genuinely asymmetric cubic profile (not symmetric about the midpoint, unlike a "tent" function). By the theorem, with $L=2$,

We compute $c_n$ by expanding $f(x) = x(2-x)^2 = 4x - 4x^2 + x^3$ and integrating termwise using

via repeated integration by parts. Let $\beta = n\pi/2$. Standard antiderivative bookkeeping (integrate by parts twice for $x^2$, three times for $x^3$) gives, after evaluating at $x=0,2$ and using $\sin(n\pi)=0$, $\cos(n\pi)=(-1)^n$:

With $2\beta = n\pi$: $\cos(2\beta) = (-1)^n$, $\sin(2\beta)=0$, so

Now $\int_0^2 x\cos(\beta x)\,dx = \left[\frac{x\sin(\beta x)}{\beta}\right]_0^2 - \frac{1}{\beta}\int_0^2\sin(\beta x)dx = 0 - \frac{1}{\beta}\left[-\frac{\cos(\beta x)}{\beta}\right]_0^2 = \frac{1}{\beta^2}\big(\cos(2\beta)-1\big) = \frac{(-1)^n-1}{\beta^2}.$ So

One further integration by parts gives $\int_0^2 x^2\cos(\beta x)dx = \left[\frac{x^2\sin(\beta x)}{\beta}\right]_0^2 - \frac{2}{\beta}\int_0^2 x\sin(\beta x)dx = 0 - \frac{2}{\beta}\cdot\frac{2(-1)^{n+1}}{\beta} = \frac{4(-1)^n}{\beta^2}.$ So

Now assemble $c_n = 4\int x\sin - 4\int x^2\sin + \int x^3\sin$:

Collect the $1/\beta$ terms: $\dfrac{8(-1)^{n+1}}{\beta} + \dfrac{16(-1)^n}{\beta} - \dfrac{8(-1)^n}{\beta} = \dfrac{-8(-1)^n+16(-1)^n-8(-1)^n}{\beta} = \dfrac{0}{\beta}=0.$ Collect the $1/\beta^3$ terms: $-\dfrac{8(-1)^n-8}{\beta^3} + \dfrac{12(-1)^n}{\beta^3} = \dfrac{-8(-1)^n+8+12(-1)^n}{\beta^3} = \dfrac{4(-1)^n+8}{\beta^3}.$ So $c_n = \dfrac{8 + 4(-1)^n}{\beta^3}$ with $\beta = n\pi/2$, i.e. $\beta^3 = n^3\pi^3/8$, giving

So $c_n = 96/(n^3\pi^3)$ for $n$ even, and $c_n = 32/(n^3\pi^3)$ for $n$ odd. As a check, $c_1 = 32/\pi^3 \approx 1.0323$, and $c_2 = 96/(8\pi^3) = 12/\pi^3 \approx 0.3873$; both positive, consistent with $f\ge 0$ on $[0,2]$. Therefore

The dominant long-time behavior ($t$ large) is governed by $n=1$:

so the profile becomes a pure half-sine hump decaying at rate $3\pi^2/4 \approx 7.402$ regardless of the asymmetry present in $f$ — a good illustration that fine initial detail (odd/even coefficient differences) washes out exponentially fast.

Example 2 — nonzero but constant boundary data, reduced to the homogeneous case. Solve

Here $\alpha^2=1$, $L=\pi$, but the boundary data is inhomogeneous, so the theorem does not apply directly to $u$. The standard device is to subtract off the (unique) steady-state solution $v(x)$ satisfying $v''=0$, $v(0)=5$, $v(\pi)=-2$: this gives $v(x) = 5 + \dfrac{-2-5}{\pi}x = 5 - \dfrac{7x}{\pi}$, a straight line interpolating the two boundary values. Let $w(x,t) = u(x,t) - v(x)$. Since $v_t = 0$ and $v_{xx}=0$,

so $w$ solves the same heat equation. Its boundary values are $w(0,t) = u(0,t)-v(0) = 5-5=0$ and $w(\pi,t)=u(\pi,t)-v(\pi) = -2-(-2)=0$: homogeneous Dirichlet conditions, exactly the hypotheses of the theorem. Its initial data is

Now apply the theorem with $L=\pi$: eigenfunctions $\sin(nx)$, eigenvalues $n^2$, and

By orthogonality of $\{\sin(nx)\}$ on $[0,\pi]$ (i.e. $\int_0^\pi \sin(mx)\sin(nx)dx = \tfrac{\pi}{2}\delta_{mn}$), the two explicit sine terms contribute directly: the $\sin x$ term contributes $c_1^{(b)} = 1$ and $c_n^{(b)}=0$ for $n\ne1$; the $3\sin2x$ term contributes $c_2^{(b)}=3$ and $0$ otherwise. It remains to expand the linear term $\tfrac{7x}{\pi}$ in a sine series on $[0,\pi]$:

A standard computation (integrate by parts once) gives $\int_0^\pi x\sin(nx)\,dx = \dfrac{\pi(-1)^{n+1}}{n}$ (using $\sin(n\pi)=0,\ \cos(n\pi)=(-1)^n$), so this term contributes

Combining,

Explicitly: $c_1 = \dfrac{14}{\pi} + 1 \approx 4.4577+1=5.4577$; $c_2 = -\dfrac{7}{\pi}+3 \approx -2.2282+3=0.7718$; $c_3 = \dfrac{14}{3\pi}\approx1.4859$; and in general $c_n = \dfrac{14(-1)^{n+1}}{\pi n}$ for $n\ge3$. Finally,

with $c_n$ as above. As $t\to\infty$, every term in the sum decays to $0$, so $u(x,t)\to v(x) = 5-\tfrac{7x}{\pi}$: the rod approaches the steady linear temperature profile fixed by the boundary values, exactly as physical intuition demands.