6.3 Separation of variables

The problem. A string of length $L$ is held fixed at both ends. Its transverse displacement $u(x, t)$ obeys the 1-D wave equation

$$\frac{\partial^2 u}{\partial t^2} \;=\; c^2\, \frac{\partial^2 u}{\partial x^2}, \qquad 0 < x < L, \quad t > 0,$$

with boundary conditions $u(0, t) = u(L, t) = 0$ for all $t$ (the ends are clamped), and initial conditions $u(x, 0) = f(x)$ (the initial shape) and $u_t(x, 0) = 0$ (the string starts at rest).

Step 1 — Substitute the product ansatz. Try

$$u(x, t) \;=\; X(x)\, T(t).$$

The partial derivatives separate cleanly:

$$\frac{\partial^2 u}{\partial t^2} \;=\; X(x)\, T''(t), \qquad \frac{\partial^2 u}{\partial x^2} \;=\; X''(x)\, T(t).$$

Substituting into the wave equation:

$$X(x)\, T''(t) \;=\; c^2\, X''(x)\, T(t).$$

Step 2 — Divide by $X(x)\, T(t)$ and separate.

$$\frac{T''(t)}{c^2\, T(t)} \;=\; \frac{X''(x)}{X(x)}.$$

The left side depends only on $t$. The right side depends only on $x$. For two functions of independent variables to be equal at all values of those variables, both must equal the same constant. Call that constant $-k^2$:

$$\frac{T''}{c^2 T} \;=\; \frac{X''}{X} \;=\; -k^2.$$

(The sign is a convention. Choosing $-k^2$ rather than $+k^2$ anticipates that the bounded-domain problem will produce oscillating spatial solutions, not exponentially growing or decaying ones. If we got the sign wrong we would discover it later — the boundary conditions would force only the trivial solution. We will see that below.)

The PDE has decoupled into two ODEs:

$$X''(x) \;+\; k^2\, X(x) \;=\; 0, \qquad T''(t) \;+\; c^2 k^2\, T(t) \;=\; 0.$$

Both are second-order linear ODEs with constant coefficients — exactly the equations we learned to solve in 5.3.

Step 3 — Solve the spatial ODE with its boundary conditions. The spatial ODE $X'' + k^2 X = 0$ has the general solution

$$X(x) \;=\; A \sin(k x) + B \cos(k x).$$

Apply the left boundary condition $X(0) = 0$ (the value $u(0, t) = 0$ forces $X(0)\, T(t) = 0$ for all $t$, so unless $T$ is identically zero we need $X(0) = 0$):

$$0 \;=\; A \sin(0) + B \cos(0) \;=\; B,$$

so $B = 0$ and $X(x) = A \sin(k x)$. Apply the right boundary condition $X(L) = 0$:

$$0 \;=\; A \sin(k L).$$

We don’t want $A = 0$ (that gives the trivial $X \equiv 0$). So we need $\sin(k L) = 0$, which forces

$$k L \;=\; n \pi, \qquad n = 1, 2, 3, \ldots$$

This is the quantisation condition. The boundary conditions have selected a discrete set of allowed $k$ values:

$$k_n \;=\; \frac{n \pi}{L}, \qquad n = 1, 2, 3, \ldots$$

with corresponding mode shapes

$$X_n(x) \;=\; \sin\!\left(\frac{n \pi x}{L}\right).$$

The amplitude $A$ is absorbed into the time-dependent coefficient at the end. Each integer $n$ labels one mode: $n = 1$ is the half-wavelength fundamental, $n = 2$ has a node at the centre, and so on. We get infinitely many modes, indexed by $n$.

Quick check on the sign of the separation constant. If we had chosen $+k^2$ instead of $-k^2$, the spatial ODE would be $X'' - k^2 X = 0$ with solution $A e^{kx} + B e^{-kx}$ — exponentials, not sinusoids. The boundary conditions $X(0) = X(L) = 0$ would force $A = B = 0$. Only the sign that gives sinusoids produces non-trivial modes. The boundary conditions chose the sign for us.

Step 4 — Solve the time ODE for each mode. For each $k_n$ the time ODE is

$$T''(t) \;+\; c^2 k_n^2\, T(t) \;=\; 0,$$

with $\omega_n \equiv c k_n = n \pi c / L$. The general solution is the simple-harmonic motion we know from 5.3:

$$T_n(t) \;=\; A_n \cos(\omega_n t) + B_n \sin(\omega_n t).$$

Step 5 — Build the general solution. Each $X_n(x)\, T_n(t)$ solves the PDE and the boundary conditions. By linearity, any sum of them also solves the PDE and the boundary conditions. The general solution is the superposition:

$$u(x, t) \;=\; \sum_{n=1}^{\infty} \sin\!\left(\frac{n \pi x}{L}\right) \bigl[\, A_n \cos(\omega_n t) + B_n \sin(\omega_n t) \,\bigr].$$

We have two coefficient sets, $\{A_n\}$ and $\{B_n\}$ — one set per piece of initial data, just as a second-order ODE in time needs two initial pieces.

Step 6 — Apply the initial conditions. From $u(x, 0) = f(x)$:

$$f(x) \;=\; \sum_{n=1}^{\infty} A_n \sin\!\left(\frac{n \pi x}{L}\right).$$

This is a Fourier sine series expansion of $f$ on $[0, L]$. The coefficients $A_n$ are extracted by projection onto the mode basis, a step we develop properly in 6.5 — Modes once we have the orthogonality relations in hand.

From $u_t(x, 0) = 0$:

$$0 \;=\; \sum_{n=1}^{\infty} \omega_n\, B_n\, \sin\!\left(\frac{n \pi x}{L}\right) \;\Longrightarrow\; B_n = 0 \text{ for all } n.$$

Step 7 — Final answer.

$$\boxed{\;u(x, t) \;=\; \sum_{n=1}^{\infty} A_n \sin\!\left(\frac{n \pi x}{L}\right) \cos\!\left(\frac{n \pi c\, t}{L}\right),\;}$$

with $\{A_n\}$ the Fourier sine-series coefficients of the initial shape $f(x)$.

Step 8 — Sanity-check.

1. Boundary conditions. Every term has $\sin(n \pi x / L)$, which vanishes at $x = 0$ and $x = L$. ✓
2. Initial shape. At $t = 0$, every cosine is 1 and the formula reduces to $u(x, 0) = \sum A_n \sin(n\pi x / L)$, the Fourier series of $f$. ✓
3. Initial velocity. Differentiate in $t$ and evaluate at $t = 0$. Every term picks up a factor of $\omega_n \sin(\omega_n \cdot 0) = 0$. So $u_t(x, 0) = 0$. ✓
4. PDE. For each $n$, $\partial_t^2 [\cos(\omega_n t)] = -\omega_n^2 \cos(\omega_n t)$ and $\partial_x^2 [\sin(k_n x)] = -k_n^2 \sin(k_n x)$. Since $\omega_n = c k_n$, the term satisfies $u_{tt} = c^2 u_{xx}$. ✓