Solving wave equation — separation of variables

The problem we are solving

WHAT we want: a function $u(x,t)$ satisfying the PDE + 2 BCs + 2 ICs. WHY separation works: the PDE is linear & homogeneous, and the BCs are homogeneous ($=0$), so sums of solutions are still solutions — that's the engine of superposition.

Step-by-step derivation FROM SCRATCH

Step 1 — Make the separation ansatz

Assume $u(x,t)=X(x)\,T(t).$ Plug into the PDE. Since $u_{tt}=X T''$ and $u_{xx}=X'' T$: $X\,T'' = c^2 X'' \,T.$

Why this step? It converts derivatives of a 2-variable function into derivatives of 1-variable functions.

Step 2 — Separate the variables

Divide both sides by $c^2 X T$: $\frac{T''}{c^2 T} = \frac{X''}{X}.$

Why this step? The left side depends only on $t$, the right side only on $x$. Two things equal for all $x,t$ can only happen if both equal the same constant. Call it $-\lambda$ (minus chosen for convenience — we'll see we need oscillation): $\frac{X''}{X}=\frac{T''}{c^2T}=-\lambda.$

This gives two ODEs: $X'' + \lambda X = 0, \qquad T'' + c^2\lambda\, T = 0.$

Step 3 — Apply BCs to the spatial ODE (the eigenvalue problem)

The BCs $u(0,t)=u(L,t)=0$ must hold for all $t$, and $T(t)\not\equiv 0$, so $X(0)=0, \qquad X(L)=0.$

We solve $X''+\lambda X=0$ and check three cases for $\lambda$ (Steel-man each!):

• $\lambda<0$ (say $\lambda=-\mu^2$): $X=Ae^{\mu x}+Be^{-\mu x}$. BCs force $A=B=0$. Only trivial solution. Rejected.
• $\lambda=0$: $X=Ax+B$. BCs force $A=B=0$. Trivial. Rejected.
• $\lambda>0$ (say $\lambda=k^2$): $X=A\cos kx + B\sin kx$.

Apply BCs to the $\lambda>0$ case:

• $X(0)=A=0 \Rightarrow A=0$.
• $X(L)=B\sin kL =0$. For a nontrivial $B\ne0$: $\sin kL=0 \Rightarrow kL=n\pi$.

So the eigenvalues and eigenfunctions are $\lambda_n=\left(\frac{n\pi}{L}\right)^2,\qquad X_n(x)=\sin\!\frac{n\pi x}{L},\quad n=1,2,3,\dots$

Step 4 — Solve the time ODE for each mode

With $\lambda_n=(n\pi/L)^2$: $T_n''+c^2\lambda_n T_n=0 \Rightarrow T_n''+\omega_n^2 T_n=0,\quad \omega_n=\frac{n\pi c}{L}.$ This is simple harmonic motion: $T_n(t)=a_n\cos\omega_n t + b_n\sin\omega_n t.$

Why this step? $\omega_n$ are the natural frequencies of the string — the harmonics you hear from a guitar.

Step 5 — Superpose (build the general solution)

Why a sum? Each $u_n=X_nT_n$ solves the (linear, homogeneous) PDE + BCs. By superposition, any sum does too. We use the freedom in $a_n,b_n$ to fit the ICs.

Step 6 — Use ICs to find the coefficients (Fourier!)

Position IC $u(x,0)=f(x)$: set $t=0$ ($\cos0=1,\sin0=0$): $f(x)=\sum_{n=1}^\infty a_n\sin\frac{n\pi x}{L}.$ This is a Fourier sine series, so by orthogonality $\int_0^L \sin\frac{m\pi x}{L}\sin\frac{n\pi x}{L}dx=\frac{L}{2}\delta_{mn}$: $\boxed{a_n=\frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx}$

Velocity IC $u_t(x,0)=g(x)$: differentiate $u$ in $t$, set $t=0$ ($\frac{d}{dt}\cos\omega_n t|_0=0$, $\frac{d}{dt}\sin\omega_n t|_0=\omega_n$): $g(x)=\sum_{n=1}^\infty b_n\,\omega_n\,\sin\frac{n\pi x}{L}\Rightarrow b_n\omega_n=\frac{2}{L}\int_0^L g(x)\sin\frac{n\pi x}{L}dx.$ So $\boxed{b_n=\frac{2}{n\pi c}\int_0^L g(x)\sin\frac{n\pi x}{L}\,dx} \quad(\text{using }\omega_n=\tfrac{n\pi c}{L}).$

Worked examples

Common mistakes (Steel-manned)

Active-recall flashcards

Connections

• Fourier sine series — how $a_n$, $b_n$ are computed; the same orthogonality engine.
• Sturm–Liouville theory — the spatial problem $X''+\lambda X=0$ is a special eigenvalue problem; guarantees real eigenvalues and orthogonal eigenfunctions.
• Heat equation — separation of variables — same spatial ODE, but time ODE is first order ($T'=-c^2\lambda T$) → decay instead of oscillation.
• d'Alembert solution of the wave equation — alternative method ($u=F(x-ct)+G(x+ct)$); equivalent to summing the modes.
• Standing waves and harmonics — physics of $\omega_n=n\pi c/L$, overtones, timbre.
• Superposition principle — why summing modes is legal.

Concept Map

Hinglish (regional understanding)