Recipe

Find the normal modes of a vibrating string

Use when: you have $u_{tt}=c^2u_{xx}$ on a finite string with both ends held fixed

Goal: every normal mode $u_n(x,t)$ and its natural frequency $\omega_n$ , ready to superpose into the general solution

Taught in Separation of variables for the vibrating string

Quick recipe

Substitute $u=X(x)T(t)$ and separate
Solve the spatial equation with the boundary conditions
Solve the time equation at that eigenvalue
Multiply the two factors and superpose over every $n$

1
Plug the product into the PDE, divide by $c^2XT$ , and set the two sides (one in $x$ alone, one in $t$ alone) equal to a shared constant $-\lambda$ .

Example

For $u_{tt}=9u_{xx}$ , dividing gives $\dfrac{T''}{9T}=\dfrac{X''}{X}=-\lambda$ , so $X''+\lambda X=0$ and $T''+9\lambda T=0$ .

Why: a pure- $x$ expression and a pure- $t$ expression agree, for every $x$ and every $t$ . That forces both to be one frozen number.
2
Impose $X(0)=X(L)=0$ on $X''+\lambda X=0$ . This is the same Dirichlet eigenvalue problem chapter 5 already solved, so its eigenvalues and eigenfunctions carry over unchanged.

Example

On $L=\pi$ , the $n$ -th eigenvalue is $\lambda_n=n^2$ with eigenfunction $X_n(x)=\sin(nx)$ ; for $n=2$ , $\lambda_2=4$ and $X_2(x)=\sin(2x)$ .

Why: the boundary condition only ever constrains $X$ . So it fixes which $\lambda$ values are allowed, before the time equation is even touched.
3
Substitute the eigenvalue into $T''+c^2\lambda T=0$ . Because this equation is second order, its solution oscillates as $T_n(t)=A_n\cos(\omega_nt)+B_n\sin(\omega_nt)$ instead of decaying.

Example

With $c=3$ and $\lambda_2=4$ : $T''+36T=0$ , so $\omega_2=6$ and $T_2(t)=A_2\cos(6t)+B_2\sin(6t)$ .

⚡ Shortcut

Skip resolving $T''+c^2\lambda_nT=0$ from scratch every time. Once $\lambda_n$ is known, read the frequency straight off $\omega_n=c\sqrt{\lambda_n}$ , which for this fixed-fixed string simplifies to $\omega_n=cn\pi/L$ .

Why: the time equation's order is set entirely by the PDE's own order in $t$ . So a second-order PDE always hands $T$ two arbitrary constants, and a solution that swings back and forth but never decays.
4
Form $u_n(x,t)=T_n(t)X_n(x)$ for the mode you need, or sum every $u_n$ together if the full general solution is required.
- only one specific mode is needed (checking a candidate, or finding one frequency) — stop after building that single $u_n$ .
- the complete solution is needed — write the infinite sum over every $n$ , leaving $A_n,B_n$ symbolic until the initial data is matched.
Why: superposition is valid because the wave equation and its boundary conditions are both linear and homogeneous. So any sum of solutions is again a solution.

Common mistakes

Writing $T'+c^2\lambda T=0$ (the heat equation's form) instead of $T''+c^2\lambda T=0$ , out of habit from a first-order time equation.
Forgetting the $c^2$ multiplying $\lambda$ in the time equation.
Reporting $\lambda_n$ itself as the natural frequency instead of $\omega_n=c\sqrt{\lambda_n}$ .
Assuming one initial condition pins down $A_n$ and $B_n$ ; a second-order time equation needs both position and velocity data.