Solve u_(tt)=a^2u_(xx) with u(0,t)=0 , u(l,t)=Asinω t…

Solve $u_{tt}=a^2u_{xx}$ with $u(0,t)=0$, $u(l,t)=A\sin\omega t$, $u(x,0)=0$, $u_t(x,0)=0$.

Step 1 · Steady-state particular solution

Try $u_p=f(x)\sin\omega t$: $f''+(\omega/a)^2f=0$ with $f(0)=0$, $f(l)=A$ gives \[ u_p(x,t) = \frac{A\sin(\omega x/a)}{\sin(\omega l/a)}\sin\omega t \qquad(\omega l/a\neq n\pi). \]

Step 2 · Transient problem

$v=u-u_p$ satisfies $v_{tt}=a^2v_{xx}$, $v(0,t)=v(l,t)=0$, $v(x,0)=0$, and $v_t(x,0)=-\partial_tu_p|_{t=0}=-\dfrac{A\omega\sin(\omega x/a)}{\sin(\omega l/a)}=:\psi(x)$.

Step 3 · Solve the transient problem

\[ v(x,t)=\sum_{n=1}^\infty C_n\sin\frac{n\pi x}{l}\sin\frac{n\pi at}{l},\qquad \frac{n\pi a}{l}C_n = \frac2l\int_0^l\psi(x)\sin\frac{n\pi x}{l}\,dx. \] Evaluating the integral (product-to-sum, using $\sin(\omega l/a\pm n\pi)=(-1)^n\sin(\omega l/a)$) gives \[ C_n = \frac{2A\omega a\,(-1)^{n+1}}{l\bigl[\omega^2-(n\pi a/l)^2\bigr]}. \]

Verification

Both boundary conditions and $u(x,0)=0$ hold termwise; differentiating and using the Fourier identity from the derivation of $C_n$ shows the transient and steady-state initial velocities cancel, giving $u_t(x,0)=0$. \checkmark

Mechanical interpretation

The first term is the forced steady-state response; the series is the free response excited by the mismatch between the initial state and the steady state. When $\omega\to\omega_n=n\pi a/l$, $\sin(\omega l/a)\to0$ and the steady-state amplitude blows up — resonance.