Solve θ_(tt)=a^2θ_(xx) with θ(0,t)=θ(l,t)=0 , θ_t(x,0)=0…
Solve $\theta_{tt}=a^2\theta_{xx}$ with $\theta(0,t)=\theta(l,t)=0$, $\theta_t(x,0)=0$, and $\theta(x,0)$ a triangular (tent) profile of peak $\theta_0$ at $x=l/2$.
Explanation of Method
Use the fixed-end series from Problem 318/140; the zero initial velocity kills the sine-in-time terms, and the Fourier sine coefficients of the triangular profile are computed piecewise.
Step 1 · Apply $\theta_t(x,0)=0$
This forces all $D_n=0$, leaving \[ \theta(x,t) = \sum_{n=1}^\infty C_n\cos\frac{n\pi at}{l}\sin\frac{n\pi x}{l}. \]
Step 2 · Compute the Fourier sine coefficients
Splitting the integral over the two linear pieces of the tent function and evaluating (via $\int x\sin\mu x\,dx=\sin\mu x/\mu^2-x\cos\mu x/\mu$): \[ C_n = \frac{8\theta_0}{n^2\pi^2}\sin\frac{n\pi}{2}, \] which vanishes for even $n$; writing $n=2k-1$, $\sin\frac{(2k-1)\pi}{2}=(-1)^{k+1}$, so \[ C_{2k-1} = \frac{8\theta_0(-1)^{k+1}}{(2k-1)^2\pi^2},\qquad C_{2k}=0. \] (Verified numerically against direct integration of the piecewise profile.)
Mechanical interpretation
Only odd modes are excited (the triangular profile is symmetric about $x=l/2$), each oscillating at $\omega_{2k-1}=(2k-1)\pi a/l$ with amplitude decaying as $1/(2k-1)^2$, so the fundamental mode dominates.
Result
The solution is: