Solve u_t=a^2u_(xx) with u(0,t)=u(l,t)=0 and u(x,0) a…
Solve $u_t=a^2u_{xx}$ with $u(0,t)=u(l,t)=0$ and $u(x,0)$ a triangular (tent) profile of peak $l/2$ at $x=l/2$.
Explanation of Method
Separation of variables gives sine eigenfunctions with exponentially decaying time factors; the Fourier sine coefficients of the tent function are computed piecewise.
Step 1 · Separate variables
$X_n=\sin\frac{n\pi x}{l}$, $\lambda_n=n^2\pi^2/l^2$, $T_n=e^{-a^2n^2\pi^2t/l^2}$, so \[ u(x,t)=\sum_{n=1}^\infty B_ne^{-a^2n^2\pi^2t/l^2}\sin\frac{n\pi x}{l}. \]
Step 2 · Compute the Fourier sine coefficients
Splitting $\varphi(x)$ into its two linear pieces and evaluating ($\int x\sin\mu x\,dx=\sin\mu x/\mu^2-x\cos\mu x/\mu$), the odd-symmetric contributions combine to give \[ B_n = \frac{4l}{n^2\pi^2}\sin\frac{n\pi}{2}, \] vanishing for even $n$; with $n=2k-1$: $B_{2k-1}=\dfrac{4l(-1)^{k+1}}{(2k-1)^2\pi^2}$ (verified numerically against direct integration).
Verification
At $t=0$ the series is the Fourier sine series of the tent function. \checkmark Each term decays as $e^{-a^2(2k-1)^2\pi^2t/l^2}\to0$, so $u\to0$ everywhere as $t\to\infty$, consistent with the zero boundary conditions.
Result
The solution is: