Solve u_t=a^2u_(xx) with u(0,t)=u(l,t)=0 and u(x,0)=x(l-x)/l^2
Solve $u_t=a^2u_{xx}$ with $u(0,t)=u(l,t)=0$ and $u(x,0)=x(l-x)/l^2$.
Explanation of Method
Same sine-eigenfunction series as Problem 326; the Fourier sine coefficients of the parabolic profile are computed via $\int x\sin\mu x\,dx$ and $\int x^2\sin\mu x\,dx$.
Step 1 · Set up the series
\[ u(x,t)=\sum_{n=1}^\infty B_ne^{-a^2n^2\pi^2t/l^2}\sin\frac{n\pi x}{l}. \]
Step 2 · Compute the Fourier sine coefficients
With $B_n=\frac{2}{l^3}\int_0^l(lx-x^2)\sin\frac{n\pi x}{l}\,dx$, evaluating the two standard integrals and combining gives \[ \int_0^l(lx-x^2)\sin\frac{n\pi x}{l}\,dx = \frac{4l^3}{n^3\pi^3}\ (n\text{ odd}),\qquad 0\ (n\text{ even}), \] so \[ B_{2k-1} = \frac{8}{(2k-1)^3\pi^3},\qquad B_{2k}=0 \] (verified numerically against direct integration).
Verification
At $t=0$ the series reproduces the known Fourier sine expansion of $x(l-x)/l^2$. \checkmark Boundary conditions hold termwise since $\sin(0)=\sin(n\pi)=0$.
Result
The solution is: