UPSC 2022 Maths Optional Paper 2 Q6a — Step-by-Step Solution

20 marks · Section B

Heat equation · PDEs · asked 3× in 13 yrs · Read the full method →

Question

Solve $u_t=u_{xx}$ , $0<x<l$ , $t>0$ , with $u(0,t)=u(l,t)=0$ and $u(x,0)=x(l-x)$ .

Standard heat-equation separation; Fourier sine series of $x(l-x)$ — only odd modes.

Setup. Standard heat equation with homogeneous Dirichlet BC. Separation of variables.

$u(x,t)=X(x)T(t)$ . Substitute: $XT'=X''T\Rightarrow T'/T=X''/X=-\lambda^2$ .

Spatial: $X''+\lambda^2 X=0$ , $X(0)=X(l)=0$ . Eigenfunctions $X_n=\sin(n\pi x/l)$ , eigenvalues $\lambda_n=n\pi/l$ .

Temporal: $T'=-\lambda_n^2 T\Rightarrow T_n=e^{-n^2\pi^2 t/l^2}$ .

u(x,t)=\sum_{n=1}^\infty B_n\sin\!\left(\dfrac{n\pi x}{l}\right)e^{-n^2\pi^2 t/l^2}.

$\sum B_n\sin(n\pi x/l)=x(l-x)$ , the Fourier sine series.

B_n=\dfrac{2}{l}\int_0^l x(l-x)\sin(n\pi x/l)\,dx.

This is the same integral computed in 2015 P1 Q7(a). The result:

\boxed{\;u(x,t)=\sum_{\substack{n=1\\n\text{ odd}}}^\infty\dfrac{8l^2}{n^3\pi^3}\sin\!\left(\dfrac{n\pi x}{l}\right)e^{-n^2\pi^2 t/l^2}.\;}