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UPSC 2021 Maths Optional Paper 2 Q6a — Step-by-Step Solution

20 marks · Section B

Wave equation · PDEs · asked 7× in 14 yrs · Read the full method →

Question

Solve $a^2 u_{xx}=u_{tt}$, $0<x<L$, $t>0$ with $u(0,t)=u(L,t)=0$, $u(x,0)=x(L-x)/4$, $u_t(x,0)=0$.

Technique

Standard separation for wave equation; Fourier sine series for IC; $u_t(x,0)=0$ kills sin-in-$t$ terms.

Solution

Step 1 — Separation of variables

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

$X''+\lambda^2 X=0,\;X(0)=X(L)=0$: $X_n=\sin(n\pi x/L)$, $\lambda_n=n\pi/L$.

$T''+a^2\lambda_n^2 T=0$: $T_n(t)=A_n\cos(a n\pi t/L)+B_n\sin(an\pi t/L)$.

Step 2 — General solution

Step 3 — Apply $u_t(x,0)=0$

$u_t(x,0)=\sum B_n(an\pi/L)\sin(n\pi x/L)=0$ for all $x$ ⇒ $B_n=0$ for all $n$.

Step 4 — Apply $u(x,0)=x(L-x)/4$

$\sum A_n\sin(n\pi x/L)=x(L-x)/4$.

Fourier sine series of $x(L-x)/4$ on $[0,L]$:

$A_n=\dfrac{2}{L}\int_0^L\dfrac{x(L-x)}{4}\sin(n\pi x/L)\,dx=\dfrac{1}{2L}\int_0^L x(L-x)\sin(n\pi x/L)\,dx$.

From 2015 P1 Q7(a) / 2022 P2 Q6(a): $\int_0^L x(L-x)\sin(n\pi x/L)\,dx=\dfrac{2L^3}{n^3\pi^3}[1-(-1)^n]$.

For $n$ odd: $=4L^3/(n^3\pi^3)$. For $n$ even: $=0$.

So $A_n=\dfrac{1}{2L}\cdot\dfrac{4L^3}{n^3\pi^3}=\dfrac{2L^2}{n^3\pi^3}$ for odd $n$, $0$ for even.

Step 5 — Final solution

Answer