← 2014 Paper 2

UPSC 2014 Maths Optional Paper 2 Q8a — Step-by-Step Solution

15 marks · Section B

Wave equation · PDEs · asked 7× in 13 yrs · Read the full method →

Question

Solve $\dfrac{\partial^{2}u}{\partial t^{2}}=\dfrac{\partial^{2}u}{\partial x^{2}}$, $0<x<1$, $t>0$, given that (i) $u(x,0)=0$, $0\le x\le 1$ (ii) $\dfrac{\partial u}{\partial t}(x,0)=x^{2}$, $0\le x\le 1$ (iii) $u(0,t)=u(1,t)=0$, for all $t$.

Technique

Standard separation of variables; sine-series expansion of initial velocity profile; the $x^{2}$ profile has $1/n$ decay (slow because $x^{2}$ extended periodically has a jump discontinuity).

Solution

Setup. Wave equation with $c=1$, length $L=1$. Standard sine-series Fourier solution.

Step 1 — General solution

Step 2 — Apply $u(x,0)=0$

$\sum A_n\sin(n\pi x)=0\Rightarrow A_n=0$ for all $n$.

Step 3 — Apply $u_t(x,0)=x^{2}$

So $B_n\cdot n\pi$ is the sine-series Fourier coefficient of $x^{2}$ on $(0,1)$:

Step 4 — Compute $I_n=\int_0^{1}x^{2}\sin(n\pi x)\,dx$

Using the standard formula $\int x^{2}\sin(ax)\,dx=-\dfrac{x^{2}\cos ax}{a}+\dfrac{2x\sin ax}{a^{2}}+\dfrac{2\cos ax}{a^{3}}+C$ with $a=n\pi$:

Evaluate:

• At $x=1$: $-\dfrac{\cos n\pi}{n\pi}+0+\dfrac{2\cos n\pi}{(n\pi)^{3}}=-\dfrac{(-1)^{n}}{n\pi}+\dfrac{2(-1)^{n}}{(n\pi)^{3}}$.
• At $x=0$: $0+0+\dfrac{2}{(n\pi)^{3}}=\dfrac{2}{(n\pi)^{3}}$.

Difference:

Cases:

• Odd $n$ ($(-1)^{n}=-1$): $I_n=\dfrac{1}{n\pi}+\dfrac{2(-2)}{(n\pi)^{3}}=\dfrac{1}{n\pi}-\dfrac{4}{(n\pi)^{3}}$.

• Even $n$ ($(-1)^{n}=1$): $I_n=-\dfrac{1}{n\pi}+0=-\dfrac{1}{n\pi}$.

Step 5 — Compute $B_n=2I_n/(n\pi)$

• Odd $n$: $B_n=\dfrac{2}{(n\pi)^{2}}-\dfrac{8}{(n\pi)^{4}}=\dfrac{2[(n\pi)^{2}-4]}{(n\pi)^{4}}$.

• Even $n$: $B_n=-\dfrac{2}{(n\pi)^{2}}$.

Step 6 — Final solution

Answer