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UPSC 2014 Maths Optional Paper 2 Q7a — Step-by-Step Solution

15 marks · Section B

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

Question

Find the deflection of a vibrating string (length $=\pi$, ends fixed, $\dfrac{\partial^{2}u}{\partial t^{2}}=\dfrac{\partial^{2}u}{\partial x^{2}}$) corresponding to zero initial velocity and initial deflection $f(x)=k(\sin x-\sin 2x)$.

Technique

Separation of variables; sine-series basis; zero initial velocity kills $B_n$; matching the initial deflection to its Fourier sine expansion (here a finite sum, so direct).

Solution

Setup. Wave equation on $(0,\pi)$ with $c=1$ (PDE $u_{tt}=u_{xx}$). Boundary $u(0,t)=u(\pi,t)=0$. ICs: $u(x,0)=k(\sin x-\sin 2x)$, $u_t(x,0)=0$.

Step 1 — General sine-series solution

Separation of variables gives the basis $\sin(nx)$ for $n=1,2,\ldots$. The general solution:

Step 2 — Apply $u_t(x,0)=0$

$u_t(x,0)=\sum_n B_n\cdot n\sin(nx)=0\quad\forall x\Rightarrow B_n=0\;\forall n$.

So $u(x,t)=\sum_{n=1}^\infty A_n\cos(nt)\sin(nx)$.

Step 3 — Apply $u(x,0)=k(\sin x-\sin 2x)$

By matching coefficients of the sine basis:

• $A_1=k$.
• $A_2=-k$.
• $A_n=0$ for $n\ge 3$.

Step 4 — Assemble the solution

Answer