UPSC 2024 Maths Optional Paper 2 Q5a — Step-by-Step Solution

10 marks · Section B

Second-order linear PDEs with constant coefficients (CF, PI) · PDEs · asked 12× in 13 yrs · Read the full method →

Question

Show that if $f$ and $g$ are arbitrary functions, then $u=f(x-kt+i\alpha y)+g(x-kt-i\alpha y)$ is a solution of

\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2},\quad\text{where }\alpha^2=1-\frac{k^2}{c^2}.

Chain-rule differentiation through $\xi=x-kt+i\alpha y$ and $\eta=x-kt-i\alpha y$ ; the condition on $\alpha$ emerges by matching coefficients.

Let $\xi=x-kt+i\alpha y$ , $\eta=x-kt-i\alpha y$ , so $u=f(\xi)+g(\eta)$ .

$x$ -derivatives:

u_{xx}=f''(\xi)+g''(\eta).

$y$ -derivatives (using $\xi_y=i\alpha$ , $\eta_y=-i\alpha$ ):

u_{yy}=(i\alpha)^2 f''(\xi)+(-i\alpha)^2 g''(\eta)=-\alpha^2(f''+g'').

$t$ -derivatives (using $\xi_t=-k$ , $\eta_t=-k$ ):

u_{tt}=k^2(f''+g'').

Combine:

u_{xx}+u_{yy}=(1-\alpha^2)(f''+g''),\qquad\frac{1}{c^2}u_{tt}=\frac{k^2}{c^2}(f''+g'').

These are equal iff $1-\alpha^2=k^2/c^2$ , i.e., $\alpha^2=1-k^2/c^2$ , which is the given condition.

\boxed{\;u=f(x-kt+i\alpha y)+g(x-kt-i\alpha y)\text{ satisfies the wave equation when }\alpha^2=1-k^2/c^2.\;}