← 2025 Paper 2

UPSC Maths 2025 Paper 2 Q6a — Solution

20 marks · Section B

Question

Solve $\dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} = 0$ for a rectangular plate subject to the boundary conditions $u(0, y) = 0$, $u(a, y) = 0$, $u(x, 0) = 0$, $u(x, b) = f(x)$.

Technique

Solve Laplace’s equation on the rectangle $[0,a]\times[0,b]$ by separation of variables: the three homogeneous edges give a sine eigenproblem in $x$, and the inhomogeneous edge $u(x,b)=f(x)$ fixes the coefficients via a half-range Fourier sine series.

Solution

Separation of variables

Seek $u(x,y) = X(x)Y(y)$. Substituting into $u_{xx}+u_{yy}=0$ and dividing by $XY$: $\frac{X''}{X} = -\frac{Y''}{Y} = -\lambda \quad(\text{separation constant}).$

So $X'' + \lambda X = 0$ and $Y'' - \lambda Y = 0$.

$x$-eigenproblem (homogeneous edges $x=0,a$)

$u(0,y)=0$ and $u(a,y)=0$ force $X(0)=X(a)=0$. A non-trivial solution of $X''+\lambda X=0$ with these conditions requires $\lambda = \lambda_n = \left(\dfrac{n\pi}{a}\right)^2$, $n=1,2,3,\dots$, with eigenfunctions $X_n(x) = \sin\frac{n\pi x}{a}.$

(Cases $\lambda \le 0$ give only the trivial solution.)

$y$-equation

With $\lambda_n = (n\pi/a)^2 > 0$, the equation $Y'' - \lambda_n Y = 0$ has general solution $Y_n(y) = A_n\cosh\frac{n\pi y}{a} + B_n\sinh\frac{n\pi y}{a}.$

Apply the homogeneous edge $u(x,0)=0\Rightarrow Y_n(0)=0\Rightarrow A_n = 0$. Hence $Y_n(y) = B_n\sinh\frac{n\pi y}{a}.$

Superposition

$u(x,y) = \sum_{n=1}^{\infty} B_n\,\sin\frac{n\pi x}{a}\,\sinh\frac{n\pi y}{a}.$

This already satisfies the three homogeneous conditions.

Fix coefficients from $u(x,b)=f(x)$

$f(x) = u(x,b) = \sum_{n=1}^{\infty}\Big(B_n\sinh\frac{n\pi b}{a}\Big)\sin\frac{n\pi x}{a}.$

This is the Fourier sine series of $f$ on $[0,a]$, so the bracketed coefficient is the sine-series coefficient: $B_n\sinh\frac{n\pi b}{a} = \frac{2}{a}\int_0^a f(x)\sin\frac{n\pi x}{a}\,dx,$ giving $B_n = \frac{2}{a\,\sinh\dfrac{n\pi b}{a}}\int_0^a f(x)\sin\frac{n\pi x}{a}\,dx.$

Final solution