← 2025 Paper 2
UPSC 2025 Maths Optional Paper 2 Q6a — Step-by-Step Solution
20 marks · Section B
Laplace equation: Dirichlet/Neumann, separation of variables · PDEs · asked 4× in 13 yrs · Read the full method →
Question
Solve ∂x2∂2u+∂y2∂2u=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]×[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 uxx+uyy=0 and dividing by XY:
XX′′=−YY′′=−λ(separation constant).
So X′′+λX=0 and Y′′−λY=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′′+λX=0 with these conditions requires λ=λn=(anπ)2, n=1,2,3,…, with eigenfunctions
Xn(x)=sinanπx.
(Cases λ≤0 give only the trivial solution.)
y-equation
With λn=(nπ/a)2>0, the equation Y′′−λnY=0 has general solution
Yn(y)=Ancoshanπy+Bnsinhanπy.
Apply the homogeneous edge u(x,0)=0⇒Yn(0)=0⇒An=0. Hence
Yn(y)=Bnsinhanπy.
Superposition
u(x,y)=n=1∑∞Bnsinanπxsinhanπy.
This already satisfies the three homogeneous conditions.
Fix coefficients from u(x,b)=f(x)
f(x)=u(x,b)=n=1∑∞(Bnsinhanπb)sinanπx.
This is the Fourier sine series of f on [0,a], so the bracketed coefficient is the sine-series coefficient:
Bnsinhanπb=a2∫0af(x)sinanπxdx,
giving
Bn=asinhanπb2∫0af(x)sinanπxdx.
Final solution
u(x,y)=n=1∑∞Bnsinanπxsinhanπy,Bn=asinh(nπb/a)2∫0af(ξ)sinanπξdξ.
Answer
u(x,y)=n=1∑∞[asinh(nπb/a)2∫0af(ξ)sinanπξdξ]sinanπxsinhanπy.