← 2024 Paper 2

UPSC 2024 Maths Optional Paper 2 Q6a — Step-by-Step Solution

20 marks · Section B

Laplace equation: Dirichlet/Neumann, separation of variables · PDEs · asked 5× in 14 yrs · Read the full method →

Question

Show that the solution of the 2D Laplace equation in the upper half-plane (y≥0y\ge 0, ϕ→0\phi\to 0 at infinity) with boundary condition ϕ(x,0)=f(x)\phi(x,0)=f(x) is

ϕ(x,y)=yπ∫−∞∞f(ξ) dξy2+(x−ξ)2.\phi(x,y)=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{f(\xi)\,d\xi}{y^2+(x-\xi)^2}.

Technique

Fourier transform in xx; solve the resulting ODE in yy; apply boundary conditions; invert via the convolution theorem to identify the Poisson kernel.

Solution

Step 1 — Fourier transform.

Let Φ(k,y)=ϕ^(k,y)\Phi(k,y)=\hat\phi(k,y), F(k)=f^(k)F(k)=\hat f(k). Transforming the PDE:

Φyy−k2Φ=0.\Phi_{yy}-k^2\Phi=0.

Step 2 — Solve the ODE in yy.

General solution: A(k)e−∣k∣y+B(k)e∣k∣yA(k)e^{-|k|y}+B(k)e^{|k|y}. The decay condition ϕ→0\phi\to 0 as y→∞y\to\infty forces B(k)=0B(k)=0:

Φ(k,y)=A(k)e−∣k∣y.\Phi(k,y)=A(k)e^{-|k|y}.

Step 3 — Boundary condition.

At y=0y=0: Φ(k,0)=A(k)=F(k)\Phi(k,0)=A(k)=F(k), so Φ(k,y)=F(k)e−∣k∣y\Phi(k,y)=F(k)e^{-|k|y}.

Step 4 — Convolution theorem.

Φ=F⋅e−∣k∣y\Phi=F\cdot e^{-|k|y} in Fourier space corresponds to ϕ=f∗Py\phi=f*P_y in physical space, where PyP_y is the inverse transform of e−∣k∣ye^{-|k|y}.

Step 5 — Compute the Poisson kernel.

Py(x)=12π∫−∞∞e−∣k∣yeikx dk=12π[1y−ix+1y+ix]=12π⋅2yy2+x2=yπ(y2+x2).P_y(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-|k|y}e^{ikx}\,dk=\frac{1}{2\pi}\left[\frac{1}{y-ix}+\frac{1}{y+ix}\right]=\frac{1}{2\pi}\cdot\frac{2y}{y^2+x^2}=\frac{y}{\pi(y^2+x^2)}.

Step 6 — Combine.

ϕ(x,y)=∫−∞∞f(ξ)⋅yπ(y2+(x−ξ)2) dξ=yπ∫−∞∞f(ξ) dξy2+(x−ξ)2.\phi(x,y)=\int_{-\infty}^{\infty}f(\xi)\cdot\frac{y}{\pi(y^2+(x-\xi)^2)}\,d\xi=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{f(\xi)\,d\xi}{y^2+(x-\xi)^2}.

Answer

  ϕ(x,y)=yπ∫−∞∞f(ξ) dξy2+(x−ξ)2.  \boxed{\;\phi(x,y)=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{f(\xi)\,d\xi}{y^2+(x-\xi)^2}.\;}
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