← 2017 Paper 2

UPSC 2017 Maths Optional Paper 2 Q5d — Step-by-Step Solution

10 marks · Section B

Laplace equation: Dirichlet/Neumann, separation of variables · PDEs · asked 4× in 13 yrs · Read the full method →

Question

Let Γ\Gamma be a closed curve in xyxy-plane and let SS denote the region bounded by the curve Γ\Gamma. Let ∂2w∂x2+∂2w∂y2=f(x,y) ∀(x,y)∈S\dfrac{\partial^2 w}{\partial x^2}+\dfrac{\partial^2 w}{\partial y^2}=f(x,y)\ \forall(x,y)\in S. If ff is prescribed at each point (x,y)(x,y) of SS and ww is prescribed on the boundary Γ\Gamma of SS, then prove that any solution w=w(x,y)w=w(x,y), satisfying these conditions, is unique.

Technique

Subtract two candidate solutions → harmonic function with zero boundary data; Green’s first identity gives ∬∣∇u∣2=0\iint|\nabla u|^2=0 → uu constant → u≡0u\equiv0. Alternative: maximum principle.

Solution

We are given the Dirichlet problem for Poisson’s equation:

∇2w=f in S,w=g on Γ,\nabla^2 w=f \text{ in } S,\qquad w=g \text{ on } \Gamma,

where ff (on SS) and gg (on Γ\Gamma) are prescribed.

Step 1 — Assume two solutions and form their difference

Suppose w1w_1 and w2w_2 both satisfy the conditions:

∇2w1=f,  ∇2w2=f in S,w1=g=w2 on Γ.\nabla^2 w_1=f,\ \ \nabla^2 w_2=f \text{ in } S,\qquad w_1=g=w_2 \text{ on } \Gamma.

Define u=w1−w2u=w_1-w_2. By linearity,

∇2u=∇2w1−∇2w2=f−f=0 in S,\nabla^2 u=\nabla^2 w_1-\nabla^2 w_2=f-f=0 \text{ in } S, u=w1−w2=g−g=0 on Γ.u=w_1-w_2=g-g=0 \text{ on } \Gamma.

Thus uu is harmonic in SS and vanishes on the boundary Γ\Gamma.

Step 2 — Energy integral (Green’s first identity)

Green’s first identity for uu over SS with boundary Γ\Gamma:

∬S(u ∇2u+∣∇u∣2) dA=∮Γu ∂u∂n ds.\iint_S \big(u\,\nabla^2 u+|\nabla u|^2\big)\,dA=\oint_\Gamma u\,\frac{\partial u}{\partial n}\,ds.

On the right, u=0u=0 on Γ\Gamma, so the boundary integral vanishes. On the left, ∇2u=0\nabla^2 u=0 in SS, so the term u ∇2uu\,\nabla^2 u vanishes. Hence

∬S∣∇u∣2 dA=0.\iint_S |\nabla u|^2\,dA=0.

Step 3 — Conclude u≡0u\equiv0

The integrand ∣∇u∣2=ux2+uy2≥0|\nabla u|^2=u_x^2+u_y^2\ge0 is continuous and non-negative, and its integral over SS is zero. Therefore

∣∇u∣2≡0 in S ⟹ ux≡0, uy≡0 ⟹ u=const in S.|\nabla u|^2\equiv0 \text{ in } S\ \Longrightarrow\ u_x\equiv0,\ u_y\equiv0\ \Longrightarrow\ u=\text{const in } S.

By continuity up to the boundary and u=0u=0 on Γ\Gamma, the constant is 00. Hence

u≡0 ⟹ w1≡w2 in S.u\equiv0\ \Longrightarrow\ w_1\equiv w_2 \text{ in } S.

Answer

  The solution w of ∇2w=f with w prescribed on Γ is unique.  \boxed{\;\text{The solution } w \text{ of } \nabla^2 w=f \text{ with } w \text{ prescribed on } \Gamma \text{ is unique.}\;}
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