Mechanical Engineering > Gate 2022 Set-1 > Quants
Solution of ∇2𝑇 = 0 in a square domain (0 < 𝑥 < 1 and 0 < 𝑦 < 1) with boundary conditions:

𝑇(𝑥, 0) = 𝑥; 𝑇(0, 𝑦) = 𝑦; 𝑇(𝑥, 1) = 1 + 𝑥; 𝑇(1, 𝑦) = 1 + 𝑦

is
A
T(x, y) = x − xy+ y
B
T(x,y)=x+y
C
T(x,y)=-x+y
D
T(x,y)=x+xy+y

Correct : b

Laplace's Equation:
Laplace's equation in two dimensions is given by:
    ∂2T/∂x2 + ∂2T/∂y2 = 0
We will test each given option by first checking if it satisfies Laplace's equation, and then if it satisfies all four boundary conditions.
Boundary Conditions:
• T(x, 0) = x
• T(0, y) = y
• T(x, 1) = 1 + x
• T(1, y) = 1 + y
Option A: T(x, y) = x − xy + y
• Check PDE:
    ∂T/∂x = 1 − y
    ∂2T/∂x2 = 0
    ∂T/∂y = −x + 1
    ∂2T/∂y2 = 0
    ∇2T = 0 + 0 = 0. The PDE is satisfied.
• Boundary Conditions:
    T(x, 0) = x − x(0) + 0 = x. (Matches)     T(0, y) = 0 − (0)y + y = y. (Matches)     T(x, 1) = x − x(1) + 1 = x − x + 1 = 1. (Does NOT match 1 + x) Since this option fails one boundary condition, it is not the correct solution.
Option B: T(x, y) = x + y
• Check PDE:
    ∂T/∂x = 1
    ∂2T/∂x2 = 0
    ∂T/∂y = 1
    ∂2T/∂y2 = 0
    ∇2T = 0 + 0 = 0. The PDE is satisfied.
• Boundary Conditions:
    T(x, 0) = x + 0 = x. (Matches T(x, 0) = x)     T(0, y) = 0 + y = y. (Matches T(0, y) = y)     T(x, 1) = x + 1. (Matches T(x, 1) = 1 + x)     T(1, y) = 1 + y. (Matches T(1, y) = 1 + y) All boundary conditions are satisfied. This option satisfies both Laplace's equation and all boundary conditions. Therefore, this is the correct solution.
For completeness, let's briefly check the PDE for the remaining options.
Option C: T(x, y) = -x + y
• PDE:
    ∂T/∂x = -1 &implies; ∂2T/∂x2 = 0
    ∂T/∂y = 1 &implies; ∂2T/∂y2 = 0
    ∇2T = 0 + 0 = 0. The PDE is satisfied.
• Boundary Conditions:
    T(x, 0) = -x + 0 = -x. (Does NOT match x) This option fails the first boundary condition.
Option D: T(x, y) = x + xy + y
• PDE:
    ∂T/∂x = 1 + y
    ∂2T/∂x2 = 0
    ∂T/∂y = x + 1
    ∂2T/∂y2 = 0
    ∇2T = 0 + 0 = 0. The PDE is satisfied.
• Check Boundary Conditions:
    T(x, 0) = x + x(0) + 0 = x. (Matches)     T(0, y) = 0 + (0)y + y = y. (Matches)     T(x, 1) = x + x(1) + 1 = x + x + 1 = 2x + 1. (Does NOT match 1 + x) This option fails one boundary condition.

Based on the analysis, only T(x, y) = x + y satisfies both Laplace's equation and all the given boundary conditions.

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