Question

Solution of \( \nabla^2 T = 0 \) in a square domain \( (0<x<1 \) and \( 0<y<1) \) with boundary conditions: \[ T(x, 0) = x; \quad T(0, y) = y; \quad T(x, 1) = 1 + x; \quad T(1, y) = 1 + y \] is

• A. \( T(x, y) = x - xy + y \)
• B. \( T(x, y) = x + y \)
• C. \( T(x, y) = x \)
• D. \( T(x, y) = x + y \)

Hint:

For boundary value problems in Laplace's equation, linear solutions often satisfy simple boundary conditions. Check boundary conditions and apply them directly to verify the solution.

Answer 1

The Correct Option is B

To solve this Laplace equation with the given boundary conditions, we use the principle of superposition for solutions to Laplace's equation. The boundary conditions suggest a simple linear solution. We assume a linear solution of the form: \[ T(x, y) = x + y. \] Step 1: Verify the boundary conditions - At \( y = 0 \), \( T(x, 0) = x + 0 = x \), which satisfies \( T(x, 0) = x \).
- At \( x = 0 \), \( T(0, y) = 0 + y = y \), which satisfies \( T(0, y) = y \).
- At \( y = 1 \), \( T(x, 1) = x + 1 \), which satisfies \( T(x, 1) = 1 + x \).
- At \( x = 1 \), \( T(1, y) = 1 + y \), which satisfies \( T(1, y) = 1 + y \).
Step 2: Verify the solution to Laplace's equation For the solution \( T(x, y) = x + y \), we compute the Laplacian: \[ \nabla^2 T = \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0 + 0 = 0. \] Thus, \( T(x, y) = x + y \) is indeed a solution to the Laplace equation. The correct answer is (B) \( T(x, y) = x + y \).