Discussion
Solution of \(\nabla^{2}T = 0\) in a square domain (\(0 \lt x \lt 1\) and \(0 \lt y \lt 1\)) with boundary conditions:
\(T(x, 0) = x;\ T(0, y) = y;\ T(x, 1) = 1 + x;\ T(1, y) = 1 + y\)
is
Answer: B
The solution is \(T(x, y) = x + y\).
It satisfies Laplace's equation, since both second derivatives are zero, and it matches every boundary condition: \(T(x, 0) = x\), \(T(0, y) = y\), \(T(x, 1) = x + 1\) and \(T(1, y) = 1 + y\). The solution of Laplace's equation with given boundary values is unique, so this is the answer.
\(x - xy + y\) gives \(T(x, 1) = 1\), not \(1 + x\). \(x + xy + y\) gives \(T(x, 1) = 2x + 1\). \(-x + y\) gives \(T(x, 0) = -x\). Each fails at least one boundary.
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