Discussion
Consider the Laplace equation \(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0\), \(0 \lt x \lt 1\), \(0 \lt y \lt 1\), with the boundary conditions
\(T(x, 0) = x\), \(T(0, y) = y\)
\(T(x, 1) = 1 + x\), \(T(1, y) = 1 + y\).
Then the value of \(T\left(\frac{1}{2}, \frac{1}{3}\right)\) is equal to
Answer: B
The value is \(T(\frac{1}{2}, \frac{1}{3}) = \frac{5}{6}\).
Try \(T(x, y) = x + y\). It is harmonic, since both second derivatives are 0, and it meets every boundary condition: \(T(x, 0) = x\), \(T(0, y) = y\), \(T(x, 1) = x + 1\) and \(T(1, y) = 1 + y\). By uniqueness of the solution of the Dirichlet problem, this is the solution.
\[ \begin{aligned} T\left(\frac{1}{2}, \frac{1}{3}\right) &= \frac{1}{2} + \frac{1}{3} \\ &= \frac{5}{6} \end{aligned} \]
\(\frac{7}{6}\) adds an extra \(\frac{1}{3}\), \(\frac{1}{6}\) is the difference \(\frac{1}{2} - \frac{1}{3}\), and \(\frac{1}{2}\) is only the x-coordinate.
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