Discussion
Let \(\Omega = \{(x,y) \in \mathbb{R}^{2} \mid x^{2} + y^{2} \lt 1\}\) be the open unit disc in \(\mathbb{R}^{2}\) with boundary \(\partial\Omega\). If \(u(x,y)\) is the solution of the Dirichlet problem
\[ \begin{aligned} u_{xx} + u_{yy} &= 0 \quad \text{in } \Omega \\ u(x,y) &= 1 - 2y^{2} \quad \text{on } \partial\Omega, \end{aligned} \]
then \(u\left(\frac{1}{2}, 0\right)\) is equal to
Answer: C
\(u\left(\frac{1}{2}, 0\right) = \) \(\frac{1}{4}\).
On the unit circle \(x^{2} + y^{2} = 1\), so the boundary data can be rewritten:
\[ 1 - 2y^{2} = (x^{2} + y^{2}) - 2y^{2} = x^{2} - y^{2} \]
The function \(x^{2} - y^{2}\) is harmonic (\(u_{xx} + u_{yy} = 2 - 2 = 0\)) and matches the boundary values, so by uniqueness of the Dirichlet problem it is the solution: \(u(x, y) = x^{2} - y^{2}\).
Then \(u\left(\frac{1}{2}, 0\right) = \frac{1}{4}\). Taking \(1 - 2y^{2}\) itself as the solution would give 1, but that function is not harmonic, so 1 is wrong; \(-1\) and \(-\frac{1}{4}\) have the wrong sign.
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