Discussion
Let \(u_{1}\) and \(u_{2}\) be solutions of the equation :
\(\Delta u = 0\) in D
\(\dfrac{\partial u}{\partial n} = g(s)\) on \(\partial D\)
where \(D = \{(x, y) \in \mathbf{R}^{2} / \frac{x^{2}}{4} + \frac{y^{2}}{9} \le 1\}\) is an ellipse with boundary \(\partial D = \{(x, y) \in \mathbf{R}^{2} /\) \(\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1\}\). If \(u_{1}(0, \frac{1}{2}) - u_{2}(0, \frac{1}{2}) = 1\). Then :
Answer: B
The answer is \(u_{1} \equiv u_{2} + 1\) in D.
Let \(w = u_{1} - u_{2}\). Then \(\Delta w = 0\) in D and \(\partial w / \partial n = 0\) on the boundary. Green's identity gives
\[ \begin{aligned} \int_{D} |\nabla w|^{2} &= \int_{\partial D} w \frac{\partial w}{\partial n} - \int_{D} w\, \Delta w \\ &= 0 \end{aligned} \]
so \(\nabla w = 0\) and \(w\) is constant: solutions of the Neumann problem are unique up to an additive constant. The given value \(w(0, \frac{1}{2}) = 1\) fixes the constant as 1.
So \(u_{1} - u_{2}\) does not change sign (C), the two are not identical (D), and they are not 'dependent' in any multiplicative sense (A).
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