Question:
Let \( \Omega = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 < 1\} \) be the open unit disc and \( \partial\Omega \) be its boundary. If \( u(x, y) \) is the solution of the following Dirichlet problem
\[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \quad \text{in } \Omega \]
\[ u(x, y) = x^2 - y^2 \quad \text{on } \partial\Omega, \]
then the value of \( 4\left(u\left(\frac{1}{2}, 0\right) - u\left(0, \frac{1}{2}\right)\right) \) is ______.
Let \( \Omega = \{(x, y) \in \mathbb{R}^2 : x^2 + y^2 < 1\} \) be the open unit disc and \( \partial\Omega \) be its boundary. If \( u(x, y) \) is the solution of the following Dirichlet problem
\[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \quad \text{in } \Omega \]
\[ u(x, y) = x^2 - y^2 \quad \text{on } \partial\Omega, \]
then the value of \( 4\left(u\left(\frac{1}{2}, 0\right) - u\left(0, \frac{1}{2}\right)\right) \) is ______.
\[ \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0 \quad \text{in } \Omega \]
\[ u(x, y) = x^2 - y^2 \quad \text{on } \partial\Omega, \]
then the value of \( 4\left(u\left(\frac{1}{2}, 0\right) - u\left(0, \frac{1}{2}\right)\right) \) is ______.
Show Hint
Check if the boundary function is already harmonic in the whole disc. If it is, it must equal the unique solution itself.
Updated On: Jul 21, 2026
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Correct Answer: 2
Solution and Explanation
Step 1: Understanding the Question.
We need to solve the Laplace equation \( u_{xx} + u_{yy} = 0 \) inside the open unit disc \( \Omega \), with the boundary value \( u = x^2 - y^2 \) fixed on the circle \( \partial\Omega \). This is a Dirichlet problem. Once we know \( u \), we plug in the two given points and combine the values.
Step 2: Check whether the boundary function is already harmonic.
A function is harmonic when it satisfies Laplace's equation everywhere, not just on the boundary. Take \( v(x,y) = x^2 - y^2 \) and find its second derivatives:
\[ \frac{\partial^2 v}{\partial x^2} = 2, \qquad \frac{\partial^2 v}{\partial y^2} = -2 \]
Adding them gives \( 2 + (-2) = 0 \), so \( v(x,y) = x^2 - y^2 \) is harmonic on the whole plane, not just on the circle.
Step 3: Use uniqueness of the Dirichlet problem.
The Dirichlet problem on a bounded region has exactly one harmonic solution for a given continuous boundary value. Here \( v(x,y) = x^2 - y^2 \) is harmonic inside \( \Omega \) and it also equals the prescribed boundary data \( x^2 - y^2 \) on \( \partial\Omega \). Since the solution is unique, we get
\[ u(x,y) = x^2 - y^2 \quad \text{for all } (x,y) \in \Omega \]
There is no need to build the Poisson integral separately, since we already found a harmonic function that matches the boundary condition exactly.
Step 4: Evaluate at the two given points.
At \( \left(\frac{1}{2}, 0\right) \):
\[ u\left(\frac{1}{2}, 0\right) = \left(\frac{1}{2}\right)^2 - 0^2 = \frac{1}{4} \]
At \( \left(0, \frac{1}{2}\right) \):
\[ u\left(0, \frac{1}{2}\right) = 0^2 - \left(\frac{1}{2}\right)^2 = -\frac{1}{4} \]
Final Answer:
\[ 4\left(u\left(\frac{1}{2},0\right) - u\left(0,\frac{1}{2}\right)\right) = 4\left(\frac{1}{4} - \left(-\frac{1}{4}\right)\right) = 4 \times \frac{1}{2} = 2 \]
\[ \boxed{2} \]
We need to solve the Laplace equation \( u_{xx} + u_{yy} = 0 \) inside the open unit disc \( \Omega \), with the boundary value \( u = x^2 - y^2 \) fixed on the circle \( \partial\Omega \). This is a Dirichlet problem. Once we know \( u \), we plug in the two given points and combine the values.
Step 2: Check whether the boundary function is already harmonic.
A function is harmonic when it satisfies Laplace's equation everywhere, not just on the boundary. Take \( v(x,y) = x^2 - y^2 \) and find its second derivatives:
\[ \frac{\partial^2 v}{\partial x^2} = 2, \qquad \frac{\partial^2 v}{\partial y^2} = -2 \]
Adding them gives \( 2 + (-2) = 0 \), so \( v(x,y) = x^2 - y^2 \) is harmonic on the whole plane, not just on the circle.
Step 3: Use uniqueness of the Dirichlet problem.
The Dirichlet problem on a bounded region has exactly one harmonic solution for a given continuous boundary value. Here \( v(x,y) = x^2 - y^2 \) is harmonic inside \( \Omega \) and it also equals the prescribed boundary data \( x^2 - y^2 \) on \( \partial\Omega \). Since the solution is unique, we get
\[ u(x,y) = x^2 - y^2 \quad \text{for all } (x,y) \in \Omega \]
There is no need to build the Poisson integral separately, since we already found a harmonic function that matches the boundary condition exactly.
Step 4: Evaluate at the two given points.
At \( \left(\frac{1}{2}, 0\right) \):
\[ u\left(\frac{1}{2}, 0\right) = \left(\frac{1}{2}\right)^2 - 0^2 = \frac{1}{4} \]
At \( \left(0, \frac{1}{2}\right) \):
\[ u\left(0, \frac{1}{2}\right) = 0^2 - \left(\frac{1}{2}\right)^2 = -\frac{1}{4} \]
Final Answer:
\[ 4\left(u\left(\frac{1}{2},0\right) - u\left(0,\frac{1}{2}\right)\right) = 4\left(\frac{1}{4} - \left(-\frac{1}{4}\right)\right) = 4 \times \frac{1}{2} = 2 \]
\[ \boxed{2} \]
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