Solve the Dirichlet problem on {x² + y² < 1} {x² + y² = 1} Au = 0 I u(x, y) = 2xy + 2y? on
Solve the Dirichlet problem on {x² + y² < 1} {x² + y² = 1} Au = 0 I u(x, y) = 2xy + 2y? on
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem 4: Solve the Dirichlet Problem
Given the Dirichlet problem:
\[
\begin{cases}
\Delta u = 0 & \text{on } \{x^2 + y^2 < 1\} \\
u(x, y) = 2xy + 2y^2 & \text{on } \{x^2 + y^2 = 1\}
\end{cases}
\]
where \( \Delta \) represents the Laplace operator.
#### Explanation:
1. The Laplace equation \(\Delta u = 0\) is a second-order partial differential equation, which in two dimensions is given by:
\[
\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}
\]
2. The problem is defined on a unit disk, which is represented by the region \(\{x^2 + y^2 < 1\}\). This means that the function \(u(x, y)\) needs to be harmonic inside this disk.
3. The boundary condition is specified on the unit circle \(\{x^2 + y^2 = 1\}\), where the function is given by \(u(x, y) = 2xy + 2y^2\). This means that the solution \(u(x, y)\) must take this form on the boundary of the disk.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b0185de-0645-4c2a-aea4-e046d61ab5cb%2F63728e40-05aa-4790-90db-431b7fea332b%2Fhjhkn09_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 4: Solve the Dirichlet Problem
Given the Dirichlet problem:
\[
\begin{cases}
\Delta u = 0 & \text{on } \{x^2 + y^2 < 1\} \\
u(x, y) = 2xy + 2y^2 & \text{on } \{x^2 + y^2 = 1\}
\end{cases}
\]
where \( \Delta \) represents the Laplace operator.
#### Explanation:
1. The Laplace equation \(\Delta u = 0\) is a second-order partial differential equation, which in two dimensions is given by:
\[
\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}
\]
2. The problem is defined on a unit disk, which is represented by the region \(\{x^2 + y^2 < 1\}\). This means that the function \(u(x, y)\) needs to be harmonic inside this disk.
3. The boundary condition is specified on the unit circle \(\{x^2 + y^2 = 1\}\), where the function is given by \(u(x, y) = 2xy + 2y^2\). This means that the solution \(u(x, y)\) must take this form on the boundary of the disk.
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