Solving Laplace's Eq with Mixed BCs

  • Context: MHB 
  • Thread starter Thread starter Dustinsfl
  • Start date Start date
  • Tags Tags
    Laplace Mixed
Click For Summary
SUMMARY

This discussion focuses on solving Laplace's equation $\nabla^2u = 0$ on a rectangular domain with mixed boundary conditions. The boundary conditions include $u_y(x,0) = 0$, $u_x(0,y) = 0$, $u(x,H) = f(x)$, and $u_x(L,y) + u(L,y) = 0$. The solution is derived using separation of variables, resulting in the form $u(x,y) = \sum_{n = 1}^{\infty}A_n\cos\lambda_nx\cosh\lambda_ny$, where the eigenvalues $\lambda_n$ are determined by the equation $\tan\lambda_n = \frac{1}{\lambda_n}$. The condition $B_n = 0$ is established due to the first boundary condition, simplifying the solution.

PREREQUISITES
  • Understanding of Laplace's equation and its applications in potential theory.
  • Familiarity with boundary conditions, specifically mixed boundary conditions.
  • Knowledge of separation of variables technique in solving partial differential equations.
  • Basic understanding of eigenvalues and eigenfunctions in the context of differential equations.
NEXT STEPS
  • Study the method of separation of variables for solving partial differential equations.
  • Explore the implications of mixed boundary conditions on solutions to Laplace's equation.
  • Investigate the properties of hyperbolic functions, particularly $\cosh$ and $\sinh$.
  • Learn about eigenvalue problems and their role in solving differential equations.
USEFUL FOR

Mathematicians, physicists, and engineers working with potential theory, particularly those dealing with boundary value problems in rectangular domains.

Dustinsfl
Messages
2,217
Reaction score
5
Solve Laplace’s equation $\nabla^2u = 0$ on the rectangle with the following boundary conditions:
$$
u_y(x,0) = 0\quad u_x(0,y) = 0\quad u(x,H) = f(x)\quad u_x(L,y) + u(L,y) = 0.
$$

How are mixed BC handled?
 
Physics news on Phys.org
Consider the boundary conditions $u_x(0,y) = 0$ and $u_x(L,y) + u(L,y) = 0$.
Therefore, if $u(x,y)$ is of the form $u(x,y) = \varphi(x)\psi(y)$, $\varphi_n(x) = A\cos\lambda_nx$ and the eigenvalues are determined by
$$
\tan\lambda_n = \frac{1}{\lambda_n}.
$$
So we have that
$$
\begin{alignat*}{3}
u(x,y) & = & \sum_{n = 1}^{\infty}A\cos\lambda_nx(B\cosh\lambda_ny + C\sinh\lambda_ny)\\
& = & \sum_{n = 1}^{\infty}\cos\lambda_nx(A_n\cosh\lambda_ny + B_n\sinh\lambda_ny)
\end{alignat*}
$$
Now because of the first boundary condition, $u_y(x,0)$, we have that $B_n = 0$.
Therefore, the solution is of the form
$$
u(x,y) = \sum_{n = 1}^{\infty}A_n\cos\lambda_nx\cosh\lambda_ny.
$$
 
Last edited:

Similar threads

  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K