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[SOLVED] Laplace mixed BC

dwsmith

Well-known member
Feb 1, 2012
1,673
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?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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
$$
u(x,y) = \sum_{n = 1}^{\infty}A_n\cos\lambda_nx\cosh\lambda_ny.
$$
 
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