# [SOLVED]Laplace mixed BC

#### dwsmith

##### Well-known member
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
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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