Laplace PDE: Dirichlet Problem on a Rectangle

Laplace PDE:

\[\Delta u = 0 \qquad \left( \Delta u = u_{xx} + u_{yy} \right)\]

Product Solution:

Assume $u(x, y) = X(x) Y(y)$.

Then

\[\Delta u = X'' Y + X Y'' = 0\] \[\frac{X''}{X} = -\frac{Y''}{Y} = c\]

$c=0$:

\[Y'' = 0 \implies Y = d_1 y + d_2\] \[X'' = 0 \implies X = c_1 x + c_2\] \[u = (d_1 y + d_2) (c_1 x + c_2)\]

$c=\lambda^2, \lambda > 0$

\[Y'' + \lambda^2 Y = 0 \implies Y = d_1 \sin(\lambda y) + d_2 \cos(\lambda y)\] \[X'' - \lambda^2 X = 0 \implies X = c_1 \sinh(\lambda x) + c_2 \cosh(\lambda x)\] \[u = \left[ d_1 \sin(\lambda y) + d_2 \cos(\lambda y) \right] \left[ (c_1 \sinh(\lambda x) + c_2 \cosh(\lambda x) \right]\]

$c=-\lambda^2, \lambda > 0$

\[Y'' - \lambda^2 Y = 0 \implies Y = d_1 \sinh(\lambda y) + d_2 \cosh(\lambda y)\] \[X'' + \lambda^2 X = 0 \implies X = c_1 \sin(\lambda x) + c_2 \cos(\lambda x)\] \[u = \left[ d_1 \sinh(\lambda y) + d_2 \cosh(\lambda y) \right] \left[ (c_1 \sin(\lambda x) + c_2 \cos(\lambda x) \right]\]

Dirichlet Boundary Problem on a Rectangle:

Non-Zero Bottom Edge:

Problem:

\[\Delta u = 0 \qquad x \in (0, L), \ y \in (0, M)\] \[u(x, 0) = f(x) \qquad u(x, M) = 0\] \[u(0, y) = 0 \qquad u(L, y) = 0\]

Considering the BC:

\[u(0, y) = u(L, y) = 0 \implies X(0) = X(L) = 0\]

$c = 0$:

\[X = c_1 x + c_2 \implies X = 0\] \[u = 0, \ \text{trivial solution}\]

$c = \lambda^2$:

\[X = c_1 \sinh(\lambda x) + c_2 \cosh(\lambda x)\] \[X(0) = c_1 \sinh(0) + c_2 \cosh(0) = c_2 = 0\] \[X(L) = c_1 \sinh(\lambda L) = 0 \implies c_1 = 0\] \[X = 0, \ u = 0, \ \text{trivial solution}\]

$c = -\lambda^2$:

\[X = c_1 \sin(\lambda x) + c_2 \cos(\lambda x)\] \[X(0) = c_1 \sin(0) + c_2 \cos(0) = c_2 = 0\] \[X(L) = c_1 \sin(\lambda L) = 0 \implies \lambda_n = \frac{n \pi}{L}\] \[X_n = \sin(\lambda_n x), \lambda_n = \frac{n \pi}{L}\] \[u_n = \left[ d_1 \sinh(\lambda_n y) + d_2 \cosh(\lambda_n y) \right] \sin(\lambda_n x) \qquad \lambda_n = \frac{n \pi}{L}\]

Considering the BC:

\[u(x, M) = 0 \implies Y(M) = 0\] \[Y = d_1 \sinh(\lambda_n y) + d_2 \cosh(\lambda_n y)\] \[Y(M) = d_1 \sinh(\lambda_n M) + d_2 \cosh(\lambda_n M) = 0\] \[d_1 = -\cosh(\lambda_n M) \qquad d_2 = \sinh(\lambda_n M) = 0\] \[Y = -\cosh(\lambda_n M) \sinh(\lambda_n y) + \sinh(\lambda_n M) \cosh(\lambda_n y)\] \[Y = \sinh \left[ \lambda_n (M - y) \right]\]

So:

\[u_n = \sinh \left[ \lambda_n (M - y) \right] \sin(\lambda_n x) \qquad \lambda_n = \frac{n \pi}{L}\] \[u = \sum_{n=1}^\infty a_n \sinh \left[ \lambda_n (M - y) \right] \sin(\lambda_n x) \qquad \lambda_n = \frac{n \pi}{L}\]

Non-Zero Top Edge:

\[u(x, 0) = 0 \qquad u(x, L) = g(x)\]

Same BC for $u(0, y) = u(L, y) = 0$, so we still have :

\[u_n = \left[ d_1 \sinh(\lambda_n y) + d_2 \cosh(\lambda_n y) \right] \sin(\lambda_n x) \qquad \lambda_n = \frac{n \pi}{L}\]

Considering the BC:

\[u(x, 0) = 0 \implies Y(0) = 0\] \[Y = d_1 \sinh(\lambda_n y) + d_2 \cosh(\lambda_n y)\] \[Y(0) = d_1 \sinh(0) + d_2 \cosh(0) = d_2 = 0\] \[Y = \sinh(\lambda_n y)\]

So:

\[u_n = \sinh(\lambda_n y) \sin(\lambda_n x) \qquad \lambda_n = \frac{n \pi}{L}\]

Non-Zero Left and Right Edges:

Just swap x and y.

$u(0, y) = h(y) \qquad u(L, y) = 0$:

\[u_n = \sinh(\gamma_n (L - x)) \sin(\gamma_n y) \qquad \gamma_n = \frac{n \pi}{M}\]

$u(0, y) = 0 \qquad u(L, y) = k(y)$:

\[u_n = \sinh(\gamma_n x) \sin(\gamma_n y) \qquad \gamma_n = \frac{n \pi}{M}\]

Generalizing:

Problem:

\[\Delta u = 0 \qquad x \in (0, L), \ y \in (0, M)\] \[u(x, 0) = f(x) \qquad u(x, M) = g(x)\] \[u(0, y) = h(y) \qquad u(L, y) = k(y)\]

By Linearity:

\[u = u^1 + u^2 + u^3 + u^4\]

Where

\[u^1(x, 0) = f(x) \qquad u^2(x, M) = g(x)\] \[u^3(0, y) = h(y) \qquad u^4(L, y) = k(y)\]

And

\[u_n^1 = \sinh \left[ \lambda_n (M - y) \right] \sin(\lambda_n x) \qquad \lambda_n = \frac{n \pi}{L}\] \[u_n^2 = \sinh(\lambda_n y) \sin(\lambda_n x)\] \[u_n^3 = \sinh(\gamma_n (L - x)) \sin(\gamma_n y) \qquad \gamma_n = \frac{n \pi}{M}\] \[u_n^4 = \sinh(\gamma_n x) \sin(\gamma_n y)\]