
# 5.4: Laplace’s Equation

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Figure $$\PageIndex{1}$$: A conducting sheet insulated from above and below.

In a square, heat-conducting sheet, insulated from above and below

$\frac{1}{k}\dfrac{\partial}{\partial t} u = \dfrac{\partial^2}{\partial x^2} u + \dfrac{\partial^2}{\partial y^2} u.$

If we are looking for a steady state solution, i.e., we take $$u(x,y,t)=u(x,y)$$ the time derivative does not contribute, and we get Laplace’s equation

$\dfrac{\partial^2}{\partial x^2} u + \dfrac{\partial^2}{\partial y^2} u=0,$

an example of an elliptic equation. Let us once again look at a square plate of size $$a\times b$$, and impose the boundary conditions

\begin{align} u(x,0) & = 0, \nonumber\\ u(a,y) & = 0, \nonumber\\ u(x,b) & = x, \nonumber\\ u(0,y) & = 0.\end{align}

(This choice is made so as to be able to evaluate Fourier series easily. It is not very realistic!) We once again separate variables,

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

and define

$\frac{X''}{X} = -\frac{Y''}{Y} = -\lambda.$

or explicitly

$X'' = -\lambda X,\;\;Y''=\lambda Y.$

With boundary conditions $$X(0)=X(a)=0$$, $$Y(0)=0$$. The 3rd boundary conditions remains to be implemented.

Once again distinguish three cases:

 $$\lambda>0$$

$$X(x) = \sin \alpha_n(x)$$, $$\alpha_n=\frac{n\pi}{a}$$, $$\lambda_n=\alpha_n^2$$. We find

\begin{align} Y(y) &= C_n\sinh \alpha_n y + D_n \cosh \alpha _n y \nonumber\\[4pt] &= C'_n\exp (\alpha_n y) + D'_n \exp(- \alpha _n y).\end{align}

Since $$Y(0)=0$$ we find $$D_n=0$$ ($$\sinh(0)=0,\cosh(0)=1$$).

 $$\lambda \leq 0$$

No solutions

So we have

$u(x,y) = \sum_{n=1}^\infty b_n \sin\alpha_n x \sinh \alpha_n y$

The one remaining boundary condition gives

$u(x,b) = x = \sum_{n=1}^\infty b_n \sin\alpha_n x \sinh \alpha_n b.$

This leads to the Fourier series of $$x$$,

\begin{align} b_n \sinh \alpha_n b &= \frac{2}{a} \int_0^a x \sin \frac{n\pi x}{a}dx \nonumber\\ &= \frac{2 a }{n\pi}(-1)^{n+1}.\end{align}

So, in short, we have

$V(x,y) = \frac{2a}{\pi} \sum_{n=1}^\infty (-1)^{n+1} \frac{\sin \frac{n\pi x}{a}\sinh \frac{n\pi y}{a}}{n \sinh \frac{n\pi b}{a}}.$

Exercise $$\PageIndex{1}$$

The dependence on $$x$$ enters through a trigonometric function, and that on $$y$$ through a hyperbolic function. Yet the differential equation is symmetric under interchange of $$x$$ and $$y$$. What happens?