5.4: Laplace’s Equation
- Page ID
- 8347
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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
\[\frac{1}{k}\dfrac{\partial}{\partial t} u = \dfrac{\partial^2}{\partial x^2} u + \dfrac{\partial^2}{\partial y^2} u. \nonumber \]
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
\[\dfrac{\partial^2}{\partial x^2} u + \dfrac{\partial^2}{\partial y^2} u=0, \nonumber \]
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
\[\begin{align} u(x,0) & = 0, \nonumber\\ u(a,y) & = 0, \nonumber\\ u(x,b) & = x, \nonumber\\ u(0,y) & = 0.\end{align} \nonumber \]
(This choice is made so as to be able to evaluate Fourier series easily. It is not very realistic!) We once again separate variables,
\[u(x,y) = X(x) Y(y), \nonumber \]
and define
\[\frac{X''}{X} = -\frac{Y''}{Y} = -\lambda. \nonumber \]
or explicitly
\[X'' = -\lambda X,\;\;Y''=\lambda Y. \nonumber \]
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
\[\begin{align} Y(y) &= C_n\sinh \alpha_n y + D_n \cosh \alpha _n y \nonumber\\[4pt] &= C'_n\exp (\alpha_n y) + D'_n \exp(- \alpha _n y).\end{align} \nonumber \]
Since \(Y(0)=0\) we find \(D_n=0\) (\(\sinh(0)=0,\cosh(0)=1\)).
| \(\lambda \leq 0\) |
No solutions
So we have
\[u(x,y) = \sum_{n=1}^\infty b_n \sin\alpha_n x \sinh \alpha_n y \nonumber \]
The one remaining boundary condition gives
\[u(x,b) = x = \sum_{n=1}^\infty b_n \sin\alpha_n x \sinh \alpha_n b. \nonumber \]
This leads to the Fourier series of \(x\),
\[\begin{align} b_n \sinh \alpha_n b &= \frac{2}{a} \int_0^a x \sin \frac{n\pi x}{a}dx \nonumber\\ &= \frac{2 a }{n\pi}(-1)^{n+1}.\end{align} \nonumber \]
So, in short, we have
\[V(x,y) = \frac{2a}{\pi} \sum_{n=1}^\infty (-1)^{n+1} \frac{\sin \frac{n\pi x}{a}\sinh \frac{n\pi y}{a}}{n \sinh \frac{n\pi b}{a}}. \nonumber \]
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?
- Answer
-
The symmetry is broken by the boundary conditions.