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[ "article:topic", "Laplace equation", "authorname:nwalet", "license:ccbyncsa", "showtoc:no" ]
Mathematics LibreTexts

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:


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

    \[\begin{align} Y(y) &= C_n\sinh \alpha_n y + D_n \cosh \alpha _n y \nonumber\\[5pt] &= C'_n\exp (\alpha_n y) + D'_n \exp(- \alpha _n y).\end{align}\]

    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?


    The symmetry is broken by the boundary conditions.