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5.4 Laplace’s equation

insulated


Figure 5.1: A conducting sheet insulated from above and below.

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

{1\over k} {∂u\over ∂t} = {{∂}^{2}u\over ∂{x}^{2}} + {{∂}^{2}u\over ∂{y}^{2}} .
(5.29)

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

{{∂}^{2}u\over ∂{x}^{2}} + {{∂}^{2}u\over ∂{y}^{2}} = 0,
(5.30)

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

\begin{eqnarray} u(x,0)& =& 0, %& \\ u(a,y)& =& 0, %& \\ u(x,b)& =& x,%& \\ u(0,y)& =& 0. %&(5.31) \\ \end{eqnarray}

(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),
(5.32)

and define

{X''\over X} = −{Y ''\over Y } = −λ.
(5.33)

Or explicitly

X'' = −λX,\kern 2.77695pt \kern 2.77695pt Y '' = λY.
(5.34)

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

Once again distinguish three cases:
λ > 0 X(x) =\mathop{ sin}\nolimits {α}_{n}(x), {α}_{n} = {nπ\over a} , {λ}_{n} = {α}_{n}^{2}. We find

\begin{eqnarray} Y (y)& =& {C}_{n}\mathop{ sinh}\nolimits {α}_{n}y + {D}_{n}\mathop{ cosh}\nolimits {α}_{n}y %& \\ & =& {C'}_{n}\mathop{ exp}\nolimits ({α}_{n}y) + {D'}_{n}\mathop{ exp}\nolimits (−{α}_{n}y).%&(5.35) \\ \end{eqnarray}

Since Y (0) = 0 we find {D}_{n} = 0 (\mathop{sinh}\nolimits (0) = 0,\mathop{cosh}\nolimits (0) = 1).
λ ≤ 0 No solutions

So we have

u(x,y) ={ \mathop{∑ }}_{n=1}^{∞}{b}_{ n}\mathop{ sin}\nolimits {α}_{n}x\mathop{sinh}\nolimits {α}_{n}y
(5.36)

The one remaining boundary condition gives

u(x,b) = x ={ \mathop{∑ }}_{n=1}^{∞}{b}_{ n}\mathop{ sin}\nolimits {α}_{n}x\mathop{sinh}\nolimits {α}_{n}b.
(5.37)

This leads to the Fourier series of x,

\begin{eqnarray}{ b}_{n}\mathop{ sinh}\nolimits {α}_{n}b& =& {2\over a}{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{0}^{a}x\mathop{sin}\nolimits {nπx\over a} dx%& \\ & =& {2a\over nπ}{(−1)}^{n+1}. %&(5.38) \\ \end{eqnarray}

So, in short, we have

V (x,y) = {2a\over π} {\mathop{∑ }}_{n=1}^{∞}{(−1)}^{n+1}{\mathop{sin}\nolimits {nπx\over a} \mathop{ sinh}\nolimits {nπy\over a} \over n\mathop{sinh}\nolimits {nπb\over a} } .
(5.39)

Question: 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.

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