Week 14

Tutorial 14: Laplace’s Equations

1. Solve the Laplace's equation for a rectangular plate
$$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0<x<a, \quad 0<y<b$$
subject to the given boundary conditions
$$\begin{array}{lll} u(0, y)=0, & u(a, y)=0, & 0<y<b \\ u(x, 0)=0, & u(x, b)=f(x) & 0<x<a \end{array}$$
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Solution
Using $u=X Y$ and $-\lambda^{2}$ as a separation constant leads to
$$\begin{gathered} X^{\prime \prime}+\lambda^{2} X=0 \\ X(0)=0 \\ X(a)=0 \end{gathered}$$
and
$$\begin{gathered} Y^{\prime \prime}-\lambda^{2} Y=0 \\ Y(0)=0 . \end{gathered}$$
Then
$$X=c_{1} \sin \frac{n \pi}{a} x \quad \text { and } \quad Y=c_{2} \sinh \frac{n \pi}{a} y$$
for $n=1,2,3, \ldots$ so that
$$u=\sum_{n=1}^{\infty} A_{n} \sin \frac{n \pi}{a} x \sinh \frac{n \pi}{a} y$$
Imposing
$$u(x, b)=f(x)=\sum_{n=1}^{\infty} A_{n} \sinh \frac{n \pi b}{a} \sin \frac{n \pi}{a} x$$
gives
$$A_{n} \sinh \frac{n \pi b}{a}=\frac{2}{a} \int_{0}^{a} f(x) \sin \frac{n \pi}{a} x d x$$
so that
$$u(x, y)=\sum_{n=1}^{\infty} A_{n} \sin \frac{n \pi}{a} x \sinh \frac{n \pi}{a} y$$
where
$$A_{n}=\frac{2}{a} \operatorname{csch} \frac{n \pi b}{a} \int_{0}^{a} f(x) \sin \frac{n \pi}{a} x d x$$

2. Solve the Laplace's equation for a rectangular plate
$$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0<x<1, \quad 0<y<1$$
subject to the given boundary conditions
$$\begin{array}{llll} u(0, y)=0, & u(1, y)=1-y, & & 0<y<1 \\ \left.\frac{\partial u}{\partial y}\right|_{y=0}=0, & \left.\frac{\partial u}{\partial y}\right|_{y=1}=0, & & 0<x<1 \end{array}$$
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Solution
Using $u=X Y$ and $\lambda^{2}$ as a separation constant leads to
$$\begin{gathered} X^{\prime \prime}+-\lambda^{2} X=0, \\ X(0)=0, \end{gathered}$$
and
$$\begin{gathered} Y^{\prime \prime}+\lambda^{2} Y=0, \\ Y^{\prime}(0)=0 \\ Y^{\prime}(1)=0 . \end{gathered}$$
Then
$$Y=c_{1} \cos n \pi y$$
for $n=0,1,2, \ldots$ and
$$X=c_{2} x \text { or } X=c_{2} \sinh n \pi x$$
for $n=1,2,3, \ldots$ so that
$$u=A_{0} x+\sum_{n=1}^{\infty} A_{n} \sinh n \pi x \cos n \pi y$$
Imposing
$$u(1, y)=1-y=A_{0}+\sum_{n=1}^{\infty} A_{n} \sinh n \pi x \cos n \pi y$$
gives
$$A_{0}=\int_{0}^{1}(1-y) d y$$
and
$$A_{n} \sinh n \pi=2 \int_{0}^{1}(1-y) \cos n \pi y=\frac{2\left[1-(-1)^{n}\right]}{n^{2} \pi^{2} \sinh n \pi}$$
for $n=1,2,3, \ldots$ so that
$$u(x, y)=\frac{1}{2} x+\frac{2}{\pi^{2}} \sum_{n=1}^{\infty} \frac{1-(-1)^{n}}{n^{2} \sinh n \pi} \sinh n \pi x \cos n \pi y$$

3. Solve the Laplace's equation for a rectangular plate
$$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0<x<\pi, \quad 0<y<\pi$$
subject to the given boundary conditions
$$\left.\frac{\partial u}{\partial x}\right|_{x=0}=u(0, y), \quad u(\pi, y)=1, \quad 0<y<\pi\\u(x, 0)=0, \quad u(x, \pi)=0, \quad 0<x<\pi$$
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Solution
Using $u=X Y$ and $\lambda^{2}$ as a separation constant leads to
$$\begin{aligned} &X^{\prime \prime}-\lambda^{2} X=0 \\ &X^{\prime}(0)=X(0) \end{aligned}$$
and
$$\begin{gathered} Y^{\prime \prime}+\lambda^{2} Y=0, \\ Y(0)=0 \\ Y(\pi)=0 \end{gathered}$$
Then
$$Y=c_{1} \sin n y \quad \text { and } \quad X=c_{2}(n \cosh n x+\sinh n x)$$
for $n=1,2,3, \ldots$ so that
$$u=\sum_{n=1}^{\infty} A_{n}(n \cosh n x+\sinh n x) \sin n y$$
Imposing
$$u(\pi, y)=1=\sum_{n=1}^{\infty} A_{n}(n \cosh n \pi+\sinh n \pi) \sin n y$$
gives
$$A_{n}(n \cosh n \pi+\sinh n \pi)=\frac{2}{\pi} \int_{0}^{\pi} \sin n y d y=\frac{2\left[1-(-1)^{n}\right]}{n \pi}$$
for $n=1,2,3, \ldots$ so that
$$u(x, y)=\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{1-(-1)^{n}}{n} \frac{n \cosh n x+\sinh n x}{n \cosh n \pi+\sinh n \pi} \sin n y$$

1. Solve the Laplace's equation for a rectangular plate
$$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0, \quad 0<x<y, \quad 0<y<1$$
subject to the given boundary conditions Answer:
$$\begin{aligned} & u(0, y)=\left.10 y \quad \frac{\partial u}{\partial x}\right|_{x=1}=-1, \quad 0<y<1 \\ & u(x, 0)=0, \quad u(x, 1)=0, \quad 0<x<y \end{aligned}$$
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Solution
This boundary-value problem has the form of Problem 2 in this section, with $a=1$ and $b=1$. Thus, the solution has the form
$$u(x, y)=\sum_{n=1}^{\infty}\left(A_{n} \cosh n \pi x+B_{n} \sinh n \pi x\right) \sin n \pi y$$
The boundary condition $u(0, y)=10 y$ implies
$$10 y=\sum_{n=1}^{\infty} A_{n} \sin n \pi y$$
and
$$A_{n}=\frac{2}{1} \int_{0}^{1} 10 y \sin n \pi y d y=\frac{20}{n \pi}(-1)^{n+1}$$
The boundary condition $u_{x}(1, y)=-1$ implies
$$-1=\sum_{n=1}^{\infty}\left(n \pi A_{n} \sinh n \pi+n \pi B_{n} \cosh n \pi\right) \sin n \pi y$$
and
$$\begin{aligned} n \pi A_{n} \sinh n \pi+n \pi B_{n} \cosh n \pi &=\frac{2}{1} \int_{0}^{1}(-\sin n \pi y) d y \\ A_{n} \sinh n \pi+B_{n} \cos n \pi &=-\frac{2}{n \pi}\left[1-(-1)^{n}\right] \\\\ \end{aligned}\\B_{n} =\frac{2}{n \pi}\left[(-1)^{n}-1\right] \operatorname{sech} n \pi-\frac{20}{n \pi}(-1)^{n+1} \tanh n \pi$$