2024-02-01 Laplace's Equation Fourier Series

We look at Laplace’s equation and solving it with separation of variables, example in Figure 2.9, and math behind Fourier series and orthogonality of harmonic functions.

General solution of Laplace’s equation in Cartesian coordinates

$${\vec{\nabla }}^{2}V=0$$ $$\frac{{\partial }^{2}V(x,y,z)}{\partial x^2}+\frac{{\partial }^{2}V(x,y,z)}{\partial y^2}+\frac{{\partial }^{2}V(x,y,z)}{\partial z^2}=0$$

Trial Solution given by

$$V(x,y,z)=X(x)Y(y)Z(z)$$

Put this in the differential Equation

$$YZ\frac{{\partial }^{2}X}{\partial x^2}+XZ\frac{{\partial }^{2}Y}{\partial y^2}+XY\frac{{\partial }^{2}Z}{\partial z^2}=0$$

Divide both sides of equation by $XYZ$

$$\frac{1}{X}\frac{{\partial }^{2}X}{\partial x^2}+\frac{1}{Y}\frac{{\partial }^{2}Y}{\partial y^2}+\frac{1}{Z}\frac{{\partial }^{2}Z}{\partial z^2}=0$$

For the sum of these terms to be 0, each term has to be constant. We assign $\alpha ,\beta$ and $\gamma$ to the terms

$${\alpha }^2+{\beta }^2-{\gamma }^2=0$$ $$\frac{1}{X}\frac{{\partial }^{2}X}{\partial x^2}={\alpha }^2$$ $$\frac{1}{Y}\frac{{\partial }^{2}Y}{\partial Y^2}={\beta }^2$$ $$\frac{1}{Z}\frac{{\partial }^{2}Z}{\partial z^2}={\gamma }^2$$

Now we get 3 separate differential equations We can use an ansatz of

$$x=e^{-i\alpha x}$$

to solve these equations for x,y and z respectively, with differing appropriate exponential components.

$$\frac{{\partial }^{2}X}{\partial x^2}=(\pm i\alpha )(\pm i\alpha )e^{\pm i\alpha x}$$ $$=-{\alpha }^2{e}^{\pm i\alpha x}$$

And generalizing this to the other 2 coordinates,

$$V(x,y,z)=e^{\pm i\alpha x}{e}^{\pm i\beta y}{e}^{\pm i\gamma z}$$

Example with Figure 2.9 from the textbook

• See powerpoint for full description
• Uses a double fourier integral over the top of the box

Orthogonality of harmonic functions

$$\int \sin ax\sin bxdx\{\begin{array}{l}\frac{\sin (a-b)x}{2(a-b)}-\frac{\sin (a+b)x}{2(a+b)}\end{array}\text{ if a}\ne \text{b}$$

Solution of integral

$$\frac{1}{2}x-\frac{1}{4a}\sin (2ax)$$

Looking at finite case

$$\underset{0}{\overset{L}{\int }}\sin \frac{n\pi x}{L}\sin \frac{m\pi x}{L}dx=\frac{\frac{\sin \left(n-\frac{m\pi x}{}\right)}{L}}{2\frac{n-L\pi }{L}}-\frac{\sin \frac{((n-m)\pi x)}{L}}{2\frac{n+m\pi }{L}}{|}_0^L$$

Evaluated:

$$\frac{\sin (n-m\pi )}{2\frac{(n-m)\pi }{L}}-\frac{\frac{\sin (n+m)\pi }{2(n-m)\pi }}{L}-\left[\frac{\sin 0}{2(n-m)\pi /L}-\frac{\sin 0}{2(n-m\pi \pi )\frac{\pi }{L}}\right]=0$$

The whole integral is 0 if n & m are both 0 or if they are unequal

$$\underset{0}{\overset{L}{\int }}\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)d=\frac{1}{2x}-\frac{1}{\frac{4n\pi }{L}}\sin \left(\frac{2n\pi }{L}\right)x{|}_0^L$$ $$=\frac{1}{2}L-\frac{1}{\frac{4n\pi }{L}}\sin n2\pi -\left[0-\frac{1}{\frac{4n\pi }{L}}\sin 0\right]$$ $$\underset{0}{\overset{L}{\int }}\sin \left(\frac{n\pi x}{L}\sin \frac{m\pi x}{L}\right)d=\frac{1}{2L}{\delta }_{n,m}$$

The ${\delta }_{n,m}$ is known as theKronecker delta. Makes life easier.

$${\delta }_{nm}=\{\begin{array}{ll}0& n\ne m\\ 1& n=m\end{array}$$

Fourier Series

Any function $F(x)$ can be represented as an infinite series of sin or cosine functions

$$F(x)=\sum _{n=1}^{\infty }{a}_n\sin \left(\frac{n\pi x}{L}\right)|\sin \left(\frac{m\pi x}{L}\right)$$ $$F(x)\sin \left(\frac{m\pi x}{L}\right)=\sum _{n=1}^{\infty }{a}_n\sin \frac{n\pi x}{L}\sin \frac{m\pi x}{L}$$

Integrate both sides from 0 → L

$$\underset{0}{\overset{L}{\int }}F(x)\sin \left(\frac{m\pi x}{L}\right)d=\underset{0}{\overset{L}{\int }}\sum _{n=1}^{\infty }{a}_n\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx$$ $$=\sum _{n=1}^{\infty }{a}_n\underset{0}{\overset{L}{\int }}\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx$$ $$=\sum _{n=1}^{\infty }{a}_n\frac{1}{2}L{\delta }_{nm}$$

Use Kronecker delta

The integral from 0 → L over $F(x)\sin (\frac{m\pi x}{L})$ is equal to

$$a_m\frac{1}{2}{\delta }_{n,m}$$

Know that the Kronecker delta can be used to represent a product of two sines like seen above.

$$a_m\frac{1}{2}L⟹a_m=\frac{2}{L}\underset{0}{\overset{L}{\int }}F(x)\sin \left(\frac{m\pi x}{L}\right)dx$$

Use boundary conditions to solve out Next time: double Fourier series

In class problem 3.15 A rectangular pipe, running parallel to the z-axis (from −∞ to +∞), has three grounded metal sides, at y = 0, y = a, and x = 0. The fourth side, at x = b, is maintained at a specified potential $V_0(y)$

(a) Develop a general formula for the potential inside the pipe.

$$\frac{{\partial }^{2}V}{\partial x^2}+\frac{{\partial }^{2}V}{\partial y^2}+\frac{{\partial }^{2}V}{\partial z^2}=0$$

Boundary Conditions

$$\{\begin{array}{l}V=0\text{ when }y=0\\ V=0\text{ when }y=a\\ V=0\text{ when }x=0\\ V=V_0(y)\text{ when }x=b\end{array}$$

Should be constant along z axis, because the $x=b$ plane runs infinitely long along the z axis

$$V(x,y,z)=X(x)Y(y)Z(z)$$ $$\frac{1}{X}\frac{{\partial }^{2}X}{\partial x^2}+\frac{1}{Y}\frac{{\partial }^{2}Y}{\partial y^2}=0$$ $$C_1+C_2=0$$ $$\frac{1}{X}\frac{{\partial }^{2}X}{\partial x^2}=C_1$$ $$\frac{1}{Y}\frac{{\partial }^{2}Y}{\partial Y^2}=C_2$$ $$X(x)=A{e}^{kx}+B{e}^{-kx}$$ $$Y(y)=C\sin ky+D\cos ky$$ $$V(x,y)=(A{e}^{kx}+B{e}^{-kx})(C\sin ky+D\cos ky)$$

Condition 1 $⟹$ $D=0$ Condition 3 $⟹$ (b) Find the potential explicitly, for the case $V_0(y)=V_0$