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Concluded discussion of wave equation

TODAY

Wish to move on to the potential equation (often called Laplace equation). In two dimension this equation can be written
\begin{displaymath}\frac{\partial^2u(x,y)}{\partial x^2}+\frac{\partial^2u(x,y)}{\partial y^2}=0\end{displaymath}
while in three dimensions we write
\begin{displaymath}\frac{\partial^2u(x,y,z)}{\partial x^2}+\frac{\partial^2u(x,y,z)}{\partial y^2}+ \frac{\partial^2u(x,y,z)}{\partial z^2}=0\end{displaymath}

Either equation may be written
\begin{displaymath}\nabla^2u=0\end{displaymath}

and we will have to learn how to express the Laplacian

 

\begin{displaymath}\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+ \frac{\partial^2}{\partial z^2}\end{displaymath}

in different ways such as in polar, cylindrical or spherical coordinates.

ORIGIN OF LAPLACE EQUATION
Often we need to be able to solve the Laplace equation as a stepping-stone to the solution to a more complicated problem:
E.g. in electrostatics one finds that in a region of space in which there are no charges the electrostatic potential satisfies

\begin{displaymath}\nabla\cdot\vec{E}=0\end{displaymath}

where
\begin{displaymath}\vec{E}=-\nabla V\end{displaymath}

giving
\begin{displaymath}\nabla\cdot\nabla V=\nabla^2 V=0\end{displaymath}

Of course, electrostatics would be uninteresting if there weren't any charges around. However, to solve Poisson's equation

\begin{displaymath}\nabla^2V=\rho\end{displaymath}

one needs to be able to solve the homogenous equation.

FUNCTIONS OF A COMPLEX VARIABLE
The 2-D Laplace equation plays a special role in the theory of complex variables. Suppose V(z) is a differentiable function of a complex variable
z=x+iy

then
\begin{displaymath}\frac{\partial V}{\partial x}=\frac{dV}{dz}\end{displaymath}

\begin{displaymath}\frac{\partial V}{\partial y}=i\frac{dV}{dz}\end{displaymath}

and
\begin{displaymath}\frac{\partial^2 V}{\partial x^2}=\frac{d^2V}{dz^2}\end{displaymath}

\begin{displaymath}\frac{\partial^2 V}{\partial y^2}=-\frac{d^2V}{dz^2}\end{displaymath}
Hence
\begin{displaymath}\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}=0\end{displaymath}


FINITE DIFFERENCES
To get a feeling for what the Laplace equation "does" let us attempt to solve it approximately by the method of finite differences.
Suppose a function is defined on the three points
\begin{displaymath}x-\Delta,x,x+\Delta\end{displaymath}

by the method of Taylor expansion

\begin{displaymath}f(x-\Delta)\approx f(x)-\Delta f^\prime+\frac{\Delta^2}{2}f^{\prime\prime}\end{displaymath}

\begin{displaymath}f(x+\Delta)\approx f(x)+\Delta f^\prime+\frac{\Delta^2}{2}f^{\prime\prime}\end{displaymath}

We can solve these equations to get approximate expressions for the derivatives
\begin{displaymath}f^\prime=\frac{1}{2\Delta}(f(x+\Delta)-f(x-\Delta))\end{displaymath}

\begin{displaymath}f^{\prime\prime}=\frac{1}{\Delta^2}(f(x+\Delta)+ f(x-\Delta)-2f(x))\end{displaymath}

If the function f satisfies the differential equation
\begin{displaymath}f^{\prime\prime}=0\end{displaymath}

we get
\begin{displaymath}f(x)=\frac{1}{2}(f(x+\Delta)+f(x-\Delta))\end{displaymath}

the value of the function in the middle is the average of the values at the ends
Similarly in two dimensions

 

\begin{displaymath}\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}\end{displaymath}

\begin{displaymath}= \frac{1}{\Delta^2}(f(x+\Delta,y)+f(x,y+\Delta)\end{displaymath}

\begin{displaymath}+f(x-\Delta,y)+f(x,y-\Delta) -4f(x))\end{displaymath}

Hence,
we can find an approximate solution to the Laplace equation by averaging over the surrounding points.
This result can be used to solve Laplace equation numerically.

EXAMPLE
Suppose we want to find the solution to the Laplace

\begin{displaymath}\nabla^2f=0\end{displaymath}

equation in an L-shaped region. To specify the problem we assume the function is known at the boundary
Let us crudely discretize the problem

\begin{displaymath}\begin{array}{ccccc} &1&&&\\ 2&3&4&&\\ 5&6&7&8&\\ 9&10&11&12&13\\ &14&15&16&\\ \end{array}\end{displaymath}

Here
1,2,4,5,7,8,9,13,14,15,16

 

are exterior points where the function is known.
3,6,10,11,12

are interior points where we wish to find f.


The approximate solution is then obtained by solving the set of equations
\begin{displaymath}f_3=\frac{1}{4}(f_1+f_2+f_4+f_6)\end{displaymath}

\begin{displaymath}f_6=\frac{1}{4}(f_3+f_5+f_7+f_{10})\end{displaymath}

\begin{displaymath}f_{10}=\frac{1}{4}(f_6+f_9+f_{11}+f_{14})\end{displaymath}

\begin{displaymath}f_{11}=\frac{1}{4}(f_7+f_{15}+f_{10}+f_{12})\end{displaymath}

\begin{displaymath}f_{12}=\frac{1}{4}(f_8+f_{11}+f_{13}+f_{16})\end{displaymath}

Of course, if one wishes to find an accurate solution it is necessary to use a finer mesh.
The generalization to three dimension is

\begin{displaymath}\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial z^2}\end{displaymath}

\begin{displaymath}= \frac{1}{\Delta^2}(f(x+\Delta,y,z)+f(x,y+\Delta,z)+f(x,y,z+\Delta)\end{displaymath}

\begin{displaymath}+f(x-\Delta,y,z)+f(x,y-\Delta,z)+f(x,y,z-\Delta) -6f(x))\end{displaymath}

LAPLACE EQUATION IN DIFFERENT COORDINATE SYSTEMS
When solving boundary value problems in more than one dimension it is often necessary to use other coordinate systems than the cartesian. It is then important to be able to express the Laplacian operator in these coordinate systems.

POLAR COORDINATES

\begin{displaymath}x=r\cos\theta\end{displaymath}

\begin{displaymath}y=r\sin\theta\end{displaymath}

Laplace's equation in this cordinate system can be shown to be:

 

\begin{displaymath}\nabla^2u(r,\theta)=\frac{\partial^2u(r,\theta)}{\partial r^2... ...+\frac{1}{r^2} \frac{\partial^2 u(r,\theta)}{\partial\theta^2}\end{displaymath}

IN CYLINDRICAL COORDINATES

\begin{displaymath}x=\rho\cos\phi\end{displaymath}

\begin{displaymath}y=\rho\sin\phi\end{displaymath}



z=z

Laplace's equation in polar coordinates can be shown to be:
\begin{displaymath}\nabla^2u(\rho,\phi,z)=\frac{\partial^2u(\rho,\phi,z)}{\parti... ...o^2}+\frac{1}{\rho}\frac{\partial u(\rho,\phi,z)}{\partial\rho}\end{displaymath}

\begin{displaymath}+\frac{1}{\rho^2} \frac{\partial^2 u(\rho,\phi,z)}{\partial\phi^2}+\frac{\partial^2 u(\rho,\phi,z)}{\partial z^2}\end{displaymath}



\begin{figure} \epsfysize=350pt \epsffile{spherco.eps} \end{figure}


Laplace's equation in spherical coordinates is:
\begin{displaymath}\nabla^2u(r,\theta,z)=\frac{1}{r^2}[\frac{\partial}{\partial r}(r^2\frac{\partial u(r\theta,\phi)} {\partial r})\end{displaymath}

\begin{displaymath}+\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}(\sin\the... ...sin^2\theta}\frac{\partial^2 u(r,\theta,\phi)}{\partial\phi^2}]\end{displaymath}


Information provided by: http://www.physics.ubc.ca