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Next: Cylindrical Coordinates, Up: Cartesian Coordinates, Previous: Separable Solutions

Analytical Functions of a Complex Variable

Solutions to

\begin{displaymath}
{\partial^{2}F\over \partial x^{2}} + {\partial^{2}F\over \partial
y^{2}} = 0,
\end{displaymath}

can be found by setting

\begin{displaymath}
F = F(z), \qquad z = x + iy.
\end{displaymath}

Here $F(z)$ is an analytic function of the complex variable $z$. Splitting the function $F(z)$ into real and imaginary parts we obtain

\begin{displaymath}
F(z) = u(x,y) + i v(x,y),
\end{displaymath}

and both $u(x,y)$ and $v(x,y)$ satisfy Laplace's equation.

Example 5. .2Consider solutions to Laplace's equation with

\begin{displaymath}
F(x,0) = {1\over 1 + x^{2}}, \qquad F(x,y) \rightarrow 0 \hbox{ as }
x,y \rightarrow \infty.
\end{displaymath}

In this case we may try

\begin{displaymath}
F(z) = {1\over 1 + z^{2}} = {1\over 1 + (x+iy)^{2}}.
\end{displaymath}

Expanding the square and expressing in terms of real and imaginary parts we obtain

\begin{displaymath}
{1\over 1 + x^{2} - y^{2} + i2xy} = {1 + x^{2} - y^{2} - i2xy\over
(1 + x^{2} - y^{2})^{2} + 4 x^{2}y^{2}}.
\end{displaymath}

Thus, we have

\begin{displaymath}
u = {1 + x^{2} - y^{2}\over (1 + x^{2} - y^{2})^{2} + 4x^{2}y^{2}}
\end{displaymath}

and

\begin{displaymath}
v = {-2xy\over (1 + x^{2} - y^{2})^{2} + 4x^{2}y^{2}}.
\end{displaymath}

Using the real part for the potential, $F$, the solution to Laplace's equation is

\begin{displaymath}
F(x,y) = {1 + x^{2} - y^{2}\over (1 + x^{2} - y^{2})^{2} + 4x^{2}y^{2}}.
\end{displaymath}

It is straightforward to show that both Laplace's equation and the boundary conditions are satisfied.
next up previous
Next: Cylindrical Coordinates, Up: Cartesian Coordinates, Previous: Separable Solutions
Prof. Alan Hood
2000-11-06