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Separable Solutions

Seek separable solutions of the form

\begin{displaymath}
F = X(x) Y(y).
\end{displaymath}

Substituting into Laplace's equation we obtain

\begin{displaymath}
{d^{2}X\over dx^{2}} Y + X {d^{2}Y\over dy^{2}} = 0.
\end{displaymath}

On dividing by $XY$ and rearranging into functions of $x$ and functions of $y$ alone, this gives
\begin{displaymath}
{1\over X}{d^{2}X\over dx^{2}} = - {1\over Y}{d^{2}Y\over dy^{2}} = K.
\end{displaymath} (5.14)

The form of the solution depends on the sign of the separation constant $K$.
  1. $K = + p^{2}$. The solution is

    \begin{displaymath}
F = \left (a_{1}e^{px} + a_{2}e^{-px}\right )\left ( b_{1}\sin py +
b_{2}\cos py \right ).
\end{displaymath}

    In general, we need to sum over all values of $p$ to obtain the most general solution.

  2. $K = - p^{2}$. In this case the solution is

    \begin{displaymath}
F = \left (a_{1}\sin px + a_{2}\cos px \right )\left (b_{1}e^{py} +
b_{2}e^{-py}\right ).
\end{displaymath}

  3. $K = 0$. The solution is somewhat simpler in this case, with linear solutions for both $X$ and $Y$.

    \begin{displaymath}
F = \left (a_{1}x + a_{2}\right )\left (b_{1}y + b_{2}\right ).
\end{displaymath}

Example 5. .1Consider an irrotational, incompressible flow in a channel of width $a$. The situation is shown in Figure 5.1

Figure 5.1: An irrotational, incompressible flow in a channel of width $a$.
\includegraphics [scale=0.7]{dummy.ps}

The velocity components are

\begin{displaymath}
v_{x} = -{\partial F\over \partial x}, \qquad v_{y}=-{\partial
F\over \partial y}.
\end{displaymath}

The boundary conditions are that there is no normal flow through the channel walls. Hence,

\begin{displaymath}
v_{y} = 0, \hbox{ on } y=0, y=a.
\end{displaymath}


\begin{displaymath}
v_{x} = h(y), \hbox{ on } x=0, \qquad v_{x}\rightarrow 0 \hbox{
as } x\rightarrow \infty,
\end{displaymath}

where $h(y)$ is a specified function of $y$. The boundary conditions in the $y$-direction imply that the solution should be oscillatory in $y$ while in the $x$-direction they imply that the variation is exponential in $x$. Therefore, we try a solution of the form, $K = p^{2}$ is taken as positive,

\begin{displaymath}
F = A e^{-px}\cos py,
\end{displaymath}


\begin{displaymath}
v_{y} = Ap e^{-px}\sin py.
\end{displaymath}

This satisfies the boundary condition at $y=0$ and the condition for large positive $x$. Hence,

\begin{displaymath}
v_{y}(x,a) = 0, \quad \Rightarrow \quad \sin pa = 0,\quad
\Rightarrow \quad p = {n\pi\over a}.
\end{displaymath}

Thus, the general solution is
\begin{displaymath}
F = \sum_{n}^{}A_{n}e^{-n\pi x/a}\cos {n\pi y\over a},
\end{displaymath} (5.15)

and

\begin{displaymath}
v_{x} = \sum_{n}^{}{n\pi \over a}A_{n}e^{-n\pi x/a}\cos {n\pi y\over a}.
\end{displaymath}

The constants in the summation are given from the condition on $v_{x}$ at $x=0$ by inverting the Fourier series as
\begin{displaymath}
{n\pi\over a}A_{n} = {2\over a}\int_{0}^{a} h(y)\cos {n\pi y\over a} dy, \quad
A_{0} = 0.
\end{displaymath} (5.16)


next up previous
Next: Analytical Functions of a Up: Cartesian Coordinates, Previous: Cartesian Coordinates,
Prof. Alan Hood
2000-11-06