Separable Solutions

Seek separable solutions of the form

$$F = X(x) Y(y).$$

Substituting into Laplace's equation we obtain

$${d^{2}X\over dx^{2}} Y + X {d^{2}Y\over dy^{2}} = 0.$$

On dividing by $XY$ and rearranging into functions of $x$ and functions of $y$ alone, this gives

$${1\over X}{d^{2}X\over dx^{2}} = - {1\over Y}{d^{2}Y\over dy^{2}} = K.$$ (5.14)

The form of the solution depends on the sign of the separation constant $K$.

1. $K = + p^{2}$. The solution is
   
   $$F = \left (a_{1}e^{px} + a_{2}e^{-px}\right )\left ( b_{1}\sin py + b_{2}\cos py \right ).$$
   
   
   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
   
   $$F = \left (a_{1}\sin px + a_{2}\cos px \right )\left (b_{1}e^{py} + b_{2}e^{-py}\right ).$$
   
   

3. $K = 0$. The solution is somewhat simpler in this case, with linear solutions for both $X$ and $Y$.
   
   $$F = \left (a_{1}x + a_{2}\right )\left (b_{1}y + b_{2}\right ).$$
   
   

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 $a$.

The velocity components are

$$v_{x} = -{\partial F\over \partial x}, \qquad v_{y}=-{\partial F\over \partial y}.$$

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

$$v_{y} = 0, \hbox{ on } y=0, y=a.$$

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

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,

$$F = A e^{-px}\cos py,$$

$$v_{y} = Ap e^{-px}\sin py.$$

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

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

Thus, the general solution is

$$F = \sum_{n}^{}A_{n}e^{-n\pi x/a}\cos {n\pi y\over a},$$ (5.15)

and

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

The constants in the summation are given from the condition on $v_{x}$ at $x=0$ by inverting the Fourier series as

$${n\pi\over a}A_{n} = {2\over a}\int_{0}^{a} h(y)\cos {n\pi y\over a} dy, \quad A_{0} = 0.$$ (5.16)