# Use Fourier's method of separation of variables to solve the boundary value problem

Use Fourier's method of separation of variables to solve the boundary value problem comprising the following PDE and BC:

PDE: $x \sin(y) u_x + \cos(y) u_y = -2 \sin(y) u$, $u = u(x,y)$

Boundary Condition: $u(x,0) = \frac{1}{x}$, for $0 \lt x, 0 \lt y \lt \frac{\pi}{2}$

ATTEMPT:

My first instinct was to divide through by $\sin(y)$ and get $x u_x + \frac{u_y}{\tan(y)} = -2u$

Then letting $u(x,y) = f(x)g(y)$ so that $u_x = f'(x)g(y)$ and $u_y = f(x)g'(y)$

Working it out I get: $$-2 = x \frac{f'(x)}{f(x)} + \frac{1}{\tan(y)} \frac{f'(y)}{f(y)}$$

And this is where I get stuck because to find the constant $= \frac{T'}{T} = \frac{X'}{X}$ is difficult with the number two affecting either side.

Am I doing this totally wrong?

• Shouldn't $\frac{f'(y)}{f(y)}$ be $\frac{g'(y)}{g(y)}$? – suresh Feb 24 '14 at 0:52

$x\sin(y)u_x+\cos(y)u_y=-2\sin(y)u$

$xu_x+\dfrac{u_y}{\tan(y)}=-2u$

$\dfrac{dy}{dt}=\dfrac{1}{\tan(y)}$ , letting $y(0)=0$ , we have $\cos(y)=e^{-t}$
$\dfrac{dx}{dt}=x$ , letting $x(0)=x_0$ , we have $x=x_0e^t=x_0\sec(y)$
$\dfrac{du}{dt}=-2u$ , letting $u(0)=f(x_0)$ , we have $u(x,y)=f(x_0)e^{-2t}=f(x\cos(y))\cos^2(y)$
$u(x,0)=\dfrac{1}{x}$ :
$f(x)=\dfrac{1}{x}$
$\therefore u(x,y)=\dfrac{\cos^2(y)}{x\cos(y)}=\dfrac{\cos(y)}{x}$