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}$


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?

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



Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=\dfrac{1}{\tan(y)}$ , 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 $u(x,y)=f(x_0)e^{-2t}=f(x\cos(y))\cos^2(y)$

$u(x,0)=\dfrac{1}{x}$ :


$\therefore u(x,y)=\dfrac{\cos^2(y)}{x\cos(y)}=\dfrac{\cos(y)}{x}$


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