$x^2\dfrac{\partial u}{\partial x}+y^2\dfrac{\partial u}{\partial y}=u^2$

Question

Help me please to solve the following PDE equation:

$x^2\dfrac{\partial u}{\partial x}+y^2\dfrac{\partial u}{\partial y}=u^2,\; \: u(x,2x)=1$

Thanks a lot!

Answer 1

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

$\dfrac{dx}{dt}=x^2$ , letting $x(1)=-1$ , we have $-\dfrac{1}{x}=t$

$\dfrac{dy}{dt}=y^2$ , we have $-\dfrac{1}{y}=t+y_0=-\dfrac{1}{x}+y_0$

$\dfrac{du}{dt}=u^2$ , we have $\dfrac{1}{u}=-t+f(y_0)=\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$ , i.e. $u(x,y)=\dfrac{1}{\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)}$

$u(x,2x)=1$ :

$\dfrac{1}{\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{2x}\right)}=1$

$\dfrac{1}{x}+f\left(\dfrac{1}{2x}\right)=1$

$f\left(\dfrac{1}{2x}\right)=1-\dfrac{1}{x}$

$f(x)=1-\dfrac{1}{\dfrac{1}{2x}}=1-2x$

$\therefore u(x,y)=\dfrac{1}{\dfrac{1}{x}+1-2\left(\dfrac{1}{x}-\dfrac{1}{y}\right)}=\dfrac{1}{1+\dfrac{2}{y}-\dfrac{1}{x}}=\dfrac{xy}{xy+2x-y}$