Method of characteristics $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + 2\frac{\partial^2 u}{\partial x \partial y}=0$

Question

I know how to solve problems with form like this (via method of characteristics): $$a(x,y) u_{x}+b(x,y)u_y=c(x,y).$$ But I got this problem: $$ \dfrac{\partial^2 u}{\partial x^2} + \dfrac{\partial^2 u}{\partial y^2} + 2\dfrac{\partial^2 u}{\partial x \partial y}=0$$ with conditions: $$ u(0,y)=0, u(x,1)=x^2. $$

My idea was to factorize the equation and I get this: $$ (\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y})(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y})u=0. $$

Probably, I can get characteristic if I solve: $$(\dfrac{\partial u}{\partial x}+\dfrac{\partial u}{\partial y})=0.$$

I get that characteristic is: $y=x+c.$

But what now? How do I continue or is there a better way to solve this equation via method of characteristics?

Thank you for any help.

Answer 1

From

$$ \left(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}\right)\left(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}\right)u=0. $$

making

$$ U = \left(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}\right)u $$

we have

$$ \left(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}\right)U=0 $$

gives

$$ U = \phi(y-x) $$

and then solving

$$ \left(\dfrac{\partial }{\partial x}+\dfrac{\partial }{\partial y}\right)u=\phi(y-x) $$

we have

$$ u(x,y) = x\phi(y-x)+\psi(y-x) $$

with the boundary conditions we have

$$ u(0,y) = 0\phi(y-0)+\psi(y-0)= \psi(y) = 0\\ u(x,1) = x\phi(1-x) = x^2 $$

or

$$ \phi(\eta) = 1-\eta $$

and finally

$$ u(x,y) = x(1 - y + x) $$