# PDE using method of characteristic strips

• Sep 21st 2011, 07:44 AM
hatsoff
PDE using method of characteristic strips
Hi guys. I'd like to solve the following:

Quote:

$\displaystyle u=xu_x+yu_y+\frac{1}{2}((u_x)^2+(u_y)^2)$,

$\displaystyle u(x,0)=\frac{1}{2}(1-x^2)$.
This corresponds to the problem $\displaystyle F=xp+yq+\frac{1}{2}(p^2+q^2)-z=0$.

Then the characteristic equations are given by

$\displaystyle \frac{dx}{dt}=x+p$, $\displaystyle \frac{dy}{dt}=y+q$, $\displaystyle \frac{dz}{dt}=p(x+p)+q(y+q)$, $\displaystyle \frac{dp}{dt}=0$, $\displaystyle \frac{dq}{dt}=0$.

And it's easy to find the curves (there are two) induced by the initial condition:

$\displaystyle \Gamma=(x(s,0),y(s,0),z(s,0),p(s,0),q(s,0))=(s,0, \frac{1}{2}(1-s^2),-s,\pm 1)$

Solving for $\displaystyle z$ we have

$\displaystyle z=-xs\pm y+\frac{1}{2}(s^2+1)$

(where the signs $\displaystyle \pm$ correspond to the signs in $\displaystyle \Gamma$).

But I don't know how to put this in closed form (i.e. in the form $\displaystyle z=z(x,y)$). It may just be a simple elementary calculus/algebra block on my part. Any help would be much appreciated!
• Sep 21st 2011, 08:00 AM
hatsoff
Re: PDE using method of characteristic strips
NEVERMIND...

I'm an idiot.... : (