We have to find the function $u(x,y)$ for the following system:
- $u_xu_y = xy$
- $u(x,y) = y+1$ for $x=y$
Using the method of characterstics I get:
$F(x,y,u,u_x,u_y) = u_xu_y-xy = 0$
Defining $p = u_x$ and $q = u_y$ we get:
$F(x,y,u,p,q) = pq-xy = 0$
I use a parametrisation for $s = 0$:
- $x(t) = y(t)$
- $y(t) = t$
- $u(t) = t+1$
We are supposed to find p and q out of the following system:
- $F(x,y,u,p,q) = 0$
- $u_t = px_t+qy_t$
By filling in what we know we get:
- $pq-t^2 = 0$
- $p+q = 1$
I am not quite sure how to find p and q out of this system.
The characteristic differential equations are:
- $x_s = F_p = q$, for $s=0 : x=t$
- $y_s = F_q = p$, for $s=0 : y=t$
- $u_s = pF_p+qF_q = 2pq$, for $s=0 : u=t+1$
- $p_s=-F_x-pF_u = y$, for $s=0 : p= ?$
- $q_s = -F_y-qF_u = x$, for $s=0 : q= ?$
Help would be much appreciated.