I am familiar with the method of characteristics. However, I have a problem that appears to only have a specified solution at a single point:
Solving for $V(x,y)$, where
$V_xy-V_y(ax+bx^2y) = -cx^2y^2-dx^2-ey^2$, $\quad V(0,0) = 0$.
and where (a,b,c,d,e) are scalar constants.
I'm not sure how to define my "data" and propagate it to execute the method. Any help would be greatly appreciated
The boundary reduced to one point doesn't define a unique solution.
For example in order to illustrate with a simplified example :
For example in the case of $a=d=e=0$ $$yV_x-bx^2yV_y=-cx^2y^2$$
$\frac{dx}{y}=\frac{dy}{-bx^2y}=\frac{dV}{-cx^2y^2}$
First characteristic equation from $\frac{dx}{y}=\frac{dy}{-bx^2y}\quad\implies\quad 3y-bx^3=c_1$
Second characteristic equation from $\frac{dy}{-bx^2y}=\frac{dV}{-cx^2y^2}\quad\implies\quad V-\frac{c}{2b}y^2=c_2$
General solution of the PDE from $c_2=F(c_1)$ : $$V(x,y)=\frac{c}{2b}y^2+F(3y-bx^3)$$ $F$ is an arfitrary function ( to be determined according to the condition ).
CONDITION : $V(0,0)=0=\frac{c}{2b}0^2+F(3*0-b*0^3)=F(0)$
This implies $F(0)=0$.
They are imfinity many functions $F$ such as $F(0)=0$. Thus they are infinity many solutions $V(x,y)=\frac{c}{2b}y^2+F(3y-bx^3)$ .