$$ x\frac{\partial u}{\partial x} +y\frac{\partial u}{\partial y}= 2u \qquad (1)$$
find, in explicit form, the solution of (1) which satisfies the condition: $$ u =x^3 \quad \text{on} \quad y=x+1 \qquad (2)$$ In which region of the $xy$-plane is $u(x, y)$ uniquely determined by the equation (1) with the condition (2)?
My working: So I used the method of characteristics to answer the first part and attained implicitly and then explicitly: $$(se^t, \,(s+1)e^t,\, s^3e^{2t})$$ $$u(x,y) =\frac{x^3}{y-x}$$
My difficulty comes in finding the domain of definition. My understanding is that it is not unique when the initial curve crosses a characteristic. I calculated the Jacobian, because when that is zero, the initial curve should be tangential to a characteristic, i.e. $$ J= e^{2t} = {(y-x)^2},$$ which equals zero when $x=y$. So is the solution unique when $y>x$ or perhaps when $y<x$?
I am not quite sure though, where to go from here? Any help would be much appreciated.
EDIT: correction made to my working. I still do not understand how to find where the solution is unique.
$$ x\frac{\partial u}{\partial x} +y\frac{\partial u}{\partial y}= 2u \qquad (1)$$ $$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{2u}$$ From $\frac{dx}{x}=\frac{dy}{y}$ : $$\frac{y}{x}=c_1$$ From $\frac{dx}{x}=\frac{du}{2u}$ : $$\frac{u}{x^2}=c_2$$ General solution of the PDE : $\quad\frac{u}{x^2}=F\left(\frac{y}{x}\right)$
$$u(x,y)=x^2F\left(\frac{y}{x}\right)$$ $F$ is an arbitrary function, to be determined according to the boundary condition.
CONDITION :
$u(x,x+1)=x^3=x^2F\left(\frac{x+1}{x}\right)$ $$F\left(\frac{x+1}{x}\right)=x$$ Let $X=\frac{x+1}{x}\quad ; \quad x=\frac{1}{X-1}$ $$F(X)=\frac{1}{X-1}$$ The function $F(X)$ is determined.
We put it into the general solution where $X=\frac{y}{x}$, thus $F(X)=\frac{1}{\frac{y}{x}-1}=\frac{x}{y-x}$
The particular solution which satisfies the boundary condition is : $$u(x,y)=\frac{x^3}{y-x}$$