While solving a Quasi-Linear PDE, I found that the general solution is given by $$u(x,y)=x+g(x^3-y^3)$$, where $g$ is an arbitrary function. Now, how can we determine the nature of solution given a set of Cauchy data? Say, we are given: i) $u=x$ on $y=4x$, ii) $u=-2y$ on $x^3-y^3=2$, iii) $u=y^2$ on $y=-x$. I wish to determine whether there exist a unique or infinite solutions in each case.
For i) since it comes to $g(-63x^3)=0$, I think the problem has infinitely many solutions. For ii) and iii), since it comes to $g(2)=-x-2y$ and $g(0)=y^2-x$ respectively, I think there exists no solutions in these cases. Am I right? Any hints? Thanks beforehand.
I believe you are correct only in case ii). In case i), the condition $g(-63x^3)=0$ for all $x\in\Bbb{R}$ implies $g(z)=0$ for all $z\in\Bbb{R}$, hence the solution $u(x,y)=x$ is unique. Finally, in case iii) your solution would be correct if $u=y^2$ on $y=x$, but not if $u=y^2$ on $y=-x$, as stated in the question. In the latter case, we have $g(2x^3)=y^2-x=x^2-x$, so that $g(z)=\left(\frac{z}{2}\right)^{2/3}-\left(\frac{z}{2}\right)^{1/3}$ and $$ u(x,y)=x+\left(\frac{x^3-y^3}{2}\right)^{2/3}-\left(\frac{x^3-y^3}{2}\right)^{1/3}. $$