Solving first order quasilinear PDEs I'm having a lot of trouble with first order quasilinear PDE problems where one has to show that there are no solutions or show that there are infinitely many. I understand how to employ the method of characteristics to find a solution when it exists, but am not able to do much other than that. 
Two specific examples are the PDE 
$$\partial_{x_1}u+3{x_1}^2\partial_{x_2}u=1$$
with the two different initial conditions $u(z,z^3)=1$ and $u(z,z^3)=z-1$ respectively.
Supposedly there are no solutions ($u\in C^1(\mathbb{R}^2)$) in the first case and infinitely many in the second case. I don't know how one is supposed to prove such things because we don't have any examples or solutions. 
My attempt in the first case was: characteristics are found to be $x(z,t)=(t+z,(t+z)^3)$, on which $u=t+1$. But these lie on the initial curve and so $u=1$, contradiction. I'm not sure if this argument is correct.
And I have no idea for the second one. The characteristics are the same but now $u=t+z-1$ on each one, which agrees with $u$ from the initial condition. I'm not sure how this helps establish that there are infinitely many solutions. 
Any help with these kind of problems would be much appreciated.
 A: $$\partial_{x_1}u+3{x_1}^2\partial_{x_2}u=1$$
Characteristic system of ODEs :
$$\frac{dx_1}{1}=\frac{dx_2}{3{x_1}^2}=\frac{du}{1}=dt$$
First characteristic equation from $\frac{dx_1}{1}=\frac{dx_2}{3 x_1^2}$ :
$$x_1^3-x_2=c_1$$
Second characteristic from $\frac{dx_1}{1}=\frac{du}{1}=dt$ :
$$u-x_1=c_2$$
General solution of the PDE on the form of implicit equation $c_1=F(c_1)$ :
$$u-x_1=F(x_1^3-x_2)$$
$F$ is an arbitrary function to be determined according to the boundary condition.
$$\boxed{u(x_1,x_2)=x_1+F(x_1^3-x_2)}$$
FIRST CASE, CONDIION $x_1=z\:;\: x_2=z^3 \:;\: u=1$
$$1=z+F(z^3-z^3)=z+F(0)\quad\implies\quad 1-z=F(0)=\text{constant}$$
This is impossible for all different $z$ with the same $F(0)$. Thus there is no solution in this case.
SECOND CASE, CONDITION $x_1=z\:;\: x_2=z^3 \:;\: u=z-1$
$$z-1=z+F(z^3-z^3)=z+F(0)\quad\implies\quad F(0)=-1$$
They are an infinity of functions $F$ such as $F(0)=-1$. Thus in this case they are an infinity of solutions.
NOTE: In both above cases the condition is specified on a characteristic curve, which is a special case. That is why they are either an infinity of solutions or no solution depending on whether the specified condition is consistent or not with the characteristic equation involved. Generally the condition is specified on a curve which doesn't coincides with a characteristic curve and then the function $F$ can be determined.
For example, if the condition is $u(z,z^3-z)=z+z^2\quad$ thus $x_1=z\:;\: x_2=z^3-z \:;\: u=z+z^2$
$z+z^2=z+F(z^3-(z^3-z))=z+F(z)\quad\implies\quad F(z)=z^2\quad$ that we put into the above general solution where $z=x_1^3-x^2$ :
$$u(x_1,x_2)=x_1+(x_1^3-x_2)^2$$
This is the unique solution if the condition is $u(z,z^3-z)=z+z^2$.
