Problem on PDE existence and uniqueness properties Consider the partial differential equation $$u_x+2xu_y=0$$
Describe the existence and  uniqueness properties for each auxiliary condition below . Include solutions possible or explain why none exists.
$(a). u(x,x^2)=2\\
(b). u(x,x^2)=e^{-x}\\
(c). u(0,y)=e^{-y}$
My attempt: Given PDE $u_x+2xu_y=0$
then $$\frac{dx}{1}=\frac{dy}{2x}=\frac{du}{0}$$
then $$2x-y-c=1, u=c_2$$
From option (a) $u(x,x^2)=2$ then $$x=t,y=t^2, u=2$$ also $$2t-t^2=c_1,2=c_2$$
Since we are not eliminate  $c_1, c_2$ pde have no slotuion
is my approach is right?
 A: $$\frac{dx}{1}=\frac{dy}{2x}=\frac{du}{0}$$
First characteristics from $\quad\frac{dx}{1}=\frac{dy}{2x}\quad\implies\quad y-x^2=c_1$.
Second characteristics from $\quad\frac{du}{0}=$finite $\quad\implies\quad u=c_2$.
General solution :
$$u(x,y)=F(y-x^2)$$
with any differentiable function $F$.
Case $(a)$ :
$u(x,x^2)=2=F(x^2-x^2)=F(0)\quad$ 
All functions $F(X)$ with condition $F(0)=2$ are convenient. There is an infinity of solutions. For example :
$u(x,y)=2$
$u(x,y)=2+(y-x^2)$
$u(x,y)=2+\sin(y-x^2)$
and so on...
Case $(b)$ :
$u(x,x^2)=e^{-x}=F(x^2-x^2)=F(0)$
It is impossible that $e^{-x}=$constant. There is no solution.
Case $(c)$ :
$u(0,y)=e^{-y}=F(y-0)=F(y)$
The function is determined : $\quad F(X)=e^{-X}\quad$ Puting it into the general solution where $X=y-x^2$ leads to a unique solution :
$u(x,y)=e^{-(y-x^2)}$
A: Caveat: this is a longer answer, but I find it easier to understand just what's going on here.
Using the method of characteristics, you examine the function $u$ along a trajectory $(x(t),y(t))$ through some point $(x_0,y_0)$ (usually in a boundary value problem you choose a convenient pair for the initial conditions, i.e. where you have boundary data) converting the partial differential equation into an ordinary differential equation. Namely, you want 
$$
x'(t)=1\\
y'(t)=2x(t)
$$
since then $\frac{d}{dt}u(x(t),y(t))=u_x+2x(t)u_y=0$ for $u$ which solve the pde. So along trajectories of the above sort, as $u(x(t),y(t))=u(x_0,y_0)$.  
Solving the coupled system, we find 
$$
x(t)=t+x_0\\
y(t)=t^2+2x_0t+y_0
$$
and eliminating the parameter $t$, we have that $u$ is constant as long as 
$$
y=(x-x_0)^2+2x_0(x-x_0)+y_0=x^2-x_0^2+y_0
$$
We know that 
$$
u(x,y)=u(x,x^2+y_0-x_0^2)=u(x_0,y_0)
$$
From here, it is obvious that the first condition is easy to satisfy. We know that the desired function is constant along the parabola $y=x^2$, this condition just fixes such a constant to be $2$.
For the second, it is clearly impossible, since we know that $u$ is constant on the parabola $y=x^2$, but $e^{-x}$ is not constant. 
For the third, we have data on the line $x=0$, so it is convenient to set $x_0=0$. Then your condition yields
$$
u(x,y)=u(0,y_0)=e^{-y_0}
$$
but our choice of characteristic is valid for any $y_0$, so what is $y_0$ when $x_0$ is zero in terms of $x$ and $y$? It is $y-x^2$. Giving us the answer
$$
u(x,y)=e^{-(y-x^2)}
$$
