I have to check solvability of this Cauchy problem:
Defining $D=\left\{(x,y)\mid y>x^{2},\, x\in\mathbb{R}\right\}$ and $a=a(x,y)$ continuous in $\overline{D}$ the problem to check is
\begin{align*} \begin{cases} a(x,y)\partial_{x}u-\partial_{y}u=-u &(x,y)\in D\\ u(x,x^{2})=g(x) &x\in\mathbb{R} \end{cases} \end{align*}
Clearly $\partial D$ is the initial curve projection and it fixes $u$ values over this curve. However, if characteristics projection intersects more than one time $\partial D$ then characteristics equation may not have solution since you are prescribing the curve in to or more points and you only need one point obtain a characteristic.
Therefore, for problem be solvable you must prove that
Tangent vector to $(x,x^{2},g(x))$ is not a scalar multiple of $(a,-1,u)$ and tangent to $(x,x^{2})$ is not a scalar multiple of $(a,-1)$.
$a(x,y)$ must be in such way that if characteristics projection intersect $\partial D$ in one point, there is no other point.
And for proving that you must give some conditions over $a(x,y)$ and $g(x)$.
You can integrate two of three characteristics ODE system \begin{align*} \begin{cases} \dot{x}=a(x,y)\\ \dot{y}=-1\\ \dot{u}=-u \end{cases} \end{align*} and therefore $y(t)=-t+c_{2}$ and $\frac{d}{dt}\ln u=-1$, then $u(t)=c_{3}e^{-t}$. Since $y(t)$ is an inyective and strictly decreasing function of $t$ for every $t\in\mathbb{R}$ we can take $y$ as the new parameter for describing characteristics, and then $t=c_{2}-y$ and $\frac{dx}{dy}=\dot{x}\vert_{t=t(y)}\frac{dt}{dy}=-a(x(y),y)$ and $u(y)=c_{3}e^{y-c_{2}}$.
Clearly over initial curve, that is $y=x^{2}$ we have $u(x^{2})=c_{3}e^{x^{2}-c_{2}}=g(x)$, and since exponential is always positive then for we have some hope for solvability we need $g(x)>0$ or $g(x)<0$ $\forall x\in\mathbb{R}$.
My question is: how, and if you can give me a hint to do this, $a(x,y)$ must be for there are only one intersection point and satisfy transversality condition? I don't find solution for this problem and is cracking me down... and I'm sure this is not so complicated...
Since I posted the question I have done some in deep thinking and I have realised that in order for my own comment on my question be realised we need to that $a(x,y)\leq 0$ over the parabola with $x<0$, for if it where positive and since $a$ is continuous in some open containing a point of the parabola then it will be positive in that open set then it must be a characteristic that crosses at least two times the parabola in that open set. The same reasoning says to me that for $x>0$ over the parabola $a(x,y)\geq 0$. And since $a$ is continuous in $\overline{D}$ then for $x=0$ over the parabola $a(x,y)=0$. That is, over parabola $a$ must be a non-positive increasing function for $x<0$ and a non-negative function for $x>0$.
Another thing is that it's not be possible that in $D$ characteristics slope change of sign and therefore them to cross from one side of the parabola, since $\dot{y}=-1\neq 0$. But we could think that if $a$ is always positive, or negative on $D$ it could cross the parabola on the other side, but this is not possible since $a$ is continuous on $\overline{D}$ it must be $a(x,y)\leq 0$ for $x\leq 0$ and $a(x,y)\geq 0$ for $x\geq 0$.
I think these are the conditions over $a(x,y)$, what do you think?