# Solvability of a first-order linear PDE Cauchy problem.

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

1. 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 $(a,-1)$.

2. $a(x,y)$ 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?

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I was thinking about some conditions on $a(x,y)$ and I realised that not only you have the problem of characteristic curves proyection intersect $\partial D$ more than one time, you also has the problem that when characteristic projection touches, the boundary it makes it with the same slope as the boundary has, that is the characteristic curve to be tangent to $\partial D$. Then $a(x,y)$ on $\partial D$, in order to characteristic projection doesn't be tangent must not be equal to the slope of $\partial D$ in that point. – elessartelkontar Sep 21 '12 at 16:15