Is this PDE $xu_x-yu_y=u$ Well posed? We have the given problem
$$xu_x-yu_y=u,\ x>0, \ y>0 \  (1)$$
$$u(x,x)=x^2, \ x>0, \ (2)$$
They ask to check if the problem is well-posed and solve it next.
I know that a problem is well-posed if:

*

*It has a solution

*The solution is unique

*A small change to PDE/side conditions produce only small changes in the solution.

Also, we can observe that the PDE is linear, first order and homogeneous if we write it in the form
$$xu_x-yu_y-u=0,\ x>0, \ y>0 \  (1)$$
$$u(x,x)=x^2, \ x>0, \ (2)$$
With characteristic curves
$$\frac{dx}{dt}=x\Leftrightarrow \frac{1}{x} \frac{dx}{dt}=1$$
$$\frac{dy}{dt}=-y \Leftrightarrow -\frac{1}{y} \frac{dy}{dt}=1 $$ and
$$\frac{du}{dt}=u \Leftrightarrow \frac{1}{u} \frac{du}{dt}=1$$
From which we obtain :$\begin{cases}
yx=c_1 \\
\frac{u}{x}=c_2 \\
\end{cases}$
Combine the forms:
$$c_2=G(c_1) \Leftrightarrow \frac{u}{x}=G(xy)\Leftrightarrow u=xG(yx)$$
Now, using the initial condition $u(x,x)=x^2$ gives
$$x^2=xG(x*x) \Leftrightarrow G(x^2)=x $$
So, how could I check if it is well-posed before solving it? And if yes!!! How? Any suggestion?
 A: The question itself is ill-posed. This is not a general boundary value problem or initial value problem. It is a specific request which you can explicitly solve. (You already solved it, actually, as noted by Kurt in comments).
A correct well-posedness question would be something like:
consider the PDE $x\partial_x u-y\partial_y u=u$ with the "boundary condition" $u(x, x)=\phi(x)$, where $\phi$ belongs to the space $X$. (This $X$ must be a topological space of functions. Typically it will be a Hilbert or Banach space). Is this problem well-posed in this space?

If you don't specify the topology, how can you say whether the solution depends continuously on the data?

For the given problem, however, there is something that remains to be done even without specifying topologies. Is the solution unique? You found one, but how can you be sure it is the only one? Most probably, this is what you are supposed to do (here I am assuming that the problem comes from an assignment).
A: To check for the existence of a unique solution to the above problem, we will need  the tangent vector at each point $(x,y)$ of the characteristic projections which is
$$\tau=(\frac{dx}{dt},\frac{dy}{dt})=(x,-y)$$
The data curve is
$$C=\{(x,y,u) \in (0,+\infty)\times(0,+\infty)\times(0,+\infty):x=t,y=t,u=t^2,t \in \mathbb{R}\}$$
The perpendicular vector in C is
$$n=(\frac{dy}{dt},-\frac{dx}{dt})=(1,-1)$$
and $$n \cdot t=x+y$$ and if $n \cdot t\neq0$, the problem is well posed.
Is my consideration right?
