Domain of definition of first order linear PDE

Question

Consider the first order PDE $$x\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}=u, \quad u(x,0)=f(x).$$

I've tried solving this PDE using the method of characteristics. The first time I did it I found that $$u(x,y)=f\left(\sqrt{x^2+y^2}\right)e^{\arctan(y/x)}.$$

When I did it again (solving the ODEs slightly differently) I found that $$u(x,y)=f\left(\sqrt{x^2+y^2}\right)e^{\arcsin(y/\sqrt{x^2+y^2})}.$$

I know that these two solutions are equivalent. However, the first one would tell me that the domain of definition is the whole of $\mathbb{R}^2$ apart from the line $x=0$ where $y/x$ blows up, but the second solution would tell me that the domain of definition is $\mathbb{R}^2\setminus\{(0,0)\}.$

Which one is correct?

Answer 1

The question is not well-posed, when you solve PDEs you found a family of solution and you fix the "correct" one (like you said) using the boundary or initial condition.

Each of the solutions are "correct" for a specific problem.

Answer 2

Citation :

the first one would tell me that the domain of definition is the whole of $\mathbb{R}^2$ apart from the line $x=0$ where $y/x$ blows up

Note : It is not asked for the domain of definition of the function $u(x,y)=\frac{y}{x}$ which is the whole of $\mathbb{R}^2$ apart from the line $x=0$ where $y/x$ blows up.

It is asked for the domain of definition of the function $u(x,y)=f\left(\sqrt{x^2+y^2}\right)e^{\arctan(y/x)}.$

$\arctan(y/x)$ doesn't blow up. Just like the second form of solution which involves $\arcsin(y/\sqrt{x^2+y^2})$, both have finite value, even on the line $x=0$ , except $(0,0)$.