Home Assignment 2 Problem 2(a)

[MikeMorris]

For this question (eqn 14) I found the characteristic curve to be $\frac{x}{y} = C$. It asks what the difference is between the cases $(x,y)\neq (0,0)$ and that the solution be continuous at (0,0). From what I understand, in the former case we can simply say the solution is any funtion of one variable, $u(x,y) = \phi (\frac{x}{y})$, on the domain $(x,y)\neq (0,0)$. In the second case, we need to pick some value for $u(0,0)$ to make it continuous, and elsewhere it's the same as the other case. Is this a sufficient explanation? It feels lacking to me. Could someone offer assistance?

[Zhiman Tang]

I think you need to find the value of u(0,0), below is my answer, but idk whether it is correct or not.
consider u(x,y) = f(x/y)
in this case, x=0 is not a problem, but y=0 is a problem
we can write u(0,y) = f(0),
to make u a contiunuos solution, we need u(0,0)=f(0) since the value of u can't jump at y=0
so u(0,0)=f(0) is the condition to make it continuos.

[Victor Ivrii]

Since integral curves are rays (straight half-lines) from $(0,0)$  the solution in the plane wit the punched out origin is any function, constant along these rays, in particular $u=f(y/x)$.

But if we want solution in the whole plane, $u$ must be continuous at $(0,0)$ and since all rays intersect there $u$ is just a constant.

[Wanying Zhang]

I wonder why the integral curves are half-lines from (0, 0)? I agree that these integral curves are straight since the equation $C = \frac{x}{y}$, but it seems no given range for either $x$ and $y$. How is the characteristic of being half determined? Any reply would be appreciated.

[Victor Ivrii]

Please all these lines tend to $(0,0)$ in one as $s\to -\infty$ ($\frac{dx}{ds}=x$, $\frac{dy}{ds}=y$) but never reach it. Please read again ODE, Chapter 7.