Uniqueness of Solution for cauchy problem I'm working on characteristics for partial differential equations and the example given is the equation $yu_x-xu_y=0$ with the condition $u|_{x=1}=y^2$. I could come up with one solution on the spot, but the book I'm using says the solution is not unique. So far I couldn't come up with another solution than $f(x,y)=x^2+y^2-1$. I understand that the methods of characteristics results in all soultions to be functions of $x^2+y^2$
 A: As you pointed out, the characteristics are concentric circles centred at the origin. The difficulty is that any such circle of radius $r < 1$ fails to intersect the line $x = 1$. Therefore, the condition $u|_{x = 1} = y$ tells us nothing about the values of $u$ on circles of radius $r < 1$!
So I claim that $u(x,y) = g(r)$ is a solution, where $g(r)$ is any function of $r = \sqrt{x^2 + y^2}$ chosen so that $u(x,y)$ is differentiable, and so that $g(r) = r^2 - 1$ for $r \geq 1$. (But there is no constraint on $g(r)$ when $r < 1$, other than the differentiability condition.)
For example, we could take $g(r) = r^2 -  1$ for all $r$. This is your example.
But we could also take:
$$ g(r) = \begin{cases} r^2 - 1 + r^2 (1 - r)^2 & \ \  r < 1 \\ r^2 - 1 & \ \ r \geq 1.\end{cases}$$
Unless I made a mistake (which is quite possible, so please check!), this choice of $g(r)$ also makes $f(x,y)$ differentiable everywhere, and it solves the PDE with $u|_{x = 1} = y$.
A: $$yu_x-xu_y=0$$
From the characteristic set of ODEs $\quad \frac{dx}{y}=\frac{dy}{-x}=\frac{du}{0}\quad$ the equations of two characteristic curves can be derived :
$u=c_1$ and$\quad xdx+ydy=0\quad\to\quad x^2+y^2=c_2\quad$ which leads to the general solution of the PDE :
$$u(x,y)=F(x^2+y^2)$$
At this point they are an infinity of solutions. The question is is there only one solution if the specified condition $u(1,y)=y^2$ is added ?
$u(1,y)=F(1+y^2)=y^2$
Let $X=1+y^2 \quad\to\quad y^2=X-1=F(X)$
So, the function $F(X)=X-1$ is determined. This is a one to one relationship. For each value of X only one value of $F(X)$ exists.
Putting this function $F(X)$ where $X=x^2+y^2$ into the above general solution gives only one particular solution of the PDE :
$$u(x,y)=X-1=(x^2+y^2)-1$$
The specified solution is sufficient to reduce the number of solutions to one solution only.
