The following PDE is given:
$xu_y-yu_x=0$ with $u(x,0)=x^2$
The following topics;
$yU_x-xU_y=1, U(x,0)=0$
Solution of the PDE $yu_x+xu_y=0$ subject to the initial condition $u(x,0) = \exp \left(-\frac{x^2}{2}\right)$
Solving $-yu_x+xu_y = u$ using method of characteristics
did not solve my troubles, because the PDE is not equal to 0, and the initial condition is different. I am stuck near the end of the problem.
I used the method of characteristics for PDE:
$x_t=-y, y_t=x, u_t=0, x(0,s)=s, y(0,s)=0, u(0,s)=s^2$
$x'(t)=-y(t)$ thus $ x''(t)=-y'(t)=-x(t)$ so $x(t)=c_{11}cos(t)+c_{12}sin(t)$.
Using $x(0,s)=s$ we get $c_{11}=s, c_{12}=0$ thus $x(t,s)=s\cos{t}$
Same procedure for $y'(t)=x(t), y''(t)=x'(t)=-y(t)$ thus $y(t,s)=s\cos{t}$
And for $u$, we get $u(t,s)=s^2$
Now, how exactly do I solve for $t$ and $s$ to write $u$ as a function of $x$ and $y$ (and not $t$ and $s$)? I cannot solve for $t$ or $s$ using $x$ and $y$ given that $x(t,s)=y(t,s)$, i.e. they are identical to each other.
That is my first question.
My second question is: if the initial condition was $u(x,0)=x, x > 0$ instead of $u(x,0)=x^2$ (so no square, and without $x>0$), do we agree that it would only change the result for $u$, i.e. for $x(t,s)$ and $y(t,s)$, we would get the same result than here.
Thank you for taking your time to help me.
Edit: As @Mattos correctly points out, $y(t)=s\sin(t)$. That changes a lot of things. Now, we can use $x^2+y^2=s^2(\cos{x}^2+\sin{x}^2)=s^2=u$
. Thus, $u(x,y)=x^2+y^2$.
Best Answer
As pointed out in the comment section by @Mattos, $y(t)=s\sin{t}$, not $s\cos{t}$. So I made a calculation error. Now, to find $u(x,y)$, we can simply use $x^2+y^2=s^2(\cos(t)^2+\sin(t)^2)=s^2=u(t,s)$. Thus $u(t,s)=x^2+y^2$.
If the initial condition is $u(x,0)=x, x > 0$ instead, we get $u(0,s)=s$ thus $u(t,s)=s$ thus $u(x,y)=\sqrt{x^2+y^2}$. If we had simply $u(x,0)=x$ (so without $x>0$),we would have $u(x,y)=\pm \sqrt{x^2+y^2}$