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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, $u(x,y)=x^2+y^2$.
EditPiAf
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    $$y_{t}(t,s) = x(t,s) \implies y(t) = \int s \sin t dt = s \color{red}{\sin t}$$ – Matthew Cassell Oct 10 '19 at 13:24
  • @Mattos uh you're right. That changes a lot of things –  Oct 10 '19 at 13:25
  • @Mattos do you agree that it would not change much if the initial condition was $u(x,0)=x, x > 0$ ? I mean, the solution is slightly different, but the procedure is the same. We would get $\sqrt{x^2+y^2}$ (and not $-\sqrt{x^2+y^2}$ because $x>0, u(x,0)=x, u(0,s) = s $) if I am correct ? –  Oct 10 '19 at 13:43
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    Well it does change much, in fact in changes the whole solution. If $u(x,0) = x$ with $x > 0$ then $$u(x,y) = \sqrt{x^2+y^2}$$ – Matthew Cassell Oct 10 '19 at 13:49
  • @Mattos Yeah, right. Thank you for your help. Can you post your answer (by stating what you said right here in the comments) so I can accept it and the question doesn't show up anymore in the unanswered section ? –  Oct 10 '19 at 13:57
  • @Ryukyu, I know this is like 2 years after you asked this, and I'm sorry for commenting here lmao, but could you pleasee explain how/why you got $c_{12} = 0$? I'm doing a similar question, and even though I get that $c_{11} = s$, I can't seem to figure out how or why to get a value for $c_12$? Especially since I too am given the condition $x(0,s) = s$. Would really appreciate if you could help! – Applesauce44 Nov 10 '21 at 13:53
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    @Applesauce44 Don't know if you got your answer, but if I recall correctly, $c_12$ is basically required to be zero so that x equals cosine and y equals sine, to respect $x'' = -y' = -x$. Only a trigonometric function can solve this kind of equations, and looking at the boundary condition, one can guess that $x = \cos$ so $c_12 = 0$ and $y = \sin$ –  Nov 10 '21 at 22:27
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    @Applesauce44 if you want to deepen your research, it's taken from here https://metaphor.ethz.ch/x/2021/hs/401-0353-00L/ex/sol03.pdf Link to the course https://metaphor.ethz.ch/x/2021/hs/401-0353-00L/ Enjoy your reading –  Nov 10 '21 at 22:29
  • AHHHHHHH I see okay okay, that makes sense, thank you. Also thanks for the links as well, I'll take a look at both! – Applesauce44 Nov 11 '21 at 03:02
  • @Applesauce44 Just wondering, are you studying at ETHZ ? –  Dec 02 '21 at 20:45
  • @Ryukyu ahaha, no, no I'm not. In a completelyyy different country. I was once looking to go there for grad school though so this is interesting lol. Are you studying there? – Applesauce44 Dec 03 '21 at 09:32
  • @Applesauce44 Previously yes. It's a very well known university, especially for grad school as you said. Was just wondering because you seem to have had the exact same exercise haha –  Dec 04 '21 at 23:38
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    @Ryukyu ohh ahaha I see I see. Yeah no so I'm taking a related class right now so we might have somewhat similar materials lol. This is really interesting to know though! ^_^ – Applesauce44 Dec 07 '21 at 17:21

1 Answers1

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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 $u(x,y)=\pm \sqrt{x^2+y^2}$