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Solve the equation $xU_x+yU_y=0$.

From Partial Differential Equations: An Introduction, 2nd Ed. (Strauss, pg. 10). There are no boundary conditions.

Solution: The PDE can be written $(x,y)\cdot\nabla U(x,y)=0$, implying that the vector $(x,y)$ is tangent to the PDE's characteristic equation. Then I can write $dy/dx=y/x$, and to solve this ODE,

$$\frac{dy}{dx}=\frac yx \\ \ln y=\ln x+c_1 \\ y=cx \\ y/x=c$$

Hence, the PDE is dependent on $c$ and I can write $U(x,y)=f(y/x)$ for some function $f$, solving the problem.

Question: Is this solution correct? The hint for the question (supplied by the professor) said I should use a substitution of variables, but this seems reasonably valid.

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    $\begingroup$ It looks fine to me. To convince yourself that it is correct you can try verifying that $U(x,y) = f(y/x)$ is indeed a solution by computing $U_x,U_y$ and substituting this into the equation. $\endgroup$ – Winther Sep 11 '15 at 18:17
  • $\begingroup$ Oh, I get it. After using the chain rule, the two terms cancel out. Thanks. $\endgroup$ – nettle Sep 11 '15 at 18:19
  • $\begingroup$ Yes, it is a solution. I wonder if $f(\frac y x)$ depends on the boundary conditions. $\endgroup$ – kleineg Sep 11 '15 at 18:23
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    $\begingroup$ @kleineg Yes, with sufficient boundary conditions, $f(y/x)$ can be determined precisely. $\endgroup$ – nettle Sep 11 '15 at 21:23
  • $\begingroup$ @Winther There is an often-missed subtlety here, which I addressed in my answer. $\endgroup$ – user147263 Sep 13 '15 at 4:57
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You did not get the most general solution of this PDE. For example, the function $$u(x,y) = \frac{x}{\sqrt{x^2+y^2}}$$ satisfies it, but cannot be written in the form $f(y/x)$. Indeed, $u(1,1)\ne u(-1,-1)$ while the ratio $y/x$ at these points is the same.

The incorrect implication in your solution is $$\frac{dy}{dx}=\frac yx \implies \ln y=\ln x+c_1$$ What you can actually conclude is $$ \ln |y|=\begin{cases}\ln x+c_1,\quad &x>0 \\ \ln (-x)+c_2,\quad &x<0 \end{cases}$$ This leads to $u(x,y)=f(y/x)$ for $x>0$ and $u(x,y)=g(y/x)$ for $x<0$, where $f$ and $g$ may well be different.

Also, $x=0$ is problematic: the PDE makes sense there but the formula for solution doesn't.

Quoting from my answer elsewhere:

I would rather use notation $u(x,y)=g(\theta)$ where $\theta$ is the polar coordinate. After all, the PDE $xu_x+yu_y = 0$ simply says that $\partial u/\partial r = 0$ in polar coordinates. Any function that is independent of radial coordinate $r$ solves the PDE. And not every such function is of the form $G(y/x)$ since this form requires $u(-x,-y)=u(x,y)$ which need not be the case.

Generally, I recommend sketching characteristic curves to see the picture more clearly. In this case, they are half-lines of the form $\{(at,bt):t>0\}$ with $a^2+b^2\ne 0$. They are not lines through the origin, because the characteristic equation makes no sense at the origin.

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  • $\begingroup$ Completely forgot about that. The substitution of variables the professor wanted was linear, though, so I can't imagine that it would fix this singularity at $0$. $\endgroup$ – nettle Sep 13 '15 at 19:20
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    $\begingroup$ The singularity at $x=0$ is an artefact of your approach; you could have just as well used $f(x/y)$ and then it's in a different place. There is a singularity at $(0,0)$ which is intrinsic to this PDE: every solution that's not identically constant does not have a limit as $(x,y)\to (0,0)$. All that said, it's quite likely that a professor will be satisfied with $u(x,y)=f(y/x)$ answer, since it demonstrates the knowledge of what's being tested... still, I wanted to point out the incompleteness of this answer. $\endgroup$ – user147263 Sep 13 '15 at 19:25

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