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.

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.