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I am trying to prove that this partial differential equation has infinitely many solutions.

$$\partial_xu + x\partial_yu = 0$$ $$u(x,0) = \cos x$$

Using the method of characteristics, I found that $u(x_,y) = \cos(\sqrt{x^2-2y})$ is a solution

I couldn't find a different solution. Does anyone have any suggestions?

Santos
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1 Answers1

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The equation is $$du\left(\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right)=0.$$ This means that $u$ is constant on trajectories of the vector field $F=\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}.$ Other than this and the initial condition $u(x,0)=\cos x$, the function $u$ does not need to satisfy anything.

The key to solving this problem lies in understanding the trajectories of the vector field $F$. Specifically, you can find an open ball $B$ which remains far away from the $x$ axis under the flow of $F$ at any time. This means that inside $B$, you can smoothly change $u$ as much as you like.

Amitai Yuval
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    @Santos, Amitai, I cannot see it. It seems that with the initial condition we can pick e.g. $u(\pi/2,0)=1$, then it seems the value of $u$ all along $x^2-y=(\pi/2)^2$ is set to one if you require continuity. Can you "move" $u$ if the ball is crossed by this curve? – Rafa Budría Apr 22 '17 at 18:59
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    @RafaBudría No, you cannot. A ball which is crossed by this curve does not remain away from the $x$ axis under the flow of $F$. You need a different ball. – Amitai Yuval Apr 22 '17 at 20:20
  • @Santos, Amitai, I think I've got it: one curve is, e.g. $x^2-2y=-1$ on wich the initial conditions doesn't determine the value for $u$. The solution with the cosine is not defined for $x^2-2y\lt0$ and I know now why. Ok, a solution can be patching with $1-x^2/2+y$ for $x^2-2y\lt0$ and another one with $1+(1/8)(x^2-2y)^2$ Thank you! – Rafa Budría Apr 22 '17 at 21:13
  • @RafaBudría Yes. Or, more generally, the trajectories are all of the form $y=\frac{x^2}{2}+c$. Consequently, the region in the plane affected by the initial condition is ${y\leq\frac{x^2}{2}}$. The values of $u$ outside this region have nothing to do with the initial condition. – Amitai Yuval Apr 22 '17 at 21:21