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Suppose $f: \mathbb{R}^2 \to \mathbb{R}$ has continuous partial derivatives and $f = a\frac{\partial f}{\partial x} + b \frac{\partial f}{\partial y}.$ Prove that $f = 0$ if $f$ is bounded.

If $a=0$ or $b=0,$ I was able to solve the problem. If $a, b \ne 0,$ let $g = (e^{-x/a} + e^{-y/b})f.$ Then $g$ is differentiable and we get $f = f + (e^{-x/a} + e^{-y/b})(ag_x + bg_y) \Rightarrow ag_x + bg_y = 0.$ Unfortunately, I have no idea how to solve this new equation. I'm not sure how I would even approach the simple PDE $g_x = g_y.$

Another approach: Suppose $f \ne 0.$ WLOG $\exists x_0 : f(x_0) > 0$ (else, we may work with $-f$). Let $D_r$ be the disk of radius $r$ centered at the origin. Since $f$ is continuous, it has a maximum $M = f(a)$ on $D_r.$ If the maximum is inside the disk, then the gradient of $f$ vanishes at $a,$ so $f(a) = 0 $ due to the given differential equation. If $r > x_0,$ then $M \ge f(x_0) > 0,$ contradiction. Thus, the maximum of $f$ is always on the boundary of $D_r$ for $r > x_0.$ But this is not a problem. We can imagine the graph of $z=f(x,y)$ approaching the planar asymptote $z=M$ ever so slowly. Maybe the shape of $f$ is something like $1 - e^{-x^2-y^2}.$ This is not a solution of course, but it has the properties of boundedness, differentiability, and always having a maximum on the boundary of $D_r.$ How do we rule out such cases? What prevents such solutions from working?

3rd idea (continuing from the 1st idea): The gradient of $g$ is always perpendicular to $v = (a,b).$ Thus, the image of $\nabla g$ is a subset of a line. Since $g$ is continuously differentiable, $\nabla g$ is continuous. By the IVT, the image of $\nabla g$ is either a line segment, ray, or the entire line. But what do we do next? $g$ is not necessarily bounded.

4th idea: If we show $f$ is harmonic, then $f$ has to be constant, and the differential equation implies the constant is $0.$ But I don't think there's any way of showing $f$ is constant from the differential equation. Afterall, $f = e^{x/a} + e^{y/b}$ is a non-harmonic solution.

Is there a simple idea I'm missing? How would you come up with it?

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    You can use the method of characteristics to show the solution to the problem is given by $$f = \exp \left( \frac{x+y}{a+b} \right) g(bx - ay)$$ with $g$ an arbitrary, differentiable function. – Matthew Cassell Aug 04 '20 at 05:27
  • @mattos What if $a=-b$? Is there a solution that does not rely on prior PDE knowledge? – Display name Aug 04 '20 at 05:28
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    Then $y$ must change accordingly and the solution is given by $$f = \exp \left( \frac{x-y}{2a} \right) g(x+y)$$ I'm sure you can get a solution that doesn't rely on prior PDE knowledge. The point of this exercise seems to be to show that the solution exhibits exponential growth and so is unbounded on $\mathbb{R^{2}}$ unless $f \equiv 0$. But how you would show this is dependent on what you have learned, the topic where this problem appeared etc. which you haven't really specified. – Matthew Cassell Aug 04 '20 at 05:35

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