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Let $f : \mathbb{R}^3 \setminus \left \{0 \right \} \to \mathbb{R}$ be a differentiable function s.t $\nabla f \neq 0$ and:

$y \frac{\partial f}{\partial x} - x \frac{\partial f}{\partial y} =0 \\ z \frac{\partial f}{\partial y} - y \frac{\partial f}{\partial z} =0$

Show there is a differentiable $g : (0, \infty) \to \mathbb{R}$ s.t $f(x,y,z) = g(||(x,y,z)||)$

I tried using spherical coordinates to show that the gradient is only determined by $r$ which I think should help, but I had some trouble with the algebra, help appreciated.

paxtibimarce
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2 Answers2

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Define $g(r,\theta,\phi) := f(r\sin\theta\cos\phi, r\sin\theta\sin\phi, r\cos\theta)$. Our assumptions say $$\frac{\partial f}{\partial x}r\sin\theta\sin\phi = \frac{\partial f}{\partial y}r\sin\theta\sin\phi$$ $$\frac{\partial f}{\partial y}r\cos\theta = \frac{\partial f}{\partial z}r\sin\theta\sin\phi$$ We have $$\frac{\partial g}{\partial \phi} = -\frac{\partial f}{\partial x}r\sin\theta\sin\phi + \frac{\partial f}{\partial y}r\sin\theta\sin\phi = 0$$ so $g$ in fact does not depend on $\phi$ and we can fix $\phi = \frac\pi2$ so $$g(r,\theta) = f(0, r\sin\theta, r\cos\theta).$$ We have $$\frac{\partial g}{\partial \theta} = \frac{\partial f}{\partial y}r\cos\theta - \frac{\partial f}{\partial z}r\sin\theta = 0$$ so $g$ doesn't depend on $\theta$ either. Therefore $g(r,\theta,\phi) = g(r)$ and is differentiable.

Taking $(x,y,z)$ from a dense subset of $\Bbb{R}^3$ we can define $$(r,\theta,\phi) = \left(\sqrt{x^2+y^2+z^2}, \arctan\frac{\sqrt{x^2+y^2}}z, \arctan\frac{y}x\right)$$ and notice $$f\left(x,y,z\right) = f(r\sin\theta\cos\phi, r\sin\theta\sin\phi, r\cos\theta) = g(r,\theta,\phi) = g(r)= g(\|(x,y,z)\|).$$ By continuity the result extends to all of $\Bbb{R}^3$.

paxtibimarce
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mechanodroid
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  • Could you elaborate on "Taking $(x,y,z)$ from a dense subset...". I don't really get why that is necessary. – paxtibimarce Feb 20 '21 at 20:42
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    @paxtibimarce Well, the denominators have to be nonzero so we can define $(r,\theta,\phi)$ as stated on the set $S = {(x,y,z) \in \Bbb{R}^3 : x \ne 0, z \ne 0}$. On $S$ we get the equality of continuous functions $f(x,y,z) = g(|(x,y,z)|)$. Since $S$ is dense in $\Bbb{R}^3$, the equality holds on all of $\Bbb{R}^3$. – mechanodroid Feb 20 '21 at 20:47
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Since I prefer a more conceptual approach, note that you can deduce from the two equations that $$\nabla f(x,y,z) = \lambda(x,y,z)(x,y,z)$$ for some function $\lambda$. Since the gradient of a differentiable function is everywhere orthogonal to the level surfaces of that function, this tells you that the function $f$ is constant on spheres centered at the origin. Thus, $f(x,y,z) = g(\|(x,y,z)\|)$ for some function $g$.

(If you prefer, you can see that the directional derivative is $0$ in any direction tangent to such a sphere, because that directional derivative is given by the dot product of the gradient and the tangent vector, and we all know that $(x,y,z)$ is the normal to the sphere centered at the origin passing through the point $(x,y,z)$.)

Ted Shifrin
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    Ratio and proportion. The first equation tells you that $(\frac{\partial f}{\partial x},\frac{\partial f}{\partial x})$ is a scalar multiple of $(x,y)$ (away from $x=0$ and $y=0$) and likewise for the second equation. But then you can take care of these cases. – Ted Shifrin Feb 20 '21 at 20:17
  • Thanks, could you elaborate on how you deduce the function is constant on spheres? – paxtibimarce Feb 20 '21 at 20:24
  • I can imagine it but I'm just not sure how to show it formally – paxtibimarce Feb 20 '21 at 20:27
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    Well, I gave the argument in the parenthetic second paragraph. The function stays constant if you move orthogonal to the position vector, which means along the sphere through that point. You know that the normal vector to a sphere centered at the origin is the position vector at every point, and there can be only one surface with that property. – Ted Shifrin Feb 20 '21 at 20:59
  • Any point $\boldsymbol{r}=(x,y,z)\ne(0,0,0)$ is joined to $(0,r,0),$ $r=\sqrt{x^2+y^2+z^2},$ by an arc of a circle, centre $O,$ parameter $t,$ on which (abusing notation): $\boldsymbol{r}\cdot\boldsymbol{r}=r^2,$ whence $\boldsymbol{r}\cdot(d\boldsymbol{r}/dt)=0;$ and excepting at most one point, $y\ne0,$ therefore $\nabla f$ is a multiple of $\boldsymbol{r};$ therefore (chain rule) $df(\boldsymbol{r})/dt=(\nabla f)\cdot(d\boldsymbol{r}/dt)=0;$ therefore $f(\boldsymbol{r})=f(0,r,0)=g(r),$ say. – Calum Gilhooley Feb 20 '21 at 21:26