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This is an unsolved exercise given in my textbook, which I am having trouble with. The exercise seems simple, but for some reason I can't solve it. Help would be very nice!

Let $o(t)$ be a real continuous positive function in $[0,\infty)$, and $F(x,y,z)=o(\|(x,y,z)\|)(x,y,z)$ ($\|\cdot \|$ means the norm, e.g. $\sqrt{x^2+y^2+z^2}$). We need to prove that $F$ has a vector potential in $\mathbb{R}^3$.

Attempt: If $G$ is the potential, we know $G_x = o(\|(x,y,z)\|)x$, so I figured $G$ could be of the form $G(x,y,z)=\int_{0}^{x} o(\|(t,y,z)\|)t dt + h(y,z)$. So now, to find $h(y,z)$ I need to solve: $$ G_y(x,y,z)=yo(\|(x,y,z)\|)=(\int_{0}^{x} o(\|(t,y,z)\|)t dt + h(y,z))_y $$ and $$ G_z(x,y,z)=zo(\|(x,y,z)\|)=(\int_{0}^{x} o(\|(t,y,z)\|)t dt + h(y,z))_z. $$ But I don't really know how to differentiate $\int_{0}^{x} o(\|(t,y,z)\|)t$ by $y$ or $z$, which gets me stuck.

Thanks for reading. Please help me figure this out!

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Unless the function $t\mapsto o(t)$ is differentiable you can't compute $G_y,\ G_z$ – Mercy King Jun 22 '12 at 3:31
@Mercy: Yeah, we aren't given that $o(t)$ is differentiable. So I'm sort of guessing my approach here is wrong but... I don't see any other approach. – ro44 Jun 22 '12 at 3:34
I think I found a simpler approach: $G(x,y,z)=(1/2)(\int_{0}^{x^2+y^2+z^2} o(\sqrt{t}) dt)$. I can't see any problem with it, but if it works I'm going to feel like an idiot. What do you guys think? – ro44 Jun 22 '12 at 3:45
Since $F$ is defined on $\mathbb{R}^3$ (which is simply connected), $F$ has a potential iff $\text{curl}F=0$. – Mercy King Jun 22 '12 at 3:46
up vote 0 down vote accepted

Since $o$ is continue in $[0,+\infty)$, let us put :$\Phi(t)=\int_0^t \tau o(\tau) d \tau $, for all $t \in[0,+\infty)$. Then we know that $\Phi$ is a class $C^1$ function on $[0,+\infty)$ and has derivative : $\Phi'(t)=t o(t)$ forall $t \in [0,+\infty)$. if we consider the function $V$ such that :$$\forall X\in \mathbb R^3 \quad V(X)= \Phi (\|X\|)$$ it is clear that $V$ in differentiable on $\mathbb R^3 \backslash \{(0,0,0)\}$ and , putting: $X=(x,y,z)$: $$\frac{\partial V}{\partial x} (X)= \frac{x}{\|X\|} \Phi'(\|X\|)=\frac{x}{\|X\|} \|X\| o(\|X\|)= x o(\|X\|) $$ With the same way we have too:$$\frac{\partial V}{\partial y} (X)= y o(\|X\|) \quad \text{and} \quad\frac{\partial V}{\partial z} (X)= z o(\|X\|)$$

Then : $\bf{Grad}$$ (V) (x,y,z) = F(x,y,z)$

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