Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
Unless the function $t\mapsto o(t)$ is differentiable you can't compute $G_y,\ G_z$ –  Mercy 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 Jun 22 '12 at 3:46
add comment

1 Answer

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)$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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