Vector field defined by constant function

Question

Let $n \geq 2$. Given $\phi:[0,\infty] \longrightarrow \mathbb{R}^n$ continuously differetiable and $F:\mathbb{R}^n \longrightarrow \mathbb{R}^n $ a vector field, Defined as $F(x) = \phi(\|x\|_2)$. Prove that $F$ is conservative if and only if $\phi$ is constant. Already proved that $\phi$ is constant implies $F$ conservative. I'm struggling with proving the other direction.

$F$ conservative $\Longrightarrow$ $\phi$ is constant.

Answer 1

Talking in terms of differential forms, if $F$ is conservative then it is closed. Thus you know that

$$0=\frac{\partial F_{i}}{\partial x_{j}}-\frac{\partial F_{j}}{\partial x_{i}}=\phi_{i}^{\prime}\frac{\partial\|\boldsymbol{x}\|_{2}}{\partial x_{j}}-\phi_{j}^{\prime}\frac{\partial\|\boldsymbol{x}\|_{2}}{\partial x_{i}}=$$

$$=\phi_{i}^{\prime}\frac{x_{j}}{\|\boldsymbol{x}\|_{2}}-\phi_{j}^{\prime}\frac{x_{i}}{\|\boldsymbol{x}\|_{2}}$$

$$\Longrightarrow x_{j}\phi_{i}^{\prime}-x_{i}\phi_{j}^{\prime}=0$$

for every $1\leq i,j\leq n$. That's true on the sphere $\|\boldsymbol{x}\|_{2}=R>0$, i.e.

$$x_{j}\phi_{i}^{\prime}\left(R\right)-x_{i}\phi_{j}^{\prime}\left(R\right)=0$$

where $\phi_{i}^{\prime}\left(R\right)$ and $\phi_{j}^{\prime}\left(R\right)$ are constants. But this is true for every value of $x_{i},x_{j}$ as long as $\|\boldsymbol{x}\|_{2}=R$ (you can argue that there are multiple choices for every $n\geq 2$), and hence you must have

$$\phi_{i}^{\prime}\left(R\right)=0$$

for every $R>0$ and $1\leq i\leq n$. Combine this to get

$$\phi^{\prime}=0 \Longrightarrow \phi=\rm const.$$

on $[0,\infty)$ (the $R=0$ case follows from continuity).

Answer 2

Let $\phi(x)=(\phi_1(x),\ldots\phi_n(x))$. Also, let $F(x_1,\ldots,x_n)=(F_1(x_1,\ldots,x_n),\ldots,F_n(x_1,\ldots,x_n))$ where we have:

$$F_i(x_1,\ldots,x_n)=\phi_i(\rho)$$

where $\rho=\|(x_1,\ldots,x_n)\|_2=\sqrt{x_1^2+\cdots+x_n^2}$. Now, if $\rho\gt 0$, we have:

$$\frac{\partial F_i}{\partial x_j}(x_1,\ldots,x_n)=\phi_i'(\rho)\frac{x_j}{\rho}$$

Assuming $F$ is conservative, which implies $\frac{\partial F_i}{\partial x_j}=\frac{\partial F_j}{\partial x_i}$, we have:

$$\phi_i'(\rho)x_j=\phi_j'(\rho)x_i$$

This also means:

$$\phi_i'(\rho)x_j=\phi_j'(\rho)(-x_i)$$

(swap $x_i$ with $-x_i$, which gives the same $\rho$).

Adding those last two equalities, we conclude $\phi_i'(\rho)x_j=0$, and as for any $\rho\gt 0$ we can choose non-zero $x_j$'s, it follows that $\phi_i'=0$ i.e. $\phi_i$ is constant. (On $(0,+\infty)$, and so by continuity on $[0,+\infty)$.)

Hence, $\phi$ is constant and $F$ is constant.