Is $ f(x):= \frac{x}{ |x|^n }$ a gradient field?

Question

for $ 2 \leq n $ let be

$ f: \mathbb{R}^n \backslash \{ 0 \} \rightarrow \mathbb{R}^n $

$ f(x):= \frac{x}{|x|^n} $ , $ x\in \mathbb{R}^n \backslash \{ 0 \} $

Is f a gradient field?

A vectorfield $ f $ ist a gradient field, if there is a function $ f= \nabla v $

$ \left[ \nabla f (x) = (\frac{ \partial f}{ \partial x_1} (x),...,\frac{ \partial f}{ \partial x_n} (x)) \right] $

So, I need to find a function $v$ so that $ f= \nabla v $ right?

If I look for the case $n=2 $ so $ f(x):= \frac{x}{|x|^2} \in \mathbb{R}^n \backslash \{ 0 \} $

let be $ \partial_j v(x)= \frac{x_j}{|x|}$

then $v(x)= ln(|x|) $ and

$$ \partial_j v(x)= \frac{1}{|x|} \frac{x_j}{|x|} = \frac{x_j}{|x^2|} $$

Is this the right idea? If yes, how can I proceed for larger $n$? If no, what would you suggest?

Appreciate any help !

Answer 1

Let's suppose that there exists some $V(x)$ such that $\frac{d}{dx}V(x)=f(x)$.

Therefore:

$$\int_a^y\frac{d}{dx}V(x)dx =\int_a^yf(x) dx$$

Applying the fundamental theorem of calculus gives:

$$V(y)=\int_a^y \frac{x}{|x|^n} dx$$ Now let's break into cases. Suppose we are only considering $x,a,y > 0$. Then, we are left with a really simple integral, namely: $$V(y)=\int_a^y \frac{x}{x^n} dx = \int_a^y \frac{1}{x^{n-1}} dx.$$

Now, let's assume we are only considering $x, a, y < 0$. This also simplifies very nicely:

$$V(y)=\int_a^y \frac{x}{(-x)^n} dx = (-1)^n\int_a^y \frac{1}{x^{n-1}} dx.$$

So it seems that we have a definition for $V$.

\begin{cases} V(x) = \frac{1}{2-n}x^{2-n} & x < 0 \\ V(x) = \frac{1}{2-n}(-1)^n x^{2-n} & x>0 \end{cases}

Answer 2

Your $f$ is a radial field and it depends only on the magnitude of your vector $r = |\vec{r}|$ so it is of the form $$f(\vec{r}) = \frac{\vec{r}}{r^n} = F(r)\hat{r}$$ where $F(r) = \frac1{r^{n-1}}$ and $\hat{r}$ is the unit radial vector.

Notice that for a scalar function $\phi : \mathbb{R}^n \to \mathbb{R}$ depending only on magnitude $r$ we have $$\nabla \phi(r) = \phi'(r)\nabla r = \phi'(r)\hat{r}$$

since $\nabla r = \hat{r}$.

Hence if $$\phi(r) = \int F(r)\,dr = -\frac1{(n-2)r^{n-2}}$$ then $f = \nabla \phi$.