# Subtracting off first coordinate to get divergence-free vector field

Let $v$ be a vector field on $\mathbb{R}^n$. Show that $v$ can be written as a sum $v=f_1\dfrac{\partial}{\partial x_1}+w$ where $w$ is a divergence-free vector field.

Suppose $v=v_1\dfrac{\partial}{\partial x_1}+v_2\dfrac{\partial}{\partial x_2}+\ldots+v_n\dfrac{\partial}{\partial x_n}$, where $v_i:\mathbb{R}^n\rightarrow\mathbb{R}$.

Then we want to choose $f_1$ so that $$v-f_1\dfrac{\partial}{\partial x_1}=(v_1-f_1)\dfrac{\partial}{\partial x_1}+v_2\dfrac{\partial}{\partial x_2}+\ldots+v_n\dfrac{\partial}{\partial x_n}$$ is divergence-free, i.e. this quantity is zero when we actually take these partial derivatives (rather than using them as just symbols).

So, take any point $p\in\mathbb{R}^n$. Then we can evaluate the real number $$q(p)=\dfrac{\partial v_1(p)}{\partial x_1}+\dfrac{\partial v_2(p)}{\partial x_2}+\ldots+\dfrac{\partial v_n(p)}{\partial x_n}$$

If I set $f_1(p)=q(p)x_1$, this yields $v-f_1\dfrac{\partial}{\partial x_1}=0$ at point $p$. Since this holds for any point $p$, we have $v-f_1\dfrac{\partial}{\partial x_1}=0$.

EDIT: This solution is currently wrong, because $q(p)$ is not a constant, so I have to differentiate using the product rule. How can I fix it?

• Looks good. Though, I would write $q(p)$ rather than just $q$ so that it is clear that $q$ is a function of $p$ as well! – Tom Nov 23 '13 at 22:07
• @Tom Just edited to incorporate your suggestion. Thanks! – JJ Beck Nov 24 '13 at 0:00
• No problem! Also, don't forget that $x_1$ is a function as well, so $f(p) = q(p)x_1(p)$ or simply $f = q\,x_1$ interpreted by standard pointwise function multiplication. – Tom Nov 24 '13 at 0:06
• @Tom Actually now I'm getting confused. If $f_1(p)=q(p)x_1(p)$, will it be true that $\dfrac{\partial f_1(p)}{\partial x_1}=q(p)$? – JJ Beck Nov 24 '13 at 0:14
• Oh... you make a good point! If you don't have $q$ constant, you'll have to use the product rule. But, if you do set $q$ constant, the divergence of $w$ will only be certain to vanish at $p$.. – Tom Nov 24 '13 at 0:17

You want $$\frac{\partial f_1}{\partial x_1}(p) = q(p) .$$ Let $$f_1(x_1,x_2,\dots,x_n) = \int_0^{x_1} q(\xi,x_2,\dots,x_n) \, d\xi .$$ Here $p = (x_1,x_2,\dots,x_n)$. All is perfectly legitimate because you are working on $\mathbb R^n$. Probably on some other manifolds you could not make this work.
• You mean integral from $0$ to $x_1$, right? – JJ Beck Nov 24 '13 at 4:49