Suppose $F$ satisfies $div F = 0$ and $curl F=0$ on all of $R^3$. Show that we can write Find $F= \nabla f$, where $ \nabla^2 f=0$ I have to solve this:
Suppose $F$ satisfies $div F = 0$ and $curl F=0$ on all of $R^3$. Show that we can write Find $F= \nabla f$, where $ \nabla^2 f=0$
Sincerely I do not even know how to start, any ideas?
 A: We have that $\operatorname{curl}F = 0$ on all of $\mathbb{R}^3$.
Then the value of a line integral along any curve $C$ only depends on the initial point and the end point of the curve, not on the curve itself. This follows because we have, assuming that $C_1, C_2$ both begin at $r_0$ and end at $r_1$:
\begin{align}
\int_{C_1}F(s)\cdot ds - \int_{C_2}F(s)\cdot ds &= \int_{C_1}F(s)\cdot ds + \int_{-C_2}F(s)\cdot ds\\
&=\oint_{C_1, C_2}F(s)\cdot ds\\
&=\iint_A\operatorname{curl}F\,dA\\
&= 0
\end{align}
where $C_1, C_2$ is the closed curve created by joining $C_1$ and $C_2$ and $A$ is the area enclosed by this closed curve. The equality of line integral and area integral follow from Stokes' Theorem.
This, however, implies that the value of any line integral is actually independant of the path taken and only depends on the start and endpoints of the line along which the function is being integrated.
Therefore, we can define
$$f(r) = \int_{r_0}^r F(s)\cdot ds$$
for an arbitrary $r_0\in\mathbb{R}^3$.
Then, by construction, we have
$$F = \nabla f$$
which, along with $\operatorname{div}F = 0$ implies that
$$0 = \operatorname{div}F = \operatorname{div}(\nabla f) = \nabla^2 f$$
as was to be shown.
