I'm doing a list of exercises, but there is one problem I can't solve. Let $$\Omega=\Big\{(x,y,z)\in\mathbb{R}^3 \ | \ \frac{x^2}{4}+\frac{y^2}{9}+\frac{z^2}{16}<1\Big\}.$$
Suppose that $\rho:\Omega\rightarrow\mathbb{R}$ is a given continuous function, and that $i:\Omega\rightarrow\mathbb{R}^3$ is $C^1$ and is also given. Also supose that $\nabla\cdot i=0$ in $\Omega$. Show that exists $$F:\Omega\rightarrow\mathbb{R}^3$$ such that $\nabla\cdot F=\rho$ and $\nabla\wedge F=i$.
PS: $\nabla\cdot F$ is the divergence and $\nabla\wedge F$ is the curl.
I know I'm supposed to use this information to construct the vector field $F$...I'm a little lost here cause I never did this before. Thanks.
Here' how it works:
Since $\nabla \cdot i = 0$ on $\Omega$, and $\Omega$ is a "nice" region in $R^3$, i.e. it is simply connected, it is basically an open ball, topologically speaking, etc., it follows that there is a vector field $G$ on $\Omega$ such that $\nabla \times G = i$. In fact, there a very many such fields. For if $\phi:\Omega \to R$ is any twice differentiable real-valued function, $\phi \in C^2(\Omega, R)$, then $\nabla \times (G + \nabla \phi) = \nabla \times G + \nabla \times \nabla \phi = \nabla \times G = i$, since $\nabla \times \nabla \phi = 0$, always. (The curl of a gradient is zero, a standard vector calculus result.) Now $\nabla \cdot (G + \nabla \phi) = \nabla \cdot G + \nabla^2 \phi$. Consider the equation $\nabla^2 \phi = \rho - \nabla \cdot G$ on $\Omega$; for $\phi$ satisfying this we have $\nabla \cdot (G + \nabla \phi) = \nabla \cdot G + \nabla^2 \phi = \nabla \cdot G + \rho - \nabla \cdot G = \rho$. Take $F = G + \nabla \phi$; then $\nabla \times F = \nabla \times G = i$ and $\nabla \cdot F = \rho$. And that does it.
Of course, there are a couple of caveats. First, though we have "constructed" the vector field $F$ on $\Omega$, it is really more of a recipe for construction rather than a construction itself. In point of fact, there are integral formulas which express $G$ in terms of $i$ and $\phi$ in terms of $\rho - \nabla \cdot G$, they won't get you very far without knowing $i$ and $\rho$ explicitly. They do however suffice to establish the existence of the vector field $G$ and the scalar field $\phi$. And in regard to this latter point, technically since we require that $\phi$ satisfy the Poisson equation $\nabla^2 \phi = \rho - \nabla \cdot G$ on $\Omega$, we need to make sure that $\rho$, $\nabla \cdot G$ and $\Omega$ are all sufficiently nice, and we need boundary conditions on $\phi$. I'm not going to to into the heavy analysis here, but I suspect, since $\Omega$ is so simple, we can specify $\phi$ on $\partial \Omega = \{(x, y, z) \in R^3 \vert \frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} = 1\}$ with quite a degree of latitude; probably taking $\phi$ to be any continuous function on $\partial \Omega$ will suffice; and probably we will need $i \in C^1(\bar{\Omega}, R^3)$ as well. Here of course we have $\bar{\Omega} = \{(x, y, z) \in R^3 \vert \frac{x^2}{4} + \frac{y^2}{9} + \frac{z^2}{16} \le 1\}$. And also $\rho \in C(\bar{\Omega}, R)$. These latter requirements to prevent things from blowing up as we approach $\partial \Omega$, which may invalidate the integral formulas for $G$, $\phi$ to which I referred.
All references, both implicit and explicit, to established results, viz., the existence of $G$ with $\nabla \times G = i$ and the solution to the Poisson equation $\nabla^2 \phi = \rho - \nabla \cdot G$ on $\Omega$, may be found by wiki'ing/googling around such topics as divergence, curl, vector calculus, Poisson's equation and so forth. There is an abundance of material available on the web.
Finally, this question/answer are deeply related to this one.
Hope that helps. Cheers.