$\varphi$ verify that $\nabla \cdot\varphi=0$ but doesn't exist $G:\Bbb R^3\to \Bbb R^3$, $\mathcal C^1$ such that $\nabla \times G=\varphi$

Question

I have this problem: Probe that $\varphi:\Bbb R^3-[0] \to \Bbb R^3, \varphi=\frac{(x,y,z)}{\vert\vert(x,y,z)\vert\vert^3}$ verify that $\nabla \cdot\varphi=0$ but does not exist $G:\Bbb R^3\to \Bbb R^3$, of class $\mathcal C^1$ such that $\nabla \times G=\varphi$

It's easy to prove that $\nabla \cdot\varphi=0$ but I have a problem with the part of there is no $G$ such that $\nabla \times G=\varphi$, if is $G=(G1,G2,G3)$ I have this conditions:

$\frac{x}{\vert\vert(x,y,z)\vert\vert^3}= \frac{\partial G3}{\partial y}-\frac{\partial G2}{\partial z}$

$\frac{y}{\vert\vert(x,y,z)\vert\vert^3}= \frac{\partial G3}{\partial x}-\frac{\partial G1}{\partial z}$

$\frac{z}{\vert\vert(x,y,z)\vert\vert^3}= \frac{\partial G2}{\partial x}-\frac{\partial G1}{\partial y}$

And with $\nabla \cdot\varphi=0$ if the lateral derivatives of $G$ coincide I can find some $G$, so something is wrong. Has something to do with the fact that $\varphi$ is not $\mathcal C^1$?

Answer 1

If $\varphi = \nabla \times G$ on $\mathbb{R}^3 - \{0\}$ where $G \in C^1(\mathbb{R}^3)$, then by the divergence theorem, since $\nabla \cdot (\nabla \times G) = 0,$

$$\int_{\partial B_R(0)} \varphi \cdot \mathbf{n} \,dS = \int_{\partial B_R(0)} \nabla \times G \cdot \mathbf{n} \,dS = \int_{B_R(0)} \nabla \cdot (\nabla \times G)\, dV = 0$$

where $B_R(0)$ is the closed ball of radius $R$ centered at the origin.

However,

$$\int_{\partial B_R(0)} \varphi \cdot \mathbf{n} \,dS = \int_0^{2\pi}\int_0^\pi\frac{R\mathbf{e_r}}{R^3} \cdot \mathbf{e}_rR^2\sin \theta \,d\theta \, d\phi = 4\pi$$

Answer 2

We note that we are asked for a $\mathcal C^1$ function

$G:\Bbb R^3 \to \Bbb R^3 \tag 1$

such that

$\nabla \times G = \varphi, \tag 2$

whilst $\varphi$ is only defined on $\Bbb R^3 \setminus \{(0, 0, 0) \}$; indeed, with

$\vec r = (x, y, z), \tag 3$

we have

$r = \Vert \vec r \Vert = (x^2 + y^2 + z^2)^{1/2}, \tag 4$

whence

$\varphi = \dfrac{(x, y, z)}{\Vert (x, y, z) \Vert^3} = r^{-3} \vec r; \tag 5$

it follows that

$\Vert \varphi \Vert^2 = r^{-6} \vec r \cdot \vec r = r^{-6} r^2 = r^{-4}, \tag 6$

so that

$\Vert \varphi \Vert = r^{-2} \tag 7$

on $\Bbb R^3 \setminus \{(0, 0, 0) \}$; now suppose a $\mathcal C^1$ function $G:\Bbb R^3 \to \Bbb R^3$ exists which satisfies (2); then both $G$ and its derivatives are $\mathcal C^0$; restricting $G$ to the closed ball $\bar B(0, \epsilon)$ of radius $\epsilon$ centered at $(0, 0, 0)$, we have, by the compactness of $\bar B(0, \epsilon)$, that the derivatives of $G$ are bounded on $\bar B(0, \epsilon)$; but this clearly contradicts (7), which in concert with

$\Vert \nabla \times G \Vert = \Vert \varphi \Vert \tag 8$

(from (2)) shows the derivatives of $G$ must become arbitrarily large on $\bar B(0, \epsilon)$; hence no such $G$ exists.

The key here is that, while $\varphi$ is only defined on $\Bbb R^3 \setminus \{(0, 0, 0) \}$, $G$ is hypothesized to be defined on all of $\Bbb R^3$.

Appendix: Derivation of

$\nabla \cdot \varphi = 0. \tag{9}$

I worked this out whilst thinking about my solution; I though it might be worth including here.

With

$\varphi = \dfrac{(x, y, z)}{\Vert (x, y, z) \Vert^3} \tag{10}$

and

$\vec r = (x, y, z), \tag{11}$

so that

$r = (x^2 + y^2 + z^2)^{1/2}, \tag{12}$

we may write

$\varphi = r^{-3} \vec r; \tag{13}$

then, in accord with the vector identity

$\nabla \cdot f \vec X = \nabla f \cdot X + f \nabla \cdot X, \tag{14}$

which holds for scalar and vector fields $f$ and $X$, respectively, we have

$\nabla \cdot \varphi = \nabla r^{-3} \cdot \vec r + r^{-3} \nabla \cdot \vec r; \tag{15}$

now,

$\nabla \cdot \vec r = 3; \tag{16}$

thus,

$\nabla \cdot \varphi = \nabla r^{-3} \cdot \vec r + 3 r^{-3} ; \tag{17}$

$\nabla r^{-3} = -3r^{-4} \nabla r; \tag{18}$

from (11) and (12),

$\nabla r = \dfrac{1}{2}(x^2 + y^2 + z^2)^{-1/2} (2x, 2y, 2z) = r^{-1}(x, y, z) = r^{-1} \vec r; \tag{19}$

$\nabla r \cdot \vec r = r^{-1} \vec r \cdot \vec r = r^{-1}r^2 = r; \tag{20}$

$\nabla \cdot \varphi = -3r^{-4} r + 3r^{-3} = 0. \tag{21}$

End: Appendix