Finding a vector field given its curl.

Question

This is the question I'm working on

Find a vector field with twice differentiable components whose curl is $\langle x,y,z \rangle$ or prove that no such field exists.

My Solution

Let $\textbf{F}=\langle M,N,P \rangle$ be the vector field which we have to find. Given $$ \nabla \times \textbf{F} = \langle x,y,z \rangle $$ Using the determinant expansion of the curl formula, we get $$ \frac{\partial P}{\partial y}-\frac{\partial N}{\partial z} = x \\ \frac{\partial P}{\partial x}-\frac{\partial M}{\partial z} = y \\ \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} = z $$

Differentiating first equation w.r.t $x$, second w.r.t $y$ and the third w.r.t $z$, we get $$ P_{xy}-N_{xz}=1 \\ P_{xy}-M_{yz}=1 \\ N_{xz}-M_{yz}=1 \\ $$ Subtracting the second from first, third from second and then adding the first and third, we get $$ M_{zy}-N_{xz}=0 \\ P_{xy}-N_{xz}=0 \\ P_{xy}-M_{yz}=0 \\ $$

Comparing the last two sets of equations yields $1=0$, a contradiction.

Hence the initial assumption that such a vector field exists is false.

A somewhat simpler argument would be to check the divergence of the curl. — Cameron L. Williams, Sep 13 '22 at 12:27
You guys didn't do divergence? $\nabla\cdot\vec{F}$? If not, a classic result is the divergence of the curl ($\nabla\cdot(\nabla\times\vec{F})$) is $0$. Your argument works just as well, too. — Cameron L. Williams, Sep 13 '22 at 12:28
https://math.stackexchange.com/questions/3903785/why-is-the-divergence-of-curl-expected-to-be-zero — Bob Dobbs, Sep 13 '22 at 12:29
@CameronWilliams it's just me. I'm self studying from a book, section by section, chapter after chapter. Divergence is in the next section. In fact, it's the last section in the book. — Abhishek A Udupa, Sep 13 '22 at 14:00
What the question describes is computing the divergence. (There is a sign error in the second component of the rotor, and that is why you had to substract — if you do it correctly you can simply sum all three equations) — Mariano Suárez-Álvarez, Sep 14 '22 at 03:36
$\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y} = z$? — ArthD21, Sep 14 '22 at 03:37

score 0 · Accepted Answer · answered Sep 14 '22 at 03:36

Your method of solving the problem is correct.

This is an example of a more general result that for a vector field, the divergence of the curl is identically zero. We can write this as $$ \nabla\cdot\nabla\times F=0. $$ This is not an equation you need to solve but an identity which is always true. It follows from the definitions of divergence and curl and is always true for any suitably differentiable $F$.

In your case, $$\nabla\cdot(x,y,z)\equiv\left({\partial x\over \partial x},...\right)=(1,1,1)$$ which means that $(x,y,z)$ cannot be the curl of a vector field.

Finding a vector field given its curl.

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