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Let $F: \mathbb{R}^3 \to \mathbb{R}^3$ such that $F$ is differenciable and $\nabla \cdot F = 0, \forall \bar{x}\in \mathbb{R}^3$.

Show that $F = \nabla \times G$, for $G: \mathbb{R}^3 \to \mathbb{R}^3$ defined as $G=(G_1, G_2, G_3)$, where $G_1(x, y, z) = \int_0^z F_2(x, y, t) dt - \int_0^y F_3(x, t, 0) dt$, $G_2(x, y, z) = -\int_0^z F_1(x, y, t) dt$ and $G_3(x,y,z) = 0$.

I've tried a proof but I'm not sure if I'm doing things well:

(Second try):

\begin{alignat*}{2} \nabla \times G & = \Big(\frac{\partial}{\partial y} G_3 - \frac{\partial}{\partial z} G_2, \frac{\partial}{\partial z} G_1 - \frac{\partial}{\partial x} G_3, \frac{\partial}{\partial x} G_2 - \frac{\partial}{\partial y} G_1 \Big)\\ & = \Big(\frac{\partial}{\partial z} \Big( \int_0^z F_1(x, y, t) dt \Big), \frac{\partial}{\partial z} \Big( \int_0^z F_2(x, y, t) dt - \int_0^y F_2(x, t, 0) dt \Big), \frac{\partial}{\partial x} \Big( -\int_0^z F_1(x, y, t) dt \Big) - \frac{\partial}{\partial y} \Big( \int_0^z F_2(x, y, t) dt - \int_0^y F_3(x, t, 0) dt \Big) \Big)\\ & = \Big(\frac{\partial}{\partial z} \Big( F_1(x, y, z^2/2) - F_1(x, y, 0) \Big), \frac{\partial}{\partial z} \Big( F_2(x, y, z^2/2) - F_2(x, y, 0) - F_2(x, y^2/2, 0) + F_2(x, 0, 0) \Big), \frac{\partial}{\partial x} \Big( -F_1(x, y, z^2/2)+F_1(x, y, 0) \Big) - \frac{\partial}{\partial y} \Big( F_2(x, y, z^2/2)-F_2(x, y, 0) - F_3(x, y^2/2, 0) + F_3(x, 0, 0) \Big) \Big)\\ & = \Big( F_1(x, y, z), F_2(x, y, z), -F_1(1, y, z^2/2)+F_1(1, y, 0) - F_2(x, 1, z^2/2)+F_2(x, 1, 0) + F_3(x, y, 0) \Big)\\ \end{alignat*}

Related articles:

Existence of $G$ for $\nabla \times G=F$?

How do I solve $F = \nabla\times G$ for $G$?

Blue Tomato
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    $\nabla\cdot(\nabla\times G)=0$ does not imply $F=\nabla\times G$ (why would it - $F$ appears nowhere in the other equation!). All $\nabla\cdot(\nabla\times G)=0$ says is that $\nabla\times G$ is some divergence-free field, not necessarily $F$ - in fact, $\nabla\cdot(\nabla\times G)=0$ is true for all (bona fide functions) $G$, not just the $G$ given in the problem, You are making this way too complicated: all you're supposed to do is calculate $\nabla\times G$ and see that you get $F$. That's what the equation $F=\nabla\times G$ means, after all. – anon Sep 21 '22 at 15:38
  • I've already tried to show that $F = \nabla \times G$, but I'm not sure how to apply hypothesis to the third entry. Any hint? – Blue Tomato Sep 21 '22 at 15:52
  • I don't know what "apply the hypothesis to the third entry" means. Did you calculate $\nabla\times G$? If so, what did you get? If not, what's stopping you? – anon Sep 21 '22 at 15:54
  • @runway44 I wrote down the second try – Blue Tomato Sep 21 '22 at 15:55
  • It is true that if $F_1$ were a constant then $\int_0^zF_1tdt$ would be $F_1z^2/2$, but $\int_0^z F_1(x,y,t)dt$ is very much not $F_1(x,y,z^2/2)$. That would be a big conceptual red flag in calc I, let alone calc III. Plus your $z^2/2$s magically disappeared with no explanation, which would be another big error, but we can leave that alone. Instead you use the fundamental theorem of calculus to differentiate the integrals. Also I think in your definition of $G_1$ the second integrand should be $F_3$ not $F_2$ (otherwise it's a typo in the original). – anon Sep 21 '22 at 16:08

1 Answers1

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So we've defined $G$ to be

$$ \begin{cases} G_1 & = \displaystyle \int_0^z F_2(x,y,t)\,\mathrm{d}t - \int_0^y F_3(x,t,0)\,\mathrm{d}t \\[5pt] G_2 & = \displaystyle -\int_0^z F_1(x,y,t)\,\mathrm{d}t \\[5pt] G_3 & = 0 \end{cases} $$

And $\nabla\times G=\displaystyle\Big(\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z}\,,\,\frac{\partial G_1}{\partial z}-\frac{\partial G_3}{\partial x}\,,\,\frac{\partial G_2}{\partial x}-\frac{\partial G_1}{\partial y}\Big)$. The first component is

$$ \frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z} \,=\, 0+\frac{\partial}{\partial z}\int_0^z F_1(x,y,t)\,\mathrm{d}t \,=\, F_1(x,y,z), $$

by the fundamental theorem of calculus. The second component is

$$ \frac{\partial G_1}{\partial z}-\frac{\partial G_3}{\partial x} \,=\, \frac{\partial}{\partial z}\left(\int_0^z F_2(x,y,t)\,\mathrm{d}t - \int_0^y F_3(x,t,0)\,\mathrm{d}t\right)-\frac{\partial }{\partial x}(0) $$

$$ =\, \big(F_2(x,y,z)-0\big)-0 ~=~ F_2(x,y,z), $$

again using the fundamental theorem. The third component is more tricky:

$$ \frac{\partial G_2}{\partial x}-\frac{\partial G_1}{\partial y} ~=$$

$$\frac{\partial}{\partial x}\left(-\int_0^z F_1(x,y,t)\,\mathrm{d}t\right) - \frac{\partial}{\partial y}\left(\int_0^z F_2(x,y,t)\,\mathrm{d}t - \int_0^y F_3(x,t,0)\,\mathrm{d}t\right) $$

$$ \int_0^z -\left(\frac{\partial F_1}{\partial x}(x,y,t)+\frac{\partial F_2}{\partial y}(x,y,t)\right)\,\mathrm{d}t +F_3(x,y,0), $$

which follows from combining the first two integrals and using the fundamental theorem of calculus on the third. We are given that $\nabla\cdot F=0$, which means we can solve for $\partial F_3/\partial z$ from

$$ \frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}=0 $$

and substitute it in for our first integrand. So the third component of $\nabla\times G$ becomes

$$ \int_0^z \frac{\partial F_3}{\partial z}(x,y,t)\,\mathrm{d}t+F_3(x,y,0) = \big(F_3(x,y,z)-F_3(x,y,0)\big)+F_3(x,y,0) $$

$$ = F_3(x,y,z) $$

and we are done.

anon
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  • I found one of many points of confusion I had: the Leibniz integral rule for differentiation under the integral sign. You interchanged the partial derivates and the integral: $-\frac{\partial}{\partial x} \int_{0}^{z} F_1(x,y,t) dt = -\int_{0}^{z} \frac{\partial F_1}{\partial x} (x,y,t) dt$. I didn't know that was possible. Thanks so much for helping me! – Blue Tomato Sep 21 '22 at 17:30