## If $\nabla \cdot \vec{F} = 0$ show that $\vec{F}=\nabla \times \int_{0}^{1} \vec{F}(tx,ty,tz)\times(tx,ty,tz)dt$.

Suppose $$\vec{F}$$ is a vector field on $$\mathbb{R}^3$$ and $$\nabla \cdot \vec{F} = 0$$. Prove that:

$$\vec{F}=\nabla \times \int_{0}^{1} \vec{F}(tx,ty,tz)\times(tx,ty,tz)dt$$.

Tried doing it, but can’t seem to get it exactly right. Can anyone give an solution with working – naturally just doing 1st component should suffice.

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## Norm of $||(x-x^*,y-y^*,z-z^*) ||\leq ||\nabla J(x,y,z)||$

I want to prove that

$$||(x-x^*,y-y^*,z-z^*) ||\leq ||\nabla J(x,y,z)||,$$

where $$J(x,y,z)=\exp(\frac{x+y+z}{2016})+\frac{x^2+2y^2+3z^2}{2}$$ and $$(x^*,y^*,z^*)$$ is the minimum of $$J$$.

Also, I have $$||(x-x^*,y-y^*,z-z^*) ||^2\leq U^t H U$$ with $$H$$ the Hessian matrix of $$J$$ and $$U=(x-x^*,y-y^*,z-z^*).$$

Could you give me some clue to complete the proof?

Thanks!

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## Does $\nabla g=\omega(\cdot) g$ imply $\nabla$ is metric w.r.t a conformal rescaling of $g$?

This is a cross-post.

Let $$E$$ be a smooth vector bundle over a manifold $$M$$, where $$\text{rank}(E) > 1,\dim M > 1$$. Suppose that $$E$$ is equipped with a metric $$g$$ and an affine connection $$\nabla$$, such that $$\nabla_X g=\omega (X) g$$ for every $$X \in \Gamma(TM)$$. (Here $$\omega$$ is a one form).

Must $$\omega$$ be closed?

Clearly, $$\nabla$$ is metric-compatible ($$\nabla g=0$$) iff $$\omega=0$$. Moreover, $$\omega=d\phi$$ is exact if and only if $$\nabla s=0$$ where $$s=e^{-\phi}g$$, i.e. $$\nabla$$ is metric w.r.t a positive conformal rescaling of $$g$$. So, an alternative formulation of the question is the following:

Suppose that $$\nabla g=\omega (\cdot) g$$ for some $$\omega \in \Omega^1(M)$$. Must $$\nabla$$ be metric w.r.t a local conformal rescalings of $$g$$?

Differentiating $$\nabla g=\omega (\cdot) g$$, we get $$R(X,Y)g=d\omega(X,Y)g$$, so if $$\nabla$$ is flat then $$\omega$$ is closed.

I required $$\text{rank}(E) > 1$$, since if the rank is $$1$$, $$\nabla g$$ can always be written as $$\omega (\cdot) g$$ for a suitable $$\omega$$, so the assumption always holds, but I think that $$d\omega=0$$ does not always hold. Maybe this can be used to construct a counter example of higher rank by taking a direct sum of line bundles.

## How to treat an equation of the form $-\Delta u=G\cdot \nabla I(u)+f(u) ?$

There are plenty of variational techniques (direct methods of calculus of variations, mountain pass type theorems, Lusternik-Schnirelmann theory) to prove the existence of solutions of a semilinear elliptic equation of the form $$-\Delta u=f(u)$$ in $$H^1_0(\Omega)$$, under suitable hypothesis on $$f:\mathbb{R}\to\mathbb{R}$$, thanks to the fact that we can see weak solutions of this problem as the stationary points of the functional: $$I:H^1_0(\Omega)\to\mathbb{R}, u\mapsto\frac{1}{2}\|u\|_{H^1_0}^2-\int_\Omega\int_0^{u(x)}f(s)\operatorname{d}s\operatorname{d}x.$$

If $$G:\Omega\rightarrow\mathbb{R}^n$$, how can we treat the equation: $$-\Delta u=G\cdot\nabla I(u)+f(u),$$ or even, if $$g:\mathbb{R}^n\to\mathbb{R}$$, the equation: $$-\Delta u=g\left(\nabla I(u)\right)+f(u)?$$

If $$n=1$$, I saw in the $$G$$-case that we can transform the equation into another semilinear elliptic equation that hasn’t the dissipative term $$u’$$, with the same trick used in Sturm-Liouville theory, and so we can bring back this problem into the realm of the previous variational problem. However, what about the $$g$$-case if $$n=1$$? What about the $$G$$-case if $$n\ge2$$? Can we say anything about the $$g$$-case if $$n\ge 2$$?

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