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.

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!

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$ ?