5

Let $I=[0,1]$.

Define:

$H_0 = L^2(I,\mathbb{R}^3)$ with inner product $\langle u,v \rangle_0 = \int_0^1 \langle u(t), v(t) \rangle \text{ dt}$

$H_1 = W^{1,2}(I, \mathbb{R}^3)$ with inner product $\langle u,v \rangle_1 = \langle u,v \rangle_0 + \langle u',v' \rangle_0$.

and the submanifold $\Omega = \{\omega \in H_1\ \mid \ \|\omega(t)\| = 1 \text{ and } w(0) = w(1) \}$ of $H$.

I have trouble showing the following "elementary" inequality between covariant and ordinary derivative.

If $A$ is an $H_1$ bounded subset of $\Omega$ then there exists a constant $C$ such that $$ \| u\|_1 \leq C\left( \| \frac{Du}{dt} \|_0 + \|u\|_0 \right) $$ for any $w \in A$ and any $u \in H_1(w^*TS^2)$.

The covariant derivative for some $u \in H_1(w^*TS^2)$ along a curve $w$ is that of the standard metric on $S^2$ and can be written as $$ \frac{Du}{\partial t} = u'(t) - \langle u'(t), w(t) \rangle w(t).$$

I've tried to keep the outline of the specific setting as brief and precise as possible, if there is anything unclear about it, feel free to comment.

What I've tried: Squaring both sides of the inequality and trying to get the $\| \|_1$ and $\| \|_0$ terms of $w$ together, since subtracting would only leave the $L^2$-norm of the ordinary derivative, which might help. When squaring I also used that $$ \langle w(t), \frac{Du}{dt}(t) \rangle = 0$$ for any $t$ since $\frac{Du}{dt}(t)$ lies in the tangential plane $S^2_{w(t)}$ and is thus orthogonal to $w(t)$. I can't get any further though and I lack (forgot) some general things to look out for when showing such inequality.

Any help would be appreciated.

Community
  • 1
Nhat
  • 1,354

1 Answers1

2

Let $n(t)=\langle u'(t),w(t)\rangle w(t)$ so $u'(t)=Du/dt(t)+n(t).$ We want to bound $\|u\|_1^2=\|u\|_0^2+\|Du/dt\|_0^2+\|n\|_0^2$ by $C(\|Du\|_0+\|u\|_0)^2$ for some $C.$ The only troublesome term is $\|n\|_0^2.$ Differentiating $\langle u,w\rangle=0$ gives $n(t)=\langle u'(t),w(t)\rangle w(t)=-\langle u(t),w'(t)\rangle w(t),$ so

$$\|n\|_0^2=\int_0^1\langle u(t),w'(t)\rangle^2dt\leq\int_0^1\| u(t)\|^2\|w'(t)\|^2dt\leq C'\|u\|_\infty^2$$ for some $C',$ using Cauchy-Schwarz and the assumption that $\|w\|_1$ is bounded. Here $\|u\|_\infty$ is the essential supremum of $\|u\|.$

Between any $0\leq a<b\leq 1,$ the change in $\|u\|^2$ is $$\int_a^b\frac{d}{dt}\|u\|^2dt=2\int_a^b\langle u(t),u'(t)\rangle dt=2\int_a^b\langle u(t),Du/dt(t)\rangle dt.$$ To justify the first expressions use a smooth approximation in $W^{1,2}.$ The second equality uses the fact that $n(t)$ is orthogonal to $u(t)$ (note $u(t)$ lies in the tangent space to $S^2$ at $w(t)$). The right-hand-side is at most $\|Du/dt(t)\|_0^2+\|u\|_0^2.$ And the average value of $\|u\|$ is $\int u(t)dt\leq \|u\|_0,$ so the essential supremum of $\|u\|$ is at most $C(\|u\|_0+\|Du/dt\|_0)$ for some $C.$

Dap
  • 25,701
  • 20
  • 52
  • Thanks for the detailed input. There are some inconsistencies in your answer, however, making it harder to follow: The definition of $n(t)$ your using for your calculations seem to vary from the one at the top. Also how is $\langle u,w \rangle = 0$? It’s the covariant derivative of $u$ that is orthogonal to $w$ isn’t it? Edit: I made a mistake looking at your calculations with $n$. The second point still stands, i will check the first one now. – Nhat Jan 26 '19 at 18:32
  • @Nhat: I've corrected a couple of missing $w$'s where I used $n(t),$ and added "note $u(t)$ lies in the tangent plane to $S^2$ at $w(t)$" - at least that's how I understand it – Dap Jan 26 '19 at 18:37
  • I’m not fluent in differential geometry enough so you may be right. Is the tangent plane at $w(t)$ the one with $w(t)$ at the origin? Or is it the plane with origin at $0$ that’s tangent to $S^2$ after translation? I feel like i’m misunderstanding the difference between tangent space and tangent plane. – Nhat Jan 26 '19 at 18:42
  • @Nhat: sorry for the confusion, I meant tangent space not plane. – Dap Jan 26 '19 at 18:42
  • So the picture i have in mind of a curve $u \in H_1(w^*TS^2)$ being a curve that follows $w$ through its tangent planes is wrong then. What a shame - but it makes sense, I just checked the definition by the author (Curve Straightening Flow by Langer / Singer) and it says $u$ is closed at $0$ which only makes sense with your interpretation. I will check your calculations when I’m at home and then mark it as answered. Tyvm! – Nhat Jan 26 '19 at 18:47