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Background:

I am reading the book Variational Problems in Geometry by Seiki Nishikawa. The main purpose of this book is to prove the existence of harmonic maps $M\to N$ between two compact Riemannian manifolds $M$ and $N$ with the target manifold being nonpositively curved. Since a harmonic map is defined as a minimizer for the energy functional $E$, the idea is to deform an existing map $u:M\to N$ along the gradient lines of $E$, which is equivalent to solving an equation: $$\frac{\partial u_t}{\partial t}=\tau(u_t)\quad\quad\text{Eq.(1)}$$ where $u_t$ is a smooth variation of some given $u_0$, and $\tau$ denotes the tension field (along this direction, $E$ decreases fastest). By embedding $N$ into $\mathbb R^q$ using the Nash embedding theorem, we can regard $N$ as a submanifold and $u$ as a vector-valued function to simplify certain statements. After this embedding, the above equation takes a new form: $$(\Delta-\partial_t)u_t=\Pi(u_t)\quad\quad\text{Eq.(2)}$$ I'll omit the definition of $\Pi(u_t)$. Just know that they are equivalent in the sense that their solutions coincide, if given identical initial conditions. The difference is:

  • Eq.(1) is intrinsic and is an equation in $\Gamma(u_t^{-1}TN)$;
  • Eq.(2) is an equation in $\mathbb R^q$.

Theorem:

One of the main theorems is the existence of a solution to any one of the equations above defined for $t\in[0,+\infty)$, given some $u_0$ as the initial condition. Suppose now we have already established:

  • local (w.r.t time) existence of a solution
  • certain estimates

The proof goes as follows:

Proof. (The part where there's no problem, so I'll give a sketch only) Since we have established the local existence, for a given $u_0$, we can find a smooth $u:M\times[0,T)\to N$ for some $T>0$ such that it solves (1) and (2) and $u(\cdot,0)=u_0$. Now suppose $T_0$ is the supremum of all such $T$. We wish to prove $T_0=+\infty$. If not, we take an increasing sequence $t_i\to T_0$. By certain estimates, we know that $\{u_{t_i}\}$ and $\{\partial_tu_{t_i}\}$ are uniformly bounded and equicontinuous, respectively in the Holder spaces $C^{2+\alpha}(M,\mathbb R^q)$ and $C^{\alpha}(M,\mathbb R^q)$, for some $0<\alpha<1$.

Proof. (The part where I am confused. I'll write exact words) By the Ascoli-Arzela theorem, there exists a subsequence $\{t_{i_k}\}$ of $\{t_i\}$ and functions$^1$ $$u(\cdot,T_0)\in C^{2+\alpha}(M,\mathbb R^q)\quad\text{and}\quad\partial_t u(\cdot,T_0)\in C^\alpha(M,\mathbb R^q)$$ such that the subsequences $$\{u(\cdot,t_{i_k})\}\quad\text{and}\quad\{\partial_tu(\cdot,t_{i_k})\}$$ respectively, converge uniformly to $u(\cdot,T_0)$ and $\partial_tu(\cdot,T_0)$, as $t_{i_k}\to T_0$. Since for each $t_{i_k}$, we have $$\partial_tu(\cdot,t_{i_k})=\tau(u(\cdot,t_{i_k}))\quad\quad\text{Eq.(3)}$$ we also get at$^2$ $T_0$ $$\partial_tu(\cdot,T_0)=\tau(u(\cdot,T_0))\quad\quad\text{Eq.(4)}$$ consequently$^3$, we see that (1) has a solution in $M\times[0,T_0]$. Using $u(\cdot,T_0)$ again as the initial condition to solve (1), we extend the solution to $M\times[0,T_0+\epsilon)$ for some $\epsilon>0$, contradicting $T_0$ being the supremum. Hence $T_0=\infty$.


Questions:

  1. How do we know that the limits do not depend on the choice of $t_i$. I think it can be argued using the uniformly boundedness and equicontinuity. Am I right?

  2. How can we go from (3) to (4)? Eq. (3) is clearly obtained by (1), which is an equation in $\Gamma(u_t^{-1}TN)$. However, (3)$\implies$(4) would require the convergence of $\partial_tu(\cdot,t_{i_k})$ to $\partial_tu(\cdot,T_0)$. But this convergence is only in $C^{\alpha}(M,\mathbb R^q)$, not in $\Gamma(u_t^{-1}TN)$. Although I think I can fix this by writing them in the form of (2) to start with.

  3. How can we conclude from (4) that (1) has a solution in $M\times[0,T_0]$. Note that $\partial_tu(\cdot,T_0)$ is so far only a notation for the limit of $\partial_tu(\cdot,t_{i_k})$, we have not proved that it is actually the (one-sided) derivative of $u$ at $t=T_0$. How can I fix this?

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$\newcommand{\R}{\mathbb{R}}\newcommand{\pa}{\partial}$Edit: The answer is now LaTeXified.

Below are my notes on this. I reworked the proof:

Proof. Let $S:=\big\{T\in[0,\infty):$ the equation has a solution in $C^{2+\alpha,1+\alpha/2}(M\times[0,T],N)\big\}$. Let $T_0:=\sup S$. By existence of local solution, $T_0>0$. We claim that $T_0=\infty$.

Suppose $T_0<\infty$. By uniqueness of solution and definition of $T_0$, we have a solution $u\in C^{2,1}(M\times[0,T_0),N)$. Take $\alpha<\alpha'<1$. By the \textit{a priori} estimate above, $u_t$ is uniformly bounded in $t$ in $C^{2+\alpha'}(M,\R^L)$.

Define $$u(x,T_0):=\int_0^{T_0}\pa_tu(x,t)\,dt+u(x,0).$$ For any sequence $t_k\nearrow T_0$, $(u_{t_k})_{k=1}^\infty$ has a subsequence that converges in $C^{2+\alpha}(M,N)$ by Arzelà–Ascoli, and its limit is necessarily $u_{T_0}$. Thus $u_{T_0}\in C^{2+\alpha}(M,N)$, and any such sequence must in fact converge to $u_{T_0}$ in $C^{2+\alpha}(M,N)$. In other words, $u\in C^{2+\alpha,0}(M\times[0,T_0],N)$, or equivalently, $t\mapsto u_t$ is continuous as a map $[0,T_0]\to C^{2+\alpha}(M,N)$.

Since $\pa_tu_t=\tau(u_t)$, we see that $t\mapsto\pa_tu_t$ has a continuous extension $[0,T_0]\to C^{\alpha/2}(M,\R^L)$. So in fact $u\in C^{2+\alpha,1+\alpha/2}(M\times[0,T_0],N)$, i.e., $T_0\in S$. Now, existence of local solution implies that $u$ can be extended to a solution on $[0,T_0+\varepsilon]$, which is a contradiction. $\square$

The a priori estimate refers to the one as given in the book: For $0<\alpha<1$, $$\sup_{t\in[0,T)}\Big(\big\|u_t\big\|_{C^{2+\alpha}(M,\R^L)}+\big\|\pa_tu_t\big\|_{C^\alpha(M,\R^L)}\Big)\leq C(M,N,f,\alpha).$$

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  • 1
    $\begingroup$ Can you change the images into LaTeX? Doing so will enable search engines to index your answer properly. $\endgroup$ – Mateusz Kwaśnicki Feb 14 at 6:56
  • $\begingroup$ @MateuszKwaśnicki The answer is now LaTeXified. $\endgroup$ – Colescu Feb 14 at 8:49

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