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I am currently reading "Variational Problems in Geometry" by Seiki Nishikawa and don't understand the last paragraph in Section 4.1 after Proposition 4.4 about the heat flow method.

Some preparation from the book:

We consider, for a map $u: M \times [0,T) \to N$, the following initial value problem of a system of nonlinear parabolic partial differential equations $$ \begin{align} &\frac{\partial u}{\partial t}(x,t) = \tau(u(x,t)), \quad (x,t) \in M \times (0,T) \\\\ &u(x,0) = f(x). \end{align} $$ Here, $T > 0$ and $f \in C^\infty(M,N)$ is a map given as the initial condition. Also we assume that $u$ is continuous in $M \times [0,T)$ and is $C^\infty$ in $M \times (0,T)$; namely, $$ u \in C^0(M \times [0,T),N) \cap C^\infty(M \times (0,T), N). $$

For a given solution to the above parabolic equation for harmonic maps, set $u_t(x) = u(x,t)$ and define $$ \begin{align} &e(u_t) = \frac{1}{2}\vert du_t \vert^2 \\\\ &E(u_t) = \int_M e(u_t) d\mu_g \\\\ &\kappa(u_t) = \frac{1}{2}\Big\vert\frac{\partial u_t}{\partial t}\Big\vert^2 \\\\ &K(u_t) = \int_M \kappa(u_t) d\mu_g \end{align} $$

Proposition 4.4 Let $u: M \times [0,T) \to N$ be a solution to the parabolic equation for harmonic maps, and set $u_t(x) = u(x,t)$. Then the following hold in $M \times (0,T)$:
(1) $E(u_t)$ is a monotone nonincreasing fucntion. Namely, $$ \frac{d}{dt} E(u_t) = -2K(u_t) \leq 0. $$ (2) If $N$ is of nonpositive curvature $K_N \leq 0$, then $$ \frac{d^2}{dt^2}E(u_t) = -2\frac{d}{dt}K(u_t) \geq 0. $$ Namely, $E(u_t)$ is a convex function and $K(u_t)$ is a monotone nonincreasing function.

Now to the paragraph in question:

Proposition 4.4 implies the following. Let $N$ be of nonpositive curvature $K_N \leq 0$. Further assume that the parabolic equation for harmonic maps has a solution $u: M \times [0,\infty) \to N$ for $T = \infty$. Noting (1) from Proposition 4.4 and $E(u_t) \geq 0$, $K(u_t) \to 0$ must hold as $t \to 0$ for $u_t(x) = u(x,t)$. Hence, $\frac{\partial u_t}{\partial t} \to 0$ must hold as $t \to 0$. However, since $u$ is a solution to the parabolic equation for harmonic maps, this implies that $\tau(u_t) \to 0$ holds as $t \to 0$. Thus, when $t \to \infty$, $u_t$ converges to $u_\infty \in C^\infty(M,N)$. Consequently, if $\tau(u_t) \to \tau(u_\infty), \tau(u_\infty) = 0$ must hold. In other words, $u_t$ converges to a harmonic map $u_\infty$.

  • Why does $K(u_t) \to 0$ if $t \to 0$ and why are we looking at $t \to 0$ in general?
  • How do we know that when $t \to \infty$, $u_t$ converges to $u_\infty \in C^\infty(M,N)$?
  • Why must $\tau(u_\infty) = 0$ if $\tau(u_t) \to \tau(u_\infty)$?
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  • $\begingroup$ Looks like a typo to me. I think it's supposed to be $t \rightarrow \infty$. $\endgroup$
    – Deane
    Commented Dec 20, 2023 at 15:34
  • $\begingroup$ Yes, that makes a lot more sense and answers the problems I had with the paragraph. $\endgroup$ Commented Dec 20, 2023 at 15:57

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As @Deane mentioned in the comments, I am quite sure now that it is a typo and should be $t \to \infty$. To answer my own questions (now with $t \to \infty$):

Why does $K(u_t) \to 0$ if $t \to \infty$?

From the definition of $K(u_t)$ we know that $K(u_t) \geq 0$ and from Proposition 4.4 we know that $K(u_t)$ is a monotone nonincreasing function. Therefore, $K(u_t) \to 0$ as $t \to \infty$.

How do we know that when $t \to \infty$, $u_t$ converges to $u_\infty \in C^\infty(M,N)$?

Since $u_\infty = u(x, \infty)$ is exactly the solution $u: M \times [0,\infty) \to N$ for $T = \infty$ of the parabolic equation for harmonic maps this makes now total sense for $t \to \infty$. I was just confused because first the paragraph stated all the results for $t \to 0$ and then it switched to $t \to \infty$.

Why must $\tau(u_\infty) = 0$ if $\tau(u_t) \to \tau(u_\infty)$?

Same thing as above. Since $\tau(u_t) \to 0$ and $\tau(u_t) \to \tau(u_\infty)$ as $t \to \infty$ it must hold that $\tau(u_\infty) = 0$ because the limit is unique.

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