Critical O(d)-equivariant biharmonic maps

Abstract

We study O(d)-equivariant biharmonic maps in the critical dimension. A major consequence of our study concerns the corresponding heat flow. More precisely, we prove that blowup occurs in the biharmonic map heat flow from $B^4(0, 1)$ into $S^4$. To our knowledge, this was the first example of blowup for the biharmonic map heat flow. Such results have been hard to prove, due to the inapplicability of the maximum principle in the biharmonic case. Furthermore, we classify the possible O(4)-equivariant biharmonic maps from $\mathbf {R}^4$ into $S^4$, and we show that there exists, in contrast to the harmonic map analogue, equivariant biharmonic maps from $B^4(0,1)$ into $S^4$ that wind around $S^4$ as many times as we wish. We believe that the ideas developed herein could be useful in the study of other higher-order parabolic equations.

Appendix

Proof

(Lemma 4) (2) $\Longrightarrow$ (1): This direction is trivial.

(1) $\Longrightarrow$ (2): We set $\widetilde{\varPhi }^0_i(s) = (-1)^{i+1} \varPhi ^0_i(-s)$ for $i \in \{1,2,3,4\}$. Observe that $\varPhi ^0$ solving (13) is equivalent to $\widetilde{\varPhi }^0$ solving (13). Therefore, after relabeling $\widetilde{\varPhi }^0$ as $\varPhi ^0$ our statement is equivalent to showing that if $s_0 \in \mathbf {R}$, $\varPhi ^0 : [s_0, \infty ) \rightarrow \mathbf {R}^4$ solves (13), and

$$\begin{aligned} \lim _{s \rightarrow \infty } \varPhi ^0_1(s) = 0, \end{aligned}$$

then

$$\begin{aligned} \lim _{s \rightarrow \infty } \varPhi ^0(s) = 0. \end{aligned}$$

It is easy to show that if $x \in C^2([s_0,\infty );\mathbf {R})$, $x(s) \rightarrow 0$ as $s \rightarrow \infty$, and $|\partial _s^2 x(s)| \le C$ for all $s \in [s_0, \infty )$ then $\partial _sx(s) \rightarrow 0$ as $s \rightarrow \infty$. We use this fact, which we call (P1), repeatedly in what follows.

First observe that there cannot exist an $s_1 \in [s_0, \infty )$ such that $|\varPhi ^0_3(s)| \ge 1$ for all $s \in [s_1, \infty )$. Indeed, if there were such an $s_1$ then eventually $\varPhi ^0_4(s)$ would be the same sign as $\varPhi ^0_3$ after which we could apply Lemma 10 and obtain a contradiction.

Therefore, if there is an $s_1 \in [s_0, \infty )$ such that $|\varPhi ^0_3(s_1)| \ge 1$ then there must be an $s_2 > s_1$ such that $|\varPhi ^0_3(s)| < 1$ for all $s \in [s_2, \infty )$, or else we could apply Lemma 10 and obtain a contradiction. Therefore, $\varPhi ^0_3$ is bounded on $[s_0, \infty )$.

Now we proceed to show, one by one, that $\lim _{s\rightarrow \infty } \varPhi ^0_i(s) = 0$ for $i \in \{2,3,4\}$. First we look at $\varPhi ^0_2$. The fact that $\varPhi ^0_1 \rightarrow 0$ as $s \rightarrow \infty$, the boundedness of $\varPhi ^0_3$, and (P1) yield

$$\begin{aligned} \lim _{s \rightarrow \infty } \varPhi ^0_2(s) = 0. \end{aligned}$$

Next, we look at $\varPhi ^0_3$. Hoping for a contradiction, we assume that $\varPhi ^0_3(s) \not \rightarrow 0$ as $s \rightarrow \infty$. From (P1) we know that $\varPhi ^0_4$ is unbounded, that is, there exists a monotone increasing sequence $\{s_i\}_{i \in \mathbf {N}} \subset [s_0, \infty )$ diverging to infinity such that $|\varPhi ^0_4(s_i)| \rightarrow \infty$. From (13) and the fact that $|(\varPhi ^0_1(s), \varPhi ^0_2(s), \varPhi ^0_3(s))| \le C$ on $[s_0,\infty )$, we have that $|\partial _s\varPhi ^0_4(s)| \le C$ on $[s_0, \infty )$. Therefore, $|\varPhi ^0_4(s)| \ge \frac{1}{2} |\varPhi ^0_4(s_i)|$ for $s \in \left[ s_i, s_i+\frac{1}{2C} |\varPhi ^0_4(s_i)|\right]$. Observe that over this interval $\varPhi ^0_4$ is non-vanishing. Therefore, there exists an $s \in [s_0,\infty )$ such that $|\varPhi ^0_3(s)| \ge 1$ and $\varPhi ^0_3(s)$ has the same sign as $\varPhi ^0_4(s) \ne 0$. Lemma 10 then yields a contradiction, hence

$$\begin{aligned} \lim _{s \rightarrow \infty } \varPhi ^0_3(s) = 0. \end{aligned}$$

Finally, we look at $\varPhi ^0_4$. Since

$$\begin{aligned} \lim _{s \rightarrow \infty } (\varPhi ^0_1(s), \varPhi ^0_2(s), \varPhi ^0_3(s)) = 0, \end{aligned}$$

from (13), we have $\partial _s\varPhi ^0_4(s) \rightarrow 0$ as $s \rightarrow \infty$. Now (P1) gives us

$$\begin{aligned} \lim _{s \rightarrow \infty } \varPhi ^0_4(s) = 0. \end{aligned}$$

$\square$

Remark 6

Observe that in the above proof we make use of Lemma 10. Our argument would be circular if Lemma 10 depended upon Lemma 4. By closely examining the proof of Lemma 10, it is clear that this is not the case.

Proof

(Lemma 6) We prove this lemma for $c_0 = \frac{99}{100}$. It is elementary to compute

$$\begin{aligned} \min _{x \in \mathbf {R}} f(x) = -\frac{1}{12} \sqrt{169+38 \sqrt{19}}. \end{aligned}$$

Since f is an odd function, we have

$$\begin{aligned} |f(y)| \le \frac{1}{12} \sqrt{169+38 \sqrt{19}} \le 2\quad \text{ for } \text{ all } y \in \mathbf {R}. \end{aligned}$$

We differentiate:

$$\begin{aligned} \partial _x Q(x;f(y)) = \frac{3 (9+21 \cos (2x)+2 f(y) \sin (2x))}{(7+3 \cos (2x))^2}. \end{aligned}$$

By periodicity, what we wish to prove is that $\partial _xQ(x;f(y)) \le c_0$ for all $x \in \left( -\frac{\pi }{2}, \frac{\pi }{2}\right]$ and $y \in \mathbf {R}$.

Firstly, $\partial _xQ\left( \frac{\pi }{2};f(y)\right) < 0$ which means we may restrict our attention to $x \in \left( -\frac{\pi }{2}, \frac{\pi }{2}\right)$. We use Weierstrass’ substitution:

$$\begin{aligned} \sin (2x) \mapsto \frac{2t}{1+t^2} \text{ and } \cos (2x) \mapsto \frac{1-t^2}{1+t^2} \quad \text{ for } t \in \mathbf {R}. \end{aligned}$$

This transforms the problem into showing that

$$\begin{aligned} \frac{3 (15+2 (\widetilde{f}-3 t) t) \left( 1+t^2\right) }{2 \left( 5+2 t^2\right) ^2} \le c_0, \end{aligned}$$

for all $t \in \mathbf {R}$ and $\widetilde{f} \in [-2,2]$. It suffices to show

$$\begin{aligned} -45+50 c_0+(-27+40 c_0) t^2+(18+8 c_0) t^4-12 (t+ t^3) \ge 0, \end{aligned}$$

(33)

for all $t \in \mathbf {R}$. Next we prove this.

We substitute $c_0 = \frac{99}{100}$ into (33) and let p be the polynomial on the left hand side of the resulting expression, that is,

$$\begin{aligned} p(t) = \frac{648}{25} t^4-12 t^3+\frac{63}{5} t^2-12 t+\frac{9}{2}. \end{aligned}$$

We calculate:

$$\begin{aligned} p'\left( \frac{2}{5}\right) < 0, \; p'\left( \frac{43}{100}\right) > 0, \quad \text{ and } \quad p''(t) > 0, \end{aligned}$$

for $t \in \mathbf {R}$. Therefore, p is convex with its unique global minimum occurring somewhere in $\left[ \frac{2}{5}, \frac{43}{100}\right]$. We use this to estimate:

$$\begin{aligned} \min _{t \in \mathbf {R}} p(t) \ge \frac{648}{25} \left( \frac{2}{5}\right) ^4-12 \left( \frac{43}{100}\right) ^3+\frac{63}{5} \left( \frac{2}{5}\right) ^2-12 \left( \frac{43}{100}\right) +\frac{9}{2} > 0. \end{aligned}$$

This is what we wished to show. $\square$