Neck analysis for biharmonic maps

Abstract

In this paper, we study the blow up of a sequence of (both extrinsic and intrinsic) biharmonic maps in dimension four with bounded energy and show that there is no neck in this process. Moreover, we apply the method to provide new proofs to the removable singularity theorem and energy identity theorem of biharmonic maps.

Appendix

Proof of Theorem 2.1

It will be convenient to assume that $\overline{u}=0$. Since u is a biharmonic map, then it satisfies the Euler–Lagrange

$$\begin{aligned} \triangle ^2u=\nabla ^3u\#\nabla u+\nabla ^2 u\#\nabla ^2 u+\nabla ^2 u\#\nabla u\#\nabla u+\nabla u\#\nabla u\#\nabla u\#\nabla u. \end{aligned}$$

Let $0<\sigma <1$ and $\sigma '=\frac{1+\sigma }{2}$, take cut-off function $\varphi \in C_0^\infty (B_{\sigma '})$ satisfying $\varphi \equiv 1$ in $B_\sigma$, $|\nabla \varphi |\le \frac{4}{1-\sigma }$.

Direct computation shows that

$$\begin{aligned} \triangle ^2(\varphi u)= & {} \triangle (\varphi \triangle u+2\nabla u \nabla \varphi +u\triangle \varphi )\\= & {} \varphi \triangle ^2u+4\nabla \triangle u\nabla \varphi +2\triangle u\triangle \varphi +4\nabla ^2u\nabla ^2\varphi +4\nabla u\nabla \triangle \varphi +u\triangle ^2\varphi \\= & {} (\nabla ^3u\#\nabla u+\nabla ^2 u\#\nabla ^2 u+\nabla ^2 u\#\nabla u\#\nabla u+\nabla u\#\nabla u\#\nabla u\#\nabla u)\varphi \\&+\,\nabla ^3 u\#\nabla \varphi +\nabla ^2u\#\nabla ^2\varphi +\nabla u\#\nabla ^3\varphi +u\nabla ^4\varphi \\= & {} (\nabla ^3(\varphi u)\#\nabla u+\nabla ^2 (\varphi u)\#\nabla ^2 u+\nabla ^2 u\#\nabla u\#\nabla (\varphi u)+\nabla u\#\nabla u\#\nabla u\#\nabla (\varphi u))\\&+\,\nabla ^3 u\#\nabla \varphi +\nabla ^2u\#\nabla ^2\varphi +\nabla u\#\nabla ^3\varphi +u\nabla ^4\varphi +\nabla ^2u\#\nabla u\#\nabla \varphi +\nabla ^2\varphi \#\nabla u\#\nabla u\\&+\,\nabla u\#\nabla u\#\nabla u\#\nabla \varphi . \end{aligned}$$

Assume first that $1<p<\frac{4}{3}$. By the standard $L^p$ theory, we have

$$\begin{aligned} \Vert \nabla ^4(\varphi u)\Vert _{L^p(B_1)}\le & {} C\left( \Vert \nabla u\Vert _{L^4(B_1)}\Vert \nabla ^3(\varphi u)\Vert _{L^{\frac{4p}{4-p}}(B_1)} +\Vert \nabla ^2 u\Vert _{L^2(B_1)}\Vert \nabla ^2(\varphi u)\Vert _{L^{\frac{4p}{4-2p}}(B_1)}\right. \\&+\,\Vert \nabla ^2 u\Vert _{L^2(B_1)}\Vert \nabla u\Vert _{L^4(B_1)}\Vert \nabla (\varphi u)\Vert _{L^{\frac{4p}{4-3p}}(B_1)}+\Vert \nabla u\Vert ^3_{L^4(B_1)}\Vert \nabla (\varphi u)\Vert _{L^{\frac{4p}{4-3p}}(B_1)} \\&+\,\frac{\Vert \nabla ^3 u\Vert _{L^p(B_{\sigma '})}}{1-\sigma }+\frac{\Vert \nabla ^2 u\Vert _{L^p(B_{\sigma '})}}{(1-\sigma )^2} +\frac{\Vert \nabla u\Vert _{L^p(B_{\sigma '})}}{(1-\sigma )^3}\\&+\,\frac{\Vert u\Vert _{L^p(B_{\sigma '})}}{(1-\sigma )^4} +\frac{\Vert \nabla ^2 u\#\nabla u\Vert _{L^p(B_{\sigma '})}}{1-\sigma }+\frac{\Vert \nabla u\#\nabla u\Vert _{L^p(B_{\sigma '})}}{(1-\sigma )^2}\\&\left. +\,\frac{1}{1-\sigma }\Vert \nabla u\#\nabla u\#\nabla u\Vert _{L^p(B_{\sigma '})} \right) , \end{aligned}$$

By the Sobolev embedding, if $\epsilon _0$ is sufficiently small, we get

$$\begin{aligned} \Vert \nabla ^4(\varphi u)\Vert _{L^p(B_1)}\le & {} C\big ( \frac{1}{1-\sigma }\Vert \nabla ^3 u\Vert _{L^p(B_{\sigma '})}\\&+\,\frac{1}{(1-\sigma )^2}\Vert \nabla ^2 u\Vert _{L^p(B_{\sigma '})} +\frac{1}{(1-\sigma )^3}\Vert \nabla u\Vert _{L^p(B_{\sigma '})}\\&+\,\frac{1}{(1-\sigma )^4}\Vert u\Vert _{L^p(B_{\sigma '})} +\frac{1}{1-\sigma }\Vert \nabla ^2 u\#\nabla u\Vert _{L^p(B_{\sigma '})}\\&+\,\frac{1}{(1-\sigma )^2}\Vert \nabla u\#\nabla u\Vert _{L^p(B_{\sigma '})}\\&+\,\frac{1}{1-\sigma }\Vert \nabla u\#\nabla u\#\nabla u\Vert _{L^p(B_{\sigma '})} \big ). \end{aligned}$$

Setting

$$\begin{aligned} \Psi _j=\sup _{0\le \sigma \le 1}(1-\sigma )^j\Vert \nabla ^j u\Vert _{L^p(B_{\sigma })} \end{aligned}$$

and noticing that $1-\sigma =2(1-\sigma ')$,$1<p<\frac{4}{3}$, we have

$$\begin{aligned} \Psi _4\le & {} C\left( \Psi _3+\Psi _2+\Psi _1+\Psi _0+\Vert \nabla ^2 u\#\nabla u\Vert _{L^p(B_1)}+\Vert \nabla u\#\nabla u\Vert _{L^p(B_1)}\right. \\&\left. +\,\Vert \nabla u\#\nabla u\#\nabla u\Vert _{L^p(B_1)}\right) \\\le & {} C\left( \Psi _3+\Psi _2+\Psi _1+\Psi _0+ \Vert \nabla ^2u\Vert _{L^2(B_1)}+\Vert \nabla u\Vert _{L^4(B_1)}\right) . \end{aligned}$$

Using the interpolation inequality as in Sect. 3.2, we get

$$\begin{aligned} \Psi _4\le & {} C\left( \Psi _0+\Vert \nabla ^2u\Vert _{L^2(B_1)}+\Vert \nabla u\Vert _{L^4(B_1)}\right) \\\le & {} C\left( \Vert \nabla ^2u\Vert _{L^2(B_1)}+\Vert \nabla u\Vert _{L^4(B_1)}\right) . \end{aligned}$$

We start with $p=\frac{16}{13}$. The above argument implies that

$$\begin{aligned} \left\| u\right\| _{W^{4,\frac{16}{13}}(B_{7/8})}\le C (\left\| \nabla ^2 u\right\| _{L^2(B_1)} +\left\| \nabla u\right\| _{L^4}(B_1)). \end{aligned}$$

The Sobolev embedding theorem implies

$$\begin{aligned} \left\| \nabla ^3 u\right\| _{L^{\frac{16}{9}}(B_{7/8})} + \left\| \nabla ^2 u\right\| _{L^{\frac{16}{5}}(B_{7/8})} +\left\| \nabla u\right\| _{L^{16}(B_{7/8})}\le C (\left\| \nabla ^2 u\right\| _{L^2(B_1)} +\left\| \nabla u\right\| _{L^4}(B_1)). \end{aligned}$$

With this, we can bound the $L^{\frac{8}{5}}$ norm of the right hand side of the Euler–Lagrange equation. The interior $L^p$ estimate then shows u is bounded in $W^{4,\frac{8}{5}}$ in $B_{3/4}$. The lemma is then proved by bootstrapping method.$\square$