Nonlinear stability of explicit self-similar solutions for the timelike extremal hypersurfaces in ${\mathbb {R}}^{1+3}$

Abstract

This paper is devoted to the study of the singularity phenomenon of timelike extremal hypersurfaces in Minkowski spacetime ${\mathbb {R}}^{1+3}$. We find that there are two explicit lightlike self-similar solutions to a graph representation of timelike extremal hypersurfaces in Minkowski spacetime ${\mathbb {R}}^{1+3}$, the geometry of them are two spheres. The linear mode unstable of those lightlike self-similar solutions for the radially symmetric membranes equation is given. After that, we show those self-similar solutions of the radially symmetric membranes equation are nonlinearly stable inside a strictly proper subset of the backward lightcone. This means that the dynamical behavior of those two spheres is as attractors. Meanwhile, we overcome the double roots case (the theorem of Poincaré can’t be used) in solving the difference equation by construction of a Newton’s polygon when we carry out the analysis of spectrum for the linear operator.

Appendix

In the appendix, we give the details on the proof of Lemma 3.5. Firstly, we recall a result of the existence of difference equation, which first established by Birkhoff and Trjitzinsk.

Proposition 5.1

(Birkhoff and Trjitzinsk [3]) The kth-order linear difference equation

$$\begin{aligned} a_0(n)u_n+a_1(n)u_{n+1}+\cdots +a_k(n)u_{n+k}=0,~~~a_0\not \equiv 0,~~a_k\not \equiv 0, \end{aligned}$$

with polynomial coefficients $a_i$ has precisely k linearly independent formal solutions of the general form

$$\begin{aligned} u_n=e^{Q(n)}n^r\sum _{i=0}^{\infty }n^{\frac{-i}{p}}\sum _{j=0}^mC_{i,j}\ln ^{j}n, \end{aligned}$$

(5.1)

where

$$\begin{aligned} Q(n)=\mu n\ln n+\sum _{s=0}^p\nu _s n^{\frac{s}{p}}, \end{aligned}$$

and $C_{i,j}$ and $\nu _s$ are coefficients, $r,p\in {\mathbb {N}}$, $\mu p\in {\mathbb {Z}}$ and $m\in {\mathbb {N}}\cap \{0\}$.

It is hard to get an exact expansion of solution to (3.23) by (5.1). So we have to use other method to analyze the asymptotic behavior of solutions to (3.23).

Let

$$\begin{aligned}&z_{n}^{(1)}=a_n,\\&z_{n}^{(2)}=a_{n+1}=z^{(1)}_{n+1},\\&z_{n}^{(3)}=a_{n+2}=z^{(2)}_{n+1},\\&z_{n}^{(4)}=a_{n+3}=z^{(3)}_{n+1}, \end{aligned}$$

then by (3.23), we have

$$\begin{aligned} z_{n+1}^{(4)}=(2-p_1(n))z^{(3)}_{n}-(1+p_2(n))z^{(1)}_{n}, \end{aligned}$$

where $z_n^{(i)}$$(i=1,2,3,4)$ depends on $\nu$ and $\kappa$.

Furthermore, let $z_n=(z^{(1)}_n,z^{(2)}_n,z^{(3)}_n,z^{(4)}_n)$, we have

$$\begin{aligned} \begin{aligned} z_{n+1}&=\left( \begin{array}{llll} 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1\\ -1-p_2(n)&{}\quad 0&{}\quad 2-p_1(n)&{}\quad 0 \end{array} \right) z_n\\&=({\mathcal {D}}+{\mathcal {T}}(n))z_n, \end{aligned} \end{aligned}$$

(5.2)

where

$$\begin{aligned} {\mathcal {D}}=\left( \begin{array}{llll} 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1\\ -1&{}\quad 0&{}\quad 2&{}\quad 0 \end{array} \right) ,~~ {\mathcal {T}}(n)=\left( \begin{array}{llll} 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0\\ -p_2(n)&{}\quad 0&{}\quad -p_1(n)&{}\quad 0 \end{array} \right) . \end{aligned}$$

Since the matrix ${\mathcal {D}}$ has eigenvalues $\lambda _1=\lambda _2=1$ and $\lambda _3=\lambda _4=-1$ (double), we should diagonalize the matrix ${\mathcal {D}}$. Direct computation shows that $\lambda _1$ has an eigenvector $\xi _1=(1,1,1,1)$, and $\lambda _3$ has an eigenvector $\xi _3=(1,-1,1,-1)$. Note that 1 and $-1$ are double eigenvalues of ${\mathcal {D}}$. We have to set $(\lambda _1E-{\mathcal {D}})\xi _2=\xi _1$, i.e.

$$\begin{aligned} \left( \begin{array}{llll} 1&{}\quad -1&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad -1&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad -1\\ 1&{}\quad 0&{}\quad -2&{}\quad 1 \end{array} \right) \left( \begin{array}{l} x_1\\ x_2\\ x_3\\ x_4 \end{array} \right) = \left( \begin{array}{l} 1\\ 1\\ 1\\ 1 \end{array} \right) , \end{aligned}$$

solving it, we have a new eigenvector $\xi _2=(0,1,2,3)$. Here E is the identity matrix.

Similarly, set $(\lambda _3E-{\mathcal {D}})\xi _4=\xi _3$, i.e.

$$\begin{aligned} \left( \begin{array}{llll} -1&{}\quad -1&{}\quad 0&{}\quad 0\\ 0&{}\quad -1&{}\quad -1&{}\quad 0\\ 0&{}\quad 0&{}\quad -1&{}\quad -1\\ 1&{}\quad 0&{}\quad -2&{}\quad -1 \end{array} \right) \left( \begin{array}{l} x_1\\ x_2\\ x_3\\ x_4 \end{array} \right) = \left( \begin{array}{l} 1\\ -1\\ 1\\ -1 \end{array} \right) , \end{aligned}$$

we get the last eigenvector $\xi _4=(0,1,-2,3)$.

Let

$$\begin{aligned} {\mathcal {P}}=\left( \begin{array}{llll} 1&{}\quad 0&{}\quad 1&{}\quad 0\\ 1&{}\quad 1&{}\quad -1&{}\quad 1\\ 1&{}\quad 2&{}\quad 1&{}\quad -2\\ 1&{}\quad 3&{}\quad -1&{}\quad 3 \end{array} \right) ,~~ {\mathcal {P}}^{-1}=\left( \begin{array}{llll} \frac{1}{2}&{}\quad \frac{3}{4}&{}\quad 0&{}\quad -\frac{1}{4}\\ -\frac{1}{4}&{}\quad -\frac{1}{4}&{}\quad \frac{1}{4}&{}\quad \frac{1}{4}\\ \frac{1}{2}&{}\quad -\frac{3}{4}&{}\quad 0&{}\quad \frac{1}{4}\\ \frac{1}{4}&{}\quad -\frac{1}{4}&{}\quad -\frac{1}{4}&{}\quad \frac{1}{4} \end{array} \right) , \end{aligned}$$

then the matrix ${\mathcal {D}}$ is transformed into Jordan matrix

$$\begin{aligned} {\mathcal {P}}^{-1}{\mathcal {D}}{\mathcal {P}}=J:=\left( \begin{array}{llll} 1&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad -1&{}\quad 1\\ 0&{}\quad 0&{}\quad 0&{}\quad -1 \end{array} \right) . \end{aligned}$$

(5.3)

So let $z_n={\mathcal {P}}y_n^{(1)}$, by (5.2), we get

$$\begin{aligned} y_{n+1}^{(1)}=(J+{\mathcal {P}}^{-1}{\mathcal {T}}(n){\mathcal {P}})y_n^{(1)}, \end{aligned}$$

(5.4)

where

$$\begin{aligned} {\mathcal {P}}^{-1}{\mathcal {T}}(n){\mathcal {P}}=J:=\left( \begin{array}{llll} \frac{p_1(n)+p_2(n)}{4}&{}\quad \frac{p_1(n)}{2}&{}\quad \frac{p_1(n)+p_2(n)}{4}&{}\quad -\frac{p_1(n)}{2}\\ -\frac{p_1(n)+p_2(n)}{4}&{}\quad -\frac{p_1(n)}{2}&{}\quad -\frac{p_1(n)+p_2(n)}{4}&{}\quad \frac{p_1(n)}{2}\\ -\frac{p_1(n)+p_2(n)}{4}&{}\quad -\frac{p_1(n)}{2}&{}\quad -\frac{p_1(n)+p_2(n)}{4}&{}\quad \frac{p_1(n)}{2}\\ -\frac{p_1(n)+p_2(n)}{4}&{}\quad -\frac{p_1(n)}{2}&{}\quad -\frac{p_1(n)+p_2(n)}{4}&{}\quad \frac{p_1(n)}{2} \end{array} \right) . \end{aligned}$$

Now our task is to transform the Jordan matrix J into a diagonal matrix with four different eigenvalues.

Lemma 5.1

There are two inverse matrices ${\mathcal {M}}_1(n)$ and ${\mathcal {M}}_2(n)$ depending on n such that

$$\begin{aligned} {\mathcal {M}}_1(n)J{\mathcal {M}}_2(n)=\left( \begin{array}{llll} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1+\frac{1}{n}&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad -1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad -(1+\frac{1}{n}) \end{array} \right) , \end{aligned}$$

(5.5)

where $n=1,2,3,\ldots$ and J is the Jordan matrix in (5.3).

Proof

This proof is based on observation. Let

$$\begin{aligned} {\mathcal {P}}_1=\left( \begin{array}{llll} 1&{}\quad 2&{}\quad 0&{}\quad 0\\ 1&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1 \end{array} \right) ,~~ {\mathcal {P}}_1^{-1}=\left( \begin{array}{llll} -1&{}\quad 2&{}\quad 0&{}\quad 0\\ 1&{}\quad -1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1 \end{array} \right) , \end{aligned}$$

and

$$\begin{aligned} {\mathcal {P}}_2=\left( \begin{array}{llll} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad -1&{}\quad -2\\ 0&{}\quad 0&{}\quad 1&{}\quad 1 \end{array} \right) ,~~ {\mathcal {P}}_2^{-1}=\left( \begin{array}{llll} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 2\\ 0&{}\quad 0&{}\quad -1&{}\quad -1 \end{array} \right) , \end{aligned}$$

then we derive

$$\begin{aligned} {\mathcal {P}}_2^{-1}{\mathcal {P}}_1^{-1}J{\mathcal {P}}_1{\mathcal {P}}_2=\left( \begin{array}{llll} 0&{}\quad -1&{}\quad 0&{}\quad 0\\ 1&{}\quad 2&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1\\ 0&{}\quad 0&{}\quad -1&{}\quad -2 \end{array} \right) . \end{aligned}$$

(5.6)

To diagonalize above matrix with four different eigenvalues, we introduce a matrix depending on n

$$\begin{aligned} {\mathcal {P}}_3(n)=\left( \begin{array}{llll} 1&{}\quad 1&{}\quad 0&{}\quad 0\\ -1&{}\quad -(1+\frac{1}{n})&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1 \end{array} \right) , \end{aligned}$$

then

$$\begin{aligned} {\mathcal {P}}_3^{-1}(n+1)=\left( \begin{array}{llll} n+2&{}\quad n+1&{}\quad 0&{}\quad 0\\ -(n+1)&{}\quad -(n+1)&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1 \end{array} \right) . \end{aligned}$$

By (5.6), direct computation shows that

$$\begin{aligned} {\mathcal {P}}_3^{-1}(n+1){\mathcal {P}}_2^{-1}{\mathcal {P}}_1^{-1}J{\mathcal {P}}_1{\mathcal {P}}_2{\mathcal {P}}_3(n)=\left( \begin{array}{llll} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1+\frac{1}{n}&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 1\\ 0&{}\quad 0&{}\quad -1&{}\quad -2 \end{array} \right) . \end{aligned}$$

(5.7)

Thus we introduce another matrix

$$\begin{aligned} {\mathcal {P}}_4(n)=\left( \begin{array}{llll} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 1&{}\quad 1\\ 0&{}\quad 0&{}\quad -1&{}\quad -(1+\frac{1}{n}) \end{array} \right) , \end{aligned}$$

then

$$\begin{aligned} {\mathcal {P}}_4^{-1}(n+1)=\left( \begin{array}{llll} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad n+2&{}\quad n+1\\ 0&{}\quad 0&{}\quad -(n+1)&{}\quad -(n+1) \end{array} \right) . \end{aligned}$$

which combining with (5.7) gives that

$$\begin{aligned} {\mathcal {M}}_1(n)J{\mathcal {M}}_2(n)=\left( \begin{array}{llll} 1&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 1+\frac{1}{n}&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad -1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad -(1+\frac{1}{n}) \end{array} \right) , \end{aligned}$$

where

$$\begin{aligned}&{\mathcal {M}}_1(n)={\mathcal {P}}_4^{-1}(n+1){\mathcal {P}}_3^{-1}(n+1){\mathcal {P}}_2^{-1}{\mathcal {P}}_1^{-1},\\&{\mathcal {M}}_2(n)={\mathcal {P}}_1{\mathcal {P}}_2{\mathcal {P}}_3(n){\mathcal {P}}_4(n). \end{aligned}$$

$\square$

We now return to the system (5.4). Set

$$\begin{aligned}&y_n^{(1)}={\mathcal {P}}_1y_n^{(2)},\\&y_n^{(2)}={\mathcal {P}}_2y_n^{(3)}, \end{aligned}$$

we derive

$$\begin{aligned} y_{n+1}^{(3)}=({\mathcal {P}}_2^{-1}{\mathcal {P}}_1^{-1}J{\mathcal {P}}_1{\mathcal {P}}_2+{\mathcal {P}}_2^{-1}{\mathcal {P}}_1^{-1}{\mathcal {P}}^{-1}{\mathcal {T}}(n){\mathcal {P}}{\mathcal {P}}_1{\mathcal {P}}_2)y_n^{(3)}. \end{aligned}$$

(5.8)

Set

$$\begin{aligned}&y_n^{(3)}={\mathcal {P}}_3(n)y_n^{(4)},~~y_{n+1}^{(3)}={\mathcal {P}}_3(n+1)y_{n+1}^{(4)}\\&y_n^{(4)}={\mathcal {P}}_4y_n^{(5)},~~y_{n+1}^{(4)}={\mathcal {P}}_4y_{n+1}^{(5)}, \end{aligned}$$

and

$$\begin{aligned}&{\mathcal {M}}_1(n)={\mathcal {P}}_4^{-1}(n+1){\mathcal {P}}_3^{-1}(n+1){\mathcal {P}}_2^{-1}{\mathcal {P}}_1^{-1},\\&{\mathcal {M}}_2(n)={\mathcal {P}}_1{\mathcal {P}}_2{\mathcal {P}}_3(n){\mathcal {P}}_4(n). \end{aligned}$$

Taking advantage of the process of proof in Lemma 5.1, we derive from (5.8) that

$$\begin{aligned} y_{n+1}^{(5)}= & {} ({\mathcal {M}}_1(n)J{\mathcal {M}}_2(n)+{\mathcal {M}}_1(n){\mathcal {P}}^{-1}{\mathcal {T}}(n){\mathcal {P}}{\mathcal {M}}_2(n))y_n^{(5)}\nonumber \\= & {} (\tilde{{\mathcal {D}}}+\tilde{{\mathcal {T}}}(n))y_n^{(5)},~~~~n\ge 1, \end{aligned}$$

(5.9)

where $\tilde{{\mathcal {D}}}:={\mathcal {M}}_1(n)J{\mathcal {M}}_2(n)$ is a diagonal matrix defined in (5.5), $\tilde{{\mathcal {T}}}(n)$ is a off-diagonal matrix, which is

$$\begin{aligned} \begin{array}{lll} &{}&{}\tilde{{\mathcal {T}}}(n)={\mathcal {M}}_1(n){\mathcal {P}}^{-1}{\mathcal {T}}(n){\mathcal {P}}{\mathcal {M}}_2(n)\\ &{}&{}\quad =\left( \begin{array}{llll} \frac{(n+4)(2p_1+p_2)}{4}&{}\quad \frac{p_1(n+\frac{16}{n}+8)+p_2(n+\frac{8}{n}+6)}{4}&{}\quad \frac{p_1(5n+2)+2(n+1)p_2}{4}&{}\quad \frac{p_1(5n^2-2n-16)+2(n^2-4)}{4n}\\ \frac{-(n+1)(5p_1+3p_2)}{4}&{}\quad \frac{-p_1(n+1)(5+\frac{8}{n})-p_2(3+\frac{4}{n})}{4}&{}\quad \frac{(n+1)(p_1+p_2)}{4}&{}\quad \frac{p_1(n^2+5n+4)+p_2(n^2+3n+2)}{4n}\\ \frac{(n+4)(p_1+p_2)}{4}&{}\quad \frac{(n+4)[(1+\frac{4}{n})p_1+(1+\frac{2}{n})p_2]}{4}&{}\quad \frac{-(n+4)(p_1+p_2)}{4}&{}\quad \frac{-(n+4)(5p_1+3p_2)}{4n}\\ \frac{-(n+1)(p_1+p_2)}{4}&{}\quad \frac{-(n+1)[(1+\frac{4}{n})p_1+(1+\frac{2}{n})p_2]}{4}&{}\quad \frac{(n+1)(p_1-3p_2)}{4}&{}\quad \frac{(n+1)(5p_1-7p_2)}{4n} \end{array} \right) , \end{array} \end{aligned}$$

(5.10)

where $p_1:=p_1(n)$ and $p_2:=p_2(n)$ defined in (3.24)–(3.25), respectively.

So it follows from (5.9) that

$$\begin{aligned} y^{(5)}_{n+1}=(\tilde{{\mathcal {D}}}(n)+\tilde{{\mathcal {T}}}(n))\cdot (\tilde{{\mathcal {D}}}(n-1)+\tilde{{\mathcal {T}}}(n-1))\cdot \ldots \cdot (\tilde{{\mathcal {D}}}(1)+\tilde{{\mathcal {T}}}(1))y_1^{(5)}, \end{aligned}$$

which combining with (5.10) that $y^{(5)}_{n+1}$ has an unbounded solution depending on n. This is coincident with the result of Birkhoff and Trjitzinsk  [3]. Thus we complete the proof of Lemma 3.5.