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Finite time blow-up of semi-linear Klein-Gordon equations
with positive initial energy in FLRW spacetimes

Makoto NAKAMURA   and  Takuma YOSHIZUMI Graduate School of Information Science and Technology, Osaka University, 1-5 Yamadaoka, Suita, Osaka 565-0871, JAPAN. E-mail: makoto.nakamura.ist@osaka-u.ac.jpGraduate School of Information Science and Technology, Osaka University, 1-5 Yamadaoka, Suita, Osaka 565-0871, JAPAN. E-mail: yoshizumi.takuma@ist.osaka-u.ac.jp
Abstract

Blowing-up solutions for semi-linear Klein-Gordon equations are considered in Friedmann-Lemaître-Robertson-Walker spacetimes. Some sufficient conditions are shown by applying the concavity method for semi-linear wave equations in the Minkowski spacetime to semi-linear Klein-Gordon equations in FLRW spacetimes.

Mathematics Subject Classification (2020): Primary 35L05; Secondary 35L71, 35Q75.

Keywords : semilinear Klein-Gordon equation, blowing-up solution, Friedmann-Lemaître-Robertson-Walker spacetime

1 Introduction

We consider the Cauchy problem of semi-linear Klein-Gordon equations in Friedmann-Lemaître-Robertson-Walker spacetimes (FLRW spacetimes for short). FLRW spacetimes are solutions of the Einstein equations with the cosmological constant under the cosmological principle. They describe the spatial expansion or contraction, and yield some important models of the universe. Let n1n\geq 1 be the spatial dimension, a()a(\cdot) be a scale-function defined on an interval [0,T0)[0,T_{0}) for some 0<T00<T_{0}\leq\infty, c>0c>0 be the speed of light. The metrics {gαβ}\{g_{\alpha\beta}\} of FLRW spacetimes are expressed by

c2(dτ)2=0α,βngαβdxαdxβ:=c2(dt)2+a(t)2j=1n(dxj)2,-c^{2}(d\tau)^{2}=\sum_{0\leq\alpha,\beta\leq n}g_{\alpha\beta}dx^{\alpha}dx^{\beta}:=-c^{2}(dt)^{2}+a(t)^{2}\sum_{j=1}^{n}(dx^{j})^{2}, (1.1)

where we have put the spatial curvature as zero, the variable τ\tau denotes the proper time, x0=tx^{0}=t is the time-variable (see e.g., [2, 3]). When a()=1a(\cdot)=1 or a(t)=eHta(t)=e^{Ht} for the Hubble constant HH\in{\mathbb{R}}, the spacetime with (1.1) reduces to the Minkowski spacetime or the de Sitter spacetime, respectively.

We recall the derivation of the Klein-Gordon equation. We denote the first and second derivatives of one variable function aa by a˙\dot{a} and a¨\ddot{a}. The Klein-Gordon equation generated by the above metric (gαβ)0α,βn(g_{\alpha\beta})_{0\leq\alpha,\beta\leq n} is given by (|g|)1α(|g|gαββu)+m2u=f(u)-(\sqrt{|g|})^{-1}\partial_{\alpha}(\sqrt{|g|}g^{\alpha\beta}\partial_{\beta}u)+m^{2}u=f(u) for the determinant g:=det(gαβ)g:=\det(g_{\alpha\beta}) and the inverse matrix (gαβ)(g^{\alpha\beta}), i.e.,

t2u+na˙atuc2a2Δu+m2c2u=c2f(u),\partial_{t}^{2}u+\frac{n\dot{a}}{a}\partial_{t}u-\frac{c^{2}}{a^{2}}\Delta u+m^{2}c^{2}u=c^{2}f(u), (1.2)

where Δ:=j=1nj2\Delta:=\sum_{j=1}^{n}\partial_{j}^{2} denotes the Laplacian.

In this paper, we consider the Cauchy problem of (1.2) given by

{t2u+na˙(t)a(t)1tuc2a(t)2Δu+m2c2u=c2f(u),u(t0,)=u0(),tu(t0,)=u1()\left\{\begin{array}[]{l}\partial_{t}^{2}u+n\dot{a}(t)a(t)^{-1}\partial_{t}u-c^{2}a(t)^{-2}\Delta u+m^{2}c^{2}u=c^{2}f(u),\\ u(t_{0},\cdot)=u_{0}(\cdot),\ \ \partial_{t}u(t_{0},\cdot)=u_{1}(\cdot)\end{array}\right. (1.3)

for (t,x)[t0,T)×n(t,x)\in[t_{0},T)\times{\mathbb{R}}^{n} with 0t0<TT00\leq t_{0}<T\leq T_{0}, where u0u_{0}, u1u_{1} are given initial data. For semi-linear terms, we assume that f:f:{\mathbb{C}}\rightarrow{\mathbb{C}} satisfies that there exists F:F:{\mathbb{C}}\rightarrow{\mathbb{R}} and ε>0\varepsilon>0 such that

tF(u)=Re{f(u)tu¯}andRe{f(u)u¯}(2+ε)F(u)\partial_{t}F(u)=\operatorname{Re}\left\{f(u)\partial_{t}\overline{u}\right\}\ \ \mbox{and}\ \ \operatorname{Re}\left\{f(u)\overline{u}\right\}\geq(2+\varepsilon)F(u) (1.4)

for any u:[t0,T)u:[t_{0},T)\rightarrow{\mathbb{C}}. We note that if we set f:f:{\mathbb{R}}\rightarrow{\mathbb{R}}, then the condition on (1.4) is described as

F(u):=0sf(ξ)𝑑ξandf(u)u(2+ε)F(u)F(u):=\int_{0}^{s}f(\xi)d\xi\ \ \mbox{and}\ \ f(u)u\geq(2+\varepsilon)F(u)

for u:[t0,T)u:[t_{0},T)\rightarrow{\mathbb{R}}, which has been considered in [8, 14]. The assumption (1.4) includes the power-type non-linearity. If we take f(u)=|u|p1uf(u)=|u|^{p-1}u for uu\in{\mathbb{C}} with p>1p>1, then we have to set εp1\varepsilon\leq p-1. If we take f(u)=|u|p1uf(u)=-|u|^{p-1}u for uu\in{\mathbb{C}} with p>1p>1, then we have to set εp1\varepsilon\geq p-1. For another example, we note that f(u)=±|u|pf(u)=\pm|u|^{p} for uu\in{\mathbb{R}} with p>1p>1, then we have to set ε=p1\varepsilon=p-1.

We denote the Lebesgue space by Lq(I)L^{q}(I) for an interval II\subset{\mathbb{R}} and 1q1\leq q\leq\infty with the norm

YLq(I):={{I|Y(t)|q𝑑t}1/qif 1q<,ess.suptI|Y(t)|ifq=.\|Y\|_{L^{q}(I)}:=\begin{cases}\left\{\int_{I}|Y(t)|^{q}dt\right\}^{1/q}&\mbox{if}\ \ 1\leq q<\infty,\\ {\mbox{ess.}\sup}_{t\in I}|Y(t)|&\mbox{if}\ \ q=\infty.\end{cases}

We denote by \|\cdot\| the L2(n)L^{2}({\mathbb{R}^{n}}) norm, and by q\|\cdot\|_{q} the Lq(n)L^{q}({\mathbb{R}^{n}}) norm for q2q\neq 2. We also denote (u,v):=nuv¯𝑑x(u,v):=\int_{\mathbb{R}^{n}}u\overline{v}dx by the L2L^{2}-inner product.

We define the energy functional

E(t):=12ut2+12c2a2u2+12m2c2u2c2nF(u)𝑑x,E(t):=\frac{1}{2}\|u_{t}\|^{2}+\frac{1}{2}c^{2}a^{-2}\|\nabla u\|^{2}+\frac{1}{2}m^{2}c^{2}\|u\|^{2}-c^{2}\int_{\mathbb{R}^{n}}F(u)dx, (1.5)

the Nehari functional

I(u):=c2a2u2+m2c2u2c2Renu¯f(u)𝑑xI(u):=c^{2}a^{-2}\|\nabla u\|^{2}+m^{2}c^{2}\|u\|^{2}-c^{2}\operatorname{Re}\int_{\mathbb{R}^{n}}\overline{u}f(u)dx (1.6)

and the unstable set

:={uC([t0,T),H1(n))C1([t0,T),L2(n));I(u)<0}\mathcal{B}:=\{u\in C([t_{0},T),H^{1}({\mathbb{R}^{n}}))\cap C^{1}([t_{0},T),L^{2}({\mathbb{R}^{n}}))\,;\,I(u)<0\} (1.7)

for the Cauchy problem of (1.3). We note that the Nehari functional I(u)I(u) depends on time. The set \mathcal{B} is invariant under our assumptions for initial data. More precisely, if I(u0)<0I(u_{0})<0 holds, then we have uu\in\mathcal{B} (see, Lemma 2.3 and Lemma 2.4, below).

Table 1: Initial data leading to high energy blowup of the problem (1.3).
Under I(u0)<0I(u_{0})<0 Progress
Case I m~2c~2ε2(ε+2)u02>E(t0)\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\|u_{0}\|^{2}>E(t_{0}) E(t0)m~2c~2ε2(ε+2)Re(u0,u1)0E(t_{0})\geq\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\operatorname{Re}(u_{0},u_{1})\geq 0 Solved in this paper
Case II m~2c~2ε2(ε+2)u02>E(t0)\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\|u_{0}\|^{2}>E(t_{0}) m~2c~2ε2(ε+2)Re(u0,u1)>E(t0)0\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\operatorname{Re}(u_{0},u_{1})>E(t_{0})\geq 0 Solved in this paper
Case III E(t0)m~2c~2ε2(ε+2)u02E(t_{0})\geq\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\|u_{0}\|^{2} m~2c~2ε2(ε+2)Re(u0,u1)>E(t0)0\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\operatorname{Re}(u_{0},u_{1})>E(t_{0})\geq 0 Solved in this paper
Case IV E(t0)m~2c~2ε2(ε+2)u02E(t_{0})\geq\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\|u_{0}\|^{2} E(t0)m~2c~2ε2(ε+2)Re(u0,u1)0E(t_{0})\geq\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\operatorname{Re}(u_{0},u_{1})\geq 0 Still open

In [9], the existence of global solutions was proved for semi-linear Klein-Gordon equations in de Sitter spacetime with n4n\leq 4. This result was extended to the case of general FLRW spacetimes in [4, 11]. Galstian and Yagdjian in [4] also proved the local well-posedness with n1n\geq 1 for semi-linear Klein-Gordon equations in FLRW spacetimes for any u0H1(n)u_{0}\in H^{1}({\mathbb{R}^{n}}), u1L2(n)u_{1}\in L^{2}({\mathbb{R}^{n}}) with a˙()>0\dot{a}(\cdot)>0.

The nonexistence of the global solution of the Klein-Gordon equation in the Minkowski spacetime (i.e., a˙0\dot{a}\equiv 0) was shown by Levine in [6] when the initial energy was negative. We also refer to the result [15], which established a sharp condition for blowup of one in the Minkowski spacetime when the initial energy had an upper bound. When the initial energy was arbitrarily high, the first result was a condition which Wang in [12] showed for blowing-up solutions corresponding to the Klein-Gordon equation in the Minkowski spacetime. This result was extended by Yang and Xu in [14].

On the other hand, for FLRW spacetimes, the case of the gauge invariant semi-linear term of the form f(u)=λ|u|p1uf(u)=\lambda|u|^{p-1}u with 1<p<1<p<\infty, λ\lambda\in{\mathbb{C}} was considered in [11] for a˙0\dot{a}\leq 0 by the concavity method for a2u2a^{2}\|u\|^{2}. Moreover, McCollum, Mwamba and Oliver in [8] showed a condition for blowing-up solutions of (1.3) for large initial data with the positive initial energy.

Our work is to solve Case I{\rm{I}}, Case II{{\rm{I}\rm{I}}} and Case III{{\rm{I}\rm{I}\rm{I}}} in Table 1 under m~>0\widetilde{m}>0 and c~>0\widetilde{c}>0 defined by (1.16), below. Case IV{{\rm{I}\rm{V}}} is still open.

We show the following theorem and corollary for blowing-up solutions of (1.3) for large initial data. Put a0:=a(0)a_{0}:=a(0) and a1:=a˙(0)a_{1}:=\dot{a}(0).

Theorem 1.1.

Let mm\in{\mathbb{R}}. Let fC(,)f\in C({\mathbb{C}},{\mathbb{C}}) be a function with (1.4) for some ε>0\varepsilon>0. Assume that t0=0t_{0}=0, and u0H1(n)u_{0}\in H^{1}({\mathbb{R}^{n}}), u1L2(n)u_{1}\in L^{2}({\mathbb{R}^{n}}) satisfy

ρ:=m2c2ε2(ε+2)u02E(0)>0\rho:=\frac{m^{2}c^{2}\varepsilon}{2(\varepsilon+2)}\|u_{0}\|^{2}-E(0)>0 (1.8)

for E()E(\cdot) given by (1.5). Assume Re(u0,u1)0\operatorname{Re}(u_{0},u_{1})\geq 0. Let 0<T00<T_{0}\leq\infty. Let aC2([0,T0),(0,))a\in C^{2}([0,T_{0}),(0,\infty)) satisfy a˙(t)0\dot{a}(t)\geq 0 and

a˙(t)2a¨(t)a(t)0\dot{a}(t)^{2}-\ddot{a}(t)a(t)\geq 0 (1.9)

for any t[0,T0)t\in[0,T_{0}). Put

T:=max{1,π2(1+na1a01)u02ε2ρ}.T:=\max\left\{1,\frac{\pi^{2}(1+na_{1}a_{0}^{-1})\|u_{0}\|^{2}}{\varepsilon^{2}\rho}\right\}. (1.10)

Then, the solution uu of (1.3) blows-up in L2(n)L^{2}({\mathbb{R}^{n}}) in finite time no later than TT if TT0T\leq T_{0}. More precisely, u\|u\|\rightarrow\infty as tTt\nearrow T_{*} for some TT_{*} with 0<TT0<T_{*}\leq T.

Remark 1.2.

Let fC(,)f\in C({\mathbb{C}},{\mathbb{C}}) satisfy the following conditions

f(0)=0,|f(s)f(v)|C|sv|(|s|p1+|v|p1)f(0)=0,\ \ \ |f(s)-f(v)|\leq C|s-v|\left(|s|^{p-1}+|v|^{p-1}\right) (1.11)

for any s,vs,v\in{\mathbb{C}} and some constant C>0C>0 independent of s,vs,v, and

1<p<{if n=1,2,1+2n2if n3.1<p<\begin{cases}\infty&\mbox{if \, }n=1,2,\\ 1+\frac{2}{n-2}&\mbox{if \, }n\geq 3.\end{cases} (1.12)

Then, we note that the condition (1.8) does not hold for sufficiently small data since ρ\rho is rewritten by

ρ\displaystyle\rho =12u1212c2a02u02m2c2ε+2u02+c2nF(u0)𝑑x\displaystyle=-\frac{1}{2}\|u_{1}\|^{2}-\frac{1}{2}c^{2}a^{-2}_{0}\|\nabla u_{0}\|^{2}-\frac{m^{2}c^{2}}{\varepsilon+2}\|u_{0}\|^{2}+c^{2}\int_{\mathbb{R}^{n}}F(u_{0})dx
12u1212c2a02u02m2c2ε+2u02+c2ε+2n|u0|p+1𝑑x\displaystyle\leq-\frac{1}{2}\|u_{1}\|^{2}-\frac{1}{2}c^{2}a^{-2}_{0}\|\nabla u_{0}\|^{2}-\frac{m^{2}c^{2}}{\varepsilon+2}\|u_{0}\|^{2}+\frac{c^{2}}{\varepsilon+2}\int_{\mathbb{R}^{n}}|u_{0}|^{p+1}dx

and we have ρ0\rho\leq 0 if u0u_{0} is sufficiently small.

Theorem 1.3.

Let mm\in{\mathbb{R}}. Let fC(,)f\in C({\mathbb{C}},{\mathbb{C}}) be a function with (1.4) for some ε>0\varepsilon>0. Let 0t0<T00\leq t_{0}<T_{0}\leq\infty. Let aC2([0,T0),(0,))a\in C^{2}([0,T_{0}),(0,\infty)) satisfy a˙(t)0\dot{a}(t)\geq 0 and (1.9) for any t[t0,T0)t\in[t_{0},T_{0}). Assume

a˙(t0)a(t0){|m|c{ε(ε+4)ε}2nifm0,<ifm=0.\frac{\dot{a}(t_{0})}{a(t_{0})}\begin{cases}\leq\frac{|m|c\left\{\sqrt{\varepsilon(\varepsilon+4)}-\varepsilon\right\}}{2n}&\mbox{if}\ \ m\neq 0,\\ <\infty&\mbox{if}\ \ m=0.\end{cases} (1.13)

Let u0H1(n)u_{0}\in H^{1}({\mathbb{R}^{n}}), u1L2(n)u_{1}\in L^{2}({\mathbb{R}^{n}}) satisfy

δ:=|m|cε2(ε+2)Re(u0,u1)E(t0)>0,\delta:=\frac{|m|c\varepsilon}{2(\varepsilon+2)}\operatorname{Re}(u_{0},u_{1})-E(t_{0})>0, (1.14)

I(u0)<0I(u_{0})<0 and Re(u0,u1)0\operatorname{Re}(u_{0},u_{1})\geq 0. Put

T:=t0+max[1,2π2(ε+4){1+na˙(t0)a(t0)1}u02ε2(ε+2)δ].T:=t_{0}+\max\left[1,\frac{2\pi^{2}(\varepsilon+4)\{1+n\dot{a}(t_{0})a(t_{0})^{-1}\}\|u_{0}\|^{2}}{\varepsilon^{2}(\varepsilon+2)\delta}\right]. (1.15)

Then, the solution uu of (1.3) blows-up in L2(n)L^{2}({\mathbb{R}^{n}}) in finite time no later than TT if TT0T\leq T_{0}. More precisely, u\|u\|\rightarrow\infty as tTt\nearrow T_{*} for some TT_{*} with t0<TTt_{0}<T_{*}\leq T.

When fC(,)f\in C({\mathbb{C}},{\mathbb{C}}) satisfies (1.11) with (1.12), we note that the condition (1.14) does not hold for sufficiently small data (see, Appendix in [8]).

Remark 1.4.

For mm\in{\mathbb{R}} and c>0c>0, we put

m~:=min{1,|m|},c~:=min{1,c}.\widetilde{m}:=\min\{1,|m|\},\ \ \widetilde{c}:=\min\{1,c\}. (1.16)

If we have

m~2c~2ε2(ε+2)u02>E(t0),\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\|u_{0}\|^{2}>E(t_{0}),

then (1.8) holds on t0=0t_{0}=0. If we have

m~2c~2ε2(ε+2)Re(u0,u1)>E(t0),\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\operatorname{Re}(u_{0},u_{1})>E(t_{0}),

then (1.14) holds. So that, in this paper, we solve Case I{\rm{I}}, Case II{{\rm{I}\rm{I}}} and Case III{{\rm{I}\rm{I}\rm{I}}} in Table 1, below.

In [8], a sufficient condition for blowing-up solutions of (1.3) was shown under I(u0)<na˙a1m~c~Re(u0,u1)<0I(u_{0})<-n\dot{a}a^{-1}\widetilde{m}\widetilde{c}\operatorname{Re}(u_{0},u_{1})<0 and m~2c~2ε3(ε+2)Re(u0,u1)>E(t0)0\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{3(\varepsilon+2)}\operatorname{Re}(u_{0},u_{1})>E(t_{0})\geq 0 with m~,c~>0\widetilde{m},\,\widetilde{c}>0. Since we have shown blowing-up solutions of (1.3) under I(u0)<0I(u_{0})<0 and m~2c~2ε2(ε+2)Re(u0,u1)>E(t0)0\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\operatorname{Re}(u_{0},u_{1})>E(t_{0})\geq 0 with m~,c~>0\widetilde{m},\,\widetilde{c}>0 in Theorem 1.3, we improve the result in [8] in terms of the upper bounds of I(u0)I(u_{0}) and E(t0)E(t_{0}). Moreover, we also have shown the blowing-up solution of (1.3) under I(u0)<0I(u_{0})<0 and m~2c~2ε2(ε+2)u0>E(t0)0\frac{\widetilde{m}^{2}\widetilde{c}^{2}\varepsilon}{2(\varepsilon+2)}\|u_{0}\|>E(t_{0})\geq 0. Blowup of (1.3) in this condition was not shown in [8].

We adapt the concavity method which was first introduced by Levine in [6, 7]. In the Minkowski spacetime, this method was employed for the function m2u2m^{2}\|u\|^{2} in [6, 7, 12, 14]. McCollum, Mwamba and Oliver in [8] also employed this method for m2u2m^{2}\|u\|^{2} in FLRW spacetimes. Instead of the function m2u2m^{2}\|u\|^{2}, we employ the concavity method for a new function θ\theta given by

θ(t):=u2+t0tn{a˙(τ)a(τ)1u2+G(τ)}𝑑τ+n(Tt)a˙(t0)a(t0)1u02\theta(t):=\|u\|^{2}+\int_{t_{0}}^{t}n\left\{\dot{a}(\tau)a(\tau)^{-1}\|u\|^{2}+G(\tau)\right\}d\tau+n(T-t)\dot{a}(t_{0})a(t_{0})^{-1}\|u_{0}\|^{2}

for t[t0,T)t\in[t_{0},T) with some T(t0,T0]T\in(t_{0},T_{0}], where

G(t):=t0t{a˙(τ)2a¨(τ)a(τ)a(τ)2}u2𝑑τ.G(t):=\int_{t_{0}}^{t}\left\{\frac{\dot{a}(\tau)^{2}-\ddot{a}(\tau)a(\tau)}{a(\tau)^{2}}\right\}\|u\|^{2}d\tau.

By the second and third terms of θ\theta, we improve the condition on I(u0)I(u_{0}), and we solve Case I{\rm{I}}. Moreover, we can consider the condition on m=0m=0 and E(t0)<0E(t_{0})<0 in Theorem 1.1 and Theorem 1.3. This yields that our results also include blowing-up solutions for semi-linear wave equations in FLRW spacetimes with the negative energy. Setting the third term of θ\theta is inspired by [5] and [13]. Gazzola and Squassina in [5], and Xu and Ding in [13] showed some conditions for blowing-up solutions of damped wave, and Klein-Gordon equations in the Minkowski spacetime.

Now, we introduce some concrete examples of the scale-function aa. For σ\sigma\in\mathbb{R} and the Hubble constant HH\in{\mathbb{R}}, we put

T0:={if(1+σ)H0,2n(1+σ)Hif(1+σ)H<0,T_{0}:=\begin{cases}\infty&\mbox{if}\ \ (1+\sigma)H\geq 0,\\ -\frac{2}{n(1+\sigma)H}&\mbox{if}\ \ (1+\sigma)H<0,\end{cases} (1.17)

and define a()a(\cdot) by

a(t):={a0{1+n(1+σ)Ht2}2/n(1+σ)ifσ1,a0exp(Ht)ifσ=1a(t):=\begin{cases}a_{0}\left\{1+\frac{n(1+\sigma)Ht}{2}\right\}^{2/n(1+\sigma)}&\mbox{if}\ \ \sigma\neq-1,\\ a_{0}\exp(Ht)&\mbox{if}\ \ \sigma=-1\end{cases} (1.18)

for 0t<T00\leq t<T_{0}. We note a0=a(0)a_{0}=a(0) and H=a˙(0)/a(0)H=\dot{a}(0)/a(0), where a˙:=da/dt\dot{a}:=da/dt. This scale-function a()a(\cdot) describes the Minkowski spacetime when H=0H=0 (namely, a()a(\cdot) is a constant a0a_{0}), the expanding space when H>0H>0 with σ1\sigma\geq-1, the blowing-up space when H>0H>0 with σ<1\sigma<-1 (the “Big-Rip” in cosmology), the contracting space when H<0H<0 with σ1\sigma\leq-1, and the vanishing space when H<0H<0 with σ>1\sigma>-1 (the “Big-Crunch” in cosmology). It describes the de Sitter spacetime when σ=1\sigma=-1 (see, e.g., [10]).

We obtain the following corollaries from the above theorems, respectively for the concrete example of aa given by (1.18).

Corollary 1.5.

Let fC(,)f\in C({\mathbb{C}},{\mathbb{C}}) be a function with (1.4) for some ε>0\varepsilon>0. Assume that t0=0t_{0}=0, mm\in{\mathbb{R}}, and u0H1(n)u_{0}\in H^{1}({\mathbb{R}^{n}}), u1L2(n)u_{1}\in L^{2}({\mathbb{R}^{n}}) satisfy (1.8) and Re(u0,u1)0\operatorname{Re}(u_{0},u_{1})\geq 0. Let T0T_{0} and aa be defined by (1.17) and (1.18). Assume one of the following conditions (i) and (ii) holds.

(i) H=0H=0, σ\sigma\in{\mathbb{R}}.

(ii) H>0H>0, σ1\sigma\geq-1.
Then, the condition on (1.9) is satisfied, and the result in Theorem 1.1 hold. Namely, the solution uu of the Cauchy problem (1.3) blows-up in finite time no later than TT defined by (1.10).

Corollary 1.6.

Let fC(,)f\in C({\mathbb{C}},{\mathbb{C}}) be a function with (1.4) for some ε>0\varepsilon>0. Assume that mm\in{\mathbb{R}}, and u0H1(n)u_{0}\in H^{1}({\mathbb{R}^{n}}), u1L2(n)u_{1}\in L^{2}({\mathbb{R}^{n}}) satisfy (1.14), I(u0)<0I(u_{0})<0 and Re(u0,u1)0\operatorname{Re}(u_{0},u_{1})\geq 0. We put

Cε:=2|m|c{ε(ε+4)ε},C_{\varepsilon}:=\frac{2}{|m|c\{\sqrt{\varepsilon(\varepsilon+4)}-\varepsilon\}}, (1.19)

for m0m\neq 0. Let T0T_{0} and aa be defined by (1.17) and (1.18). Assume one of the following conditions (i), (ii), (iii) and (iv) holds.

(i) mm\in{\mathbb{R}}, H=0H=0, σ\sigma\in{\mathbb{R}}, t0=0t_{0}=0.

(ii) m=0m=0, H>0H>0, σ1\sigma\geq-1, t0=0t_{0}=0.

(iii) m0m\neq 0, 0<H1/nCε0<H\leq 1/nC_{\varepsilon}, σ1\sigma\geq-1, t0=0t_{0}=0.

(iv) m0m\neq 0, H>1/nCεH>1/nC_{\varepsilon}, σ>1\sigma>-1, t0=2Cε/(1+σ)2/n(1+σ)H(>0)t_{0}=2C_{\varepsilon}/(1+\sigma)-2/n(1+\sigma)H(>0)
Then, the conditions (1.9) and (1.13) are satisfied, and the result in Theorem 1.3 hold. Namely, the solution uu of the Cauchy problem (1.3) blows-up in finite time no later than TT defined by (1.15).

In Corollaries 1.5 and 1.6, the case H=0H=0 reduces to the semi-linear Klein-Gordon equation in the Minkowski spacetime, which was extensively studied (see, e.g., [1, 6, 12, 13, 14, 15] on blowing-up solutions, and the references therein). We included this case to compare it with the case H0H\neq 0.

This paper is organized as follows. In Section 2, we collect fundamental properties on some ordinary differential equation of second order, the variant space I(u)<0I(u)<0, the energy EE and the concavity function. In Section 3.1, 3.2, 3.3 and 3.4, we prove Theorem 1.1, Theorem 1.3, Corollary 1.5 and Corollary 1.6.

2 Preliminaries

We prepare several lemmas to prove the results in the previous section. We start with the following fundamental statement for E(t)E(t) and I(u)I(u) defined by (1.5) and (1.6), respectively.

Lemma 2.1.

Let T>0T>0, a0>0a_{0}>0, a˙0\dot{a}\geq 0, u0H1(n)u_{0}\in H^{1}({\mathbb{R}^{n}}) and u1L2(n)u_{1}\in L^{2}({\mathbb{R}^{n}}). If uu is the solutions to the Cauchy problem (1.3), then the following results hold for any t[t0,T)t\in[t_{0},T).

(1)

E(t)12ut2+I(u)ε+2+ε2(ε+2)(c2a2u2+m2c2u2).E(t)\geq\frac{1}{2}\|u_{t}\|^{2}+\frac{I(u)}{\varepsilon+2}+\frac{\varepsilon}{2(\varepsilon+2)}\left(c^{2}a^{-2}\|\nabla u\|^{2}+m^{2}c^{2}\|u\|^{2}\right). (2.1)

(2)

E(t)+t0t(na˙a1ut2+c2a˙a3u2)𝑑τ=E(t0)E(t)+\int_{t_{0}}^{t}\left(n\dot{a}a^{-1}\|u_{t}\|^{2}+c^{2}\dot{a}a^{-3}\|\nabla u\|^{2}\right)d\tau=E(t_{0}) (2.2)

and E(t)E(t0)E(t)\leq E(t_{0}).

(3)

Re(u,utt)+na˙a1Re(u,ut)+I(u)=0.\operatorname{Re}(u,u_{tt})+n\dot{a}a^{-1}\operatorname{Re}(u,u_{t})+I(u)=0. (2.3)
Proof.

(1) By (1.4), (1.5) and (1.6), we have

c2a2u2+m2c2u2I(u)ε+2\displaystyle\frac{c^{2}a^{-2}\|\nabla u\|^{2}+m^{2}c^{2}\|u\|^{2}-I(u)}{\varepsilon+2} =c2ε+2Renu¯f(u)𝑑x\displaystyle=\frac{c^{2}}{\varepsilon+2}\operatorname{Re}\int_{\mathbb{R}^{n}}\overline{u}f(u)dx
c2nF(u)𝑑x\displaystyle\geq c^{2}\int_{\mathbb{R}^{n}}F(u)dx
=12ut2+12c2a2u2+12m2c2u2E(t).\displaystyle=\frac{1}{2}\|u_{t}\|^{2}+\frac{1}{2}c^{2}a^{-2}\|\nabla u\|^{2}+\frac{1}{2}m^{2}c^{2}\|u\|^{2}-E(t).

This yields the inequality of (2.1).

(2) Multiplying ut¯\overline{u_{t}} to the first equation of (1.3), integrating it for xnx\in{\mathbb{R}^{n}}, taking its real part and integrating it for [t0,t][t_{0},t], we obtain the equation of (2.2). By a0>0a_{0}>0, a˙0\dot{a}\geq 0 and (2.2), we have E(t)E(t0)E(t)\leq E(t_{0}).

(3) Multiplying u¯\overline{u} to the first equation of (1.3), integrating it for xnx\in{\mathbb{R}^{n}} and taking its real part, we obtain the equation of (2.3). ∎

Next, we set a lemma with the fundamental statement for the ordinary differential inequality.

Lemma 2.2.

Let T>0T>0, γC1([t0,T),)\gamma\in C^{1}([t_{0},T),{\mathbb{R}}), and hC1([t0,T),)h\in C^{1}([t_{0},T),{\mathbb{R}}). Assume that h(t0)0h(t_{0})\geq 0 and

h(t)+γ(t)h(t)>0h^{\prime}(t)+\gamma^{\prime}(t)h(t)>0 (2.4)

for any t[t0,T)t\in[t_{0},T). Then, h(t)>0h(t)>0 for any t(t0,T)t\in(t_{0},T).

Proof.

Multiplying eγ(t)>0e^{\gamma(t)}>0 to the both sides of (2.4), and noting (h+γh)eγ=(heγ)(h^{\prime}+\gamma^{\prime}h)e^{\gamma}=(he^{\gamma})^{\prime}, we have

{h(t)eγ(t)}>0\left\{h(t)e^{\gamma(t)}\right\}^{\prime}>0

for any t[t0,T)t\in[t_{0},T). Integrating its both sides for [t0,t)[t_{0},t), we obtain

h(t)eγ(t)\displaystyle h(t)e^{\gamma(t)} >h(t0)eγ(t0)\displaystyle>h(t_{0})e^{\gamma(t_{0})}
0\displaystyle\geq 0

for any t[t0,T)t\in[t_{0},T) by h(t0)0h(t_{0})\geq 0. This yields h(t)>0h(t)>0 for any t(t0,T)t\in(t_{0},T). ∎

Here, we put

L(t):=u(t)2L(t):=\|u(t)\|^{2} (2.5)

for t[t0,T)t\in[t_{0},T). We prove that the set \mathcal{B} is invariant under our assumptions.

Lemma 2.3.

Let 0t0<T0\leq t_{0}<T, a(t0)>0a(t_{0})>0 and a˙0\dot{a}\geq 0. Let u0H1(n)u_{0}\in H^{1}({\mathbb{R}^{n}}), u1L2(n)u_{1}\in L^{2}({\mathbb{R}^{n}}) satisfy Re(u0,u1)0\operatorname{Re}(u_{0},u_{1})\geq 0 and (1.8). Then, I(u0)<0I(u_{0})<0, and the solution uu of (1.3) belongs to \mathcal{B}.

Proof.

Firstly, we prove I(u0)<0I(u_{0})<0. By (1.8) and (2.1), we have

12u12+I(u0)ε+2+ε2(ε+2)c2a(t0)2u02<0.\frac{1}{2}\|u_{1}\|^{2}+\frac{I(u_{0})}{\varepsilon+2}+\frac{\varepsilon}{2(\varepsilon+2)}c^{2}a(t_{0})^{-2}\|\nabla u_{0}\|^{2}<0.

This yields I(u0)<0I(u_{0})<0.

Next, we prove I(u(t))<0I(u(t))<0 for any t>t0t>t_{0}. When m=0m=0, we obtain E(t0)<0E(t_{0})<0 by (1.8). Since we have E(t)E(t0)E(t)\leq E(t_{0}) by (2.2) and a˙0\dot{a}\geq 0, we obtain E(t)<0E(t)<0. Since we have E(t)I(u)/(ε+2)E(t)\geq I(u)/(\varepsilon+2) by (2.1), we obtain I(u)<0I(u)<0. Next, we consider the case of m0m\neq 0. If I(u(t))<0I(u(t))<0 for any t[t0,T)t\in[t_{0},T) does not hold, then there exists a first t(t0,T)t_{*}\in(t_{0},T) such that I(u(t))=0I(u(t_{*}))=0 by I(u0)<0I(u_{0})<0 and the continuity of I(u)I(u) on [t0,T)[t_{0},T). Here, we have

L=2Re(u,ut),L(t0)=2Re(u0,u1)0L^{\prime}=2\operatorname{Re}(u,u_{t}),\ \ L^{\prime}(t_{0})=2\operatorname{Re}(u_{0},u_{1})\geq 0 (2.6)

and

L′′=2ut2+2Re(u,utt),L^{\prime\prime}=2\|u_{t}\|^{2}+2\operatorname{Re}(u,u_{tt}), (2.7)

where LL has been defined by (2.5). By (2.3), (2.6) and (2.7), we obtain

L′′\displaystyle L^{\prime\prime} =2{ut2I(u)na˙a1Re(u,ut)}\displaystyle=2\left\{\|u_{t}\|^{2}-I(u)-n\dot{a}a^{-1}\operatorname{Re}(u,u_{t})\right\}
=2{ut2I(u)}na˙a1L.\displaystyle=2\left\{\|u_{t}\|^{2}-I(u)\right\}-n\dot{a}a^{-1}L^{\prime}.

This yields

L′′+na˙a1L=2{ut2I(u)}.L^{\prime\prime}+n\dot{a}a^{-1}L^{\prime}=2\left\{\|u_{t}\|^{2}-I(u)\right\}. (2.8)

Noting that I(u)<0I(u)<0 for t[t0,t)t\in[t_{0},t_{*}), we have

L′′+na˙a1L>0L^{\prime\prime}+n\dot{a}a^{-1}L^{\prime}>0

for t[t0,t)t\in[t_{0},t_{*}). Replacing TT, γ(t)\gamma(t), h(t)h(t) in Lemma 2.2 with tt_{*}, nloga(t)n\log{a(t)}, L(t)L^{\prime}(t), we obtain L(t)>0L^{\prime}(t)>0 for any t(t0,t)t\in(t_{0},t_{*}) by L(t0)0L^{\prime}(t_{0})\geq 0 in (2.6). Thus, this yields L(t)>L(t0)L(t_{*})>L(t_{0}) since LL is continuous on [t0,t][t_{0},t_{*}]. By L(t)>L(t0)L(t_{*})>L(t_{0}) and (1.8), we have

L(t)>2(ε+2)m2c2εE(t0).L(t_{*})>\frac{2(\varepsilon+2)}{m^{2}c^{2}\varepsilon}E(t_{0}). (2.9)

In contrast, we have

E(t)\displaystyle E(t_{*}) 12ut(t)2+ε2(ε+2)(c2a(t)2u(t)2+m2c2u(t)2)\displaystyle\geq\frac{1}{2}\|u_{t}(t_{*})\|^{2}+\frac{\varepsilon}{2(\varepsilon+2)}\left(c^{2}a(t_{*})^{-2}\|\nabla u(t_{*})\|^{2}+m^{2}c^{2}\|u(t_{*})\|^{2}\right)
m2c2ε2(ε+2)L(t)\displaystyle\geq\frac{m^{2}c^{2}\varepsilon}{2(\varepsilon+2)}L(t_{*})

by (2.1) and I(u(t))=0I(u(t_{*}))=0. Since we have E(t0)E(t)E(t_{0})\geq E(t_{*}) by (2) of Lemma 2.1, we have

E(t0)m2c2ε2(ε+2)L(t).E(t_{0})\geq\frac{m^{2}c^{2}\varepsilon}{2(\varepsilon+2)}L(t_{*}).

This leads to a contradiction to (2.9). Thus, we have proved I(u(t))<0I(u(t))<0 for any t[t0,T)t\in[t_{0},T). This yields uu\in\mathcal{B}. ∎

Lemma 2.4.

Let a>0a>0, a˙0\dot{a}\geq 0 and (1.9). Assume that there exists t0[0,T)t_{0}\in[0,T) such that (1.13) holds. Assume that u0H1(n)u_{0}\in H^{1}({\mathbb{R}^{n}}), u1L2(n)u_{1}\in L^{2}({\mathbb{R}^{n}}) satisfy I(u0)<0I(u_{0})<0 and (1.14). Then, the solution uu of (1.3) belongs to \mathcal{B}.

Proof.

When m=0m=0, the proof is similar to Lemma 2.3. We consider the case of m0m\neq 0 in the following. Putting

H(t):=L(t)4(ε+2)|m|cεE(t0)=2Re(u,ut)4(ε+2)|m|cεE(t0),H(t):=L^{\prime}(t)-\frac{4(\varepsilon+2)}{|m|c\varepsilon}E(t_{0})=2\operatorname{Re}(u,u_{t})-\frac{4(\varepsilon+2)}{|m|c\varepsilon}E(t_{0}),

we have

H\displaystyle H^{\prime} =L′′\displaystyle=L^{\prime\prime}
=2{ut2I(u)}na˙a1L\displaystyle=2\left\{\|u_{t}\|^{2}-I(u)\right\}-n\dot{a}a^{-1}L^{\prime}
(ε+4)ut2+εm2c2u22(ε+2)E(t)na˙a1L\displaystyle\geq\left(\varepsilon+4\right)\|u_{t}\|^{2}+\varepsilon m^{2}c^{2}\|u\|^{2}-2(\varepsilon+2)E(t)-n\dot{a}a^{-1}L^{\prime}
2|m|cε(ε+4)uut2(ε+2)E(t0)na˙a1L\displaystyle\geq 2|m|c\sqrt{\varepsilon(\varepsilon+4)}\|u\|\|u_{t}\|-2(\varepsilon+2)E(t_{0})-n\dot{a}a^{-1}L^{\prime}
|m|cε(ε+4)|Re(u,ut)|2(ε+2)E(t0)na˙a1L\displaystyle\geq|m|c\sqrt{\varepsilon(\varepsilon+4)}|\operatorname{Re}(u,u_{t})|-2(\varepsilon+2)E(t_{0})-n\dot{a}a^{-1}L^{\prime}
={|m|cε(ε+4)2na˙a1}|Re(u,ut)|2(ε+2)E(t0)\displaystyle=\left\{|m|c\sqrt{\varepsilon(\varepsilon+4)}-2n\dot{a}a^{-1}\right\}|\operatorname{Re}(u,u_{t})|-2(\varepsilon+2)E(t_{0})
|m|cεRe(u,ut)2(ε+2)E(t0)\displaystyle\geq|m|c\varepsilon\operatorname{Re}(u,u_{t})-2(\varepsilon+2)E(t_{0})
=|m|cε2H\displaystyle=\frac{|m|c\varepsilon}{2}H (2.10)

for any t[t0,T)t\in[t_{0},T), by (2.8), (2.1), x2+y22xyx^{2}+y^{2}\geq 2xy for any x,yx,y\in{\mathbb{R}} and E(t)E(t0)E(t)\leq E(t_{0}) for any t[t0,T)t\in[t_{0},T), where we have used |m|cε(ε+4)2na˙a1|m|cε|m|c\sqrt{\varepsilon(\varepsilon+4)}-2n\dot{a}a^{-1}\geq|m|c\varepsilon by (1.13) and ddta˙a10\frac{d}{dt}\dot{a}a^{-1}\leq 0 derived from (1.9). Since we have H(t0)>0H(t_{0})>0 by (1.14), we obtain H(t)>0H(t)>0 for any t[t0,T)t\in[t_{0},T) by putting h:=Hh:=H and γ:=|m|cεt/2\gamma:=-|m|c\varepsilon t/2, in Lemma 2.2. This yields L′′>0L^{\prime\prime}>0 for any t[t0,T)t\in[t_{0},T). Thus, we have

Re(u,ut)=L(t)2\displaystyle\operatorname{Re}(u,u_{t})=\frac{L^{\prime}(t)}{2} >L(t0)2=Re(u0,u1)\displaystyle>\frac{L^{\prime}(t_{0})}{2}=\operatorname{Re}(u_{0},u_{1}) (2.11)
>2(ε+2)|m|cεE(t0)\displaystyle>\frac{2(\varepsilon+2)}{|m|c\varepsilon}E(t_{0}) (2.12)

for any t[t0,T)t\in[t_{0},T) by L′′>0L^{\prime\prime}>0 and (1.14). Here, if I(u(t))<0I(u(t))<0 for any t[0,T)t\in[0,T) does not hold, then there exists a first t(t0,T)t_{*}\in(t_{0},T) such that I(u(t))=0I(u(t_{*}))=0 by I(u0)<0I(u_{0})<0 and the continuity of I(u)I(u) on [t0,T)[t_{0},T). Then, we have

E(t0)\displaystyle E(t_{0}) E(t)\displaystyle\geq E(t_{*})
12ut(t)2+ε2(ε+2)(c2a(t)2u(t)+m2c2u(t)2)\displaystyle\geq\frac{1}{2}\|u_{t}(t_{*})\|^{2}+\frac{\varepsilon}{2(\varepsilon+2)}(c^{2}a(t_{*})^{-2}\|\nabla u(t_{*})\|+m^{2}c^{2}\|u(t_{*})\|^{2})
ε2(ε+2)ut(t)2+m2c2ε2(ε+2)u(t)2\displaystyle\geq\frac{\varepsilon}{2(\varepsilon+2)}\|u_{t}(t_{*})\|^{2}+\frac{m^{2}c^{2}\varepsilon}{2(\varepsilon+2)}\|u(t_{*})\|^{2}
|m|cε2(ε+2)u(t)ut(t)\displaystyle\geq\frac{|m|c\varepsilon}{2(\varepsilon+2)}\|u(t_{*})\|\|u_{t}(t_{*})\|
|m|cε2(ε+2)Re(u(t),ut(t))\displaystyle\geq\frac{|m|c\varepsilon}{2(\varepsilon+2)}\operatorname{Re}(u(t_{*}),u_{t}(t_{*}))

by E(t0)E(t)E(t_{0})\geq E(t) for any t[t0,T)t\in[t_{0},T), (2.1) and I(u(t))=0I(u(t_{*}))=0. This leads a contradiction to (2.12). Thus, we have proved I(u(t))<0I(u(t))<0 for any t[t0,T)t\in[t_{0},T). This yields uu\in\mathcal{B}. ∎

Lemma 2.5.

Let 0t0<T0\leq t_{0}<T, a(t0)>0a(t_{0})>0 and a˙0\dot{a}\geq 0. Let u0H1(n)u_{0}\in H^{1}({\mathbb{R}^{n}}), u1L2(n)u_{1}\in L^{2}({\mathbb{R}^{n}}) satisfy Re(u0,u1)0\operatorname{Re}(u_{0},u_{1})\geq 0 and (1.8). Let uu be the solution of (1.3). Then,

L(t)>0andm2c2εL(t)2(ε+2)E(t0)>2(ε+2)ρL^{\prime}(t)>0\ \ \mbox{and}\ \ m^{2}c^{2}\varepsilon L(t)-2(\varepsilon+2)E(t_{0})>2(\varepsilon+2)\rho (2.13)

holds for any t(t0,T)t\in(t_{0},T), where ρ\rho is defined by (1.8).

Proof.

By Lemma 2.3, we have I(u(t))<0I(u(t))<0 for any t[t0,T)t\in[t_{0},T). By (2.8) and I(u)<0I(u)<0, we have L′′(t)+na˙(t)a(t)1L(t)>0L^{\prime\prime}(t)+n\dot{a}(t)a(t)^{-1}L^{\prime}(t)>0 for any [t0,T)[t_{0},T). Using Lemma 2.2, we obtain L(t)>0L^{\prime}(t)>0 for any t(t0,T)t\in(t_{0},T) by L(t0)=2Re(u0,u1)0L^{\prime}(t_{0})=2\operatorname{Re}(u_{0},u_{1})\geq 0. By L>0L^{\prime}>0 and (1.8), we have

m2c2εL(t)2(ε+2)E(t0)\displaystyle m^{2}c^{2}\varepsilon L(t)-2(\varepsilon+2)E(t_{0}) >m2c2εL(t0)2(ε+2)E(t0)\displaystyle>m^{2}c^{2}\varepsilon L(t_{0})-2(\varepsilon+2)E(t_{0})
=2(ε+2)ρ\displaystyle=2(\varepsilon+2)\rho

for any t(t0,T)t\in(t_{0},T). ∎

In the following, we show the key lemma in order to use the concavity method.

Lemma 2.6.

Let κ\kappa, AA and BB be positive constants. Let 0t0<T0\leq t_{0}<T satisfy

T>t0+π2(2κ+1)B8κ2A.T>t_{0}+\frac{\pi^{2}(2\kappa+1)B}{8\kappa^{2}A}. (2.14)

We consider the following differential inequality ;

{y′′(t)κAy(t)1+1/κ,y(t0)=y0,y(t0)=y1\left\{\begin{array}[]{l}y^{\prime\prime}(t)\leq-\kappa Ay(t)^{1+1/\kappa},\\ y(t_{0})=y_{0},\ \ y^{\prime}(t_{0})=y_{1}\end{array}\right. (2.15)

for any t[t0,T)t\in[t_{0},T). If y0y_{0} and y1y_{1} satisfy

{B(Tt0)}κy0,y10,\left\{B(T-t_{0})\right\}^{-\kappa}\leq y_{0},\ \ y_{1}\leq 0, (2.16)

then there exists T(t0,T)T_{*}\in(t_{0},T) such that the solution yC2([t0,T),)y\in C^{2}([t_{0},T),{\mathbb{R}}) of (2.15) satisfies y0y\searrow 0 as tTt\nearrow T_{*}.

Proof.

Assume y(t)>0y(t)>0 for any t[t0,T)t\in[t_{0},T). We show a contradiction. We note that we have y′′<0y^{\prime\prime}<0 by y>0y>0 and (2.15), and we also have y(t)<0y^{\prime}(t)<0 for any t(t0,T)t\in(t_{0},T) by y′′<0y^{\prime\prime}<0 and y10y_{1}\leq 0. Multiplying y<0y^{\prime}<0 to the both sides of the first inequality in (2.15), we obtain

12{(y)2}κ2A2κ+1{y2+1/κ}.\frac{1}{2}\left\{(y^{\prime})^{2}\right\}^{\prime}\geq-\frac{\kappa^{2}A}{2\kappa+1}\left\{y^{2+1/\kappa}\right\}^{\prime}.

This yields

(y)2\displaystyle(y^{\prime})^{2} y12+Iy02+1/κIy2+1/κ\displaystyle\geq y_{1}^{2}+{\rm{I}}y_{0}^{2+1/\kappa}-{\rm{I}}y^{2+1/\kappa}
>y12+Iy02+1/κIy01/κy2\displaystyle>y_{1}^{2}+{\rm{I}}y_{0}^{2+1/\kappa}-{\rm{I}}y_{0}^{1/\kappa}y^{2} (2.17)

for t(t0,T)t\in(t_{0},T) by 0<y<y00<y<y_{0}, where we have put

I:=2κ2A2κ+1.{\rm{I}}:=\frac{2\kappa^{2}A}{2\kappa+1}.

Noting that the right hand side of (2) is positive by y0>yy_{0}>y, we have

y<(IIIIIy2)1/2y^{\prime}<-\left({{\rm{I}\rm{I}}}-{{\rm{I}\rm{I}\rm{I}}}y^{2}\right)^{1/2}

by y<0y^{\prime}<0, where we have put

II:=y12+Iy02+1/κ,III:=Iy01/κ.{{\rm{I}\rm{I}}}:=y_{1}^{2}+{\rm{I}}y_{0}^{2+1/\kappa},\ \ {{\rm{I}\rm{I}\rm{I}}}:={\rm{I}}y_{0}^{1/\kappa}.

This yields

y(IIIIIy2)1/2<1.\frac{y^{\prime}}{\left({{\rm{I}\rm{I}}}-{{\rm{I}\rm{I}\rm{I}}}y^{2}\right)^{1/2}}<-1. (2.18)

Putting z:=(III/II)1/2yz:=({{\rm{I}\rm{I}\rm{I}}}/{{\rm{I}\rm{I}}})^{1/2}y and integrating the both sides of (2.18) on [t0,t][t_{0},t] with t0<t<Tt_{0}<t<T, we have z>0z>0 and

t0tIII1/2ddτSin1z(τ)𝑑τ<(tt0),\int_{t_{0}}^{t}{{\rm{I}\rm{I}\rm{I}}}^{-1/2}\frac{d}{d\tau}\mathrm{Sin}^{-1}z(\tau)d\tau<-(t-t_{0}), (2.19)

where Sin1z()\mathrm{Sin}^{-1}z(\cdot) is the inverse to z()=sinφz(\cdot)=\sin\varphi for φ[π/2,π/2]\varphi\in[-\pi/2,\pi/2]. This yields

III1/2(tt0)Sin1z(t0)<Sin1z(t)<0{{\rm{I}\rm{I}\rm{I}}}^{1/2}(t-t_{0})-\mathrm{Sin}^{-1}z(t_{0})<-\mathrm{Sin}^{-1}z(t)<0

by z>0z>0 and we have

|III1/2(tt0)Sin1z(t0)|<|Sin1z(t)|<π2\left|{{\rm{I}\rm{I}\rm{I}}}^{1/2}(t-t_{0})-\mathrm{Sin}^{-1}z(t_{0})\right|<\left|\mathrm{Sin}^{-1}z(t)\right|<\frac{\pi}{2} (2.20)

by Sin1z()[π/2,π/2]\mathrm{Sin}^{-1}z(\cdot)\in[-\pi/2,\pi/2]. We have

Sin1z(t)<III1/2(tt0)+Sin1z(t0).\mathrm{Sin}^{-1}z(t)<-{{\rm{I}\rm{I}\rm{I}}}^{1/2}(t-t_{0})+\mathrm{Sin}^{-1}z(t_{0}).

by (2.19). This yields

z(t)<sin(III1/2(tt0)Sin1z(t0))z(t)<-\sin\left({{\rm{I}\rm{I}\rm{I}}}^{1/2}(t-t_{0})-\mathrm{Sin}^{-1}z(t_{0})\right)

by noting (2.20). Putting

T~:=t0+Sin1z(t0)III1/2,\widetilde{T}_{*}:=t_{0}+\frac{\mathrm{Sin}^{-1}z(t_{0})}{{{\rm{I}\rm{I}\rm{I}}}^{1/2}}, (2.21)

where we note T~>t0\widetilde{T}_{*}>t_{0} by

Sin1z(t0)=Sin1{(Iy02+1/κy12+Iy02+1/κ)1/2}[0,π/2],\mathrm{Sin}^{-1}z(t_{0})=\mathrm{Sin}^{-1}\left\{\left(\frac{{\rm{I}}y_{0}^{2+1/\kappa}}{y_{1}^{2}+{\rm{I}}y_{0}^{2+1/\kappa}}\right)^{1/2}\right\}\ \ \in[0,\pi/2], (2.22)

there exists T(t0,T~]T_{*}\in(t_{0},\widetilde{T}_{*}] such that z0z\searrow 0 as tTt\nearrow T_{*}, which yields y0y\searrow 0 as tTt\nearrow T_{*}. Since we have

T~t0\displaystyle\widetilde{T}_{*}-t_{0} =I1/2y01/2κSin1{(Iy02+1/κy12+Iy02+1/κ)1/2}\displaystyle={\rm{I}^{-1/2}}y_{0}^{-1/2\kappa}\mathrm{Sin}^{-1}\left\{\left(\frac{{\rm{I}}y_{0}^{2+1/\kappa}}{y_{1}^{2}+{\rm{I}}y_{0}^{2+1/\kappa}}\right)^{1/2}\right\}
πI1/2y01/2κ2\displaystyle\leq\frac{\pi{\rm{I}^{-1/2}}y_{0}^{-1/2\kappa}}{2}
={π2(2κ+1)8κ2A}1/2y01/2κ\displaystyle=\left\{\frac{\pi^{2}(2\kappa+1)}{8\kappa^{2}A}\right\}^{1/2}y_{0}^{-1/2\kappa}
{π2(2κ+1)8κ2A}1/2B1/2(Tt0)1/2\displaystyle\leq\left\{\frac{\pi^{2}(2\kappa+1)}{8\kappa^{2}A}\right\}^{1/2}B^{1/2}(T-t_{0})^{1/2}
<(Tt0)1/2(Tt0)1/2=Tt0\displaystyle<(T-t_{0})^{1/2}(T-t_{0})^{1/2}=T-t_{0}

by (2.21), (2.22), (2.16) and (2.14), we have TT~<TT_{*}\leq\widetilde{T}_{*}<T. ∎

3 Proof of theorems and corollaries

We prove Theorem 1.1, Theorem 1.3, Corollary 1.5 and Corollary 1.6 in this section.

3.1 Proof of Theorem 1.1

Let t0=0t_{0}=0. We remember that we have put a0:=a(0)a_{0}:=a(0), a1:=a˙(0)a_{1}:=\dot{a}(0) and

T:=max{1,π2(1+na1a01)u02ε2ρ}.T:=\max\left\{1,\frac{\pi^{2}(1+na_{1}a_{0}^{-1})\|u_{0}\|^{2}}{\varepsilon^{2}\rho}\right\}.

We consider the Cauchy problem of (1.3) for (t,x)[0,T)×n(t,x)\in[0,T)\times{\mathbb{R}^{n}}. For any t[0,T)t\in[0,T), we put

G(t):=0t{a˙(τ)2a¨(τ)a(τ)a(τ)2}u(τ)2𝑑τG(t):=\int_{0}^{t}\left\{\frac{\dot{a}(\tau)^{2}-\ddot{a}(\tau)a(\tau)}{a(\tau)^{2}}\right\}\|u(\tau)\|^{2}d\tau (3.1)

and

θ(t):=u(t)2+0tn{a˙(τ)a(τ)1u(τ)2+G(τ)}𝑑τ+n(Tt)a1a01u02.\theta(t):=\|u(t)\|^{2}+\int_{0}^{t}n\left\{\dot{a}(\tau)a(\tau)^{-1}\|u(\tau)\|^{2}+G(\tau)\right\}d\tau+n(T-t)a_{1}a^{-1}_{0}\|u_{0}\|^{2}. (3.2)

We note that θ()>0\theta(\cdot)>0 since we have G()0G(\cdot)\geq 0 by (1.9). By (3.2), we have

θ\displaystyle\theta^{\prime} =2Re(u,ut)+na˙a1u2na1a01u02+nG\displaystyle=2\operatorname{Re}(u,u_{t})+n\dot{a}a^{-1}\|u\|^{2}-na_{1}a^{-1}_{0}\|u_{0}\|^{2}+nG
=2Re(u,ut)+0tτ(na˙a1u2)dτ+nG\displaystyle=2\operatorname{Re}(u,u_{t})+\int_{0}^{t}\partial_{\tau}\left(n\dot{a}a^{-1}\|u\|^{2}\right)d\tau+nG
=2Re(u,ut)+2Re0tna˙a1(u,ut)𝑑τ\displaystyle=2\operatorname{Re}(u,u_{t})+2\operatorname{Re}\int_{0}^{t}n\dot{a}a^{-1}(u,u_{t})d\tau (3.3)

and

θ′′\displaystyle\theta^{\prime\prime} =2Re(u,utt)+2na˙a1Re(u,ut)+2ut2\displaystyle=2\operatorname{Re}(u,u_{tt})+2n\dot{a}a^{-1}\operatorname{Re}(u,u_{t})+2\|u_{t}\|^{2}
=2{ut2I(u)}\displaystyle=2\left\{\|u_{t}\|^{2}-I(u)\right\} (3.4)

by (2.3). Since we have I(u)<0I(u)<0 by Lemma 2.3, we obtain θ′′>0\theta^{\prime\prime}>0, which yields

θ(t)>θ(0)=2Re(u0,u1)0\theta^{\prime}(t)>\theta^{\prime}(0)=2\operatorname{Re}(u_{0},u_{1})\geq 0 (3.5)

for any t(0,T)t\in(0,T) by (3.1) and the assumption Re(u0,u1)0\operatorname{Re}(u_{0},u_{1})\geq 0. By (3.1), we have

(θ(t))24=η(t)+{θ(t)0tnG(τ)𝑑τ(Tt)na1a01u02}{ut2+0tna˙(τ)a(τ)1ut2𝑑τ},\frac{\left(\theta^{\prime}(t)\right)^{2}}{4}=-\eta(t)+\left\{\theta(t)-\int_{0}^{t}nG(\tau)d\tau-(T-t)na_{1}a_{0}^{-1}\|u_{0}\|^{2}\right\}\\ \cdot\left\{\|u_{t}\|^{2}+\int_{0}^{t}n\dot{a}(\tau)a(\tau)^{-1}\|u_{t}\|^{2}d\tau\right\}, (3.6)

where we have put

η(t)\displaystyle\eta(t) :={u2+0tna˙(τ)a(τ)1u2𝑑τ}{ut2+0tna˙(τ)a(τ)1ut2𝑑τ}\displaystyle:=\left\{\|u\|^{2}+\int_{0}^{t}n\dot{a}(\tau)a(\tau)^{-1}\|u\|^{2}d\tau\right\}\left\{\|u_{t}\|^{2}+\int_{0}^{t}n\dot{a}(\tau)a(\tau)^{-1}\|u_{t}\|^{2}d\tau\right\}
{Re(u,ut)+Re0tna˙(τ)a(τ)1(u,ut)𝑑τ}2\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\left\{\operatorname{Re}(u,u_{t})+\operatorname{Re}\int_{0}^{t}n\dot{a}(\tau)a(\tau)^{-1}(u,u_{t})d\tau\right\}^{2}
=u2ut2{Re(u,ut)}2\displaystyle=\|u\|^{2}\|u_{t}\|^{2}-\left\{\operatorname{Re}(u,u_{t})\right\}^{2}
+(0tna˙a1u2𝑑τ)(0tna˙a1ut2𝑑τ){Re0tna˙a1(u,ut)𝑑τ}2\displaystyle\ \ \ \ \ \ +\left(\int_{0}^{t}n\dot{a}a^{-1}\|u\|^{2}d\tau\right)\left(\int_{0}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau\right)-\left\{\operatorname{Re}\int_{0}^{t}n\dot{a}a^{-1}(u,u_{t})d\tau\right\}^{2}
+u20tna˙a1ut2𝑑τ+ut20tna˙a1u2𝑑τ\displaystyle\ \ \ \ \ \ +\|u\|^{2}\int_{0}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau+\|u_{t}\|^{2}\int_{0}^{t}n\dot{a}a^{-1}\|u\|^{2}d\tau
2{Re(u,ut)}{Re0tna˙a1(u,ut)𝑑τ}.\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2\left\{\operatorname{Re}(u,u_{t})\right\}\left\{\operatorname{Re}\int_{0}^{t}n\dot{a}a^{-1}(u,u_{t})d\tau\right\}. (3.7)

We note that we have

Re(u,ut)n|uut¯|𝑑xuut\operatorname{Re}(u,u_{t})\leq\int_{\mathbb{R}^{n}}|u\overline{u_{t}}|dx\leq\|u\|\|u_{t}\| (3.8)

and

|Re0tna˙a1(u,ut)𝑑τ|2\displaystyle\left|\operatorname{Re}\int_{0}^{t}n\dot{a}a^{-1}(u,u_{t})d\tau\right|^{2} (0tna˙a1uut𝑑τ)2\displaystyle\leq\left(\int_{0}^{t}n\dot{a}a^{-1}\|u\|\|u_{t}\|d\tau\right)^{2}
(0tna˙a1u2𝑑τ)(0tna˙a1ut2𝑑τ)\displaystyle\leq\left(\int_{0}^{t}n\dot{a}a^{-1}\|u\|^{2}d\tau\right)\left(\int_{0}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau\right) (3.9)

by the Hölder inequality. Using (3.8) and (3.1), we have

{Re(u,ut)}\displaystyle\biggl{\{}\operatorname{Re}(u,u_{t})\biggr{\}} {Re0tna˙a1(u,ut)𝑑τ}\displaystyle\left\{\operatorname{Re}\int_{0}^{t}n\dot{a}a^{-1}(u,u_{t})d\tau\right\}
uut(0tna˙a1u2𝑑τ)1/2(0tna˙a1ut2𝑑τ)1/2\displaystyle\leq\|u\|\|u_{t}\|\left(\int_{0}^{t}n\dot{a}a^{-1}\|u\|^{2}d\tau\right)^{1/2}\left(\int_{0}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau\right)^{1/2}
12(u20tna˙a1ut2𝑑τ+ut20tna˙a1u2𝑑τ),\displaystyle\leq\frac{1}{2}\left(\|u\|^{2}\int_{0}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau+\|u_{t}\|^{2}\int_{0}^{t}n\dot{a}a^{-1}\|u\|^{2}d\tau\right), (3.10)

where we have used xy(x2+y2)/2xy\leq(x^{2}+y^{2})/2 for any x,yx,y\in{\mathbb{R}} at the last inequality. By (3.1), (3.8), (3.1) and (3.1), we have

η(t)0\eta(t)\geq 0 (3.11)

for any t[0,T)t\in[0,T). So that, we have

{θ(t)}24θ(t){ut2+0tna˙(τ)a(τ)1ut2𝑑τ}\frac{\left\{\theta^{\prime}(t)\right\}^{2}}{4}\leq\theta(t)\left\{\|u_{t}\|^{2}+\int_{0}^{t}n\dot{a}(\tau)a(\tau)^{-1}\|u_{t}\|^{2}d\tau\right\} (3.12)

by a˙()0\dot{a}(\cdot)\geq 0, G()0G(\cdot)\geq 0 and (3.11). By (3.12) and (3.1), we obtain

θ(t)θ′′(t)κ~+34{θ(t)}2\displaystyle\theta(t)\theta^{\prime\prime}(t)-\frac{\widetilde{\kappa}+3}{4}\left\{\theta^{\prime}(t)\right\}^{2} θ(t){θ′′(t)(κ~+3)(ut2+0tna˙a1ut2𝑑τ)}\displaystyle\geq\theta(t)\left\{\theta^{\prime\prime}(t)-(\widetilde{\kappa}+3)\left(\|u_{t}\|^{2}+\int_{0}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau\right)\right\}
=θ(t)ζ(t)\displaystyle=\theta(t)\zeta(t) (3.13)

for any κ~>0\widetilde{\kappa}>0, where we have put

ζ(t):=(κ~+1)ut22I(u)(κ~+3)0tna˙(τ)a(τ)1ut2𝑑τ.\zeta(t):=-(\widetilde{\kappa}+1)\|u_{t}\|^{2}-2I(u)-(\widetilde{\kappa}+3)\int_{0}^{t}n\dot{a}(\tau)a(\tau)^{-1}\|u_{t}\|^{2}d\tau. (3.14)

Here, let

κ~=ε+1.\widetilde{\kappa}=\varepsilon+1.

By (2.1), a˙0\dot{a}\geq 0, (2.2) and Lemma 2.5, we have

ζ(t)\displaystyle\zeta(t) ε(c2a2u2+m2c2u2)2(ε+2)E(t)(ε+4)0tna˙a1ut2𝑑τ\displaystyle\geq\varepsilon(c^{2}a^{-2}\|\nabla u\|^{2}+m^{2}c^{2}\|u\|^{2})-2(\varepsilon+2)E(t)-(\varepsilon+4)\int_{0}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau
=2(ε+2){E(t)+0tna˙a1ut2𝑑τ}\displaystyle=-2(\varepsilon+2)\left\{E(t)+\int_{0}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau\right\}
+ε0tna˙a1ut2𝑑τ+ε(c2a2u2+m2c2u2)\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +\varepsilon\int_{0}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau+\varepsilon\left(c^{2}a^{-2}\|\nabla u\|^{2}+m^{2}c^{2}\|u\|^{2}\right)
2(ε+2){E(t)+0tna˙a1ut2𝑑τ}+m2c2εu2\displaystyle\geq-2(\varepsilon+2)\left\{E(t)+\int_{0}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau\right\}+m^{2}c^{2}\varepsilon\|u\|^{2}
=2(ε+2){E(0)0tc2a˙a3u2𝑑τ}+m2c2εu2\displaystyle=-2(\varepsilon+2)\left\{E(0)-\int_{0}^{t}c^{2}\dot{a}a^{-3}\|\nabla u\|^{2}d\tau\right\}+m^{2}c^{2}\varepsilon\|u\|^{2}
2(ε+2)E(0)+m2c2εu2\displaystyle\geq-2(\varepsilon+2)E(0)+m^{2}c^{2}\varepsilon\|u\|^{2}
=2(ε+2)E(0)+m2c2εL(t)\displaystyle=-2(\varepsilon+2)E(0)+m^{2}c^{2}\varepsilon L(t)
2(ε+2)E(0)+m2c2εL(0)\displaystyle\geq-2(\varepsilon+2)E(0)+m^{2}c^{2}\varepsilon L(0)
=2(ε+2)ρ\displaystyle=2(\varepsilon+2)\rho (3.15)

for any t[0,T)t\in[0,T), where ρ\rho and LL are defined by (1.8) and (2.5), and we have used L(t)L(0)L(t)\geq L(0) by the first inequality L(t)>0L^{\prime}(t)>0 for any t(0,T)t\in(0,T) in (2.13). So that, we have

θ(t)θ′′(t)ε+44{θ(t)}22(ε+2)ρθ(t)\theta(t)\theta^{\prime\prime}(t)-\frac{\varepsilon+4}{4}\left\{\theta^{\prime}(t)\right\}^{2}\geq 2(\varepsilon+2)\rho\theta(t) (3.16)

for any t[0,T)t\in[0,T) by (3.1), (3.1) and θ>0\theta>0. Putting κ:=ε/4\kappa:=\varepsilon/4, we obtain

{θκ}′′(t)\displaystyle\left\{\theta^{-\kappa}\right\}^{\prime\prime}(t) =κθ(t)κ2[θ(t)θ′′(t)ε+44{θ(t)}2]\displaystyle=-\kappa\theta(t)^{-\kappa-2}\left[\theta(t)\theta^{\prime\prime}(t)-\frac{\varepsilon+4}{4}\left\{\theta^{\prime}(t)\right\}^{2}\right]
2κ(ε+2)ρθ(t)κ1\displaystyle\leq-2\kappa(\varepsilon+2)\rho\theta(t)^{-\kappa-1}
=2κ(ε+2)ρ{θ(t)κ}1+1/κ\displaystyle=-2\kappa(\varepsilon+2)\rho\left\{\theta(t)^{-\kappa}\right\}^{1+1/\kappa} (3.17)

by θ>0\theta>0 and (3.16). Putting

y(t):=θ(t)κy(t):=\theta(t)^{-\kappa}

and A:=2(ε+2)ρA:=2(\varepsilon+2)\rho, we have

y′′(t)κAy(t)1+1/κy^{\prime\prime}(t)\leq-\kappa Ay(t)^{1+1/\kappa} (3.18)

for any t[0,T)t\in[0,T) by (3.1). Since we have

0<θ(0)=u02+na1a01Tu02(1+na1a01)Tu020<\theta(0)=\|u_{0}\|^{2}+na_{1}a_{0}^{-1}T\|u_{0}\|^{2}\leq(1+na_{1}a_{0}^{-1})T\|u_{0}\|^{2}

by T1T\geq 1, we obtain

y0(BT)κ,y_{0}\geq(BT)^{-\kappa}, (3.19)

where we have put y0:=y(0)y_{0}:=y(0) and B:=(1+na1a01)u02B:=(1+na_{1}a_{0}^{-1})\|u_{0}\|^{2}. Since we have

{θκ}(0)=κθ(0)θκ1(0)0\left\{\theta^{-\kappa}\right\}^{\prime}(0)=-\kappa\theta^{\prime}(0)\theta^{-\kappa-1}(0)\leq 0

by θ(0)0\theta^{\prime}(0)\geq 0 and θ(0)>0\theta(0)>0, we obtain

y1:=y(0)0.y_{1}:=y^{\prime}(0)\leq 0. (3.20)

Moreover, we note that we have

π2(2κ+1)B8κ2A=π2(1+na1a01)u022ε2ρ<T\frac{\pi^{2}(2\kappa+1)B}{8\kappa^{2}A}=\frac{\pi^{2}(1+na_{1}a_{0}^{-1})\|u_{0}\|^{2}}{2\varepsilon^{2}\rho}<T (3.21)

by κ=ε/4\kappa=\varepsilon/4 and (1.10). Since TT is defined by (1.10), and y=θκy=\theta^{-\kappa} satisfy (2.14), (2.15) and (2.16) by (3.21), (3.18), (3.19) and (3.20), there exists T(0,T)T_{*}\in(0,T) such that y(t)=θ(t)κ0y(t)=\theta(t)^{-\kappa}\searrow 0 as tTt\nearrow T_{*} by Lemma 2.6. From this fact, we obtain the following result as required in this theorem.

Claim 3.1.

u(t)\|u(t)\|\nearrow\infty as tTt\nearrow T_{*}.

Proof.

Since we have

G(t)\displaystyle G(t) =0t{a˙(τ)2a¨(τ)a(τ)a(τ)2}u(τ)2𝑑τ\displaystyle=\int_{0}^{t}\left\{\frac{\dot{a}(\tau)^{2}-\ddot{a}(\tau)a(\tau)}{a(\tau)^{2}}\right\}\|u(\tau)\|^{2}d\tau
u(t)20tddτ(a˙a)𝑑τ\displaystyle\leq\|u(t)\|^{2}\int_{0}^{t}\frac{d}{d\tau}\left(-\frac{\dot{a}}{a}\right)d\tau
{a1a01a˙(t)a(t)1}u(t)2,\displaystyle\leq\{a_{1}a_{0}^{-1}-\dot{a}(t)a(t)^{-1}\}\|u(t)\|^{2},

where we have used (1.9) and u2\|u\|^{2} is non-decreasing by Lemma 2.5, we obtain

θ(t)\displaystyle\theta(t) =u(t)2+0tn{a˙(τ)a(τ)1u(τ)2+G(τ)}𝑑τ+n(Tt)a1a01u02\displaystyle=\|u(t)\|^{2}+\int_{0}^{t}n\left\{\dot{a}(\tau)a(\tau)^{-1}\|u(\tau)\|^{2}+G(\tau)\right\}d\tau+n(T-t)a_{1}a^{-1}_{0}\|u_{0}\|^{2}
u(t)2+0tna1a01u(τ)2𝑑τ+n(Tt)a1a01u02\displaystyle\leq\|u(t)\|^{2}+\int_{0}^{t}na_{1}a_{0}^{-1}\|u(\tau)\|^{2}d\tau+n(T-t)a_{1}a^{-1}_{0}\|u_{0}\|^{2}
u(t)2+na1a01tu(t)2+n(Tt)a1a01u02.\displaystyle\leq\|u(t)\|^{2}+na_{1}a_{0}^{-1}t\|u(t)\|^{2}+n(T-t)a_{1}a^{-1}_{0}\|u_{0}\|^{2}.

Since θ\theta\nearrow\infty as tTt\nearrow T_{*}, we have the claim. ∎

3.2 Proof of Theorem 1.3

We remember the definition of TT by (1.15). We define GG, θ\theta by

G(t):=t0t{a˙(τ)2a¨(τ)a(τ)a(τ)2}u(τ)2𝑑τG(t):=\int_{t_{0}}^{t}\left\{\frac{\dot{a}(\tau)^{2}-\ddot{a}(\tau)a(\tau)}{a(\tau)^{2}}\right\}\|u(\tau)\|^{2}d\tau

and

θ(t):=u(t)2+t0tn{a˙(τ)a(τ)1u(τ)2+G(τ)}𝑑τ+n(Tt)a˙(t0)a(t0)1u02\theta(t):=\|u(t)\|^{2}+\int_{t_{0}}^{t}n\left\{\dot{a}(\tau)a(\tau)^{-1}\|u(\tau)\|^{2}+G(\tau)\right\}d\tau\\ +n(T-t)\dot{a}(t_{0})a(t_{0})^{-1}\|u_{0}\|^{2} (3.22)

for any t[t0,T)t\in[t_{0},T) by the similar definition of (3.1) and (3.2). Then, the similar argument holds from (3.1) to (3.12), where we have replaced [0,T)[0,T) with [t0,T)[t_{0},T). Similarly to the derivation of (3.1), we obtain

θ(t)θ′′(t)κ~+34{θ(t)}2\displaystyle\theta(t)\theta^{\prime\prime}(t)-\frac{\widetilde{\kappa}+3}{4}\left\{\theta^{\prime}(t)\right\}^{2} θ(t)[θ′′(t)(κ~+3){ut2+t0tna˙(τ)a(τ)1ut2𝑑τ}]\displaystyle\geq\theta(t)\left[\theta^{\prime\prime}(t)-(\widetilde{\kappa}+3)\left\{\|u_{t}\|^{2}+\int_{t_{0}}^{t}n\dot{a}(\tau)a(\tau)^{-1}\|u_{t}\|^{2}d\tau\right\}\right]
=θ(t)ζ(t)\displaystyle=\theta(t)\zeta(t) (3.23)

for any κ~>0\widetilde{\kappa}>0, where we have put

ζ(t):=(κ~+1)ut22I(u)(κ~+3)t0tna˙(τ)a(τ)1ut2𝑑τ.\zeta(t):=-(\widetilde{\kappa}+1)\|u_{t}\|^{2}-2I(u)-(\widetilde{\kappa}+3)\int_{t_{0}}^{t}n\dot{a}(\tau)a(\tau)^{-1}\|u_{t}\|^{2}d\tau.

Here, let

κ~=ε2+1.\widetilde{\kappa}=\frac{\varepsilon}{2}+1.

By (2.1), (2.2), 2ε+1κ~=3ε/2>02\varepsilon+1-\widetilde{\kappa}=3\varepsilon/2>0 and ε+1κ~=ε/2\varepsilon+1-\widetilde{\kappa}=\varepsilon/2, we have

ζ(t)\displaystyle\zeta(t) (κ~+1)ut2+(ε+2)ut2+ε(c2a2u2+m2c2u2)\displaystyle\geq-(\widetilde{\kappa}+1)\|u_{t}\|^{2}+(\varepsilon+2)\|u_{t}\|^{2}+\varepsilon(c^{2}a^{-2}\|\nabla u\|^{2}+m^{2}c^{2}\|u\|^{2})
2(ε+2)E(t)(κ~+3)t0tna˙a1ut2𝑑τ\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -2(\varepsilon+2)E(t)-(\widetilde{\kappa}+3)\int_{t_{0}}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau
=(ε+1κ~)ut2+ε(c2a2u2+m2c2u2)\displaystyle=(\varepsilon+1-\widetilde{\kappa})\|u_{t}\|^{2}+\varepsilon(c^{2}a^{-2}\|\nabla u\|^{2}+m^{2}c^{2}\|u\|^{2})
2(ε+2){E(t)+t0tna˙a1ut2𝑑τ}+(2ε+1κ~)t0tna˙a1ut2𝑑τ\displaystyle\ \ \ \ \ \ \ \ \ -2(\varepsilon+2)\left\{E(t)+\int_{t_{0}}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau\right\}+(2\varepsilon+1-\widetilde{\kappa})\int_{t_{0}}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau
=(ε+1κ~)ut2+ε(c2a2u2+m2c2u2)\displaystyle=(\varepsilon+1-\widetilde{\kappa})\|u_{t}\|^{2}+\varepsilon(c^{2}a^{-2}\|\nabla u\|^{2}+m^{2}c^{2}\|u\|^{2})
2(ε+2){E(t0)t0tc2a˙a3u2𝑑τ}+(2ε+1κ~)t0tna˙a1ut2𝑑τ\displaystyle\ \ \ \ \ \ \ \ \ -2(\varepsilon+2)\left\{E(t_{0})-\int_{t_{0}}^{t}c^{2}\dot{a}a^{-3}\|\nabla u\|^{2}d\tau\right\}+(2\varepsilon+1-\widetilde{\kappa})\int_{t_{0}}^{t}n\dot{a}a^{-1}\|u_{t}\|^{2}d\tau
(ε+1κ~)ut2+m2c2εu22(ε+2)E(t0)\displaystyle\geq(\varepsilon+1-\widetilde{\kappa})\|u_{t}\|^{2}+m^{2}c^{2}\varepsilon\|u\|^{2}-2(\varepsilon+2)E(t_{0})
=ε2ut2+m2c2εu22(ε+2)E(t0)\displaystyle=\frac{\varepsilon}{2}\|u_{t}\|^{2}+m^{2}c^{2}\varepsilon\|u\|^{2}-2(\varepsilon+2)E(t_{0}) (3.24)

for any t[t0,T)t\in[t_{0},T). By (3.2), x2+y22xyx^{2}+y^{2}\geq 2xy for x,yx,y\in{\mathbb{R}}, (2.11) and (1.14), we have

ζ(t)\displaystyle\zeta(t) ε2ut2+m2c2εu22(ε+2)E(t0)\displaystyle\geq\frac{\varepsilon}{2}\|u_{t}\|^{2}+m^{2}c^{2}\varepsilon\|u\|^{2}-2(\varepsilon+2)E(t_{0})
ε2(ut2+m2c2u2)2(ε+2)E(t0)\displaystyle\geq\frac{\varepsilon}{2}(\|u_{t}\|^{2}+m^{2}c^{2}\|u\|^{2})-2(\varepsilon+2)E(t_{0})
|m|cεuut2(ε+2)E(t0)\displaystyle\geq|m|c\varepsilon\|u\|\|u_{t}\|-2(\varepsilon+2)E(t_{0})
|m|cεRe(u,ut)2(ε+2)E(t0)\displaystyle\geq|m|c\varepsilon\operatorname{Re}(u,u_{t})-2(\varepsilon+2)E(t_{0})
|m|cεRe(u0,u1)2(ε+2)E(t0)\displaystyle\geq|m|c\varepsilon\operatorname{Re}(u_{0},u_{1})-2(\varepsilon+2)E(t_{0})
=2(ε+2)δ\displaystyle=2(\varepsilon+2)\delta (3.25)

for any t[t0,T)t\in[t_{0},T). By (3.2) and (3.2), we have

θ(t)θ′′(t)κ~+34{θ(t)}22(ε+2)δθ(t)\theta(t)\theta^{\prime\prime}(t)-\frac{\widetilde{\kappa}+3}{4}\left\{\theta^{\prime}(t)\right\}^{2}\geq 2(\varepsilon+2)\delta\theta(t) (3.26)

for any t[t0,T)t\in[t_{0},T). Putting κ:=(κ~1)/4\kappa:=(\widetilde{\kappa}-1)/4, we obtain

{θκ}′′(t)\displaystyle\left\{\theta^{-\kappa}\right\}^{\prime\prime}(t) =κθκ2(t)[θ(t)θ′′(t)κ~+34{θ(t)}2]\displaystyle=-\kappa\theta^{-\kappa-2}(t)\left[\theta(t)\theta^{\prime\prime}(t)-\frac{\widetilde{\kappa}+3}{4}\left\{\theta^{\prime}(t)\right\}^{2}\right]
2κ(ε+2)δθκ1(t)\displaystyle\leq-2\kappa(\varepsilon+2)\delta\theta^{-\kappa-1}(t) (3.27)

by θ>0\theta>0 and (3.26). Putting

y(t):=θ(t)κy(t):=\theta(t)^{-\kappa}

and A:=2(ε+2)δA:=2(\varepsilon+2)\delta, we have

y′′(t)2κ(ε+2)δθκ1(t)=κAy(t)1+1/κy^{\prime\prime}(t)\leq-2\kappa(\varepsilon+2)\delta\theta^{-\kappa-1}(t)=-\kappa Ay(t)^{1+1/\kappa} (3.28)

for any t[t0,T)t\in[t_{0},T) by (3.2). We note that we have

0<y(t0)1/κ=θ(t0){1+na˙(t0)a(t0)1}(Tt0)u02=B(Tt0)0<y(t_{0})^{-1/\kappa}=\theta(t_{0})\leq\left\{1+n\dot{a}(t_{0})a(t_{0})^{-1}\right\}(T-t_{0})\|u_{0}\|^{2}=B(T-t_{0})

by (3.22) and Tt01T-t_{0}\geq 1, and

y(t0)={θκ}(t0)=κθ(t0)θ(t0)κ10,y^{\prime}(t_{0})=\left\{\theta^{-\kappa}\right\}^{\prime}(t_{0})=-\kappa\theta^{\prime}(t_{0})\theta(t_{0})^{-\kappa-1}\leq 0,

by θ(t0)0\theta^{\prime}(t_{0})\geq 0 in (3.5), where we have put B:={1+na˙(t0)a(t0)1}u02B:=\left\{1+n\dot{a}(t_{0})a(t_{0})^{-1}\right\}\|u_{0}\|^{2}. This yields

y0{B(Tt0)}κ,y10,y_{0}\geq\left\{B(T-t_{0})\right\}^{-\kappa},\ \ y_{1}\leq 0, (3.29)

where we have put y0:=y(t0)y_{0}:=y(t_{0}), y1:=y(t0)y_{1}:=y^{\prime}(t_{0}). Moreover, we note that we have

π2(2κ+1)B8κ2A=π2(ε+4){1+na˙(t0)a(t0)1}u02ε2(ε+2)δ<Tt0\frac{\pi^{2}(2\kappa+1)B}{8\kappa^{2}A}=\frac{\pi^{2}(\varepsilon+4)\left\{1+n\dot{a}(t_{0})a(t_{0})^{-1}\right\}\|u_{0}\|^{2}}{\varepsilon^{2}(\varepsilon+2)\delta}<T-t_{0} (3.30)

by (1.15). Since TT is defined by (1.15), and y=θκy=\theta^{-\kappa} satisfy (2.14), (2.15) and (2.16) by (3.30), (3.28) and (3.29), there exists T(t0,T)T_{*}\in(t_{0},T) such that y(t)=θ(t)κ0y(t)=\theta(t)^{-\kappa}\searrow 0 as tTt\nearrow T_{*} by Lemma 2.6. For u\|u\|\nearrow\infty as tTt\nearrow T_{*}, we use the similar argument for Claim 3.1.

3.3 Proof of Corollary 1.5

We check the assumptions a˙0\dot{a}\geq 0 and (1.9). By (1.18), we have

a˙(t)a(t)=H{1+n(1+σ)Ht2}1,\frac{\dot{a}(t)}{a(t)}=H\left\{1+\frac{n(1+\sigma)Ht}{2}\right\}^{-1}, (3.31)
a¨(t)a(t)=H2{1n(1+σ)2}{1+n(1+σ)Ht2}2.\frac{\ddot{a}(t)}{a(t)}=H^{2}\left\{1-\frac{n(1+\sigma)}{2}\right\}\left\{1+\frac{n(1+\sigma)Ht}{2}\right\}^{-2}.

This yields

a˙(t)2a¨(t)a(t)a(t)2=n(1+σ)H22{1+n(1+σ)Ht2}2.\frac{\dot{a}(t)^{2}-\ddot{a}(t)a(t)}{a(t)^{2}}=\frac{n(1+\sigma)H^{2}}{2}\left\{1+\frac{n(1+\sigma)Ht}{2}\right\}^{-2}.

So that, we note that the assumption (1.9) holds if and only if σ1\sigma\geq-1 under H0H\neq 0, or σ\sigma\in{\mathbb{R}} under H=0H=0. Under σ1\sigma\geq-1 and H0H\neq 0, we note that a˙0\dot{a}\geq 0 holds if and only if H>0H>0 by (3.31), a>0a>0 and 1+n(1+σ)Ht/2>01+n(1+\sigma)Ht/2>0. Under σ\sigma\in{\mathbb{R}} and H=0H=0, we note that a˙=0\dot{a}=0 holds. So that, the assumptions a˙0\dot{a}\geq 0 and (1.9) hold, and the required result follows from Theorem 1.1.

3.4 Proof of Corollary 1.6

In the proof of Corollary 1.5, we have shown the assumptions a˙0\dot{a}\geq 0 and (1.9) hold if and only if H=0H=0, σ\sigma\in{\mathbb{R}}, or H>0H>0, σ1\sigma\geq-1 hold. So that, we check the assumption (1.13) in the following. Firstly, let t0=0t_{0}=0. Then, we have

a˙(t0)a(t0)=a˙(0)a(0)=H\frac{\dot{a}(t_{0})}{a(t_{0})}=\frac{\dot{a}(0)}{a(0)}=H

by (3.31). So that, we have

(1.13)H{1nCεifm0,<ifm=0,\eqref{t_0-condition}\iff H\begin{cases}\leq\frac{1}{nC_{\varepsilon}}&\mbox{if}\ \ m\neq 0,\\ <\infty&\mbox{if}\ \ m=0,\end{cases} (3.32)

where CεC_{\varepsilon} is defined by (1.19). By H=0H=0, σ\sigma\in{\mathbb{R}}, or H>0H>0, σ1\sigma\geq-1, and (3.32), we have checked the cases of (i), (ii) and (iii) in Corollary 1.6 satisfy all assumptions of Theorem 1.3.

Next, let t0>0t_{0}>0. Since the cases of m=0m=0 or H1/nCεH\leq 1/nC_{\varepsilon} have been considered when t0=0t_{0}=0, we consider the case of m0m\neq 0 and H>1/nCεH>1/nC_{\varepsilon}. Then, we have

(1.13)a˙(t0)a(t0)=H{1+n(1+σ)Ht02}11nCε\eqref{t_0-condition}\iff\frac{\dot{a}(t_{0})}{a(t_{0})}=H\left\{1+\frac{n(1+\sigma)Ht_{0}}{2}\right\}^{-1}\leq\frac{1}{nC_{\varepsilon}} (3.33)

by (3.31). Noting that the last inequality in (3.33) does not hold when σ=1\sigma=-1 and H>1/nCεH>1/nC_{\varepsilon}, we have

(1.13)t02Cε1+σ2n(1+σ)H>0andσ>1\eqref{t_0-condition}\iff t_{0}\geq\frac{2C_{\varepsilon}}{1+\sigma}-\frac{2}{n(1+\sigma)H}>0\ \ \mbox{and}\ \ \sigma>-1

under H>1/nCεH>1/nC_{\varepsilon}. So that, we have checked the case of (iv) satisfies all assumptions of Theorem 1.3. Since all assumptions of Theorem 1.3 hold in the cases of (i), (ii), (iii) and (iv), the required result follows from Theorem 1.3.

Acknowledgments. This work was supported by JSPS KAKENHI Grant Numbers 16H03940, 17KK0082, 22K18671, and by JST SPRING Grant Number JPMJSP2138.

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