# Blow-Up of Solutions to a Novel Two-Component Rod System

Authors: Shengrong Liu; Zhengguang Guo; Weiming Wang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We consider a novel two-component rod system which is closely connected to the shallow water theory. The present work is mainly concerned with the blow-up mechanism of strong solutions; we establish new conditions in view of some special classes of initial value to guarantee finite time blow-up of solutions.


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## Body

## 1. Introduction

We consider the following variation of the two-component rod system:(1)ut-uxxt+3uux-σ(2uxuxx+uuxxx)+π(ρ)ρx=0,nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnt>0,x∈𝕊,π(ρ)t+(π(ρ)u)x=0,t>0,x∈𝕊.
If we introduce a momentum y=u-uxx, the previous system possesses the following form:
(2)yt+σyxu+2σyux-3(σ-1)uux+π(ρ)ρx=0,nnnnnnnnnnnnnnnnnnnnnnnnnnnt>0,x∈𝕊,π(ρ)t+(π(ρ)u)x=0,t>0,x∈𝕊,
where
(3)π(ρ)=ρ-μ(ρ)=ρ-∫01‍ρdx,
with u(x,t) and ρ(x,t) depending on a space variable x∈𝕊=ℝ/ℤ and a time variable t. It is obvious that system (1) for π(ρ)=0 reduces to the rod equation studied in [1–12]. Particularly, for π(ρ)=0 and σ=1, (1) becomes the celebrated Camassa-Holm equation which was investigated by many authors [13–21]. It reduces to the two-component rod system for μ(ρ)=0 studied in [22]. Moreover, system (1) reduces to the two-component Camassa-Holm (CH2) equation for μ(ρ)=0 and σ=1 which was investigated in [23–30]. The two-component rod system which includes both velocity and density variables in the dynamics possesses the following form:
(4)ut-uxxt-Aux+3uux-σ(2uxuxx+uuxxx)+ρρx=0,gggggggggggggggggggggggggggggt>0,x∈ℝ,ρt+(ρu)x=0,t>0,x∈ℝ.
We note that the geometric structure of (1) is different from (4) with the incorporation of the term μ(ρ); this restricts our discussion only on the unit circle. Furthermore, the discussion of this work shows that system (1) possesses wave breaking phenomenon which is described with different blow-up criteria, while system (4) admits not only breaking wave solutions but also global in time solutions. However, we do not know whether the global solutions of (1) exist or not for the time being. It is the structure of (1) that breaks some properties which previously holds for (4). It is also known that for (4)
(5)ddt∫𝕊‍ρdx=-∫𝕊‍(ρu)xdx=0,
which follows that ∫𝕊‍ρdx is an invariant with respect to time. Particularly, if ρ has zero mean initially, then the solution ρ to (4) will preserve zero mean for all time. This motivates us to introduce the term π(ρ) since it always has zero mean. On the other hand, we consider the system (1) in the spaces Hs×Hs-1/ℝ for s>5/2 on the circle, where Hs=Hs(𝕊) denotes the L2-Sobolev space of regularity, and denote by Hs-1/ℝ the space Hs-1 with two functions being identified if they differ by a constant. The basic idea of this variation is the decomposition of ρ∈Hs-1 into π(ρ) and μ(ρ), where μ(ρ)∈ℝ is independent on variable x, π(ρ) belongs to H^s-1, a subspace of Hs-1 containing all zero mean functions. The main purpose of our work is to investigate formation of singularities of solutions to (1) where the conservation laws play crucial roles. The idea is motivated by Guo's recent works [28, 31].The rest of this paper is organized as follows. In Section2, we recall the local well-posedness theorem and show some auxiliary results which will be used in the sequel. In Section 3, the detailed blow-up criteria are established via various initial conditions and as a byproduct, the blow-up rate is presented.


## 2. Preliminaries

In this section, we would like to list some useful results for later use. We now provide the framework in which we shall reformulate (1). Let
(6)G(x)=cosh(x-[x]-(1/2))2sinh(1/2),x∈ℝ.
Then (1-∂x2)-1f=G*f for all f∈L2(𝕊) and G*y=u,  where * is the spatial convolution. With this in hand, we rewrite (1) as follows:
(7)ut+σuux+∂x(G*(3-σ2u2+σ2ux2+12π(ρ)2))=0,2222222222222222222222222222222…t>0,x∈𝕊,π(ρ)t+(π(ρ)u)x=0,t>0,x∈𝕊.
Firstly, we recall the elementary result on the local well-posedness theorem for system (7) which was shown in [32] from the geometric formulation.Theorem 1.
Lets>5/2. There is an open neighborhood U containing (0,π(0))∈Hs×Hs-1/ℝ such that for any (u0,π(ρ0))∈U there exists a maximal T>0 and a unique solution (u,π(ρ)) to the initial value problem for system (7) with
(8)(u,π(ρ))∈C([0,T);Hs×Hs-1ℝ)∩C1([0,T);Hs-1×Hs-2ℝ),(u,π(ρ))(0)=(u0,π(ρ0)). Moreover, the solution depends continuously on the initial data, that is, the mapping
(9)(u0,π(ρ0))↦(u,π(ρ)):U⟶C([0,T);Hs×Hs-1ℝ)∩C1([0,T);Hs-1×Hs-2ℝ)
is continuous.We now introduce two characteristics for (7)
(10)∂q1∂t=u(q1,t),0<t<T,x∈ℝ,q1(x,0)=x,x∈ℝ,(11)∂q2∂t=σu(q2,t),0<t<T,x∈ℝ,q2(x,0)=x,x∈ℝ,
where u denotes the first component of the solution to (7) with certain initial data and T is the lifespan of the solution, then qi(i=1,2) is a diffeomorphism of the line. These two characteristics are both increasing diffeomorphisms of ℝ. A direct calculation shows
(12)q1,tx(x,t)=ux(q1(x,t),t)q1,x(x,t),q2,tx(x,t)=σux(q2(x,t),t)q2,x(x,t).
Thus, we have
(13)q1,x(x,t)=exp(∫0t‍ux(q1,s)ds)>0,(x,t)∈ℝ×[0,T),(14)q2,x(x,t)=exp(∫0t‍σux(q2,s)ds)>0,q2,x(x,t)=expq2,xv(x,t)∈ℝ×[0,T).
This is usually called the particle trajectory method, and it is important in the discussion of blow-up phenomena. We are now in a position to state the following.Lemma 2.
LetX0=(u0,π(ρ0))∈Hs×Hs-1/ℝ,s>5/2, and let T be the maximal existence time of the solution X=(u,π(ρ)) to (7) with the initial data X0. Then for all (x,t)∈𝕊×[0,T), we have
(15)π(ρ(q1,t))q1,x(x,t)=π(ρ0(x)).Proof.
Differentiating the left-hand side of (15) with respect to t, we obtain
(16)ddt[π(ρ(q1,t))q1,x(x,t)]=[πt(ρ(q1,t))+π(ρ)xq1,t]q1,x(x,t)11111+π(ρ(q1,t))q1,xt(x,t)=[πt(ρ(q1,t))+π(ρ)xu(q1,t)]q1,x(x,t)ssssss+π(ρ(q1,t))ux(q1,t)q1,x(x,t)=-(π(ρ)u)xq1,x(x,t)+π(ρ)xu(q1,t)q1,x(x,t)ssssss+π(ρ(q1,t))ux(q1,t)q1,x(x,t)=0,
where we have used q1 defined in (10) and the second equation of system (7). Then π(ρ(q1,t))q1,x(x,t) is independent on time t. Now we choose t=0, due to (13) we know q1,x(x,0)=1. Therefore, this lemma is easily proved.The application of (15) leads to the following result.Theorem 3.
LetX0=(u0,π(ρ0))∈Hs×Hs-1/ℝ,s>5/2, and let T be the maximal existence time of the solution X=(u,π(ρ)) to (7) with the initial data X0. If there exists M>0 such that
(17)∥ux(·,t)∥L∞+∥π(ρ)x(·,t)∥L∞≤M,
then Hs×Hs-1-norm of X does not blow up on [0,T).Proof.
Applying the operatorΛs where Λ:=(1-∂x2)1/2 to the first equation in (7), and multiplying by Λsu, then integrating over 𝕊, we obtain
(18)ddt∥u∥Hs2=-2(σuux,u)s-2(u,f1(u))s-2(u,f2(u))s,
where
(19)f1(u)=∂x(G*3-σ2u2),f2(u)=12∂x(G*(σux2+π(ρ)2)).
From the proof in [25], we get
(20)|(σuux,u)s|≤c∥ux∥L∞∥u∥Hs2,|(u,f1(u))s|≤c∥ux∥L∞∥u∥Hs2.
For the third term on the right-hand side of (13), we estimate it as follows:
(21)|(u,f2(u))s|≤∥f2(u)∥Hs∥u∥Hs≤c(∥ux2∥Hs-1+∥π(ρ)2∥Hs-1)∥u∥Hs≤c(∥ux∥L∞∥ux∥Hs-1ddds+∥π(ρ)∥L∞∥π(ρ)∥Hs-1)∥u∥Hs≤c(∥ux∥L∞+∥π(ρ)∥L∞)×(∥u∥Hs2+∥π(ρ)∥Hs-12).
Combining the previous inequalities, we can get
(22)ddt∥u∥Hs2≤c(∥ux∥L∞+∥π(ρ)∥L∞)×(∥u∥Hs2+∥π(ρ)∥Hs-12),
where the constant c may be different from instance to instance. We do similar estimates for the second component by applying the operator Λs-1 to the second equation in (7), and multiplying by Λs-1π(ρ), then integrating over 𝕊 to obtain
(23)ddt∥u∥Hs2=-2(π(ρ)xu,π(ρ))s-1-2(π(ρ)ux,π(ρ))s-1.
Similarly, we have
(24)|(π(ρ)xu,π(ρ))s-1|≤c(∥ux∥L∞+∥π(ρ)x∥L∞)(∥u∥Hs2+∥π(ρ)∥Hs-12),|(π(ρ)ux,π(ρ))s-1|≤c(∥ux∥L∞+∥π(ρ)∥L∞)(∥u∥Hs2+∥π(ρ)∥Hs-12).
Therefore, it follows that
(25)ddt∥π(ρ)∥Hs-12≤c(∥ux∥L∞+∥π(ρ)x∥L∞+∥π(ρ)∥L∞)×(∥u∥Hs2+∥π(ρ)∥Hs-12).
By (22) and (25) we have
(26)ddt(∥u∥Hs2+∥π(ρ)∥Hs-12)≤c(∥ux∥L∞+∥π(ρ)x∥L∞)×(∥u∥Hs2+∥π(ρ)∥Hs-12).
It is easy to obtain by Gronwall's inequality, Lemma 2, and (13) that
(27)∥u∥Hs2+∥π(ρ)∥Hs-12≤exp(cMt)(∥u0∥Hs2+∥π(ρ0)∥Hs-12).
Then the Hs×Hs-1-norm of X does not blow up on [0,T).Next we present the precise blow-up scenario for sufficiently regular solutions to (7).Theorem 4.
LetX0=(u0,π(ρ0))∈Hs×Hs-1/ℝ,s>5/2, and let T be the maximal existence time of the solution X=(u,π(ρ)) to (7) with the initial data X0. Then the solution X blows up in finite time if and only if
(28)limt→Tinfx∈𝕊{ux(x,t)}=-∞orlimt→Tsup{∥ρx(·,t)∥L∞}=+∞.Proof.
Multiplying the first equation in (2) by y and integrating by parts, we get
(29)∫𝕊‍yytdx+σ∫𝕊‍yyxudx+2σ∫𝕊‍y2uxdx-3(σ-1)2∫𝕊‍y(u2)xdx+∫𝕊‍yπ(ρ)π(ρ)xdx=0.
It then follows that
(30)12ddt∫𝕊‍y2dx=-3σ2∫𝕊‍y2uxdx-3(σ-1)∫𝕊‍uuxuxxdxss+12∫𝕊‍π(ρ)2yxdx=-3σ2∫𝕊‍y2uxdx-3(σ-1)∫𝕊‍uuxuxxdxSS+12∫𝕊‍π(ρ)2uxdx-12∫𝕊‍π(ρ)2uxxxdx.
Differentiating the first equation in (2) with respect to x, multiplying by yx, and integrating by parts yield
(31)12ddt∫𝕊‍yx2dx=-5σ2∫𝕊‍uxyx2dx+σ∫𝕊‍y2uxdxss+3(σ-1)2∫𝕊‍(u2)xxyxdxss+∫𝕊‍uxxx(12π(ρ)x2ffffffffm+π(ρ)π(ρ)xx-12π(ρ)2)dx.
Combining (30) and (31) together, we obtain
(32)ddt∫𝕊‍(y2+yx2)dx=∫𝕊‍(π(ρ)2-5σyx2-σy2)uxdx+3(σ-1)∫𝕊‍ux(2ux2+5uxx2)dx+2∫𝕊‍uxxx(π(ρ)x2+π(ρ)π(ρ)xx-π(ρ)2)dx.
Similar arguments made on the second equation in (2) yield
(33)ddt∫𝕊‍π(ρ)2dx=-∫𝕊‍π(ρ)2uxdx,ddt∫𝕊‍π(ρ)x2dx=-3∫𝕊‍π(ρ)x2uxdxcc+∫𝕊‍π(ρ)2uxxxdx,ddt∫𝕊‍π(ρ)xx2dx=-5∫𝕊‍π(ρ)xx2uxdxdv-∫𝕊‍(2π(ρ)π(ρ)xx-3π(ρ)x2)uxxxdx.
It follows by combining (32) and (33) that
(34)ddt∫𝕊‍(y2+yx2+π(ρ)2+π(ρ)x2+π(ρ)xx2)dx=-∫𝕊‍ux[σy2+5σyx2+3π(ρ)x2+5π(ρ)xx2FFFFFFFFFF+3(1-σ)(2ux2+5uxx2)]dx+∫𝕊‍uxxx(5π(ρ)x2-π(ρ)2)dx.
Assume that there exist M1>0 and M2>0, such that
(35)ux(x,t)≥-M1,(x,t)∈𝕊×[0,T),∥π(ρ)x∥L∞=∥ρx∥L∞≤M2,∀t∈[0,T).
By Lemma 2, we have
(36)∥π(ρ)∥L∞≤eM1T∥π(ρ0)∥L∞≤ceM1T∥ρ0∥L∞.
It then follows that
(37)ddt∫𝕊‍(y2+yx2+π(ρ)2+π(ρ)x2+π(ρ)xx2)dx≤cM1∫𝕊‍(y2+yx2+π(ρ)x2+π(ρ)xx2)dx+cM2∫𝕊‍(uxxx2+π(ρ)2+π(ρ)x2)dx≤c∫𝕊‍(y2+yx2+π(ρ)2+π(ρ)x2+π(ρ)xx2)dx.
It is easy to obtain by Gronwall's inequality that
(38)∥u∥H32+∥π(ρ)∥H22≤∥y∥H12+∥π(ρ)∥H22≤ect(∥y0∥H12+∥π(ρ0)∥H22).
Sobolev imbedding, (38), and Theorem 3 ensure that the solution X does not blow up in finite time.
On the other hand, due to the Sobolev imbedding theorem, we observe thatux(x,t)→-∞ or ∥ρx(·,t)∥L∞→+∞ will lead to blow-up of solutions.After the local well-posedness of strong solutions (see Theorem1) is established, a natural question is whether this local solution can exist globally. If the solution only exists in finite time, what induces the blow-up phenomenon? On the other hand, finding sufficient conditions to guarantee the finite time singularities or global existence is very interesting, especially for sufficient conditions added on certain initial data. The following results will give some positive answers.


## 3. Blow-Up

In this section, we pay more attention to the formation of singularities for strong solutions to our system. It will show that wave breaking is one way that singularities arise in smooth solutions. We start this section with the following lemmas.Lemma 5.
LetX0=(u0,π(ρ0))∈Hs×Hs-1/ℝ, s≥5/2, and let T be the maximal existence time of the solution X=(u,π(ρ)) to (7) with the initial data X0. Then we obtain the following conservation laws:
(39)E1=∫𝕊‍(u2+ux2+π(ρ)2)dx,E2=∫𝕊‍(u3+σuux2+uπ(ρ)2)dx.Proof.
The proofs are direct consequences of the energy method; similar ones are given in [28] for the two-component Camassa-Holm equations; we refer the readers to [28] for the details.Remark 6.
The conservation ofE1 guarantees the uniform bound of u(x,t), then Theorem 4 is also interpreted as wave breaking.Lemma 7 (see [10]).
For allf∈H1(𝕊), the following inequality holds:
(40)G*(f2+fx22)(x)≥C0f2(x),
with
(41)C0=12+arctan(sinh(1/2))2sinh(1/2)+2arctan(sinh(1/2))sinh2(1/2)≈0.869.
Moreover, C0 is the optimal constant obtained by the function
(42)f0=1+arctan(sinh(x-[x]-(1/2)))sinh(x-[x]-(1/2))1+arctan(sinh(1/2))sinh(1/2).Lemma 8 (see [33]).
For allf∈H1(𝕊), the following inequality holds:
(43)maxx∈[0,1]f2(x)≤C1∥f∥H1(𝕊)2,
where
(44)C1=e1/2+e-1/22(e1/2-e-1/2).
Moreover, C1 is the minimum value, so in this sense, C1 is the optimal constant which is obtained by the associated Green's function
(45)G(x)=cosh(x-[x]-(1/2))2sinh(1/2).Lemma 9 (see [33]).
For any functionf∈H2(𝕊), the following inequality holds:
(46)∥f(x)∥L∞(𝕊)2≤(∫𝕊‍f(x)dx)2+∥f(x)∥H1(𝕊)2.Lemma 10 (see [34]).
Assumef(x)∈Hs(𝕊), s>2. If ∫𝕊‍f(x)dx=0, then
(47)∥f(x)∥L∞(𝕊)2≤112∫𝕊‍fx2(x)dx,∫𝕊‍f2(x)dx≤112∫𝕊‍fx2(x)dx,∫𝕊‍f2(x)fx2(x)dx≤112(∫𝕊‍fx2(x)dx)2.It is now to state our result.Theorem 11.
LetX0=(u0,π(ρ0))∈Hs×Hs-1/ℝ, s≥5/2, and let T be the maximal existence time of solution X=(u,π(ρ)) to (7) with the initial data X0. If the following inequality holds:
(48)(1)∫𝕊‍u0x3dx<-2E1(0)κσ,if0<σ<1,(2)∫𝕊‍u0x3dx<-3(3-σ)(1-C0)σC1E13(0)+3σM*2E12(0),f<-3(3-σ)(1-C0)σC1E13(0)+3σif1≤σ<3,
where E1(0) is the initial value of E1 in Lemma 5, and constant κ is determined later. Then the corresponding solution X blows up in finite time.Proof.
Differentiating the first equation in system (7) with respect to x, we obtain
(49)uxt+σ(ux2+uuxx)+∂x2(G*(3-σ2u2+σ2ux2+12π(ρ)2))=0.
Applying the relation (1-∂x2)-1f=G*f to (49) gives
(50)uxt+σ2ux2+σuuxx+G*(3-σ2u2+σ2ux2+12π(ρ)2)-3-σ2u2-12π(ρ)2=0.
If 0<σ<1,multiplying (50) with ux2 and integrating by parts subsequently, we obtain
(51)13ddt∫𝕊‍ux3dx=-σ6∫𝕊‍ux4dx-∫𝕊‍ux2(G*(3-σ2u2+σ2ux2+12π(ρ)2))dxvv+∫𝕊‍ux2(3-σ2u2+12π(ρ)2)dx=-σ6∫𝕊‍ux4dx-σ∫𝕊‍ux2(G*(u2+ux22))dx-3-3σ2∫𝕊‍ux2(G*u2)dxhh-12∫𝕊‍ux2(G*π(ρ)2)dx+∫𝕊‍ux2(3-σ2u2+12π(ρ)2)dx≤-σ6∫𝕊‍ux4dx+3-σ-2σC02∫𝕊‍ux2u2dx+12∫𝕊‍ux2π(ρ)2dx,
where we have used the following identity:
(52)∫𝕊‍ux2uuxxdx=-13∫𝕊‍ux4dx.
Since 3-3σ>0, and
(53)∫𝕊‍ux2(G*π(ρ)2)dx≥0,∫𝕊‍ux2(G*u2)dx≥0,G*(u2+ux22)(x)≥C0u2(x),
where C0≈0.869<1.
In the following, we estimate the three terms on the right-hand side of (51) one by one. The Cauchy-Schwarz inequality implies that
(54)|∫𝕊‍ux3dx|≤(∫𝕊‍ux4dx)1/2(∫𝕊‍ux2dx)1/2,
hence
(55)|∫𝕊‍ux4dx|≥1∫𝕊‍ux2dx(∫𝕊‍ux3dx)2≥1E1(0)(∫𝕊‍ux3dx)2.
Using Lemma 8 and the invariant property of E1, we have
(56)∫𝕊‍ux2u2dx≤∥u∥L∞2|∫𝕊‍ux2dx|≤C1E12(0).
Note the fact that
(57)∫𝕊‍π(ρ)dx=∫𝕊‍ρ(x,t)dx-∫𝕊‍(∫𝕊‍ρ(x,t)dx)dx=0.
This implies that for any t∈[0,T), π(ρ) has at least a zero point ηt, that is, π(ρ(ηt,t))=0. Therefore, we have
(58)π(ρ)=∫ηtx‍π(ρ(s,t))sds,x∈[ηt,ηt+1].
Suppose that the solution does not blow-up in finite time, it follows that there exists a constant M*>0 such that ux(x,t)>-M* and ∥ρx(·,t)∥L∞(𝕊) is bounded by M*; thus
(59)|π(ρ)|=|∫ηtx‍π(ρ(s,t))sds|=|∫ηtx‍ρs(s,t)ds|≤M*.
Then
(60)|∫𝕊‍ux2π(ρ)2dx|≤M*2|∫𝕊‍ux2dx|≤M*2E1(0).
By (55)–(60), we have
(61)ddt∫𝕊‍ux3dx≤-σ2∫𝕊‍ux4dx+3(3-σ-2σC0)2∫𝕊‍ux2u2dxxx+32∫𝕊‍ux2π(ρ)2dx≤-σ2E1(0)(∫𝕊‍ux3dx)2fff+3(3-σ-2σC0)2C1E12(0)+32M*2E1(0).
For convenience, we denote
(62)φ(t)=∫𝕊‍ux3dx,κ=3(3-σ-2σC0)2C1E12(0)+32M*2E1(0),
where we noticed the fact that
(63)3-σ-2σC0>0.
That is,
(64)dφ(t)dt≤-σ2E1(0)φ2(t)+κ.
Note that if the initial quantity satisfies
(65)φ(0)<-2E1(0)κσ,
the standard argument on the Riccati type inequality and the initial hypothesis ensure that there exists a finite time T such that
(66)limt→T∫𝕊‍ux3(x,t)dx=-∞.
Since
(67)∫𝕊‍ux3dx≥infux(x,t)∫𝕊‍ux2dx>infux(x,t)E1(0).
This implies that
(68)limt→Tinfux(x,t)=-∞.
Then it contradicts the assumption ux(x,t)>-M*. By Theorem 4, we know that the solution must blow up in finite time.
If1≤σ<3, we get
(69)ddt∫𝕊‍ux3dx=-σ2∫𝕊‍ux4dx-3(3-σ)2∫𝕊‍ux2(G*(u2+ux22))dxfd-9(σ-1)4∫𝕊‍ux2(G*ux2)dxss-12∫𝕊‍ux2(G*π(ρ)2)dxss+3∫𝕊‍ux2(3-σ2u2+12π(ρ)2)dx≤-σ2∫𝕊‍ux4dx+3(3-σ)(1-C0)2∫𝕊‍ux2u2dxss+32∫𝕊‍ux2π(ρ)2dx.
We also denote that
(70)φ(t)=∫𝕊‍ux3dx.
It follows that
(71)dφ(t)dt≤-σ2E1(0)φ2(t)+3(3-σ)(1-C0)2C1E12(0)ss+32M*2E1(0),
where we note 3(3-σ)(1-C0)/2>0. Similar to case (1), it is easy to see that under the corresponding condition (2) of our theorem and the previous arguments, blow-up phenomenon occurs. We complete the proof.As we know, the key issue for partial differential equations lies in the estimates. In the following results, we apply different strategies to derive suitable bounds for solution, then blow-up phenomenon occurs while some special initial values are involved. Precisely, we show the following.Theorem 12.
Suppose thatX0=(u0,π(ρ0))∈Hs×Hs-1/ℝ,s≥5/2, u0 does not vanish identically. If ∫𝕊‍u0(x)dx=0, and one of the following conditions is satisfied
(72)(1)σ∈(3sinh(1/2)6+sinh(1/2),3),E1(0)>C2,(2)σ∈(0,3sinh(1/2)6+sinh(1/2)],φ(0)<-(3-σ4σ-32sinh(1/2))E13(0)+3σM*2E12(0),
for some constant C2 and function φ(t) is defined as previously, then the corresponding solution to initial data X0 of (7) blows up in finite time.Proof.
Differentiating both sides of the first equation of (7) with respect to variable x, we obtain
(73)uxt+σ(ux2+uuxx)+∂x2(G*(3-σ2u2+σ2ux2+12π(ρ)2))=0.
Applying the relation ∂x2(G*f)=G*f-f to (73) gives
(74)uxt+σ2ux2+σuuxx+G*(3-σ2u2+σ2ux2+12π(ρ)2)-3-σ2u2-12π(ρ)2=0.
Multiplying (74) with ux2 and integrating by parts subsequently, we obtain
(75)ddt∫𝕊‍ux3dx=-σ2∫𝕊‍ux4dx+3(3-σ)2∫𝕊‍ux2u2dxdd+32∫𝕊‍ux2π(ρ)2dxdd-3∫𝕊‍ux2(G*(3-σ2u2222222222222222+σ2ux2+12π(ρ)2))dx,
where we have used the following identity:
(76)∫𝕊‍ux2uuxxdx=-13∫𝕊‍ux4dx.
On the other hand, we know that ∫𝕊‍u(x,t)dx=0 in view of the hypothesis, and the following inequality holds
(77)12sinh(1/2)≤G(x)≤cosh(1/2)2sinh(1/2),
where
(78)cosh(1/2)2sinh(1/2)≈1.082>1.
Using Lemma 10 and (77), we obtain
(79)ddt∫𝕊‍ux3dx≤-σ2∫𝕊‍ux4dx+3-σ8(∫𝕊‍ux2dx)2+32∫𝕊‍ux2π(ρ)2dxdd-3σ2∫𝕊‍ux2(G*ux2)dx≤-σ2∫𝕊‍ux4dx+3-σ8(∫𝕊‍ux2dx)2cc+32∫𝕊‍ux2π(ρ)2dx-3σ4sinh(1/2)(∫𝕊‍ux2dx)2=-σ2∫𝕊‍ux4dx-(3σ4sinh(1/2)-3-σ8)(∫𝕊‍ux2dx)2dd+32∫𝕊‍ux2π(ρ)2dx.
In view of (60), we obtain that
(80)ddt∫𝕊‍ux3dx≤-σ2∫𝕊‍ux4dx-(3σ4sinh(1/2)-3-σ8)×(∫𝕊‍ux2dx)2+32M*2E1(0).
If 3sinh(1/2)/(6+sinh(1/2))<σ<3, it is easy to know that
(81)3σ4sinh(1/2)-3-σ8>0.
If there exists a constant C2 such that the initial energy E1(0)>C2, then there is some δ>0 such that
(82)32M*2E1(0)≤δ(∫𝕊‍ux2dx)2,3σ4sinh(1/2)-3-σ8-δ>0.
On the other hand, we have by Lemma 10(83)∫𝕊‍ux2dx≥1213∥u∥H1(𝕊)2.
Therefore, the previous inequality yields
(84)ddt∫𝕊‍ux3dx≤-σ2∫𝕊‍ux4dx-144169(3σ4sinh(1/2)-3-σ8-δ)∥u∥H1(𝕊)2.
In view of Hölder's inequality, there holds
(85)∫𝕊‍ux4dx≥(∫𝕊‍ux3dx)4/3.
For simplicity of notations, we denote by φ(t) and μ the following quantities:
(86)∫𝕊‍ux3dx,144169(3σ4sinh(1/2)-3-σ8-δ),
respectively. Therefore we have
(87)dφ(t)dt≤-σ2φ4/3(t)-μ∥u∥H1(𝕊)4.
First, we can easily get φ(t)≤φ(0)-μ∥u∥H1(𝕊)4t, and it is not difficult to find that there exists a t0≥0 such that φ(t0)<0. Then for t>t0, we get
(88)dφ(t)dt≤-σ2φ4/3(t),withφ(t0)<0.
Solving this inequality yields
(89)φ(t)≤(φ-1/3(t0)+σ6(t-t0))-3,
which approaches -∞ as t arrives at T0=-(6/σ)φ-1/3(t0)+t0; that is, there exists a time T≤-(6/σ)φ-1/3(t0)+t0 such that
(90)limt→T∫𝕊‍ux3dx=-∞.
Therefore, it follows that
(91)limt→Tinfx∈ℝux(x,t)=-∞.
Then it contradicts the assumption ux(x,t)>-M*. By blow-up scenario, we know that the solution must blow up in finite time.
If0<σ≤3sinh(1/2)/(6+sinh(1/2)), then we have
(92)ddt∫𝕊‍ux3dx≤-σ2∫𝕊‍ux4dxn+(3-σ8-3σ4sinh(1/2))f×(∫𝕊‍ux2dx)2+32M*2E1(0),
where
(93)3-σ8-3σ4sinh(1/2)≥0.
We also use φ(t) as previously to get
(94)dφ(t)dt≤-σ2E1(0)φ2(t)sx+(3-σ8-3σ4sinh(1/2))E12(0)+32M*2E1(0).
Similar arguments to (64) in Theorem 11 and condition (2) guarantee the finite time blow-up of solution to (7).The zero mean ofu(x,t) in the previous theorem can be substituted by E2(0)=0; blow-up still occurs with the aid of different estimate from (83).Theorem 13.
Assume thatX0=(u0,π(ρ0))∈Hs×Hs-1/ℝ, s≥5/2, u0 does not vanish identically, ∫𝕊‍(u03+σu0u0x2+u0π(ρ0)2)dx=0, if one of the following conditions is satisfied:
(95)(1)σ∈(3sinh(1/2)2+sinh(1/2),3),E1(0)>C3,(2)σ∈(0,3sinh(1/2)2+sinh(1/2)],φ(0)<-(3(3-σ)4σ-32sinh(1/2))E13(0)+3σM*2E12(0),
for some constant C3, then the corresponding solution to (7) blows up in finite time.Proof.
By assumption and the invariance property ofE2, we have
(96)∫𝕊‍u(u2+σux2+π(ρ)2)dx=∫𝕊‍(u3+σuux2+uπ(ρ)2)dx=0.
Therefore, u(x,t) must change sign, so there exists at least one zero point on 𝕊. Then for each t∈[0,T), suppose that there is a ξt∈[0,1] such that u(ξt,t)=0, for x∈𝕊 we have
(97)u2(x,t)=(∫ξtx‍uxdx)2≤(x-ξt)∫ξtx‍ux2dx,x∈[ξt,ξt+12].
Thus, the previous relation and an integration by parts yield
(98)∫ξtξt+1/2‍u2ux2dx≤∫ξtξt+1/2‍(x-ξt)ux2(∫ξtx‍ux2dz)dx=∫ξtξt+1/2‍(x-ξt)(∫ξtx‍ux2dz)d(∫ξtx‍ux2dz)=14(∫ξtξt+1/2‍ux2dz)2sz-12∫ξtξt+1/2‍(∫ξtx‍ux2dz)2dx≤14(∫ξtξt+1/2‍ux2dz)2.
Doing a similar estimate on [ξt+1/2,ξt+1], we obtain
(99)∫𝕊‍u2ux2dx≤14(∫𝕊‍ux2dx)2.
In view of (97), we also have
(100)∥u(x,t)∥L∞(𝕊)2≤12∫𝕊‍ux2dx.
Let
(101)φ(t)=∫𝕊‍ux3dx,t≥0,
we obtain
(102)dφ(t)dt=-σ2∫𝕊‍ux4dx+3(3-σ)2∫𝕊‍ux2u2dx+32∫𝕊‍ux2π(ρ)2dx-3∫𝕊‍ux2(G*(3-σ2u2+σ2ux2+12π(ρ)2))dx≤-σ2∫𝕊‍ux4dx-14(3σsinh(1/2)-3(3-σ)2)×(∫𝕊‍ux2dx)2+32∫𝕊‍ux2π(ρ)2dx,
where we use the fact 1/2sinh(1/2)≤G(x)≤cosh(1/2)/2sinh(1/2), then
(103)∫𝕊‍ux2(G*ux2)dx≤∥G*ux2∥L∞(𝕊)≤12sinh(1/2)(∫𝕊‍ux2dx)2.
For σ∈(3sinh(1/2)/(2+sinh(1/2)),3),  we have
(104)3σsinh(1/2)-3(3-σ)2≥0.
If there is a constant C3 such that the initial energy E1(0)>C3, there exists some δ>0 such that
(105)32M*2E1(0)≤δ4(∫𝕊‍ux2dx)2,3σsinh(1/2)-3(3-σ)2-δ>0.
On the other hand, we have ∫𝕊‍ux2dx≥(2/3)∥u∥H1(𝕊)2. Therefore, the previous inequality yields
(106)dφ(t)dt≤-σ2∫𝕊‍ux4dx-19(3σsinh(1/2)-3(3-σ)2-δ)×∥u∥H1(𝕊)4≤-σ2φ4/3(t)-19(3σsinh(1/2)-3(3-σ)2-δ)×∥u∥H1(𝕊)4.
The remaining part is very close to Theorem 12, so we omit it. So the solution must blow up in finite time.
Forσ∈(0,3sinh(1/2)/(2+sinh(1/2))], then
(107)dφ(t)dt≤-σ2E1(0)φ2(t)cc+(3(3-σ)8-3σ4sinh(1/2))E12(0)ss+32M*2E1(0),
where 3(3-σ)/8-3σ/4sinh(1/2)≥0. Note that if the initial quantity satisfies
(108)φ(0)<-(3(3-σ)4σ-32sinh(1/2))E13(0)+3σM*2E12(0),
then condition (2) can conclude that the solution to (7) goes to -∞ in finite time. This completes the proof.Theorem 14.
Suppose thatX0=(u0,π(ρ0))∈Hs×Hs-1/ℝ,s≥5/2, X=(u,π(ρ)) is the solution to system (7) with the initial data X0. If there is some point x0∈𝕊 such that one of the following conditions is satisfied:
(109)(1)u0′(x0)<-3-σ-2σC0σC1E1(0),0<σ<1,(2)u0′(x0)<-3-σσ(1-C0)C1E1(0),1≤σ<3,
where the constants C0 and C1 are given in Lemmas 7 and 8, E1(0) is the initial value of E1, then the solution X must blow up in finite time.Proof.
Differentiating the first equation in system (7) with respect to x and noticing that ∂x2(G*f)=G*f-f, we have
(110)uxt+σ2ux2+σuuxx+G*(3-σ2u2+σ2ux2+12π(ρ)2)-3-σ2u2-12π(ρ)2=0.
When 0<σ<1, this equation, in combination with (11), yields
(111)ddtux(q2(x,t),t)=(uxt+σuuxx)(q2,t)=(-σ2ux2+3-σ2u2+12π(ρ)2ccc-G*(3-σ2u2+σ2ux2+12π(ρ)2))(q2,t)≤(-σ2ux2+3-σ2u2+12π(ρ)2ccc-σG*(u2+ux22)-3-3σ2G*u2)(q2,t).
Note that
(112)∫𝕊‍π(ρ)dx=0.
We can deduce that there exists at least one point x0 such that π(ρ(q2(x0,t),t))=0 for t∈[0,T). Let us consider this problem at (q2(x0,t),t). For convenience, we denote ux(q2(x0,t),t)=m(t). Then we have
(113)dm(t)dt≤-σ2m2(t)+3-σ2u2-σG*(u2+ux22)≤-σ2m2(t)+3-σ-2σC02C1E1(0).
Using the notation ϱ=(3-σ-2σC0/σ)C1E1(0), where 3-σ-2σC0>0, we have
(114)dm(t)dt≤-σ2(m2(t)-ϱ).
In view of the initial condition, it is not difficult to obtain
(115)dm(t)dt≤σ(δ-1)2m2(t),
with 0<δ<1 determined by δm2(0)=ϱ. Then, by using the standard arguments for this type of inequality and our hypothesis, it is easy to conclude that the lifespan of the solution is finite; that is, blow-up phenomenon occurs.
When1≤σ<3, then
(116)dm(t)dt≤-σ2ux2+3-σ2u2-3-σ2G*(u2+ux22)xx-3σ-34(G*ux2)≤-σ2ux2+3-σ2(1-C0)u2≤-σ2m2(t)+3-σ2(1-C0)C1E1(0),
where (3-σ)(1-C0)>0. Similar to the arguments about (113), under the condition (2), the corresponding solution X blows up in finite time. We complete the proof.Theorem 15.
Suppose thatX0=(u0,π(ρ0))∈Hs×Hs-1/ℝ,s≥5/2, X=(u,π(ρ)) is the solution to system (7) with the initial data X0. If there holds one of the following conditions for some point x0∈𝕊(117)(1)u0′(x0)<-3-σ-2σC0σK(0),0<σ<1,(2)u0′(x0)<-3-σσ(1-C0)K(0),1≤σ<3,
where the constant C0 is the best constant given by Lemma 7 and
(118)K(0)=(∫𝕊‍u0(x)dx)2+∫𝕊‍(u02+u0x2+π(ρ0)2)dx.
Let T be the maximal existence time of the corresponding solution to (7) with the initial data X0. Then T is finite.Proof.
We easily know that∫𝕊‍u(x,t)dx is also an invariant with respect to time. The result follows by using Lemma 9 that
(119)∥u(x)∥L∞(𝕊)2≤(∫𝕊‍u(x)dx)2+∥u(x)∥H1(𝕊)2<(∫𝕊‍u0(x)dx)2+E0=K0,
in (113) instead of ∥u(x)∥L∞(𝕊)2≤C1∥u(x)∥H1(𝕊)2≤C1E1(0).In the following, as a byproduct, we examine the blow-up rate while the solution blows up in finite time.Theorem 16.
Assume thatX0=(u0,π(ρ0))∈Hs×Hs-1/ℝ,s≥5/2, X=(u,π(ρ)) is the corresponding solution with the initial data X0. If there holds the condition of Theorem 14, then we have the following description:
(120)limt→T(T-t)m(t)=-2σ,
where σ>0 and m(t) is defined in Theorem 14.Proof.
The conclusion follows from the theory of ordinary differential equations to inequality (113). Indeed, when 0<σ<1, by (113) we have
(121)dm(t)dt≤-σ2m2(t)+3-σ-2σC02C1E1(0).
In view of Lemma 8 and the conservation of E1, there holds for all t∈[0,T) that
(122)|dm(t)dt+σ2m2(t)|≤σϱ2.
It follows that
(123)-σϱ2≤dm(t)dt+σ2m2(t)≤σϱ2,a.e.on(0,T).
Since limt→Tm(t)=-∞ by Theorem 14, it implies that for any ε∈(0,σ/2) there exists a t0 such that m2(t)>σϱ/2ε for all t∈[t0,T). Therefore,
(124)-m2(t)ε≤dm(t)dt+σ2m2(t)≤m2(t)ε,
that is,
(125)-σ2-ε≤1m2(t)dm(t)dt≤-σ2+ε.
Direct integration from t to T gives
(126)-σ2-ε≤1(T-t)m(t)≤-σ2+ε,
and the arbitrariness of ε leads to our result.
When1≤σ<3, by (116) we obtain
(127)dm(t)dt≤-σ2m2(t)+σω2,
where ω=((3-σ)/σ)(1-C0)C1E1(0). Similar to the arguments for 0<σ<1, we can get the same result.



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*Source: 730860-2013-07-17.xml*