Blow-up of solutions for the dissipative Dullin–Gottwald–Holm equation with arbitrary coefficients

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Blow-up of solutions for the dissipative Dullin–Gottwald–Holm equation with arbitrary coefficients Emil Novruzov Department of Mathematics, Gebze Technical University, Turkey Received 1 February 2016; revised 27 March 2016

Abstract We study the Cauchy problem for the weakly dissipative Dullin–Gottwald–Holm equation describing unidirectional propagation of surface waves in a shallow water regime:   ut − α 2 uxxt + c0 ux + 3uux + γ uxxx + λ u − α 2 uxx = α 2 (2ux uxx + uuxxx ). In the present paper we demonstrate the simple conditions on the initial data that lead to blow-up of the solution in finite time. © 2016 Published by Elsevier Inc. Keywords: Weakly dissipative Dullin–Gottwald–Holm equation; Blow-up

1. Introduction In the paper we consider the following Cauchy problem   ut − α 2 uxxt + c0 ux + 3uux + γ uxxx + λ u − α 2 uxx = α 2 (2ux uxx + uuxxx )

(1)

u (0, x) = u0 (x) ,

(2)

E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jde.2016.03.034 0022-0396/© 2016 Published by Elsevier Inc.

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where the constants α 2 (α > 0) and cγ0 are squares of length scales, c0 ≥ 0 is the linear wave speed for undisturbed water resting at spatial infinity and u (t, x) stands for the fluid velocity. The equation under consideration is the so-called Dullin–Gottwald–Holm equation with weakly dissipative term λ (u − uxx ). Dullin–Gottwald–Holm (DGH) equation describing the unidirectional propagation of surface waves in a shallow water regime was derived by the method of asymptotic analysis and a near-identity normal form transformation from water wave theory, combining the linear dispersive of the KdV equation with the nonlinear dispersion of the Camassa–Holm equation. It is completely integrable and its travelling wave solutions contain both the KdV solitons and the CH peakons as limiting cases [6,7,19]. Note that this is especially interesting since the governing equations for water waves admit regular travelling waves that resemble those of the KdV equation, as well as limiting waves — the waves of greatest height — that present a peak at the wave crest, like the CH-peakons (see the discussion in [8,14]). When α = 0, DGH equation becomes the well-known KdV equation. Bourgain [4] proved that solutions to the KdV equation are global as long as the initial data is square integrable. Another remarkable property is that it is integrable and the solitary waves are orbitally stable [2]. For α = 1 and γ = 0, Eq. (1) becomes the Camassa–Holm equation [6,7] ut − utxx + kux + 3uux = 2ux uxx + uuxxx ,

t > 0, x ∈ R

where k is a dispersive coefficient related to the critical shallow water speed. The details concerning the hydrodynamical relevance of Camassa–Holm equation were mathematically rigorously described in [10], where, in addition, authors investigate in what sense model under consideration gives us insight into the wave breaking phenomenon. Alternative derivations of Camassa–Holm equation as a equation for geodesic flow on the diffeomorphism group of the circle were presented by Constantin and Kolev [9] and Ionescu-Kruse [23]. The equation has bi-Hamiltonian structure [20] and is completely integrable [1,3,7,11,12]. Note that local well-posedness for the initial datum u0 (x) ∈ H s with s > 3/2 was proved by several authors (see, for example, [15, 26,28]). For the initial data with lower regularity, we refer to papers [5] and [29]. When k = 0, Camassa–Holm equation possesses a solitary wave with discontinuous first derivatives [6], which is named “peakon” (travelling wave solutions with a corner at their peak). More importantly, the peakons are orbitally stable [17,25], which means that the shape of the peakons is stable so that these wave patterns are physically recognizable. Wave breaking for a large class of initial data has been established in [15,16,26,35,36] and in the recent paper [24], where, in particular, new and direct proof for the result from [27] on the necessary and sufficient condition for wave breaking was presented. It is also worth mentioning that from the point of view of theory of water waves the fact that solutions that originate from smooth localized initial data can develop singularities only in the form of breaking waves, as proved in the paper [13], is especially interesting. The infinite propagation speed for the Camassa–Holm equation for k = 0 was investigated in [18,21, 22]. Also, the wide range of problems for CH equation with non-zero dispersion coefficient was considered in [26,33,37]. In particular, certain conditions on the initial datum to guarantee that the corresponding solution exists globally or blows up in finite time were established. The local well-posedness, global existence and blow-up profile of the Cauchy problem for DGH equation were considered in [19,30,33,34,38,39]. In the paper [31], the following dissipative equation was considered ut − uxxt + k(u − uxx )x + 3uux + λ (u − uxx ) = 2ux uxx + uuxxx .

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Under the certain integral condition on the initial data a new blow-up condition was presented. The same equation for arbitrary (not necessarily positive) solution has been considered in [32]. Authors demonstrated the simple conditions on the initial data that lead to blow-up of the solution in finite time or guarantee that the solution exists globally and investigated the propagation speed for  the equation under consideration. In particular, it is proved that under condition  2 λ + u0 H 1 < u0 (x0 ) − u0x (x0 ) the corresponding solution blows up in finite time. The obtained condition is simple and convenient, in particular, it doesn’t require a condition of the changing of the sign of the potential y0 = u0 − u0xx in the point x0 ; also, this condition does not contain the power function depending on λ, u0 H 1 , etc. It is easy to see that in the mentioned paper α = 1 and c0 + αγ2 = 0. It is worth to note that this case has some additional properties   which is not true for c0 + αγ2 > 0. For example, we do not have the identity connecting the sign of potential y = u − α 2 uxx at time t with the sign of y0 , if c0 + αγ2 is not equal to 0. In the present paper we would like to extend the obtained in [32] “blow-up” result to the  dissipative DGH equation with arbitrary positive α and c0 + αγ2 ≥ 0. So, we will prove the following result.   Theorem 1. Suppose u0 ∈ H s (R), s  3 and c0 + αγ2 ≥ 0. Also, for some point x0 , assume    1 1 1 u0 H 1 + 2α − 14 c0 + 3 αγ2 − 2αγ 2 for α ≤ 1 and that 2α (u0 (x0 ) − αu0x (x0 )) ≥ λ + √ α 2       γ 1 α 1 1 √ u0 H 1 + 1 c0 + 3 γ2 + γ2 − 1 ≥ − + c − αu ≥ λ + (u (x ) (x )) 0 0 0x 0 0 2 2α 4α 2α 4α α α α 2 for α > 1. Then the corresponding solution u (t, x) to Eq. (1) blows up in finite time. Our paper is organized as follows. In Section 2, we recall several useful results which are crucial in the proof of the new blow-up result. We give this proof in Section 3. 2. Preliminaries In this section, we recall several useful results in order to pursue our goal. Note that, the local well-posedness of the Cauchy problem for (1) with initial data u0 (x) ∈ H s (R), s > 32 and λ = 0 can be obtained by applying the Kato’s theorem (see, for example, [33]). It is easy to see that the same result holds for equation (1). The proof of this fact repeats almost word for word the corresponding proof for the case λ = 0, that is why we omit the further details and present corresponding result directly. (Throughout this paper, for 1  p  ∞, the norm in the Lebesgue space Lp (R) will be written by ·Lp , while ·H s , s > 0, will stand for the norm in the classical Sobolev spaces H s (R).) Lemma 1. Given u0 (x) ∈ H s (R), s > 32 , there exist a maximal T = T (u0 , α, c0 , γ , λ) > 0 and a unique solution u to Eq. (1), such that     u = u (·, u0 (x)) ∈ C [0, T ) ; H s (R) ∩ C 1 [0, T ) ; H s−1 (R) . Moreover, the solution depends continuously on the initial   data, i.e. the mapping u0 → u (·, u0 ) : H s (R) → C ([0, T ) ; H s (R)) ∩ C 1 [0, T ) ; H s−1 (R) is continuous and the maximal time of existence T > 0 can be chosen to be independent of s.

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The following lemma gives necessary and sufficient condition for the blow-up of the solution. Lemma 2. Given u0 (x) ∈ H s (R), s > 32 , the solution u = u (·, u0 ) of Eq. (1) blows up in a finite time T > 0 if and only if   lim inf inf [ux (x, t)] → −∞.

t→T

x∈R

Next, multiplying both sides of (1) by u, then integrating by parts on R, by simple calculation, we get 1 d 2 dt



 (u2 + α 2 u2x )dη + λ

R

(u2 + α 2 u2x )dη = 0. R

Thus, we finally can formulate the following lemma which will be used in the sequel. Lemma 3. Let u0 ∈ H s (R), s  3, and let T > 0 be the maximal existence time of the corresponding solution u to Eq. (1). Then we have 

(u2 + α 2 u2x )dη = e−2λt

R

 (u20 + α 2 u20x )dη, ∀ t ∈ [0, T ) .

(3)

R

3. Proof of the main result We are now ready to prove Theorem 1. Proof of Theorem 1. We consider only case α ≤ 1, because for α > 1 the proof is similar. First, it is convenient to rewrite the equation in the following form γ ut − α 2 uxxt + 2kux − 2 (u − α 2 uxx )x + 3uux + α   λ u − α 2 uxx = α 2 (2ux uxx + uuxxx ) , where c0 +

γ α2

(4)

= 2k ≥ 0. Let v = u + k. Then v satisfies the equation

 γ    vt − α 2 vxxt + − 2 − k (v − α 2 vxx )x + 3vvx + λ v − α 2 vxx − λk = α 2 (2vx vxx + vvxxx ) . α We introduce the following notation: y = v − α 2 vxx and q(x  0 ,t)

E (t) = −∞

1 ξ e α y (ξ, t) dξ, α

where q (x0 , t) solution of the following problem (see [27] and [33], for example)

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qt = u (q (x, t) , t) −

5

γ , α2

(5)

q (x, 0) = x.

(6)

Then, differentiating E (t) we get dE (t) = dt

q(x  0 ,t) −∞





1 ξ 1 q t,x0 e α yt (ξ, t) dξ + e α y (q (x0 , t) , t) qt (x0 , t). α α

(7)

From Eq. (1) one can get the equation for y yt = −yx v − 2yvx + k1 yx − λy + λk, where k1 = k + q(x  0 ,t) −∞

γ . α2

(8)

Therefore, integration by parts and Eq. (8) yield the following identities

1 ξ e α yt (ξ, t) dξ = α

q(x  0 ,t) −∞

 1 ξ e α − (yv)x − yvx + k1 yx − λy + λk dξ = α

q(x  0 ,t)  q(x0 ,t) 1 ξ q(t,x0 ) 1 1 ξ 1 ξ 2 2 2 − yv e α + + e α yvdξ − e α v − α vx α α α 2α −∞ −∞ −∞

q(x  0 ,t) −∞

q(x q(x q(x  0 ,t)  0 ,t)  0 ,t)  ξ ξ ξ 1 1 ξ 2 1 λk k1 2 2 e α yx dξ − λye α dξ + e α dξ = e α v − α vx dξ + 2α α α α α −∞

−∞

−∞

q(x  0 ,t)   ξ q(x0 ,t) ξ ξ q(x0 ,t) 1 1 1 2 ξ q(x0 ,t) α − yve + 2 e α v 2 + α 2 vx2 dξ − vx ve α + − v eα −∞ −∞ −∞ α 2α α −∞

q(x q(x  0 ,t)  0 ,t)   q(x0 ,t)  ξ ξ 1 ξ 2 1 1 2 2 2 − 2 v e α dξ + 2 e α v 2 − α 2 vx2 dξ + e α v − α vx 2α 2α 2α −∞ −∞

 k1 ξ q(x0 ,t) 1 k1 − e α y λ+ −∞ α α α 1 α2

q(x  0 ,t)

e −∞

ξ α



−∞

q(x  0 ,t) −∞

λk e ydξ + α ξ α

q(x  0 ,t)

ξ

e α dξ = −∞

 q(x   0 ,t) ξ ξ q(x0 ,t) 1 2 2 k1 1 e α ydξ − vx ve α + λ+ v + α vx dξ − −∞ 2 α α 2

−∞

ξ q(x0 ,t) ξ q(x0 ,t) α ξ 2 q(x0 ,t) 1 − y (v − k1 ) e α + λk e α . e α vx −∞ −∞ 2 α −∞

(9)

We need to estimate from below the integral from the right-hand side of the previous equality

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6 q(x  0 ,t)

e

ξ α

v2 +

α2

−∞

2

⎜1 vx2 dξ = ⎝ 2

⎛ ⎜1 ⎝ 2 ⎛ ⎜1 ⎝ 2 ⎛ ⎜1 ⎝ 2

⎛



q(x  0 ,t)

q(x  0 ,t)

e v 2 dξ + −∞

ξ

e α v 2 dξ + −∞ q(x  0 ,t)

ξ α

ξ

e α v 2 dξ + −∞

1 2

α 2

1 2

⎞   ⎟ e v 2 + α 2 vx2 dξ ⎠ ≥

q(x  0 ,t) −∞

q(x  0 ,t)

ξ α

⎞

ξ ⎟ 2αe α vvx dξ ⎠ ≥

−∞

⎞   ξ ⎟ e α v 2 dξ ⎠ ≥

q(x  0 ,t)

x

−∞

⎞ q(x 0 ,t)      ξ ξ α ξ ⎟ q(x0 ,t) 1 e α v 2 dx + e α v 2 − e α v 2 dξ ⎠ ≥ −∞ 2 2

q(x  0 ,t) −∞

−∞

  α ξ  2  q(x0 ,t) α q x0 ,t 2 = e α v (q (x0 , t) , t) . eα v −∞ 2 2

(10)

Then, inserting (9) and (10) into (7) and bearing in the mind (5) (it is easy to see that u (q (x, t) , t) − αγ2 = v (q (x, t) , t) − k1 and consequently, qt = v (q (x, t) , t) − k1 ) we have

 q(x  0 ,t) ξ dE (t) 1 k1 e α ydξ ≥ + λ+ dt α α −∞









1 q x0 ,t 1 q t,x0 e α y (q (x0 , t) , t) qt (q (x0 , t) , t) − e α y (q (x0 , t) , t) (v (q (x0 , t) , t) − k1 ) α α 1 α2

q(x  0 ,t)

ξ

eα −∞

v2 +

 ξ q(x0 ,t) ξ q(x0 ,t) α2 2 α ξ q(x0 ,t) + e α vx2 + λk e α ≥ vx dξ − vx ve α −∞ −∞ 2 2 −∞ 



q x0 ,t 1 q(t,x0 ) 2 e α v (q (x0 , t) , t) − e α vx (q (x0 , t) , t) v (q (x0 , t) , t) 2α   ξ q(x0 ,t) α q x0 ,t 2 α vx (q (x0 , t) , t) + λk e α = + e −∞ 2     q x0 ,t 1 q x0 ,t e α (v (q (x0 , t) , t) − αvx (q (x0 , t) , t))2 + λke α . 2α

Note that condition of the theorem yields 

1 1 1 1 −1 (u0 (x0 , 0) − αu0x (x0 , 0)) ≥ λ + √ u0  + k + k1 2α 2 α α 2 or, by equality v0 = u0 + k,

(11)

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1 1 (v0 (x0 , 0) − αv0x (x0 , 0)) ≥ λ + √ u0  + 2α α 2  



1 1 1 + k + k1 − 1 > 0. 2 2α α

(12)

Hence, integration by parts gives that E (0) > 0: 1 E (0) = α 1 ( α

x0

x0

1 e y0 (x) dξ = α ξ α

−∞

x0

ξ α

e v0 (ξ ) dξ + α

−∞

−∞

x0

x0

  ξ e α v0 − α 2 v0xx (ξ ) dξ =

−∞

1 e v0x (ξ ) dξ − α e v0x (x0 )) = ( α ξ α

x0 α

2

x0

ξ

e α v0 (ξ ) dξ −

−∞

x0

ξ

x0

x0

e α v0 dξ + αe α v0 (x0 ) − α 2 e α v0x (x0 )) = e α (v0 (x0 ) −

− −∞

αv0x (x0 )) ≥ 2αe

x0 α



 

1 1 1 1 + k + k1 −1 > 0. λ + √ u0 H 1 + 2 2α α α 2

(13)

By similar way we obtain 1 E (t) = α 1 ( α

q(x  0 ,t)

q(x  0 ,t) −∞

1 e y (ξ, t) dξ = α q(x  0 ,t)

ξ α

e vdξ + α −∞

1 ( α

q(x  0 ,t)

  ξ e α v − α 2 vxx dξ =

ξ α

−∞

ξ α

e vx dξ − αe

  q x0 ,t α

vx (q (x0 , t) , t)) =

−∞

q(x  0 ,t)

ξ α

q(x  0 ,t)

e vdξ − −∞

ξ

e α vdξ + αe

  q x0 ,t α

v (t, q (t, x0 )) −

−∞

αeq(x0 ,t) vx (q (x0 , t) , t)) = e

  q x0 ,t α

(v (q (x0 , t) , t) − αvx (q (x0 , t) , t)).

Thus, from (11) and (14) we obtain

 q(x  0 ,t)   ξ dE (t) 1 1 q x0 ,t k1 α α e ydξ ≥ + λ+ e (v (q (x0 , t) , t) − αvx (q (x0 , t) , t))2 + dt α α 2α −∞

λke Therefore,

  q x0 ,t α

.

(14)

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     q x0 ,t dE (t) k1 1 − q x0 ,t ≥ e α (E (t))2 − λ + E (t) + λke α . dt 2α α Multiplying the obtained inequality by e− e

−

  q x0 ,t α

  q x0 ,t α

we get

   dE (t) k1 −q(x0 ,t) 1 −2 q x0 ,t 2 α E (t) − λ + E (t) + λk. ≥ e e dt 2α α −

  q x0 ,t α

Adding the expression −qt (x0 , t) e E (t) to the right-hand side of the inequality and (v(q (x0 , t) , t) − k1 ) e−q(x0 ,t) E (t) to the left-hand side of the inequality, by virtue of equality qt = v − k1 we have 











q x0 ,t q x0 ,t dE (t) 1 −2 q x0 ,t α −qt (x0 , t) e− α E (t) + e− α E (t)2 ≥ e dt 2α

     q x0 ,t k1 − q x0 ,t − λ+ e α E (t) − (v (q (x0 , t) , t) − k1 ) e− α E (t) + λk. α

Therefore, 

  q x ,t d 1 − α0 E (t) ≥ e (v (q (x0 , t) , t) − αvx (q (x0 , t) , t))2 − dt 2α 

    q x0 ,t k1 − q x0 ,t e α E (t) − (v (q (x0 , t) , t) − k1 ) e− α E (t) + λk. λ+ α Then by (14) we obtain that d 1 (v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) ≥ (v (q (x0 , t) , t) − αvx (q (x0 , t) , t))2 − dt 2α 



1 − 1 (v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) + λk. λ + v (q (x0 , t) , t) + k1 α Hence, d (v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) ≥ (v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) × dt 1 ( (v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) − λ − 2α 1 v (q (x0 , t) , t) − k1 ( − 1)) + λk. α

(15)

d So, by (15), inequality dt (v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) > 0 holds as long as v (q (x0 , t) , t) − αvx (q (x0 , t) , t) > 0 and



1 1 (v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) ≥ λ + v (q (x0 , t) , t) + k1 −1 . 2α α

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However, in view of Lemma 3, 1 1 max (1, α) −λt 1 uL∞ ≤ √ uH 1 ≤ √ e u0 H 1 = √ e−λt u0 H 1 min α) (1, 2 2 α 2 and using (12) we obtain that

 1 1 1 1 + + (v0 (x0 , 0) − αv0x (x0 , 0)) ≥ λ + √ u0 H 1 + k 2α 2 2α α 2   



1 1 1 1 − 1 ≥ λ + u0 (x0 ) + k + + k1 −1 = k1 α 2 2α α  



1 1 k1 − 1 ≥ λ + v0 (x0 ) + k1 − 1 > 0. α α 1 (According to condition of Theorem 1, for α > 1 we have 2α (u (x ) − αu (x )) ≥ λ +       0 0  0x  0 γ γ α 1 1 α 1 1 √ u0 H 1 + k 1 − √ u0 H 1 + − 1 = λ + + 3 − 1 ≥ + k c + 1 0 2 2 2α α 4α 2α α α 2  2  γ 1 k 1 . It implies that 2α c0 + α 2 = − 2α − 4α (v0 (x0 , 0) − αv0x (x0 , 0)) ≥ λ + √α u0 H 1 + k +   2   1 k1 α1 − 1 ≥ 0 and 2α (v0 (x0 , 0) − αv0x (x0 , 0)) ≥ λ + v0 (x0 ) + k1 α1 − 1 .) Taking into account that v (q (x0 , t) , t) − αvx (q (x0 , t) , t) is a continuous function, by virtue d of (15) we have dt (v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) > 0 for t ∈ [0, t0 ) with some t0 > 0. Therefore,

 

1 1 1 1 1 + + k1 −1 (v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) ≥ λ + √ u0 H 1 + k 2α 2 2α α α 2 and by Lemma 3 we have that

 

1 1 1 1 1 + + k1 −1 ≥ (v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) ≥ λ + √ u0 H 1 + k 2α 2 2α α α 2  



1 1 1 − 1 ≥ λ + v (q (x0 , t) , t) + k1 −1 λ + √ u (t)H 1 + k + k1 α α 2 for t ∈ [0, t0 ). Consequently,

d dt

(v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) > 0 for all t ≥ t0 . Therefore,

 

1 1 1 1 1 + + k1 −1 (v (q (x0 , t) , t) − αvx (q (x0 , t) , t)) ≥ λ + √ u0 H 1 + k 2α 2 2α α α 2 for all t which belong to lifespan of the solution. However, the identity (3) implies that v (q (x0 , t) , t) = u (q (x0 , t) , t) + k → k as t → ∞ and therefore, there exists T0 such that √



2 1 1 u0 H 1 + k + 2k1 −ux (q (x0 , t) , t) ≥ 2λ + −1 α 2 α and

(16)

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1 1 − ux (q (x0 , t) , t) − λ = − vx (q (x0 , t) , t) − λ > 2 2 

1 1 1 − 1 for t > T0 . √ u0 H 1 + k + k1 4 α α 2

(17)

Thus, from (16)–(17) we obtain that

 1 2 1 − ux (q (x0 , t) , t) − λux (q (x0 , t) , t) = ux (q (x0 , t) , t) − ux (q (x0 , t) , t) − λ = 2 2

 1 − |ux (q (x0 , t) , t)| − ux (q (x0 , t) , t) − λ < 2  √ 

2 1 1 u0 H 1 + k + 2k1 − 2λ + −1 × α 2 α



1 1 1 (18) −1 = −C1 for t > T0 . √ u0 H 1 + k + k1 4 α α 2 −1  1 − |x| Meanwhile, note that if p (x) = 2α e α , then 1 − α 2 ∂x2 f = p ∗ f for all f ∈ L2 (R) and p ∗ y = u, where we denote the convolution by ∗. Using the last identity, we can rewrite (1) as follows: ut + uux −

 γ α2 2  γ  2 u + ∂ p ∗ u + + c + u u + λu = 0. x x 0 2 x α2 α2

(19)

Differentiating (19) with respect to the special variable, in view of identity α 2 ∂x2 p ∗f = p ∗f −f and c0 + γ = 2k, we have

  γ  1 u2 2ku 1 α2 utx + u − 2 uxx = − u2x + 2 + 2 − 2 p ∗ u2 + u2x + 2ku − λux . 2 2 α α α α It follows from the definition of q(x, t) and (20) that d ux (q (x0 , t) , t) = utx (q (x0 , t) , t) + qt (x0 , t) uxx (q (x0 , t) , t) = dt γ utx (q (x0 , t) , t) + (u (q (x0 , t) , t) − 2 )uxx (q (x0 , t) , t) = α 1 2 1 2 2ku − ux (q (x0 , t) , t) + 2 u (q (x0 , t) , t) + 2 − 2 α α

 2 1 α p ∗ u2 + u2x + 2ku (q (x0 , t) , t) − λux (q (x0 , t) , t) ≤ 2 α2 1 2ku 1 − u2x (q (x0 , t) , t) + 2 u2 (q (x0 , t) , t) + 2 2 α α 1 − 2 2kp ∗ u (q (x0 , t) , t) − λux (q (x0 , t) , t) ≤ α

(20)

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1 2 1 u (q (x0 , t) , t) − u2x (q (x0 , t) , t) + 2 2 α 1 1 2k |p ∗ u (q (x0 , t) , t)| + 2 2ku − λux (q (x0 , t) , t) . 2 α α

11

(21)

Next, using that |p ∗ u (q (x0 , t) , t)| ≤ uL∞ ≤ 1 1 √ uH 1 ≤ √ e−λt u0 H 1 (see Lemma 3) 2 α 2 and applying (18) and (21) we get d 1 4 ux (q (x0 , t) , t) < 2 u2 (q (x0 , t) , t) + √ ke−λt u0 H 1 − C1 . 3 dt α α 2

(22)

Further, for sufficiently large t , by Lemma 3, we get 1 2 4 1 u (q (x0 , t) , t) + √ ke−λt u0 H 1 < C1 . 2 α2 α3 2

(23)

Therefore, according to (22) and (23), there exists T1 such that for t > T1 > T0 d 1 ux (q (x0 , t) , t) < − C1 . dt 2 Thus, we arrive at 1 ux (q (x0 , t) , t) < ux (q (x0 , T1 ) , T1 ) − C1 (t − T1 ) . 2 That means that we can choose T2 > T1 such that 1 −λux (q (x0 , t) , t) < u2x (q (x0 , t) , t) 8

(24)

and (by (3)) 1 2 1 1 1 u (q (x0 , t) , t) + 2 2k |p ∗ u (q (x0 , t) , t)| + 2 2ku < u2L∞ + 4k uL∞ < 2 2 2α α α 1 4 1 (25) √ e−2λt u0 2H 1 + √ ke−λt u0 H 1 < u2x (q (x0 , t) , t) 8 2α 2 α 2 for t > T2 . Then, substituting (24) and (25) into (21) we obtain

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12

d 1 1 ux (q (x0 , t) , t) = 2 u2 (q (x0 , t) , t) − u2x (q (x0 , t) , t) + dt 2 α 1 1 2k |p ∗ u (q (x0 , t) , t)| + 2 2ku − λux (q (x0 , t) , t) < α2 α 1 1 1 1 2 u (q (x0 , t) , t) − u2x (q (x0 , t) , t) + u2x (q (x0 , t) , t) < − u2x (q (x0 , t) , t) 8 x 2 8 4 for t > T2 . So that, −

d dt



1 ux (q (x0 , t) , t)

 <−

1 4

and by integration we get 0<−

1 1 1 <− − (t − T2 ) ux (q (x0 , t) , t) ux (q (x0 , T2 ) , T2 ) 4

for t > T2 . The left-hand side of the last inequality holds since by (16) we know that ux (q (x0 , t) , t) < 0 for t > T0 . Finally, if we suppose that solution exists globally then for sufficiently large t ∗ > T2 0<−

 1 1 1 <− − t ∗ − T2 < 0. ∗ ∗ ux (q (x0 , t ) , t ) ux (q (x0 , T2 ) , T2 ) 4

This contradiction shows that the solution blows up in finite time.

2

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