# Blowup Phenomena for a Modified Dullin-Gottwald-Holm Shallow Water System

Authors: Panpan Zhai; Zhengguang Guo; Weiming Wang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We discuss blowup phenomena for a modified two-component Dullin-Gottwald-Holm shallow water system. In this paper, some new blowup criteria of strong solutions involving the density and suitable integral form of the momentum are established.


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## Body

## 1. Introduction

We consider the following two-component DGH type system:(1)yt+2ωux+uyx+2yux+γuxxx+gρρ¯x=0,y=u-α2uxx,ρt+(ρu)x=0,
where u=u(x,t), (x,t)∈(ℝ,ℝ+) denotes the velocity field, g is the downward constant acceleration of gravity in applications to shallow water waves, and ρ=(1-∂x2)(ρ¯-ρ¯0), where ρ¯0 is taken to be a constant. It is obvious that if ρ≡0, then (1) reduces to the well-known Dullin-Gottwald-Holm equation [1] (DGH equation for short). There are some contributions to DGH equation concerning the well-posedness, scattering problem, blowup phenomenon, and so forth; see, for example, [2–5] and references therein. We find that (1) is expressed in terms of an averaged filtered density component ρ¯ in analogy to the relation between momentum and velocity by setting ρ=(1-∂x2)(ρ¯-ρ¯0) and the velocity component u. The idea is actually from the recent work [6]. Our modification breaks the structure of DGH2 system derived by following Ivanov’s approach [7] by the authors in [8]. The motivation of current research is stated as follows. From geometric point of view, (1) is the model for geodesic motion on the semidirect product Lie group of diffeomorphisms acting on densities, with respect to the H1-norm of velocity u and the H1-norm on filtered density. From a physical point of view, (1) admits wave breaking phenomena in finite time which attracts researchers’ interest. We also find that the H1-norm of (u,ρ¯) is conserved with respect to time variable. This makes further different discussions on the singularities, unlike those for the DGH2 system or two-component Camassa-Holm system, possible. In the previous works [9–11] on the two-component Camassa-Holm equation and its modified version, blowup conditions were established in view of the negativity of initial velocity slope at some point; basically, the initial integral form of momentum is never involved. That is why we consider this kind of blowup condition in this paper. Precisely, we show the solutions blowup in finite time provided that the initial density and momentum satisfy certain sign conditions. To our knowledge, less results exist yet for the formation of singularities of (1) although the approaches we applied here are standard. The methods in previous works cannot be moved to this model parallelly. For convenience, let v=ρ¯-ρ¯0 and Λ=(1-α2∂x2)-1; then the operator Λ can be expressed by its associated Green’s function G(x)=(1/2α)e-|x/α| with
(2)Λf(x)=G*f(x)=∫ℝ‍G(x-y)f(y)dy.
Using this identity, system (1) takes an equivalent form of a quasilinear evolution equation of hyperbolic type as follows:
(3)ut+ux(u-γα2)=-∂xG*(u2+α22ux2+(2ω+γα2)u+g2v2-g2vx2),vt+uvx=-G*((uxvx)x+uxv).The current paper is based on some results on the Camassa-Holm equation [12–19] and its two-component generalizations [20–27]. We investigate further formation of singularities of solutions to (3) with the case of g=1 and α>0, just for simplicity mathematically. This paper is organized as follows. In Section 2, we recall some preliminary results on the well-posedness and blowup scenario. In Section 3, the detailed blowup conditions are presented.


## 2. Preliminaries

In this section, for completeness, we recall some elementary results and skip their proofs since they are not the main concern of this work. For convenience, in what follows, we letλ=-γ/α2 and 2κ=2ω+γ/α2.We can apply Kato’s theory [28] to establish the following local well-posedness theorem for (3).Theorem 1.
Assume an initial data(u0,v0)∈Hs×Hs-1,s≥5/2. Then there exists a maximal T=T(∥u0,v0∥Hs×Hs-1)>0 and a unique solution
(4)(u,v)∈C([0,T);Hs×Hs-1)∩C1([0,T);Hs-1×Hs-2)
of system (3). Moreover, the solution (u,v) depends continuously on the initial value (u0,v0), and the maximal time of existence T>0 is independent of s.The proof of Theorem1 is similar to the one in [11]. Moreover, using the techniques in [11], one can get the criterion for finite time wave breaking to (3) as follows.Theorem 2.
Let(u0,v0)∈Hs×Hs-1 with s≥5/2, and let T>0 be the maximal time of existence of the solution (u,v) to (3) with initial data (u0,v0). Then the corresponding solution (u,v) blowsup in finite time if and only if
(5)limt↑T{infx∈ℝux(t,x)}=-∞.Lemma 3 (see [29]).
Assume that a differentiable functiony(t) satisfies
(6)y′(t)≤-Cy2(t)+K
with constants C,K>0. If the initial datum y(0)=y0<-K/C, then the solution to (9) goes to -∞ before t tends to 1/(-Cy0+K/y0).Lemma 4 (see [19]).
Suppose thatΨ(t) is twice continuously differential satisfying
(7)Ψ′′(t)≥C0Ψ′(t)Ψ(t),t>0,C0>0,Ψ(t)>0,Ψ′(t)>0.
Then Ψ blowsup in finite time. Moreover the blowup time can be estimated in terms of the initial datum as
(8)T≤max{2C0Ψ(0),Ψ(0)Ψ′(0)}.We also need to introduce the standard particle trajectory method for later use. Consider now the following two initial value problems:(9)q1,t=u(t,q1)+λ,t∈[0,T),q1(0,x)=x,x∈ℝ,(10)q2,t=u(t,q2),t∈[0,T),q2(0,x)=x,x∈ℝ,
where u∈C1([0,T),Hs-1) is the first component of the solution (u,v) to system (3) with initial data (u0,v0)∈Hs×Hs-1(s≥5/2), and T>0 is the maximal time of existence. By direct computation, we have
(11)qi,tx(t,x)=ux(t,qi(t,x))qi,x(t,x),i=1,2.
Then,
(12)qi,x(t,x)=exp(∫0t‍ux(τ,qi(τ,x))dτ)>0,t>0,x∈ℝ,
which means that qi(t,·): ℝ→ℝ is a diffeomorphism of the line for every t∈[0,T). Consequently, the L∞-norm of any function v(t,·) is preserved under the family of the diffeomorphisms qi(t,·); that is,
(13)∥v(t,·)∥L∞=∥v(t,q1(t,·))∥L∞=∥v(t,q2(t,·))∥L∞,t∈[0,T).
Similarly,
(14)infx∈ℝv(t,x)=infx∈ℝv(t,q1(t,x))=infx∈ℝv(t,q2(t,x)),t∈[0,T),supx∈ℝv(t,x)=supx∈ℝv(t,q1(t,x))=supx∈ℝv(t,q2(t,x)),t∈[0,T).


## 3. Blowup Phenomenon

In this section, we show that blowup phenomenon is the only one way that singularity arises in smooth solutions. We start this section with the following useful lemma.Lemma 5.
LetX0=(u0,v0)∈Hs×Hs-1, s≥2. T is assumed to be the maximal existence time of the solution X=(u,v) to system (3) corresponding to the initial data X0. Then for all t∈[0,T), one has the following conservation law:
(15)E(t)=∫ℝ‍(u2+α2ux2+v2+vx2)dx.Proof.
We will prove thatE(t) is a conserved quantity with respect to time variable. Here we use the classical energy method. Multiplying the first equation in (3) by u(x,t) and integrating by parts, we obtain
(16)∫ℝ‍uutdx+∫ℝ‍α2uxuxtdx=-∫ℝ‍uvx(v-vxx)dx.
Similarly, we have the following inequality for the second equation (3):
(17)∫ℝ‍vvt+vxvxtdx=∫ℝ‍uvx(v-vxx)dx.
This implies that
(18)∫ℝ‍(uut+α2uxuxx+vvt+vxvxx)dx=0.
Thus, we have
(19)ddt∫ℝ‍(u2+α2ux2+v2+vx2)dx=2∫ℝ‍(uut+α2uxuxx+vvt+vxvxx)dx=0.
This completes the proof.Using this conservation law, we obtain(20)∥u(·,t)∥L∞(ℝ)2+∥v(·,t)∥L∞(ℝ)2≤12α∥u(·,t)∥Hα1(ℝ)2+12∥v(·,t)∥H1(ℝ)2≤C1E(0),
where
(21)C1=max{12α,12}.Theorem 6.
Suppose thatX0=(u0,v0)∈Hs×Hs-1, s≥5/2, ρ0(x0)=y0(x0)+κ=0, and the initial data satisfies the following conditions:(i)
ρ0(x)≥0on(-∞,x0) and ρ0(x)≤0on(x0,∞),(ii)
∫-∞x0‍eξ/α(y0(ξ)+κ)dξ>0 and ∫-∞x0‍e-ξ/α(y0(ξ)+κ)dξ<0,for some point x0∈ℝ. Then the solution to system (3) with the initial value X0 blowsup in finite time.Proof.
Differentiating the first equation of (3) with respect to x, we obtain
(22)uxt+ux2+uuxx+λuxx+∂x2G*(u2+α22ux2+2κu+12v2-12vx2)=0.
Applying the relation ∂x2(G*f)=(1/α2)(G*f-f) yields
(23)uxt+ux2+uuxx+λuxx+1α2G*(u2+α22ux2+2κu+12v2-12vx2)-1α2(u2+α22ux2+2κu+12v2-12vx2)=0.
From (23) we have
(24)ddtux(q1(x0,t),t)=(uxt+uuxx+λuxx)(q1(x0,t),t)=-ux2(q1(x0,t),t)-1α2G=*(u2+α22ux2+2κu+12v2-12vx2)(q1(x0,t),t)=+1α2(u2+α22ux2+2κu+12v2-12vx2)(q1(x0,t),t)≤-12ux2(q1(x0,t),t)+12α2(u+κ)2(q1(x0,t),t)=+12α2(v2-vx2)(q1(x0,t),t)=-12α2(G*(v2-vx2))(q1(x0,t),t),
where we used the fact proved in [30] that
(25)G*((u+κ)2+α22ux2)≥12(u+κ)2.
In order to arrive at our result, we need the following three claims.
Claim  1.
y
(
q
1
(
x
0
,
t
)
,
t
)
)
+
κ
=
0 for all t in its lifespan; q1 is defined in (9).
It is worth noting the equivalent form of the first equation in (3) in what follows:
(26)yt+uyx+2yux+λyx+2κux+ρvx=0.
From the previous equation, we can get
(27)ddt((y(q1(x,t),t)+κ)q1,x2(x,t))=(yt+uyx+2yux+λyx+2κux)(q1(x,t),t)q1,x2(x,t)=-ρ(q1(x,t),t)vx(q1(x,t),t)q1,x2(x,t).
Since q2(x,·) defined by (10) is a diffeomorphism of the line for any t∈[0,T), so there exists an x3(t)∈ℝ such that
(28)q2(t,x3(t))=q1(t,x0),t∈[0,t).
When t=0, we have
(29)x3(0)=q2(0,x3(0))=q1(0,x0)=x0.
Now we prove that ρ(t,q1(t,x0))=0. It is easy to get
(30)ddtρ(t,q2(t,x3(t)))=-(ρux)(t,q2(t,x3(t))).
Since
(31)ρ0(x0)=0,
integrating the previous equation, we can obtain
(32)ρ(t,q2(t,x3(t)))=ρ(0,q2(0,x3(0)))e-∫0t‍ux(τ,q2(τ,x3(τ)))dτ=ρ0(x0)e-∫0t‍ux(τ,q2(τ,x3(τ)))dτ=0;
thus we have
(33)ρ(t,q1(t,x0))=ρ(t,q2(t,x3(t)))=0.
So we can get
(34)ddt((y(q1(x0,t),t)+κ)q1,x2(x0,t))=-ρ(q1(x0,t),t)vx(q1(x0,t),t)q1,x2(x0,t)=0;
then we have(35)y(q1(x0,t),t)+κ=y0(x0)+κ=0.
Our claim is proved.
Claim  2. For any fixed t, vx2(x,t)-v2(x,t)≤vx2(q1(x0,t),t)-v2(q1(x0,t),t) for all x∈ℝ. For any fixed t, if x≤q1(x0,t), then
(36)vx2(x,t)-v2(x,t)=-(∫-∞q1(x0,t)‍eξρ(ξ,t)dξ-∫xq1(x0,t)‍eξρ(ξ,t)dξ)=×(∫q1(x0,t)∞‍e-ξρ(ξ,t)dξ+∫xq1(x0,t)‍e-ξρ(ξ,t)dξ)=vx2(q1(x0,t),t)-v2(q1(x0,t),t)=-∫-∞x‍eξρ(ξ,t)dξ∫xq1(x0,t)‍e-ξρ(ξ,t)dξ=+∫xq1(x0,t)‍eξρ(ξ,t)dξ∫q1(x0,t)∞‍e-ξρ(ξ,t)dξ≤vx2(q1(x0,t),t)-v2(q1(x0,t),t),
where the condition (i) is used. Similarly, for x≥q1(x0,t), we also have
(37)vx2(x,t)-v2(x,t)≤vx2(q1(x0,t),t)-v2(q1(x0,t),t).
So Claim 2 is proved. Consequently, we can obtain
(38)(G*(v2-vx2))(q1(x0,t),t)=12α∫ℝ‍e-|q1(x0,t)-ξ|/α(v2-vx2)(ξ,t)dξ≥12α∫ℝ‍e-|q1(x0,t)-ξ|/α(v2-vx2)(q1(x0,t),t)dξ=(v2-vx2)(q1(x0,t),t).
Thus, one can get
(39)ddtux(q1(x0,t),t)≤-12ux2(q1(x0,t),t)+12α2(u+κ)2(q1(x0,t),t).
Claim  3.
(
u
+
κ
)
2
(
q
1
(
x
0
,
t
)
,
t
)
<
α
2
u
x
2
(
q
1
(
x
0
,
t
)
,
t
) for all t≥0. Furthermore, ux(q1(x0,t),t)<0 is strictly decreasing.
Suppose that there exists at0 such that (u+κ)2(q1(x0,t),t)<α2ux2(q1(x0,t),t) on [0,t0) and (u+κ)2(q1(x0,t0),t0)=α2ux2(q1(x0,t0),t0). From the expression of u(x,t) in terms of y(x,t), we can rewrite u(x,t)+κ and ux(x,t) as follows:
(40)u(x,t)+κ=12αe-x/α∫-∞x‍eξ/α(y(ξ,t)+κ)dξ+12αex/α∫x∞‍e-ξ/α(y(ξ,t)+κ)dξ,ux(x,t)=-12α2e-x/α∫-∞x‍eξ/α(y(ξ,t)+κ)dξ+12α2ex/α∫x∞‍e-ξ/α(y(ξ,t)+κ)dξ.
Letting
(41)I(t)=e-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/α(y(ξ,t)+κ)dξ,II(t)=eq1(x0,t)/α∫q1(x0,t)∞‍e-ξ/α(y(ξ,t)+κ)dξ,
then
(42)dI(t)dt=-1α(u(q1(x0,t),t)+λ)e-q1(x0,t)/α×∫-∞q1(x0,t)‍eξ/α(y(ξ,t)+κ)dξ+e-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/αyt(ξ,t)dξ.
Integrating by parts, the first term of (42) yields
(43)-1α(u(q1(x0,t),t)+λ)e-q1(x0,t)/α×∫-∞q1(x0,t)‍eξ/α(y(ξ,t)+κ)dξ=(αuux-u2-κu)(q1(x0,t),t)-1αe-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/αλ(y(ξ,t)+κ)dξ.
For the second term of (42), we have the following equation in the view of Claim 1:
(44)e-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/αyt(ξ,t)dξ=-e-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/α(((y+κ)u)x+12(u2-α2ux2)x=+λ(y+κ)x+κux+ρvx12)dξ=(α22ux2-12u2-κu)(q1(x0,t),t)=-e-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/αρvxdξ=+1αe-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/α((y+κ)u+12(u2-α2ux2)+λ(y+κ)+κu12)dξ=(α22ux2-12u2-κu+12(vx2-v2))(q1(x0,t),t)=+12αe-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/α(v2-vx2)dξ=+1αe-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/α(32u2-α22ux2-α2uuxx+λ(y+κ)+2κu32u2-α22)dξ=(α22ux2-αuux-κu+12(vx2-v2))(q1(x0,t),t)=+12αe-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/α(v2-vx2)dξ=+1αe-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/α(u2+α22ux2+λ(y+κ)+2κuα22)dξ.
Here we have used
(45)-αe-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/α(uuxx)(ξ,t)dξ=-α(uux)(q1(x0,t),t)+12u2(q1(x0,t),t)=+αe-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/α(ux2-12α2u2)(ξ,t)dξ.
Combining the previous equations together, and with the help of (38), (42) reads as
(46)dI(t)dt=(α22ux2-u2-2κu)(q1(x0,t),t)+1αe-q1(x0,t)/α∫-∞q1(x0,t)‍eξ/α(u2+α22ux2+2κu)dξ+12(vx2-v2)(q1(x0,t),t)+12αe-q1(x0,t)/α×∫-∞q1(x0,t)‍eξ/α(v2-vx2)dξ=(α22ux2-(u+κ)2)(q1(x0,t),t)+1αe-q1(x0,t)/α×∫-∞q1(x0,t)‍eξ/α(α22ux2+(u+κ)2)dξ+12(vx2-v2)(q1(x0,t),t)+12αe-q1(x0,t)/α×∫-∞q1(x0,t)‍eξ/α(v2-vx2)dξ≥12(α2ux2-(u+κ)2)(q1(x0,t),t)>0,on[0,t0),
where Claim 2 and the inequality [30]
(47)∫-∞x‍eξ/α((u+κ)2+α22uξ2)(ξ,t)dξ≥α2ex/α(u+κ)2
have been used. From the continuity property, we have
(48)e-q1(x0,t0)/α∫-∞q1(x0,t0)‍eξ/α(y(ξ,t0)+κ)dξ>e-x0/α∫-∞x0‍eξ/α(y0(ξ)+κ)dξ>0.
Similarly,
(49)dII(t)dt≤12((u+κ)2-α2ux2)(q1(x0,t),t)<0,on[0,t0).
Thus, by continuity property,
(50)eq1(x0,t0)/α∫-∞q1(x0,t0)‍e-ξ/α(y(ξ,t0)+κ)dξ<ex0/α∫-∞x0‍e-ξ/α(y0(ξ)+κ)dξ<0.
Summarizing (48) and (50), we obtain
(51)α2ux2(q1(x0,t0),t0)-(u+κ)2(q1(x0,t0),t0)=-1α2∫-∞q1(x0,t0)‍eξ/α(y(ξ,t0)+κ)dξ=×∫q1(x0,t0)∞‍e-ξ/α(y(ξ,t0)+κ)dξ>-1α2∫-∞x0‍eξ/α(y0(ξ)+κ)dξ∫x0∞‍e-ξ/α(y0(ξ)+κ)dξ=α2u0x2(x0)-(u0+κ)2(x0)>0.
That is a contradiction. On the other hand, from the expression of ux(x,t) in terms of y(x,t), we can easily get that ux(q1(x0,t),t)<0. So we complete the proof of Claim 3.
Furthermore, due to (46) and (49), we can obtain
(52)ddt(α2ux2-(u+κ)2)(q1(x0,t),t)=-1α2ddt(∫-∞q1(x0,t)‍eξ/α(y(ξ,t)+κ)dξ×∫q1(x0,t)∞‍e-ξ/α(y(ξ,t)+κ)dξ)≥-12α2(α2ux2-(u+κ)2)(q1(x0,t),t)eq1(x0,t)/α≥×∫-∞x‍e-ξ/α(y(ξ,t)+κ)dξ≥+12α2(α2ux2-(u+κ)2)(q1(x0,t),t)e-q1(x0,t)/α≥×∫x∞‍eξ/α(y(ξ,t)+κ)dξ=-ux(q1(x0,t),t)(α2ux2-(u+κ)2)(q1(x0,t),t).
Integrating (39) and then substituting it into the previous inequality, we have
(53)ddt(α2ux2-(u+κ)2)(q1(x0,t),t)≥12α2(α2ux2-(u+κ)2)(q1(x0,t),t)≥×(∫0t‍(α2ux2-(u+κ)2)(q1(x0,τ),τ)dτ-2α2u0x(x0)∫0t‍).
Let Ψ(t)=∫0t‍(α2ux2-(u+κ)2)(q1(x0,τ),τ)dτ-2α2u0x(x0); then we can complete the proof with the help of Lemma 4.Remark 7.
We note that if the condition (i) is replaced by the following one:(i′)
ρ0(x)≤0on(-∞,x0) and ρ0(x)≥0on(x0,∞),then Claim 2 also holds; that is, the theorem always holds with anyone of (i) and (i').As a corollary of Theorem6, we have the following.Theorem 8.
Suppose thatX0=(u0,v0)∈Hs×Hs-1,s≥5/2, and the initial data satisfies the following conditions:(i)
ρ0(x)≥0  on(-∞,x0) and ρ0(x)≤0  on(x0,∞)(orρ0(x)≤0  on(-∞,x0) and ρ0(x)≥0  on(x0,∞)),(ii)
u0′(x0)≤-(C1E0+|κ|)/α,for some point x0∈ℝ. Then the solution to system (3) with the initial value X0 blowsup in finite time.Proof.
As shown in Theorem6, condition (i) guarantees that vx2(x,t)+v2(x,t)≤(vx2-v2)(q1(x0,t),t) for all x∈ℝ. Then,
(54)ddtux(q1(x0,t),t)≤-12ux2(q1(x0,t),t)≤+12α2(u+κ)2(q1(x0,t),t)≤-12ux2(q1(x0,t),t)≤+12α2(C1E0+|κ|)2:=-12ux2(q1(x0,t),t)+K2,
where K>0 is a constant. By setting φ(t)=ux(q1(x0,t),t), we obtain
(55)dφdt=-12φ2+K2.
Applying Lemma 3, we have
(56)limt↑Tφ(t)=-∞withT=1-(1/2)φ0-(K2/φ0),
when
(57)φ0<-2K=-C1E0+|κ|α.
This completes the proof.



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