AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 851476 10.1155/2013/851476 851476 Research Article On the Cauchy Problem for a Class of Weakly Dissipative One-Dimensional Shallow Water Equations http://orcid.org/0000-0002-7691-1412 Xu Jingjing Jiang Zaihong Zheng Sining Department of Mathematics Zhejiang Normal University Jinhua 321004 China zjnu.edu.cn 2013 1 10 2013 2013 21 06 2013 22 08 2013 2013 Copyright © 2013 Jingjing Xu and Zaihong Jiang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We investigate a more general family of one-dimensional shallow water equations with a weakly dissipative term. First, we establish blow-up criteria for this family of equations. Then, global existence of the solution is also proved. Finally, we discuss the infinite propagation speed of this family of equations.

1. Introduction

Recently, in , the following one-dimensional shallow water equations were studied: (1)yt+auxy+buyx=0,t>0,x, where u(x,t) and y(x,t)=(1-x2)u(x,t). A detailed description of the corresponding strong solution with the initial data u0 was also given by them in .

When a=θ-1,  b=θ, and λ=0, (1) reduces to θ-equation which is studied by Ni and Zhou in .

When a=2, b=1, and λ=0, (1) reduces to the Camassa-Holm equation, which was derived physically by Camassa and Holm in  (found earlier by Fuchssteiner and Fokas  as a bi-Hamiltonian generalization of the KdV equation) by approximating directly the Hamiltonian for Euler’s equations in the shallow water region with u(x,t) representing the free surface above a flat bottom. The Camassa-Holm equation is completely integrable and has infinite conservation laws. Local well-posedness for the initial datum u0(x)Hs with s>3/2 was proved in [5, 6]. One of the remarkable features of Camassa-Holm equation is the presence of breaking waves and global solutions. Necessary and sufficient condition for wave breaking was established by Mckean  in 1998. A new and direct proof was also given in . The solitary waves of Camassa-Holm equation are peaked solitons. The orbital stability of the peakons was shown by Constantin and Strauss in  (see also ). The property of propagation speed of solutions to the Camassa-Holm equation, which was presented by Himonas and his collaborators in their work is worthy of being mentioned here .

The Degasperis-Procesi equation  and b-family equation  are the special cases with a=3, b=1, and b=1, respectively. There have been extensive studies on the two equations, (cf. [14, 15]).

In this paper, we consider the following weakly dissipative one-dimensional shallow water equation: (2)yt+auxy+buyx+λy=0,t>0,x, where λy=λ(u-uxx) is the weakly dissipative term.

It is worth pointing out that many works have been done for related equations which have a weakly dissipative term (cf. ).

The paper is organized as follows. In Section 2, we establish the local well-posedness of the initial-value problem associated with (2) and present the precise blow-up scenario. Some blow-up results are given in Section 3. In Section 4, we establish a sufficient condition added on the initial data to guarantee global existence. We will consider the infinite propagation speed in Section 5.

2. Local Well-Posedness and Blow-Up Scenario

In this section, we first establish the local well-posedness of (2) by using Kato’s theory. Then, we provide the precise blow-up scenario for solutions to (2).

System (2) is equivalent to the following system: (3)ut+buux+xG*(a2u2+3b-a2ux2)+λu=0, where G(x)=(1/2)e-|x|, * means doing convolution.

Theorem 1.

Given u0Hs(), s>3/2, then there exist a T and a unique solution u to (2) such that (4)u(x,t)C([0,T);Hs())C1([0,T);Hs-1()).

To make the paper concise, we would like to omit the detailed proof, since one can find similar ones for these types of equations in .

3. Blow-Up Phenomenon

In this section, we will give some conditions to guarantee the finite time blowup. Motivated by Mckean’s deep observation for the Camassa-Holm equation , we can consider the similar particle trajectory as (5)qt=bu(q,t),0<t<T,x,q(x,0)=x,x, where T is the lifespan of the solution: then q is a diffeomorphism of the line. Taking derivative (5) with respect to x, we obtain (6)dqtdx=qtx=bux(q,t)qx,t(0,T). Therefore (7)qx(x,t)=exp{b0tux(q,s)ds},qx(x,0)=1. Hence, from (2), the following identity can be proved: (8)y(q)qxa/b=y0(x)e-λt. In fact, direct calculation yields (9)ddt(y(q)qxa/b)=[yt(q)+bu(q,t)yx(q)+aux(q,t)y(q)]qxa/b=-λyqxa/b.

Motivated by , we give the following theorem.

Theorem 2.

Let a-2b>0, b>0: suppose that u0H2(), and there exists a x0 such that y0(x0)=(1-x2)u0(x0)=0, (10)y00(0),for  x(-,x0),y00(0),for  x(x0,),(11)e-x0-x0eξy0(ξ)dξ>2λb,ex0x0e-ξy0(ξ)dξ<-2λb. Then the corresponding solution u(x,t) to (2) with u0 as the initial datum blows up in finite time.

Proof.

Suppose that the solution exists globally. From (8) and initial condition (10), we have y(q(x0,t),t)=0 and (12)y(q(x,t),t)0(0),for  x(-,q(x0,t)),y(q(x,t),t)0(0),for  x(q(x0,t),), for all t0. Due to u(x,t)=G*y(x,t), we can write u(x,t) and ux(x,t) as (13)u(x,t)=12e-x-xeξy(ξ,t)dξ+12exxe-ξy(ξ,t)dξ,ux(x,t)=-12e-x-xeξy(ξ,t)dξ+12exxe-ξy(ξ,t)dξ. Therefore, (14)ux2(x,t)-u2(x,t)=--xeξy(ξ,t)dξxe-ξy(ξ,t)dξ, for all t>0.

By direct calculation, for xq(x0,t), we have (15)ux2(x,t)-u2(x,t)=--xeξy(ξ,t)dξxe-ξy(ξ,t)dξ=-(-q(x0,t)eξy(ξ,t)dξ-xq(x0,t)eξy(ξ,t)dξ)×(q(x0,t)e-ξy(ξ,t)dξ+xq(x0,t)e-ξy(ξ,t)dξ)=ux2(q(x0,t),t)-u2(q(x0,t),t)--xeξy(ξ,t)dξxq(x0,t)e-ξy(ξ,t)dξ+xq(x0,t)eξy(ξ,t)dξq(x0,t)e-ξy(ξ,t)dξux2(q(x0,t),t)-u2(q(x0,t),t). Similarly, for xq(x0,t), we have (16)ux2(x,t)-u2(x,t)ux2(q(x0,t),t)-u2(q(x0,t),t). So for any fixed t, combination of (15) and (16), we obtain (17)ux2(x,t)-u2(x,t)ux2(q(x0,t),t)-u2(q(x0,t),t), for all x.

From the expression of ux(x,t) in terms of y(x,t), differentiating ux(q(x0,t),t) with respect to t, we have (18)tux(q(x0,t),t)=uxt(q(x0,t),t)+uxx(q(x0,t),t)qt(q(x0,t),t)=a2u2(q(x0,t),t)+a-b2ux2(q(x0,t),t)-λux(q(x0,t),t)-G*(a2u2(x,t)+3b-a2ux2(x,t))=G*(a2u2(q(x0,t),t)+a-b2ux2(q(x0,t),t)-a2u2(x,t)-3b-a2ux2(x,t))-λux(q(x0,t),t)=G*(a-2b2(u2(q(x0,t),t)-ux2(q(x0,t),t)-u2(x,t)+ux2(x,t))a-2b2)+G*(bu2(q(x0,t),t)-b2ux2(q(x0,t),t)-bu2(x,t)-b2ux2(x,t))-λux(q(x0,t),t)b2u2(q(x0,t),t)-b2ux2(q(x0,t),t)-λux(q(x0,t),t), where we have used (17), and the inequality G*(u2(x,t)+(1/2)ux2(x,t))(1/2)u2. In addition, we also used the equation utx+uuxx-(a/2)u2-((b-a)/2)ux2+G*((a/2)u2+((3b-a)/2)ux2)+λux=0, which is obtained by differentiating equation (3).

For (11), we know that (19)(u0x(x0)+λb)2-(u0(x0)+λb)2=-e-x0-x0eξy0(ξ)dξ×(ex0x0e-ξy0(ξ)dξ+2λb)>0,(u0x(x0)+λb)2-(u0(x0)-λb)2=-ex0x0e-ξy0(ξ)dξ×(e-x0-x0eξy0(ξ)dξ-2λb)>0.

Claim. u x ( q ( x 0 , t ) , t ) < 0 is decreasing. (u(q(x0,t),t)+λ/b)2<(ux(q(x0,t),t)+λ/b)2 and (u(q(x0,t),t)-λ/b)2<(ux(q(x0,t),t)+λ/b)2, for all t0.

Suppose that there exists a t0 such that (u(q(x0,t),t)+λ/b)2<(ux(q(x0,t),t)+λ/b)2 and (u(q(x0,t),t)-λ/b)2<(ux(q(x0,t),t)+λ/b)2 on [0,t0); then (u(q(x0,t0),t0)+λ/b)2=(ux(q(x0,t0),t0)+λ/b)2 or (u(q(x0,t0),t0)-λ/b)2=(ux(q(x0,t0),t0)+λ/b)2.

Now, let (20)I(t)=12e-q(x0,t)-q(x0,t)eξy(ξ,t)dξ,II(t)=12eq(x0,t)q(x0,t)e-ξy(ξ,t)dξ. Firstly, differentiating I(t), we have (21)dI(t)dt=-b2u(q(x0,t),t)e-q(x0,t)-q(x0,t)eξy(ξ,t)dξ+12e-q(x0,t)-q(x0,t)eξyt(ξ,t)dξ=b2u(ux-u)(q(x0,t),t)-12e-q(x0,t)×-q(x0,t)eξ(a-2b2buyx+2buxy+a-2b2(u2-ux2)x+λy)dξb2u(ux-u)(q(x0,t),t)+b4(u2+ux2-2uux)×(q(x0,t),t)-λ2(u-ux)(q(x0,t),t)=b4(ux2-u2)(q(x0,t),t)-λ2(u-ux)(q(x0,t),t)=b4(ux(q(x0,t),t)+λb)2-b4(u(q(x0,t),t)+λb)2>0,on[0,t0). Secondly, by the same argument, we obtain (22)dII(t)dt=b2u(q(x0,t),t)eq(x0,t)q(x0,t)e-ξy(ξ,t)dξ+12eq(x0,t)q(x0,t)e-ξyt(ξ,t)dξ=b2u(ux+u)(q(x0,t),t)-12eq(x0,t)×q(x0,t)e-ξ(a-2b2buyx+2buxy+  a-2b2(u2-ux2)x+λy)dξb2u(ux+u)(q(x0,t),t)-b4(u2+ux2+2uux)(q(x0,t),t)-λ2(ux+u)(q(x0,t),t)=-b4(ux2-u2)(q(x0,t),t)-λ2(ux+u)(q(x0,t),t)=-b4(ux(q(x0,t),t)+λb)2+b4(u(q(x0,t),t)-λb)2<0,on[0,t0). Therefore, it follows from (21), (22), and the continuity property of ODEs that (23)(ux(q(x0,t),t)+λb)2-(u(q(x0,t),t)+λb)2=-4I(t)(II(t)+λb)>-4I(0)(II(0)+λb)>0,(ux(q(x0,t),t)+λb)2-(u(q(x0,t),t)-λb)2=-4(I(t)-λb)II(t)>-4(I(0)-λb)II(0)>0, for all t>0. This implies that t0 can be extended to the infinity.

Moreover, using (21) and (22) again, we have the following equation for [2(ux+λ/b)2-(u+λ/b)2-(u-λ/b)2](q(x0,t),t): (24)ddt[2(ux+λb)2-(u+λb)2-(u-λb)2](q(x0,t),t)=-4ddt[I(t)(II(t)+λb)]-4ddt[(I(t)-λb)II(t)]-b[(ux+λb)2-(u+λb)2](q(x0,t),t)(II(t)+λb)+b[(ux+λb)2-(u-λb)2](q(x0,t),t)I(t)-b[(ux+λb)2-(u+λb)2](q(x0,t),t)II(t)+b[(ux+λb)2-(u-λb)2](q(x0,t),t)(I(t)-λb)=b(-λb[2(ux+λb)2-(u+λb)2-(u-λb)2]  ×(q(x0,t),t)-ux(q(x0,t),t)×(2(ux(q(x0,t),t)+λb)2)  +2(u+λb)2II(t)-2(u-λb)2I(t)[2(ux+λb)2-(u+λb)2-(u-λb)2])=b((u-λb)2-(ux(q(x0,t),t)+λb)×[2(ux+λb)2-(u+λb)2-(u-λb)2]×(q(x0,t),t)-ux(q(x0,t),t)×[(u+λb)2+(u-λb)2](q(x0,t),t)+2(u+λb)2II(t)-2(u-λb)2I(t))-b(ux(q(x0,t),t)+λb)×[2(ux+λb)2-(u+λb)2-(u-λb)2](q(x0,t),t), where we use ux(q(x0,t),t)=-I(t)+II(t).

Now, recalling (18), we have (25)tux(q(x0,t),t)(q(x0,t),t)b2u2(q(x0,t),t)-b2ux2(q(x0,t),t)-λux=b4[(u+λb)2+(u-λb)2-2(ux+λb)2]×(q(x0,t),t).

Putting (25) into (24), it yields (26)ddt[2(ux+λb)2-(u+λb)2-(u-λb)2](q(x0,t),t)b24[2(ux+λb)2-(u+λb)2-(u-λb)2]×(q(x0,t),t)×(0t[2(ux+λb)2-(u+λb)2-(u-λb)2]×(q(x0,τ),τ)dτ-4u0x(x0)-4λb0t[2(ux+λb)2-(u+λb)2-(u-λb)2]). Before finishing the proof, we need the following technical lemma.

Lemma  3 (see ). Suppose that Ψ(t) is twice continuously differential satisfying(27)Ψ′′(t)C0Ψ(t)Ψ(t),t>0,C0>0,Ψ(t)>0,Ψ(t)>0.Then ψ(t) blows up in finite time. Moreover the blow-up time can be estimated in terms of the initial datum as(28)Tmax{2C0Ψ(0),Ψ(0)Ψ(0)}.

Let Ψ(t)=0t[2(ux+λ/b)2-(u+λ/b)2-(u-λ/b)2](q(x0,τ),τ)dτ-4u0x(x0)-4λ/b; then (26) is an equation of type (27) with C0=b2/4. The proof is complete by applying Lemma 3.

Remark 4.

When b=1, Theorem 2 reduces to the result in .

Theorem 5.

Let a=2b>0. Suppose that u0H2() and there exists a x0 such that y0(x0)=(1-x2)u0(x0)=0, (29)e-x0-x0eξy0(ξ)dξ>2λb,ex0x0e-ξy0(ξ)dξ<-2λb. Then the corresponding solution u(x,t) to (2) with u0 as the initial datum blows up in finite time.

Proof.

We easily obtain (30)utx+buuxx-a2u2+b2ux2+G*(a2u2+b2ux2)+λux=0. Differentiating ux at the point (q(x0,t),t) with respect to t, we get (31)ddtux(q(x0,t),t)b2u2(q(x0,t),t)-b2ux2(q(x0,t),t)-λux(q(x0,t),t).

Process of the proof is similar to Theorem 2. Thus to be concise, we omit the detailed proof.

When a=2b>0,λ=0, using u~(x,t)=bu(x,t), (2) can be reformulated into (32)u~x+u~xxt+3u~u~x-2u~xu~xx-u~u~xxx=0, which is the well-known Camassa-Holm equation. Meanwhile, we also find that the condition in Theorem 5 can be reformulated into (33)-x0eξy~0(ξ)dξ>0,x0e-ξy~0(ξ)dξ<0, which is one of the sufficient conditions to guarantee blow-up add-on initial data for the Camassa-Holm equation.

So, we show the necessary and sufficient condition for the special case a=2b>0 and λ=0 in the following theorem.

Theorem 6.

When a=2b>0 and λ=0, then the nonlinear wave equation (2) breaks if and only if some portion of the positive part of y0(x) lies to the left of some portion of its negative part.

Proof.

As studied in , when a=2b>0 and λ=0, rewriting (2) yields (34)ux+uxxt+3buux-2buxuxx-buuxxx=0.

Recalling Mckean’s theorem in , (32) breaks if and only if some portion of the positive part of y~0(x,t)=(1-x2)u~0 lies to the left of some portion of its negative part.

So (34) breaks if and only if some portion of the positive part of by0(x,t)=b(1-x2)u0 lies to the left of some portion of its negative part.

This completes the proof.

Remark 7.

Mckean’s theorem  is for the special case a=2, b=1. Condition a=2b here is more general. However, the necessary and sufficient condition for (2) is still a challenging problem for us at present.

4. Global Existence

Now, let us try to find a condition for global existence. Unfortunately, When a2b, like the Degasperis-Procesi equation , only the following easy one can be proved at present.

Theorem 8.

Suppose that u0H3(), and y0=(1-x2)u0 is one sign. Then the corresponding solution to (2) exists globally.

Proof.

We can assume that y00. It is sufficient to prove that ux(x,t) has a lower and upper bound for all t. In fact, (35)ux(x,t)=-12e-x-xeξy(ξ,t)dξ+12exxe-ξy(ξ,t)dξ. Therefore, we have (36)ux(x,t)-12e-x-xeξy(ξ,t)dξ-12-xy(ξ,t)dξ-12-y(ξ,t)dξ=-12-y0(ξ,t)dξ,ux(x,t)12exxe-ξy(ξ,t)dξ12xy(ξ,t)dξ12-y(ξ,t)dξ=12-y0(ξ,t)dξ. This completes the proof.

5. Infinite Propagation Speed

In this section, we will give a more detailed description on the corresponding strong solution u(x,t) to (2) in its life span with initial data u0(x) being compactly supported. The main theorem reads as follows.

Theorem 9.

Let 0<a3b. Assume that for some T0 and s5/2, uC([0,T);Hs()) is a strong solution of (2). If u0(x)=u(x,0) has compact support [a,c], then for t(0,T), one has (37)u(x,t)={L(t)e-x,for  x>q(c,t),l(t)e-x,for  x<q(a,t), where L(t) and l(t) denote continuous nonvanishing functions, with L(t)>0 and l(t)<0 for t(0,T). Furthermore, L(t) is strictly increasing function, while l(t) is strictly decreasing function.

Proof.

Since u0 has compact support in x in [a,c], from (8), so does y(,t) has compact support in x in [q(a,t),q(c,t)] in its lifespan. Hence the following functions are well-defined:(38)E(t)=exy(x,t)dx,F(t)=e-xy(x,t)dx, with (39)E0=exy0(x)dx=0,F0=e-xy0(x)dx=0. Thus, for x>q(c,t), we obtain (40)u(x,t)=12e-|x|*y(x,t)=12e-xq(a,t)q(b,t)eτy(τ,t)dτ=12e-xE(t). Similarly, for x<q(a,t), we have (41)u(x,t)=12e-|x|*y(x,t)=12exq(a,t)q(b,t)e-τy(τ,t)dτ=12exF(t). Hence, as consequences of (40) and (41), we get (42)u(x,t)=-ux(x,t)=uxx(x,t)=12e-xE(t),as  x>q(c,t),u(x,t)=ux(x,t)=uxx(x,t)=12exF(t),as  x<q(a,t). On the other hand, (43)dE(t)dt=exyt(x,t)dx. It is easy to get (44)yt=-buux+b(uux)xx-x(a2u2+3b-a2ux2)-λu+λuxx. Putting the identity (44) into dE(t)/dt, we have (45)dE(t)dt=ex(-buux+b(uux)xx-x(a2u2+3b-a2ux2))dx+ex(-λu+λuxx)dx=ex(a2u2+3b-a2ux2)dx, where we have used (42).

Therefore, in the lifespan of the solution, we get (46)E(t)=0tex(a2u2+3b-a2ux2)(x,τ)dxdτ>0. By the same argument, one can check that the following identity for F(t) is true: (47)F(t)=-0te-x(a2u2+3b-a2ux2)(x,τ)dxdτ<0.

In order to complete the proof, it is sufficient to let L(t)=(1/2)E(t) and l(t)=(1/2)F(t), respectively.

Acknowledgments

This work is partially supported by Zhejiang Innovation Project (T200905), ZJNSF (Grant no. R6090109), and NSFC (Grant no. 10971197 and 11101376).

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