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Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013 Article ID 851476 7 pageshttpdxdoiorg1011552013851476
Research ArticleOn the Cauchy Problem for a Class of Weakly DissipativeOne-Dimensional Shallow Water Equations
Jingjing Xu and Zaihong Jiang
Department of Mathematics Zhejiang Normal University Jinhua 321004 China
Correspondence should be addressed to Zaihong Jiang jzhong790405gmailcom
Received 21 June 2013 Accepted 22 August 2013
Academic Editor Sining Zheng
Copyright copy 2013 J Xu and Z Jiang This is an open access article distributed under the Creative Commons Attribution Licensewhich 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 establishblow-up criteria for this family of equations Then global existence of the solution is also proved Finally we discuss the infinitepropagation speed of this family of equations
1 Introduction
Recently in [1] the following one-dimensional shallow waterequations were studied
119910119905+ 119886119906119909119910 + 119887119906119910
119909= 0 119905 gt 0 119909 isin R (1)
where 119906(119909 119905) isin R and 119910(119909 119905) = (1 minus 1205972
119909)119906(119909 119905) A detailed
description of the corresponding strong solution with theinitial data 119906
0was also given by them in [1]
When 119886 = 120579 minus 1 119887 = 120579 and 120582 = 0 (1) reduces to 120579-equation which is studied by Ni and Zhou in [2]
When 119886 = 2 119887 = 1 and 120582 = 0 (1) reduces to theCamassa-Holm equation which was derived physically byCamassa and Holm in [3] (found earlier by Fuchssteinerand Fokas [4] as a bi-Hamiltonian generalization of the KdVequation) by approximating directly the Hamiltonian forEulerrsquos equations in the shallow water region with 119906(119909 119905) rep-resenting the free surface above a flat bottom The Camassa-Holm equation is completely integrable and has infiniteconservation laws Local well-posedness for the initial datum1199060(119909) isin 119867
119904 with 119904 gt 32 was proved in [5 6] One ofthe remarkable features of Camassa-Holm equation is thepresence of breaking waves and global solutions Necessaryand sufficient condition for wave breaking was establishedby Mckean [7] in 1998 A new and direct proof was alsogiven in [8] The solitary waves of Camassa-Holm equationare peaked solitons The orbital stability of the peakons was
shown by Constantin and Strauss in [9] (see also [10]) Theproperty of propagation speed of solutions to the Camassa-Holm equation which was presented by Himonas and hiscollaborators in their work is worthy of beingmentioned here[11]
The Degasperis-Procesi equation [12] and b-family equa-tion [13] are the special cases with 119886 = 3 119887 = 1 and 119887 = 1respectively There have been extensive studies on the twoequations (cf [14 15])
In this paper we consider the followingweakly dissipativeone-dimensional shallow water equation
119910119905+ 119886119906119909119910 + 119887119906119910
119909+ 120582119910 = 0 119905 gt 0 119909 isin R (2)
where 120582119910 = 120582(119906 minus 119906119909119909
) is the weakly dissipative termIt is worth pointing out that many works have been done
for related equationswhich have aweakly dissipative term (cf[16ndash19])
The paper is organized as follows In Section 2 weestablish the local well-posedness of the initial-value problemassociated with (2) and present the precise blow-up scenarioSome blow-up results are given in Section 3 In Section 4we establish a sufficient condition added on the initial datato guarantee global existence We will consider the infinitepropagation speed in Section 5
2 Abstract and Applied Analysis
2 Local Well-Posedness and Blow-Up Scenario
In this section we first establish the local well-posedness of(2) by using Katorsquos theoryThen we provide the precise blow-up scenario for solutions to (2)
System (2) is equivalent to the following system
119906119905+ 119887119906119906
119909+ 120597119909119866 lowast (
119886
21199062+
3119887 minus 119886
21199062
119909) + 120582119906 = 0 (3)
where 119866(119909) = (12)119890minus|119909| lowast means doing convolution
Theorem 1 Given 1199060
isin 119867119904(R) 119904 gt 32 then there exist a 119879
and a unique solution 119906 to (2) such that
119906 (119909 119905) isin 119862 ([0 119879) 119867119904(R)) cap 119862
1([0 119879) 119867
119904minus1(R)) (4)
To make the paper concise we would like to omit thedetailed proof since one can find similar ones for these typesof equations in [5]
3 Blow-Up Phenomenon
In this section we will give some conditions to guarantee thefinite time blowup Motivated by Mckeanrsquos deep observationfor the Camassa-Holm equation [7] we can consider thesimilar particle trajectory as
119902119905= 119887119906 (119902 119905) 0 lt 119905 lt 119879 119909 isin R
119902 (119909 0) = 119909 119909 isin R
(5)
where 119879 is the lifespan of the solution then 119902 is a diffeomor-phism of the line Taking derivative (5) with respect to 119909 weobtain
119889119902119905
119889119909= 119902119905119909
= 119887119906119909(119902 119905) 119902
119909 119905 isin (0 119879) (6)
Therefore
119902119909(119909 119905) = exp119887int
119905
0
119906119909(119902 119904) 119889119904 119902
119909(119909 0) = 1 (7)
Hence from (2) the following identity can be proved
119910 (119902) 119902119886119887
119909= 1199100(119909) 119890minus120582119905
(8)
In fact direct calculation yields119889
119889119905(119910 (119902) 119902
119886119887
119909) = [119910
119905(119902) + 119887119906 (119902 119905) 119910
119909(119902)
+119886119906119909(119902 119905) 119910 (119902)] 119902
119886119887
119909= minus120582119910119902
119886119887
119909
(9)
Motivated by [19] we give the following theorem
Theorem 2 Let 119886 minus 2119887 gt 0 119887 gt 0 suppose that 1199060isin 1198672(R)
and there exists a1199090isin R such that119910
0(1199090) = (1minus120597
2
119909)1199060(1199090) = 0
1199100ge 0 ( equiv 0) for 119909 isin (minusinfin 119909
0)
1199100le 0 ( equiv 0) for 119909 isin (119909
0infin)
(10)
119890minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt
2120582
119887 119890
1199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt minus
2120582
119887
(11)
Then the corresponding solution 119906(119909 119905) to (2) with 1199060as the
initial datum blows up in finite time
Proof Suppose that the solution exists globally From (8) andinitial condition (10) we have 119910(119902(119909
0 119905) 119905) = 0 and
119910 (119902 (119909 119905) 119905) ge 0 ( equiv 0) for 119909 isin (minusinfin 119902 (1199090 119905))
119910 (119902 (119909 119905) 119905) le 0 ( equiv 0) for 119909 isin (119902 (1199090 119905) infin)
(12)
for all 119905 ge 0 Due to 119906(119909 119905) = 119866 lowast 119910(119909 119905) we can write 119906(119909 119905)
and 119906119909(119909 119905) as
119906 (119909 119905) =1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 +
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
119906119909(119909 119905) = minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 +
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
(13)
Therefore
1199062
119909(119909 119905) minus 119906
2(119909 119905) = minusint
119909
minusinfin
119890120585119910 (120585 119905) 119889120585int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
(14)
for all 119905 gt 0By direct calculation for 119909 le 119902(119909
0 119905) we have
1199062
119909(119909 119905) minus 119906
2(119909 119905)
= minusint
119909
minusinfin
119890120585119910 (120585 119905) 119889120585int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
= minus(int
119902(1199090 119905)
minusinfin
119890120585119910 (120585 119905) 119889120585 minus int
119902(1199090 119905)
119909
119890120585119910 (120585 119905) 119889120585)
times (int
infin
119902(1199090119905)
119890minus120585
119910 (120585 119905) 119889120585 + int
119902(1199090 119905)
119909
119890minus120585
119910 (120585 119905) 119889120585)
= 1199062
119909(119902 (1199090 119905) 119905) minus 119906
2(119902 (1199090 119905) 119905)
minus int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585int
119902(1199090 119905)
119909
119890minus120585
119910 (120585 119905) 119889120585
+ int
119902(1199090 119905)
119909
119890120585119910 (120585 119905) 119889120585int
infin
119902(1199090119905)
119890minus120585
119910 (120585 119905) 119889120585
le 1199062
119909(119902 (1199090 119905) 119905) minus 119906
2(119902 (1199090 119905) 119905)
(15)
Similarly for 119909 ge 119902(1199090 119905) we have
1199062
119909(119909 119905) minus 119906
2(119909 119905) le 119906
2
119909(119902 (1199090 119905) 119905) minus 119906
2(119902 (1199090 119905) 119905)
(16)
So for any fixed 119905 combination of (15) and (16) we obtain
1199062
119909(119909 119905) minus 119906
2(119909 119905) le 119906
2
119909(119902 (1199090 119905) 119905) minus 119906
2(119902 (1199090 119905) 119905)
(17)
for all 119909 isin R
Abstract and Applied Analysis 3
From the expression of 119906119909(119909 119905) in terms of 119910(119909 119905)
differentiating 119906119909(119902(1199090 119905) 119905) with respect to 119905 we have
120597119905119906119909(119902 (1199090 119905) 119905)
= 119906119909119905
(119902 (1199090 119905) 119905) + 119906
119909119909(119902 (1199090 119905) 119905) 119902
119905(119902 (1199090 119905) 119905)
=119886
21199062(119902 (1199090 119905) 119905) +
119886 minus 119887
21199062
119909(119902 (1199090 119905) 119905)
minus 120582119906119909(119902 (1199090 119905) 119905) minus 119866 lowast (
119886
21199062(119909 119905) +
3119887 minus 119886
21199062
119909(119909 119905))
= 119866 lowast (119886
21199062(119902 (1199090 119905) 119905) +
119886 minus 119887
21199062
119909(119902 (1199090 119905) 119905)
minus119886
21199062(119909 119905) minus
3119887 minus 119886
21199062
119909(119909 119905)) minus 120582119906
119909(119902 (1199090 119905) 119905)
= 119866 lowast (119886 minus 2119887
2(1199062(119902 (1199090 119905) 119905) minus 119906
2
119909(119902 (1199090 119905) 119905)
minus1199062(119909 119905) + 119906
2
119909(119909 119905)))
+ 119866 lowast (1198871199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905)
minus1198871199062(119909 119905) minus
119887
21199062
119909(119909 119905)) minus 120582119906
119909(119902 (1199090 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909(119902 (1199090 119905) 119905)
(18)
where we have used (17) and the inequality 119866 lowast (1199062(119909 119905) +
(12)1199062
119909(119909 119905)) ge (12)119906
2 In addition we also used theequation 119906
119905119909+119906119906119909119909
minus (1198862)1199062minus ((119887minus119886)2)119906
2
119909+119866lowast((1198862)119906
2+
((3119887minus119886)2)1199062
119909)+120582119906119909= 0 which is obtained by differentiating
equation (3)For (11) we know that
(1199060119909
(1199090) +
120582
119887)
2
minus (1199060(1199090) +
120582
119887)
2
= minus119890minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 times (119890
1199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 +
2120582
119887) gt 0
(1199060119909
(1199090) +
120582
119887)
2
minus (1199060(1199090) minus
120582
119887)
2
= minus1198901199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 times (119890
minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 minus
2120582
119887) gt 0
(19)
Claim 119906119909(119902(1199090 119905) 119905) lt 0 is decreasing (119906(119902(119909
0 119905) 119905) +
120582119887)2
lt (119906119909(119902(1199090 119905) 119905) + 120582119887)
2 and (119906(119902(1199090 119905) 119905) minus 120582119887)
2lt
(119906119909(119902(1199090 119905) 119905) + 120582119887)
2 for all 119905 ge 0Suppose that there exists a 119905
0such that (119906(119902(119909
0 119905) 119905) +
120582119887)2
lt (119906119909(119902(1199090 119905) 119905) + 120582119887)
2 and (119906(119902(1199090 119905) 119905) minus 120582119887)
2lt
(119906119909(119902(1199090 119905) 119905) + 120582119887)
2 on [0 1199050) then (119906(119902(119909
0 1199050) 1199050) +
120582119887)2= (119906119909(119902(1199090 1199050) 1199050) + 120582119887)
2 or (119906(119902(1199090 1199050) 1199050) minus 120582119887)
2=
(119906119909(119902(1199090 1199050) 1199050) + 120582119887)
2Now let
119868 (119905) =1
2119890minus119902(1199090 119905) int
119902(1199090 119905)
minusinfin
119890120585119910 (120585 119905) 119889120585
119868119868 (119905) =1
2119890119902(1199090 119905) int
infin
119902(1199090 119905)
119890minus120585
119910 (120585 119905) 119889120585
(20)
Firstly differentiating 119868(119905) we have
119889119868 (119905)
119889119905= minus
119887
2119906 (119902 (119909
0 119905) 119905) 119890
minus119902(1199090 119905) int
119902(1199090 119905)
minusinfin
119890120585119910 (120585 119905) 119889120585
+1
2119890minus119902(1199090 119905) int
119902(1199090 119905)
minusinfin
119890120585119910119905(120585 119905) 119889120585
=119887
2119906 (119906119909minus 119906) (119902 (119909
0 119905) 119905) minus
1
2119890minus119902(1199090 119905)
times int
119902(1199090 119905)
minusinfin
119890120585(119887119906119910119909+ 2119887119906
119909119910
+119886 minus 2119887
2(1199062minus 1199062
119909)119909+ 120582119910)119889120585
ge119887
2119906 (119906119909minus 119906) (119902 (119909
0 119905) 119905) +
119887
4(1199062+ 1199062
119909minus 2119906119906
119909)
times (119902 (1199090 119905) 119905) minus
120582
2(119906 minus 119906
119909) (119902 (119909
0 119905) 119905)
=119887
4(1199062
119909minus 1199062) (119902 (119909
0 119905) 119905) minus
120582
2(119906 minus 119906
119909) (119902 (119909
0 119905) 119905)
=119887
4(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
minus119887
4(119906 (119902 (119909
0 119905) 119905) +
120582
119887)
2
gt 0 on [0 1199050)
(21)
Secondly by the same argument we obtain
119889119868119868 (119905)
119889119905=
119887
2119906 (119902 (119909
0 119905) 119905) 119890
119902(1199090 119905) int
infin
119902(1199090 119905)
119890minus120585
119910 (120585 119905) 119889120585
+1
2119890119902(1199090 119905) int
infin
119902(1199090 119905)
119890minus120585
119910119905(120585 119905) 119889120585
=119887
2119906 (119906119909+ 119906) (119902 (119909
0 119905) 119905) minus
1
2119890119902(1199090 119905)
times int
infin
119902(1199090 119905)
119890minus120585
(119887119906119910119909+ 2119887119906
119909119910
+119886 minus 2119887
2(1199062minus 1199062
119909)119909+ 120582119910)119889120585
4 Abstract and Applied Analysis
le119887
2119906 (119906119909+ 119906) (119902 (119909
0 119905) 119905)
minus119887
4(1199062+ 1199062
119909+ 2119906119906
119909) (119902 (119909
0 119905) 119905)
minus120582
2(119906119909+ 119906) (119902 (119909
0 119905) 119905)
= minus119887
4(1199062
119909minus 1199062) (119902 (119909
0 119905) 119905)
minus120582
2(119906119909+ 119906) (119902 (119909
0 119905) 119905)
= minus119887
4(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
+119887
4(119906 (119902 (119909
0 119905) 119905) minus
120582
119887)
2
lt 0 on [0 1199050)
(22)
Therefore it follows from (21) (22) and the continuityproperty of ODEs that
(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
minus (119906 (119902 (1199090 119905) 119905) +
120582
119887)
2
= minus4119868 (119905) (119868119868 (119905) +120582
119887) gt minus4119868 (0) (119868119868 (0) +
120582
119887) gt 0
(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
minus (119906 (119902 (1199090 119905) 119905) minus
120582
119887)
2
= minus4(119868 (119905) minus120582
119887) 119868119868 (119905) gt minus4(119868 (0) minus
120582
119887) 119868119868 (0) gt 0
(23)
for all 119905 gt 0 This implies that 1199050can be extended to the
infinityMoreover using (21) and (22) again we have the following
equation for [2(119906119909+120582119887)
2minus(119906+120582119887)
2minus(119906 minus 120582119887)
2](119902(1199090 119905) 119905)
119889
119889119905[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
= minus4119889
119889119905[119868 (119905) (119868119868 (119905) +
120582
119887)] minus 4
119889
119889119905[(119868 (119905) minus
120582
119887) 119868119868 (119905)]
ge minus119887 [(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
] (119902 (1199090 119905) 119905) (119868119868 (119905) +
120582
119887)
+ 119887 [(119906119909+
120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905) 119868 (119905)
minus 119887 [(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
] (119902 (1199090 119905) 119905) 119868119868 (119905)
+ 119887 [(119906119909+
120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905) (119868 (119905) minus
120582
119887)
= 119887(minus120582
119887[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905) minus 119906
119909(119902 (1199090 119905) 119905)
times (2(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
)
+2(119906 +120582
119887)
2
119868119868 (119905) minus 2(119906 minus120582
119887)
2
119868 (119905))
= 119887(minus (119906119909(119902 (1199090 119905) 119905) +
120582
119887)
times [2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905) minus 119906
119909(119902 (1199090 119905) 119905)
times [(119906 +120582
119887)
2
+ (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
+2(119906 +120582
119887)
2
119868119868 (119905) minus 2(119906 minus120582
119887)
2
119868 (119905))
ge minus119887(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
times [2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
(24)
where we use 119906119909(119902(1199090 119905) 119905) = minus119868(119905) + 119868119868(119905)
Now recalling (18) we have
120597119905119906119909(119902 (1199090 119905) 119905) (119902 (119909
0 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909
=119887
4[(119906 +
120582
119887)
2
+ (119906 minus120582
119887)
2
minus 2(119906119909+
120582
119887)
2
]
times (119902 (1199090 119905) 119905)
(25)
Putting (25) into (24) it yields
119889
119889119905[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
ge1198872
4[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905)
times (int
119905
0
[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 120591) 120591) 119889120591 minus 4119906
0119909(1199090) minus
4120582
119887)
(26)
Abstract and Applied Analysis 5
Before finishing the proof we need the following technicallemma
Lemma 3 (see [15]) Suppose that Ψ(119905) is twice continuouslydifferential satisfying
Ψ10158401015840(119905) ge 119862
0Ψ1015840(119905) Ψ (119905) 119905 gt 0 119862
0gt 0
Ψ (119905) gt 0 Ψ1015840(119905) gt 0
(27)
Then 120595(119905) blows up in finite time Moreover the blow-up timecan be estimated in terms of the initial datum as
119879 le max2
1198620Ψ (0)
Ψ (0)
Ψ1015840 (0) (28)
Let Ψ(119905) = int119905
0[2(119906119909
+ 120582119887)2
minus (119906 + 120582119887)2
minus (119906 minus
120582119887)2](119902(1199090 120591) 120591)119889120591 minus 4119906
0119909(1199090) minus 4120582119887 then (26) is an
equation of type (27) with 1198620
= 11988724 The proof is complete
by applying Lemma 3
Remark 4 When 119887 = 1 Theorem 2 reduces to the result in[19]
Theorem 5 Let 119886 = 2119887 gt 0 Suppose that 1199060
isin 1198672(R) and
there exists a 1199090isin R such that 119910
0(1199090) = (1 minus 120597
2
119909)1199060(1199090) = 0
119890minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt
2120582
119887 119890
1199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt minus
2120582
119887
(29)
Then the corresponding solution 119906(119909 119905) to (2) with 1199060as the
initial datum blows up in finite time
Proof We easily obtain
119906119905119909
+ 119887119906119906119909119909
minus119886
21199062+
119887
21199062
119909+ 119866 lowast (
119886
21199062+
119887
21199062
119909) + 120582119906
119909= 0
(30)
Differentiating 119906119909at the point (119902(119909
0 119905) 119905)with respect to 119905 we
get
119889
119889119905119906119909(119902 (1199090 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909(119902 (1199090 119905) 119905)
(31)
Process of the proof is similar to Theorem 2 Thus to beconcise we omit the detailed proof
When 119886 = 2119887 gt 0 120582 = 0 using (119909 119905) = 119887119906(119909 119905) (2) canbe reformulated into
119909+ 119909119909119905
+ 3119909minus 2119909119909119909
minus 119909119909119909
= 0 (32)
which is the well-known Camassa-Holm equation Mean-while we also find that the condition in Theorem 5 can bereformulated into
int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt 0 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt 0 (33)
which is one of the sufficient conditions to guarantee blow-upadd-on initial data for the Camassa-Holm equation
So we show the necessary and sufficient condition for thespecial case 119886 = 2119887 gt 0 and 120582 = 0 in the following theorem
Theorem 6 When 119886 = 2119887 gt 0 and 120582 = 0 then the nonlinearwave equation (2) breaks if and only if some portion of thepositive part of 119910
0(119909) lies to the left of some portion of its
negative part
Proof As studied in [1] when 119886 = 2119887 gt 0 and120582 = 0 rewriting(2) yields
119906119909+ 119906119909119909119905
+ 3119887119906119906119909minus 2119887119906
119909119906119909119909
minus 119887119906119906119909119909119909
= 0 (34)
RecallingMckeanrsquos theorem in [7] (32) breaks if and onlyif some portion of the positive part of 119910
0(119909 119905) = (1 minus 120597
2
119909)0
lies to the left of some portion of its negative partSo (34) breaks if and only if some portion of the positive
part of 1198871199100(119909 119905) = 119887(1 minus 120597
2
119909)1199060lies to the left of some portion
of its negative partThis completes the proof
Remark 7 Mckeanrsquos theorem [7] is for the special case 119886 = 2119887 = 1 Condition 119886 = 2119887 here is more general However thenecessary and sufficient condition for (2) is still a challengingproblem for us at present
4 Global Existence
Now let us try to find a condition for global existence Unfor-tunately When 119886 = 2119887 like the Degasperis-Procesi equation[12] only the following easy one can be proved at present
Theorem 8 Suppose that 1199060isin 1198673(R) and 119910
0= (1 minus 120597
2
119909)1199060is
one sign Then the corresponding solution to (2) exists globally
Proof We can assume that 1199100ge 0 It is sufficient to prove that
119906119909(119909 119905) has a lower and upper bound for all 119905 In fact
119906119909(119909 119905) = minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 +
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
(35)
Therefore we have
119906119909(119909 119905) ge minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 ge minus
1
2int
119909
minusinfin
119910 (120585 119905) 119889120585
ge minus1
2int
infin
minusinfin
119910 (120585 119905) 119889120585 = minus1
2int
infin
minusinfin
1199100(120585 119905) 119889120585
119906119909(119909 119905) le
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585 le1
2int
infin
119909
119910 (120585 119905) 119889120585
le1
2int
infin
minusinfin
119910 (120585 119905) 119889120585 =1
2int
infin
minusinfin
1199100(120585 119905) 119889120585
(36)
This completes the proof
6 Abstract and Applied Analysis
5 Infinite Propagation Speed
In this section we will give a more detailed description onthe corresponding strong solution 119906(119909 119905) to (2) in its life spanwith initial data 119906
0(119909) being compactly supported The main
theorem reads as follows
Theorem 9 Let 0 lt 119886 le 3119887 Assume that for some 119879 ge 0 and119904 ge 52 119906 isin 119862([0 119879)119867
119904(R)) is a strong solution of (2) If
1199060(119909) = 119906(119909 0) has compact support [ac] then for 119905 isin (0 119879)
one has
119906 (119909 119905) = 119871 (119905) 119890
minus119909 for 119909 gt 119902 (119888 119905)
119897 (119905) 119890minus119909
for 119909 lt 119902 (119886 119905) (37)
where 119871(119905) and 119897(119905) denote continuous nonvanishing functionswith 119871(119905) gt 0 and 119897(119905) lt 0 for 119905 isin (0 119879) Furthermore 119871(119905)is strictly increasing function while 119897(119905) is strictly decreasingfunction
Proof Since 1199060has compact support in 119909 in [119886 119888] from (8)
so does 119910( 119905) has compact support in 119909 in [119902(119886 119905) 119902(119888 119905)] inits lifespan Hence the following functions are well-defined
119864 (119905) = intR
119890119909119910 (119909 119905) 119889119909 119865 (119905) = int
R
119890minus119909
119910 (119909 119905) 119889119909 (38)
with
1198640= int
R
1198901199091199100(119909) 119889119909 = 0 119865
0= int
R
119890minus119909
1199100(119909) 119889119909 = 0
(39)
Thus for 119909 gt 119902(119888 119905) we obtain
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905)
=1
2119890minus119909
int
119902(119887119905)
119902(119886119905)
119890120591119910 (120591 119905) 119889120591 =
1
2119890minus119909
119864 (119905)
(40)
Similarly for 119909 lt 119902(119886 119905) we have
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905) =1
2119890119909int
119902(119887119905)
119902(119886119905)
119890minus120591
119910 (120591 119905) 119889120591
=1
2119890119909119865 (119905)
(41)
Hence as consequences of (40) and (41) we get
119906 (119909 119905) = minus119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890minus119909
119864 (119905)
as 119909 gt 119902 (119888 119905)
119906 (119909 119905) = 119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890119909119865 (119905)
as 119909 lt 119902 (119886 119905)
(42)
On the other hand
119889119864 (119905)
119889119905= int
R
119890119909119910119905(119909 119905) 119889119909 (43)
It is easy to get
119910119905= minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909)
minus 120582119906 + 120582119906119909119909
(44)
Putting the identity (44) into 119889119864(119905)119889119905 we have
119889119864 (119905)
119889119905
= intR
119890119909(minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909))119889119909
+ intR
119890119909(minus120582119906 + 120582119906
119909119909) 119889119909
= intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909)119889119909
(45)
where we have used (42)Therefore in the lifespan of the solution we get
119864 (119905) = int
119905
0
intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 gt 0 (46)
By the same argument one can check that the followingidentity for 119865(119905) is true
119865 (119905) = minusint
119905
0
intR
119890minus119909
(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 lt 0
(47)
In order to complete the proof it is sufficient to let 119871(119905) =
(12)119864(119905) and 119897(119905) = (12)119865(119905) respectively
Acknowledgments
This work is partially supported by Zhejiang InnovationProject (T200905) ZJNSF (Grant no R6090109) and NSFC(Grant no 10971197 and 11101376)
References
[1] Z Jiang and S Hakkaev ldquoWave breaking and propagationspeed for a class of one-dimensional shallow water equationsrdquoAbstract and Applied Analysis vol 2011 Article ID 647368 15pages 2011
[2] L Ni and Y Zhou ldquoWave breaking and propagation speed fora class of nonlocal dispersive 120579-equationsrdquo Nonlinear AnalysisReal World Applications vol 12 no 1 pp 592ndash600 2011
[3] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[4] B Fuchssteiner and A S Fokas ldquoSymplectic structures theirBacklund transformations and hereditary symmetriesrdquo PhysicaD vol 4 no 1 pp 47ndash66 1981-1982
[5] A Constantin and J Escher ldquoWell-posedness global existenceand blowup phenomena for a periodic quasi-linear hyperbolicequationrdquo Communications on Pure and Applied Mathematicsvol 51 no 5 pp 475ndash504 1998
Abstract and Applied Analysis 7
[6] Y A Li and P J Olver ldquoWell-posedness and blow-up solutionsfor an integrable nonlinearly dispersive model wave equationrdquoJournal of Differential Equations vol 162 no 1 pp 27ndash63 2000
[7] H P McKean ldquoBreakdown of a shallow water equationrdquo TheAsian Journal of Mathematics vol 2 no 4 pp 867ndash874 1998
[8] Z Jiang L Ni and Y Zhou ldquoWave breaking of the Camassa-Holm equationrdquo Journal of Nonlinear Science vol 22 no 2 pp235ndash245 2012
[9] A Constantin and W A Strauss ldquoStability of peakonsrdquo Com-munications on Pure and AppliedMathematics vol 53 no 5 pp603ndash610 2000
[10] Y Zhou ldquoStability of solitary waves for a rod equationrdquo ChaosSolitons and Fractals vol 21 no 4 pp 977ndash981 2004
[11] A A Himonas G Misiołek G Ponce and Y Zhou ldquoPersis-tence properties and unique continuation of solutions of theCamassa-Holm equationrdquo Communications in MathematicalPhysics vol 271 no 2 pp 511ndash522 2007
[12] A Degasperis and M Procesi ldquoAsymptotic integrabilityrdquo inSymmetry and PerturbationTheory ADegasperis andGGaetaEds pp 23ndash37 World Scientific Singapore 1999
[13] D D Holm and M F Staley ldquoNonlinear balance and exchangeof stability of dynamics of solitons peakons rampscliffs andleftons in a 1 + 1 nonlinear evolutionary PDErdquo Physics LettersA vol 308 no 5-6 pp 437ndash444 2003
[14] Y Zhou ldquoBlow-up phenomenon for the integrable Degasperis-Procesi equationrdquoPhysics Letters A vol 328 no 2-3 pp 157ndash1622004
[15] Y Zhou ldquoOn solutions to the Holm-Staley 119887-family of equa-tionsrdquo Nonlinearity vol 23 no 2 pp 369ndash381 2010
[16] Z Guo ldquoBlow up global existence and infinite propagationspeed for the weakly dissipative Camassa-Holm equationrdquoJournal ofMathematical Physics vol 49 no 3 Article ID 0335162008
[17] Z Guo ldquoSome properties of solutions to the weakly dissipativeDegasperis-Procesi equationrdquo Journal of Differential Equationsvol 246 no 11 pp 4332ndash4344 2009
[18] W Niu and S Zhang ldquoBlow-up phenomena and globalexistence for the nonuniform weakly dissipative 119887-equationrdquoJournal of Mathematical Analysis and Applications vol 374 no1 pp 166ndash177 2011
[19] M Zhu andZ Jiang ldquoSomeproperties of solutions to theweaklydissipative 119887-family equationrdquo Nonlinear Analysis Real WorldApplications vol 13 no 1 pp 158ndash167 2012
Submit your manuscripts athttpwwwhindawicom
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International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Abstract and Applied Analysis
2 Local Well-Posedness and Blow-Up Scenario
In this section we first establish the local well-posedness of(2) by using Katorsquos theoryThen we provide the precise blow-up scenario for solutions to (2)
System (2) is equivalent to the following system
119906119905+ 119887119906119906
119909+ 120597119909119866 lowast (
119886
21199062+
3119887 minus 119886
21199062
119909) + 120582119906 = 0 (3)
where 119866(119909) = (12)119890minus|119909| lowast means doing convolution
Theorem 1 Given 1199060
isin 119867119904(R) 119904 gt 32 then there exist a 119879
and a unique solution 119906 to (2) such that
119906 (119909 119905) isin 119862 ([0 119879) 119867119904(R)) cap 119862
1([0 119879) 119867
119904minus1(R)) (4)
To make the paper concise we would like to omit thedetailed proof since one can find similar ones for these typesof equations in [5]
3 Blow-Up Phenomenon
In this section we will give some conditions to guarantee thefinite time blowup Motivated by Mckeanrsquos deep observationfor the Camassa-Holm equation [7] we can consider thesimilar particle trajectory as
119902119905= 119887119906 (119902 119905) 0 lt 119905 lt 119879 119909 isin R
119902 (119909 0) = 119909 119909 isin R
(5)
where 119879 is the lifespan of the solution then 119902 is a diffeomor-phism of the line Taking derivative (5) with respect to 119909 weobtain
119889119902119905
119889119909= 119902119905119909
= 119887119906119909(119902 119905) 119902
119909 119905 isin (0 119879) (6)
Therefore
119902119909(119909 119905) = exp119887int
119905
0
119906119909(119902 119904) 119889119904 119902
119909(119909 0) = 1 (7)
Hence from (2) the following identity can be proved
119910 (119902) 119902119886119887
119909= 1199100(119909) 119890minus120582119905
(8)
In fact direct calculation yields119889
119889119905(119910 (119902) 119902
119886119887
119909) = [119910
119905(119902) + 119887119906 (119902 119905) 119910
119909(119902)
+119886119906119909(119902 119905) 119910 (119902)] 119902
119886119887
119909= minus120582119910119902
119886119887
119909
(9)
Motivated by [19] we give the following theorem
Theorem 2 Let 119886 minus 2119887 gt 0 119887 gt 0 suppose that 1199060isin 1198672(R)
and there exists a1199090isin R such that119910
0(1199090) = (1minus120597
2
119909)1199060(1199090) = 0
1199100ge 0 ( equiv 0) for 119909 isin (minusinfin 119909
0)
1199100le 0 ( equiv 0) for 119909 isin (119909
0infin)
(10)
119890minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt
2120582
119887 119890
1199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt minus
2120582
119887
(11)
Then the corresponding solution 119906(119909 119905) to (2) with 1199060as the
initial datum blows up in finite time
Proof Suppose that the solution exists globally From (8) andinitial condition (10) we have 119910(119902(119909
0 119905) 119905) = 0 and
119910 (119902 (119909 119905) 119905) ge 0 ( equiv 0) for 119909 isin (minusinfin 119902 (1199090 119905))
119910 (119902 (119909 119905) 119905) le 0 ( equiv 0) for 119909 isin (119902 (1199090 119905) infin)
(12)
for all 119905 ge 0 Due to 119906(119909 119905) = 119866 lowast 119910(119909 119905) we can write 119906(119909 119905)
and 119906119909(119909 119905) as
119906 (119909 119905) =1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 +
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
119906119909(119909 119905) = minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 +
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
(13)
Therefore
1199062
119909(119909 119905) minus 119906
2(119909 119905) = minusint
119909
minusinfin
119890120585119910 (120585 119905) 119889120585int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
(14)
for all 119905 gt 0By direct calculation for 119909 le 119902(119909
0 119905) we have
1199062
119909(119909 119905) minus 119906
2(119909 119905)
= minusint
119909
minusinfin
119890120585119910 (120585 119905) 119889120585int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
= minus(int
119902(1199090 119905)
minusinfin
119890120585119910 (120585 119905) 119889120585 minus int
119902(1199090 119905)
119909
119890120585119910 (120585 119905) 119889120585)
times (int
infin
119902(1199090119905)
119890minus120585
119910 (120585 119905) 119889120585 + int
119902(1199090 119905)
119909
119890minus120585
119910 (120585 119905) 119889120585)
= 1199062
119909(119902 (1199090 119905) 119905) minus 119906
2(119902 (1199090 119905) 119905)
minus int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585int
119902(1199090 119905)
119909
119890minus120585
119910 (120585 119905) 119889120585
+ int
119902(1199090 119905)
119909
119890120585119910 (120585 119905) 119889120585int
infin
119902(1199090119905)
119890minus120585
119910 (120585 119905) 119889120585
le 1199062
119909(119902 (1199090 119905) 119905) minus 119906
2(119902 (1199090 119905) 119905)
(15)
Similarly for 119909 ge 119902(1199090 119905) we have
1199062
119909(119909 119905) minus 119906
2(119909 119905) le 119906
2
119909(119902 (1199090 119905) 119905) minus 119906
2(119902 (1199090 119905) 119905)
(16)
So for any fixed 119905 combination of (15) and (16) we obtain
1199062
119909(119909 119905) minus 119906
2(119909 119905) le 119906
2
119909(119902 (1199090 119905) 119905) minus 119906
2(119902 (1199090 119905) 119905)
(17)
for all 119909 isin R
Abstract and Applied Analysis 3
From the expression of 119906119909(119909 119905) in terms of 119910(119909 119905)
differentiating 119906119909(119902(1199090 119905) 119905) with respect to 119905 we have
120597119905119906119909(119902 (1199090 119905) 119905)
= 119906119909119905
(119902 (1199090 119905) 119905) + 119906
119909119909(119902 (1199090 119905) 119905) 119902
119905(119902 (1199090 119905) 119905)
=119886
21199062(119902 (1199090 119905) 119905) +
119886 minus 119887
21199062
119909(119902 (1199090 119905) 119905)
minus 120582119906119909(119902 (1199090 119905) 119905) minus 119866 lowast (
119886
21199062(119909 119905) +
3119887 minus 119886
21199062
119909(119909 119905))
= 119866 lowast (119886
21199062(119902 (1199090 119905) 119905) +
119886 minus 119887
21199062
119909(119902 (1199090 119905) 119905)
minus119886
21199062(119909 119905) minus
3119887 minus 119886
21199062
119909(119909 119905)) minus 120582119906
119909(119902 (1199090 119905) 119905)
= 119866 lowast (119886 minus 2119887
2(1199062(119902 (1199090 119905) 119905) minus 119906
2
119909(119902 (1199090 119905) 119905)
minus1199062(119909 119905) + 119906
2
119909(119909 119905)))
+ 119866 lowast (1198871199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905)
minus1198871199062(119909 119905) minus
119887
21199062
119909(119909 119905)) minus 120582119906
119909(119902 (1199090 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909(119902 (1199090 119905) 119905)
(18)
where we have used (17) and the inequality 119866 lowast (1199062(119909 119905) +
(12)1199062
119909(119909 119905)) ge (12)119906
2 In addition we also used theequation 119906
119905119909+119906119906119909119909
minus (1198862)1199062minus ((119887minus119886)2)119906
2
119909+119866lowast((1198862)119906
2+
((3119887minus119886)2)1199062
119909)+120582119906119909= 0 which is obtained by differentiating
equation (3)For (11) we know that
(1199060119909
(1199090) +
120582
119887)
2
minus (1199060(1199090) +
120582
119887)
2
= minus119890minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 times (119890
1199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 +
2120582
119887) gt 0
(1199060119909
(1199090) +
120582
119887)
2
minus (1199060(1199090) minus
120582
119887)
2
= minus1198901199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 times (119890
minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 minus
2120582
119887) gt 0
(19)
Claim 119906119909(119902(1199090 119905) 119905) lt 0 is decreasing (119906(119902(119909
0 119905) 119905) +
120582119887)2
lt (119906119909(119902(1199090 119905) 119905) + 120582119887)
2 and (119906(119902(1199090 119905) 119905) minus 120582119887)
2lt
(119906119909(119902(1199090 119905) 119905) + 120582119887)
2 for all 119905 ge 0Suppose that there exists a 119905
0such that (119906(119902(119909
0 119905) 119905) +
120582119887)2
lt (119906119909(119902(1199090 119905) 119905) + 120582119887)
2 and (119906(119902(1199090 119905) 119905) minus 120582119887)
2lt
(119906119909(119902(1199090 119905) 119905) + 120582119887)
2 on [0 1199050) then (119906(119902(119909
0 1199050) 1199050) +
120582119887)2= (119906119909(119902(1199090 1199050) 1199050) + 120582119887)
2 or (119906(119902(1199090 1199050) 1199050) minus 120582119887)
2=
(119906119909(119902(1199090 1199050) 1199050) + 120582119887)
2Now let
119868 (119905) =1
2119890minus119902(1199090 119905) int
119902(1199090 119905)
minusinfin
119890120585119910 (120585 119905) 119889120585
119868119868 (119905) =1
2119890119902(1199090 119905) int
infin
119902(1199090 119905)
119890minus120585
119910 (120585 119905) 119889120585
(20)
Firstly differentiating 119868(119905) we have
119889119868 (119905)
119889119905= minus
119887
2119906 (119902 (119909
0 119905) 119905) 119890
minus119902(1199090 119905) int
119902(1199090 119905)
minusinfin
119890120585119910 (120585 119905) 119889120585
+1
2119890minus119902(1199090 119905) int
119902(1199090 119905)
minusinfin
119890120585119910119905(120585 119905) 119889120585
=119887
2119906 (119906119909minus 119906) (119902 (119909
0 119905) 119905) minus
1
2119890minus119902(1199090 119905)
times int
119902(1199090 119905)
minusinfin
119890120585(119887119906119910119909+ 2119887119906
119909119910
+119886 minus 2119887
2(1199062minus 1199062
119909)119909+ 120582119910)119889120585
ge119887
2119906 (119906119909minus 119906) (119902 (119909
0 119905) 119905) +
119887
4(1199062+ 1199062
119909minus 2119906119906
119909)
times (119902 (1199090 119905) 119905) minus
120582
2(119906 minus 119906
119909) (119902 (119909
0 119905) 119905)
=119887
4(1199062
119909minus 1199062) (119902 (119909
0 119905) 119905) minus
120582
2(119906 minus 119906
119909) (119902 (119909
0 119905) 119905)
=119887
4(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
minus119887
4(119906 (119902 (119909
0 119905) 119905) +
120582
119887)
2
gt 0 on [0 1199050)
(21)
Secondly by the same argument we obtain
119889119868119868 (119905)
119889119905=
119887
2119906 (119902 (119909
0 119905) 119905) 119890
119902(1199090 119905) int
infin
119902(1199090 119905)
119890minus120585
119910 (120585 119905) 119889120585
+1
2119890119902(1199090 119905) int
infin
119902(1199090 119905)
119890minus120585
119910119905(120585 119905) 119889120585
=119887
2119906 (119906119909+ 119906) (119902 (119909
0 119905) 119905) minus
1
2119890119902(1199090 119905)
times int
infin
119902(1199090 119905)
119890minus120585
(119887119906119910119909+ 2119887119906
119909119910
+119886 minus 2119887
2(1199062minus 1199062
119909)119909+ 120582119910)119889120585
4 Abstract and Applied Analysis
le119887
2119906 (119906119909+ 119906) (119902 (119909
0 119905) 119905)
minus119887
4(1199062+ 1199062
119909+ 2119906119906
119909) (119902 (119909
0 119905) 119905)
minus120582
2(119906119909+ 119906) (119902 (119909
0 119905) 119905)
= minus119887
4(1199062
119909minus 1199062) (119902 (119909
0 119905) 119905)
minus120582
2(119906119909+ 119906) (119902 (119909
0 119905) 119905)
= minus119887
4(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
+119887
4(119906 (119902 (119909
0 119905) 119905) minus
120582
119887)
2
lt 0 on [0 1199050)
(22)
Therefore it follows from (21) (22) and the continuityproperty of ODEs that
(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
minus (119906 (119902 (1199090 119905) 119905) +
120582
119887)
2
= minus4119868 (119905) (119868119868 (119905) +120582
119887) gt minus4119868 (0) (119868119868 (0) +
120582
119887) gt 0
(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
minus (119906 (119902 (1199090 119905) 119905) minus
120582
119887)
2
= minus4(119868 (119905) minus120582
119887) 119868119868 (119905) gt minus4(119868 (0) minus
120582
119887) 119868119868 (0) gt 0
(23)
for all 119905 gt 0 This implies that 1199050can be extended to the
infinityMoreover using (21) and (22) again we have the following
equation for [2(119906119909+120582119887)
2minus(119906+120582119887)
2minus(119906 minus 120582119887)
2](119902(1199090 119905) 119905)
119889
119889119905[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
= minus4119889
119889119905[119868 (119905) (119868119868 (119905) +
120582
119887)] minus 4
119889
119889119905[(119868 (119905) minus
120582
119887) 119868119868 (119905)]
ge minus119887 [(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
] (119902 (1199090 119905) 119905) (119868119868 (119905) +
120582
119887)
+ 119887 [(119906119909+
120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905) 119868 (119905)
minus 119887 [(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
] (119902 (1199090 119905) 119905) 119868119868 (119905)
+ 119887 [(119906119909+
120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905) (119868 (119905) minus
120582
119887)
= 119887(minus120582
119887[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905) minus 119906
119909(119902 (1199090 119905) 119905)
times (2(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
)
+2(119906 +120582
119887)
2
119868119868 (119905) minus 2(119906 minus120582
119887)
2
119868 (119905))
= 119887(minus (119906119909(119902 (1199090 119905) 119905) +
120582
119887)
times [2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905) minus 119906
119909(119902 (1199090 119905) 119905)
times [(119906 +120582
119887)
2
+ (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
+2(119906 +120582
119887)
2
119868119868 (119905) minus 2(119906 minus120582
119887)
2
119868 (119905))
ge minus119887(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
times [2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
(24)
where we use 119906119909(119902(1199090 119905) 119905) = minus119868(119905) + 119868119868(119905)
Now recalling (18) we have
120597119905119906119909(119902 (1199090 119905) 119905) (119902 (119909
0 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909
=119887
4[(119906 +
120582
119887)
2
+ (119906 minus120582
119887)
2
minus 2(119906119909+
120582
119887)
2
]
times (119902 (1199090 119905) 119905)
(25)
Putting (25) into (24) it yields
119889
119889119905[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
ge1198872
4[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905)
times (int
119905
0
[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 120591) 120591) 119889120591 minus 4119906
0119909(1199090) minus
4120582
119887)
(26)
Abstract and Applied Analysis 5
Before finishing the proof we need the following technicallemma
Lemma 3 (see [15]) Suppose that Ψ(119905) is twice continuouslydifferential satisfying
Ψ10158401015840(119905) ge 119862
0Ψ1015840(119905) Ψ (119905) 119905 gt 0 119862
0gt 0
Ψ (119905) gt 0 Ψ1015840(119905) gt 0
(27)
Then 120595(119905) blows up in finite time Moreover the blow-up timecan be estimated in terms of the initial datum as
119879 le max2
1198620Ψ (0)
Ψ (0)
Ψ1015840 (0) (28)
Let Ψ(119905) = int119905
0[2(119906119909
+ 120582119887)2
minus (119906 + 120582119887)2
minus (119906 minus
120582119887)2](119902(1199090 120591) 120591)119889120591 minus 4119906
0119909(1199090) minus 4120582119887 then (26) is an
equation of type (27) with 1198620
= 11988724 The proof is complete
by applying Lemma 3
Remark 4 When 119887 = 1 Theorem 2 reduces to the result in[19]
Theorem 5 Let 119886 = 2119887 gt 0 Suppose that 1199060
isin 1198672(R) and
there exists a 1199090isin R such that 119910
0(1199090) = (1 minus 120597
2
119909)1199060(1199090) = 0
119890minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt
2120582
119887 119890
1199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt minus
2120582
119887
(29)
Then the corresponding solution 119906(119909 119905) to (2) with 1199060as the
initial datum blows up in finite time
Proof We easily obtain
119906119905119909
+ 119887119906119906119909119909
minus119886
21199062+
119887
21199062
119909+ 119866 lowast (
119886
21199062+
119887
21199062
119909) + 120582119906
119909= 0
(30)
Differentiating 119906119909at the point (119902(119909
0 119905) 119905)with respect to 119905 we
get
119889
119889119905119906119909(119902 (1199090 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909(119902 (1199090 119905) 119905)
(31)
Process of the proof is similar to Theorem 2 Thus to beconcise we omit the detailed proof
When 119886 = 2119887 gt 0 120582 = 0 using (119909 119905) = 119887119906(119909 119905) (2) canbe reformulated into
119909+ 119909119909119905
+ 3119909minus 2119909119909119909
minus 119909119909119909
= 0 (32)
which is the well-known Camassa-Holm equation Mean-while we also find that the condition in Theorem 5 can bereformulated into
int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt 0 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt 0 (33)
which is one of the sufficient conditions to guarantee blow-upadd-on initial data for the Camassa-Holm equation
So we show the necessary and sufficient condition for thespecial case 119886 = 2119887 gt 0 and 120582 = 0 in the following theorem
Theorem 6 When 119886 = 2119887 gt 0 and 120582 = 0 then the nonlinearwave equation (2) breaks if and only if some portion of thepositive part of 119910
0(119909) lies to the left of some portion of its
negative part
Proof As studied in [1] when 119886 = 2119887 gt 0 and120582 = 0 rewriting(2) yields
119906119909+ 119906119909119909119905
+ 3119887119906119906119909minus 2119887119906
119909119906119909119909
minus 119887119906119906119909119909119909
= 0 (34)
RecallingMckeanrsquos theorem in [7] (32) breaks if and onlyif some portion of the positive part of 119910
0(119909 119905) = (1 minus 120597
2
119909)0
lies to the left of some portion of its negative partSo (34) breaks if and only if some portion of the positive
part of 1198871199100(119909 119905) = 119887(1 minus 120597
2
119909)1199060lies to the left of some portion
of its negative partThis completes the proof
Remark 7 Mckeanrsquos theorem [7] is for the special case 119886 = 2119887 = 1 Condition 119886 = 2119887 here is more general However thenecessary and sufficient condition for (2) is still a challengingproblem for us at present
4 Global Existence
Now let us try to find a condition for global existence Unfor-tunately When 119886 = 2119887 like the Degasperis-Procesi equation[12] only the following easy one can be proved at present
Theorem 8 Suppose that 1199060isin 1198673(R) and 119910
0= (1 minus 120597
2
119909)1199060is
one sign Then the corresponding solution to (2) exists globally
Proof We can assume that 1199100ge 0 It is sufficient to prove that
119906119909(119909 119905) has a lower and upper bound for all 119905 In fact
119906119909(119909 119905) = minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 +
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
(35)
Therefore we have
119906119909(119909 119905) ge minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 ge minus
1
2int
119909
minusinfin
119910 (120585 119905) 119889120585
ge minus1
2int
infin
minusinfin
119910 (120585 119905) 119889120585 = minus1
2int
infin
minusinfin
1199100(120585 119905) 119889120585
119906119909(119909 119905) le
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585 le1
2int
infin
119909
119910 (120585 119905) 119889120585
le1
2int
infin
minusinfin
119910 (120585 119905) 119889120585 =1
2int
infin
minusinfin
1199100(120585 119905) 119889120585
(36)
This completes the proof
6 Abstract and Applied Analysis
5 Infinite Propagation Speed
In this section we will give a more detailed description onthe corresponding strong solution 119906(119909 119905) to (2) in its life spanwith initial data 119906
0(119909) being compactly supported The main
theorem reads as follows
Theorem 9 Let 0 lt 119886 le 3119887 Assume that for some 119879 ge 0 and119904 ge 52 119906 isin 119862([0 119879)119867
119904(R)) is a strong solution of (2) If
1199060(119909) = 119906(119909 0) has compact support [ac] then for 119905 isin (0 119879)
one has
119906 (119909 119905) = 119871 (119905) 119890
minus119909 for 119909 gt 119902 (119888 119905)
119897 (119905) 119890minus119909
for 119909 lt 119902 (119886 119905) (37)
where 119871(119905) and 119897(119905) denote continuous nonvanishing functionswith 119871(119905) gt 0 and 119897(119905) lt 0 for 119905 isin (0 119879) Furthermore 119871(119905)is strictly increasing function while 119897(119905) is strictly decreasingfunction
Proof Since 1199060has compact support in 119909 in [119886 119888] from (8)
so does 119910( 119905) has compact support in 119909 in [119902(119886 119905) 119902(119888 119905)] inits lifespan Hence the following functions are well-defined
119864 (119905) = intR
119890119909119910 (119909 119905) 119889119909 119865 (119905) = int
R
119890minus119909
119910 (119909 119905) 119889119909 (38)
with
1198640= int
R
1198901199091199100(119909) 119889119909 = 0 119865
0= int
R
119890minus119909
1199100(119909) 119889119909 = 0
(39)
Thus for 119909 gt 119902(119888 119905) we obtain
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905)
=1
2119890minus119909
int
119902(119887119905)
119902(119886119905)
119890120591119910 (120591 119905) 119889120591 =
1
2119890minus119909
119864 (119905)
(40)
Similarly for 119909 lt 119902(119886 119905) we have
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905) =1
2119890119909int
119902(119887119905)
119902(119886119905)
119890minus120591
119910 (120591 119905) 119889120591
=1
2119890119909119865 (119905)
(41)
Hence as consequences of (40) and (41) we get
119906 (119909 119905) = minus119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890minus119909
119864 (119905)
as 119909 gt 119902 (119888 119905)
119906 (119909 119905) = 119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890119909119865 (119905)
as 119909 lt 119902 (119886 119905)
(42)
On the other hand
119889119864 (119905)
119889119905= int
R
119890119909119910119905(119909 119905) 119889119909 (43)
It is easy to get
119910119905= minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909)
minus 120582119906 + 120582119906119909119909
(44)
Putting the identity (44) into 119889119864(119905)119889119905 we have
119889119864 (119905)
119889119905
= intR
119890119909(minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909))119889119909
+ intR
119890119909(minus120582119906 + 120582119906
119909119909) 119889119909
= intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909)119889119909
(45)
where we have used (42)Therefore in the lifespan of the solution we get
119864 (119905) = int
119905
0
intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 gt 0 (46)
By the same argument one can check that the followingidentity for 119865(119905) is true
119865 (119905) = minusint
119905
0
intR
119890minus119909
(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 lt 0
(47)
In order to complete the proof it is sufficient to let 119871(119905) =
(12)119864(119905) and 119897(119905) = (12)119865(119905) respectively
Acknowledgments
This work is partially supported by Zhejiang InnovationProject (T200905) ZJNSF (Grant no R6090109) and NSFC(Grant no 10971197 and 11101376)
References
[1] Z Jiang and S Hakkaev ldquoWave breaking and propagationspeed for a class of one-dimensional shallow water equationsrdquoAbstract and Applied Analysis vol 2011 Article ID 647368 15pages 2011
[2] L Ni and Y Zhou ldquoWave breaking and propagation speed fora class of nonlocal dispersive 120579-equationsrdquo Nonlinear AnalysisReal World Applications vol 12 no 1 pp 592ndash600 2011
[3] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[4] B Fuchssteiner and A S Fokas ldquoSymplectic structures theirBacklund transformations and hereditary symmetriesrdquo PhysicaD vol 4 no 1 pp 47ndash66 1981-1982
[5] A Constantin and J Escher ldquoWell-posedness global existenceand blowup phenomena for a periodic quasi-linear hyperbolicequationrdquo Communications on Pure and Applied Mathematicsvol 51 no 5 pp 475ndash504 1998
Abstract and Applied Analysis 7
[6] Y A Li and P J Olver ldquoWell-posedness and blow-up solutionsfor an integrable nonlinearly dispersive model wave equationrdquoJournal of Differential Equations vol 162 no 1 pp 27ndash63 2000
[7] H P McKean ldquoBreakdown of a shallow water equationrdquo TheAsian Journal of Mathematics vol 2 no 4 pp 867ndash874 1998
[8] Z Jiang L Ni and Y Zhou ldquoWave breaking of the Camassa-Holm equationrdquo Journal of Nonlinear Science vol 22 no 2 pp235ndash245 2012
[9] A Constantin and W A Strauss ldquoStability of peakonsrdquo Com-munications on Pure and AppliedMathematics vol 53 no 5 pp603ndash610 2000
[10] Y Zhou ldquoStability of solitary waves for a rod equationrdquo ChaosSolitons and Fractals vol 21 no 4 pp 977ndash981 2004
[11] A A Himonas G Misiołek G Ponce and Y Zhou ldquoPersis-tence properties and unique continuation of solutions of theCamassa-Holm equationrdquo Communications in MathematicalPhysics vol 271 no 2 pp 511ndash522 2007
[12] A Degasperis and M Procesi ldquoAsymptotic integrabilityrdquo inSymmetry and PerturbationTheory ADegasperis andGGaetaEds pp 23ndash37 World Scientific Singapore 1999
[13] D D Holm and M F Staley ldquoNonlinear balance and exchangeof stability of dynamics of solitons peakons rampscliffs andleftons in a 1 + 1 nonlinear evolutionary PDErdquo Physics LettersA vol 308 no 5-6 pp 437ndash444 2003
[14] Y Zhou ldquoBlow-up phenomenon for the integrable Degasperis-Procesi equationrdquoPhysics Letters A vol 328 no 2-3 pp 157ndash1622004
[15] Y Zhou ldquoOn solutions to the Holm-Staley 119887-family of equa-tionsrdquo Nonlinearity vol 23 no 2 pp 369ndash381 2010
[16] Z Guo ldquoBlow up global existence and infinite propagationspeed for the weakly dissipative Camassa-Holm equationrdquoJournal ofMathematical Physics vol 49 no 3 Article ID 0335162008
[17] Z Guo ldquoSome properties of solutions to the weakly dissipativeDegasperis-Procesi equationrdquo Journal of Differential Equationsvol 246 no 11 pp 4332ndash4344 2009
[18] W Niu and S Zhang ldquoBlow-up phenomena and globalexistence for the nonuniform weakly dissipative 119887-equationrdquoJournal of Mathematical Analysis and Applications vol 374 no1 pp 166ndash177 2011
[19] M Zhu andZ Jiang ldquoSomeproperties of solutions to theweaklydissipative 119887-family equationrdquo Nonlinear Analysis Real WorldApplications vol 13 no 1 pp 158ndash167 2012
Submit your manuscripts athttpwwwhindawicom
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 3
From the expression of 119906119909(119909 119905) in terms of 119910(119909 119905)
differentiating 119906119909(119902(1199090 119905) 119905) with respect to 119905 we have
120597119905119906119909(119902 (1199090 119905) 119905)
= 119906119909119905
(119902 (1199090 119905) 119905) + 119906
119909119909(119902 (1199090 119905) 119905) 119902
119905(119902 (1199090 119905) 119905)
=119886
21199062(119902 (1199090 119905) 119905) +
119886 minus 119887
21199062
119909(119902 (1199090 119905) 119905)
minus 120582119906119909(119902 (1199090 119905) 119905) minus 119866 lowast (
119886
21199062(119909 119905) +
3119887 minus 119886
21199062
119909(119909 119905))
= 119866 lowast (119886
21199062(119902 (1199090 119905) 119905) +
119886 minus 119887
21199062
119909(119902 (1199090 119905) 119905)
minus119886
21199062(119909 119905) minus
3119887 minus 119886
21199062
119909(119909 119905)) minus 120582119906
119909(119902 (1199090 119905) 119905)
= 119866 lowast (119886 minus 2119887
2(1199062(119902 (1199090 119905) 119905) minus 119906
2
119909(119902 (1199090 119905) 119905)
minus1199062(119909 119905) + 119906
2
119909(119909 119905)))
+ 119866 lowast (1198871199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905)
minus1198871199062(119909 119905) minus
119887
21199062
119909(119909 119905)) minus 120582119906
119909(119902 (1199090 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909(119902 (1199090 119905) 119905)
(18)
where we have used (17) and the inequality 119866 lowast (1199062(119909 119905) +
(12)1199062
119909(119909 119905)) ge (12)119906
2 In addition we also used theequation 119906
119905119909+119906119906119909119909
minus (1198862)1199062minus ((119887minus119886)2)119906
2
119909+119866lowast((1198862)119906
2+
((3119887minus119886)2)1199062
119909)+120582119906119909= 0 which is obtained by differentiating
equation (3)For (11) we know that
(1199060119909
(1199090) +
120582
119887)
2
minus (1199060(1199090) +
120582
119887)
2
= minus119890minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 times (119890
1199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 +
2120582
119887) gt 0
(1199060119909
(1199090) +
120582
119887)
2
minus (1199060(1199090) minus
120582
119887)
2
= minus1198901199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 times (119890
minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 minus
2120582
119887) gt 0
(19)
Claim 119906119909(119902(1199090 119905) 119905) lt 0 is decreasing (119906(119902(119909
0 119905) 119905) +
120582119887)2
lt (119906119909(119902(1199090 119905) 119905) + 120582119887)
2 and (119906(119902(1199090 119905) 119905) minus 120582119887)
2lt
(119906119909(119902(1199090 119905) 119905) + 120582119887)
2 for all 119905 ge 0Suppose that there exists a 119905
0such that (119906(119902(119909
0 119905) 119905) +
120582119887)2
lt (119906119909(119902(1199090 119905) 119905) + 120582119887)
2 and (119906(119902(1199090 119905) 119905) minus 120582119887)
2lt
(119906119909(119902(1199090 119905) 119905) + 120582119887)
2 on [0 1199050) then (119906(119902(119909
0 1199050) 1199050) +
120582119887)2= (119906119909(119902(1199090 1199050) 1199050) + 120582119887)
2 or (119906(119902(1199090 1199050) 1199050) minus 120582119887)
2=
(119906119909(119902(1199090 1199050) 1199050) + 120582119887)
2Now let
119868 (119905) =1
2119890minus119902(1199090 119905) int
119902(1199090 119905)
minusinfin
119890120585119910 (120585 119905) 119889120585
119868119868 (119905) =1
2119890119902(1199090 119905) int
infin
119902(1199090 119905)
119890minus120585
119910 (120585 119905) 119889120585
(20)
Firstly differentiating 119868(119905) we have
119889119868 (119905)
119889119905= minus
119887
2119906 (119902 (119909
0 119905) 119905) 119890
minus119902(1199090 119905) int
119902(1199090 119905)
minusinfin
119890120585119910 (120585 119905) 119889120585
+1
2119890minus119902(1199090 119905) int
119902(1199090 119905)
minusinfin
119890120585119910119905(120585 119905) 119889120585
=119887
2119906 (119906119909minus 119906) (119902 (119909
0 119905) 119905) minus
1
2119890minus119902(1199090 119905)
times int
119902(1199090 119905)
minusinfin
119890120585(119887119906119910119909+ 2119887119906
119909119910
+119886 minus 2119887
2(1199062minus 1199062
119909)119909+ 120582119910)119889120585
ge119887
2119906 (119906119909minus 119906) (119902 (119909
0 119905) 119905) +
119887
4(1199062+ 1199062
119909minus 2119906119906
119909)
times (119902 (1199090 119905) 119905) minus
120582
2(119906 minus 119906
119909) (119902 (119909
0 119905) 119905)
=119887
4(1199062
119909minus 1199062) (119902 (119909
0 119905) 119905) minus
120582
2(119906 minus 119906
119909) (119902 (119909
0 119905) 119905)
=119887
4(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
minus119887
4(119906 (119902 (119909
0 119905) 119905) +
120582
119887)
2
gt 0 on [0 1199050)
(21)
Secondly by the same argument we obtain
119889119868119868 (119905)
119889119905=
119887
2119906 (119902 (119909
0 119905) 119905) 119890
119902(1199090 119905) int
infin
119902(1199090 119905)
119890minus120585
119910 (120585 119905) 119889120585
+1
2119890119902(1199090 119905) int
infin
119902(1199090 119905)
119890minus120585
119910119905(120585 119905) 119889120585
=119887
2119906 (119906119909+ 119906) (119902 (119909
0 119905) 119905) minus
1
2119890119902(1199090 119905)
times int
infin
119902(1199090 119905)
119890minus120585
(119887119906119910119909+ 2119887119906
119909119910
+119886 minus 2119887
2(1199062minus 1199062
119909)119909+ 120582119910)119889120585
4 Abstract and Applied Analysis
le119887
2119906 (119906119909+ 119906) (119902 (119909
0 119905) 119905)
minus119887
4(1199062+ 1199062
119909+ 2119906119906
119909) (119902 (119909
0 119905) 119905)
minus120582
2(119906119909+ 119906) (119902 (119909
0 119905) 119905)
= minus119887
4(1199062
119909minus 1199062) (119902 (119909
0 119905) 119905)
minus120582
2(119906119909+ 119906) (119902 (119909
0 119905) 119905)
= minus119887
4(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
+119887
4(119906 (119902 (119909
0 119905) 119905) minus
120582
119887)
2
lt 0 on [0 1199050)
(22)
Therefore it follows from (21) (22) and the continuityproperty of ODEs that
(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
minus (119906 (119902 (1199090 119905) 119905) +
120582
119887)
2
= minus4119868 (119905) (119868119868 (119905) +120582
119887) gt minus4119868 (0) (119868119868 (0) +
120582
119887) gt 0
(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
minus (119906 (119902 (1199090 119905) 119905) minus
120582
119887)
2
= minus4(119868 (119905) minus120582
119887) 119868119868 (119905) gt minus4(119868 (0) minus
120582
119887) 119868119868 (0) gt 0
(23)
for all 119905 gt 0 This implies that 1199050can be extended to the
infinityMoreover using (21) and (22) again we have the following
equation for [2(119906119909+120582119887)
2minus(119906+120582119887)
2minus(119906 minus 120582119887)
2](119902(1199090 119905) 119905)
119889
119889119905[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
= minus4119889
119889119905[119868 (119905) (119868119868 (119905) +
120582
119887)] minus 4
119889
119889119905[(119868 (119905) minus
120582
119887) 119868119868 (119905)]
ge minus119887 [(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
] (119902 (1199090 119905) 119905) (119868119868 (119905) +
120582
119887)
+ 119887 [(119906119909+
120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905) 119868 (119905)
minus 119887 [(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
] (119902 (1199090 119905) 119905) 119868119868 (119905)
+ 119887 [(119906119909+
120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905) (119868 (119905) minus
120582
119887)
= 119887(minus120582
119887[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905) minus 119906
119909(119902 (1199090 119905) 119905)
times (2(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
)
+2(119906 +120582
119887)
2
119868119868 (119905) minus 2(119906 minus120582
119887)
2
119868 (119905))
= 119887(minus (119906119909(119902 (1199090 119905) 119905) +
120582
119887)
times [2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905) minus 119906
119909(119902 (1199090 119905) 119905)
times [(119906 +120582
119887)
2
+ (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
+2(119906 +120582
119887)
2
119868119868 (119905) minus 2(119906 minus120582
119887)
2
119868 (119905))
ge minus119887(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
times [2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
(24)
where we use 119906119909(119902(1199090 119905) 119905) = minus119868(119905) + 119868119868(119905)
Now recalling (18) we have
120597119905119906119909(119902 (1199090 119905) 119905) (119902 (119909
0 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909
=119887
4[(119906 +
120582
119887)
2
+ (119906 minus120582
119887)
2
minus 2(119906119909+
120582
119887)
2
]
times (119902 (1199090 119905) 119905)
(25)
Putting (25) into (24) it yields
119889
119889119905[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
ge1198872
4[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905)
times (int
119905
0
[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 120591) 120591) 119889120591 minus 4119906
0119909(1199090) minus
4120582
119887)
(26)
Abstract and Applied Analysis 5
Before finishing the proof we need the following technicallemma
Lemma 3 (see [15]) Suppose that Ψ(119905) is twice continuouslydifferential satisfying
Ψ10158401015840(119905) ge 119862
0Ψ1015840(119905) Ψ (119905) 119905 gt 0 119862
0gt 0
Ψ (119905) gt 0 Ψ1015840(119905) gt 0
(27)
Then 120595(119905) blows up in finite time Moreover the blow-up timecan be estimated in terms of the initial datum as
119879 le max2
1198620Ψ (0)
Ψ (0)
Ψ1015840 (0) (28)
Let Ψ(119905) = int119905
0[2(119906119909
+ 120582119887)2
minus (119906 + 120582119887)2
minus (119906 minus
120582119887)2](119902(1199090 120591) 120591)119889120591 minus 4119906
0119909(1199090) minus 4120582119887 then (26) is an
equation of type (27) with 1198620
= 11988724 The proof is complete
by applying Lemma 3
Remark 4 When 119887 = 1 Theorem 2 reduces to the result in[19]
Theorem 5 Let 119886 = 2119887 gt 0 Suppose that 1199060
isin 1198672(R) and
there exists a 1199090isin R such that 119910
0(1199090) = (1 minus 120597
2
119909)1199060(1199090) = 0
119890minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt
2120582
119887 119890
1199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt minus
2120582
119887
(29)
Then the corresponding solution 119906(119909 119905) to (2) with 1199060as the
initial datum blows up in finite time
Proof We easily obtain
119906119905119909
+ 119887119906119906119909119909
minus119886
21199062+
119887
21199062
119909+ 119866 lowast (
119886
21199062+
119887
21199062
119909) + 120582119906
119909= 0
(30)
Differentiating 119906119909at the point (119902(119909
0 119905) 119905)with respect to 119905 we
get
119889
119889119905119906119909(119902 (1199090 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909(119902 (1199090 119905) 119905)
(31)
Process of the proof is similar to Theorem 2 Thus to beconcise we omit the detailed proof
When 119886 = 2119887 gt 0 120582 = 0 using (119909 119905) = 119887119906(119909 119905) (2) canbe reformulated into
119909+ 119909119909119905
+ 3119909minus 2119909119909119909
minus 119909119909119909
= 0 (32)
which is the well-known Camassa-Holm equation Mean-while we also find that the condition in Theorem 5 can bereformulated into
int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt 0 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt 0 (33)
which is one of the sufficient conditions to guarantee blow-upadd-on initial data for the Camassa-Holm equation
So we show the necessary and sufficient condition for thespecial case 119886 = 2119887 gt 0 and 120582 = 0 in the following theorem
Theorem 6 When 119886 = 2119887 gt 0 and 120582 = 0 then the nonlinearwave equation (2) breaks if and only if some portion of thepositive part of 119910
0(119909) lies to the left of some portion of its
negative part
Proof As studied in [1] when 119886 = 2119887 gt 0 and120582 = 0 rewriting(2) yields
119906119909+ 119906119909119909119905
+ 3119887119906119906119909minus 2119887119906
119909119906119909119909
minus 119887119906119906119909119909119909
= 0 (34)
RecallingMckeanrsquos theorem in [7] (32) breaks if and onlyif some portion of the positive part of 119910
0(119909 119905) = (1 minus 120597
2
119909)0
lies to the left of some portion of its negative partSo (34) breaks if and only if some portion of the positive
part of 1198871199100(119909 119905) = 119887(1 minus 120597
2
119909)1199060lies to the left of some portion
of its negative partThis completes the proof
Remark 7 Mckeanrsquos theorem [7] is for the special case 119886 = 2119887 = 1 Condition 119886 = 2119887 here is more general However thenecessary and sufficient condition for (2) is still a challengingproblem for us at present
4 Global Existence
Now let us try to find a condition for global existence Unfor-tunately When 119886 = 2119887 like the Degasperis-Procesi equation[12] only the following easy one can be proved at present
Theorem 8 Suppose that 1199060isin 1198673(R) and 119910
0= (1 minus 120597
2
119909)1199060is
one sign Then the corresponding solution to (2) exists globally
Proof We can assume that 1199100ge 0 It is sufficient to prove that
119906119909(119909 119905) has a lower and upper bound for all 119905 In fact
119906119909(119909 119905) = minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 +
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
(35)
Therefore we have
119906119909(119909 119905) ge minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 ge minus
1
2int
119909
minusinfin
119910 (120585 119905) 119889120585
ge minus1
2int
infin
minusinfin
119910 (120585 119905) 119889120585 = minus1
2int
infin
minusinfin
1199100(120585 119905) 119889120585
119906119909(119909 119905) le
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585 le1
2int
infin
119909
119910 (120585 119905) 119889120585
le1
2int
infin
minusinfin
119910 (120585 119905) 119889120585 =1
2int
infin
minusinfin
1199100(120585 119905) 119889120585
(36)
This completes the proof
6 Abstract and Applied Analysis
5 Infinite Propagation Speed
In this section we will give a more detailed description onthe corresponding strong solution 119906(119909 119905) to (2) in its life spanwith initial data 119906
0(119909) being compactly supported The main
theorem reads as follows
Theorem 9 Let 0 lt 119886 le 3119887 Assume that for some 119879 ge 0 and119904 ge 52 119906 isin 119862([0 119879)119867
119904(R)) is a strong solution of (2) If
1199060(119909) = 119906(119909 0) has compact support [ac] then for 119905 isin (0 119879)
one has
119906 (119909 119905) = 119871 (119905) 119890
minus119909 for 119909 gt 119902 (119888 119905)
119897 (119905) 119890minus119909
for 119909 lt 119902 (119886 119905) (37)
where 119871(119905) and 119897(119905) denote continuous nonvanishing functionswith 119871(119905) gt 0 and 119897(119905) lt 0 for 119905 isin (0 119879) Furthermore 119871(119905)is strictly increasing function while 119897(119905) is strictly decreasingfunction
Proof Since 1199060has compact support in 119909 in [119886 119888] from (8)
so does 119910( 119905) has compact support in 119909 in [119902(119886 119905) 119902(119888 119905)] inits lifespan Hence the following functions are well-defined
119864 (119905) = intR
119890119909119910 (119909 119905) 119889119909 119865 (119905) = int
R
119890minus119909
119910 (119909 119905) 119889119909 (38)
with
1198640= int
R
1198901199091199100(119909) 119889119909 = 0 119865
0= int
R
119890minus119909
1199100(119909) 119889119909 = 0
(39)
Thus for 119909 gt 119902(119888 119905) we obtain
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905)
=1
2119890minus119909
int
119902(119887119905)
119902(119886119905)
119890120591119910 (120591 119905) 119889120591 =
1
2119890minus119909
119864 (119905)
(40)
Similarly for 119909 lt 119902(119886 119905) we have
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905) =1
2119890119909int
119902(119887119905)
119902(119886119905)
119890minus120591
119910 (120591 119905) 119889120591
=1
2119890119909119865 (119905)
(41)
Hence as consequences of (40) and (41) we get
119906 (119909 119905) = minus119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890minus119909
119864 (119905)
as 119909 gt 119902 (119888 119905)
119906 (119909 119905) = 119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890119909119865 (119905)
as 119909 lt 119902 (119886 119905)
(42)
On the other hand
119889119864 (119905)
119889119905= int
R
119890119909119910119905(119909 119905) 119889119909 (43)
It is easy to get
119910119905= minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909)
minus 120582119906 + 120582119906119909119909
(44)
Putting the identity (44) into 119889119864(119905)119889119905 we have
119889119864 (119905)
119889119905
= intR
119890119909(minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909))119889119909
+ intR
119890119909(minus120582119906 + 120582119906
119909119909) 119889119909
= intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909)119889119909
(45)
where we have used (42)Therefore in the lifespan of the solution we get
119864 (119905) = int
119905
0
intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 gt 0 (46)
By the same argument one can check that the followingidentity for 119865(119905) is true
119865 (119905) = minusint
119905
0
intR
119890minus119909
(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 lt 0
(47)
In order to complete the proof it is sufficient to let 119871(119905) =
(12)119864(119905) and 119897(119905) = (12)119865(119905) respectively
Acknowledgments
This work is partially supported by Zhejiang InnovationProject (T200905) ZJNSF (Grant no R6090109) and NSFC(Grant no 10971197 and 11101376)
References
[1] Z Jiang and S Hakkaev ldquoWave breaking and propagationspeed for a class of one-dimensional shallow water equationsrdquoAbstract and Applied Analysis vol 2011 Article ID 647368 15pages 2011
[2] L Ni and Y Zhou ldquoWave breaking and propagation speed fora class of nonlocal dispersive 120579-equationsrdquo Nonlinear AnalysisReal World Applications vol 12 no 1 pp 592ndash600 2011
[3] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[4] B Fuchssteiner and A S Fokas ldquoSymplectic structures theirBacklund transformations and hereditary symmetriesrdquo PhysicaD vol 4 no 1 pp 47ndash66 1981-1982
[5] A Constantin and J Escher ldquoWell-posedness global existenceand blowup phenomena for a periodic quasi-linear hyperbolicequationrdquo Communications on Pure and Applied Mathematicsvol 51 no 5 pp 475ndash504 1998
Abstract and Applied Analysis 7
[6] Y A Li and P J Olver ldquoWell-posedness and blow-up solutionsfor an integrable nonlinearly dispersive model wave equationrdquoJournal of Differential Equations vol 162 no 1 pp 27ndash63 2000
[7] H P McKean ldquoBreakdown of a shallow water equationrdquo TheAsian Journal of Mathematics vol 2 no 4 pp 867ndash874 1998
[8] Z Jiang L Ni and Y Zhou ldquoWave breaking of the Camassa-Holm equationrdquo Journal of Nonlinear Science vol 22 no 2 pp235ndash245 2012
[9] A Constantin and W A Strauss ldquoStability of peakonsrdquo Com-munications on Pure and AppliedMathematics vol 53 no 5 pp603ndash610 2000
[10] Y Zhou ldquoStability of solitary waves for a rod equationrdquo ChaosSolitons and Fractals vol 21 no 4 pp 977ndash981 2004
[11] A A Himonas G Misiołek G Ponce and Y Zhou ldquoPersis-tence properties and unique continuation of solutions of theCamassa-Holm equationrdquo Communications in MathematicalPhysics vol 271 no 2 pp 511ndash522 2007
[12] A Degasperis and M Procesi ldquoAsymptotic integrabilityrdquo inSymmetry and PerturbationTheory ADegasperis andGGaetaEds pp 23ndash37 World Scientific Singapore 1999
[13] D D Holm and M F Staley ldquoNonlinear balance and exchangeof stability of dynamics of solitons peakons rampscliffs andleftons in a 1 + 1 nonlinear evolutionary PDErdquo Physics LettersA vol 308 no 5-6 pp 437ndash444 2003
[14] Y Zhou ldquoBlow-up phenomenon for the integrable Degasperis-Procesi equationrdquoPhysics Letters A vol 328 no 2-3 pp 157ndash1622004
[15] Y Zhou ldquoOn solutions to the Holm-Staley 119887-family of equa-tionsrdquo Nonlinearity vol 23 no 2 pp 369ndash381 2010
[16] Z Guo ldquoBlow up global existence and infinite propagationspeed for the weakly dissipative Camassa-Holm equationrdquoJournal ofMathematical Physics vol 49 no 3 Article ID 0335162008
[17] Z Guo ldquoSome properties of solutions to the weakly dissipativeDegasperis-Procesi equationrdquo Journal of Differential Equationsvol 246 no 11 pp 4332ndash4344 2009
[18] W Niu and S Zhang ldquoBlow-up phenomena and globalexistence for the nonuniform weakly dissipative 119887-equationrdquoJournal of Mathematical Analysis and Applications vol 374 no1 pp 166ndash177 2011
[19] M Zhu andZ Jiang ldquoSomeproperties of solutions to theweaklydissipative 119887-family equationrdquo Nonlinear Analysis Real WorldApplications vol 13 no 1 pp 158ndash167 2012
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Differential EquationsInternational Journal of
Volume 2014
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Journal of
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Abstract and Applied Analysis
le119887
2119906 (119906119909+ 119906) (119902 (119909
0 119905) 119905)
minus119887
4(1199062+ 1199062
119909+ 2119906119906
119909) (119902 (119909
0 119905) 119905)
minus120582
2(119906119909+ 119906) (119902 (119909
0 119905) 119905)
= minus119887
4(1199062
119909minus 1199062) (119902 (119909
0 119905) 119905)
minus120582
2(119906119909+ 119906) (119902 (119909
0 119905) 119905)
= minus119887
4(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
+119887
4(119906 (119902 (119909
0 119905) 119905) minus
120582
119887)
2
lt 0 on [0 1199050)
(22)
Therefore it follows from (21) (22) and the continuityproperty of ODEs that
(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
minus (119906 (119902 (1199090 119905) 119905) +
120582
119887)
2
= minus4119868 (119905) (119868119868 (119905) +120582
119887) gt minus4119868 (0) (119868119868 (0) +
120582
119887) gt 0
(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
minus (119906 (119902 (1199090 119905) 119905) minus
120582
119887)
2
= minus4(119868 (119905) minus120582
119887) 119868119868 (119905) gt minus4(119868 (0) minus
120582
119887) 119868119868 (0) gt 0
(23)
for all 119905 gt 0 This implies that 1199050can be extended to the
infinityMoreover using (21) and (22) again we have the following
equation for [2(119906119909+120582119887)
2minus(119906+120582119887)
2minus(119906 minus 120582119887)
2](119902(1199090 119905) 119905)
119889
119889119905[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
= minus4119889
119889119905[119868 (119905) (119868119868 (119905) +
120582
119887)] minus 4
119889
119889119905[(119868 (119905) minus
120582
119887) 119868119868 (119905)]
ge minus119887 [(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
] (119902 (1199090 119905) 119905) (119868119868 (119905) +
120582
119887)
+ 119887 [(119906119909+
120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905) 119868 (119905)
minus 119887 [(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
] (119902 (1199090 119905) 119905) 119868119868 (119905)
+ 119887 [(119906119909+
120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905) (119868 (119905) minus
120582
119887)
= 119887(minus120582
119887[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905) minus 119906
119909(119902 (1199090 119905) 119905)
times (2(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
2
)
+2(119906 +120582
119887)
2
119868119868 (119905) minus 2(119906 minus120582
119887)
2
119868 (119905))
= 119887(minus (119906119909(119902 (1199090 119905) 119905) +
120582
119887)
times [2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905) minus 119906
119909(119902 (1199090 119905) 119905)
times [(119906 +120582
119887)
2
+ (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
+2(119906 +120582
119887)
2
119868119868 (119905) minus 2(119906 minus120582
119887)
2
119868 (119905))
ge minus119887(119906119909(119902 (1199090 119905) 119905) +
120582
119887)
times [2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
(24)
where we use 119906119909(119902(1199090 119905) 119905) = minus119868(119905) + 119868119868(119905)
Now recalling (18) we have
120597119905119906119909(119902 (1199090 119905) 119905) (119902 (119909
0 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909
=119887
4[(119906 +
120582
119887)
2
+ (119906 minus120582
119887)
2
minus 2(119906119909+
120582
119887)
2
]
times (119902 (1199090 119905) 119905)
(25)
Putting (25) into (24) it yields
119889
119889119905[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
] (119902 (1199090 119905) 119905)
ge1198872
4[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 119905) 119905)
times (int
119905
0
[2(119906119909+
120582
119887)
2
minus (119906 +120582
119887)
2
minus (119906 minus120582
119887)
2
]
times (119902 (1199090 120591) 120591) 119889120591 minus 4119906
0119909(1199090) minus
4120582
119887)
(26)
Abstract and Applied Analysis 5
Before finishing the proof we need the following technicallemma
Lemma 3 (see [15]) Suppose that Ψ(119905) is twice continuouslydifferential satisfying
Ψ10158401015840(119905) ge 119862
0Ψ1015840(119905) Ψ (119905) 119905 gt 0 119862
0gt 0
Ψ (119905) gt 0 Ψ1015840(119905) gt 0
(27)
Then 120595(119905) blows up in finite time Moreover the blow-up timecan be estimated in terms of the initial datum as
119879 le max2
1198620Ψ (0)
Ψ (0)
Ψ1015840 (0) (28)
Let Ψ(119905) = int119905
0[2(119906119909
+ 120582119887)2
minus (119906 + 120582119887)2
minus (119906 minus
120582119887)2](119902(1199090 120591) 120591)119889120591 minus 4119906
0119909(1199090) minus 4120582119887 then (26) is an
equation of type (27) with 1198620
= 11988724 The proof is complete
by applying Lemma 3
Remark 4 When 119887 = 1 Theorem 2 reduces to the result in[19]
Theorem 5 Let 119886 = 2119887 gt 0 Suppose that 1199060
isin 1198672(R) and
there exists a 1199090isin R such that 119910
0(1199090) = (1 minus 120597
2
119909)1199060(1199090) = 0
119890minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt
2120582
119887 119890
1199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt minus
2120582
119887
(29)
Then the corresponding solution 119906(119909 119905) to (2) with 1199060as the
initial datum blows up in finite time
Proof We easily obtain
119906119905119909
+ 119887119906119906119909119909
minus119886
21199062+
119887
21199062
119909+ 119866 lowast (
119886
21199062+
119887
21199062
119909) + 120582119906
119909= 0
(30)
Differentiating 119906119909at the point (119902(119909
0 119905) 119905)with respect to 119905 we
get
119889
119889119905119906119909(119902 (1199090 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909(119902 (1199090 119905) 119905)
(31)
Process of the proof is similar to Theorem 2 Thus to beconcise we omit the detailed proof
When 119886 = 2119887 gt 0 120582 = 0 using (119909 119905) = 119887119906(119909 119905) (2) canbe reformulated into
119909+ 119909119909119905
+ 3119909minus 2119909119909119909
minus 119909119909119909
= 0 (32)
which is the well-known Camassa-Holm equation Mean-while we also find that the condition in Theorem 5 can bereformulated into
int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt 0 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt 0 (33)
which is one of the sufficient conditions to guarantee blow-upadd-on initial data for the Camassa-Holm equation
So we show the necessary and sufficient condition for thespecial case 119886 = 2119887 gt 0 and 120582 = 0 in the following theorem
Theorem 6 When 119886 = 2119887 gt 0 and 120582 = 0 then the nonlinearwave equation (2) breaks if and only if some portion of thepositive part of 119910
0(119909) lies to the left of some portion of its
negative part
Proof As studied in [1] when 119886 = 2119887 gt 0 and120582 = 0 rewriting(2) yields
119906119909+ 119906119909119909119905
+ 3119887119906119906119909minus 2119887119906
119909119906119909119909
minus 119887119906119906119909119909119909
= 0 (34)
RecallingMckeanrsquos theorem in [7] (32) breaks if and onlyif some portion of the positive part of 119910
0(119909 119905) = (1 minus 120597
2
119909)0
lies to the left of some portion of its negative partSo (34) breaks if and only if some portion of the positive
part of 1198871199100(119909 119905) = 119887(1 minus 120597
2
119909)1199060lies to the left of some portion
of its negative partThis completes the proof
Remark 7 Mckeanrsquos theorem [7] is for the special case 119886 = 2119887 = 1 Condition 119886 = 2119887 here is more general However thenecessary and sufficient condition for (2) is still a challengingproblem for us at present
4 Global Existence
Now let us try to find a condition for global existence Unfor-tunately When 119886 = 2119887 like the Degasperis-Procesi equation[12] only the following easy one can be proved at present
Theorem 8 Suppose that 1199060isin 1198673(R) and 119910
0= (1 minus 120597
2
119909)1199060is
one sign Then the corresponding solution to (2) exists globally
Proof We can assume that 1199100ge 0 It is sufficient to prove that
119906119909(119909 119905) has a lower and upper bound for all 119905 In fact
119906119909(119909 119905) = minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 +
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
(35)
Therefore we have
119906119909(119909 119905) ge minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 ge minus
1
2int
119909
minusinfin
119910 (120585 119905) 119889120585
ge minus1
2int
infin
minusinfin
119910 (120585 119905) 119889120585 = minus1
2int
infin
minusinfin
1199100(120585 119905) 119889120585
119906119909(119909 119905) le
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585 le1
2int
infin
119909
119910 (120585 119905) 119889120585
le1
2int
infin
minusinfin
119910 (120585 119905) 119889120585 =1
2int
infin
minusinfin
1199100(120585 119905) 119889120585
(36)
This completes the proof
6 Abstract and Applied Analysis
5 Infinite Propagation Speed
In this section we will give a more detailed description onthe corresponding strong solution 119906(119909 119905) to (2) in its life spanwith initial data 119906
0(119909) being compactly supported The main
theorem reads as follows
Theorem 9 Let 0 lt 119886 le 3119887 Assume that for some 119879 ge 0 and119904 ge 52 119906 isin 119862([0 119879)119867
119904(R)) is a strong solution of (2) If
1199060(119909) = 119906(119909 0) has compact support [ac] then for 119905 isin (0 119879)
one has
119906 (119909 119905) = 119871 (119905) 119890
minus119909 for 119909 gt 119902 (119888 119905)
119897 (119905) 119890minus119909
for 119909 lt 119902 (119886 119905) (37)
where 119871(119905) and 119897(119905) denote continuous nonvanishing functionswith 119871(119905) gt 0 and 119897(119905) lt 0 for 119905 isin (0 119879) Furthermore 119871(119905)is strictly increasing function while 119897(119905) is strictly decreasingfunction
Proof Since 1199060has compact support in 119909 in [119886 119888] from (8)
so does 119910( 119905) has compact support in 119909 in [119902(119886 119905) 119902(119888 119905)] inits lifespan Hence the following functions are well-defined
119864 (119905) = intR
119890119909119910 (119909 119905) 119889119909 119865 (119905) = int
R
119890minus119909
119910 (119909 119905) 119889119909 (38)
with
1198640= int
R
1198901199091199100(119909) 119889119909 = 0 119865
0= int
R
119890minus119909
1199100(119909) 119889119909 = 0
(39)
Thus for 119909 gt 119902(119888 119905) we obtain
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905)
=1
2119890minus119909
int
119902(119887119905)
119902(119886119905)
119890120591119910 (120591 119905) 119889120591 =
1
2119890minus119909
119864 (119905)
(40)
Similarly for 119909 lt 119902(119886 119905) we have
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905) =1
2119890119909int
119902(119887119905)
119902(119886119905)
119890minus120591
119910 (120591 119905) 119889120591
=1
2119890119909119865 (119905)
(41)
Hence as consequences of (40) and (41) we get
119906 (119909 119905) = minus119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890minus119909
119864 (119905)
as 119909 gt 119902 (119888 119905)
119906 (119909 119905) = 119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890119909119865 (119905)
as 119909 lt 119902 (119886 119905)
(42)
On the other hand
119889119864 (119905)
119889119905= int
R
119890119909119910119905(119909 119905) 119889119909 (43)
It is easy to get
119910119905= minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909)
minus 120582119906 + 120582119906119909119909
(44)
Putting the identity (44) into 119889119864(119905)119889119905 we have
119889119864 (119905)
119889119905
= intR
119890119909(minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909))119889119909
+ intR
119890119909(minus120582119906 + 120582119906
119909119909) 119889119909
= intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909)119889119909
(45)
where we have used (42)Therefore in the lifespan of the solution we get
119864 (119905) = int
119905
0
intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 gt 0 (46)
By the same argument one can check that the followingidentity for 119865(119905) is true
119865 (119905) = minusint
119905
0
intR
119890minus119909
(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 lt 0
(47)
In order to complete the proof it is sufficient to let 119871(119905) =
(12)119864(119905) and 119897(119905) = (12)119865(119905) respectively
Acknowledgments
This work is partially supported by Zhejiang InnovationProject (T200905) ZJNSF (Grant no R6090109) and NSFC(Grant no 10971197 and 11101376)
References
[1] Z Jiang and S Hakkaev ldquoWave breaking and propagationspeed for a class of one-dimensional shallow water equationsrdquoAbstract and Applied Analysis vol 2011 Article ID 647368 15pages 2011
[2] L Ni and Y Zhou ldquoWave breaking and propagation speed fora class of nonlocal dispersive 120579-equationsrdquo Nonlinear AnalysisReal World Applications vol 12 no 1 pp 592ndash600 2011
[3] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[4] B Fuchssteiner and A S Fokas ldquoSymplectic structures theirBacklund transformations and hereditary symmetriesrdquo PhysicaD vol 4 no 1 pp 47ndash66 1981-1982
[5] A Constantin and J Escher ldquoWell-posedness global existenceand blowup phenomena for a periodic quasi-linear hyperbolicequationrdquo Communications on Pure and Applied Mathematicsvol 51 no 5 pp 475ndash504 1998
Abstract and Applied Analysis 7
[6] Y A Li and P J Olver ldquoWell-posedness and blow-up solutionsfor an integrable nonlinearly dispersive model wave equationrdquoJournal of Differential Equations vol 162 no 1 pp 27ndash63 2000
[7] H P McKean ldquoBreakdown of a shallow water equationrdquo TheAsian Journal of Mathematics vol 2 no 4 pp 867ndash874 1998
[8] Z Jiang L Ni and Y Zhou ldquoWave breaking of the Camassa-Holm equationrdquo Journal of Nonlinear Science vol 22 no 2 pp235ndash245 2012
[9] A Constantin and W A Strauss ldquoStability of peakonsrdquo Com-munications on Pure and AppliedMathematics vol 53 no 5 pp603ndash610 2000
[10] Y Zhou ldquoStability of solitary waves for a rod equationrdquo ChaosSolitons and Fractals vol 21 no 4 pp 977ndash981 2004
[11] A A Himonas G Misiołek G Ponce and Y Zhou ldquoPersis-tence properties and unique continuation of solutions of theCamassa-Holm equationrdquo Communications in MathematicalPhysics vol 271 no 2 pp 511ndash522 2007
[12] A Degasperis and M Procesi ldquoAsymptotic integrabilityrdquo inSymmetry and PerturbationTheory ADegasperis andGGaetaEds pp 23ndash37 World Scientific Singapore 1999
[13] D D Holm and M F Staley ldquoNonlinear balance and exchangeof stability of dynamics of solitons peakons rampscliffs andleftons in a 1 + 1 nonlinear evolutionary PDErdquo Physics LettersA vol 308 no 5-6 pp 437ndash444 2003
[14] Y Zhou ldquoBlow-up phenomenon for the integrable Degasperis-Procesi equationrdquoPhysics Letters A vol 328 no 2-3 pp 157ndash1622004
[15] Y Zhou ldquoOn solutions to the Holm-Staley 119887-family of equa-tionsrdquo Nonlinearity vol 23 no 2 pp 369ndash381 2010
[16] Z Guo ldquoBlow up global existence and infinite propagationspeed for the weakly dissipative Camassa-Holm equationrdquoJournal ofMathematical Physics vol 49 no 3 Article ID 0335162008
[17] Z Guo ldquoSome properties of solutions to the weakly dissipativeDegasperis-Procesi equationrdquo Journal of Differential Equationsvol 246 no 11 pp 4332ndash4344 2009
[18] W Niu and S Zhang ldquoBlow-up phenomena and globalexistence for the nonuniform weakly dissipative 119887-equationrdquoJournal of Mathematical Analysis and Applications vol 374 no1 pp 166ndash177 2011
[19] M Zhu andZ Jiang ldquoSomeproperties of solutions to theweaklydissipative 119887-family equationrdquo Nonlinear Analysis Real WorldApplications vol 13 no 1 pp 158ndash167 2012
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 5
Before finishing the proof we need the following technicallemma
Lemma 3 (see [15]) Suppose that Ψ(119905) is twice continuouslydifferential satisfying
Ψ10158401015840(119905) ge 119862
0Ψ1015840(119905) Ψ (119905) 119905 gt 0 119862
0gt 0
Ψ (119905) gt 0 Ψ1015840(119905) gt 0
(27)
Then 120595(119905) blows up in finite time Moreover the blow-up timecan be estimated in terms of the initial datum as
119879 le max2
1198620Ψ (0)
Ψ (0)
Ψ1015840 (0) (28)
Let Ψ(119905) = int119905
0[2(119906119909
+ 120582119887)2
minus (119906 + 120582119887)2
minus (119906 minus
120582119887)2](119902(1199090 120591) 120591)119889120591 minus 4119906
0119909(1199090) minus 4120582119887 then (26) is an
equation of type (27) with 1198620
= 11988724 The proof is complete
by applying Lemma 3
Remark 4 When 119887 = 1 Theorem 2 reduces to the result in[19]
Theorem 5 Let 119886 = 2119887 gt 0 Suppose that 1199060
isin 1198672(R) and
there exists a 1199090isin R such that 119910
0(1199090) = (1 minus 120597
2
119909)1199060(1199090) = 0
119890minus1199090 int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt
2120582
119887 119890
1199090 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt minus
2120582
119887
(29)
Then the corresponding solution 119906(119909 119905) to (2) with 1199060as the
initial datum blows up in finite time
Proof We easily obtain
119906119905119909
+ 119887119906119906119909119909
minus119886
21199062+
119887
21199062
119909+ 119866 lowast (
119886
21199062+
119887
21199062
119909) + 120582119906
119909= 0
(30)
Differentiating 119906119909at the point (119902(119909
0 119905) 119905)with respect to 119905 we
get
119889
119889119905119906119909(119902 (1199090 119905) 119905)
le119887
21199062(119902 (1199090 119905) 119905) minus
119887
21199062
119909(119902 (1199090 119905) 119905) minus 120582119906
119909(119902 (1199090 119905) 119905)
(31)
Process of the proof is similar to Theorem 2 Thus to beconcise we omit the detailed proof
When 119886 = 2119887 gt 0 120582 = 0 using (119909 119905) = 119887119906(119909 119905) (2) canbe reformulated into
119909+ 119909119909119905
+ 3119909minus 2119909119909119909
minus 119909119909119909
= 0 (32)
which is the well-known Camassa-Holm equation Mean-while we also find that the condition in Theorem 5 can bereformulated into
int
1199090
minusinfin
1198901205851199100(120585) 119889120585 gt 0 int
infin
1199090
119890minus120585
1199100(120585) 119889120585 lt 0 (33)
which is one of the sufficient conditions to guarantee blow-upadd-on initial data for the Camassa-Holm equation
So we show the necessary and sufficient condition for thespecial case 119886 = 2119887 gt 0 and 120582 = 0 in the following theorem
Theorem 6 When 119886 = 2119887 gt 0 and 120582 = 0 then the nonlinearwave equation (2) breaks if and only if some portion of thepositive part of 119910
0(119909) lies to the left of some portion of its
negative part
Proof As studied in [1] when 119886 = 2119887 gt 0 and120582 = 0 rewriting(2) yields
119906119909+ 119906119909119909119905
+ 3119887119906119906119909minus 2119887119906
119909119906119909119909
minus 119887119906119906119909119909119909
= 0 (34)
RecallingMckeanrsquos theorem in [7] (32) breaks if and onlyif some portion of the positive part of 119910
0(119909 119905) = (1 minus 120597
2
119909)0
lies to the left of some portion of its negative partSo (34) breaks if and only if some portion of the positive
part of 1198871199100(119909 119905) = 119887(1 minus 120597
2
119909)1199060lies to the left of some portion
of its negative partThis completes the proof
Remark 7 Mckeanrsquos theorem [7] is for the special case 119886 = 2119887 = 1 Condition 119886 = 2119887 here is more general However thenecessary and sufficient condition for (2) is still a challengingproblem for us at present
4 Global Existence
Now let us try to find a condition for global existence Unfor-tunately When 119886 = 2119887 like the Degasperis-Procesi equation[12] only the following easy one can be proved at present
Theorem 8 Suppose that 1199060isin 1198673(R) and 119910
0= (1 minus 120597
2
119909)1199060is
one sign Then the corresponding solution to (2) exists globally
Proof We can assume that 1199100ge 0 It is sufficient to prove that
119906119909(119909 119905) has a lower and upper bound for all 119905 In fact
119906119909(119909 119905) = minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 +
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585
(35)
Therefore we have
119906119909(119909 119905) ge minus
1
2119890minus119909
int
119909
minusinfin
119890120585119910 (120585 119905) 119889120585 ge minus
1
2int
119909
minusinfin
119910 (120585 119905) 119889120585
ge minus1
2int
infin
minusinfin
119910 (120585 119905) 119889120585 = minus1
2int
infin
minusinfin
1199100(120585 119905) 119889120585
119906119909(119909 119905) le
1
2119890119909int
infin
119909
119890minus120585
119910 (120585 119905) 119889120585 le1
2int
infin
119909
119910 (120585 119905) 119889120585
le1
2int
infin
minusinfin
119910 (120585 119905) 119889120585 =1
2int
infin
minusinfin
1199100(120585 119905) 119889120585
(36)
This completes the proof
6 Abstract and Applied Analysis
5 Infinite Propagation Speed
In this section we will give a more detailed description onthe corresponding strong solution 119906(119909 119905) to (2) in its life spanwith initial data 119906
0(119909) being compactly supported The main
theorem reads as follows
Theorem 9 Let 0 lt 119886 le 3119887 Assume that for some 119879 ge 0 and119904 ge 52 119906 isin 119862([0 119879)119867
119904(R)) is a strong solution of (2) If
1199060(119909) = 119906(119909 0) has compact support [ac] then for 119905 isin (0 119879)
one has
119906 (119909 119905) = 119871 (119905) 119890
minus119909 for 119909 gt 119902 (119888 119905)
119897 (119905) 119890minus119909
for 119909 lt 119902 (119886 119905) (37)
where 119871(119905) and 119897(119905) denote continuous nonvanishing functionswith 119871(119905) gt 0 and 119897(119905) lt 0 for 119905 isin (0 119879) Furthermore 119871(119905)is strictly increasing function while 119897(119905) is strictly decreasingfunction
Proof Since 1199060has compact support in 119909 in [119886 119888] from (8)
so does 119910( 119905) has compact support in 119909 in [119902(119886 119905) 119902(119888 119905)] inits lifespan Hence the following functions are well-defined
119864 (119905) = intR
119890119909119910 (119909 119905) 119889119909 119865 (119905) = int
R
119890minus119909
119910 (119909 119905) 119889119909 (38)
with
1198640= int
R
1198901199091199100(119909) 119889119909 = 0 119865
0= int
R
119890minus119909
1199100(119909) 119889119909 = 0
(39)
Thus for 119909 gt 119902(119888 119905) we obtain
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905)
=1
2119890minus119909
int
119902(119887119905)
119902(119886119905)
119890120591119910 (120591 119905) 119889120591 =
1
2119890minus119909
119864 (119905)
(40)
Similarly for 119909 lt 119902(119886 119905) we have
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905) =1
2119890119909int
119902(119887119905)
119902(119886119905)
119890minus120591
119910 (120591 119905) 119889120591
=1
2119890119909119865 (119905)
(41)
Hence as consequences of (40) and (41) we get
119906 (119909 119905) = minus119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890minus119909
119864 (119905)
as 119909 gt 119902 (119888 119905)
119906 (119909 119905) = 119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890119909119865 (119905)
as 119909 lt 119902 (119886 119905)
(42)
On the other hand
119889119864 (119905)
119889119905= int
R
119890119909119910119905(119909 119905) 119889119909 (43)
It is easy to get
119910119905= minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909)
minus 120582119906 + 120582119906119909119909
(44)
Putting the identity (44) into 119889119864(119905)119889119905 we have
119889119864 (119905)
119889119905
= intR
119890119909(minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909))119889119909
+ intR
119890119909(minus120582119906 + 120582119906
119909119909) 119889119909
= intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909)119889119909
(45)
where we have used (42)Therefore in the lifespan of the solution we get
119864 (119905) = int
119905
0
intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 gt 0 (46)
By the same argument one can check that the followingidentity for 119865(119905) is true
119865 (119905) = minusint
119905
0
intR
119890minus119909
(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 lt 0
(47)
In order to complete the proof it is sufficient to let 119871(119905) =
(12)119864(119905) and 119897(119905) = (12)119865(119905) respectively
Acknowledgments
This work is partially supported by Zhejiang InnovationProject (T200905) ZJNSF (Grant no R6090109) and NSFC(Grant no 10971197 and 11101376)
References
[1] Z Jiang and S Hakkaev ldquoWave breaking and propagationspeed for a class of one-dimensional shallow water equationsrdquoAbstract and Applied Analysis vol 2011 Article ID 647368 15pages 2011
[2] L Ni and Y Zhou ldquoWave breaking and propagation speed fora class of nonlocal dispersive 120579-equationsrdquo Nonlinear AnalysisReal World Applications vol 12 no 1 pp 592ndash600 2011
[3] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[4] B Fuchssteiner and A S Fokas ldquoSymplectic structures theirBacklund transformations and hereditary symmetriesrdquo PhysicaD vol 4 no 1 pp 47ndash66 1981-1982
[5] A Constantin and J Escher ldquoWell-posedness global existenceand blowup phenomena for a periodic quasi-linear hyperbolicequationrdquo Communications on Pure and Applied Mathematicsvol 51 no 5 pp 475ndash504 1998
Abstract and Applied Analysis 7
[6] Y A Li and P J Olver ldquoWell-posedness and blow-up solutionsfor an integrable nonlinearly dispersive model wave equationrdquoJournal of Differential Equations vol 162 no 1 pp 27ndash63 2000
[7] H P McKean ldquoBreakdown of a shallow water equationrdquo TheAsian Journal of Mathematics vol 2 no 4 pp 867ndash874 1998
[8] Z Jiang L Ni and Y Zhou ldquoWave breaking of the Camassa-Holm equationrdquo Journal of Nonlinear Science vol 22 no 2 pp235ndash245 2012
[9] A Constantin and W A Strauss ldquoStability of peakonsrdquo Com-munications on Pure and AppliedMathematics vol 53 no 5 pp603ndash610 2000
[10] Y Zhou ldquoStability of solitary waves for a rod equationrdquo ChaosSolitons and Fractals vol 21 no 4 pp 977ndash981 2004
[11] A A Himonas G Misiołek G Ponce and Y Zhou ldquoPersis-tence properties and unique continuation of solutions of theCamassa-Holm equationrdquo Communications in MathematicalPhysics vol 271 no 2 pp 511ndash522 2007
[12] A Degasperis and M Procesi ldquoAsymptotic integrabilityrdquo inSymmetry and PerturbationTheory ADegasperis andGGaetaEds pp 23ndash37 World Scientific Singapore 1999
[13] D D Holm and M F Staley ldquoNonlinear balance and exchangeof stability of dynamics of solitons peakons rampscliffs andleftons in a 1 + 1 nonlinear evolutionary PDErdquo Physics LettersA vol 308 no 5-6 pp 437ndash444 2003
[14] Y Zhou ldquoBlow-up phenomenon for the integrable Degasperis-Procesi equationrdquoPhysics Letters A vol 328 no 2-3 pp 157ndash1622004
[15] Y Zhou ldquoOn solutions to the Holm-Staley 119887-family of equa-tionsrdquo Nonlinearity vol 23 no 2 pp 369ndash381 2010
[16] Z Guo ldquoBlow up global existence and infinite propagationspeed for the weakly dissipative Camassa-Holm equationrdquoJournal ofMathematical Physics vol 49 no 3 Article ID 0335162008
[17] Z Guo ldquoSome properties of solutions to the weakly dissipativeDegasperis-Procesi equationrdquo Journal of Differential Equationsvol 246 no 11 pp 4332ndash4344 2009
[18] W Niu and S Zhang ldquoBlow-up phenomena and globalexistence for the nonuniform weakly dissipative 119887-equationrdquoJournal of Mathematical Analysis and Applications vol 374 no1 pp 166ndash177 2011
[19] M Zhu andZ Jiang ldquoSomeproperties of solutions to theweaklydissipative 119887-family equationrdquo Nonlinear Analysis Real WorldApplications vol 13 no 1 pp 158ndash167 2012
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Abstract and Applied Analysis
5 Infinite Propagation Speed
In this section we will give a more detailed description onthe corresponding strong solution 119906(119909 119905) to (2) in its life spanwith initial data 119906
0(119909) being compactly supported The main
theorem reads as follows
Theorem 9 Let 0 lt 119886 le 3119887 Assume that for some 119879 ge 0 and119904 ge 52 119906 isin 119862([0 119879)119867
119904(R)) is a strong solution of (2) If
1199060(119909) = 119906(119909 0) has compact support [ac] then for 119905 isin (0 119879)
one has
119906 (119909 119905) = 119871 (119905) 119890
minus119909 for 119909 gt 119902 (119888 119905)
119897 (119905) 119890minus119909
for 119909 lt 119902 (119886 119905) (37)
where 119871(119905) and 119897(119905) denote continuous nonvanishing functionswith 119871(119905) gt 0 and 119897(119905) lt 0 for 119905 isin (0 119879) Furthermore 119871(119905)is strictly increasing function while 119897(119905) is strictly decreasingfunction
Proof Since 1199060has compact support in 119909 in [119886 119888] from (8)
so does 119910( 119905) has compact support in 119909 in [119902(119886 119905) 119902(119888 119905)] inits lifespan Hence the following functions are well-defined
119864 (119905) = intR
119890119909119910 (119909 119905) 119889119909 119865 (119905) = int
R
119890minus119909
119910 (119909 119905) 119889119909 (38)
with
1198640= int
R
1198901199091199100(119909) 119889119909 = 0 119865
0= int
R
119890minus119909
1199100(119909) 119889119909 = 0
(39)
Thus for 119909 gt 119902(119888 119905) we obtain
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905)
=1
2119890minus119909
int
119902(119887119905)
119902(119886119905)
119890120591119910 (120591 119905) 119889120591 =
1
2119890minus119909
119864 (119905)
(40)
Similarly for 119909 lt 119902(119886 119905) we have
119906 (119909 119905) =1
2119890minus|119909|
lowast 119910 (119909 119905) =1
2119890119909int
119902(119887119905)
119902(119886119905)
119890minus120591
119910 (120591 119905) 119889120591
=1
2119890119909119865 (119905)
(41)
Hence as consequences of (40) and (41) we get
119906 (119909 119905) = minus119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890minus119909
119864 (119905)
as 119909 gt 119902 (119888 119905)
119906 (119909 119905) = 119906119909(119909 119905) = 119906
119909119909(119909 119905) =
1
2119890119909119865 (119905)
as 119909 lt 119902 (119886 119905)
(42)
On the other hand
119889119864 (119905)
119889119905= int
R
119890119909119910119905(119909 119905) 119889119909 (43)
It is easy to get
119910119905= minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909)
minus 120582119906 + 120582119906119909119909
(44)
Putting the identity (44) into 119889119864(119905)119889119905 we have
119889119864 (119905)
119889119905
= intR
119890119909(minus119887119906119906
119909+ 119887(119906119906
119909)119909119909
minus 120597119909(119886
21199062+
3119887 minus 119886
21199062
119909))119889119909
+ intR
119890119909(minus120582119906 + 120582119906
119909119909) 119889119909
= intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909)119889119909
(45)
where we have used (42)Therefore in the lifespan of the solution we get
119864 (119905) = int
119905
0
intR
119890119909(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 gt 0 (46)
By the same argument one can check that the followingidentity for 119865(119905) is true
119865 (119905) = minusint
119905
0
intR
119890minus119909
(119886
21199062+
3119887 minus 119886
21199062
119909) (119909 120591) 119889119909 119889120591 lt 0
(47)
In order to complete the proof it is sufficient to let 119871(119905) =
(12)119864(119905) and 119897(119905) = (12)119865(119905) respectively
Acknowledgments
This work is partially supported by Zhejiang InnovationProject (T200905) ZJNSF (Grant no R6090109) and NSFC(Grant no 10971197 and 11101376)
References
[1] Z Jiang and S Hakkaev ldquoWave breaking and propagationspeed for a class of one-dimensional shallow water equationsrdquoAbstract and Applied Analysis vol 2011 Article ID 647368 15pages 2011
[2] L Ni and Y Zhou ldquoWave breaking and propagation speed fora class of nonlocal dispersive 120579-equationsrdquo Nonlinear AnalysisReal World Applications vol 12 no 1 pp 592ndash600 2011
[3] R Camassa and D D Holm ldquoAn integrable shallow waterequation with peaked solitonsrdquo Physical Review Letters vol 71no 11 pp 1661ndash1664 1993
[4] B Fuchssteiner and A S Fokas ldquoSymplectic structures theirBacklund transformations and hereditary symmetriesrdquo PhysicaD vol 4 no 1 pp 47ndash66 1981-1982
[5] A Constantin and J Escher ldquoWell-posedness global existenceand blowup phenomena for a periodic quasi-linear hyperbolicequationrdquo Communications on Pure and Applied Mathematicsvol 51 no 5 pp 475ndash504 1998
Abstract and Applied Analysis 7
[6] Y A Li and P J Olver ldquoWell-posedness and blow-up solutionsfor an integrable nonlinearly dispersive model wave equationrdquoJournal of Differential Equations vol 162 no 1 pp 27ndash63 2000
[7] H P McKean ldquoBreakdown of a shallow water equationrdquo TheAsian Journal of Mathematics vol 2 no 4 pp 867ndash874 1998
[8] Z Jiang L Ni and Y Zhou ldquoWave breaking of the Camassa-Holm equationrdquo Journal of Nonlinear Science vol 22 no 2 pp235ndash245 2012
[9] A Constantin and W A Strauss ldquoStability of peakonsrdquo Com-munications on Pure and AppliedMathematics vol 53 no 5 pp603ndash610 2000
[10] Y Zhou ldquoStability of solitary waves for a rod equationrdquo ChaosSolitons and Fractals vol 21 no 4 pp 977ndash981 2004
[11] A A Himonas G Misiołek G Ponce and Y Zhou ldquoPersis-tence properties and unique continuation of solutions of theCamassa-Holm equationrdquo Communications in MathematicalPhysics vol 271 no 2 pp 511ndash522 2007
[12] A Degasperis and M Procesi ldquoAsymptotic integrabilityrdquo inSymmetry and PerturbationTheory ADegasperis andGGaetaEds pp 23ndash37 World Scientific Singapore 1999
[13] D D Holm and M F Staley ldquoNonlinear balance and exchangeof stability of dynamics of solitons peakons rampscliffs andleftons in a 1 + 1 nonlinear evolutionary PDErdquo Physics LettersA vol 308 no 5-6 pp 437ndash444 2003
[14] Y Zhou ldquoBlow-up phenomenon for the integrable Degasperis-Procesi equationrdquoPhysics Letters A vol 328 no 2-3 pp 157ndash1622004
[15] Y Zhou ldquoOn solutions to the Holm-Staley 119887-family of equa-tionsrdquo Nonlinearity vol 23 no 2 pp 369ndash381 2010
[16] Z Guo ldquoBlow up global existence and infinite propagationspeed for the weakly dissipative Camassa-Holm equationrdquoJournal ofMathematical Physics vol 49 no 3 Article ID 0335162008
[17] Z Guo ldquoSome properties of solutions to the weakly dissipativeDegasperis-Procesi equationrdquo Journal of Differential Equationsvol 246 no 11 pp 4332ndash4344 2009
[18] W Niu and S Zhang ldquoBlow-up phenomena and globalexistence for the nonuniform weakly dissipative 119887-equationrdquoJournal of Mathematical Analysis and Applications vol 374 no1 pp 166ndash177 2011
[19] M Zhu andZ Jiang ldquoSomeproperties of solutions to theweaklydissipative 119887-family equationrdquo Nonlinear Analysis Real WorldApplications vol 13 no 1 pp 158ndash167 2012
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Abstract and Applied Analysis 7
[6] Y A Li and P J Olver ldquoWell-posedness and blow-up solutionsfor an integrable nonlinearly dispersive model wave equationrdquoJournal of Differential Equations vol 162 no 1 pp 27ndash63 2000
[7] H P McKean ldquoBreakdown of a shallow water equationrdquo TheAsian Journal of Mathematics vol 2 no 4 pp 867ndash874 1998
[8] Z Jiang L Ni and Y Zhou ldquoWave breaking of the Camassa-Holm equationrdquo Journal of Nonlinear Science vol 22 no 2 pp235ndash245 2012
[9] A Constantin and W A Strauss ldquoStability of peakonsrdquo Com-munications on Pure and AppliedMathematics vol 53 no 5 pp603ndash610 2000
[10] Y Zhou ldquoStability of solitary waves for a rod equationrdquo ChaosSolitons and Fractals vol 21 no 4 pp 977ndash981 2004
[11] A A Himonas G Misiołek G Ponce and Y Zhou ldquoPersis-tence properties and unique continuation of solutions of theCamassa-Holm equationrdquo Communications in MathematicalPhysics vol 271 no 2 pp 511ndash522 2007
[12] A Degasperis and M Procesi ldquoAsymptotic integrabilityrdquo inSymmetry and PerturbationTheory ADegasperis andGGaetaEds pp 23ndash37 World Scientific Singapore 1999
[13] D D Holm and M F Staley ldquoNonlinear balance and exchangeof stability of dynamics of solitons peakons rampscliffs andleftons in a 1 + 1 nonlinear evolutionary PDErdquo Physics LettersA vol 308 no 5-6 pp 437ndash444 2003
[14] Y Zhou ldquoBlow-up phenomenon for the integrable Degasperis-Procesi equationrdquoPhysics Letters A vol 328 no 2-3 pp 157ndash1622004
[15] Y Zhou ldquoOn solutions to the Holm-Staley 119887-family of equa-tionsrdquo Nonlinearity vol 23 no 2 pp 369ndash381 2010
[16] Z Guo ldquoBlow up global existence and infinite propagationspeed for the weakly dissipative Camassa-Holm equationrdquoJournal ofMathematical Physics vol 49 no 3 Article ID 0335162008
[17] Z Guo ldquoSome properties of solutions to the weakly dissipativeDegasperis-Procesi equationrdquo Journal of Differential Equationsvol 246 no 11 pp 4332ndash4344 2009
[18] W Niu and S Zhang ldquoBlow-up phenomena and globalexistence for the nonuniform weakly dissipative 119887-equationrdquoJournal of Mathematical Analysis and Applications vol 374 no1 pp 166ndash177 2011
[19] M Zhu andZ Jiang ldquoSomeproperties of solutions to theweaklydissipative 119887-family equationrdquo Nonlinear Analysis Real WorldApplications vol 13 no 1 pp 158ndash167 2012
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of