# The Uniqueness of Strong Solutions for the Camassa-Holm Equation

Authors: Meng Wu; Chong Lai
Journal: Journal of Function Spaces and Applications
 (2013)

---

## Abstract

Assume that there exists a strong solution of the Camassa-Holm equation and the initial value of the solution belongs to the Sobolev spaceH1(R). We provide a new proof of the uniqueness of the strong solution
for the equation.


---

## Body

## 1. Introduction

The integrable Camassa-Holm model [1]
(1)ut-utxx+3uux=2uxuxx+uuxxx
has been investigated by many scholars. Equation (1) has peaked solitary wave solutions, which takes the form ce-|x-ct|,c∈R. The existence and uniqueness of the global weak solutions for (1) have been given by Constantin and Escher [2] and Constantin and Molinet [3] in which the m=u0-u0xx is a positive (or negative) Radon measure. The local well-posedness of strong solutions for the Camassa-Holm model and its various generalized forms are provided in [4–8]. For the initial value u0 satisfying u0-u0xx≥0 or u0-u0xx≤0, it is shown in [9] that the Camassa-Holm equation has unique global strong solutions in the Sobolev space Hs(R) with s>3/2. If the initial data satisfy certain conditions, we know that the local strong solutions blow up in finite time [10, 11]. It means that the slope of the solution becomes unbounded while the solution itself remains bounded. For other techniques to obtain the dynamic properties for the Camassa-Holm equation and other related shallow water equations, the reader is referred to [12–16] and the references therein.We consider the equivalent form of the Cauchy problem for (1)
(2)ut+12(u2)x+∂xPu(t,x)=0,u(0,x)=u0,
where Pu(t,x)=Λ-2(u2+(1/2)ux2)=(1/2)∫-∞∞‍e|x-y|(u2(t,y)+(1/2)uy2(t,y))dy. If (1) has a suitable smooth strong solution, we have the conservation law
(3)∫-∞∞‍(u2(t,x)+ux2(t,x))dx00=∫-∞∞‍(u2(0,x)+ux2(0,x))dx,
which derives
(4)∥u∥L∞(R)≤∥u∥H1(R)=∥u0∥H1(R).The objective of this work is to give a new proof of the uniqueness for the solutions of the Camassa-Holm equation (1). Firstly, we establish the following inequality:
(5)∥u(t,·)-v(t,·)∥L1(R)00≤cect(∫-∞∞‍|u0(x)-v0(x)|dx+∫-∞∞‍|u0x2-v0x2|dx),
where t∈[0,T0), functions u and v are two local or global strong solutions of problem (2) with initial data u(0,·)=u0∈H1(R) and v(0,·)=v0∈H1(R), respectively. Constant c depends on ∥u0∥H1(R), ∥v0∥H1(R), and the maximum existence time T0. Secondly, from (5), we immediately arrive at the goal of the uniqueness. Here we state that the approach to establish (5) is the device of doubling variables which was presented in Kruzkov's paper [17].This paper is organized as follows. Several lemmas are given in Section2, while the proofs of the main results are established in Section 3.


## 2. Notations and Several Lemmas

SetξT=[0,T]×R for an arbitrary T>0. The space of all infinitely differentiable functions f(t,x) with compact support in [0,T]×R is denoted by C0∞(ξT). We define ρ(σ) as a function which is infinitely differentiable on (-∞,+∞) such that ρ(σ)≥0, ρ(σ)=0 for |σ|≥1 and ∫-∞∞‍ρ(σ)dσ=1. For any number h>0, we let ρh(σ)=ρ(h-1σ)/h. Then we have that ρh(σ) is a function in C∞(-∞,∞) and
(6)ρh(σ)≥0,ρh(σ)=0if|σ|≥h,|ρh(σ)|≤ch,∫-∞∞‍ρh(σ)=1.
Assume that the function v(x) is locally integrable in (-∞,∞). We define the approximation of function v(x) as
(7)vh(x)=1h∫-∞∞‍ρ(x-yh)v(y)dy,h>0.We callx0 a Lebesgue point of function v(x) if
(8)limh→01h∫|x-x0|≤h|v(x)-v(x0)|dx=0.
At any Lebesgue point x0 of the function v(x), we have limh→0vh(x0)=v(x0). Since the set of points which are not Lebesgue points of v(x) has measured zero, we get vh(x)→v(x) as h→0 almost everywhere.We introduce notation connected with the concept of a characteristic cone. For anyM0>0, we define N>supt∈[0,∞)∥u∥L∞(R)<∞. Let ℧ designate the cone {(t,x):|x|≤M0-Nt,0<t<T0=min(T,M0N-1)}. We let Sτ designate the cross section of the cone ℧ by the plane t=τ,τ∈[0,T0].LetHr={x:|x|≤r}, where r>0.Lemma 1 (see [17]).
Let the functionv(t,x) be bounded and measurable in cylinder Ω=[0,T]×Hr. If δ∈(0,min[r,T]) and any number h∈(0,δ), then the function
(9)Vh=1h2∭∫|(t-τ)/2|≤h,δ≤(t+τ)/2≤T-δ,|(x-y)/2|≤h,|(x+y)/2|≤r-δ|v(t,x)-v(τ,y)|dxdtdydτ
satisfies limh→0Vh=0.Lemma 2 (see [17]).
If the function|∂F(u)/∂u| is bounded, then the function H(u,v)=sign(u-v)(F(u)-F(v)) satisfies the Lipschitz condition in u and v, respectively.Lemma 3.
Letu0∈H1(R). It holds that the function Qu(t,x)=∂xPu(t,x)=∂xΛ-2(u2+(1/2)ux2) satisfies
(10)∥Pu(t,x)∥L∞(R)<∞,∥Pu(t,x)∥L1(R)<∞,∥Pu(t,x)∥L2(R)<∞,∥Qu(t,x)∥L∞(R)<∞,∥Qu(t,x)∥L1(R)<∞,∥Qu(t,x)∥L2(R)<∞.The proof of Lemma3 can be found in [13, 15] (see [13, Lemma 5.1]).Lemma 4.
Letu be the strong solution of problem (2), f(t,x)∈C0∞(ξT), and f(0,x)=0. Then
(11)∬ξT{|u-k|ft+sign(u-k)12[u2-k2]fx00000-sign(u-k)Qu(t,x)f12[u2-k2]}dxdt=0,
where k is an arbitrary constant.Proof.
LetΦ(u) be an arbitrary twice smooth function on the line -∞<u<∞. We multiply (2) by the function Φ′(u)f(t,x), where f(t,x)∈C0∞(ξT). Integrating over ξT and transferring the derivatives with respect to t and x to the test function f, for any constant k, we obtain
(12)∬ξT{Φ(u)ft+[∫ku‍Φ′(z)zdz]fx00000-Φ′(u)Qu(t,x)f[∫ku‍Φ′(z)zdz]}dxdt=0,
in which we have used ∫-∞∞‍[∫ku‍Φ′(z)zdz]fxdx=-∫-∞∞‍[fΦ′(u)uux]dx.
We have(13)∫-∞∞[∫ku‍Φ′(z)zdz]fxdx00=∫-∞∞[∫ku(z2-k2)Φ′′(z)dzΦ′(u)(12u2-12k2)000000000-12∫ku(z2-k2)Φ′′(z)dz]fxdx.
Let Φh(u) be an approximation of the function |u-k| and set Φ(u)=Φh(u). Using the properties of the sign(u-k), (12), and (13) and sending h→0, we have
(14)∬ξT{|u-k|ft+sign(u-k)12[u2-k2]fx00000-sign(u-k)Qu(t,x)f12}dxdt=0,
which completes the proof.In fact, the proof of (11) can also be found in [17].Lemma 5.
Assumeu and v are two strong solutions of problem (2). It has
(15)|∫-∞∞‍sign(u-v)[Qu(t,x)-Qv(t,x)]fdx|00≤c∫-∞∞‍(|u02-v02|+|u0x2-v0x2|+|u-v|)dx.Proof.
We have(16)∫-∞∞‍(ux2-vx2)dx=∫-∞∞‍[(u02+u0x2)-(v02+v0x2)]dx000000000000000000-∫-∞∞‍(u2-v2)dx,|∫-∞∞‍Λ-2(u2+12ux2-v2-12vx2)0000×∂x[sign[u(t,x)-v(t,x)]f(t,x)]dx∫-∞∞‍|00=|12∬-∞∞e-|x-y|(u2+12uy2-v2-12vy2)dy0000×∂x[sign[u(t,x)-v(t,x)]f(t,x)]dx12∫-∞∞‍|00≤12∫-∞∞|∫-∞∞(u2+12uy2-v2-12vy2)dy|00000×e-|x-y||∂x[sign[u(t,x)-v(t,x)]f(t,x)]|dx00≤12|∫-∞∞(u2+12uy2-v2-12vy2)dy|0000×∫-∞∞|∂x[sign[u(t,x)-v(t,x)]f(t,x)]|dx00≤c|∫-∞∞(u2+12uy2-v2-12vy2)dy|00≤c(∫-∞∞|u02-v02|dx+∫-∞∞‍|u0x2-v0x2|dx0000000+∫-∞∞‍|u-v|dx),
in which we have used the Fubini theorem, ∥u∥L∞≤∥u0∥H1(R) and ∥v∥L∞≤∥v0∥H1(R). The proof is completed.


## 3. Main Results

Theorem 6.
Letu and v be two local or global strong solutions of problem (2) with initial data u(0,·)=u0∈H1(R) and v(0,·)=v0∈H1(R), respectively. Let T0 be the maximum existence time of solutions u and v. For any t∈[0,T0), it holds that
(17)∥u(t,·)-v(t,·)∥L1(R)00≤ectc(1+T0)0000×(∫-∞∞‍|u0(x)-v0(x)|dx+∫-∞∞‍|u0x2-v0x2|dx),
where c depends on ∥u0∥H1(R) and ∥v0∥H1(R).From Theorem6, we immediately obtain the uniqueness result.Theorem 7.
Letu(t,x) be a strong solution of (1) with u0∈H1(R), and let T0 be the maximum existence time of solution u. Then any strong solution of (1) is unique.Proof of Theorem6.
For an arbitraryT>0, set ξT=[0,T]×R. Let f(t,x)∈C0∞(ξT). We assume that f(t,x)=0 outside some cylinder
(18)⨄={(t,x)}=[δ,T-2δ]×Hr-2δ,0<2δ≤min(T,r).
We define
(19)g=f(t+τ2,x+y2)ρh(t-τ2)ρh(x-y2)=f(⋯)λh(*),
where (⋯)=((t+τ)/2,(x+y)/2) and (*)=((t-τ)/2,(x-y)/2). The function ρ(σ) is defined in (6). Note that
(20)gt+gτ=ft(⋯)λh(*),gx+gy=fx(⋯)λh(*).
Takingk=v(τ,y) and assuming f(t,x)=0 outside the cylinder ⨄, from Lemma 4, we have
(21)∭∫ξT×ξT{(u2(t,x)2-v2(τ,y)2)|u(t,x)-v(τ,y)|gt+sign(u(t,x)-v(τ,y))0000000000×(u2(t,x)2-v2(τ,y)2)gx0000000000+sign(u(t,x)-v(τ,y))0000000000×Qu(t,x)g(u2(t,x)2-v2(τ,y)2)}dxdtdydτ=0.
Similarly, it has
(22)∭∫ξT×ξT{(v2(τ,y)2-u2(t,x)2)|v(τ,y)-u(t,x)gτ|+sign(v(τ,y)-u(t,x))0000000000×(v2(τ,y)2-u2(t,x)2)gy0000000000+sign(v(τ,y)-u(t,x))0000000000×Qv(τ,y)g(v2(τ,y)2-u2(t,x)2)+sign(v(τ,y)-u(t,x))}dxdtdydτ=0,
from which we obtain
(23)0≤∭∫ξT×ξT{(v2(τ,y)2-u2(t,x)2)×(gx+gy)|u(t,x)-v(τ,y)|(gt+gτ)00000000000000+sign(u(t,x)-v(τ,y))00000000000000×(u2(t,x)2-v2(τ,y)2)00000000000000×(gx+gy)(v2(τ,y)2-u2(t,x)2)}dxdtdydτ000+|∭∫ξT×ξTsign(u(t,x)-v(t,x))00000000000×(Qu(t,x)-Qv(τ,y))gdxdtdydτ∭∫ξT×ξTsign|00=∭∫ξT×ξT(I1+I2+I3)dxdtdydτ.
We will show that(24)0≤∬ξT{(u2(t,x)2)|u(t,x)-v(t,x)|ft+sign(u(t,x)-v(t,x))000000000×(u2(t,x)2-v2(t,x)2)fx}dxdt00+|∬ξTsign(u(t,x)-v(t,x))00000×[Qu(t,x)-Qv(t,x)]fdxdt∬ξT|.
We note that the first two terms in the integrand of (23) can be represented in the form
(25)Yh=Y(t,x,τ,y,u(t,x),v(τ,y))λh(*).
Since ∥u∥L∞≤∥u0∥H1(R) and ∥v∥L∞≤∥v0∥H1(R), from Lemma 2, we know Yh satisfies the Lipschitz condition in u and v, respectively. By the choice of g, we have Yh=0 outside the region
(26){(t,x;τ,y)}={|x-y|2δ≤t+τ2≤T-2δ,|t-τ|2≤h,0000000000000|x+y|2≤r-2δ,|x-y|2≤h},∭∫ξT×ξTYhdxdtdydτ00=∭∫ξT×ξT[Y(t,x,τ,y,u(t,x),v(τ,y))0000000000000-Y(t,x,t,x,u(t,x),v(t,x))(t,x,τ,y,u(t,x),v(τ,y))]0000000×λh(*)dxdtdydτ0000+∭∫ξT×ξTY(t,x,t,x,u(t,x),v(t,x))00000000×λh(*)dxdtdydτ00=J1(h)+J2.
Considering the estimate |λ(*)|≤c/h2 and the expression of function Yh, we have
(27)|J1(h)|00≤c[∫∭|(t-τ)/2|≤h,δ≤((t+τ)/2)≤T-δ,|(x-y)/2|≤h,|(x+y)/2|≤r-δh+1h20000000×∭∫|(t-τ)/2|≤h,δ≤(t+τ)/2≤T-δ,|(x-y)/2|≤h,|(x+y)/2|≤r-δ|(τ,y)v(t,x)-v(τ,y)|dxdtdydτ,1h2∫∭|(t-τ)/2|≤h,δ≤((t+τ)/2)≤T-δ,|(x-y)/2|≤h,|(x+y)/2|≤r-δ],
where the constant c does not depend on h. Using Lemma 1, we obtain J1(h)→0 as h→0. The integral J2 does not depend on h. In fact, substituting t=α, (t-τ)/2=β, x=η, (x-y)/2=ξ and noting that
(28)∫-hh‍∫-∞∞‍λh(β,ξ)dξdβ=1,
we have
(29)J2=22∬ξTYh(α,η,α,η,u(α,η),v(α,η))0000×{∫-hh‍∫-∞∞‍λh(β,ξ)dξdβ}dηdα=4∬ξTYh(t,x,t,x,u(t,x),v(t,x))dxdt.
Hence
(30)limh→0∭∫ξT×ξTYhdxdtdydτ00=4∬ξTYh(t,x,t,x,u(t,x),v(t,x))dxdt.
Since
(31)I3=sign(u(t,x)-v(τ,y))(Qu(t,x)-Qv(τ,y))fλh(*),∭∫ξT×ξTI3dxdtdydτ00=∭∫ξT×ξT[I3(t,x,τ,y)-I3(t,x,t,x)]000000000×λh(*)dxdtdydτ0000+∭∫ξT×ξTI3(t,x,t,x)λh(*)dxdtdydτ00=K1(h)+K2,
we obtain
(32)|K1(h)|00≤c(∭|(t-τ)/2|≤h,δ≤(t+τ)/2≤T-δ,|(x-y)/2|≤h,|(x+y)/2|≤r-δh+1h200000000×∭∫|(t-τ)/2|≤h,δ≤(t+τ)/2≤T-δ,|(x-y)/2|≤h,|(x+y)/2|≤r-δ|Qv(t,x)-Qv(τ,y)|dxdtdydτ∭|(t-τ)/2|≤h,δ≤(t+τ)/2≤T-δ,|(x-y)/2|≤h,|(x+y)/2|≤r-δ|Qv(t,x)).
By Lemmas 1 and 3, we have K1(h)→0 as h→0. Using (28), we have
(33)K2=22∬ξTI3(α,η,α,η,u(α,η),v(α,η))00000×{∫-hh‍∫-∞∞‍λh(β,ξ)dξdβ}dηdα=4∬ξTI3(t,x,t,x,u(t,x),v(t,x))dxdt=4∬ξTsign(u(t,x)-v(t,x))0000×(Qu(t,x)-Qv(t,x))f(t,x)dxdt.
From (30) and (33), we prove that inequality (24) holds.
Let(34)ω(t)=∫-∞∞‍|u(t,x)-v(t,x)|dx.
We define
(35)θh=∫-∞σ‍ρh(σ)dσ(θh′(σ)=ρh(σ)≥0)
and choose two numbers τ1 and τ2∈(0,T0), τ1<τ2. In (24), we choose
(36)f=[θh(t-τ1)-θh(t-τ2)]χ(t,x),000000000h<min(τ1,T0-τ2),
where
(37)χ(t,x)=χε(t,x)=1-θε(|x|+Nt-M0+ε),ε>0.
We note that function χ(t,x)=0 outside the cone ℧ and f(t,x)=0 outside the set ⨄. For (t,x)∈℧, we have the relations
(38)0=χt+N|χx|≥χt+Nχx.
Applying (24) and (35)–(38), we have the inequality
(39)0≤∬ξT0{[ρh(t-τ1)-ρh(t-τ2)]00000×χε|u(t,x)-v(t,x)|}dxdt+|∬ξT0[θh(t-τ1)-θh(t-τ2)]00000×[Qu(t,x)-Qv(t,x)]B(t,x)χ(t,x)dxdt∬ξT0[θh(t-τ1)-θh(t-τ2)]|,
where B(t,x)=sign[u(t,x)-v(t,x)].
From (39), we obtain
(40)0≤∬ξT0{[ρh(t-τ1)-ρh(t-τ2)]00000×χε|u(t,x)-v(t,x)|}dxdt+∫0T0‍(θh(t-τ1)-θh(t-τ2))000×|∫-∞∞‍[Qu(t,x)-Qv(t,x)]B(t,x)χ(t,x)dx|dt.
Using Lemma 5, we have
(41)0≤∬ξT0{[ρh(t-τ1)-ρh(t-τ2)]00000×χε|u(t,x)-v(t,x)|}dxdt+c∫0T0‍(θh(t-τ1)-θh(t-τ2))00000×(G+∫-∞∞‍|u-v|dx)dt,
where G=∫-∞∞‍|u02-v02|dx+∫-∞∞‍|u0x2-v0x2|dx and c is defined in Lemma 5.
Lettingε→0 in (41) and sending M0→∞, we have
(42)0≤∫0T0‍{∫-∞∞‍[ρh(t-τ1)-ρh(t-τ2)]0000×∫-∞∞‍|u(t,x)-v(t,x)|dx}dt+c∫0T0‍(θh(t-τ1)-θh(t-τ2))0000×(G+∫-∞∞‍|u-v|dx)dt.
By the properties of the functionρh(σ) for h≤min(τ1,T0-τ1), we have
(43)|∫0T0‍ρh(t-τ1)ω(t)dt-ω(τ1)|00=|∫0T0‍ρh(t-τ1)[ω(t)-ω(τ1)]dt|00≤c1h∫τ1-hτ1+h‍|ω(t)-ω(τ1)|dt⟶0ash⟶0,
where c is independent of h.
Set(44)L(τ1)=∫0T0‍θh(t-τ1)μ(t)dt=∫0T0‍∫-∞t-τ1‍ρh(σ)dσω(t)dt.
Using the similar proof of (43), we get
(45)L′(τ1)=-∫0T0‍ρh(t-τ1)ω(t)dt⟶-ω(τ1)ash⟶0,
from which we obtain
(46)L(τ1)⟶L(0)-∫0τ1‍ω(σ)dσash⟶0.
Similarly, we have
(47)L(τ2)⟶L(0)-∫0τ2‍ω(σ)dσash⟶0.
Then, we get
(48)L(τ1)-L(τ2)⟶∫τ1τ2‍ω(σ)dσash⟶0.
Furthermore, if h→0, we have
(49)∫0T0‍(θh(t-τ1)-θh(t-τ2))Gdt⟶G∫τ1τ2‍dt00=(∫-∞∞‍|u02-v02|dx+∫-∞∞‍|u0x2-v0x2|dx)(τ2-τ1).
Letτ1→0 and τ2→t, and note that
(50)|u(τ1,x)-v(τ1,x)|≤|u(τ1,x)-u0(x)||u(τ1,x)-v(τ1,x)|≤+|v(τ1,x)-v0(x)||u(τ1,x)-v(τ1,x)|≤+|u0(x)-v0(x)|.
Thus, from (42), (43), (48), (49), and (50), for any t∈[0,T0], we have
(51)∫-∞∞|u(t,x)-v(t,x)|dx00≤∫-∞∞|u(0,x)-v(0,x)|dx0000+cT0(∫-∞∞|u02-v02|dx+∫-∞∞|u0x2-v0x2|dx)0000+∫0t∫-∞∞|u-v|dx,
from which we complete the proof of Theorem 6 by using the Gronwall inequality.



---

*Source: 409760-2013-06-10.xml*