# The Local Stability of Solutions for a Nonlinear Equation

Authors: Haibo Yan; Ls Yong
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

The approach of Kruzkov’s device of doubling the variables is applied to establish the local stability of strong solutions for a nonlinear partial differential equation in the spaceL1(R) by assuming that the initial value only lies in the space L1(R)∩L∞(R).


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## Body

## 1. Introduction and Main Results

Coclite and Karlsen [1] studied the well-posedness for the nonlinear equation
(1)vt-vtxx+4h′(v)vx=h′′′(v)vx3+3h′′(v)vxvxx+h′(v)vxxx,
where function h(v) satisfies
(2)|h′(v)|≤c|v|,|h(v)|≤c|v|2,
or
(3)|h′(v)|≤c,|h(v)|≤c|v|,
where c is a positive constant. The existence of entropy weak solutions and several other dynamic properties for (1) are investigated in [1].Consider the following partial differential equation:(4)vt-vtxx+mh′(v)vx=h′′′(v)vx3+3h′′(v)vxvxx+h′(v)vxxx,
where m is a positive constant and h(v)∈C3. If m=4 and h(v)=v2/2, (4) becomes the Degasperis-Procesi equation (see [2–12]). If h(v)=v3, we note that h(v) does not satisfy conditions (2) and (3). Recently, Wu [13] established the local well-posedness of strong solutions for (4) in the Sobolev space C([0,T);Hs(R))∩C1([0,T);Hs-1(R)) provided that h(v)=v3 and the initial value lies in the space Hs(R) with s>3/2. The objective of this work is to study (4) in the case h(v)=v3. The local stability of strong solutions for this case is established in the space L1(R). We think that this stability result is a new conclusion for (4).Whenh(v)=v3, the Cauchy problem of (4) takes the form
(5)vt-vtxx+3mv2vx=6vx3+18vvxvxx+3v2vxxx,v(0,x)=v0(x),
which is equivalent to
(6)vt+3v2vx+(m-1)(1-∂x2)-1∂x(v3)=0,v(0,x)=v0(x),
where the operator Λ-2g=(1-∂x2)-1g=(1/2)∫-∞∞‍e-|x-y|g(y)dy for any g∈L∞(R) or g∈Lp0(R) with a parameter p0 satisfying 1≤p0<∞.Using the approach of Kruzkov’s device of doubling the variables in [14], we obtain the following result.Theorem 1.
Letv1(t,x) and v2(t,x) be two strong solutions of problem (5) or (6) with initial data v1(0,x)=v10(x)∈L1(R)∩L∞(R) and v2(0,x)=v20(x)∈L1(R)∩L∞(R). Then, for an arbitrary T>0,
(7)∫-∞∞|v1(t,x)-v2(t,x)|dx≤ect∫-∞∞|v10(x)-v20(x)|dx,∫-∞∞|v1(t,x)-v2(t,x)|dx≤ect∫-∞∞|v10(x)(x)|,t∈[0,T]
holds, where c is a constant depending on ∥v10∥L1(R), ∥v10∥L∞(R), ∥v20∥L1(R), ∥v20∥L∞(R), and T.This paper is organized as follows. Several lemmas are given in Section2 and the proof of Theorem 1 is completed in Section 3.


## 2. Several Lemmas

Lemma 1 (see [13]).
Assume that initial valuev0=v(0,x)∈L2(R). Then the solution of problem (5) with m>0 satisfies
(8)∫RK1Kdx=∫R1+ξ2m+ξ2|v^(ξ)|2dξ=∫R1+ξ2m+ξ2|v^0(ξ)|2dξ,
where K1=v-∂xx2v and K=(m-∂xx2)-1v. Moreover, there exist two constants c1>0 and c2>0 depending only on m such that
(9)c1∥v0∥L2(R)≤c1∥v∥L2(R)≤c2∥v0∥L2(R).
In fact, if v∈L1(R)∩L∞(R), we know that v∈L2(R).Lemma 2 (see [13]).
Assume thatv0∈L2(R). Then
(10)∥v(t,x)∥L∞≤∥v0∥L∞ect,∀t∈[0,∞),
where c>0 is a constant independent of t.Lemma 3.
LetJv=(m-1)(1-∂x2)-1∂x(v3) and v0∈L1(R)∩L∞(R). Consider that
(11)∥Jv∥L∞(R)≤|m-1|∥v0∥L∞(R)3ect,t∈[0,∞)
holds, where c>0 is a constant independent of t.Proof.
Since(12)|Λ-2∂x(v3)|=|12∂x∫-∞∞e-|x-y|v3(t,y)dy|=|-12e-x∫-∞xeyv3(t,y)dy+12ex∫x∞e-yv3(t,y)dy|≤12∫-∞∞e-|x-y||v3(t,y)|dy,
we complete the proof by using Lemma 2 and ∫-∞∞‍e|x-y|dy=2.LetζT=[0,T]×R for an arbitrary T>0. C0∞(ζT) is the space of all infinitely differentiable functions with compact support in ζT. We define θ(ω) as a function which is infinitely differentiable on (-∞,+∞) such that θ(ω)≥0, θ(ω)=0 for |ω|≥1, and ∫-∞∞‍θ(ω)dω=1. For any number k>0, we let θk(ω)=(θ(k-1ω)/k). Then we have that θk(ω) is a function in C∞(-∞,∞) and
(13)θk(ω)≥0,θk(ω)=0if|ω|≥k,|θk(ω)|≤ck,∫-∞∞θk(ω)dω=1,
where c is a positive constant. Assume that the function u(x) is locally integrable in (-∞,∞). We define the approximation of function u(x) as
(14)uk(x)=1k∫-∞∞θ(x-yk)u(y)dy,k>0.We know that a pointx0 is defined as a Lebesgue point of function u(x) if
(15)lim⁡k→01k∫|x-x0|≤k|u(x)-u(x0)|dx=0.
For the Lebesgue point x0 of the function u(x), we get lim⁡k→0⁡uk(x0)=u(x0). Since the measure for the set of points which are not the Lebesgue points of u(x) is zero, we have uk(x)→u(x) as k→0 almost everywhere.For anyN>0, let R0>sup⁡t∈[0,∞)⁡∥v∥L∞(R)2<∞. Let ℶ represent the cone {(t,x):|x|≤N-R0t, 0<t<T0=min⁡(T,NR0-1)}. We let Sτ represent the cross section of the cone ℶ by the plane t=τ, τ∈[0,T0]. Let Kr={x:|x|≤r}, where r>0.Lemma 4 (see [14]).
Let the functionu(t,x) be bounded and measurable in cylinder Ω=[0,T]×Kr. If δ∈(0,min⁡[r,T]) and h∈(0,δ), then the function(16)Uk=1k2⨌|(t-τ)/2|≤k,δ≤(t+τ)/2≤T-δ,|(x-y)/2|≤k,|(x+y)/2|≤r-δ⁡|u(t,x)-u(τ,y)|dxdtdydτ
satisfieslim⁡k→0⁡Uk=0.Lemma 5 (see [14]).
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 6.
Letv be the strong solution of problem (6) and ϕ(t,x)∈C0∞(ζT). Then
(17)∬ζT{|v-α|ϕt+sign⁡(v-k)(v3-α3)ϕx-sign⁡(v-α)Jv(t,x)ϕ|v-α|ϕt+sign⁡(v-k)(v3-α3)ϕx}dxdt=0,
where α is an arbitrary constant.Proof.
LetΦ(v) be an arbitrary twice smooth function on the line -∞<v<∞. We multiply the first equation of problem (6) by the function Φ′(v)ϕ(t,x), where ϕ(t,x)∈C0∞(ζT). Integrating over ζT and transferring the derivatives with respect to t and x to the test function ϕ, for any constant α, we obtain
(18)∬ζT{[∫αv]Φ(v)ϕt+[∫αvΦ′(z)3z2dz]ϕx-Φ′(v)Jv(t,x)ϕ[∫αv]}dxdt=0,
in which we have used ∫-∞∞‍[∫αv‍Φ′(z)3z2dz]ϕxdx=-∫-∞∞‍[ϕΦ′(v)3v2vx]dx.
Integration by parts yields(19)∫-∞∞[∫αv‍Φ′(z)3z2dz]ϕxdx=∫-∞∞[-∫αv(z3-α3)Φ′′(z)dz(v3-α3)Φ′(v)-∫αv(z3-α3)Φ′′(z)dz]ϕxdx.
Let Φk(v) be an approximation of the function |v-α| and set Φ(v)=Φk(v). Using the properties of the sign⁡(v-α), (18), and (19) and sending k→0, we have
(20)∬ζT{|v-α|ϕt+sign⁡(v-α)[v3-α3]ϕx-sign⁡(v-α)Jv(t,x)ϕ|v-α|ϕt+sign⁡(v-α)[v3-α3]ϕx}dxdt=0,
which completes the proof.We note that the proof of (17) can also be found in [14].Lemma 7.
Assume thatv1(t,x) and v2(t,x) are two strong solutions of problem (6) associated with the initial data v10=v1(0,x)∈L1(R)∩L∞(R) and v20=v2(0,x)∈L1(R)∩L∞(R). For any ϕ∈C0∞(ζT),
(21)|∫-∞∞sign⁡(v1-v2)[Jv1(t,x)-Jv2(t,x)]ϕdx|≤c∫-∞∞|v1-v2|dx
holds, where c depends on ∥v10∥L1(R), ∥v10∥L∞(R), ∥v20∥L1(R), ∥v20∥L∞(R), T, and ϕ.Proof.
We obtain(22)Λ-2[v13(t,x)-v23(t,x)]=12∫-∞∞e-|x-y|[v13(t,y)-v23(t,y)]dy,(23)|Λ-2∂x[v13(t,x)-v23(t,x)]|=|-12e-x∫-∞xey[v13(t,y)-v23(t,y)]dy+12ex∫x∞e-y[v13(t,y)-v23(t,y)]dy|≤c∫-∞∞e-|x-y||v13(t,y)-v23(t,y)|dy≤c∫-∞∞e-|x-y||v1(t,y)-v2(t,y)|dy,
in which we have used Lemma 2. Using (23) and the Fubini theorem completes the proof.


## 3. Proof of Theorem1

Here we state that the techniques used in this paper to establish the local stability of solutions for problem (6) come from the methods of Kruzkov’s device of doubling the variables presented in Kruzkov’s paper [14].Proof of Theorem1.
For an arbitraryT>0, set ζT=[0,T]×R. Let f(t,x)∈C0∞(ζT). We assume that f(t,x)=0 outside the cylinder
(24)Θ={(t,x)}=[δ,T-2δ]×Hr-2δ,0<2δ≤min⁡(T,r).
We define
(25)h=f(t+τ2,x+y2)θk(t-τ2)θk(x-y2)=f(⋯)λk(*),
where (⋯)=((t+τ)/2,(x+y)/2) and (*)=((t-τ)/2,(x-y)/2). The function θk(ω) is defined in (13). Note that
(26)ht+hτ=ft(⋯)λk(*),hx+hy=fx(⋯)λk(*).
Takingv=v1(t,x) and α=v2(τ,y) and assuming that f(t,x)=0 outside the cylinder Θ, from Lemma 6, we have
(27)⨌ζT×ζT⁡{-sign⁡(v1(t,x)-v2(τ,y))Jv1(t,x)h|v1(t,x)-v2(τ,y)|ht+sign⁡(v1(t,x)-v2(τ,y))×(v13(t,x)-v23(τ,y))hx-sign⁡(v1(t,x)-v2(τ,y))×Jv1(t,x)h}dxdtdydτ=0.
Similarly, it has
(28)⨌ζT×ζT⁡{-sign⁡(v2(τ,y)-v1(t,x))Jv2(τ,y)|v2(τ,y)-v1(t,x)|gτ+sign⁡(v2(τ,y)-v1(t,x))×(v23(τ,y)-v13(t,x))hy-sign⁡(v2(τ,y)-v1(t,x))×Jv2(τ,y)h}dxdtdydτ=0,
from which we obtain
(29)0≤⨌ζT×ζT⁡|v1(t,x)-v2(τ,y)|(ht+hτ)dxdtdydτ+⨌ζT×ζT⁡sign⁡(v1(t,x)-v2(τ,y))×(v13(t,x)-v23(τ,y))×[hx+hy]dxdtdydτ+|⨌ζT×ζT⁡sign⁡(v1(t,x)-v2(τ,y))×(Jv1(t,x)-Jv2(τ,y))hdxdtdydτ∭∫ζT×ζTsign(v1(t,x)-v2(τ,y))|=⨌ζT×ζT⁡(I1+I2)dxdtdydτ+|⨌ζT×ζT⁡I3dxdtdydτ|.
We claim that(30)0≤∬ζT|v1(t,x)-v2(t,x)|ft+sign⁡(v1(t,x)-v2(t,x))×(v13(t,x)-v23(t,x))fxdxdt+|∬ζTsign⁡(v1(t,x)-v2(t,x))×[Jv1(t,x)-Jv2(t,x)]fdxdt∬ζTsign(v1(t,x)-v2(t,x))|.
We note that the first two terms in the integrand of (29) can be represented in the form
(31)Hk=H(t,x,τ,y,v1(t,x),v2(τ,y))λk(*).
From Lemma 5, we know that Hk satisfies the Lipschitz condition in v1 and v2, respectively. By the choice of h, we have Hk=0 outside the region as follows:
(32){(t,x;τ,y)}={|x-y|2δ≤t+τ2≤T-2δ,|t-τ|2≤k,|x+y|2≤r-2δ,|x-y|2≤k},⨌ζT×ζT⁡Hkdxdtdydτ=⨌ζT×ζT⁡[H(t,x,τ,y,v1(t,x),v2(τ,y))-H(t,x,t,x,v1(t,x),v2(t,x))]×λk(*)dxdtdydτ+⨌ζT×ζT⁡H(t,x,t,x,v1(t,x),v2(t,x))×λk(*)dxdtdydτ=B1(k)+B2.
Considering the estimate |λ(*)|≤c/k2 and the expression of function Hk, we have(33)|B1(k)|≤c[k+1k2⨌|(t-τ)/2|≤k,δ≤(t+τ)/2≤T-δ,|(x-y)/2|≤k,|(x+y)/2|≤r-δ⁡|v2(t,x)-v2(τ,y)|dxdtdydτ],k+1k2
where the constantc does not depend on k. Using Lemma 4, we obtain B1(k)→0 as k→0. The integral B2 does not depend on k. In fact, substituting t=α1, (t-τ)/2=β, x=η, (x-y)/2=ξ and noting that
(34)∫-kk∫-∞∞λk(β,ξ)dξdβ=1,
we have
(35)B2=22∬ζTHk(α1,η,α1,η,v1(α1,η),v2(α1,η))×{∫-kk∫-∞∞λk(β,ξ)dξdβ}dηdα1=4∬ζTH(t,x,t,x,v1(t,x),v2(t,x))dxdt.
Hence
(36)lim⁡k→0⨌ζT×ζT⁡Hkdxdtdydτ=4∬ζTH(t,x,t,x,v1(t,x),v2(t,x))dxdt.
Since
(37)I3=sign⁡(v1(t,x)-v2(τ,y))×(Jv1(t,x)-Jv2(τ,y))fλk(*),(38)⨌ζT×ζT⁡I3dxdtdydτ=⨌ζT×ζT⁡[I3(t,x,τ,y)-I3(t,x,t,x)]dxdtdydτ+⨌ζT×ζT⁡I3(t,x,t,x)dxdtdydτ=C1(h)+C2,
we obtain(39)|C1(k)|≤c(k+1k2⨌|(t-τ)/2|≤k,δ≤(t+τ)/2≤T-δ,|(x-y)/2|≤k,|(x+y)/2|≤r-δ⁡|Jv2(t,x)-Jv2(τ,y)|dxdtdydτ).
By Lemmas3 and 4, we have C1(h)→0 as k→0. Using (34), we have
(40)C2=22∬ζTI3(α1,η,α1,η,v1(α1,η),v2(α1,η))×{∫-hh∫-∞∞λh(β,ξ)dξdβ}dηdα1=4∬ζTI3(t,x,t,x,v1(t,x),v2(t,x))dxdt=4∬ζTsign⁡(v1(t,x)-v2(t,x))×(Jv1(t,x)-Jv2(t,x))f(t,x)dxdt.
From (36) and (38)–(40), we prove that inequality (30) holds.
Let(41)B(t)=∫-∞∞|v1(t,x)-v2(t,x)|dx.
We define
(42)ρk(σ)=∫-∞σθk(ω)dω,(ρk′(σ)=θk(σ)≥0)
and choose the two numbers τ1 and τ2∈(0,T0), τ1<τ2. In (30), we choose
(43)f=[ρk(t-τ1)-ρk(t-τ2)]χ(t,x),k<min⁡(τ1,T0-τ2),
where
(44)χ(t,x)=χɛ(t,x)=1-ρɛ(|x|+R0t-N+ɛ),ɛ>0.
When ɛ is sufficiently small, we note that function χ(t,x)=0 outside the cone ℶ and f(t,x)=0 outside the set Θ. For (t,x)∈Θ, we have
(45)0=χt+R0|χx|≥χt+R0χx.
Applying (30) and (42)–(45) and suitably choosing large R0, we have the inequality
(46)0≤∬ξT0{[θk(t-τ1)-θk(t-τ2)]×χɛ|v1(t,x)-v2(t,x)|}dxdt+|∬ξT0[ρk(t-τ1)-ρk(t-τ2)][Jv1(t,x)-Jv2(t,x)]×E(t,x)χ(t,x)dxdt∬ξT0[ρk(t-τ1)-ρk(t-τ2),Jv1(t,x)-Jv2(t,x)]|,
where E(t,x)=sign⁡[v1(t,x)-v2(t,x)].
From (46), we obtain
(47)0≤∬ξT0{[θk(t-τ1)-θk(t-τ2)]×χɛ|v1(t,x)-v2(t,x)|}dxdt+∫0T0(ρk(t-τ1)-ρk(t-τ2))×|∫-∞∞[Jv1(t,x)-Jv2(t,x)]E(t,x)χ(t,x)dx|dt.
Using Lemma 7, we have
(48)0≤∬ξT0{[θk(t-τ1)-θk(t-τ2)]×χɛ|v1(t,x)-v2(t,x)|}dxdt+c∫0T0(ρk(t-τ1)-ρk(t-τ2))×∫-∞∞|v1-v2|dxdt,
where c>0 is a constant as described in (21).
Lettingɛ→0 in (48) and sending N→∞, we have
(49)0≤∬ξT0{[θk(t-τ1)-θk(t-τ2)]×|v1(t,x)-v2(t,x)|}dxdt+c∫0T0(ρk(t-τ1)-ρk(t-τ2))×∫-∞∞|v1-v2|dxdt.
By the properties of the function θk(ω) for k≤min⁡(τ1,T0-τ1), we have
(50)|∫0T0θk(t-τ1)B(t)dt-B(τ1)|=|∫0T0θk(t-τ1)[B(t)-B(τ1)]dt|≤c1k∫τ1-kτ1+k|B(t)-B(τ1)|dt⟶0ask⟶0,
where c is independent of h.
Set(51)P(τ1)=∫0T0ρk(t-τ1)B(t)dt=∫0T0∫-∞t-τ1θk(σ)dσB(t)dt.
Using the similar proof of (50), we get
(52)P′(τ1)=-∫0T0θk(t-τ1)B(t)dt⟶-B(τ1)ask⟶0,
from which we obtain
(53)P(τ1)⟶P(0)-∫0τ1B(σ)dσask⟶0.
Similarly, we have
(54)P(τ2)⟶P(0)-∫0τ2B(σ)dσask⟶0.
Then, we get
(55)P(τ1)-P(τ2)⟶∫τ1τ2B(σ)dσask⟶0.
Lettingτ1→0 and τ2→t, from (49), (50), and (55), for any t∈[0,T0], we have
(56)∫-∞∞|v1(t,x)-v2(t,x)|dx≤∫-∞∞|v1(0,x)-v2(0,x)|dx+c∫0t∫-∞∞|v1-v2|dx,
where c depends on ∥v10∥L1(R), ∥v10∥L∞(R), ∥v20∥L1(R), ∥v20∥L∞(R), and T. Using the Gronwall inequality and (56) completes the proof of Theorem 1.



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