# Local Well-Posedness to the Cauchy Problem for an Equation of the Nagumo Type

Authors: Vladimir Lizarazo; Richard De la cruz; Julio Lizarazo
Journal: The Scientific World Journal
 (2022)

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## Abstract

In this paper, we show the local well-posedness for the Cauchy problem for the equation of the Nagumo type in this equation (1) in the Sobolev spacesHsℝ. If D>0, the local well-posedness is given for s>1/2 and for s>3/2 if D=0.


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## Body

## 1. Introduction

In this paper, we show the local well-posedness for the following Cauchy problem:(1)ut=Duxx−uu−αu−1−ϵunux,x∈ℝ,t>0,ux,0=ψx,where D>0 is a constant diffusion coefficient, α∈0,1/2 and ϵ>0 is a small positive quantity. In [1], the equation (1) was used to model chemotaxis (see equation (55) in [1]). Organisms which use chemotaxis to locate food sources include amoebae of the cellular slime mold Dictyostelium discoideum, and the motile bacterium Escherichia coli [1]. Therefore, u=ux,t models the population density, n is a positive integer, and α is a parameter which determines the minimal required density for a population to be able to survive (for normalized population density, i.e., such that u=1 is the maximum sustainable population). Balasuriya and Gottwald [1] studied the wave speed of travelling waves for the equation (1). Also, they have the numerical evidence for the wave speed of travelling waves for the equation (1). Other results related to the equation (1) can be found in [2].Whenϵ=0, the equation (1) is called a Nagumo equation or bistable equation [3–7] in which case the model describes an active pulse transmission line simulating a nerve axon.Also, we can see the equation (1) as a generalized viscous Burgers equation with a source term. Dix [8] proved local well-posedness of the viscous Burgers equation with a source term using a contraction mapping argument. Moreover, for the classical Burgers equation (without viscosity) is well known that classical solutions cannot exits for all time, but weak global solutions can be established [9]. In addition, the uniqueness of the weak solution depends on some entropy condition. Observe that when D=0, the equation (1) is a generalized Burgers equation (without viscosity) and nonlinear source term. Therefore, from the mathematical viewpoint, the case D=0 is very interesting to study the existence and uniqueness of classical solution.In this paper, we show the local well-posedness for the Cauchy problem to the equation of the Nagumo type (1) in the Sobolev spaces Hsℝ for s>1/2 if D>0, and for s>3/2 if D=0. Our proof of local well-posedness is based on the results given in [10–12]. We use the Banach fixed point in a suitable complete space to guarantee the existence of local solutions to the problem (1) with D>0. The Banach fixed point technique has been widely used to show existence and uniqueness of solutions to differential equations in Banach spaces (for instance, see [10–14] for more details). When D=0, we use the parabolic regularization method to show local well-posedness for the Cauchy problem (1) (e.g., [12,15]).We will use the following notation:ℝ for the real numbers; Sℝ for the Schwartz’s space usual; f^ denotes the Fourier transform of f; the inverse Fourier transform will be denoted by ∨; by Hsℝ, s∈ℝ, the set of all f∈S′ℝ such that 1+ξ2s/2f^∈L2ℝ. Hsℝ is called the Sobolev space and it is a Hilbert space with respect to the inner product f,gs=∫ℝ1+ξ2sf^ξg^ξ¯dξ; CI;X for the space of all continuous functions on an interval I into the Banach space X; if I is compact, CI;X is seen as a Banach space with the sup norm; CwI;X for the space of all weakly continuous functions on an interval I into Banach space X; Cw1I;X for the space of all weakly differentiable functions on an interval I into Banach space X. We also denote by Vt=etD∂x2−αId, t≥0, the semigroup in Hsℝ generated by the operator tQ where Q=D∂x2−αId, i.e.,(2)Vtf=e−tDξ2+αf^∨,forf∈Hsℝ,t≥0,Vtt≥0 is a C0-semigroup of contractions in Hsℝ, s∈ℝ. Moreover, ux,t=Vtψx is the unique solution to the linear problem associated with (1), i.e., ux,t=Vtψx is the unique solution to the following problem.(3)ut=Duxx−αu,x∈ℝ,t>0,ux,0=ψx.Proposition 1.
Letψ∈Hsℝ, s∈ℝ, λ≥0, D>0 and t>0. Then, there exists a constant Cλ, depending only on λ, such that(4)Vtψs+λ≤Cλ1+λ2Dtλ1/2ψs.In particular,Vtψ∈Sℝ for all t>0.When there is no risk of confusion, we will use the notationsut for ux,t, ϕ for ϕx, and Fu=ϵunux+u3−α+1u2.


## 2. Local Well-Posedness of the Problem (1) with D>0

In this section, we use the Banach fixed point in a suitable complete metric space to show the existence of local solutions for integral equation (9) in Sobolev space Hsℝ for s>1/2. In addition, the uniqueness of the solution and continuous dependence are established.Proposition 2.
Lets>1/2 be fixed. Then, Fu is a continuous map from Hsℝ into Hs−1ℝ and satisfies the estimates as follows:(5)Fu−Fws−1≤Lsus,wsu−ws,for all u,v∈Hsℝ, where Ls⋅,⋅ is a continuous function, nondecreasing with respect to each of their arguments. In particular,(6)Fus−1≤Lsus,0us.Proof.
Observe thatFu=ϵ/n+1un+1x+u3−α+1u2. Then, as Hsℝ is a Banach algebra for s>1/2, we have the following:(7)Fu−Fws−1≤ϵn+1∂xun+1−wn+1s−1+u3−w3s−1+α+1u2−w2s−1≤ϵn+1un+1−wn+1s+u3−w3s+α+1u2−w2s=Lsus,wsu−ws,where(8)Lsus,ws=ϵn+1∑k=0nusn−kwsk+∑k=02usn−kwsk+α+1us+ws.
The following result is to prove the existence of solutions. The proof is based in standard arguments [10,11]. We only present a sketch of proof.Proposition 3.
LetD>0 be fixed, s>1/2, ψ∈Hsℝ, and Vt is defined by (2). Then, there exists T=Tψs,M>0 and a unique function u∈C0,T;Hsℝ satisfying the following integral equation:(9)u⋅,t=Vtψ⋅−∫0tVt−τFu⋅,τdτ.Sketch of proof.
LetM,T>0 be fixed, but arbitrary. Consider the following:(10)XM,T,ψ=u∈C0,T;Hsℝ:sup0,Tut−Vtψs≤M,which is a complete metric space with distance du,v=sup0,Tut−vts. Define on the space XM,T,ψ the following map:(11)Agt≔Vtψ−∫0tVt−τFgτdτ.
We have the following:(1)
Ifg∈XM,T,ψ then Ag∈XM,T,ψ(2)
We can chooseT˜>0 sufficiently small such that AXM,T˜,ψ⊂XM,T˜,ψ(3)
There existsT¯∈0,T˜ such that A is a contraction on XM,T¯,ψ
So,A has a unique fixed point u in XM,T¯,ψ which satisfies the integral equation (40) where T¯=Tψs,M>0.Proposition 4.
The problem (1) is equivalent to the integral equation (40). More precisely, if s>1/2 and u∈C0,T;Hsℝ∩C10,T;Hs−2ℝ is a solution of (1), then u satisfies the integral equation (40). Conversely, if s>1/2 and u∈C0,T;Hsℝ is a solution of (40) then u∈C10,T;Hs−2ℝ and satisfies (1).proof.
Assume thatu∈C0,T;Hsℝ∩C10,T;Hs−2ℝ is a solution of (1). Then, d/dτVt−τuτ=−Vt−τFuτ, 0<τ<t. So, u satisfies the integral equation (40). Conversely, assume that u∈C0,T;Hsℝ is a solution of (40). For t>0, let ηt:=−∫0tVt−τFuτdτ. Then, for h>0 arbitrary,(12)ηt+h−ηth−Qηt+Futs−2≤∫0tVt−τVh−1h−QFuτs−2dτ+1h∫tt+hVt+h−τFuτ−Futs−2dτ.
However,(13)Vt−τVh−1h−QFuτs−2≤C12Dt−τ+12Dt−τ1/2Vh−1h−QFuτs−2≤C12Dt−τ+12Dt−τ1/2KFuτs−2≤C12Dt−τ+12Dt−τ1/2KFuτs−1≤C12Dt−τ+12Dt−τ1/2KLssupτ∈0,Tuτs,0supτ∈0,Tuτs,and the right hand side of (57) is a integrable function of τ in 0,t. Thus, using the dominated convergence theorem, we have as follows:(14)limh⟶0+∫0tVt−τVh−1h−QFuτs−2dτ=0.
Now, from the mean value theorem for integrals, there exists a valuec on the interval t,t+h such that(15)1h∫tt+hVt+h−τFuτ−Futs−2dτ=Vt+h−cFuc−Futs−2=Vh−c∗Fut+c∗−Futs−2,and therefore, limh⟶0+1/h∫tt+hVt+h−τFuτ−Futs−2dτ=0.
After,∂t+ηt=Qηt−Fut in Hs−2ℝ. where ∂t+ is the right derivative. In similar way, we can conclude that the left derivative is ∂t−ηt=Qηt−Fut in Hs−2ℝ. So, η∈C10,T;Hs−2ℝ and ∂tηt=Qηt−Fut. As Vtψx is the solution of the linear problem (3), we conclude that ut=Vtψ+ηt∈C10,T;Hs−2ℝ and satisfies (1).Lemma 1.
Supposeβ>0, γ>0, β+γ>1, a≥0, b≥0, u is nonnegative and tγ−1ut is locally integrable on 0,T. If(16)ut≤a+b∫0tt−sβ−1sγ−1usds,i.e., in 0,T, then(17)ut≤aEβ,γbΓβ1/νt,where ν=β+γ−1>0, Eβ,γs=∑m=0∞cmsmν with c0=1 and cm+1/cm=Γmν+γ/Γmν+γ+β for m≥0.The proof of this lemma is given in Lemma 7.1.2 in [16].Proposition 5.
Letψ,ϕ∈Hsℝ and u,v∈C0,T;Hsℝ be the corresponding solutions of equation (9). If s>1/2, then(18)ut−vts≤Kψ−ϕs,where K=E1/2,1bΓ1/22T, b=LsL,LC12DT+1/2D and L=maxsup0,Tus,sup0,Tvs (here E1/2,1 is given by previous lemma).proof.
Letψ,ϕ,u and v as in the statement of the proposition. Let s>1/2. From (9) we have as follows:(19)ut−vt=Vtψ−ϕ−∫0tVt−τFuτ−Fvτdτ.
By Propositions1 and 2, we obtain the following:(20)ut−vts≤ψ−ϕs+∫0tVt−τFuτ−Fvτs−1dτ≤ψ−ϕs+C1∫0t1+12Dt−τ1/2Fuτ−Fvτs−1dτ≤ψ−ϕs+C12DT+12D∫0tt−τ−1/2Fuτ−Fvτs−1dτ≤ψ−ϕs+LsL,LC12DT+12D∫0tt−τ−1/2uτ−vτsdτ,where L=maxsup0,Tus,sup0,Tvs. Let b=LsL,LC12DT+1/2D. Observe that E1/2,1bΓ1/22T is finite. In fact, E1/2,1bΓ1/22T=∑m=0∞am where am=cmb2πTm/2. Therefore, am+1/am=cm+1/cmb2πT1/2 and from Lemma 1 we have that(21)cm+1cm=Γm/2+1Γm+1/2+1=mΓm/2m+1Γm+1/2.
As for allx>0, Γx=2πxx−1/2e−xeθx/12x with 0<θx<1, we have that 1<eθx<e and em/6θm/2−6/m+1θm+1/2 is bounded for m≥1. From (21), we obtain as follows:(22)cm+1cm=mm+1m/2m/2−1m+1/2m+1/2−1e−1/2e6/mθm/2−6/m+1θm+1/2⟶0,asm⟶∞.
So,limm⟶∞am+1/am=0 and E1/2,1bΓ1/22T=∑m=0∞am, with am=cmb2πTm/2, is convergent.Proposition 6.
Lets>1/2. Then, the map ψ↦u is continuous in the following sense: if ψn⟶ψ in Hsℝ and un∈C0,Tn;Hsℝ, where Tn=Tψns,M>0, are the corresponding solutions (of the problem (1) with un0=ψn). Let T∈0,T∞. Then, there exists a positive integer N=ND,ψ,T such that Tn≥T for all n≥N and(23)limn⟶∞sup0,Tunt−uts=0.proof.
AsTn=Tψns,M>0 is a continuous function a ψns, then there exists N∈ℕ such that T∗≤Tn for all n≥N. Let T=minT∗,T1,T2,…,TN−1. Therefore, un is defined on 0,T for all n. It follows that u∈XM,T,ψn for all n and satisfies unts≤ψns+M≤γ+M where γ=supnψns. Therefore, supt∈0,Tunts≤γ+M for all n and supt∈0,Tuts≤γ+M. Now, similar to the proof of the previous proposition, we have as follows:(24)unt−uts≤ψn−ψs+Lsγ+M,γ+MC12DT+12D∫0tt−τ−1/2unτ−uτsdτ.
Letb=Lsγ+M,γ+MC12DT+1/2D. Thus, E1/2,1bΓ1/22T is finite (where E1/2,1 is given in Lemma 1) and we have as follows:(25)unt−uts≤ψn−ψsE1/2,1bΓ122T,for allt∈0,T.
This finishes the proof.
Finally, from Propositions3, 5 and 6, we can summarize in the following theorem:Theorem 1.
Lets>1/2. The problem (1) is locally well-posed in Hsℝ.


## 3. Local Well-Posedness of the Problem (1) with D = 0

In this section, we show the local well-posedness of the problem (1) with D=0 using a priori estimate and the parabolic regularization method, the so-called vanishing viscosity method (for more details see [12]).Lemma 2.
Letηt, at and bt be real valued positive continuous functions defined on 0,T⊆0,+∞. Let Gr and Hr be positive continuous functions for r≥0, with G strictly increasing and H nondecreasing. Define At=sup0≤s≤tas and Bt=sup0≤s≤tbs. Then, the inequality(26)Gηt≤at+bt∫0tHητdτ,0≤t<T,implies the inequality(27)ηt≤G−1Ω−1ΩAt+tBt,0≤t≤T∗≤T,where Ωr=∫ϵrdζ/HG−1ζ, ϵ>0,r>0, and T∗=supτ∈0,T:ΩAτ+τBτΩlimr⟶∞Gr.proof.
This is a particular case of the theorem given in [17] [pp. 78].Proposition 7.
Lets>3/2 be fixed. Then, Fu satisfies the estimate(28)u−w,Fu−Fw0≤L0us,wsu−w02,for all u,w∈Hsℝ, where L0x,y=ϵx∑k=0n−1xkyn−1−k+ϵn/2yn+x2+xy+y2+α+1x+y.proof.
We definequ,w=∑k=0n−1ukwn−1−k. As s>3/2 thus Hsℝ and Hs−1ℝ are Banach algebras. Moreover, we have that Hsℝ⟶Hs−1ℝ and Hsℝ⟶L∞ℝ. Thus, using the Cauchy–Schwartz inequality, we have as follows:(29)u−wFu−Fw0≤ϵu−wun−wnux0+ϵu−wwnu−wx0+u−wu3−w30+α+1u−wu2−w20≤ϵu−w0un−wnux0+ϵ2u−w2xwn0+u−w0u3−w30+α+1u−w0u2−w20≤ϵuxL∞qu,wL∞+ϵ2u−w2wnx0+u2+uw+w2L∞+α+1u+wL∞u−w02≤ϵuxs−1qu,ws−1+ϵ2wnxL∞+u2+uw+w2s−1+α+1u+ws−1u−w02≤ϵusqu,ws+ϵn2wn−1wxL∞+u2+uw+w2s−1+α+1u+ws−1u−w02≤ϵusqu,ws+ϵn2wn−1s−1wxs−1+u2+uw+w2s+α+1u+wsu−w02≤L0us,vsu−w02.Lemma 3.
(T. Kato). Let r≥1 and s>3/2 be fixed and h,v are real valued functions. Then, there exists a constant C=Cr,s such that(30)v,h∂xvr≤C∂xhs−1vr2+∂xhr−1vrvs.In particular,v,h∂xvs≤C∂xhs−1vs2.proof.
See Lemma A.5. in [13].Theorem 2.
Lets>3/2 be fixed. For D>0, consider the initial value problem (1) with initial data + and let uD∈C0,Ts∗;Hsℝ be the corresponding solution of (1) for some Ts∗>0. Then, there exists a Ts=Tsψ>0, depending on ψs, such that uD can be extended to the interval 0,Tsψ, and there is a function ρ∈C0,Tsψ;0,+∞ such that ρ0=ψs2 and uDts2≤ρt,for allt∈0,Tsψ.proof.
Using the inner product inHsℝ and Lemma 3 we have that(31)∂tuDs2=2uD∂tuDs=2uDD∂x2uD−uDuD−αuD−1−ϵuDn∂xuDs=2uDD∂x2uDs+uD−uDuD−αuD−1s+uD−ϵuDn∂xuDs=2−D∂xuDs2−uDuDuD−αuD−1s−ϵuDuDn∂xuDs≤2CsuDs4+α+1uDs3+ϵ∂xuDns−1uDs2≤2CsuDs4+α+1uDs3+ϵuDsn+2,for allt∈0,Ts∗D,ψ.
Then,uDts2≤ρt for all t∈0,T∗, where ρ∈C0,T∗;0,+∞ is the maximally extended solution of the following problem.(32)dρtdt=2Csρ2+α+1ρ3/2+ϵρn+2/2,ρ0=ψs2.
Forn≥2, from the problem (32) we obtain as follows:(33)dρtdt=2Csρ2dt+α+1ρ3/2t+ϵρn+2/2t≤2Csα+ϵ+2+4ρn+2/2t,and integrating from 0 to t we have as follows:(34)ρt≤ψs2+2Csα+ϵ+2t+8Cs∫0tρn+2/2τdτ,0≤t≤T<T∗.
From Lemma2 with at=ψs2+2Csα+ϵ+2t, bt=8Cs, Gr=r and Hr=rn+2/2 we have the following bound:(35)ρt≤ψs2+2Csα+ϵ+2t−n/2−4nCst−2/n,for t≤T∗, where T∗=supt∈0,+∞:ψs2+2Csα+ϵ+2t−n/2≥4nCst. Observe that T∗>0, since the function Φt=ψs2+2Csα+ϵ+2t−n/2−4nCst is strictly decreasing for t≥0, Φ0=1/ψsn and there exists an unique tr∈0,+∞ such that Φtr=0. Therefore, we can choose Tsψ such that 0<Tsψ<T∗. Moreover, the function Wt=Φt−2/n is increasing on 0,Tsψ, and therefore we have that(36)ρt≤ψs2+2Csα+ϵ+2Tsψ−n/2−4nCsTsψ−2/n,for all t∈0,Tsψ.
For the casen=1, from (32) we have that(37)ρt≤ψs2+2Csα+ϵ+2t+8Cs∫0tρ2τdτ,0≤t≤T<T∗,and from Lemma 2 we have ρt≤1/ψs2+2Csα+ϵ+2t−1−8Cst, for 0≤t≤T∗ where T∗=−ψs2+ψs4+4α+ϵ+2/16Csα+ϵ+2. So, we can choose Tsψ such that 0<Tsψ<T∗, and therefore, we conclude that(38)ρt≤1ψs2+2Csα+ϵ+2Tsψ−1−8CsTsψ,for all t∈0,Tsψ.
As,uDts2≤ρt for all t∈0,T∗, and since ρt and T∗ do not depend on D, the usual extension method shows that we must have Ts∗D,ψ≥Tsψ for all D>0, where Tsψ is any positive number satisfying 0<Tsψ<T∗.Theorem 3.
Lets>3/2 be fixed. If ψ∈Hsℝ, then there exists a Ts=Tsψ and a function u0∈Cw0,Ts;Hsℝ∩Cw10,Ts;Hs−2ℝ such that u00=ψ, and u0 satisfies (1) with D=0, in the weak sense, i.e.,(39)ddtu0tφs−2=−αu0−ϵu0n∂xu0−u03+α+1u02φs−2,for all φ∈Hs−2ℝ and t∈0,Ts.Moreover,u0s2≤ρt for all t∈0,Ts, where ρt is as in Theorem 2.proof.
LetTs=Tsψ be as in Theorem 2. Now, we will split the proof into four steps:Step 1.
First we will show thatuDtD>0 is a net which converges to a function u0∈C0,Ts;L2ℝ in the L2− norm, uniformly over 0,Ts.
LetD1,D2∈0,+∞. Then,(40)12ddtuD1−uD202=uD1−uD2ddtduD1−uD20=uD1−uD2D1∂x2uD1−D2∂x2uD20−uD1−uD2uD13−uD230+α+1uD1−uD2uD12−uD220−αuD1−uD2uD1−uD20−ϵuD1−uD2uD1n∂xuD1−uD2n∂xuD20=−D1∂xuD1−uD202−D1−D2∂xuD1−uD2∂xuD20−uD1−uD2uD13−uD230+α+1uD1−uD2uD12−uD220−αuD1−uD202−ϵuD1−uD2uD1n∂xuD1−uD2n∂xuD20≤D1−D2∂xuD1−uD2∂xuD20+uD1−uD2uD13−uD230+α+1uD1−uD2uD12−uD220+ϵuD1−uD2uD1n∂xuD1−uD2n∂xuD20,for allt∈0,Ts.
LetM=supt∈0,Tsρt, where ρ is the function defined in the proof of Theorem 2. We bound separately each term on the right-hand side of (40) as follows:
In order to bound the first term, we have as follows:(41)D1−D2∂xuD1−uD2∂xuD20≤D1−D2∂xuD1−uD20∂xuD20≤D1−D2∂xuD10+∂xuD20∂xuD20≤2M2D1−D2.
We can bound the second term by the following:(42)uD1−uD2uD13−uD230≤uD1−uD202uD102+uD10uD20+uD202≤uD1−uD202uD1s2+uD1suD2s+uD2s2≤3M2uD1−uD202.
The third term is bounded byuD1−uD2uD12−uD220≤2MuD1−uD202.
Finally, we have(43)uD1−uD2uD1n∂xuD1−uD2n∂xuD20=1n+1uD1−uD2∂xuD1n+1−uD2n+10=1n+1∂xuD1−uD2uD1n+1−uD2n+10=1n+1∂xuD1−uD2uD1−uD2quD1,uD20=12n+1uD1−uD22∂xquD1,uD20≤12n+1uD1−uD202∂xquD1,uD2L∞,(where q is defined in the proof of Proposition 7). From Theorem 2, we obtain as follows:(44)∂xquD1,uD2L∞≤∂xquD1,uD2s≤ρtn/2≤Mn,for all t∈0,Ts.
Therefore, from the above bounds, we have as follows:(45)12ddtuD1−uD202≤2M2D1−D2+3M2+2α+1M+Mn2n+1uD1−uD202.
Applying Gronwall’s inequality to the last relation, we show that there is a constantC>0 satisfying uD1t−uD2t02≤CD1−D2 for all t∈0,Ts, and since L2ℝ is complete there exists the limit u0t=limD⟶0uDt in L2ℝ uniformly with respect to t∈0,Ts, i.e.(46)limD⟶0+supt∈0,TsuDt−u0t0=0,and so u0∈C0,Ts;L2ℝ.Step 2.
Now we show thatu0∈Hsℝ. Let t∈0,Ts. Since uD⟶u0 in L2ℝ, as D⟶0+, then there exists a subsequence Dnj such that(47)limj⟶∞uDnj^t,ξ=u0^t,ξ,ξ−a.e..
We obtain by Fatou’s Lemma as follows:(48)u0s2=∫ℝ1+ξ2su0^2dξ≤liminfj⟶∞∫ℝ1+ξ2suDnj^2dξ≤ρt.Step 3.
We must show thatuD⇀u0 in Hsℝ for all t∈0,Ts as D⟶0+.
First of all, we will show thatuDtD>0 is a weak Cauchy net in Hsℝ, uniformly with respect to t∈0,Ts. In fact, given φ∈Hsℝ and ϵ>0, choosing φϵ∈Hsℝ such that φ−φϵs≤ϵ, then(49)uD1t−uD2tφs=uD1t−uD2tφ−φϵs+uD1t−uD2tφϵs≤uD1t−uD2tsφ−φϵs+uD1t−uD2t1−∂x2sφϵ0≤2Mϵ+uD1t−uD2t01−∂x2sφϵ0≤2Mϵ+CD1−D2,for allt∈0,Tsψ,and therefore, we have limD1,D2⟶0+supt∈0,TsuD1t−uD2tφs=0.
Thus, we have thatuD⇀u0 for all t∈0,T, i.e.,(50)limD⟶0+supt∈0,TsuDt−u0tφs=0,for all φ∈Hsℝ. Moreover, since the convergence is uniform for all φ∈Hsℝ, we can conclude that u0∈Cw0,Ts;Hsℝ.Step 4.
Finally, we show thatu0∈Cw10,Ts;Hs−2ℝ.
Letφ∈Hs−2ℝ. Then,(51)uDtφs−2=ψφs−2+∫0tD∂x2uDτ−uDτuDτ−αuDτ−1−ϵuDτn∂xuDτφs−2dτ,for all t∈0,Ts. Since uD⟶u0 in L2ℝ and uD⇀u0 in Hsℝ, we have ∂xuD⇀∂xu0 in Hs−1ℝ and ∂x2uD⇀∂x2u0 in Hs−2ℝ uniformly on 0,Ts. Observe that if r>1/2, fn⇀f in Hrℝ and gn⇀g in Hrℝ then fngn⇀fg in Hrℝ. After, we have(52)uDuD−αuD−1⇀u0u0−αu0−1inHsℝ,uDn∂xuD⇀u0n∂xu0inHs−1ℝ,uniformly on 0,Ts. Thereby, taking the limit as D⟶0+ in (51), we obtain as follows:(53)u0tφs−2=ψφs−2+∫0t−u0u0−αu0−1−ϵu0n∂xu0φs−2dτ.Corollary 1.
Letu0 be as in the preceding theorem, then u0∈AC0,Ts;Hs−2ℝ.proof.
Sincet∈0,Tsψ↦uu−αu−1+ϵunux is weakly continuous in Hs−2ℝ and the Sobolev space is separable, then applying the Bochner–Pettis theorem, it is a strongly measurable function in Hs−2ℝ. Therefore,(54)∫0tuu−αu−1+ϵunuxdτ,exists as a Bochner integral. So, from (53) we conclude that(55)u0t=ψ+∫0tuu−αu−1+ϵunuxdτ,and therefore, u0∈AC0,Ts;Hs−2ℝ.Theorem 4.
Lets>3/2 and T>0 be fixed, ψj∈Hsℝ, j=1,2, and vj∈C0,T;L2ℝ∩Cw0,T;Hsℝ∩AC0,T;Hs−2ℝ two weak sense solutions to (1) with D=0 such that vj0=ψj, j=1,2. Then,(56)v1t−v2t0≤ψ1−ψ20etL0R,R+α,where L0 is as in the Proposition 7 and(57)R=maxsupt∈0,Tv1ts,supt∈0,Tv2ts.proof.
Letwt=v1t−v2t. Since s>3/2, we have s−2>−1/2>1−s and 1−s>−s, and also Hs−2ℝ⟶H−sℝ. Using the fact that wt is real valued we have(58)wt+hwt+h0−wtwt0h=wt+h−wthwt+h0+wt+h−wthwt0=wt+h−wthwt+hs+wt+h−wthwts,where t∈0,T is fixed, h is such that t+h∈0,T, and ⋅⋅s is the Hs duality bracket. As t∈0,T↦wt∈Hsℝ is bounded and(59)∂twt=limh⟶0wt+h−wth=−v1tv1t−αv1t−1−v2tv2t−αv2t−1−ϵv1tn∂xv1t−v2tn∂xv2t,exists in the norm of Hs−2ℝ⟶H−sℝ, from (58) and (59) we have as follows:(60)∂twt02=−2v1tv1t−αv1t−1−v2tv2t−αv2t−1+ϵv1tn∂xv1t−ϵv2tn∂xv2twts=−2v1tv1t−αv1t−1−v2tv2t−αv2t−1+ϵv1tn∂xv1t−ϵv2tn∂xv2twt0=−2v1t3−α+1v1t2−ϵv1tn∂xv1t−v2t3−α+1v2t2−ϵv2tn∂xv2t+αv1t−v2twt0.
From Proposition7 and (60) we have as follows:(61)∂twt02≤L0R,R+αv1t−v2t02,where R is given by (57). Applying Gronwall’s inequality to (61), and we have proved the theorem.Theorem 5.
Letψ∈Hsℝ with s>3/2. Then, there exists a Ts=Tsψ>0 and a unique u0∈C0,Ts;Hsℝ such that(62)∂tu0t+u0tu0t−αu0t−1+u0tn∂xu0=0,u00=ψ.proof.
From the previous results, there exists a unique solution of (62) in the class described in Theorem 4. Now, we will show that u0∈C0,Ts;Hsℝ.
Letφ∈Hsℝ be such that φs=1. Then, we have u0φs≤u0ts≤ρt1/2, for all φ∈Hsℝ and for all t∈0,Ts. Additionally,(63)ψφs=limt⟶0+u0φs=liminft⟶0+u0φs≤liminft⟶0+u0ts≤limsupt⟶0+u0ts≤limsupt⟶0+ρt1/2=ψs,for all φ∈Hsℝ. As we have ψs=supφs=1ψφs, then taking supremum over φs=1 in (63) we have liminft⟶0+u0ts=limsupt⟶0+u0ts=ψs, i.e., the limit of u0ts exists as t⟶0+ and limt⟶0+u0ts=ψs. Since ut⟶ψ weakly in Hsℝ as t⟶0+, it follows that limt⟶0+u0t=ψ in the norm of Hsℝ. Let t′∈0,Ts be fixed. Then, there exists T˜>0, with Ts−t′>T˜, and a unique v∈Cw0,T˜;Hsℝ∩Cw10,T˜;Hs−2ℝ satisfying ∂tvt+vtvt−αvt−1+vtn∂xv=0, with v0=ut′. We have noticed that the uniqueness of solutions implies that vt=u0t+t′, for t∈0,T˜. Since v is continuous from the right at t=0, then u0 is continuous from the right at t=t′. Now, let t′∈0,T be fixed. Observe that the following problem(64)−∂twt+wtwt−αwt−1−wtn∂xw=0,w0=u˜t′,has a unique solution wt,x=u0t′−t,−x with u˜t′,x=u0t′,−x, because the equation in problem (64) is similar to the equation in problem (1) with D=0 and it is easy to show similar results to those obtained for problem (1) with D=0, specially the uniqueness results.
In particular, for the problem (64) there are results analogous to Theorems 3 and 4. Therefore, since w is continuous from the right at t=0, then u0 is continuous from the left at t′. So, u0∈C0,Ts;Hsℝ. Moreover, we have u0u0−αu0−1+ϵu0n∂xu0∈C0,T;Hs−2ℝ. From (55) we also conclude that u0∈C10,T;Hs−2ℝ and that it is the unique strong solution of (1) with D=0.Theorem 6.
Lets>3/2, ψ∈Hs+1ℝ and uD∈C0,Ts+1ψ;Hs+1ℝ∩C10,Ts+1ψ;Hs−1ℝ be the corresponding solution of the problem (1) for D≥0, defined in the interval 0,Ts+1ψ which is independent of D. Then, uD can be extended, if necessary, to the interval 0,Ts, with ψ viewed as an element of Hsℝ.proof.
Applying (30) with r=s+1 to obtain(65)ddtuDs+12=2uD∂uDts+1=2−D∂xuDs+12−uDuDuD−αuD−1s+1−ϵuDuDn∂xuDs+1≤2CsuD3s+1uDs+1+α+1uD2s+1uDs+1+uDsnuDs+12.
Now, from inequality (3.12) and Theorem 3.2 in [14], we obtain as follows:(66)uD2s+1≤2uDL∞uDs+1≤uDsuDs+1,(67)uD3s+1≤uDL∞uD2s+1+uD2L∞uDs+1≤2uDL∞2uDs+1+uL∞2uDs+1=3uDL∞2uDs+1≤CuDs2uDs+1.
Therefore, using (66), (67) in (65), we have as follows:(68)ddtuDts+12≤2C′uDts2+α+1uDts+uDtsnuDts+12,and integrating from 0 to t, we have as follows:(69)uDts+12≤ψs+12+2C′∫0tuDτs2+α+1uDτs+uDτsnuDτs+12dτ,and applying the Gronwall’s inequality to obtain(70)uDts+12≤ψs+12exp2C′∫0tuDτs2+α+1uDτs+uDτsndτ.
Observe that on the right-hand side of (70) is well-defined for t∈0,Tsψ and therefore we can extend (if necessary) u=ut to 0,Tsψ as a solution in Hs+1ℝ. Thus, we conclude that Tsψ≤Ts+1ψ. So, uD∈C0,Tsψ;Hs+1ℝ for D>0. From (70) also we have that(71)uDts+12≤ψs+12exp2C′M2+α+1M+MnTsψ.
Observe that the last inequality is independent ofD>0 and since uD weakly converges and uniformly to u0 in Hs+1ℝ, then we have u0∈C0,Tsψ;Hs+1ℝ.
Following Lemma 5 in [15] we have the next lemma.Lemma 4.
Lets>3/2. For ψ∈Hsℝ and τ>0, we define(72)ψτ=exp−τ1−∂x2s/2ψ=ψ^⋅exp−τ1+⋅2s/2∨.Then,limτ⟶0+ψτ−ψs=0, and there exists a constant C=Cs such that(73)ψτs+1≤C1+1τs2/s1/2ψs,ψτ−ψθ0≤τ−θψs.Moreover,limτ⟶0+ψτ−ψs=0 uniformly on compact subsets of Hsℝ.proof.
Notice thatψτ−ψs2=∫ℝ1+ξ2se−τ1+ξ2s/2−12ψ^ξ2dξ. Then, using Lebesque’s dominated convergence theorem, we obtain limτ⟶0+ψτ−ψs=0. Now, to prove the uniformity on compact subsets, it is enough to show that ψn⟶ψ in Hsℝ implies limτ⟶0+ψnτ−ψns=0 uniformly for n=1,2,…, since sequential compactness is equivalent to compactness in metric spaces. Thus, observe that(74)ψnτ−ψτs2=∫ℝ1+ξ2se−2τ1+ξ2s/2ψ^nξ−ψ^ξ2dξ≤∫ℝ1+ξ2sψ^nξ−ψ^ξ2dξ=ψn−ψs2.
Letϵ>0 be given and choose N such that if n≥N, then ψn−ψs<1/3ϵ. Thus, for τ0>0 small enough that 0<τ<τ0 we have(75)ψnτ−ψn<ϵ,for 1≤n≤N. Now, if n≥N then we have(76)ψnτ−ψns≤ψnτ−ψτs+ψτ−ψs+ψ−ψns≤ψn−ψs+ψτ−ψs+ψ−ψns<ϵ.
Hence (75) holds for all n.
On the other hand, we have(77)ψτs+12=∫ℝ1+ξ2s+1e−2τ1+ξ2s/2ψ^ξ2dξ≤1+maxξ∈ℝξ2e−2τ1+ξ2s/2ψs2≤C1+1τs2/s.
Finally, using the mean value theorem,e−τ1+ξ2s/2−e−τ1+ξ2s/2≤τ−θ1+ξ2s/2, and then we have(78)ψτ−ψθ02=∫ℝψτx−ψθx2dx=∫ℝe−τ1+ξ2s/2−e−τ1+ξ2s/22ψ^ξ2dξ≤τ−θ2ψs2.
The proof is complete.Proposition 8.
Lets>3/2, ψ∈Hsℝ, ψτ (for τ>0) be as in the preceding lemma. If u0τ is solution of the problem (62) with u0τ0=ψτ, for all τ>0, then there are constants C=Cs,ψs,T>0 and η=ηs∈0,1 such that(79)u0τ−u0θs2≤Cψτ−ψθs2+τ1−η,for τ sufficiently small and 0≤θ≤2τ.proof.
Letτ0>0 be such that u0τt is well-define in 0,T for all 0<τ≤τ0. Then,(80a)12∂tu0τ−u0θs2=u0τ−u0θu0τu0τ−αu0τ−1−u0θu0θ−αu0θ−1s+ϵu0τ−u0θu0τn∂xu0τ−u0θn∂xu0θs≤u0τ−u0θu0τu0τ−αu0τ−1−u0θu0θ−αu0θ−1s,(80b)+ϵu0τ−u0θu0θn∂xu0τ−u0θs,(80c)+ϵu0τ−u0θu0τn−u0θn∂xu0τs.
Now, the right-hand side of the inequality (3) will be estimated.
First, we will estimate (80a). Applying the Cauchy–Schwartz inequality to (80a) we have(81)u0τ−u0θu0τu0τ−αu0τ−1−u0θu0θ−αu0θ−1s≤L˜u0τ,u0θu0τ−u0θs2≤C˜u0τ−u0θs2,where L˜x,y=x2+xy+y2+α+1x+y+α and C˜=L˜M,M.
Now, we will estimate (80b). Observe that(82)u0τ−u0θu0θn∂xu0τ−u0θs≤C∂xu0θns−1u0τ−u0θs2≤Cu0θsnu0τ−u0θs2≤CMnu0τ−u0θs2.
Finally, we estimate (80c). As s>3/2, there is s0 such that 3/2<s0+1<s. From the Cauchy–Schwartz inequality, we obtain(83)u0τ−u0θu0τn−u0θn∂xu0τs≤u0τ−u0θsu0τn−u0θn∂xu0τs≤u0τ−u0θsCu0τn−u0θnsu0τs+u0τn−u0θns0u0τs+1.
Now, we will estimate each term on the right-hand side of the last inequality. First, observe that(84)u0τn−u0θnsu0τs≤u0τ−u0θsqu0τ,u0θsu0τs≤CMu0τ−u0θs,where q is defined in the proof of Proposition 7.
We also estimateu0τn−u0θns0u0τs+1. From Lemma 4 and the inequality (71), we have(85)u0τs+1≤ψτs+1eCTsM2+α+1M+Mn≤cψsτ−1/s,for all τ≤τ0.
From Lemma4, we have(86)u0τn−u0θns0≤qu0τ,u0θs0u0τ−u0θs0≤Cu0τ−u0θsϱu0τ−u0θ01−ϱ≤2CMϱu0τ−u0θ01−ϱ,where ϱ=s0/s. To estimate the term u0τ−u0θ0, observe that(87)∂tu0τ−u0θ02=2u0τ−u0θu0τu0τ−αu0τ−1−u0θu0θ−αu0θ−10+2ϵu0τ−u0θu0τn∂xu0τ−u0θn∂xu0θ0≤C1u0τ−u0θ02+1n+1q˜u0τ,u0θL∞u0τ−u0θ02≤Cu0τ−u0θ02,where q˜u,v=∑j=0nujvn−j, and from Gronwall inequality we have as follows:(88)u0τ−u0θ02≤Cψτ−ψθ02.
From Lemma4,(89)u0τn−u0θns0u0τs+1≤Cψτ−ψθ01−s0/sψsτ−1/s≤Cτ−θ1−s0/sψs1−s0/sψsτ−1/s≤Cψs2−s0/sτ1−s0+1/s,for 0≤θ≤τ. Therefore, we have(90)u0τ−u0θsu0τn−u0θns0u0τs+1≤2MCψs2−s0/sτ1−s0+1/s,and the term (80c) is bounded for(91)u0τ−u0θu0τn−u0θn∂xu0τs≤Cu0τ−u0θs2+τ1−s0+1/s.
Of the bounds that were found for (80a), (80b), (80c) we conclude that(92)∂tu0τ−u0θs2≤C˜˜u0τ−u0θs2+τ1−s0+1/s,and using Gronwall inequality, we obtain (79).
The following corollary follows immediately from Proposition8 and Lemma 4.Corollary 2.
LetF be a compact subset in Hsℝ. Suppose that ψ∈F, ψτ and u0τ are defined as in the preceding result. Then, u0τ converges uniformly to u0, for all t∈0,T, as τ⟶0+.Theorem 7.
The mapψ↦u0 is continuous in the following sense: let ψj∈Hsℝ, j=1,2,3,… such that ψj⟶ψ in Hsℝ and u0,j∈C0,Ts,j;Hsℝ∩C10,Ts,j;Hs−2ℝ are the corresponding solutions of the problem (62) with initial condition u0,j0=ψj. Let T∈0,Ts,∞. Then, there exists a positive integer N0=N0s,ψ such that Ts,j≥T for all j≥N0 and(93)limj⟶∞sup0,Tu0,jt−u0ts=0.proof.
Considerψ∈Hsℝ and let ψjj∈ℕ be a sequence in Hsℝ such that converges to ψ. Suppose that u0, u0,j, u0τ, u0,jτ are the corresponding solutions of (62) with initial values ψ, ψj, ψτ, ψjτ, respectively. As ψj⟶ψ, from Corollary 2 we have that u0,jτ converges uniformly to u0,j and u0τ converges uniformly to u0, as τ⟶0+. Thus, given ϵ>0, for τ sufficiently small, we have(94)u0,j−u0s=u0,j−u0,jτ+u0τ−u0+u0,jτ−u0τs<2ϵ+u0,jτ−u0τs,for all t∈0,T. Now, we will show that u0,jτ−u0τs converges uniformly to zero, as j⟶∞. In fact, from Lemma 3 and the Cauchy–Schwartz inequality, we have as follows:(95)∂tu0,jτ−u0τs2=2u0,jτ−u0τu0,jτu0,jτ−αu0,jτ−1−u0τu0τ−αu0τ−1s+2ϵu0,jτ−u0τu0,jτn∂xu0,jτ−u0τn∂xu0τs≤CL˜u0,jτs,u0τs+u0τs+1n+qu0,jτs,u0τsu0,jτ−u0τs2.
Applying Gronwall inequality to the last relation and the fact thatψjτ−ψτs≤ψj−ψs, we have(96)u0,jτ−u0τs2≤ψj−ψs2expc∫0TL˜u0,jτs,u0τs+u0τs+1n+qu0,jτs,u0τsdτ.
Therefore, forj sufficiently large, we have u0,j−u0s≤ϵ, for all t∈0,T.
Finally, the results obtained above can be summarized as follows:Theorem 8.
Lets>3/2. For D=0, the problem (1) is locally well-posed in Hsℝ.



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