# Global Existence of Solutions to the Fowler Equation in a Neighbourhood of Travelling-Waves

Authors: Afaf Bouharguane
Journal: International Journal of Differential Equations
 (2011)

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## Abstract

We investigate a fractional diffusion/anti-diffusion equation proposed by Andrew C. Fowler to describe the dynamics of sand dunes sheared by a fluid flow. In this paper, we prove the global-in-time well-posedness in the neighbourhood of travelling-waves solutions of the Fowler equation.


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## Body

## 1. Introduction

The study of mechanisms that allow the formation of structures such as sand dunes and ripples at the bottom of a fluid flow plays a crucial role in the understanding of coastal dynamics. The modeling of these phenomena is particularly complex since we must not only solve the Navier-Stokes or Saint-Venant equations with equation for sediment transport, but also take into account the evolution of the bottom. Instead of solving the whole system fluid flow, free surface and free bottom, nonlocal models of fluid flow interacting with the bottom were introduced in [1, 2]. Among these models, we are interested in the following nonlocal conservation law [1, 3]:
(1.1)∂tu(t,x)+∂x(u22)(t,x)+I[u(t,⋅)](x)-∂xx2u(t,x)=0,t∈(0,T),x∈R,u(0,x)=u0(x),x∈R,
where T is any given positive time, u=u(t,x) represents the dune height (see Figure 1), and ℐ is a nonlocal operator defined as follows: for any Schwartz function φ∈𝒮(ℝ) and any x∈ℝ,
(1.2)I[φ](x):=∫0+∞|ξ|-1/3φ′′(x-ξ)dξ.Figure 1
Domain considered for the Fowler model:h is the depth water, η the free surface, and u the sea bottom.Equation (1.1) is valid for a river flow over an erodible bottom u(t,x) with slow variation and describes both accretion and erosion phenomena [4]. See [4, 5] for numerical results on this equation.The nonlocal termℐ can be seen as a fractional power of order 2/3 of the Laplacian with the bad sign. Indeed, it has been proved [4]
(1.3)F(I[φ]-φ′′)(ξ)=ψI(ξ)Fφ(ξ),
where(1.4)ψI(ξ)=4π2ξ2-aI|ξ|4/3+ibIξ|ξ|1/3,
with aℐ,bℐ positive constants, ℱ denotes the Fourier transform defined in (1.7), and Γ denotes the Euler function. One simple way to establish this fact is the derivation of a new formula for the operator ℐ, see Proposition 2.5.Remark 1.1.
For causal functions (i.e.,φ(x)=0 for x<0), this operator is up to a multiplicative constant, the Riemann-Liouville fractional derivative operator which is defined as follows [6]:
(1.5)1Γ(2/3)∫0+∞φ′′(x-ξ)|ξ|1/3dξ=d-2/3dx-2/3φ′′(x)=d4/3dx4/3φ(x).Therefore, the Fowler model has two antagonistic terms: a usual diffusion and a nonlocal fractional anti-diffusive term of lower order. This remarkable feature enabled to apply this model for signal processing. Indeed, the diffusion is used to reduce the noise whereas the nonlocal anti-diffusion is used to enhance the contrast [7].Recently, some results regarding this equation have been obtained, namely, existence of travelling-wavesuϕ(t,x)=ϕ(x-ct) where ϕ∈Cb1(ℝ) and c∈ℝ represents wave velocity, the global well-posedness for L2-initial data, the failure of the maximum principle, and the local-in-time well-posedness in a subspace of Cb1 [4, 8]. Notice that the travelling-waves are not necessarily of solitary type (see [8]) and therefore may not belong to L2(ℝ), the space where a global well-posedness result is available. In [8], the authors prove local well-posedness in a subspace of Cb1(ℝ) but fail to obtain global existence.To prove the existence of travelling-waves solutions of the Fowler equation, the authors used the implicit function theorem on suitable Banach spaces [8]. Much work has been devoted to investigate existence, uniqueness, and regularity of travelling-waves for integral differential equations, see for instance [9] and references therein.An interesting topic is to know if the shape of this travelling-wave is maintained when it is perturbed. This raises the question of the stability of travelling-waves. But before interesting ourselves in this problem, we have to show first the global existence of perturbations around these travelling-waves. Hence in this paper, we prove the global well-posedness in anL2-neighbourhood of a regular travelling-wave, namely u=uϕ+v. To prove this result, we consider the following Cauchy problem:
(1.6)∂tv(t,x)+∂x(v22+uϕv)(t,x)+I[v(t,⋅)](x)-∂xx2v(t,x)=0,t∈(0,T),x∈R,v(0,x)=v0(x),x∈R,
where v0∈L2(ℝ) is an initial perturbation and T is any given positive time.To prove the existence and uniqueness results, we begin by introducing the notion ofmild solution (see Definition 2.1) based on Duhamel’s formula (2.1), in which the kernel K of ℐ-∂xx2 appears. The use of this formula allows to prove the local-in-time existence with the help of a contracting fixed point theorem. The global existence is obtained thanks to an energy estimate (4.68). This approach is classical: we refer for instance to [4, 10].The plan of this paper is organised as follows. In the next section, we define the notion of mild solution to (1.6) and we give some properties on the kernel K of ℐ-∂xx2 that will be needed in the sequel. Sections 3 and 4 are, respectively, devoted to the proof of the uniqueness and the existence of a mild solution for (1.6). Section 5 contains the proof of the regularity of the solution.Notations
(i) The norm of a measurable functionf∈Lp(ℝ) is written ∥f∥Lp(ℝ)p=∫ℝ|f(x)|pdx for 1≤p<∞.
(ii) We denote byℱ the Fourier transform of f which is defined by the following: for allξ∈ℝ(1.7)Ff(ξ):=∫Re-2iπxξf(x)dx,
and ℱ-1 denotes the inverse of Fourier transform.
(iii) The Schwartz space of rapidly decreasing functions onℝ is denoted by 𝒮(ℝ).
(iv) We writeCk(ℝ)={f:ℝ→ℂ;f,f′,…,f(k)arecontinousonℝ}.
(v) We denote byCb(ℝ) the space of all bounded continuous real-valued functions on ℝ with the norm ∥·∥L∞=supℝ|f|.
(vi) We write for anyT>0,
(1.8)C1,2((0,T]×R):={u∈C((0,T]×R);∂tu,∂xu,∂xx2u∈C((0,T]×R)}.
(vii) We denote by𝒟(U) the space of test functions on U and 𝒟′(U) denotes the distribution space.Here is our main result.Theorem 1.2.
LetT>0andv0∈L2(ℝ).  There exists a unique mild solutionv∈L∞((0,T);L2(ℝ)) of (1.6) (see Definition 2.1) which satisfies
(1.9)v∈C([0,T];L2(R)),v(0,⋅)=v0almosteverywhere.
Moreover, if ϕ∈Cb2(ℝ) then v∈C1,2((0,T]×ℝ) and satisfies
(1.10)∂tv+∂x(v22+uϕv)+I[v]-∂xx2v=0,
on (0,T]×ℝ, in the classical sense or equivalently, u=uϕ+v is a classical solution of (1.1).


## 2. Duhamel Formula and Main Properties ofK

Definition 2.1.
LetT>0 and v0∈L2(ℝ). We say that v∈L∞((0,T);L2(ℝ)) is a mild solution to (1.6) if for any t∈(0,T):
(2.1)v(t,⋅)=K(t,⋅)*v0-∫0t∂xK(t-s,⋅)*(v22+uϕv)(s,⋅)ds,
where K(t,x)=ℱ-1(e-tψℐ(·))(x) is the kernel of the operator ℐ-∂xx2, and ψℐ is defined in (1.4).The expression (2.1) is the Duhamel formula and is obtained using the spatial Fourier transform.Proposition 2.2 (main properties ofK, [4]).
The kernelK satisfies (1)
∀t>0, K(t,·)∈L1(ℝ) and K∈C∞((0,∞)×ℝ),(2)
∀s,t>0,K(s,·)*K(t,·)=K(s+t,·),∀u0∈L2(ℝ),limt→0K(t,·)*u0=u0inL2(ℝ),(3)
∀T>0,∃K0 such that ∀t∈(0,T],||∂xK(t,·)||L2(ℝ)≤K0t-3/4,(4)
∀T>0,∃K1 such that ∀t∈(0,T],||∂xK(t,·)||L1(ℝ)≤K1t-1/2.Remark 2.3.
An interesting property for the kernelK is the non-positivity (see Figure 2), and the main consequence of this feature is the failure of maximum principle [4]. We use again this property to show that the constant solutions of the Fowler equation are unstable [11].Figure 2
Evolution of the kernelK for t=0.1 (red) and t=0.5s (blue).Remark 2.4.
Using Plancherel formula, we have for anyv0∈L2(ℝ) and anyt∈(0,T](2.2)‖K(t,⋅)*v0‖L2(R)≤eα0t‖v0‖L2(R),
where α0=-minRe(ψℐ)>0.Proposition 2.5 (integral formula forℐ).
For allφ∈𝒮(ℝ) and all x∈ℝ,
(2.3)I[φ](x)=49∫-∞0φ(x+z)-φ(x)-φ′(x)z|z|7/3dz.Proof.
The proof is based on simple integrating by parts. The regularity and the rapidly decreasing ofφ ensure the validity of the computations that follow. We have
(2.4)∫0+∞φ′′(x-ξ)|ξ|-1/3dξ=∫0+∞ddξ(φ′(x)-φ′(x-ξ))|ξ|-1/3dξ=13∫0+∞|ξ|-4/3(φ′(x)-φ′(x-ξ))dξ=13∫0+∞|ξ|-4/3ddξ(φ′(x)ξ+φ(x-ξ)-φ(x))dξ=49∫0+∞φ(x-ξ)-φ(x)+φ′(x)ξ|ξ|7/3dξ=49∫-∞0φ(x+ξ)-φ(x)-φ′(x)ξ|ξ|7/3dξ.
There is no boundary term at infinity (resp., at zero) because φ is a rapidly decreasing function on ℝ (resp., φ is smooth).Remark 2.6.
Using integral formula (2.3), [4, 8] proved that
(2.5)F(I[φ])(ξ)=4π2Γ(23)|ξ|4/3(-12+i32sgn⁡(ξ))Fφ(ξ).
Notice that ℱ(ℐ[φ])(ξ)=4π2Γ(2/3)(iξ)4/3 which is consistent with Remark 1.1: up to a multiplicative constant ℐ[φ] isd4/3φ/dx4/3.Proposition 2.7.
Lets∈ℝ and φ∈Hs(ℝ). Then ℐ[φ]∈Hs-4/3(ℝ) and we have
(2.6)‖I[φ]‖Hs-4/3(R)≤4π2Γ(23)‖φ‖Hs(R).Proof.
For alls∈ℝ and all φ∈Hs(ℝ), we have, using Remark 2.6(2.7)‖I[φ]‖Hs-4/3(R)=(∫R(1+|ξ|2)s-4/3|F(I[φ])(ξ)|2dξ)1/2=4π2Γ(23)(∫R(1+|ξ|2)s-4/3|12-isgn⁡(ξ)32||ξ|8/3|F(φ)(ξ)|2dξ)1/2=4π2Γ(23)(∫R(|ξ|21+|ξ|2)4/3(1+|ξ|2)s|F(φ)(ξ)|2dξ)1/2≤4π2Γ(23)[∫R(1+|ξ|2)s|F(φ)(ξ)|2dξ]1/2=4π2Γ(23)‖φ‖Hs(R).Remark 2.8.
From the previous proposition, we deduce that for alls∈ℝ and all φ∈Hs(ℝ), ℐ[φ]∈Hs-4/3(ℝ). In particular, using the Sobolev embedding H2/3(ℝ)↪Cb(ℝ)∩L2(ℝ), we deduce that ℐ:H2(ℝ)→Cb(ℝ)∩L2(ℝ) is a bounded linear operator.Proposition 2.9 (Duhamel formula (2.1) is well defined).
LetT>0, v0∈L2(ℝ) and w∈L∞((0,T);L1(ℝ))+L∞((0,T);L2(ℝ)). Then, the function
(2.8)v:t∈(0,T]⟶K(t,⋅)*v0-∫0t∂xK(t-s,⋅)*w(s,⋅)ds
is well defined and belongs to C([0,T];L2(ℝ)) (being extended at t=0 by the value v(0,·)=v0).Proof.
From Proposition2.2, it is easy to see that v is well defined and that for any t∈(0,T], v(t,·)∈L2(ℝ). Indeed, ∀t>0,∂xK(t,·)∈L1(ℝ)∩L2(ℝ) so by Young inequalities ∂xK(t,·)*w(t,·) exists, and using the estimates on the gradient (item 3 and 4 of Proposition 2.2) we deduce that v is well defined and v(t,·)∈L2(ℝ).
Let us prove the continuity ofv. By the second item of Proposition 2.2, we have that the function t∈(0,T]→K(t,·)*v0 is continuous and it is extended continuously up to t=0 by the value v(0,·)=v0. We define the function
(2.9)F:t∈[0,T]⟶∫0t∂xK(t-s,⋅)*w(s,⋅)ds.
Now, we are going to prove that F is uniformly continuous. For any h>0, Young inequalities imply
(2.10)‖F(t+h,⋅)-F(t,⋅)‖L2≤∫0t‖∂xK(t+h-s,⋅)-∂xK(t-s,⋅)‖Lids‖w‖L∞((0,T);Lj)+∫tt+h‖∂xK(t+h-s,⋅)‖Lids‖w‖L∞((0,T);Lj),
where i,j∈ℕ* are such that i+j=3. Since ∂xK(t,·)=ℱ-1(ξ→2iπξe-tψℐ(ξ)), the dominated convergence theorem implies that
(2.11)‖∂xK(t-s+h,⋅)-∂xK(t-s,⋅)‖Li(R)⟶0,ash⟶0.
Moreover, using the estimates on the gradient (items 3 and 4 of Proposition 2.2), we have the following inequality:
(2.12)∫tt+h‖∂xK(t-s+h,⋅)‖Lj(R)ds≤cjhαj,
where cj is a positive constant and αj=1/2ifj=1,1/4ifj=2.
From (2.10), we obtain that ∥F(t+h,·)-F(t,·)∥L2(ℝ)→0,ash→0. Hence, the function F is continuous and this completes the proof of the continuity of v.Remark 2.10.
Using the semi group property of the kernelK, we have for all t0∈(0,T) and all t∈[0,T-t0], [4]
(2.13)v(t+t0,⋅)=K(t,⋅)*v(t0,⋅)-∫0t∂xK(t-s,⋅)*w(t0+s,⋅)ds.


## 3. Uniqueness of a Solution

Let us first establish the following Lemma.Lemma 3.1.
LetT>0 and v0∈L2(ℝ). For i=1,2, let wi∈L∞((0,T);L1(ℝ))∪L∞((0,T);L2(ℝ)) and define vi as in Proposition 2.9 by
(3.1)vi(t,⋅)=K(t,⋅)*v0-∫0t∂xK(t-s,⋅)*wi(s,⋅)ds.
Then,
(3.2)‖v1-v2‖C([0,T];L2(R))≤{4K0T1/4‖w1-w2‖L∞((0,T);L1(R))ifwi∈L∞((0,T);L1(R)),2K1T‖w1-w2‖L∞((0,T);L2(R))ifwi∈L∞((0,T);L2(R)).Proof.
For allt∈[0,T], we have
(3.3)v1(t,⋅)-v2(t,⋅)=-∫0t∂xK(t-s,⋅)*(w1-w2)(s,⋅)ds.
Hence with the help of Young inequalities, we get
(3.4)‖v1(t,⋅)-v2(t,⋅)‖L2(R)≤{∫0t‖∂xK(t-s,⋅)‖L2(R)‖(w1-w2)(s,⋅)‖L1(R)dsifwi∈L∞((0,T);L1(R)),∫0t‖∂xK(t-s,⋅)‖L1(R)‖(w1-w2)(s,⋅)‖L2(R)dsifwi∈L∞((0,T);L2(R)).
It then follows that
(3.5)‖v1(t,⋅)-v2(t,⋅)‖L2(R)≤{∫0t‖∂xK(t-s,⋅)‖L2(R)ds‖w1-w2‖L∞((0,T);L1(R))ifwi∈L∞((0,T);L1(R)),∫0t‖∂xK(t-s,⋅)‖L1(R)ds‖w1-w2‖L∞((0,T);L2(R))ifwi∈L∞((0,T);L2(R)).
Using again the estimates of the gradient of K (see Proposition 2.2), we conclude the proof of this Lemma.Proposition 3.2.
LetT>0 and v0∈L2(ℝ). There exists at most one v∈L∞((0,T);L2(ℝ)) which is a mild solution to (1.6).Proof.
Letv1,v2∈L∞((0,T);L2(ℝ)) be two mild solutions to (1.6) and t∈[0,T]. Using the previous Lemma, we get
(3.6)‖v1-v2‖C([0,t];L2(R))≤2K0t1/4‖v12-v22‖L∞((0,t);L1(R))+2K1t‖uϕv1-uϕv2‖L∞((0,t);L2(R)).
Since,
(3.7)‖v12-v22‖L∞((0,t);L1(R))≤M‖v1-v2‖C([0,t];L2(R)),
with M=∥v1∥C([0,T];L2(ℝ))+∥v2∥C([0,T];L2(ℝ)), then
(3.8)‖v1-v2‖C([0,t];L2(R))≤(2MK0t1/4+2K1t1/2‖uϕ‖Cb1(R))‖v1-v2‖C([0,t];L2(R)).
Therefore, v1=v2 on [0,t] for any t∈(0,T] satisfying 2MK0t1/4+2K1t1/2∥uϕ∥Cb1(ℝ)<1. Since v1 and v2 are continuous with values in L2(ℝ), we have that v1=v2 on [0,T*] where T* is the positive solution of the following:
(3.9)2MK0t1/4+2K1t1/2‖uϕ‖Cb1(R)=1,
that is, T*=(4M2K02+8K1∥uϕ∥Cb1(ℝ)(-2MK0+4M2K02+8K1∥uϕ∥Cb1(ℝ))/4K1∥uϕ∥Cb1(ℝ))4. To prove that v1=v2 on [0,T], let us define
(3.10)t0:=sup⁡{t∈[0,T]s.tv1=v2[0,t]},
and we assume that t0<T. By continuity of v1 and v2, we have that v1(t0,·)=v2(t0,·). Using the semigroup property, see Remark 2.10, we deduce that v1(t0+·,·)=v2(t0+·,·) are mild solutions to (1.6) with the same initial data v1(t0,·)=v2(t0,·) which implies, from the first step of this proof, that v1(t,·)=v2(t,·) for t∈[t0,T*+t0]. Finally, we get a contradiction with the definition of t0 and we infer that t0=T. This completes the proof of this proposition.


## 4. Global-In-Time Existence of a Mild Solution

Proposition 4.1 (local-in-time existence).
Letv0∈L2(ℝ). There exists T*>0 that only depends on ∥v0∥L2(ℝ) and ||uϕ||Cb1(ℝ) such that (1.6) admits a unique mild solution v∈C([0,T*];L2(ℝ))∩C((0,T*];H1(ℝ)). Moreover, v satisfies
(4.1)sup⁡t∈(0,T*]t1/2‖∂xv(t,⋅)‖L2(R)<+∞.Proof.
Forv∈C([0,T];L2(ℝ))∩C((0,T];H1(ℝ)), we consider the following norm:
(4.2)‖|v|‖:=‖v‖C([0,T];L2(R))+sup⁡t∈(0,T]t1/2‖∂xv(t,⋅)‖L2(R),
and we define the affine space
(4.3)X:={v∈C([0,T];L2(R))∩C((0,T];H1(R))s.t.v(0,⋅)=v0,‖|v|‖<+∞}.
It is readily seen that X endowed with the distance induced by the norm ∥|·∥| is a complete metric space. For v∈X, we define the function
(4.4)Θv:t∈[0,T]⟶K(t,⋅)*v0-12∫0t∂xK(t-s,⋅)*v2(s,⋅)ds-∫0t∂xK(t-s,⋅)*uϕv(s,⋅)ds.
From Proposition 2.9, Θv∈C([0,T];L2(ℝ)) and satisfies Θv(0,·)=v0. Step 1 (Θv∈X).
Since(4.5)∂x(K(t,⋅)*v0)=∂xK(t,⋅)*v0=F-1(ξ⟼2iπξe-tψI(ξ)Fv0(ξ)),
the dominated convergence theorem implies that for any t0>0,
(4.6)∫R4π2|ξ|2|e-tψI(ξ)-e-t0ψI(ξ)|2|Fv0(ξ)|2dξ⟶0,ast⟶t0.
Therefore, the function t>0→(ξ↦2iπξe-tψℐ(ξ)ℱv0(ξ))∈L2(ℝ) is continuous and since ℱ is an isometry of L2, we deduce that t>0→∂xK(t,·)*v0∈L2(ℝ) is continuous. We have then established that t>0→K(t,·)*v0∈H1(ℝ) is continuous. Moreover, from Proposition 2.2, we have
(4.7)‖∂xK(t,⋅)*v0‖L2(R)≤K1t-1/2‖v0‖L2(R).
Let w denote the function
(4.8)w(t,⋅)=12∫0t∂xK(t-s,⋅)*v2(s,⋅)ds+∫0t∂xK(t-s,⋅)*uϕv(s,⋅)ds.
Let us now prove that w∈C((0,T];H1(ℝ)). We first have
(4.9)∂xw(t,⋅)=∫0t∂xK(t-s,⋅)*v∂xv(s,⋅)ds+∫0t∂xK(t-s,⋅)*∂x(uϕv)(s,⋅)ds.
Using Young inequalities and Proposition 2.2, we get
(4.10)‖∂xw(t,⋅)‖L2(R)≤∫0t‖∂xK(t-s,⋅)*v∂xv(s,⋅)‖L2(R)ds+∫0t‖∂xK(t-s,⋅)*∂x(uϕv)(s,⋅)‖L2(R)ds≤∫0t‖∂xK(t-s,⋅)‖L2(R)‖v∂xv(s,⋅)‖L1(R)ds+∫0t‖∂xK(t-s,⋅)‖L1(R)‖∂x(uϕv)(s,⋅)‖L2(R)ds≤‖v‖C([0,T];L2(R))∫0tK0(t-s)-3/4s-1/2dssup⁡s∈(0,T]s1/2‖∂xv(s,.)‖L2(R)+∫0tK1(t-s)-1/2s-1/2dssup⁡s∈(0,T]s1/2‖∂x(uϕv)(s,⋅)‖L2(R).
We then obtain
(4.11)‖∂xw(t,⋅)‖L2(R)≤K0I‖v‖C([0,T];L2(R))T-1/4sup⁡s∈(0,T]s1/2‖∂xv(s,⋅)‖L2(R)+K1Jsup⁡s∈(0,T]s1/2‖∂x(uϕv)(s,⋅)‖L2(R),
where I=B(1/2,1/4) and J=B(1/2,1/2)=π, B being the beta function defined by
(4.12)B(x,y):=∫01tx-1(1-t)y-1dt.
As ∥|v∥|<∞ then
(4.13)sup⁡s∈(0,T]s1/2‖∂xv(s,⋅)‖L2(R)<∞,sup⁡s∈(0,T]s1/2‖∂x(uϕv)(s,⋅)‖L2(R)<∞.
We then deduce that ∂xw(t,·) is in L2 and so ∂xΘv(t,·)∈L2(ℝ) for all t∈(0,T].
Let us now prove that∂xw is continuous on (0,T] with values in L2. For δ>0 and t∈(0,T], we define
(4.14)(∂xw)δ(t,⋅):=∫0t∂xK(t-s,⋅)*1{s>δ}(v∂xv)(s,⋅)ds+∫0t∂xK(t-s,⋅)*1{s>δ}∂x(uϕv)(s,⋅)ds.
Since 1{s>δ}(v∂xv)(s,·)∈L∞([0,T];L1(ℝ)) and 1{s>δ}∂x(uϕv)(s,·)∈L∞([0,T];L2(ℝ)) then Proposition 2.9 implies that (∂xw)δ:[0,T]→L2(ℝ) is continuous. Moreover, we have for any t∈(0,T] and δ≤t,
(4.15)‖∂xw(t,⋅)-(∂xw)δ(t,⋅)‖L2≤K0∫0δ(t-s)-3/4s-1/2ds‖v‖C([0,T];L2)sup⁡s∈(0,T]s1/2‖∂xv(s,⋅)‖L2+K1∫0δ(t-s)-1/2s-1/2dssup⁡s∈(0,T]s1/2‖∂x(uϕv)(s,⋅)‖L2.
It then follows that
(4.16)sup⁡t∈(0,T]‖∂xw(t,⋅)-(∂xw)δ(t,⋅)‖L2(R)⟶0asδ⟶0.
We next infer that ∂xw∈C((0,T];L2(ℝ)) because it is a local uniform limit of continuous functions. Hence, we have established that Θv∈C([0,T];L2(ℝ))∩C((0,T];H1(ℝ)). To prove that Θv∈X, it remains to show that ∥|Θv∥|<+∞. Using (4.7) and (4.11), we have
(4.17)sup⁡t∈(0,T]t1/2‖∂xΘv(t,⋅)‖L2≤K1‖v0‖L2+K0IT1/4sup⁡s∈(0,T]s1/2‖∂xv(s,⋅)‖L2‖v‖C([0,T];L2)+K1JT1/2sup⁡s∈(0,T]s1/2‖∂x(uϕv)(s,⋅)‖L2.
Finally, we have Θ:X→X.Step 2.
We begin by considering a ball ofX of radius R centered at the origin
(4.18)BR:={v∈X/|‖v‖|≤R},
where R>||v0||L2(ℝ)+K1||v0||L2(ℝ). Take v∈BR and let us now prove that Θ maps BR into itself. We have
(4.19)‖Θv(t,⋅)‖L2(R)≤‖K(t,⋅)*v0‖L2(R)+∫0t‖∂xK(t-s,⋅)*(v22+uϕv)(s,⋅)‖L2(R)ds.
By Remark 2.4, we get
(4.20)‖K(t,⋅)*v0‖L2(R)≤eα0T‖v0‖L2(R),
where α0=-minRe(ψℐ)>0. Moreover, since ||v2||L∞((0,T);L1(ℝ))=||v||L∞((0,T);L2(ℝ))2 and with the help of Proposition 2.2, we get
(4.21)‖Θv(t,⋅)‖L2(R)≤eα0T‖v0‖L2(R)+2K0T1/4R2+2K1T1/2‖uϕ‖Cb1(R)R.
Using (4.17) and (4.21), we deduce that
(4.22)|‖Θv‖|≤eα0T‖v0‖L2(R)+K1‖v0‖L2(R)+(2+I)K0T1/4R2+(2+J)RK1T1/2‖uϕ‖Cb1(R)+K1J‖uϕ‖Cb1(R)RT.
Therefore, for T>0 sufficiently small such that
(4.23)eα0T‖v0‖L2(R)+K1‖v0‖L2(R)+(2+I)K0T1/4R2+(2+J)RK1T1/2‖uϕ‖Cb1(R)+K1J‖uϕ‖Cb1(R)RT≤R,
we get that |∥Θv∥|≤R.
To finish with, we are going to prove thatΘ is a contraction.
Forv,w∈BR, we have for anyt∈(0,T)(4.24)‖Θv(t,⋅)-Θw(t,⋅)‖L2(R)≤12∫0t‖∂xK(t-s,⋅)‖L2(R)‖(v2-w2)(s,⋅)‖L1(R)ds+∫0t‖∂xK(t-s,⋅)‖L1(R)‖uϕ(v-w)(s,⋅)‖L2(R)ds≤2K0t1/4‖v2-w2‖C([0,T];L1(R))+2K1t1/2‖uϕ‖Cb1(R)‖v-w‖C([0,T];L2(R)),
and since,
(4.25)‖v2-w2‖C([0,T];L1(R))≤(‖v‖C([0,T];L2(R))+‖w‖C([0,T];L2(R)))‖v-w‖C([0,T];L2(R))≤2R‖v-w‖C([0,T];L2(R)),
we get
(4.26)‖Θv(t,⋅)-Θw(t,⋅)‖L2(R)≤(4RK0t1/4+2K1t1/2‖uϕ‖Cb1(R))‖v-w‖C([0,T];L2(R)).
Moreover
(4.27)‖∂x(Θv-Θw)(t,⋅)‖L2(R)≤12∫0t‖∂xK(t-s,⋅)*∂x(v2-w2)(s,⋅)‖L2(R)ds+∫0t‖∂xK(t-s,⋅)*∂x(uϕ(v-w))(s,⋅)‖L2(R)ds≤K0It-1/4sup⁡s∈(0,T]s1/2‖(v∂xv-w∂xw)(s,⋅)‖L1(R)+K1Jsup⁡s∈(0,T]s1/2‖∂x(uϕ(v-w))(s,⋅)‖L2(R).
And since
(4.28)‖(v∂xv-w∂xw)(t,⋅)‖L1≤‖∂xw(t,⋅)‖L2‖(v-w)(t,⋅)‖L2+‖v(t,⋅)‖L2‖∂x(v-w)(t,⋅)‖L2,
then
(4.29)t1/2‖(v∂xv-w∂xw)(t,⋅)‖L1≤‖(v-w)(t,⋅)‖L2|‖w‖|+|‖v‖|t1/2‖∂x(v-w)(t,⋅)‖L2≤2R|‖v-w‖|.
Therefore, we obtain
(4.30)‖∂x(Θv-Θw)(t,⋅)‖L2(R)≤2K0It-1/4R|‖v-w‖|+K1J‖uϕ‖Cb1(R)T1/2|‖v-w‖|+K1J‖uϕ‖Cb1(R)|‖v-w‖|.
Finally, using (4.26) and (4.30), we get
(4.31)|‖Θv-Θw‖|≤[(2+I)2RK0T1/4+(2+J)‖uϕ‖Cb1(R)K1T1/2+K1JT‖uϕ‖Cb1(R)]‖|v-w‖|.Step 3 (conclusion).
For anyT*>0 sufficiently small such that (4.23) holds true and
(4.32)(2+I)2RK0T*1/4+(2+J)‖uϕ‖Cb1(R)K1T*1/2+K1JT*‖uϕ‖Cb1(R)<1,Θ is a contraction from BR into itself. The Banach fixed point theorem then implies that Θ admits a unique fixed point v∈C([0,T*];L2(ℝ))∩C((0,T*];H1(ℝ)) which is a mild solution to (1.6).Lemma 4.2 (regularityH2 of v(t,·)).
Letv0∈L2(ℝ) and ϕ∈Cb2(ℝ). There exists T*′>0 that only depends on ∥v0∥L2(ℝ) and ||uϕ||Cb2(ℝ) such that (1.6) admits a unique mild solution v∈C([0,T*′];L2(ℝ))∩C((0,T*′];H2(ℝ)). Moreover, v satisfies
(4.33)sup⁡t∈(0,T*′]t1/2‖∂xv(t,⋅)‖L2(R)<+∞,sup⁡t∈(0,T*′]t‖∂xx2v(t,⋅)‖L2(R)<+∞.Proof.
To prove this result, we use again a contracting fixed point theorem. But this time, it is the gradient of the solutionv which is searched as a fixed point.
From Proposition4.1, there exists T*>0 which depends on ∥v0∥L2(ℝ) and ∥uϕ∥C1(ℝ) such that v∈C([0,T*];L2(ℝ))∩C((0,T*];H1(ℝ)) is a mild solution to (1.6). Since v∈C((0,T*];H1(ℝ)), we can consider the gradient of v(t,·) for any t∈(0,T*]. Let then t0∈(0,T*) and T*′∈(0,T*-t0]. We consider the same complete metric space X defined in the proof of Proposition 4.1 and we take the norm |∥·∥| defined in (4.2):
(4.34)X:={w∈C([0,T*′];L2(R))∩C((0,T*′];H1(R))s.t.w(0,⋅)=w0,|‖w‖|<+∞},
with the initial data w0=∂xv(t0,·).
We now wish to apply the fixed point theorem at the following function:(4.35)Θw:t∈[0,T*′]⟶K(t,⋅)*w0-∫0t∂xK(t-s,⋅)*(v¯w)(s,⋅)ds-∫0t∂xK(t-s,⋅)*(∂x(uϕ)v¯)(s,⋅)ds-∫0t∂xK(t-s,⋅)*(uϕw)(s,⋅)ds,
where v¯(t,·):=v(t0+t,·). First, we leave the reader to verify that Θ maps X into itself. The proof is similar to the one given in Proposition 4.1.
For anyw∈X, we have from Young inequalities and Remark 2.4(4.36)‖Θw(t,⋅)‖L2(R)≤eα0T*′‖w0‖L2(R)+‖v¯‖C([t0,T*′];H1(R))|‖w‖|∫0t‖∂xK(t-s,⋅)‖L2(R)ds+‖uϕ‖Cb1(R)‖v¯‖C([t0,T*′];H1(R))∫0t‖∂xK(t-s,⋅)‖L1(R)ds+‖uϕ‖Cb1(R)|‖w‖|∫0t‖∂xK(t-s,⋅)‖L1(R)ds,
and from Proposition 2.2, we get
(4.37)‖Θw(t,⋅)‖L2≤eα0T*′‖w0‖L2+4K0T*′1/4‖v¯‖C([t0,T*′];H1)|‖w‖|+2K1T*′1/2‖uϕ‖Cb1‖v¯‖C([t0,T*′];H1)+2K1T*′1/2‖uϕ‖Cb1|‖w‖|.
Differentiating Θv(t,·) with respect to the space variable, we obtain
(4.38)∂xΘv(t,⋅)=∂xK(t,⋅)*w0-∫0t∂xK(t-s,⋅)*∂x(v¯w)(s,⋅)ds-∫0t∂xK(t-s,⋅)*∂x(∂x(uϕ)v¯)(s,⋅)ds-∫0t∂xK(t-s,⋅)*∂x(uϕw)(s,⋅)ds,
and developing, we get
(4.39)∂xΘv(t,⋅)=∂xK(t,⋅)*w0-∫0t∂xK(t-s,⋅)*[w∂xv¯+v¯∂xw](s,⋅)ds-∫0t∂xK(t-s,⋅)*[∂x2(uϕ)v¯+∂x(uϕ)∂xv¯](s,⋅)ds-∫0t∂xK(t-s,⋅)*[∂x(uϕ)w+uϕ∂xw](s,⋅)ds.
Now, from Young inequalities, we have
(4.40)‖∂xΘv(t,⋅)‖L2≤‖∂xK(t,⋅)‖L1‖w0‖L2+∫0t‖∂xK(t-s,⋅)‖L2‖w∂xv¯(s,⋅)‖L1ds+∫0t‖∂xK(t-s,⋅)‖L2‖v¯∂xw(s,⋅)‖L1ds+∫0t‖∂xK(t-s,⋅)‖L1[‖∂x2(uϕ)v¯(s,⋅)‖L2+‖∂x(uϕ)∂xv¯(s,⋅)‖L2]ds+∫0t‖∂xK(t-s,⋅)‖L1[‖∂x(uϕ)w(s,⋅)‖L2+‖uϕ∂xw(s,⋅)‖L2]ds.
Finally, from Proposition 2.2, we obtain
(4.41)‖∂xΘv(t,⋅)‖L2≤t-1/2K1‖w0‖L2+4t1/4K0‖v¯‖C([t0;T*′];H1)|‖w‖|+∫0tK0(t-s)-3/4s-1/2ds‖v¯‖C([t0;T*′];H1)sup⁡s∈(0,T*′]s1/2‖∂xw(s,⋅)‖L2+4K1t1/2‖uϕ‖Cb2‖v¯‖C([t0;T*′];H1)+2K1t1/2‖uϕ‖Cb1|‖w‖|+∫0tK1(t-s)-1/2s-1/2ds‖uϕ‖Cb2sup⁡s∈(0,T*′]s1/2‖∂xw(s,⋅)‖L2.
In other words, we have for allt∈(0,T*′](4.42)t1/2‖∂xΘv(t,⋅)‖L2≤K1‖w0‖L2+4T*′3/4K0‖v¯‖C([t0;T*′];H1)|‖w‖|+K0IT*′1/4‖v¯‖C([t0;T*′];H1)‖|w|‖+4K1T*′‖uϕ‖Cb2‖v¯‖C([t0;T*′];H1)+2K1πT*′‖uϕ‖Cb1|‖w‖|+K1T*′1/2‖uϕ‖Cb2|‖w‖|,
where I=B(1/2,1/4). Hence, using (4.37) and (4.42), we get
(4.43)|‖Θw‖|≤eα0T*′‖w0‖L2(R)+K1‖w0‖L2(R)+2K1‖uϕ‖Cb2(R)‖v¯‖C([t0;T*′];H1(R))(2T*′+T*′1/2)+C|‖w‖|(T*′1/4+T*′1/2+T*′3/4+T*′),
for some positive constant C which depends on K0,K1,∥v¯∥C([t0;T*′];H1(ℝ)) and ∥uϕ∥Cb2(ℝ).
We next leave reader to verify that: for anyw1,w2∈X,
(4.44)‖|Θw1-Θw2|‖≤C′(T*′1/4+T*′1/2+T*′3/4+T*′)|‖w1-w2‖|,
where C′ is a positive constant which depends on K0,K1,∥v¯∥C([t0;T*′];H1(ℝ)) and ∥uϕ∥Cb2(ℝ).
Then, ifT*′>0 satisfies
(4.45)eα0T*′‖w0‖L2(R)+K1‖w0‖L2(R)+2K1‖uϕ‖Cb1(R)‖v¯‖C([t0;T′];H1(R))(2T*′+T*′1/2)+CR(T*′1/4+T*′1/2+T*′3/4+T*′)≤R,C′(T*′1/4+T*′1/2+T*′3/4+T*)<1,Θ:BR(X)→BR(X) is a contraction, where BR(X) is ball of X of radius R centered at the origin. Using a contracting point fixed theorem, there exists a unique fixed point, which we denote by w. But it is easy to see that Θ∂xv¯=∂xv¯ taking into account the space derived from the Duhamel formulation (2.1). Thanks to a uniqueness argument, we deduce that w=∂xv¯ and thus that v∈C((0,T*′];H2(ℝ)), which completes the proof of this lemma.Let us now prove the global-in-time existence of mild solutionv.Proposition 4.3 (global-in-time existence).
Letv0∈L2(ℝ), ϕ∈Cb2(ℝ) and T>0. Then, there exists a (unique) mild solution v∈C([0,T];L2(ℝ))∩C((0,T];H2(ℝ)) to (1.6). Moreover, v satisfies the PDE (1.6) in the distribution sense.Proof.
Step 1 (v is a distribution solution).
Taking the Fourier transform with respect to the space variable in (2.1), we get for all t∈[0,T] and all ξ∈ℝ,
(4.46)F(v(t,⋅))(ξ)=e-tψI(ξ)Fv0(ξ)-∫0tiπξe-(t-s)ψI(ξ)F(v2(s,⋅))(ξ)ds-∫0t2iπξe-(t-s)ψI(ξ)F(uϕv(s,⋅))(ξ)ds.
Define
(4.47)G(t,ξ)=-∫0t2iπξe-(t-s)ψI(ξ)F(v22+uϕv)(s,⋅)(ξ)ds.
Classical results on ODE imply that G is differentiable with respect to the time with
(4.48)∂tG(t,ξ)+ψI(ξ)G(t,ξ)=-iπξF(v2(t,⋅))(ξ)-2iπξF((uϕv)(t,⋅))(ξ)=-F(∂x(v22)(t,⋅))(ξ)-F(∂x(uϕv)(t,⋅))(ξ).
Let us now prove that all terms in (4.48) are continuous with values in L2. Since, v∈C((0,T];H1(ℝ)) then ∂x(v2),∂x(uϕv)∈C((0,T];L2(ℝ)). We thus deduce that ℱ(∂x(v2/2)) and ℱ(∂x(uϕv)) are continuous with values in L2(ℝ). Moreover, (4.46) implies that
(4.49)ψIG(t,⋅)=ψI(F(v(t,⋅))-e-tψIFv0),
and so ψℐG(t,·) is continuous with values in L2. Indeed,
(4.50)∫R|ψI(ξ)G(t,ξ)|2dξ=∫-11|ψI(ξ)G(t,ξ)|2dξ+∫R∖(-1,1)|ψ(ξ)G(t,ξ)|2dξ≤sup⁡ξ∈[-1,1]|ψI(ξ)|2‖G(t,⋅)‖L2(R)2+C∫R∖(-1,1)|ξ2G(t,ξ)|2dξ≤sup⁡ξ∈[-1,1]|ψI(ξ)|2‖G(t,⋅)‖L2(R)2+C∫R∖(-1,1)|F(∂xx2v(t,⋅))-ξ2e-tψI(ξ)Fv0|2dξ≤sup⁡ξ∈[-1,1]|ψI(ξ)|2‖G(t,⋅)‖L2(R)2+C̃‖v(t,⋅)‖H2(R)2+C̃‖v0‖L2(R)2+C‖v(t,⋅)‖H2‖v0‖L2<∞,
because ψℐ behaves at infinity as |·|2. C,C̃ are two positive constants. Hence, we have that the function t→ψℐG(t,·)∈L2(ℝ,ℂ) is continuous. Finally, we have proved that all the terms in (4.48) are continuous with values in L2. Therefore, from (4.48), we get that G∈C1([0,T];L2(ℝ,ℂ)) and then
(4.51)ddt(G(t,⋅))+ψIG(t,⋅)=-F(∂x(v22)(t,⋅))-F(∂x(uϕv)(t,⋅)).
Moreover, t∈[0,T]→e-tψℐℱv0∈L2(ℝ,ℂ) is C1 with
(4.52)ddt(e-tψIFv0)+ψIe-tψIFv0=0.
From (4.46), we infer that ℱv is C1 on [0,T] with values in L2 with
(4.53)ddtF(v(t,⋅))=-ψIF(v(t,⋅))-F(∂x(v22)(t,⋅))-F(∂x(uϕv)(t,⋅)).
Since ℱ is an isometry of L2, we deduce that v∈C1([0,T];L2(ℝ)) and by (1.3), we get
(4.54)ddt(v(t,⋅))=-F-1(ψIF(v(t,⋅)))-∂x(v22)(t,⋅)-∂x(uϕv)(t,⋅)=-I[v(t,⋅)]+∂xx2v(t,⋅)-∂x(v22)(t,⋅)-∂x(uϕv)(t,⋅).
We are now going to prove that v satisfies the PDE (1.6) in the distribution sense. Let us note
(4.55)w(t,⋅):=-I[v(t,⋅)]+∂xx2v(t,⋅)-∂x(v22)(t,⋅)-∂x(uϕv)(t,⋅),
and let us show that
(4.56)∂tv=winD′((0,T)×R).
By definition, we have for any φ∈𝒟(0,T) and ψ∈𝒟(ℝ):
(4.57)〈∂tv,φψ〉=-∫0T∫Rv(t,x)dφdtψ(x)dtdx=-∫0T(∫Rv(t,x)ψ(x)dx)dφ(t)dtdt.
Therefore, it is enough to prove that
(4.58)∫0T(∫Rw(t,x)ψ(x)dx)φ(t)dt=-∫0T(∫Rv(t,x)ψ(x)dx)φ′(t)dt,
that is,
(4.59)ddt∫Rv(t,x)ψ(x)dx=∫Rw(t,x)ψ(x)dx,
in the sense of 𝒟′(0,T). But by (4.54), we have that the function
(4.60)t∈(0,T)⟼∫Rv(t,x)ψ(x)dx∈R
is C1 and
(4.61)ddt∫Rv(t,x)ψ(x)dx=∫Rw(t,x)ψ(x)dx
in the classical sense, which proves that the mild solution v is a distribution solution of (1.6).Step 2 (priori estimate).
By Step1, we have
(4.62)∂tv+∂x(v22+uϕv)+I[v]-∂xx2v=0
in the distribution sense. Therefore, multiplying this equality by v and integrating with respect to the space variable, we get:
(4.63)∫Rvtvdx+∫R(I[v]-vxx)vdx+∫R(uϕv)xvdx=0,
because the nonlinear term is zero. Indeed, integrating by parts, we have
(4.64)∫R∂x(v22)vdx=-∫Rv22∂xvdx=-12∫R∂x(v22)vdx.
There is no boundary term from the infinity because for all t∈(0,T], v(t,·)∈H2(ℝ). Using (1.3) and the fact that ∫ℝ(ℐ[v]-∂xx2v)vdx is real, we get
(4.65)∫R(I[v]-∂xx2v)vdx=∫RF-1(ψIFv)vdx=∫RψI|Fv|2dξ=∫RRe(ψI)|Fv|2dξ.
Moreover, since uϕv∈H1(ℝ) we have
(4.66)∫R(uϕv)xvdx=-∫Ruϕvvxdx=-∫Ruϕ(v22)xdx=∫R(∂xuϕ)v22dx.
Using (4.63), (4.65), and (4.66), we obtain
(4.67)12ddt‖v(t,⋅)‖L22≤(α0+Cϕ)‖v(t,⋅)‖L22,
where α0=-minRe(ψℐ)>0 and Cϕ=(1/2)||uϕ||Cb1. Finally, we get for all t∈[0,T] the following estimate:
(4.68)‖v(t,⋅)‖L2(R)≤e(α0+Cϕ)t‖v0‖L2(R).Step 3 (global-in-time existence).
Up to this point, we know thanks to Proposition4.1 and Lemma 4.2 that there exists T*=T*(||v0||L2(ℝ),||uϕ||Cb2(ℝ))>0 such that v∈C([0,T*];L2(ℝ))∩C((0,T*];H2(ℝ)) is a mild solution of (1.6) on (0,T*]. Let us define
(4.69)t0:=sup⁡{t>0/thereexistsamildsolutionof(1.6)on⁡(0,t)withinitialconditionv0}.
To prove the global-in-time existence of a mild solution, we have to prove that t0≥T, where T is any positive constant. Assume by contradiction that t0<T. With again the help of Proposition 4.1, there exists T*′>0 such that for any initial data w0 that satisfy
(4.70)‖w0‖L2(R)≤e(α0+Cϕ)t0‖v0‖L2(R),
it exists a mild solution w on (0,T*′]. Using (4.68), we have that w0:=v(t0-T*′/2,·) satisfies (4.70). Therefore, using an argument of uniqueness, we deduce that v(t0-T*′/2+t,·)=w(t,·) for all t∈[0,T*′/2). To finish with, we define ṽ by ṽ=v on [0,t0) and ṽ(t0-T*′/2+t,·)=w(t,·) for t∈[T*′/2,T*′]. Hence, ṽ is a mild solution on [0,t0+T*′/2] with initial datum v0, which gives us a contradiction.


## 5. Regularity of the Solution

This section is devoted to the proof of the existence of classical solutionsv to (1.6).Proposition 5.1 (Solution in the classical sense).
Letv0∈L2(ℝ), ϕ∈Cb2(ℝ) and T>0. The unique mild solution v∈C([0,T];L2(ℝ))∩C((0,T];H2(ℝ)) of (1.6) belongs to C1,2((0,T]×ℝ) and satisfies
(5.1)∂tv+∂x(v22+uϕv)+I[v]-∂xx2v=0,
on (0,T]×ℝ in the classical sense.Proof.
Step 1 (C2-regularity in space).
Let us take anyt0∈(0,T] as initial time and let T′∈(0,T-t0]. Differentiating the Duhamel formulation (2.1) two times with respect to the space, we get for any t∈[0,T′],
(5.2)∂xx2v(t+t0,⋅)=K(t,⋅)*∂xx2v(t0,⋅)-∫0t∂xK(t-s,⋅)*(u1+u2)(t0+s,⋅)ds,
where u1:=(∂xv)2+v∂xx2v and u2:=v∂x2uϕ+2∂xuϕ∂xv+uϕ∂xx2v. Since v∈C((0,T];H2(ℝ)) then u2∈C((0,T];L2(ℝ)) and from the Sobolev embedding H2(ℝ)↪Cb1(ℝ), we get that u1∈C((0,T];L1(ℝ)∩L2(ℝ)). Let us now define the following functions:
(5.3)Fi(t,x):=∫0t∂xK(t-s,⋅)*ui(t0+s,⋅)(x)ds,fori=1,2.
For allx,y∈ℝ, we have thanks to Cauchy-Schwartz inequality
(5.4)|∂xK(t-s,⋅)*ui(t0+s,⋅)(x)-∂xK(t-s,⋅)*ui(t0+s,⋅)(y)|≤∫R|∂xK(t-s,z)||ui(t0+s,x-z)-ui(t0+s,y-z)|dz≤‖T(x-y)(ui(t0+s,⋅))-ui(s+t0,⋅)‖L2(R)‖∂xK(t-s,⋅)‖L2(R),
where 𝒯zφ denotes the translated function x→φ(x+z).
Therefore, for allx,y∈ℝ and all t∈[0,T′], we deduce that
(5.5)|Fi(t,x)-Fi(t,y)|≤∫0tK0(t-s)-3/4‖T(x-y)(ui(t0+s,⋅))-ui(t0+s,⋅)‖L2(R)ds≤4K0T′1/4sup⁡s∈[0,T′]‖T(x-y)(u¯i(s,⋅))-u¯i(s,⋅)‖L2(R),
where u¯i(s,·)=ui(t0+s,·). Then, u¯i is uniformly continuous with values in L2 as a continuous function on a compact set [0,T′]. Therefore, for any ϵ>0, there exists a finite sequence 0=s0<s1<⋯<sN=T′ such that for any s∈[0,T′], there exists j∈{0,…,N-1} such that
(5.6)‖u¯i(s,⋅)-u¯i(sj,⋅)‖L2(R)≤ϵ.
Therefore, using (5.5) we have
(5.7)|Fi(t,x)-Fi(t,y)|≤4K0T′1/4sup⁡s∈[0,T′]‖T(x-y)(u¯i(s,⋅))-T(x-y)(u¯i(sj,⋅))‖L2+4K0T′1/4{‖T(x-y)(u¯i(sj,⋅))-u¯i(sj,⋅)‖L2+sup⁡s∈[0,T]‖u¯i(s,⋅)-u¯i(sj,⋅)‖L2}.
And since ∥𝒯(x-y)(u¯i(s,·))-𝒯(x-y)(u¯i(sj,·))∥L2(ℝ)=∥u¯i(s,·)-u¯i(sj,·)∥L2(ℝ), we get
(5.8)|Fi(t,x)-Fi(t,y)|≤4K0T′1/4{‖T(x-y)(u¯i(sj,⋅))-u¯i(sj,⋅)‖L2(R)+2ϵ}.
And since the translated function is continuous in L2(ℝ), we have
(5.9)‖T(x-y)(u¯i(sj,⋅))-u¯i(sj,⋅)‖L2(R)⟶0,
as (x-y)→0. Hence,
(5.10)limsup⁡(x-y)→0|Fi(t,x)-Fi(t,y)|≤2ϵ.
Taking the infimum with respect toϵ>0, we infer that Fi is continuous with respect to the variable x. Moreover, arguing as the proof of Proposition 2.9, we get that Fi∈C([0,T′];L2(ℝ)). From classical results, we then deduce that Fi is continuous with respect to the couple (t,x) on [0,T′]×ℝ.
Moreover, sincev(t0,·)∈H2(ℝ), we can easily check that (t,x)→K(t,·)*∂xx2v(t0,·)(x) is continuous on (0,T]×ℝ. Finally, we get that ∂xx2v∈C([t0,T]×ℝ) and since t0 is arbitrary in (0,T], we conclude that ∂xx2v∈C((0,T]×ℝ).Step 2 (C1-regularity in time).
From Proposition4.3, we know that the terms ∂tv and -∂x(v2/2+uϕv)+∂xx2v-ℐ[v] have the same regularity. Moreover, by the Step 1 of this proposition, we have that ∂xx2v∈C((0,T]×ℝ), and from Sobolev embeddings and Remark 2.8, we deduce that ∂x(v2/2+uϕv) and ℐ[v] belong to C((0,T]×ℝ). Finally, we obtain that ∂tv∈C((0,T]×ℝ) and thus v∈C1,2((0,T]×ℝ). The proof of this proposition is now complete.



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*Source: 408704-2011-11-24.xml*