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Existence, Regularity, and Compactness Properties in the α-Norm for Some Partial Functional Integrodifferential Equations with Finite Delay

Written By

Boubacar Diao, Khalil Ezzinbi and Mamadou Sy

Submitted: April 1st, 2019 Reviewed: June 17th, 2019 Published: May 13th, 2020

DOI: 10.5772/intechopen.88090

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Abstract

The objective, in this work, is to study the alpha-norm, the existence, the continuity dependence in initial data, the regularity, and the compactness of solutions of mild solution for some semi-linear partial functional integrodifferential equations in abstract Banach space. Our main tools are the fractional power of linear operator theory and the operator resolvent theory. We suppose that the linear part has a resolvent operator in the sense of Grimmer. The nonlinear part is assumed to be continuous with respect to a fractional power of the linear part in the second variable. An application is provided to illustrate our results.

Keywords

  • integrodifferential
  • mild solution
  • resolvent operator
  • fractional power operator

1. Introduction

We consider, in this manuscript, partial functional equations of retarded type with deviating arguments in terms of involving spatial partial derivatives in the following form [1]:

dutdt=Aut+0tBtsusds+Ftut for t0,u0=φCα=Cr0DAα],E1

where A is the infinitesimal generator of an analytic semigroup Ttt0 on a Banach space X. Bt is a closed linear operator with domain DBtDA time-independent. For 0<α<1, Aα is the fractional power of A which will be precise in the sequel. The domain DAα is endowed with the norm xα=Aαx called α norm. Cα is the Banach space Cr0DAα of continuous functions from r0 to DAα endowed with the following norm:

ϕα=suprθ0ϕθα for ϕCα.

F:R+×CαX is a continuous function, and as usual, the history function utCα is defined by

utθ=ut+θ for θr0.

As a model for this class, one may take the following Lotka-Volterra equation:

utxt=2utxx2+0thts2usxx2ds+r0gtut+θxx for t0andx0π,ut0=utπ=0 for t0,uθx=u0θx for θr0andx0π.E2

Here u0:r0×0πR,g:R+×RR and h:R+R are appropriate functions.

In the particular case where α=0, many results are obtained in the literature under various hypotheses concerning A, B, and F (see, for instance, [2, 3, 4, 5, 6] and the references therein). For example, in [7], Ezzinbi et al. investigated the existence and regularity of solutions of the following equation:

dutdt=Aut+0tBtsusds+Ftut for t0,u0=φCr0X,E3

The authors obtained also the uniqueness and the representation of solutions via a variation of constant formula, and other properties of the resolvent operator were studied. In [8], Ezzinbi et al. studied a local existence and regularity of Eq. (3). To achieve their goal, the authors used the variation of constant formula, the theory of resolvent operator, and the principle contraction method. Ezzinbi et al. in [9] studied the local existence and global continuation for Eq. (3). Recall that the resolvent operator plays an important role in solving Eq. (3); in the weak and strict sense, it replaces the role of the c0 semigroup theory. For more details in this topic, here are the papers of Chen and Grimmer [2], Hannsgen [10], Smart [11], Miller [12, 13], and Miller and Wheeler [14, 15]. In the case where the nonlinear part involves spatial derivative, the above obtained results become invalid. To overcome this difficulty, we shall restrict our problem in a Banach space YαX, to obtain our main results for Eq. (1).

Considering the case where B=0, Travis and Webb in [16] obtained results on the existence, stability, regularity, and compactness of Eq. (1). To achieve their goal, the authors assumed that A is the infinitesimal generator of a compact analytic semigroup and F is only continuous with respect to a fractional power of A in the second variable. The present paper is motivated by the paper of Travis and Webb in [16].

The paper is organized as follows. In Section 2, we recall some fundamental properties of the resolvent operator and fractional powers of closed operators. The global existence, uniqueness, and continuous dependence with respect to the initial data are studied in Section 3. In Section 4, we study the local existence and bowing up phenomena. In Section 5 we prove, under some conditions, the regularity of the mild solutions. And finally, we illustrate our main results in Section 6 by examining an example.

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2. Fractional power of closed operators and resolvent operator for integrodifferential equations

We shall write Y for DA endowed with the graph norm xY=x+Ax,Yα for DAα and LYαX will denote the space of bounded linear operators from Yα to X, and for Y0=X, we write LX with norm .LX. We also frequently use the Laplace transform of f which is denoted by f. If we assume that A generates an analytic semigroup and, without loss of generality, that 0ϱA, then one can define the fractional power Aα for 0<α<1, as a closed linear operator on its domain Yα with its inverse Aα given by

Aα=1Γα0tα1Ttdt,

where Γ is the gamma function

Γα=0tα1etdt.

We have the following known results.

Theorem 2.1. [17] The following properties are true.

  1. Yα=DAα is a Banach space with the norm xα=Aαx for xYα.

  2. Aα is a closed linear operator with domain Yα=ImAα and Aα=Aα1.

  3. Aα is a bounded linear operator in X.

  4. If 0<αβ then DAβDAα. Moreover the injection is compact if Tt is compact for t>0.

Definition 2.2. [18] A family of bounded linear operators Rtt0 in X is called resolvent operator for the homogeneous equation of Eq. (3) if:

  1. R0=I and RtM1expσt for some M11 and σR.

  2. For all xX, tRtx is continuous for t0.

  3. RtLY for t0. For xY, R.xC1R+XCR+Y, and for t0 we have

Rtx=ARtx+0tBtsRsxds=RtAx+0tRtsBsxds.E4

What follows is we assume the hypothesis taken from [1] which implies the existence of an analytic resolvent operator Rtt0.

(V1) A generates an analytic semigroup on X. Btt0 is a closed operator on X with domain at least DA a.e t0 with Btx strongly measurable for each xY and BtxbtxY,forbLloc10 with bλ absolutely convergent for Reλ>0.

(V2) ρλ=λI+ABλ1 exists as a bounded operator on X which is analytic for λΛ=λC:argλ<π/2+δ, where 0<δ<π/2. In Λ if λϵ>0, there exists M=Mϵ>0 so that ρλM/λ.

(V3) λLX for λΛ and is analytic from Λ to LX. BλLYX and BλρλLYX for λΛ. Given ϵ>0, there exists a positive constant M=Mε so that λx+BλρλxM/λxY for xY and λΛ with λε and Bλ0 as λ in Λ. In addition, λxMλnx for some n>0,λΛ with λε. Further, there exists DDA2 which is dense in Y such that AD and BλD are contained in Y and BλxY is bounded for each xD and λΛ with λϵ.

Theorem 2.3. [1] Assume that conditions (V1)–(V3) are satisfied. Then there exists an analytic resolvent operator Rtt0. Moreover, there exist positive constants N,Nα such that RtNandAαRtNαtα for t>0 and 0α<1.

We take the following hypothesis.

(H0) The semigroup Ttt0 is compact for t>0.

Theorem 2.4. [19] Under the conditions (V1)–(V3) and (H0), the corresponding resolvent operator Rtt0 is compact for t>0.

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3. Global existence, uniqueness, and continuous dependence with respect to the initial data

Definition 3.1. A function u:0bYα is called a strict solution of Eq. (1), if:

  1. tut is continuously differentiable on 0b.

  2. utY for t0b.

  3. u satisfies Eq. (1) on 0b.

Definition 3.2. A continuous function u:0bYα is called a mild solution of Eq. (1) if

ut=Rtφ0+0tRtsFsusds for t0b,u0=φCα.E5

Now to obtain our first result, we take the following assumption.

(H1) There exists a constant LF>0 such that

Ftφ1F(tφ2)LFφ1φ2α for t0andφ1,φ2Cα.

Theorem 3.3. Assume that (V1)–(V3) and (H1) hold. Then for φCα, Eq. (1) has a unique mild solution which is defined for all t0.

Proof. Let a>0. For φCα, we define the set by

=yC0aYα:y0=φ0.

The set is a closed subset of C0aYα where C0aYα is the space of continuous functions from 0a to Yα equipped with the uniform norm topology

yα=sup0taytα for yC0aYα.

For y, we introduce the extension y¯ of y on ra defined by y¯t=yt for t0a and y¯t=φt for tr0. We consider the operator Γ defined on by

Γyt=Rtφ0+0tRtsFsy¯sds for t0a.

We claim that Γ. In fact for y, we have Γy0=φ0, and by continuity of F and Rtx for xX, we deduce that Γy. In order to obtain our result, we apply the strict contraction principle. In fact, let u,v and t0a. Then

ΓuΓvt=0tRtsFsu¯sFsv¯sds.

Using the α norm, we have

AαΓuΓvtLFNα0t1tsαu¯sv¯sαdsLFNα0t1tsαsup0τauτvταdsLFNα0adssαuvα.

Now we choose a such that

LFNα0adssα<1.

Then Γ is a strict contraction on , and it has a unique fixed point y which is the unique mild solution of Eq. (1) on 0a. To extend the solution of Eq. (1) in a2a, we show that the following equation has a unique mild solution:

ddtzt=Azt+atBtszsds+Ftzt for ta2a,za=yaCraYα.E6

Notice that the solution of Eq. (6) is given by

zt=Rtaza+atRtsFszsds for ta2a.

Let z¯ be the function defined by z¯t=zt for ta2a and z¯t=yt for tra. Consider now again the set defined by

=zCa2aYα:za=ya,

provided with the induced topological norm. We define the operator Γa on by

Γazt=Rtaza+atRtsFsz¯sds for ta2a.

We have Γaza=ya and Γaz is continuous. Then it follows that Γa. Moreover, for u,v, one has

AαΓauΓavtLFNαat1tsαu¯sv¯sαds.

Since u¯=v¯=φ in r0, we deduce that

AαΓauΓavLFNα0adssαuvα.

Then we deduce that Γa has a unique fixed point in which extends the solution y in a2a. Proceeding inductively, y is uniquely and continuously extended to nan+1a for all n1, and this ends the proof.

Now we show the continuous dependence of the mild solutions with respect to the initial data.

Theorem 3.4. Assume that (V1)–(V3) and (H1) hold. Then the mild solution u.φ of Eq. (1) defines a continuous Lipschitz operator Ut,t0 in Cα by Utφ=ut.φ. That is, Utφ is continuous from 0 to Cα for each fixed φCα. Moreover there exist a real number δ and a scalar function P such that for t0 and φ1,φ2Cα we have

Utφ1Utφ2Pδeδtφ1φ2α.E7

Proof. We use the gamma formula

Γ1αkα1=0ekssαds,

where k>0 (see [20], p. 265). The continuity is obvious that the map tut.φ is continuous. Now, let φ1,φ2Cα. If we pose wt=utφ1utφ2, then we have

wtαM1eσtφ1φ2α+LFNα0teσtstsαwsαds.E8

Let δ a real number be such that

σδ<0andM1maxeδr1LFΓ1αδσα1<1.

We define the function P by

Pδ=M1M3M41M1M4LFΓ1αδσα11

where

M3=maxeδr1,M4=maxeδr1.

Fix t¯>0 and let E=sup0st¯eδsws. If 0τt¯, then from Eq. (8), we have

eδτwταM1eσδτφ1φ2α+LFNα0τeσδτsτsαeδswsαdsM1φ1φ2α+LFM1EΓ1αδσα1.E9

If rτ0, we have

eδτwταφ1φ2αM3.E10

Therefore, Eqs. (9) and (10) imply that

suprτt¯eδτwταM1M3φ1φ2Cα+LFM1EΓ1αδσα1.E11

For 0tt¯, we have

eδtwtα=suprθ0eδθeδt+θwt+θαM4suprθ0eδt+θwt+θαM4suprτt¯eδτwτα.E12

Then from Eqs. (11) and (12), we deduce that for 0tt¯

eδtwtαM1M3M4φ1φ2α+LFM1M3EΓ1αδσα1,

which implies that

EM1M3M4φ1φ2α+LFM1M4EΓ1αδσα1.

Then the result follows.

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4. Local existence, blowing up phenomena, and the compactness of the flow

We start by generalizing a result, obtained in [19] in the case of the usual norm on Xα=0, in the case where α0. We take the following assumption.

(H2) BtLXβX for some 0<β<1, a.e t0 and Btxbtxβ for xXβ, with bLlocq0 where q>1/1β.

Theorem 4.1. Assume that (V1)–(V3) and (H2) hold. Then for any a>0, there exists a positive constant M=Ma such that for xX we have

AαRt+hxRhRtxM0hdssαx for 0h<ta.

Proof. Let a>0 and xX. Then

ddtRt+hx=ARt+hx+0t+hBt+hsRsxds=ARt+hx+0tBtsRs+hxds+tt+hBt+hsRsxds.

We deduce that Rt+hx satisfies the equation of the form

ddtyt=Ayt+0tBtsysds+ft.

Then by the variation o constante formula, it follows that

Rt+hx=RtRhx+0tRtsss+hBs+huRuxduds=RhRtx+0hRhs0tBuRs+tuxduds.

Which yields that

Rt+hxRhRtx=0hRhs0tBuRs+tuxduds.

Taking the αnorm, we obtain that

AαRt+hxRhRtxNα0h1hsα0tBuRs+tuxdudsNα0h1hsα0tbuAβRt+suxdudsNαNβ0hdshsα0tbutuβxdu.

Let p be such that 1/q+1/p=1, so p<1/β. Then it follows that

AαRt+hxRhRtxNαNβbLq0auβLp0a0hdssαx.

And the proof is complete.

The local existence result is given by the following Theorem.

Theorem 4.2. Suppose that (V1)–(V3), (H0), and (H2) hold. Moreover, assume that F defined from J×Ω into X is continuous where J×Ω is an open set in R+×Cα. Then for each φΩ, Eq. (1) has at least one mild solution which is defined on some interval 0b.

Proof. Let φΩ. For any real ζJ and p>0, we define the following sets:

Iζ=t:0tζandHp=ϕCα:ϕαp.

For ϕHp, we choose ζ and p such that tϕ+φIζ×Hp and HpΩ. By continuity of F, there exists N10 such that Ftϕ+φN1 for tϕ in Iζ×Hp. We consider φ¯CrζYα as the function defined by φ¯t=Rtφ0 for tIζ and φ¯0=φ. Suppose that p¯<p and choose 0<b<ζ such that

NαN10bdssα<p¯andφ¯tφαpp¯ for tIb.

Let K0=ηCrbYα:η0=0and ηtαp¯ for 0tb. Then we have Ftφ¯t+ηtN1 for 0tb and ηK0, since ηt+φ¯tφαp. Consider the mapping S from K0 to CrbYα defined by 0=0

t=0tRtsFsηs+φ¯sds for 0tb.E13

Notice that finding a fixed point of S in K0 is equivalent to finding a mild solution of Eq. (1) in K0. Furthermore, S is a mapping from K0 to K0, since if ηK0 we have 0=0 and

tα0tAαRtsF(sηs+φ¯s)ds.

Then

tαNαN10tdstsαNαN10bdssα<p¯

which implies that SK0K0. We claim that t:ηK0} is compact in Yα for fixed trb. In fact, let β be such that 0<αβ<1. The above estimate show that Aβt:ηK0 is bounded in X. Since Aαβ is compact operator, we infer that AαβAβt:ηK0 is compact in X, hence t:ηK0 is compact in Yα. Next, we show that t:ηK0 is equicontinuous. The equicontinuity of t:ηK0 at t=0 follows from the above estimation of t. Now let 0<t0<tb with t0 be fixed. Then we have

Aαtt00t0AαRtsRtt0Rt0sFsηs+φ¯sds+AαRtt0I0t0Rt0sF(sηs+φ¯s)ds+t0tAαRtsF(sηs+φ¯s)ds.E14

Using Theorem 4.1, it follows that

Aαtt0t0N1M0tt0dssα+Rtt0IAα0t0Rt0sF(sηs+φ¯s)ds+ NαN10tt01sαds.

As the set t0:ηK0 is compact in Yα, we have

limtt0+tt0α=0uniformlyinηK0.

We obtain the same results by taking t0 be fixed with 0<t<t0b. Then we claim that limtt0tt0α=0 uniformly in ηK0 which means that t:ηK0 is equicontinuous. Then by Ascoli-Arzela theorem, :ηK0 is relatively compact in K0. Finally, we prove that S is continuous. Since F is continuous, given ε>0, there exists δ>0, such that

sup0sbηsη̂sα<δ implies that Fsηs+φ¯sF(sη̂s+φ¯s)<ε.

Then for 0tb, we have

tSη̂tαNα0t1tsαFsηs+φ¯sF(sη̂s+φ¯s)dsNαε0tdssα.

This yields the continuity of S, and using Schauder’s fixed point theorem, we deduce that S has a fixed point. Then the proof of the theorem is complete.

The following result gives the blowing up phenomena of the mild solution in finite times.

Theorem 4.3. Assume that (V1)–(V3), (H0), and (H2) hold and F is a continuous and bounded mapping. Then for each φCα, Eq. (1) has a mild solution u.φ on a maximal interval of existence rbφ. Moreover if bφ<, then lim¯tbφutφα=+.

Proof. Let u.φ be the mild solution of Eq. (1) defined on 0b. Similar arguments used in the local existence result can be used for the existence of b1>b and a function u.ub defined from bb1 to Yα satisfying

utub.φ=Rtubφ+btRtsFsusds for tbb1.

By a similar proceeding, we show that the mild solution u.φ can be extended to a maximal interval of existence rbφ. Assume that bφ<+ and lim¯tbφutφα<+. There exists N2>0 such that FsusN2, for s0bφ. We claim that u.φ is uniformly continuous. In fact, let 0<htt+h<bφ. Then

ut+hut=Rt+hRtφ0+0tRt+hsRtsFsusds+tt+hRt+hsFsusds.

By continuity of AαRt, we claim that AαRt+hRtφ0 is uniformly continuous on each compact set. Moreover, Theorem 4.1 implies that AαRt+hsRtsFsus0 uniformly in t when h0. In fact we have

0tRt+hsRtsFsusαds0tRt+hsRhRtsFsusαds+RhIAα0tRtsF(sus)ds

Then using Theorem 4.1, we obtain that

0tRt+hsRtsFsusαdsbφN2M0hdssα+RhIAα0tRtsF(sus)ds.

We claim that the set {Aα0tRtsFsusds:t[0,bφ)} is relatively compact. In fact, let tnn0 be a sequence of 0bφ. Then there exist a subsequence tnkk and a real number t0 such that tnkt0. Using the dominated convergence theorem, we deduce that

0tnkAαRtnksFsusds0t0AαRt0sFsusds.

This implies that {Aα0tRtsFsusds:t[0,bφ)} is relatively compact. Now using Banach-Steinhaus’ theorem, we deduce that

RhIAα0tRtsFsusds0

uniformly when h0 with respect to t0bφ. Moreover we have

tt+hRt+hsFsusdsαN2Nα0hdssα.

Consequently ut+hutα0ash0 uniformly in t0bφ. If h0, that is, for tt0, we have

utut0=RtRt0φ00tRt0sRt0tRtsFsusdsRt0tI0tRtsFsusdstt0Rt0sFsusds,

one can show similar results by using the same reasoning. This implies that u.φ is uniformly continuous. Therefore limtbφutφexistsinYα. And consequently, u.φ can be extended to bφ which contradicts the maximality of 0bφ.

The next result gives the global existence of the mild solutions under weak conditions of F. To achieve our goal, we introduce a following necessary result which is a consequence of Lemma 7.1.1 given in ([21], p. 197, Exo 4).

Lemma 4.4. [21] Let α,a,b0,β<1 and 0<d<. Also assume that v is nonnegative and locally integrable on 0d with

vtatα+b0tvstsβds for t0d.

Then there exists a constant M2=M2abαβd< such that vtM2/tα on 0d.

Theorem 4.5. Assume that (V1)–(V3), (H0), and (H2) hold and F is a completely continuous function on R+×Cα. Moreover suppose that there exist continuous nonnegative functions f1 and f2 such that Ftφf1tφα+f2t for φCα and t0. Then Eq. (1) has a mild solution which is defined for t0.

Proof. Let 0bφ be the maximal interval of existence of a mild solution u.φ. Assume that bφ<+. By Theorem 4.3 we have lim¯ttφutφα=+. Recall that the solution of Eq. (1) is given by u0=φ and

utφ=Rtφ0+0tRtsFsus.φds for t0bφ.

Then taking the αnorm, we obtain

utφαRtφ0α+k2Nα0bφdssαds+k1Nα0t1tsαus.φαds,

where k1=max0tbφf1t and k2=max0tbφf2t. Then we deduce that

utφαNφ0α+k1Nα0t1tsαsuprτsuτφαds+k2Nα0bφdssα.E15

Now we claim that the function

t0t1tsαsuprτsuτφαds,

is nondecreasing. In fact, let 0t1t2. Then

0t11t1sαsuprτsuτφαds=0t11sαsuprτt1suτφαds0t21sαsuprτt2suτφαds=0t21t2sαsuprτsuτφαds

which yields the result. Then it follows from Eq. (15) that

suprstusφαNφ0α+k2Nα0bφdssαds+k1Nα0t1tsαsuprτsuτφαds.

Then using Lemma 4.4, we deduce that u.φ is bounded in 0bφ. Then we obtain that lim¯tbφutφα<, which contradicts our hypothesis. Then the mild solution is global.

We focus now to the compactness of the flow defined by the mild solutions.

Theorem 4.6. Assume that (V1)–(V3) and (H0)–(H2) hold. Then the flow Ut defined from Cα to Cα by Utφ=ut.φ is compact for t>r, where ut.φ denotes the mild solution starting from φ.

Proof. We use Ascoli-Arzela’s theorem. Let E=φγ:γΓ be a bounded subset of Cα and let t>r be fixed, but arbitrary. We will prove that UtE¯ is compact. It follows from (H1) and inequality Eq. (7) that there exists N5 such that

F(tutφγ)N2utφγ)+Ft0=N5 for γΓ.

For each γΓ, we define fγCα by fγ=ut.φγ. We show now that for fixed θr0, the set fγθ:γΓ is precompact in Yα. For any γΓ, we have

fγθ=Rt+θφγ0+0t+θRt+θsFsus.φds.

As Rt is compact for t>0, we need only to prove that the set

0t+θRt+θsF(sus(.φγ))ds:γΓ

is compact. Also we have

μRε0t+θεRt+θεsF(sus(.φγ))ds:γΓ=0,

where μ is the measure of non-compactness. Moreover, using Theorem 4.1, we have

Aα0t+θεRt+θεsRεRt+θεsF(sus(.φγ))ds0t+θεRt+θsRεRt+θεsF(sus(.φ))αdsN5M0εdssα0asε0.

We deduce that

μ0t+θεRt+θsF(sus(.φγ))ds:γΓ=0.

On the other hand, for 0<αβ<1, we have

Aβt+θεt+θRt+θsF(sus(.φγ))dst+θεt+θRt+θsF(sus(.φγ))βdsNβN5t+θεt+θdst+θsβ=NβN50εdssβ0asε0.

Thus Aβt+θεt+θRt+θsF(sus(.φγ))ds:γΓ is a bounded subset of X. The precompactness in Yα now follows from the compactness of Aβ:XYα. Then the set UtEθ:rθ0 is precompacted in Yα. We prove that the family fγ:γΓ is equicontinuous. Let γ in Γ,0<ε<tr, and rθ̂θ0 with θ̂ be fixed and h=θθ̂. Then

Aαfγh+θ̂fγθ̂Rt+θ̂+hRt+θ̂φγ0α+0t+θ̂AαRt+θ̂+hsRhRt+θ̂sF(sus.φγds+RhIAα0t+θ̂Rt+θ̂sF(sus(.φγ))ds+t+θ̂t+θ̂+hAαRt+θ̂+hsF(sus(.φγ))ds.

Then it follows that

Aαfγh+θ̂fγθ̂Rt+θ̂+hRt+θ̂Aαφγ0+MN5t+θ̂0hdssα+RhIAα0t+θ̂Rt+θ̂sF(sus(.φγ))ds+N5Nα0hdssα.

Using the compactness of the set Aα0t+θRt+θsF(sus(.φγ))ds:γΓ and the continuity of tRtx for xX, the right side of the above inequality can be made sufficiently small for h>0 small enough. Then we conclude that fγ:γΓ is equicontinuous. Consequently, by Ascoli-Arzela’s theorem, we conclude that the set Utφ:φE is compact, which means that the operator Ut is compact for t>r.

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5. Regularity of the mild solutions

We define the set Cα1 by Cα1=C1r0Yα as the set of continuously differentiable functions from r0 to Yα. We assume the following hypothesis.

(H3) F is continuously differentiable, and the partial derivatives DtF and DφF are locally Lipschitz in the classical sense with respect to the second argument.

Theorem 5.1. Assume that (V1)–(V3), (H1), and (H3) hold. Let φ in Cα1 be such that φ0Y and φ̇0=0+F0φ. Then the corresponding mild solution u becomes a strict solution of Eq. (1).

Proof. Let a>0. Take φCα1 such that φ0Y and φ̇0=0+F0φ, and let u be the mild solution of Eq. (1) which is defined on [0,+[. Consider the following equation:

vt=Rtφ̇0+0tRtsDtFsus+DφF(sus)vsds+0tRtsBsφ0ds for t0,v0=φ̇.E16

Using the strict contraction principle, we can show that there exists a unique continuous function v solution in 0a of Eq. (16). We introduce the function w defined by

wt=φ0+0tvsdsift0,φtifrt0.

Then it follows

wt=φ+0tvs ds for t0a.

Consequently, the maps twt and t0tRtsFswsds are continuously differentiable, and the following formula holds

ddt0tRtsFswsds=RtF0w0+0tRtsDtFsws+DφFswsvsds=RtF0φ+0tRtsDtFsws+DφFswsvsds.

This implies that

0tRsF0φds=0tRtsFswsds0t0sRsτDtFτwτ+DφFτwτvτds.

On the other hand, from equality Eq. (4), we have

0tRs0ds=Rtφ0φ00t0sRsτBτφ0ds.

We rewrite w as follows:

wt= φ00tRs0ds+0tRsF0φds+0t0sRsτDtFτuτ+DφFτuτvτds+0t0sRsτBτφ0ds.

Then it follows that

wt= Rtφ0+0tRtsFswsds+0t0sRsτ(DτFτuτDτF(τwτ))ds+0t0sDφFτuτvτDφFτwτvτds.

We deduce, for t0a, that

utwtα0tAαRts(F(sus)F(sws))ds+0t0sAαRsτ(DτF(τuτ)DτF(τwτ))ds+0t0sAαRsτ(DφF(τuτ)DφF(τwτ))vτds.E17

The set H=usws:s0a is compact in Cα. Since the partial derivatives of F are locally Lipschitz with respect to the second argument, it is well-known that they are globally Lipschitz on H. Then we deduce that

utwtαNαha0t1tsαuswsαdsNαha0t1tsαsup0τauτwταds,

where ha=LFNα+aNαLipDtF+aNαLipDφF, with LipDφF and LipDtF the Lipschitz constant of DφF and DtF, respectively, which implies that

uwαNαha0adssαuwα.

If we choose a such that

Nαha0adssα<1,

then u=w in 0a. Now we will prove that u=w in 0+. Assume that there exists t0>0 such that ut0wt0. Let t1=inft>0:utwt>0. By continuity, one has ut=wt for tt1, and there exists ε>0 such that utwt>0 for tt1t1+ε. Then it follows that for tt1t1+ε,

utwtαNαhε0εdssαsupετt1+εuτwτα.

Now choosing ε such that

Nαhε0εdssα<1,

then u=w in t1t1+ε which gives a contradiction. Consequently, ut=wt for t0. We conclude that tut from 0+ to Yα and tFtut from 0+×Cα to X are continuously differentiable. Thus, we claim that u is a strict solution of Eq. (1) on 0+ [22, 23, 24, 25, 26, 27, 28, 29, 30, 31].

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6. Application

For illustration, we propose to study the model Eq. (2) given in the Introduction. We recall that this is defined by

twtx=2x2wtx+0thts2x2wsxds+r0gtxwt+θx for t0andx0π,wt0=wtπ=0 for t0,wθx=w0θx for θr0andx0π,E18

where w0:r0×0πR, g:R+×RR and h:R+R+ are appropriate functions. To study this equation, we choose X=L20π, with its usual norm .. We define the operator A:Y=DAXX by

Aw=w with  domainDA=H20πH010π,

and Btx=htAxX,for0,xY. For α=1/2, we define Y1/2=DA1/21/2 where x1/2=A1/2x for each xY1/2. We define C1/2 = Cr0Y1/2 equipped with norm and the functions u and φ and F by ut=wtx,φθx=w0θx for a.e x0π and θr0,t0, and finally

Ftφx=r0gtxφθx for a.ex0πandφC1/2.

Then Eq. (18) takes the abstract form

dutdt=Aut+0tBtsusds+Ftut for t0,u0=φC1/2=Cr0DA1/2],E19

The A is a closed operator and generates an analytic compact semigroup Ttt0 on X. Thus, there exists δ in 0π/2 and M0 such that Λ=λC:argλ<π2+δ0 is contained in ρA, the resolvent set of A, and RλA<M/λ for λΛ. The operator Bt is closed and for xY, BtxhtxY. The operator A has a discrete spectrum, the eigenvalues are n2, and the corresponding normalized eigenvectors are enx=2πsinnx,n=1,2,. Moreover the following formula holds:

  1. Au=n=1n2uenenuDA.

  2. A1/2u=n=11nuenen for uX.

  3. A1/2u=n=1nuenen for uDA1/2=uX:n=11nuenenX.

One also has the following result.

Lemma 6.1 [16] LetφY1/2.Thenφis absolutely continuous,φ'Xand

φ'=A12φ.

We assume the following assumptions.

(H4) The scalar function h.L10 and satisfies g1λ=1+hλ0(h the Laplace transform of h) and λg11λΛ for λΛ. Further, hλ0 as λ, for λΛ and hλ1=λn.

(H5) The function g:R+×RR is continuous and Lipschitz with respect to the second variable.

By assumption (H4), the operator ρλ=λI+g1λA1=g11λλg11λI+A1 exists as a bounded operator on X, which is analytic in Λ and satisfies ρλ<M/λ. On the other hand, for xX, we have

λx=AλI+g1λA1x=A+λg11λIλg11λIλI+g1λA1x=g11λIλg11λλg11λI+A1x.

Since λg11λλg11λI+A1 is bounded because g11λΛ, then λx has the growth properties of g11λ which tends to 1 if λ goes to infinity. Then we deduce that λLX. Moreover, it is analytic from Λ to LX. Now, for xY, one has

λx=g11λλg11λI+A1AxandBλρλx=hλρλAx.

Then it follows that

λxM/λxYandBλρλM/λxY.

We deduce that λLYX, Bλ=hλALYX, and BλρλLYX. Considering D=C00π, we see that the conditions (V1)–(V3) and (H0) are verified. Hence the homogeneous linear equation of Eq. (18) has an analytic compact resolvent operator Rtt0. The function F is continuous in the first variable from the fact that g is continuous in the first variable. Moreover from Lemma 6.1 and the continuity of g, we deduce that F is continuous with respect to the second argument. This yields the continuity of F in R+×C1/2. In addition, by assumption (H5) we deduce that

Ftφ1Ftφ2rLfφ1φ2C1/2.

Then F is a continuous globally Lipschitz function with respect to the second argument. We obtain the following important result.

Proposition 6.2. Suppose that the assumptions (H4)–(H5) hold. Then Eq. (19) has a mild solution which is defined for t0.

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Written By

Boubacar Diao, Khalil Ezzinbi and Mamadou Sy

Submitted: April 1st, 2019 Reviewed: June 17th, 2019 Published: May 13th, 2020