# General Decay of the Moore–Gibson–Thompson Equation with Viscoelastic Memory of Type II

Authors: Salah Boulaaras; Abdelbaki Choucha; Andrea Scapellato
Journal: Journal of Function Spaces
 (2022)

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## Abstract

This study deals with the general decay of solutions of a new class of Moore–Gibson–Thompson equation with respect to the memory kernel of type II. By using the energy method in the Fourier space, we establish the main results.


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## Body

## 1. Introduction and Preliminaries

In this work, we are interested in the study of the general decay of solutions of the following problem:(1)auttt+βutt−γ2Δu−bΔut+∫0tht−σΔutσdσ=0,ux,0=u0x,utx,0=u1x,uttx,0=u2x,x∈ℝn,where a,b,β>0 are physical parameters and γ is the speed of sound. The convolution term ∫0tht−sΔutsds reflects the memory type-II effect of materials due to viscoelasticity, where h is the relaxation function.Moore–Gibson–Thompson (MGT) equation is based on the modeling of high amplitude sound waves. There has been quite a bit of work in this area of research due to a wide range of applications such as medical and industrial use of high intensity ultrasound in lithotripsy, heat therapy, ultrasonic cleaning, etc. Classical models of nonlinear acoustics include the Kuznetsov equation, the Westervelt equation, and the Kokhlov–Zabulotskaya–Kuznetsov equation. An in-depth study of linear models is a good starting point for a better understanding of the approximate behaviors of nonlinear models. Indeed, even in the linear case, the works [1] have shown a rich dynamic. In [2], Marchand et al. presented a detailed analysis of this equation. Using a quasi-abstract group approach and refined spectral analysis, they establish the well posedness of the problem and define the accumulation point of eigenvalues. Kaltenbacher et al. [3] also studied the fully nonlinear version of the MGT equation and established the global well posedness and the exponential decay for the nonlinear equation under consideration.In the case of a third-order equation, the modeling of memory effects is quite complex. The memory term may affectu, or ut, or even a combination of both. Accordingly, the corresponding models are called as memory of type I, type II, and type III. This classification raises a fundamental question on the impact of the memory terms on the stability properties of the corresponding evolutions. We mention that the memory has some stabilizing effects (see, for instance, [4, 5]). On the contrary, a quantification of the stability results via decay rates shows that the memory may destroy the stability properties of a stable system.In this paper, we are interested in studying the dynamics that results from the memory kernel of type II (see [6]) in our problem (1).Natural questions can be asked based on the study of the viscoelastic memory and integral condition system (see [7–10]): could the addition of the memory kernel of type II harm the stability of this kind of problem in any way? If the answer to the question in the wave condition with friction damping are relatively simple, it is not easy to answer in the case of memory kernel of type II that we present below, especially in Fourier space. This work is part of the effort to understand this the MGT problem and memory kernel of type II.The coupling between the Fourier law of heat condition and different systems has been considered by many authors and there are many results. For example, Bresse system (Bresse–Fourier) has been studied in [11], the Bresse system coupled with the Cattaneo law of heat condition (Bresse–Cattaneo) has been studied in [12], and Timoshenko system with past history has been considered in [13]. For further details, we refer the reader to the following papers [14, 15].We mention also that several results related to viscoelasticity have been obtained by using the theory of Lie symmetries. Precisely, thanks to the symmetry reduction obtained by means of the classical Lie symmetries, it was possible to obtain exact solutions of considerable interest. For further details in this direction, we refer the reader to [16].Based on all last mentioned works, especially [6, 10], we would like to prove the general decay result in the Fourier space to problem (1). To the best of our knowledge, this is one of the first works that deals with the MGT problem with viscoelastic memory kernel of type II in the Fourier space.The paper has the following structure. In Section1, we put our assumptions and preliminaries that will be employed in our main decay result. In Section 2, by using the energy method in the Fourier space, we construct the Lyapunov functional and find the estimate for the Fourier image. Section 3 is devoted to the conclusion.To prove our main result, we need the following hypotheses and lemmas.(H1)h∈C1ℝ+,ℝ+∩Cℝ+,ℝ+ is a nonincreasing function; there exists a positive constant κ satisfying(2)h′t≤−κht,t≥0.(H2) There exist positive numbersθ and λ∈γ2/b,β/a such that λ/θ<κ and(3)H∞≤b−γ2λmin2λ2+θ,1,where H∞=∫0∞hsds>0 and Ht=∫0thsds.(H3)(4)β−γ2ab>0.Throughout this paper, we usec,C,Ci,i=1,2, to denote a generic positive constant.Lemma 1.
Suppose that (3) holds. Then, for allω∈Lloc2ℝ+,ℂ, we have(5)∫0tht−σωt−ωσdσ2≤ch∘ωt,∀t≥0,and(6)∫0th′t−σωt−ωσdσ2≤−h0h′∘ωt,∀t≥0,where(7)h∘ωt≔∫0tht−σωt−ωσ2dσ.Lemma 2.
For anyk≥0andc>0, there exist a constantC>0such that, for allt≥0, the following estimate holds:(8)∫ξ2≤1ξke−cξ2tdξ≤C1+t−k+n/2,ξ∈ℝn.


## 2. The Energy Method in the Fourier Space

In this section, we get the decay estimate of the Fourier image of the solution for system (1). By using Plancherel’s theorem together with some integral estimates such as (2), this method will allow us to give the decay rate of the solution in the energy space. To this goal, we use the energy method in the Fourier space and construct appropriate Lyapunov functionals. Finally, we prove our main result.Applying the Fourier transform to (1), we get the following problem:(9)au^ttt+βu^tt+γ2ξ2u^+bξ2u^t−ξ2∫0tht−σu^tσdσ=0,ξ∈ℝn,t>0u^ξ,0=u^0ξ,u^tξ,0=u^1ξ,u^ttξ,0=u^2ξ.Lemma 3.
Suppose that (3)–(4) hold. Letu^ξ,tbe the solution of (9). Then, the energy functionalE^t, given by(10)E^ξ,t=E^t=a2u^tt+λu^t2+ξ2γ22λu^t+λu^2+ξ2b−γ2λ−Htu^t2+λ2β−aλu^t2+12ξ2h∘u^tt,satisfies(11)E^′t≤−ξ2λC1u^t2+C2ξ2h′∘u^tt,where C1=b−γ2/λ−Ht1+θ/2>0, C2=κ0/2κ, and κ0=κ−λ/θ>0.Proof.
Firstly, multiplying (9) by u^¯tt and taking the real part, we obtain(12)Reau^tttu^¯tt+βu^ttu^¯tt+γ2ξ2u^u^¯tt+bξ2u^tu^¯tt−ξ2∫0tht−σu^tσu^¯ttdσ=ddtE^1t+R^1t=0.
Then, we have(13)aReu^tttu^¯tt=a2ddtu^tt2,βReu^ttu^¯tt=βu^tt2,bξ2Reu^tu^¯tt=b2ξ2ddtu^t2γ2ξ2Reu^u^¯tt=γ2ξ2Reddtu^u^¯t−γ2ξ2u^t2,and(14)−ξ2Re∫0tht−σu^tσu^¯ttdσ=ξ2Re∫0tht−σu^tt−u^tσu^¯ttdσ−ξ2HtReu^ttu^¯tt=ξ212ddt∫0tht−σu^tt−u^tσ2dσ−Ht2ξ2ddtu^t2−ξ212∫0th′t−σu^tt−u^tσ2dσ=ξ212ddth∘u^tt−ξ212h′∘u^tt−Ht2ξ2ddtu^t2=ξ212ddth∘u^tt−ξ212h′∘u^tt−12ξ2ddtHtu^t2+12ξ2htu^t2.
Substituting (13) and (14) into (12), we obtain(15)E^1t=a2u^tt2+b2ξ2u^t2+γ2ξ2Reu^u^¯t−Ht2ξ2u^t2+ξ212h°u^ttR^1t=βu^tt2−γ2ξ2u^t2−ξ212h′∘u^tt+12ξ2htu^t2.
Secondly, multiplying (9) by u^¯t and taking the real part, we obtain(16)Reau^tttu^¯t+βu^ttu^¯t+γ2ξ2u^u^¯t+bξ2u^tu^¯t−ξ2∫0tht−σu^tσu^¯tdσ=ddtE^2t+R^2t=0.
We have(17)aReu^tttu^¯t=aReddtu^ttu^¯t−au^tt2,βReu^ttu^¯t=β2ddtu^t2,bξ2Reu^tu^¯t=bξ2u^t2,γ2ξ2Reu^u^¯t=γ22ξ2ddtu^2,and(18)−ξ2Re∫0tht−σu^tσu^¯tdσ=ξ2Re∫0tht−σu^tt−u^tσu^¯tdσ−ξ2Htu^t2.
Substituting (17) and (18) into (16), we obtain(19)E^2t=aReu^ttu^¯t+β2u^t2+γ22ξ2u^2,R^2t=−au^tt2+bξ2u^t2+|ξ|2Re∫0tht−σu^tt−u^tσu^¯tdσ−ξ2Htu^t2.
At this point, by (4), we have β−γ2a/b>0, and by (3), we select λ such that γ2/b<λ<β/a.
LetE^t be the energy functional. Then, we have(20)ddtE^t+R^t=0,E^t=E^1t+λE^2t,R^t=R^1t+λR^2t.
By (15) and (19), we have(21)E^t=E^1t+λE^2t=a2u^tt2+b2ξ2u^t2+γ2ξ2Reu^u^¯t−Ht2ξ2u^t2+ξ212h∘u^tt+aλReu^ttu^¯t+β2λu^t2+γ2λ2ξ2u^2=a2u^tt+λu^t2+12ξ2b−γ2λ−Htu^t2+γ22λξ2u^t+λu^2+λ2β−aλu^t2+ξ212h∘u^tt,(22)R^t=R^1t+λR^2t=β−aλu^tt2−γ2ξ2u^t2−ξ212h′∘u^tt+12ξ2htu^t2+bλξ2u^t2+λξ2Re∫0tht−σu^tt−u^tσu^¯tdσ−λξ2Htu^t2.
By (3), we have β−aλ>0, and using Young’s inequality, we obtain(23)R^t=λξ2b−γ2λ−Htu^t2−ξ212h′∘u^tt−12ξ2htu^t2+λξ2Re∫0tht−σu^tt−u^tσu^¯tdσ≥λξ2b−γ2λ−Htu^t2−ξ212h′∘u^tt−λξ2Htθ2u^t2−ξ2λ2θh′∘u^tt=λξ2b−γ2λ−Ht1+θ2u^t2−ξ212h∘u^tt−ξ2λ2θh∘u^tt.
By using the fact thatλ/θ<κ and under supposition (2), we have(24)h′t≤−κht⇒κht≤−h′t.
Thus, (23) becomes(25)R^t≥λξ2b−γ2λ−Ht1+θ2u^t2−ξ2κ02h∘u^tt,where κ0=κ−λ/θ>0,
Finally, by settingC1=b−γ2/λ−Ht1+θ/2>0 and C2=κ0/2κ, we find (11).Now, to achieve our goal, we need the following lemmas.Lemma 4.
Suppose that (3)-(4) hold. Then, the functional(26)D1t≔aReu^tt+λu^tu^¯t+λu^¯+β−aλReu^tu^¯t+λu^¯,satisfies(27)D1′t≤−γ22λξ2u^t+λu^2+c1+ξ2u^t2+cu^tt+λu^t2+cξ2h°u^tt.Proof.
DifferentiatingD1 and by using (9), we obtain(28)D1′t=aReu^ttt+λu^ttu^¯t+λu^¯+au^tt+λu^t2+β−aλReu^ttu^¯t+λu^¯+β−aλReu^tu^¯tt+λu^¯t=β−aλReu^tu^¯tt+λu^¯t︸I1+au^tt+λu^t2+ξ2Re−bu^tu^¯t+λu^¯︸I2+ξ2Re−γ2u^u^¯t+λu^¯︸I3+ξ2Re∫0tht−σu^tσdσu^¯t+λu^¯︸I4.
In what follows, we estimate the termsIi, i=1,…,4, that appear in the right-hand side of (28); using Young’s inequality, we obtain(29)I1≤β−aλ2u^t2+β−aλ2u^tt+λu^t2,(30)I2+I3=−bξ2Reu^tu^¯t+λu^¯−γ2λξ2Reλu^u^¯t+λu^¯=−γ2λξ2u^t+λu^2−b−γ2λξ2Reu^tu^¯t+λu^¯≤−γ2λξ2u^t+λu^2+b−γ2λ2ξ214ε1u^t2+ε1ξ2u^t+λu^2,(31)I4=−ξ2Re∫0tht−σu^tσ−u^tσdσu^¯t+λu^¯+ξ2HtReu^tu^¯t+λu^¯≤c4ε1ξ2h°u^tt+H2t4ε1ξ2u^t2+2ε1ξ2u^t+λu^2.
By lettingε1=γ2/6λ, we obtain (27).Lemma 5.
The function,(32)D2t≔−aReu^tt+λu^t∫0tht−σu^¯tt−u^¯tσdσ,satisfies, for any ε,ε2>0,(33)D2′t≤−aHt−3ε2u^tt+λu^t2+c1+ξ2u^t2+εu^t+λu^2+c1+1ε2+1ε1+ξ2h∘u^tt−cε2h′∘u^tt.Proof.
DifferentiatingD2 and by using (9), we obtain(34)D2′t=−aReu^ttt+λu^tt∫0tht−σu^¯tt−u^¯tσdσ−aReu^tt+λu^t∫0tht−σu^¯tt−u^¯tσdσ−aHtReu^tt+λu^tu^¯ttt=−β−aλReu^tt∫0th′t−σu^¯tt−u^¯tσdσ︸J1+ξ2Re−bu^t∫0tht−σu^¯tt−u^¯tσdσ︸J2+ξ2Re−γ2u^∫0tht−σu^¯tt−u^¯tσdσ︸J3+ξ2Re∫0tht−σu^tσdσ∫0tht−σu^¯tt−u^¯tσdσ︸J4−aReu^tt+λu^t∫0th′t−σu^¯tt−u^¯tσdσ︸J5−aHtu^tt+λu^t2+aHtλReu^tt+λu^tu^¯t︸J6.
Similarly, we estimate the termsJi, i=1,…,6, that appear in the right-hand side of (34). Using Young’s inequality, we obtain(35)J1=−β−aλReu^tt+λu^t∫0tht−σu^¯tt−u^¯tσdσ+β−aλλReu^t∫0tht−σu^¯tt−u^¯tσdσ≤β−aλ2c4ε2+β−aλc2h∘u^tt+ε2u^tt+λu^t2+β−aλ2u^t2≤c1+1ε2h∘u^tt+ε2u^tt+λu^t2+cu^t2,(36)J2+J3=−γ2λξ2Reu^t+λu^∫0tht−σu^¯tt−u^¯tσdσ−b−γ2λξ2Reu^t∫0tht−σu^¯tt−u^¯tσdσ≤εξ2u^t+λu^2+c1+1εξ2h∘u^tt+b−γ2λ2ξ212u^t2,and(37)J4=−ξ2Re∫0tht−σu^tσ−u^tσdσ∫0tht−σu^¯tt−u^¯tσdσ+ξ2HtReu^t∫0tht−σu^¯tt−u^¯tσdσ≤cξ2h∘u^tt+H2t2ξ2u^t2.
Similarly, we have(38)J5=−aReu^tt+λu^t∫0th′t−σu^¯tt−u^¯tσdσ≤ε2u^tt+λu^t2−c4ε2h′∘u^tt,and(39)J6=aHtλReu^tt+λu^tu^¯t≤ε2u^tt+λu^t2+aHtλ24ε2u^t2.By substituting (35)–(39) into (34), we obtain (33).At this stage, we define the functional(40)Kt≔N1+ξ2E^t+N1ξ2D1t+N2ξ2D2t,where N,N1, and N2 are positive constants to be properly chosen later.Lemma 6.
There existμi,t0>0,i=1,…,4, such that the functionalKtgiven by (40) satisfies(41)K′t≤−μ1ξ2E^t+μ21+ξ22h°∇ut,t>t0,and(42)μ31+ξ2E^t≤Kt≤μ41+ξ2E^t.Proof.
Since the functionh is a positive and continuous, for all t0>0, we have(43)Ht=∫0thσdσ≥∫0t0hσdσ≔h0,∀t≥t0.
Firstly, by differentiating (40) and using (11), (27), and (33), we have(44)K′t≤−ξ2N2aHt−3ε2−cN1u^tt+λu^t2−ξ2γ22λξ2N1−εN2u^t+λu^2−ξ21+ξ2C1N−cN1−cN2u^t2+ξ21+ξ2cN1+c1+1ε2+1εN2h∘u^tt+ξ21+ξ2C2N−cε2N2h′∘u^tt.
By setting(45)ε=γ2ξ2N14λN2,ε2=ah06,we obtain(46)K′t≤−ξ2N2ah02−cN1u^tt+λu^t2.−ξ2γ24λξ2N1u^t+λu^2−ξ21+ξ2C1N−cN1−cN2u^t2+1+ξ22cN1+cN2h°u^tt+ξ21+ξ2C2N−cN2h′°u^tt.
Next, we fixedN1 and chose N2 large enough such that(47)N2ah02−cN1>0.
Hence, we arrive at(48)K′t≤−ξ2α0u^tt+λu^t2−ξ4α1u^t+λu^2−ξ21+ξ2C1N−cu^t2+c1+ξ22h∘u^tt+ξ21+ξ2C2N−ch′∘u^tt.
Secondly, we have(49)Kt−N1+ξ2E^t=N1ξ2D1t+N2ξ2D2t≤aN1ξ2Reu^tt+λu^tu^¯t+λu^¯+N1ξ2β−aλReu^tu^¯t+λu^¯+aN2ξ2Reu^tt+λu^t∫0tht−σu^¯tt−u^¯tσdσ.
Using Young’s inequality and Lemma1, we find(50)Kt−N1+ξ2E^t≤12N1ξ2u^tt+λu^t2+12N1ξ2u^t+λu^2+12N1ξ2u^t2+12N1ξ2u^t+λu^2+a2N2ξ2u^tt+λu^t2+c2N2h∘u^tt≤cξ2u^tt+λu^t2+u^t+λu^2+u^t2+h∘u^tt≤c1+ξ2u^tt+λu^t2+ξ2u^t+λu^2+1+ξ2u^t2+ξ2h∘u^tt≤c1+ξ2E^t.
Hence, we obtain(51)N−c1+ξ2E^t≤Kt≤N+c1+ξ2E^t.
Now, we chooseN large enough such that(52)N−c>0,C1N−c>0,C2N−c>0,and exploiting (10), estimates (48) and (51), respectively, give(53)K′t≤−μ1ξ2E^t+μ21+ξ22h∘u^tt,∀t≥t0.and(54)μ31+ξ2E^t≤Kt≤μ41+ξ2E^t.for some μi>0,i=1,…,4.Theorem 1.
Suppose (3)-(4) hold. Then, there exist positive constantsd1andd2such that the energy functional given by (10) satisfies(55)E^t≤d1E^0e−d2ρξt,∀t≥0,where ρξ=ξ2/1+ξ2.Proof.
From (53) and (2), we have(56)K′t≤−μ1ξ2E^t−μ2κ1+ξ22h′°u^tt≤−μ1ξ2E^t−c1+ξ22E^′t≤−μ1ξ2E^t−c1+ξ2E^′t,∀t≥t0.
Hence,(57)Kt+c1+ξ2E^t′≤−μ1ξ2E^t,∀t≥t0.
By exploiting (54), it can easily be shown that(58)K1t≔Kt+c1+ξ2E^t∼1+ξ2E^t.
Consequently, for somed2>0, we find(59)K1′t≤−d2ξ21+ξ2K1t,∀t≥t0.
By integrating of (59) over t0,t, we obtain(60)K1t≤K1t0e−d2ξ2/1+ξ2t−t0,∀t≥t0.
Hence, by invoking (54), (58), and (60), the continuity, and the boundedness of E, we establish (55).Now, we state and prove the following result.Theorem 2.
Letkbe a nonnegative integer, and suppose thatU^ξ,02is bounded. Then,U=utt+λut,∇ut+λu,∇utTsatisfies; for allt≥0, the decay estimate is(61)∇kUξ,t2≤CU011+t−n/4−k/2+Ce−ct∇kU02.Proof.
From (10), let(62)E^∗t=E^∗ξ,t≔12u^tt+λu^t2+ξ2u^t+λu^2+1+ξ2u^t2,and we have U^ξ,t2∼E^∗t and E^∗t≤cE^t,∀t≥0; then, by applying the Plancherel theorem and exploiting (55), we find(63)∇kUξ,t22=∫ℝnξ2kU^ξ,t2dξ≤∫ℝnξ2kE^∗ξ,tdξ≤c∫ℝnξ2ke−d2ρξtU^ξ,02dξ≤c∫ξ2≤1ξ2ke−d2ρξtU^ξ,02dξ︸M1+c∫ξ2≥1ξ2ke−d2ρξtU^ξ,02dξ︸M2.
Now, we estimateM1,M2, the low-frequency part ξ2≤1, and the high-frequency part ξ2≥1.
Firstly, we haveρξ≥1/2ξ2, for ξ2≤1. Then,(64)M1≤c∫ξ2≤1ξ2ke−d2/2ξ2tU^ξ,02dξ≤csupξ2≤1U^ξ,02∫ξ2≤1ξ2ke−d2/2ξ2tdξ,and, by using Lemma 2, we obtain(65)M1≤csupξ2≤1U^ξ,021+t−k−n/2≤cU0121+t−k−n/2.
Secondly, we haveρξ≥1/2, for ξ2≥1. Then,(66)M2≤c∫ξ2≥1ξ2ke−d2/2tU^ξ,02dξ≤ce−d2/2t∇kUx,022,∀t≥0.
Substituting (65) and (66) into (63), we find (61).


## 3. Conclusion

The aim of this work is the study of the general decay estimate of solutions of a new class of Moore–Gibson–Thompson (MGT) equation with respect to the memory kernel of type II by using the energy method in Fourier space. MGT equation is a nonlinear partial differential equation that arises in hydrodynamics and some physical applications. In this paper, we examined the different mechanism resulting from the memory kernel of type II (see [6]), which dictates the emergence of the memory term in the system in the framework of Fourier space.In the next work, we will try to use the same method with MGT equations, but in light of a generalization of assumption (3). Precisely, we will assume that κ is a positive number and ϑ is a function that fulfills the following conditions:(67)∃κ>0,0<κ≤ϑt≤ϑ0,h′t≤−ϑtht,∀t≥0.Also, we will study the same problem, but with the memory of the type III. In the future, we will try to use the same method with Boussinesq and Hall-MHD equations which are nonlinear partial differential equations that arises in hydrodynamics and some physical applications (see, for example, [17–19]).



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