# Pointwise Estimates of Solutions for the Viscous Cahn-Hilliard Equation with Inertial Term

Authors: Nianying Li; Li Yin; Honglian You
Journal: Journal of Function Spaces
 (2020)

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## Abstract

In this paper, we study the pointwise estimates of solutions to the viscous Cahn-Hilliard equation with the inertial term in multidimensions. We use Green’s function method. Our approach is based on a detailed analysis on the Green’s function of the linear system. And we get the solution’sLp convergence rate.


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## Body

## 1. Introduction

In this paper, we study the pointwise estimates of the solutionρx,t to the Cauchy problem:
(1)ηρtt+ρt+Δ2ρ−kΔρt−Δfρ=0,x,t∈ℝn×0,∞,ρx,tt=0=ρ0x,ρtx,tt=0=ρ1x,where n≥4. fρ is the intrinsic chemical potential which is smooth in the small neighborhood of the origin, and fρ=Oρ1+θ when ρ≤1 and θ is a positive integer. When η=0, Eq. (1) is the well-known Cahn-Hilliard equation. When η≠0, ηρtt is the inertial term. When k≠0, −kΔρt is the viscous term. Without loss of generality, we let η=1 and k=1.The classical Cahn-Hilliard equation was proposed in the sixties by Cahn and Hilliard which describes the phase separation in materials science, and it has been widely studied. The reader may see references ([1–6]) and the related references therein. The Cahn-Hilliard equations with inertial term model nonequilibrium decompositions caused by deep supercooling in certain glasses. As we know, the well-known Cahn-Hilliard equation is a parabolic equation, but the Cahn-Hilliard equation with the inertial term is a hyperbolic equation with relaxation which brings many mathematical difficulties to study. For which, without smallness assumption on initial data, [7] got the global existence of the classical solution. [8] obtained the global existence and the optimal decay rate of the classical solution by the Fourier splitting method. Wang and Wu [9] obtained the global existence and optimal decay rate of the classical solution by long wave-short wave method. Li and Mi [10] got the pointwise estimates and the Lp (1≤p≤∞) convergence rate of the solution by Green’s function method. Some other works on the Cahn-Hilliard equation with the inertial term can be seen in [11–13].For viscous Cahn-Hilliard equation, [14] discussed the large time behavior of solutions when the dimension n≤5. For the viscous Cahn-Hilliard equation with the inertial term, it describes the early stages of spinodal decomposition in certain glasses (see [15–16]). And for which, [17] established the existence of families of exponential attractors and inertial manifolds; [18] studied the long time dynamic of the system in three-dimensional. In this paper, we are interested in the viscous Cahn-Hilliard equation with the inertial term. Under the smallness assumption on initial data, based on the detailed analysis of the Green’s function, we get the pointwise estimates of solutions. From the representation of the symbol value to the Green’s function for the linear problem of Eq. (1), we also find that the decay rate mainly depends on the lower-frequency part, i.e., the long wave part. It is shown that the solution’s decay rate is the same as [10]. Our study bases on Section 4 in [9].To the best of our knowledge, this is the first time to obtain the pointwise estimates of the solution to Eq. (1).Throughout this paper, C denotes the generic positive constants.Wm,p=Wm,pℝnm∈ℤ+,p∈1,∞ denote the usual Lebesgue space with norms ·Lp and the usual Sobolev space with its norm
(2)fWm,p=∑∣α∣=0m∂xαfLp.In particular, we useWm,2=Hm.The main result can be stated as following Theorem1:Theorem 1.
Ifρ0Hs+1∩W1,1+ρ1Hs∩L1≤ε, s≥maxn/2+5,2n, and for any multi-index β, β<s−n/2, there exists a constant d>n/2, such that
(3)Dxβρ0x+Dxβ+2ρ0x+Dxβρ1x≤CE1+x4−d,then for ∣β∣<n, the solution to Eq. (1) has the following estimates:
(4)Dxβρx,t≤C1+t−n+β/4Bdx2,t,where ε and E is sufficiently small positive constants, m=maxa∣a∈ℤ,a≤m, Bdx,t=1+x2/1+t−d.Corollary 2.
Under the assumptions of Theorem1, for p≥1, β<n, we have that
(5)Dxβρx,tLp≤C1+t−n/41−1/p−β/4.Remark 3.
We get the same decay rate of the solution as [10].Remark 4.
Our study bases on [9, 10], where the spacial dimension n≥4. Then in this paper, we have the same assumptions for the spacial dimension.


## 2. The Green Function

We first study the Green’s function to Eq. (1)which satisfies
(6)Gtt+Gt+Δ2G−ΔGt=0,x,t∈ℝn×0,∞,Gx,tt=0=0,Gtx,tt=0=δx.We apply the Fourier transformf^ξ,t=∫ℝnfx,te−ix·ξdx and the inverse Fourier transform F−1f^x,t=2π−2n∫ℝnf^ξ,teix·ξdξ.By applying the Fourier transform with respect to the variablex, we get
(7)∂t2+∂t1+ξ2+ξ4G^ξ,t=0,G^ξ,tt=0=0,G^tξ,tt=0=1,the symbol of which is
(8)ν2+ν1+ξ2+ξ4=0.Here,ν and ξ correspond to ∂t and Dxj=1/i∂xj, j=1,2,⋯,n. By a direct calculation, we have
(9)ν=ν±ξ=−1+ξ2±1+ξ22−4ξ42,(10)G^ξ,t=1+ξ22−4ξ4−1/2eν+ξt−eν−ξt.By Duhamel’s principle, we get the solution of the nonlinear problem (1)
(11)ρx,t=Gt∗ρ0+ρ1−Δρ0+∂tG∗ρ0+∫0tGt−τ∗Δfρτdτ.Now we decomposeG^ξ,t=G^+ξ,t+G^−ξ,t, where
(12)G^±ξ,t=±ν0ξeν±ξt,ν0ξ=1+ξ22−4ξ4−1/2.Let(13)Γ1ξ=1,ξ≤s,0,ξ>2s,Γ3ξ=1,ξ>R+1,0,ξ≤Rand Γ2ξ=1−Γ1ξ−Γ3ξ be smooth cut-off functions, where s,R>0, 2s<R.Set(14)G^i±ξ,t=ΓiξG^±ξ,t,i=1,2,3.Now we estimate the Green’s functionGx,t.


### 2.1. Lower-Frequency Part

First, we give the following Lemma.Lemma 5.
Ify^ξ,t has compact support in the variable ξ, N is a positive integer, and there is a constant u>0, such that
(15)Dξβξαy^ξ,t≤Cξα−β++ξαtβ/41+tξ4ke−uξ4t,for any multi-indexes α,β with ∣β∣≤4N, then
(16)Dxαyx,t≤Ct−n+α/4BNx2,t,where k is any fixed positive number, s+=max0,s.The proof of Lemma (9) can be seen in [10].For∣ξ∣ is sufficiently small, from (9) and the Taylor expansion, we have
(17)ν+ξ=−ξ4+Oξ6,ν−ξ=−1−ξ2+ξ4+Oξ6,ν0ξ=1+Oξ2,eν+ξt=e−ξ4t1+Oξ8t,eν−ξt=e−te−ξ2teξ4t1+Oξ6t.Then(18)G^+ξ,t=ν0ξeν+ξt=1+Oξ21+Oξ6te−ξ4t=1+Oξ2+Oξ6te−ξ4t.SinceG∧±ξ,t are smooth functions to variable ξ near ∣ξ∣=0, we obtain that when ∣β∣≤4N,
(19)DξβξαG^1+ξ,t=∑μ1+μ2=ββ!μ1!μ2!Dξμ1χ1ξDξμ2ξαG^+ξ,t≤Cξα−μ2+ξαtμ2/41+tξ43/4μ2e−ξ4t≤Cξα−β++ξαtβ/41+tξ43/4βe−ξ4t.By Lemma (9), we get
(20)DxαG1+x,t≤Ct−n+α/4BNx2,t.ForG^−ξ,t, we have
(21)G^−ξ,t=−ν0ξeλ−ξt=−1+Oξ21+Oξ6te−te−ξ2teξ4t=−1+Oξ2+Oξ6te−te−ξ2teξ4t.Then, we have(22)DξβξαG^1−ξ,t=∑μ1+μ2+μ3=ββ!μ1!μ2!μ3!Dξμ1χ1ξDξμ2ξαDξμ3G^−ξ,t≤Cξα−β++ξαtβ/41+tξ23/4β+1e−1/2−ξ2+ξ4t≤Cξα−β++ξαtβ/41+tξ43/4βe−1/2−ξ4te−r/2t,where r>0.By Lemma (9), we get
(23)DxαG1−x,t≤Ce−r/2tt−n+α/2BNx2,t≤Ct−n+∣α∣/4BNx2,t.From (20)–(23), we have the following proposition:Proposition 6.
For sufficiently smalls, we have
(24)DxαG1x,t≤Ct−n+α/4BNx2,t.


### 2.2. Middle-Frequency Part

We can get the following proposition.Proposition 7.
For fixeds and R, there exist positive numbers m and C such that
(25)DxαG2x,t≤Ce−mtBNx2,t.
The proof of Proposition7 is similar to that of Proposition 12 in [10], so we omit it.


### 2.3. Higher-Frequency Part

For∣ξ∣ is large enough, we have
(26)1+ξ22−4ξ4=ξ22ξ−2+ξ−4−3=3iξ2−13ξ−2+Oξ−4,ν0ξ=11+ξ22−4ξ4=ξ−212ξ−2+ξ−4−3=33iξ−2+Oξ−4.Then, we have(27)G^3+ξ,t=ν0ξeν+ξt=13iξ−2+Oξ−4e−1/2−1/21−3iξ2−3i/3ξ−2+Oξ−4t=e−1/2−1/21−3iξ2t∑j=0kBj+tξ−2j+Bk+1+tOξ−2k+1G^3−ξ,t=−ν0ξeν−ξt=−13iξ−2+Oξ−4e−1/2−1/21+3iξ2+3i/3ξ−2+Oξ−4t=−e−1/2−1/21+3iξ2t∑j=0kBj+tξ−2j+Bk+1+tOξ−2k+1,where Bj±t are polynomials in t with degree no more than j.Let(28)R^+ξ,t=Γ3ξe−1/2−1/21−3iξ2t∑j=0∣α∣+n+1/2Bj+tξ−2j,R^−ξ,t=−Γ3ξe−1/2−1/21+3iξ2t∑j=0∣α∣+n+1/2Bj−tξ−2j,R^ξ,t=R^+ξ,t+R^−ξ,t.Because(29)x2γDxαG3±−R±x,t≤∫ℝnDξ2γξαG^3±−R^±dξ≤Ce−bt,taking ∣γ∣=0 or ∣γ∣=2N, we get the following proposition:Proposition 8.
ForR being sufficiently large, we have
(30)DxαG3±−R±x,t≤Ce−mtBNx2,t,where m>0.Combining Proposition6–8, we obtain the following estimate of the Green’s function:Proposition 9.
For any multi-indexα, we have
(31)DxαG−Rx,t≤C1+t−n+α/4BNx2,t.


## 2.1. Lower-Frequency Part

First, we give the following Lemma.Lemma 5.
Ify^ξ,t has compact support in the variable ξ, N is a positive integer, and there is a constant u>0, such that
(15)Dξβξαy^ξ,t≤Cξα−β++ξαtβ/41+tξ4ke−uξ4t,for any multi-indexes α,β with ∣β∣≤4N, then
(16)Dxαyx,t≤Ct−n+α/4BNx2,t,where k is any fixed positive number, s+=max0,s.The proof of Lemma (9) can be seen in [10].For∣ξ∣ is sufficiently small, from (9) and the Taylor expansion, we have
(17)ν+ξ=−ξ4+Oξ6,ν−ξ=−1−ξ2+ξ4+Oξ6,ν0ξ=1+Oξ2,eν+ξt=e−ξ4t1+Oξ8t,eν−ξt=e−te−ξ2teξ4t1+Oξ6t.Then(18)G^+ξ,t=ν0ξeν+ξt=1+Oξ21+Oξ6te−ξ4t=1+Oξ2+Oξ6te−ξ4t.SinceG∧±ξ,t are smooth functions to variable ξ near ∣ξ∣=0, we obtain that when ∣β∣≤4N,
(19)DξβξαG^1+ξ,t=∑μ1+μ2=ββ!μ1!μ2!Dξμ1χ1ξDξμ2ξαG^+ξ,t≤Cξα−μ2+ξαtμ2/41+tξ43/4μ2e−ξ4t≤Cξα−β++ξαtβ/41+tξ43/4βe−ξ4t.By Lemma (9), we get
(20)DxαG1+x,t≤Ct−n+α/4BNx2,t.ForG^−ξ,t, we have
(21)G^−ξ,t=−ν0ξeλ−ξt=−1+Oξ21+Oξ6te−te−ξ2teξ4t=−1+Oξ2+Oξ6te−te−ξ2teξ4t.Then, we have(22)DξβξαG^1−ξ,t=∑μ1+μ2+μ3=ββ!μ1!μ2!μ3!Dξμ1χ1ξDξμ2ξαDξμ3G^−ξ,t≤Cξα−β++ξαtβ/41+tξ23/4β+1e−1/2−ξ2+ξ4t≤Cξα−β++ξαtβ/41+tξ43/4βe−1/2−ξ4te−r/2t,where r>0.By Lemma (9), we get
(23)DxαG1−x,t≤Ce−r/2tt−n+α/2BNx2,t≤Ct−n+∣α∣/4BNx2,t.From (20)–(23), we have the following proposition:Proposition 6.
For sufficiently smalls, we have
(24)DxαG1x,t≤Ct−n+α/4BNx2,t.


## 2.2. Middle-Frequency Part

We can get the following proposition.Proposition 7.
For fixeds and R, there exist positive numbers m and C such that
(25)DxαG2x,t≤Ce−mtBNx2,t.
The proof of Proposition7 is similar to that of Proposition 12 in [10], so we omit it.


## 2.3. Higher-Frequency Part

For∣ξ∣ is large enough, we have
(26)1+ξ22−4ξ4=ξ22ξ−2+ξ−4−3=3iξ2−13ξ−2+Oξ−4,ν0ξ=11+ξ22−4ξ4=ξ−212ξ−2+ξ−4−3=33iξ−2+Oξ−4.Then, we have(27)G^3+ξ,t=ν0ξeν+ξt=13iξ−2+Oξ−4e−1/2−1/21−3iξ2−3i/3ξ−2+Oξ−4t=e−1/2−1/21−3iξ2t∑j=0kBj+tξ−2j+Bk+1+tOξ−2k+1G^3−ξ,t=−ν0ξeν−ξt=−13iξ−2+Oξ−4e−1/2−1/21+3iξ2+3i/3ξ−2+Oξ−4t=−e−1/2−1/21+3iξ2t∑j=0kBj+tξ−2j+Bk+1+tOξ−2k+1,where Bj±t are polynomials in t with degree no more than j.Let(28)R^+ξ,t=Γ3ξe−1/2−1/21−3iξ2t∑j=0∣α∣+n+1/2Bj+tξ−2j,R^−ξ,t=−Γ3ξe−1/2−1/21+3iξ2t∑j=0∣α∣+n+1/2Bj−tξ−2j,R^ξ,t=R^+ξ,t+R^−ξ,t.Because(29)x2γDxαG3±−R±x,t≤∫ℝnDξ2γξαG^3±−R^±dξ≤Ce−bt,taking ∣γ∣=0 or ∣γ∣=2N, we get the following proposition:Proposition 8.
ForR being sufficiently large, we have
(30)DxαG3±−R±x,t≤Ce−mtBNx2,t,where m>0.Combining Proposition6–8, we obtain the following estimate of the Green’s function:Proposition 9.
For any multi-indexα, we have
(31)DxαG−Rx,t≤C1+t−n+α/4BNx2,t.


## 3. The Proof of Theorem 1

In this section, we shall give the pointwise estimates of the solution to the problem (1). From (3), we have
(32)ρx,t=Φ1−Φ2+Φ3.where
(33)Φ1=Gt∗ρ0+ρ1,Φ2=Gt∗Δρ0,Φ3=∂tG∗ρ0+∫0tGt−τ∗Δfρτdτ.ForΦ1 and Φ3, we have the following proposition:Proposition 10.
(34)DxαΦ1≤CE1+t−n+α/4Bdx2,t,DxαΦ3≤CΘtθ+11+t−n+α/4Bdx2,t,
whereα<n and n≥4, ψx,t=1+tn+α/4Bdx2,t−1,(35)Θt=supx,τ∈ℝn×0,t,α<nDxαρx,τψx,τ.
The proof of the above proposition is similar to proposition 4.1-4.2 in [10], so we omit it.Next, we give a Lemma which is important to estimateΦ2 and has been proved in [10].Lemma 11.
Ifa,b>n/2, c=mina,b, we have
(36)∫ℝn1+x−y41+t−a1+y4−bdy≤C1+x41+t−c.
We write(37)Φ2=G∗Δρ0=G−R∗Δρ0+R∗Δρ0≕S1+S2.
Making use of (31), Lemma 11 and (3), we have
(38)DxαS1≤C1+t−n+α/4BNx2,t∗Δρ0x≤CE1+t−n+α/4∫ℝnBNx−y2,t1+y4−ddy≤CE1+t−n+α/4Bdx2,t,where d>n/2.
From the definition ofRx,t, we have
(39)xγDxαRx,t≤C∫ℝnDξγξαR^ξ,tdξ≤Ce−mt,taking γ=0 or γ=4N, we obtain
(40)DxαRx,t≤Ce−mtBNx2,t,then, we get
(41)DxαRx,t∗ρi−1x≤Ce−mtBNx2,t∗1+x4−d≤Ce−mtBrx2,t,i=1,2.
Thus, we obtain(42)DxαS2≤Ce−mtBdx2,t.Together with (38) and (42), we obtain the following result:Proposition 12.
Ifα<n, then
(43)DxαΦ2≤CE1+t−n+α/4Bdx2,t.Combining Proposition10–12, we have the following result:Proposition 13.
(44)Dxαρx,t≤CE+Θtθ+11+t−n+α/4Bdx2,t,where ∣α∣<n.
By the smallness of E and the continuity ofΘt, we have
(45)Dxαρx,t≤CE1+t−n+α/4Bdx2,t.
Thus, we complete the proof of Theorem1.



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