# The Local Strong and Weak Solutions for a Generalized Pseudoparabolic Equation

Authors: Nan Li
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The Cauchy problem for a nonlinear generalized pseudoparabolic equation is investigated. The well-posedness of local strong solutions for the problem is established in the Sobolev spaceC([0,T);Hs(R))⋂C1([0,T);Hs-1(R)) with s>3/2, while the existence of local weak solutions is proved in the space Hs(R) with 1≤s≤3/2. Further, under certain assumptions of the nonlinear terms in the equation, it is shown that there exists a unique global strong solution to the problem in the space C([0,∞);Hs(R))⋂C1([0,∞);Hs-1(R)) with s≥2.


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## Body

## 1. Introduction

Davis [1] investigated the pseudoparabolic equation(1.1)ut(t,x)=∂∂xφ(ux)+αutxx,
where the constant α≥0, the function φ∈C2(-∞,∞), φ(0)=0 and φ′(ξ)>0, and the subscripts x and t indicate partial derivatives. Equation (1.1) arises from the study of shearing flows of incompressible simple fluids. The quantity φ(ux)+αutx is viewed as an approximation to the stress functional during such a flow. Much attention has been given to this approximation when the function φ is linear (see [2, 3]). The existence and uniqueness of the global weak solution of the initial value problem for (1.1) were established in [1].Recently, Chen and Xue [4] investigated the Cauchy problem for the nonlinear generalized pseudoparabolic equation(1.2)ut-αutxx-λuxx+γux+f(u)x=∂∂xφ(ux)+g(u)-αg(u)xx,x∈R,t>0,
where u(t,x) is an unknown function, α>0, λ≥0, γ is a real number, f(s), φ(s), and g(s) denote given nonlinear functions. The well-posedness of global strong solution in a Sobolev space, the global classical solution and its asymptotic behavior are studied in [4] in which several key assumptions are imposed on the functions φ(s) and g(s). In fact, various dynamic properties for many special cases of (1.2) have been established in [5–7]. For example, when φ(s)=g(s)=0, (1.2) becomes the generalized regularized long wave Burger equation.Motivated by the works in [1, 4], we study the problem(1.3)ut-αutxx=∂∂xφ(ux)+βu2muxx,x∈R,t>0,u(0,x)=u0(x),x∈R,
where α>0 and β≥0, m is a nature number, φ(s) is a given function, and u0(x) is a given initial value function. Here we should address that (1.2) does not include the first equation of problem (1.3) due to the term βu2muxx. Letting β=0, the first equation of problem (1.3) reduces to (1.1).The objectives of this work are threefold. The first objective is to establish the local well-posedness of system (1.3) in the space C([0,T);Hs(R))⋂C1([0,T);Hs-1(R)) with s>3/2. We should address that the Sobolev index s≥2 is required to guarantee the local well-posedness of (1.1) and (1.2) in the works of Davis [1] and Chen and Xue [4]. The second aim is to study the existence of local weak solutions for system (1.3). The third aim is to discuss the well-posedness of the global strong solution for problem (1.3). Under the assumptions of the function φ(s) and the initial value u0(x) similar to those presented in [1, 4], problem (1.3) is shown to have a unique global solution in the space C([0,∞);Hs(R))⋂C1([0,∞);Hs-1(R)).The organization of this paper is as follows. The well-posedness of local strong solutions for problem (1.3) is investigated in Section 2, and the existence of local weak solutions is established in Section 3. Section 4 deals with the well-posedness of the global strong solution.


## 2. Local Well-Posedness

LetLp=Lp(R)  (1≤p<+∞) be the space of all measurable functions h such that ∥h∥Lpp=∫R|h(t,x)|pdx<∞. We define L∞=L∞(R) with the standard norm ∥h∥L∞=infm(e)=0supx∈R∖e|h(t,x)|. For any real number s, Hs=Hs(R) denotes the Sobolev space with the norm defined by(2.1)∥h∥Hs=(∫R(1+|ξ|2)s|ĥ(t,ξ)|2dξ)1/2<∞,
where ĥ(t,ξ)=∫Re-ixξh(t,x)dx.ForT>0 and nonnegative number s, C([0,T);Hs(R)) denotes the Frechet space of all continuous Hs-valued functions on [0,T). We set Λ=(1-∂x2)1/2. For simplicity, throughout this paper, we let c denote any positive constants.The local well-posedness theorem is stated as follows.Theorem 2.1.
Provided thats≥3/2,  u0∈Hs(R),  φ is a polynomial of order  N with  φ(0)=0. Then problem (1.3) admits a unique local solution:
(2.2)u(t,x)∈C([0,T);Hs(R))⋂C1([0,T);Hs-1(R)).Proof.
In fact, the first equation of problem (1.3) is equivalent to the equation
(2.3)ut=Λ-2(∂∂xφ(ux)+βu2muxx),
which leads to
(2.4)u=u0+∫0tΛ-2(∂∂xφ(ux)+βu2muxx)dτ.
Suppose that both u and v are in the closed ball BM0(0) of radius M0>1 about the zero function in C([0,T];Hs(R)) and A is the operator in the right-hand side of (2.4), for fixed t∈[0,T], we get
(2.5)‖∫0tΛ-2(φ(ux)x+βu2muxx)dt-∫0tΛ-2(φ(vx)x+βv2mvxx)dt‖Hs≤T(sup⁡0≤t≤T‖φ(ux)-φ(vx)‖Hs-1+sup⁡0≤t≤T‖u2muxx-v2mvxx‖Hs-2).
The algebraic property of Hs0(R) with s0>1/2 (see [8–10]) and s>3/2 derives that
(2.6)‖uxj-vxj‖Hs-1=‖(ux-vx)(uxj-1+uxj-2vx+⋯+uxvxj-2+vxj-1)‖Hs-1≤‖ux-vx∥Hs-1∥(uxj-1+uxj-2vx+⋯+uxvxj-2+vxj-1)‖Hs-1≤c‖ux-vx‖Hs-1∑i=0j-1‖ux‖Hs-1j-1-i‖vx‖Hs-1i≤cM0j-1∥u-v∥Hs,∥φ(ux)-φ(vx)∥Hs-1≤c∑j=0N∥(ux)j-(vx)j∥Hs-1≤cM0N-1∥u-v∥Hs.
Using u2muxx=∂x[u2mux]-2mu2m-1(ux)2 and v2mvxx=∂x[v2mvx]-2mv2m-1(vx)2, we get
(2.7)‖u2muxx-v2mvxx‖Hs-2≤∥∂x[u2mux-v2mvx]∥Hs-2+c∥u2m-1ux2-v2m-1vx2∥Hs-2≤c‖(u2m-v2m)vx+u2m(ux-vx)‖Hs-1+c‖u2m-1(ux2-vx2)+(u2m-1-v2m-1)vx2‖Hs-1≤c(∥(u2m-v2m)vx∥Hs-1+‖u2m(ux-vx)‖Hs-1+‖u2m-1(ux2-vx2)‖Hs-1+‖(u2m-1-v2m-1)vx2‖Hs-1)≤cM02m‖u-v‖Hs,
in which s>3/2 is used.
From (2.5)–(2.7), we obtain
(2.8)‖Au-Av‖Hs≤‖u-v‖Hs,
where θ=max(cTM0N-1,cTM02m) and c is independent of T. Choosing T sufficiently small such that θ<1, we know that A is a contractive mapping. Applying the above inequality and (2.4) yields
(2.9)‖Au‖Hs≤‖u0‖Hs+θ‖u‖Hs.
Choosing T sufficiently small such that θM0+∥u0∥Hs<M0, we know that A maps BM0(0) to itself. It follows from the contractive mapping principle that the mapping A has a unique fixed point u in BM0(0). This completes the proof of Theorem 2.1.


## 3. Existence of Local Weak Solutions

In this section, we assume thatφ(η)=η2N+1 where N is a nature number. In order to establish the existence of local weak solution, we need the following lemmas.Lemma 3.1 (see Kato and Ponce [8]).
Ifr≥0, then  Hr⋂L∞ is an algebra. Moreover,
(3.1)∥uv∥r≤c(∥u∥L∞∥v∥r+∥u∥r∥v∥L∞),
where c is a constant depending only on r.Lemma 3.2 (see Kato and Ponce [8]).
Letr>0. If  u∈Hr⋂W1,∞ and  v∈Hr-1⋂L∞, then
(3.2)∥[Λr,u]v∥L2≤c(∥∂xu∥L∞∥Λr-1v∥L2+∥Λru∥L2∥v∥L∞).Lemma 3.3.
Lets≥3/2,  φ(ux)=ux2N+1, and the function  u(t,x) is a solution of problem (1.3) and the initial data  u0(x)∈Hs. Then the following results hold.
Forq∈(0,s-1], there is a constant c such that
(3.3)∫R(Λq+1u)2dx≤∫R[(Λq+1u0)2]dx+c∫0t‖u‖Hq+12(‖ux‖L∞2N+‖ux‖L∞2‖u‖L∞2m-2+‖ux‖L∞‖u‖L∞2m-1+‖u‖L∞2m)dτ.
Forq∈[0,s-1], there is a constant c such that
(3.4)‖ut‖Hq≤c‖u‖Hq+1(‖u‖L∞2m-1+‖ux‖L∞2N-1).Proof.
Forq∈(0,s-1], applying (Λqu)Λq to both sides of the first equation of system (1.3) and integrating with respect to x by parts, we have the identity
(3.5)12ddt∫R[(Λqu)2+α(Λqux)2]dx=∫R(Λqu)Λq(φ(ux)x)dx+β∫RΛquΛq[u2muxx]dx.
We will estimate the two terms on the right-hand side of (3.5), respectively. For the first term, by using the Cauchy-Schwartz inequality and Lemmas 3.1 and 3.2, we have
(3.6)|∫R(Λqu)Λq(φ(ux)x)dx|=|∫R(Λqux)Λq(φ(ux))dx|≤c‖Λqux‖L2‖Λq(ux)2N+1‖L2≤c‖u‖Hq+12‖ux‖L∞2N.
For the second term, we have
(3.7)∫RΛquΛq[u2muxx]dx=∫RΛquΛq[(u2mux)x-2mu2m-1ux2]dx=∫RΛquxΛq(u2mux)dx-2m∫RΛquΛq[u2m-1ux2]dx=K1+K2.
For K1, applying Lemma 3.1 derives
(3.8)|K1|≤c∥u∥Hq+12(∥u∥L∞2m+∥ux∥L∞∥u∥L∞2m-1).
For K2, we get
(3.9)|K2|≤c∥u∥Hq∥u2m-1ux2∥Hq≤c∥u∥Hq(∥u2m-1ux∥L∞∥ux∥Hq+∥u2m-1ux∥Hq∥ux∥L∞)≤c∥u∥Hq+12(∥ux∥L∞∥u∥L∞2m-1+∥ux∥L∞2∥u∥L∞2m-2).
It follows from (3.5)–(3.9) that there exists a constant c such that
(3.10)12ddt∫R[(Λqu)2+(Λqux)2]dx≤c∥u∥Hq+12(∥ux∥L∞2N+∥ux∥L∞2∥u∥L∞2m-2+∥ux∥L∞∥u∥L∞2m-1+∥u∥L∞2m).
Integrating both sides of the above inequality with respect to t results in inequality (3.3).
To estimate the norm ofut, we apply the operator (1-∂x2)-1 to both sides of the first equation of system (1.3) to obtain the equation
(3.11)ut=Λ-2(∂∂xφ(ux)+βu2muxx).
Applying (Λqut)Λq to both sides of (3.11) for q∈[0,s-1] gives rise to
(3.12)∫R(Λqut)2dx=∫R(Λqut)Λq-2[∂xφ(ux)+u2muxx]dτ.
For the right-hand of (3.12), we have
(3.13)|∫R(Λqut)(1-∂x2)-1Λq∂xφ(ux)dx|≤c∥ut∥Hq(∫R(1+ξ2)q-1[∫R[ux2N̂(ξ-η)ux̂(η)]dη]2)1/2≤c∥ut∥Hq∥u∥H1∥u∥Hq+1∥ux∥L∞2N-1,|∫R(Λqut)(1-∂x2)-1Λq∂x(u2mux)dx|≤c∥ut∥Hq(∫R(1+ξ2)q-1[∫R[u2m̂(ξ-η)ux̂(η)]dη]2)1/2≤c∥ut∥Hq∥u∥H1∥u∥Hq+1∥u∥L∞2m-1,|∫R(Λqut)(1-∂x2)-1Λq(u2m-1ux2)dx|≤c∥ut∥Hq(∫R(1+ξ2)q-1[∫R[u2m-1ux̂(ξ-η)ux̂(η)]dη]2)1/2≤c∥ut∥Hq∥u∥H1∥u∥Hq+1∥u∥L∞2m-1.
Applying (3.13) into (3.12) yields the inequality
(3.14)∥ut∥Hq≤c∥u∥H1∥u∥Hq+1(∥u∥L∞2m-1+∥ux∥L∞2N-1)
for a constant c>0. This completes the proof of Lemma 3.3.Lemma 3.4.
Ifu(t,x) is a solution of problem (1.3),  α>0,  φ(η)=η2N+1, then
(3.15)∥u∥L∞≤c∥u∥H1(R)≤c∥u0∥H1(R),
where c is a constant.Proof.
Multiplying both sides of the first equation of (1.3) by u(t,x) and integrating with respect to x over R, we have
(3.16)12ddt∫R[u(t,x)2+αux(t,x)2]dx=∫Rφ(ux)xu(t,x)dx+β∫Ru2m+1uxxdx.
Since(3.17)∫Rφ(ux)xu(t,x)dx+β∫Ru2m+1uxxdx=-∫Rux2N+2dx-β(2m+1)∫Ru2mux2dx<0,
we derive that
(3.18)12ddt∫R[u(t,x)2+αux(t,x)2]dx<0,
which results in
(3.19)∫R[u(t,x)2+αux(t,x)2]<∫R[u(0,x)2+αux(0,x)2]≤c∥u0∥H12.
From (3.19), we know that (3.15) holds. This completes the proof.Defining(3.20)ϕ(x)={e1/(x2-1),|x|<1,0,|x|≥1
and setting ϕɛ(x)=ɛ-1/4ϕ(ɛ-1/4x) with 0<ɛ<1/4 and uɛ0=ϕɛ⋆u0, we know that uɛ0∈C∞ for any u0∈Hs(R) and s>0.It follows from Theorem2.1 that for each ɛ the Cauchy problem(3.21)ut-utxx=∂∂xφ(ux)+βu2muxx,u(0,x)=uɛ0(x),x∈R
has a unique solution uɛ(t,x)∈C∞([0,T);H∞).Lemma 3.5.
Under the assumptions of problem (3.21), the following estimates hold for any  ɛ with  0<ɛ<1/4,  u0∈Hs(R) and  s>0:(3.22)∥uɛ0x∥L∞≤c1∥u0x∥L∞,∥uɛ0∥Hq≤c1,ifq≤s,
where c1 is a constant independent of ɛ.Proof.
Using the definition ofuɛ0 and uɛ0x results in the conclusion of the lemma.Lemma 3.6.
Suppose thatu0(x)∈Hs(R) with  s∈[1,3/2] such that  ∥u0x∥L∞<∞. Let  uɛ0 be defined as in system (3.21) and let  φ(η)=η2N+1. Then there exist two positive constants  T and  c, independent of  ɛ, such that the solution  uɛ of problem (3.21) satisfies  ∥uɛx∥L∞≤c for any  t∈[0,T).Proof.
Using notationu=uɛ and differentiating both sides of the first equation of problem (3.11) with respect to x give rise to
(3.23)utx=-φ(ux)-βu2mux+Λ-2[φ(ux)+βu2mux-2mβ(u2m-1ux)x].
Letting p>0 be an integer and multiplying the above equation by (ux)2p+1 and then integrating the resulting equation with respect to x yield the equality
(3.24)12p+2ddt∫R(ux)2p+2dx=-∫Rφ(ux)ux2p+1dx-β∫Ru2mux2p+2dx+∫RJux2p+1dx,
where
(3.25)J=Λ-2[φ(ux)+βu2mux-2mβ(u2m-1ux)x].
Applying the Hölder’s inequality to (3.24) and noting Lemmas 3.4 and 3.5, we obtain
(3.26)12p+2ddt∫R(ux)2p+2dx≤c‖ux‖L∞2N∫R|ux|2p+2dx+c∫Rux2p+2dx+(∫R|J|2p+2dx)1/(2p+2)(ux2p+2dx)2(p+1)/2(p+2)
or
(3.27)ddt(∫R(ux)2(p+2)dx)1/(2p+2)≤c∥ux∥L∞2N(∫Rux2p+2dx)1/(2p+2)+c(∫Rux2p+2dx)1/(2p+2)+(∫R|J|2p+2dx)1/(2p+2).
Since∥f∥Lp→∥f∥L∞ as p→∞ for any f∈L∞⋂L2, integrating both sides of the inequality (3.27) with respect to t and taking the limit as p→∞ result in the estimate
(3.28)∥ux∥L∞≤∥u0x∥L∞+∫0tc(∥ux∥L∞+∥ux∥L∞2N+1+∥J∥L∞)dτ.
Using the algebra property of Hs0(R) with s0>1/2 yields (∥uɛ∥H1/2+ means that there exists a sufficiently small δ>0 such that ∥uɛ∥1/2+=∥uɛ∥H1/2+δ)
(3.29)∥J∥L∞≤c∥J∥H1/2+≤c∥Λ-2[φ(ux)+βu2mux-2mβ(u2m-1ux)x]∥H1/2+≤c(∥φ(ux)∥H0+∥u∥H1+∥u2m-1ux∥H0)≤c(∥ux∥2N∥u∥H1+∥u∥H1+∥u∥L∞2m-1∥u∥H1)≤c(∥ux∥2N+1),
in which Lemmas 3.4 and 3.5 are used. From (3.28) and (3.29), one has
(3.30)∥ux∥L∞≤∥u0x∥L∞+c∫0t[1+∥ux∥L∞+∥ux∥L∞2N+∥ux∥L∞2N+1]dτ.
From Lemma3.5, it follows from the contraction mapping principle that there is a T>0 such that the equation
(3.31)∥W∥L∞=∥u0x∥L∞+c∫0t[1+∥W∥L∞+∥W∥L∞2N+∥W∥L∞2N+1]dτ
has a unique solution W∈C[0,T]. Using the result presented on page  51 in [11] yields that there are constants T>0 and c>0 independent of ɛ such that ∥ux∥L∞≤∥W(t)∥L∞≤c for arbitrary t∈[0,T], which leads to the conclusion of Lemma 3.6.Using Lemmas3.3–3.6, notation uɛ=u and Gronwall’s inequality result in the inequalities
(3.32)∥uɛ∥Hq≤CTeCT,∥uɛt∥Hr≤CTeCT,
where q∈(0,s], r∈(0,s-1](1≤s≤3/2) and CT depends on T. It follows from the Aubin’s compactness theorem that there is a subsequence of {uɛ}, denoted by {uɛn}, such that {uɛn} and their temporal derivatives {uɛnt} are weakly convergent to a function u(t,x) and its derivative ut in L2([0,T],Hs) and L2([0,T],Hs-1), respectively. Moreover, for any real number R1>0, {uɛn} is convergent to the function u strongly in the space L2([0,T],Hq(-R1,R1)) and {uɛnt} converges to ut strongly in the space L2([0,T],Hr(-R1,R1)) for r∈[0,s-1]. Thus, we can prove the existence of a weak solution to (1.3).Theorem 3.7.
Suppose thatu0(x)∈Hs with  1≤s≤3/2,  ∥u0x∥L∞<∞ and  φ(η)=η2N+1. Then there exists a  T>0 such that (1.3) subject to initial value  u0(x) has a weak solution  u(t,x)∈L2([0,T],Hs) in the sense of distribution and  ux∈L∞([0,T]×R).Proof.
From Lemma3.6, we know that {uɛnx}(ɛn→0) is bounded in the space L∞. Thus, the sequences {uɛn}, {uɛnx}, {uɛnx2}, and {uɛnx2N+1} are weakly convergent to u, ux, ux2, and ux2N+1 in L2[0,T],Hr(-R,R) for any r∈[0,s-1), separately. Therefore, u satisfies the equation
(3.33)∫0T∫Ru(gt-gxxt)dxdt=∫0T∫R[ux2N+1gx+βu2muxgx-2mβu2m-1ux2g]dxdt,
with u(0,x)=u0(x) and g∈C0∞. Since X=L1([0,T]×R) is a separable Banach space and {uɛnx} is a bounded sequence in the dual space X*=L∞([0,T]×R) of X, there exists a subsequence of {uɛnx}, still denoted by {uɛnx}, weakly star convergent to a function v in L∞([0,T]×R). It derives from the weakly convergence of {uɛnx} to ux in L2([0,T]×R) that ux=v almost everywhere. Thus, we obtain ux∈L∞([0,T]×R).


## 4. Well-Posedness of Global Solutions

Lemma 4.1.
Ifu(t,x) is a solution of problem (1.3),  α>0,  φ(η)=η2N+1, then
(4.1)∥ux∥L∞≤A1/2,
where
(4.2)A=∫R[1+αα(u0′(x))2+(1+α)(u0′′(x))2]dx.Proof.
Multiplying each side of the first equation of problem (1.3) by uxx and integrating over [0,t]×R yields
(4.3)∫0t∫R(uxx2φ′(ux)+u2muxx2+α2∂∂t(uxx2))dxdt=∫0t∫Rutuxxdxdt.
Integrating the right-hand side of the above identity by parts and using ux(±∞)=0, we get
(4.4)2∫0t∫Rutuxxdxdt=∫R[u0′(x)]2dx-∫Rux2(t,x)dx.
From (4.3), (4.4) and the assumption of this lemma, we have
(4.5)α∥uxx∥L22+∥ux∥L2≤∫R[(u0′(x))2+α(u0′′(x))2]dx,
from which we obtain (4.1).Theorem 4.2.
Suppose thats≥2,  u0∈Hs(R),  φ(ux)=ux2N+1 with positive integer  N. Then problem (1.3) has a unique global solution:
(4.6)u(t,x)∈C([0,∞);Hs(R))⋂C1([0,∞);Hs-1(R)).Proof.
Using the Gronwall inequality and Lemma3.3 and choosing s=q+1, we have
(4.7)∥u∥Hs≤c∥u0∥Hse∫0t(∥ux∥L∞2N+∥ux∥L∞2∥u∥L∞2m-2+∥ux∥L∞∥u∥L∞2m-1+∥u∥L∞2m)dτ.
From Lemma 4.1, we have
(4.8)∥ux∥≤A1/2=(∫R[1+αα(u0′(x))2+(1+α)(u0′′(x))2]dx)1/2≤c∥u0∥H2(R).
Using (4.7) and (4.8) derives
(4.9)∥u∥Hs≤c∥u0∥Hsect,
which completes the proof of Theorem 4.2.



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