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Global well-posedness of nonlinear wave equation with weak and strong damping terms and logarithmic source term

  • Wei Lian and Runzhang Xu
Published/Copyright: July 4, 2019

Abstract

The main goal of this work is to investigate the initial boundary value problem of nonlinear wave equation with weak and strong damping terms and logarithmic term at three different initial energy levels, i.e., subcritical energy E(0) < d, critical initial energy E(0) = d and the arbitrary high initial energy E(0) > 0 (ω = 0). Firstly, we prove the local existence of weak solution by using contraction mapping principle. And in the framework of potential well, we show the global existence, energy decay and, unlike the power type nonlinearity, infinite time blow up of the solution with sub-critical initial energy. Then we parallelly extend all the conclusions for the subcritical case to the critical case by scaling technique. Besides, a high energy infinite time blow up result is established.

1 Introduction and main results

In this paper, we study initial boundary value problem of nonlinear wave equation with weak and strong damping terms and logarithmic source term

uttΔuωΔut+μut=uln|u|,(x,t)Ω×(0,), (1.1)
u(x,t)=0,xΩ,t0, (1.2)
u(x,0)=u0(x),ut(x,0)=u1(x),xΩ, (1.3)

where Ω ⊂ ℝn (n ≥ 1) is a bounded domain with a smooth boundary Ω,

ω0,μ>ωλ1, (1.4)

λ1 being the first eigenvalue of the operator −Δ under homogeneous Dirichlet boundary conditions.

The undamped hyperbolic equation

uttΔu=f(u), (1.5)

was introduced by D’Alembert [1] to model the propagation of waves along vibrating elastic string. By introducing the potential well, the global existence and finite time blow up of solution to (1.5) with E(0) < d were proved by Payne and Sattinger in [2], [3] respectively.

The nonlinear wave equation with linear weak damping term was considered by Levine and Serrin [4] in abstract form

Putt+A(u)+Q(t,ut)=F(u) (1.6)

and they proved the nonexistence of global solution for the negative initial energy, i.e., E(0) < 0. Later, Pucci and Serrin [5] extended this results to E(0) < D1, where D1 is positive. Vitillaro [6] studied the similar problem replacing the linear weak damping term by the nonlinear one

uttΔu+b|ut|m2ut=c|u|p2u (1.7)

and derived the conditions of initial data leading to finite time blow up of the solution for E(0) ≤ E1.

For the initial boundary value problem of classical strongly damped wave equation

uttΔuΔut=f(u), (1.8)

Webb [7] gave the local existence uniqueness, global existence and the asymptotic behavior of the solution. For the initial boundary value problem of strongly damped semilinear wave equation

uttωΔutΔu+ϕ(u)=f, (1.9)

Pata and Squassina [8] proved the existence of the universal attractor, in the presence of a quite general nonlinearity of critical growth. Moreover, they obtained the asymptotic behavior of the solutions in dependence of the damping coefficient.

For the wave equation with both linear weak and strong damping terms, we can directly go to [9] for the most recent progress. Gazzola and Squassina [9] proved that the initial boundary value problem of the weak and strong damping hyperbolic equation

uttΔu+μutωΔut=|u|p2u, (1.10)

has a unique local solution, and the global existence and nonexistence results were also proved for E(0) ≤ d. Also, the finite time blow up of solution with high energy E(0) > d (ω = 0, μ > 0) was obtained.

The logarithmic nonlinearity is of much interest in physics, since it appears naturally in inflation cosmology and super symmetric filed theories, quantum mechanics and nuclear physics [10], [11]. Haraux and Cazenave[12] proved the existence and uniqueness of solution for the Cauchy problem for the nonlinear Schrödinger equation

iut+Δu+Vu+kulog|u|2=0inRN×R+ (1.11)

and for the nonlinear Klein-Gordon equation

uttΔu=kulog|u|2inR3×R+. (1.12)

Górka [13] obtained that the initial boundary value problem of logarithmic Klein-Gordon equation

uttuxx+u=εulog|u|2,xO,t[0,T),u(x,0)=u0(x),ut(x,0)=u1(x),xO,u(x,t)=0,xO,t[0,T), (1.13)

where 𝓞 is a finite interval V = [a, b], admits global weak solutions for all (u0, u1) ∈ H01 × L2 and ε ∈ [0, 1] in one-dimensional case.

The weak damping with logarithmic wave equation

uttΔu+ut+u+|u|2u=uln|u|2 (1.14)

was introduced by Hiramatsu [14] to model the dynamics of Q-ball in theoretical physics. Then its initial boundary value problem was considered by Han in [15]

uttΔu+u+ut+|u|2u=uln|u|2,xΩ,t[0,T),u(x,0)=u0(x),ut(x,0)=u1(x),xΩ,u(x,t)=0,xΩ,t[0,T) (1.15)

and the global existence of weak solutions was proved, for all (u0, u1) ∈ H01 × L2 in ℝ3. By constructing an appropriate Lyapunov function, Zhang and Liu [16] obtained the exponential decay estimates of energy with E(0) < d for all (u0, u1) ∈ H01 × L2, I(u0) > 0.

Al-Gharabli and Messaoudi [17] proved the global existence and the exponential decay of solutions of the following plate equation

utt+Δ2u+u+ut=kuln|u|,xΩ,t>0,u(x,t)=uv(x,t)=0,xΩ,t0,u(x,0)=u0(x),ut(x,0)=u1(x),xΩ, (1.16)

for all (u0, u1) ∈ H02 (Ω) × L2(Ω) and 0 < E(0) < d, I(u0) > 0. Later, Al-Gharabli and Messaoudi [18] considered a general damping h(ut) instead of a linear ut one considered in [17], where h is a function having a polynomial growth near the origin. They established the global existence and the general decay of solutions for all (u0, u1) ∈ H02 (Ω) × L2(Ω), 0 < E(0) < d and I(u0) > 0.

As shown in the previous works, the dynamical behaviors of solutions are quite different when with presence of different nonlinearities, i.e., power type and logarithmic type. To be specific, with the presence of logarithmic term, the finite time blow up of solution don’t occur anymore which means the Nehari manifold in the initial energy space H01 no longer plays a role as a threshold separating global and non-global existence of solution. And instead in this article the results show that Nehari manifold can be viewed as a threshold which indicates the decay or infinite time blow up of solutions. In order to investigate and describe the dynamical behavior of solution that strongly relied on the initial data, we focus on the logarithmic term in three initial data levels, i.e., subcritical energy E(0) < d, critical initial energy E(0) = d and the arbitrary high initial energy E(0) > 0 (ω = 0).

The present paper is organized as follows. Section 2 presents some notations and preliminaries. In section 3, we state our main results. Section 4 prove the local existence of solution. Section 5 prove the global existence, asymptotic behavior and infinite time blow up of solution for E(0) < d. In section 6, the global existence, asymptotic behavior and blowup of solution for E(0) = d is proved. At last, in Section 7, we prove the infinite time blowup result for E(0) > 0 (ω = 0).

2 Notations and primary lemmas

In this section, we present some preliminaries to prove the main results. We denote the inner product and the norm on H01 (Ω) by (⋅,⋅) and ∥∇⋅∥, respectively. The symbol ∥⋅∥ will indicate the norm on L2(Ω). Moreover, we denote by 〈⋅,⋅〉 the duality pairing between H−1(Ω) and H01 (Ω). For any v, w H01 (Ω), we have

(v,w)=ωΩvw+μΩvw

and

v=(v,v)12.

By (1.4), ∥⋅∥* is an equivalent eccentric module over H01 (Ω).

First, for problem (1.1)-(1.3) we introduce the energy functional

E(t)=12ut2+12u212Ωu2ln|u|dx+14u2, (2.1)

the potential energy functional

J(u)=12u212Ωu2ln|u|dx+14u2 (2.2)

and the Nehari functional

I(u)=u2Ωu2ln|u|dx. (2.3)

By a direct computation,

J(u)=12I(u)+14u2. (2.4)

By I(u) we define the potential well (stable set)

W={uH01(Ω)I(u)>0}{0}, (2.5)

the outer space of potential well (unstable set)

V={uH01(Ω)I(u)<0} (2.6)

and the Nehari manifold

N={uH01(Ω){0}I(u)=0}. (2.7)

The depth of potential well is defined as

d=infuNJ(u). (2.8)

On the other hand, as the difference between two types of nonlinearities, assumptions on the power type nonlinearity don’t work on the logarithmic one. Consequently, introducing the logarithmic Sobolev inequality and revisiting the corresponding estimates is a necessity to handle logarithmic nonlinear term u ln ∣u∣. The following logarithmic Sobolev inequality was introduced by [20], Chapter 8.14 (also see [19] for a different proof).

Lemma 2.1

[19, 20] If uH1(ℝn) and a > 0. Then

2Rn|u(x)|2ln(|u(x)|u)dx+n(1+lna)u2a2πRn|u(x)|2dx.

For u H01 (Ω), we can define u(x) = 0 for x ∈ ℝnΩ. Then uH1(ℝn), that is to say, for a general domain Ω, we have following logarithmic Sobolev inequality,

2Ω|u(x)|2ln(|u(x)|u)dx+n(1+lna)u2a2πΩ|u(x)|2dx, (2.9)

where u H01 (Ω) and a > 0.

Lemma 2.2

Assume that u H01 (Ω)∖{0}. Then we have

  1. limλ0 J(λu) = 0 and limλ+ J(λu) = −∞;

  2. There exists a unique λ* = λ*(u) such that ddλJ(λu)|λ=λ=0,whereλ=expu2Ωu2ln|u|dxu2

  3. J(λu) is increasing on the interval (0, λ*), decreasing on the interval (λ*, +∞) and attains its maximum unique λ* ∈ (0, +∞) such that

    I(λu)>0,0<λ<λ,=0,λ=λ,<0,λ<λ<+.

Proof

  1. We get

    J(λu)=λ22u2λ22Ωu2ln|λu|dx+λ24u2=λ22u2λ22Ωu2ln|u|dxλ22ln|λ|u2+λ24u2=λ22u2Ωu2ln|u|dxln|λ|u2+12u2, (2.10)

    which tells limλ0 J(λu) = 0 and limλ+ J(λu) = −∞.

  2. A direct computation on (2.10) yields

    ddλJ(λu)=λu2λΩu2ln|u|dxλln|λ|u2=λu2Ωu2ln|λu|dx, (2.11)

    which means ddλJ(λu)|λ=λ=0,whereλ=expu2Ωu2ln|u|dxu2 .

  3. The conclusion (iii) directly follows from

    I(λu)=λ2u2λ2Ωu2ln|u|dxλ2ln|λ|u2=λdJ(λu)dλ

and conclusion (ii).□

Lemma 2.3

Let u H01 (Ω) and r:=(2π)n4en2.

  1. If 0 < ∥u∥ ≤ r, then I(u) ≥ 0;

  2. If I(u) < 0, thenu∥ > r;

  3. If I(u) = 0 andu∥ = 0, thenu∥ ≥ r.

Proof

  1. Using the logarithmic Sobolev inequality (2.9), for a > 0, we get

    I(u)=u2Ωu2ln|u|dx=u2Ωu2ln|u|u+lnudx1a22πu2+n(1+lna)2u2u2lnu. (2.12)

    Taking a=2π, we gain

    I(u)(n(2+ln(2π))4lnu)u2. (2.13)

    If 0 < ∥u∥ ≤ r, then n(2+ln(2π))4 ≥ ln ∥u∥, which gives I(u) ≥ 0.

  2. From (2.13) and I(u) < 0, we can see that

    (n(2+ln(2π))4lnu)u2<0,

    which means

    u>(2π)n4en2=r.

    1. This conclusion is similar to the proof of (ii).

Lemma 2.4

The depth of potential well d in (2.8) satisfies

d14(2π)n2en. (2.14)

Proof

From the definition of d in (2.8) and Lemma 2.3 in (iii), it follows that u ∈ 𝓝. As a result, we obtain

0=I(u)(n(2+ln(2π))4lnu)u2, (2.15)

which implies

u(2π)n4en2. (2.16)

By virtue of (2.4), I(u) = 0 and (2.16), we obtain

J(u)=12I(u)+14u2=14u214(2π)n2en, (2.17)

which gives (2.14).□

Lemma 2.5

Let u be a solution of problem (1.1)-(1.3), then E(t) is a non-increasing function with respect to t.

Proof

Multiplying Eq. (1.1) by ut and integrating it over Ω × [s, t), we can obtain

E(t)+stut(τ)2dτ=E(s). (2.18)

Thus, the proof is completed.□

Definition 2.6

Function u = u(x, t) is called a weak solution of problem (1.1)-(1.3) over Ω × [0, T], if

uC([0,T],H01(Ω))C1([0,T],L2(Ω))C2([0,T],H1(Ω)),

utL2([0, T], H01 (Ω)) and there holds

utt(t),η+Ωu(t)ηdx+ωΩut(t)ηdx+μΩut(t)ηdx=Ωu(t)ln|u(t)|ηdx,ηH01(Ω),a.e.t[0,T), (2.19)

where u(x, 0) = u0(x) in H01 (Ω), ut(x, 0) = u1(x) in L2(Ω).

3 Main results

In this section, we state our main results about problem (1.1)-(1.3).

Theorem 3.1

Let u0(x) ∈ H01 (Ω), u1(x) ∈ L2(Ω) and (1.4) holds. Then there exists a time T > 0 such that problem (1.1)-(1.3) admits a unique weak solution u on [0, T] satifying

uC([0,T],H01(Ω))C1([0,T],L2(Ω))C2([0,T],H1(Ω)).

Theorem 3.2

Let u0(x) ∈ H01 (Ω), u1(x) ∈ L2(Ω) and (1.4) holds. Assume that E(0) < d, then when u0(x) ∈ W, problem (1.1)-(1.3) admits a global weak solution u(t) ∈ L(0, +∞; H01 (Ω)) with ut(t) ∈ L(0, +∞;L2(Ω)) for 0 ≤ t < ∞.

Theorem 3.3

Let u0(x) ∈ H01 (Ω), u1(x) ∈ L2(Ω) and (1.4) holds. Assume that E(0) < d, then when u0(x) ∈ W, there exist two positive constants Ĉ and ξ independent of t such that

0<E(t)C^eξt,t0.

Theorem 3.4

Let u0(x) ∈ H01 (Ω), u1(x) ∈ L2(Ω) and (1.4) holds. Assume that E(0) < d, then when u0(x) ∈ V, the existence time of solution to problem (1.1)-(1.3) is infinite and

limt+u(t)2=+.

Theorem 3.5

Let u0(x) ∈ H01 (Ω), u1(x) ∈ L2(Ω) and (1.4) holds. Assume that E(0) = d, then when u0(x) ∈ W, problem (1.1)-(1.3) admits a global weak solution u(t) ∈ L(0, +∞; H01 (Ω)) with utL(0, +∞;L2(Ω)) for 0 ≤ t < ∞.

Theorem 3.6

Let u0(x) ∈ H01 (Ω), u1(x) ∈ L2(Ω) and (1.4) holds. Assume that E(0) = d, then when u0(x) ∈ W, there exist two positive constants Ĉ and ξ independent of t such that

0<E(t)C^eξt,t0.

Theorem 3.7

Let u0(x) ∈ H01 (Ω), u1(x) ∈ L2(Ω) and (1.4) holds. Assume that E(0) = d, then when u0(x) ∈ V, the existence time of solution to problem (1.1)-(1.3) is infinite and

limt+u(t)2=+.

Theorem 3.8

Let u0(x) ∈ H01 (Ω), u1(x) ∈ L2(Ω). Assume that ω = 0, μ ≥ 0 and u0(x), u1(x) satisfy that

  1. u02 > 4E(0) > 0,

  2. (u0(x), u1(x)) > 0,

  3. u0(x) ∈ V,

then the existence time of solution to problem (1.1)-(1.3) is infinite and

limt+u(t)2=+.

Remark 3.9

When E(0) ≤ d, we derive the global well-posedness and the decay of solution provided u0(x) ∈ W. However when E(0) > 0 and u0(x) ∈ W, it is still open that whether global solution exists.

4 Local existence

For a fixed T > 0, consider the space

H=C([0,T],H01(Ω))C1([0,T],L2(Ω)),

endowed with the norm

uH2=maxt[0,T](u(t)2+ut(t)2). (4.1)

4.1 Proof of Theorem 3.1

To establish the uniqueness and existence of local solution to problem (1.1)-(1.3), we first prove the following lemma.

Lemma 4.1

For any T > 0 and u ∈ 𝓗, there exists a unique

vC([0,T],H01(Ω))C1([0,T],L2(Ω))C2([0,T],H1(Ω)) (4.2)

such that vtL2([0, T], H01 (Ω)) solves the linear problem

vttΔvωΔvt+μvt=uln|u|,xΩ,t[0,T),v(x,0)=u0(x),vt(x,0)=u1(x),xΩ,v(x,t)=0,xΩ,t[0,T), (4.3)

where u0(x) and u1(x) are defined in (1.3).

Proof

(4.3) is an inhomogenous linear problem for v, carrying a good energy structure. Therefore it suffices to show that the forcing term is L2, so that one can apply a standard Galerkin method to extract a solution. So we estimate Ω(u ln ∣u∣)2dx. First by a direct calculation and sobolev inequality, we have

Ω(uln|u|)2dx={xΩ;|u(x)|1}uln|u|2dx+{xΩ;|u(x)|>1}uln|u|2dxe2|Ω|+n222{xΩ;|u(x)|>1}u2nn2dxe2|Ω|+n222c2nn2u2nn2, (4.4)

where c* is the best constant of the Sobolev embedding H01(Ω)L2nn2(Ω). Next we claim the uniqueness. Arguing by contradiction, we suppose that there exist two solutions v and w such that (4.3) hold. Then by subtracting the obtained two equations and testing with vt-wt, we can derive

v(t)w(t)2+vt(t)wt(t)2+20tvt(τ)wt(τ)2dτ=0,

which directly says wv. So we complete the proof.□

Take (u0(x),u1(x)) satisfying (1.4), set R2 = 2(∥∇ u02 + ∥u12) and consider

UT={uH:u(x,0)=u0(x),ut(x,0)=u1(x),uHR},T>0. (4.5)

Again Lemma 4.1 simplifies that for ∀ u ∈ 𝓤T, there exists v = Φ(u) such that v is the unique solution to problem (4.3). Next we prove that for a suitable T > 0, Φ is a contractive map satisfying Φ(𝓤T) ⊂ 𝓤T.

First we can conclude that

v(t)2+v˙(t)2+20tvt(τ)2dτ=u02+u12+20tΩu(τ)ln|u(τ)|vt(τ)dxdτ. (4.6)

where v = Φ(u) is the corresponding solution, to problem (4.3) for fixed u ∈ 𝓤T.

Similar to the arguments of (4.6), we can derive that

20tΩu(τ)ln|u(τ)|vt(τ)dxdτ0t(u(τ)ln|u(τ)|2+v˙h(τ)2)dτ. (4.7)

Hence, (4.7) becomes

20tΩu(τ)ln|u(τ)|vt(τ)dxdτcTR2n(n2)+20tv˙h(τ)2dτ, (4.8)

for all t ∈ (0, T]. Combining (4.6) with (4.7) and taking the maximum over [0, T] gives

vH212R2+cTR2n(n2). (4.9)

Choosing T sufficiently small, we get ∥v𝓗R, which shows that Φ(𝓤T) ⊂ 𝓤T.

Now taking w1 and w2 in 𝓤T, subtracting the two equations (4.3) for v1 = Φ(w1) and v2 = Φ(w2), and setting v = v1v2 we obtain for all η H01 (Ω) and a.e. t ∈ [0, T]

vtt(t),η+Ωv(t)ηdx+ωΩvt(t)ηdx+μΩvt(t)ηdx=Ωw1(t)ln|w1(t)|w2(t)ln|w2(t)|ηdx=Ωξ(t)(w1(t)w2(t))ηdx, (4.10)

where

0ξ(t)(ln|w1(t)+w2(t)|+1)(w1(t)+w2(t))ln|w1(t)+w2(t)|.

Therefore, taking η = vt in (4.10) and arguing as above, gives

Φ(w1)Φ(w2)H2=vH2cR4n(n2)Tw1w2H2δw1w2H2

for some δ < 1 as long as T is sufficiently small. So by the Contraction Mapping Principle, we can conclude that problem (1.1)-(1.3) admits a unique solution.

5 Sub-critical initial energy

5.1 Global existence for E(0) < d

We first prove the invariant set W under the flow of problem (1.1)-(1.3).

Lemma 5.1

(Invariant set W when E(0) < d) Let u0(x) ∈ H01 (Ω) and u1(x) ∈ L2(Ω). If E(0) < d and u0(x) ∈ W, then we have

u(t)W (5.1)

and

u2<4dfor allt[0,T). (5.2)

Proof

Let u(t) be the weak solution of problem (1.1)-(1.3) with E(0) < d, u0(x) ∈ W and T be the maximum existence time of u(t). From Lemma 2.5 we have

E(t)E(0)<dfor allt[0,T), (5.3)

namely,

12ut(t)2+J(u(t))=E(t)<dt[0,T). (5.4)

Arguing by contradiction, we suppose that there exists the first t0 ∈ (0, T) such that I(u(t0)) = 0 and I(u(t)) > 0 for 0 ≤ t < t0, i.e.,

u(t0)2=Ωu(t0)2ln|u(t0)|dx. (5.5)

From the definition of 𝓝, then we have u(t0) ∈ 𝓝, which implies that J(u(t0)) ≥ d. From (2.8) and the definition of E(t), it holds that

E(t0)=12ut(t0)2+J(u(t0))d,

which contradicts (5.3). Then u(t) ∈ W for all t ∈ [0, T), which together with (5.4) and (2.17) gives

d>E(t)=12ut(t)2+J(u)=12ut(t)2+12I(u)+14u214u2.

Thus, the proof of Lemma 5.1 is completed.□

Now let us turn to prove the existence of global solution to problem (1.1)-(1.3) with E(0) < d.

Proof of Theorem 3.2

From Lemma 5.1 it implies u(t) ∈ W for any t ∈ [0, T). By (2.18), we get

12ut2+12u2+14u2E(0)+Ωu2ln|u|dx, (5.6)

which together with the Logarithmic Sobolev inequality leads to

ut2+(1a22π)u2+n(1+lna)+12u2C1+u2lnu. (5.7)

Here and in the sequel we denote by Ci > 0, i = 1, 2, 3 as constants. Choosing 0 < a < 2π that

1a22π>0,

we get

ut2+u2+u2C2(1+u2lnu2). (5.8)

Recalling Lemma 5.1, we have

u2<4d.

Hence, from inequality (5.8) it follows that

ut2+u2+u2C3,

where C3 > 0 is a constant independent of t. Therefore similar as the proof of Lemma 4.1 we know that problem (1.1)-(1.3) admits a global weak solution.□

5.2 Asymptotic behavior for E(0) < d

Next, we can now prove the asymptotic behavior of the solution to problem (1.1)-(1.3), which relies on the construction of a Lyapunov functional by performing a suitable modification of the energy.

Proof of Theorem 3.3

First, we define

L(t):=E(t)+εΩutudx+εω2u2, (5.9)

where ε > 0 will be chosen later. From Lemma 5.1 and (2.1)-(2.3) it implies that

0<E(t),t0. (5.10)

It is easy to see that L(t) and E(t) are equivalent in the sense that there exist two positive constants β1 > 0 and β2 > 0 depending on ε such that

β1E(t)L(t)β2E(t)fort0. (5.11)

Taking the derivative of L(t) with respect to time yields

dL(t)dt=ωut2μut2+εut2εu2+εΩu2ln|u|dxεμΩutudx, (5.12)

together with Eq.(1.1). Now, we estimate the last term on the right hand side of (5.12) as follows. By using Young’s inequality, we obtain,

Ωutudx14δut2+δu2,δ>0. (5.13)

Substituting (5.13) into (5.12) with (2.1) gives that

dL(t)dtMεE(t)ωut2+Mε2+εμ+εμ4δut2+Mε4+εμδu2+εM21u2+ε1M2Ωu2ln|u|dx,M>0. (5.14)

By logarithmic Sobolev inequality, we have

dL(t)dtMεE(t)ωut2+Mε2+εμ+εμ4δut2+Mε4+εμδ+ε1M2(lnun2(1+lna))u2+ε1M2(a22π1)u2. (5.15)

Recalling (2.1), (2.4) and E(t) ≤ E(0) < d, we get

lnu2<ln(4J(u))<ln4d. (5.16)

Now, choosing M < 2, and ε small enough such that

Mε2+εμ+εμ4δ<0.

Then inequality (5.15) becomes

dL(t)dtMε4+εμδ+ε21M2ln(4d)n(1+lna)u2MεE(t)+ε1M2(a22π1)u2. (5.17)

Since 0 < M < 2 and J(u) ≤ E(0) < d, by taking

e2αna22π,

where α=ln(4d)+M+μδ2Mn, and taking δ > 0 small enough such that

a22π10 (5.18)

and

Mε4+εμδ+ε21M2ln(4d)n(1+lna)0, (5.19)

then we have

dL(t)dtMεE(t),t>0. (5.20)

Further, by virtue of (5.11), let ξ=Mεβ2, (5.20) becomes

dL(t)dtξL(t),t>0, (5.21)

namely,

L(t)Ceξt,t>0, (5.22)

which together with (5.11) shows

0<E(t)C^eξt,t>0. (5.23)

This completes the proof.□

5.3 Blow-up for E(0) < d

To prove the blow-up results for problem (1.1)-(1.3) with E(0) < d, we first give the invariant set V. Similar to Lemma 5.1, we also have the following lemma.

Lemma 5.2

(Invariant set V when E(0) < d) Let u0(x) ∈ H01 (Ω) and u1(x) ∈ L2(Ω). If E(0) < d and u0(x) ∈ V, then we have

u(t)V (5.24)

and

u2>4dfor allt[0,T). (5.25)

Now we turn to show that the blow up in infinite time for problem (1.1)-(1.3) with E(0) < d.

Proof of Theorem 3.4

Let u(x, t) be a weak solution of problem (1.1)-(1.3) with E(0) < d, I(u0) < 0. Arguing by contradiction, we suppose that the solution u is global. Then for any T > 0 we may consider the auxiliary function G : [0, T] → ℝ+ defined by

G(t):=u2+0tu(τ)2dτ+(Tt)u02. (5.26)

Since G(t) > 0 is positive over [0, T] and continuous, there exists a ρ > 0 (independent of the choice of T) such that

G(t)ρ for allt[0,T]. (5.27)

Furthermore,

G(t)=2Ωuutdx+u2u02=2Ωuutdx+20t(u(τ)ut(τ))dτ (5.28)

and

G(t)=2utt,u+2ut2+2(u,ut)for a.e.t[0,T]. (5.29)

Testing Eq. (1.1) with u and plugging the result into (5.29) we obtain

G(t)=2ut2u2+Ωu2ln|u|dxfor a.e.t[0,T]. (5.30)

Therefore, we get

G(t)G(t)(G(t))2=2G(t)ut2u2+Ωu2ln|u|dx+4η(t)(G(t)(Tt)u02)ut2+0tut(τ)2dτ,

where

η(t)=u2+0tu(τ)2dτut2+0tut(τ)2dτΩuutdx+0t(u(τ),ut(τ))dτ2.

By Schwarz inequality, we can get

u2ut2Ωuutdx2,0tu(τ)2dτ0tut(τ)2dτ0t(u(τ),ut(τ))dτ2

and

Ωuutdx0t(u(τ),ut(τ))dτu0tu(τ)2dτ12ut0tu(τ)2dτ12.

These three inequalities entail η(t) ≥ 0 for any t ∈ [0, T]. As a consequence, we reach the following differential inequality

G(t)G(t)(G(t))2G(t)ξ(t)for a.e.t[0,T], (5.31)

where ξ:[0, T] → ℝ+ is the map defined by

ξ(t):=2ut22I(u)40tut(τ)2dτ. (5.32)

From Lemma 2.5, Lemma 5.2, (2.4) and E(0) < d, we know

ξ(t)=4J(u(t))4E(t)2I(u(t))40tut(τ)2dτ=u24E(t)40tut(τ)2dτu24E(0)>u24d>0. (5.33)

Since E(0) < d, there exists δ > 0 (independent of T) such that

ξ(t)δfor allt0. (5.34)

Hence, by (5.27) and (5.34), it follows that

G(t)G(t)(G(t))2ρδ>0for a.e.t[0,T],

which means G(t)G″(t) − (G′(t))2 > 0. On the other hand, by directly calculation, we can see that

(lnG(t))=G(t)G(t) (5.35)

and

(lnG(t))=G(t)G(t)(G(t))2G2(t)>0. (5.36)

By (5.36), we know that (ln G(t))′ is increasing with respect to t, using this fact, integrating (5.35) from t0 to t, we have

(lnG(t))=(lnG(t0))+t0tG(t)G(t)(G(t))2G2(t)dτ

and

ln|G(t)|ln|G(t0)|=t0t(lnG(τ))dτ=t0tGG(τ)dτG(t0)G(t0)(tt0),

where 0 ≤ t0t. Then

G(t)G(t0)exp(G(t0)G(t0)(tt0)).

Since G(0) = 0 and G′(0) > 0, we can take t0 sufficiently small such that G′(t0) > 0 and G(t0) > 0. Then for sufficiently large t,

u2=G(t)G(t)G(t0)G(t)G(t)exp(G(t)G(t0)(tt0))Ce2t,

i.e.,

limt+G(t)=+.

This shows that the weak solution u to problem (1.1)-(1.3) blows up at +∞.□

6 Critical initial energy

Lemma 6.1

If E(0) = d, then there exists a t0 ∈ (0, T) such that

0t0uτ2dτ>0. (6.1)

Proof

Arguing by contradiction we suppose that 0tuτ2dτ0 for 0 ≤ t < T, which gives ∥ut∥ ≡ 0 for 0 ≤ t < T. So we can get u(t) ≡ u0(x), which contradicts the assumption of Lemma 6.1.□

6.1 Global existence for critical initial energy

Lemma 6.2

Let u0(x) ∈ H01 (Ω) and u1(x) ∈ L2(Ω). If E(0) = d and u0(x) ∈ W, then we have

u(t)W. (6.2)

Proof

We prove that u(t) ∈ W for 0 ≤ t < T. Arguing by contradiction we suppose that there exists a t0 ∈ (0, T) such that I(u(t0)) = 0 and ∥u(t0)∥ ≠ 0, which says J(u(t0) ≥ d. Hence, by

12ut(t0)2+J(u(t0))+0t0uτ2dτ=E(0)=d,

we get 0t0uτ2dτ=0 and ∥ut2 = 0 for 0 ≤ tt0, which implies dudt=0 and u(x, t) = u0(x) for xΩ, 0 ≤ tt0. Thus we can conclude I(u(t0)) = I(u0) > 0, which contradicts I(u(t0)) = 0.□

Now we turn to prove the existence of global solution to problem (1.1)-(1.3) with E(0) = d.

Proof of Theorem 3.5

First Theorem 3.1 gives the existence of the local solution. From Lemma 6.1 we can get (6.1). By (2.18) and E(0) = d we obtain

E(t0)=d0tuτ2dτ<d.

Moreover, from Lemma 6.2 it follows u(t0) ∈ W. Let v(t) = u(t + t0) and t ≥ 0, then v(t) is a solution of problem (1.1)-(1.3), which combining Theorem 3.2 says that T = +∞.□

6.2 Asymptotic behavior for E(0) = d

Next, we can now prove the asymptotic behavior of the solution to problem (1.1)-(1.3) with E(0) = d.

Proof of Theorem 3.6

From Theorem 3.5 it implies that there exists a t0 > 0 such that E(t0) < d, I(u(t0)) > 0 or ∥u(t0)∥ = 0, which together with Theorem 3.3 says,

0<E(t)C^eξ(tt0),tt0

and

0<E(t)C^eξt,tt0,

where Ĉ* = Ĉeξt0.□

6.3 Infinite time blow up for the critical initial energy

Lemma 6.3

Let u0(x) ∈ H01 (Ω) and u1(x) ∈ L2(Ω). If E(0) = d and u0(x) ∈ V, then we have

u(t)V. (6.3)

Proof

Arguing by contradiction, we suppose that there exists a t0 ∈ (0, T) such that

I(u)<0,0<t<t0

and

I(u(t0))=0,

which together with (2.8) gives J(u(t0)) ≥ d. The remainder of this proof is similar to the proof of Lemma 6.2.□

Now we turn to show that the blow up in infinite time of the solution to problem (1.1)-(1.3) with E(0) = d.

Proof of Theorem 3.7

First Theorem 3.1 gives the existence of the local solution. By the similar arguments as those in the proof of Theorem 3.5, together with Lemma 6.1 and Lemma 6.3. we can set E(t0) < d and I(u(t0)) < 0. The remainder of the proof is the same as Theorem 3.4.□

7 Infinite time blow up for E(0) > 0 (ω = 0)

We first prove the following lemmas to obtain that the unstable set V is invariant with respect to t under the flow of problem (1.1)-(1.3) with E(0) > 0, ω = 0 and μ ≥ 0.

Lemma 7.1

Let δ ∈ ℝ, T > 0 and let h be a Lipschitzian function over [0, T). Assume that h(0) ≥ 0, h′(t) + δ h(t) > 0 a.e. t ∈ (0, T). Then h(t) > 0, ∀ t ∈ (0, T).

Proof

Differentiating eδth(t) gives

eδth(t)dt=δeδth(t)+eδth(t)=eδt(h(t)+δh(t))>0, (7.1)

combining with h(0) ≥ 0, we know eδth(t) > e0h(0) ≥ 0. In other words, h(t) > 0 for all t ∈ (0, T).□

Lemma 7.2

Let u0(x) ∈ H01 (Ω), u1(x) ∈ L2(Ω), ω = 0, μ ≥ 0 and (u0, u1) ≥ 0, then the map {t ↦ ∥u(t)∥2} is strictly increasing as long as u(t) ∈ V, where u(t) is the solution of problem (1.1)-(1.3).

Proof

Let

H(t):=u2, (7.2)

then we have

H(t)=2(u,ut)

and

H(t)=2utt,u+2ut2.

Notice that, for any t ∈ (0, T) we have

utt,u=ddt(u,ut)ut2. (7.3)

And multiplying Eq. (1.1) by u yields

utt,u+u2+μ(ut,u)=(uln|u|,u). (7.4)

Substituting (7.3) into (7.4) gives

ddt((u,ut)+μ2u2)=ut2I(u), (7.5)

which shows

d2dt2u2+μddtu2=2ut22I(u). (7.6)

From u(t) ∈ V, (7.6) implies

H(t)+H(t)>0,t[0,T).

Obviously, from Lemma 7.1 and H′(0) = (u0(x), u1(x)) ≥ 0. We conclude that the map {t ↦ ∥u(t)∥2} is strictly increasing for all t ∈ (0, T).□

In the following, we show the invariance of the unstable set V under the flow of problem (1.1)-(1.3) with E(0) > 0.

Lemma 7.3

(Invariant set V when E(0) > 0) Let u0(x) ∈ H01 (Ω), u1(x) ∈ L2(Ω) and u be a weak solution of problem (1.1)-(1.3) with the maximum existence time interval [0, T), T ≤ +∞. Assume that E(0) > 0, ω = 0 and the initial data satisfy

u02>4E(0), (7.7)

then the solution u(t) of problem (1.1)-(1.3) with E(0) > 0 belongs to V, provided that u0(x) ∈ V.

Proof

We prove u(t) ∈ V for t ∈ [0, T). Arguing by contradiction, we suppose that t0 ∈ (0, T) is the first time such that

I(u(t0))=0 (7.8)

and

I(u(t)<0fort[0,t0).

Hence from Lemma 7.2 it follows that the map

tu(t)2

is strictly increasing on the interval [0, t0), which together with (7.2) gives that

H(t)=u(t)2>u02>4E(0),t(0,t0). (7.9)

Moreover, from the continuity of u(t) in t, we obtain

u(t0)2>4E(0). (7.10)

Recalling (2.1) and (2.18), we have

E(0)=E(t)+μ0tuτ2dτ=12ut2+12u212Ωuln|u|dx+14u2+μ0tuτ2dτ=12ut2+12I(u)+14u2+μ0tuτ2dτ, (7.11)

which together with (7.8), μ ≥ 0 shows that

E(0)E(t0)+μ0t0uτ2dτ12ut(t0)2+12I(u(t0))+14u(t0)2=12ut(t0)2+14u(t0)214u(t0)2. (7.12)

Obviously (7.12) contradicts (7.10). So the proof is completed.□

In the end we present the infinite time blow up result of the solution to problem (1.1)-(1.3) with E(0) > 0.

Proof of Theorem 3.8

Recalling the auxiliary function G(t) in (5.26), similarly as the proof of Theorem 3.4, we have

G(t)G(t)(G(t))2G(t)G(t)4ut2+0tuτ(τ)2dτG(t)2ut22I(u)40tuτ(τ)2dτ (7.13)

and

ξ(t):=2ut22I(u)40tuτ(τ)2dτ. (7.14)

From (7.11) and (2.1)-(2.3), we can deduce (7.14) to

ξ(t)=u24E(0). (7.15)

At this point, (7.7) and (7.15) indicate that

ξ(t)>σ>0. (7.16)

Therefore by (7.13)-(7.16) and (5.27), we have

G(t)G(t)(G(t))2>ρσ>0,t[0,T).

This tells us that

limt+G(t)=+.

The remainder of the proof is similar to that of Theorem 3.4.□

Acknowledgement

This work was supported by the National Natural Science Foundation of China (11871017), the China Postdoctoral Science Foundation(2013M540270), the Fundamental Research Funds for the Central Universities. And the authors are grateful for the insightful comments and constructive suggestions made by the referee, and also the support and contribution from Dr. Yuxuan Chen and Miss Longqi Shang.

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Received: 2018-12-09
Accepted: 2019-01-18
Published Online: 2019-07-04

© 2020 Wei Lian and Runzhang Xu, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 Public License.

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