NONEXISTENCE OF SOLUTIONS TO CAUCHY PROBLEMS FOR ...

Electronic Journal of Differential Equations, Vol. 2016 (2016), No. 20, pp. 1–14. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu

NONEXISTENCE OF SOLUTIONS TO CAUCHY PROBLEMS FOR FRACTIONAL TIME SEMI-LINEAR PSEUDO-HYPERBOLIC SYSTEMS SALEM ABDELMALEK, MAHA BAJNEED, KHALED SIOUD

Abstract. We study Cauchy problems time fractional semi-linear pseudohyperbolic equations and systems. Using the method of nonlinear capacity, we show that there are no solutions for certain nonlinearities and initial data. Our work complements the work by Aliev and col. [1, 6, 7].

1. Introduction In this article, we study Cauchy problems for time fractional pseudo-hyperbolic equations and systems. We start by considering the time fractional equation β α utt + η(−∆)k utt + (−∆)` u + ξ(−∆)r D0|t u + γD0|t u = f (u),

(1.1)

for x ∈ RN , t > 0, supplemented with the initial data u(x, 0) = u0 (x),

ut (x, 0) = u1 (x),

x ∈ RN ,

(1.2)

and with η, ξ, γ ≥ 0

for r, k ∈ N ∪ {0}, ` ∈ N, 0 < β ≤ α ≤ 1,

(1.3)

α D0|t

∆ is the Laplacian and is the left-sided Riemann-Liouville fractional derivative of order α. The aim of this paper is to show, using the method of nonlinear capacity proposed by Pokhozhaev in 1997 [18] and developed successfully and jointly with Mitidieri [15, 16, 17], that under certain conditions, there are no solutions to (1.1)-(1.2). For the non fractional case α = β = 1, Lions’ monograph [13] considered equation (1.1) in the case where η = 0 and f (u) = −|u|p u. A step forward was achieved by [10, 15, 20] where they considered the absence of global solutions for the case where η = ξ = 0 and f (u) = |u|p−1 u or f (u) = ±|u|p . Kato [10] showed that for ` = 1, ξ = 0, and 1 < p < 1 + N2 , problem (1.1)-(1.2) admits no global solution under a certain condition on the initial data. A further study of John [9] considered the case ` = 1, ξ = 0, η = 0, γ = 0 and f (u) = |u|p for u close to zero. This was generalized to ` ∈ N, ξ = η = 0, γ > 0, α = 1, by Zhang [20] and Kirane and 2010 Mathematics Subject Classification. 80A23, 65N21, 26A33, 45J05, 34K37, 42A16. Key words and phrases. Semi-linear pseudo-hyperbolic equation; non-existence of solutions; non-linear capacity method. c

2016 Texas State University. Submitted June 25, 2015. Published January 12, 2016. 1

2

S. ABDELMALEK, M. BAJNEED, K. SIOUD

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Qafsaoui [12]. The two studies proved that the critical exponent for this case is in fact p = 1 + N2 . The existence of global solutions of problem (1.1)-(1.3) for the non-fractional case α = β = 1, η, ξ, γ ≥ 0, r, k ∈ N ∪ {0} and ` ∈ N was achieved by Aliev and Kazymov [5]. Recently, by using the method of the test function Aliev and col. [1, 6, 7] established sufficient conditions for the nonexistence of global solutions of problem (1.1)-(1.3) for the non-fractional case α = β = 1: Aliev and Lichaei [6] considered the case α = β = 1 and η, ξ, γ > 0 for r = k ∈ N ∪ {0}, ` ∈ N, and f (u) ≥ C|u|p . Aliev and Kazymov [1] examined the case 1 p α = β = 1, k = 0, r = 0, ` ∈ N and f (u) = (1+|x| 2 )s |u| . Aliev and Mamedov [7] treated the non existence of global solutions of a semilinear hyperbolic equation with an anisotopic elliptic part (α = β = 1, k = r = 0), utt + εut +

N X

(−1)`k Dx2`kk u = f (u),

f (u) ≥ c|u|p .

k=1

Our work will complement the results of [6] for r, k ∈ N ∪ {0}, η, ξ, γ ≥ 0 and ` ∈ N and extend it to the time-fractional case 0 < β ≤ α < 1, using the test function method. In the second part of this paper, we study the Cauchy problem for the timefractional pseudo-hyperbolic system β1 α1 utt + η1 (−∆)k1 utt + (−∆)`1 u + ξ1 (−∆)r1 D0|t u + γ1 D0|t u = f (v) = |v|p β2 α2 vtt + η2 (−∆)k2 vtt + (−∆)`2 v + ξ2 (−∆)r2 D0|t v + γ2 D0|t v = g(u) = |u|q

(1.4)

posed in Q∞ := RN × (0, ∞), subject to the initial conditions u(x, 0) = u0 (x), ut (x, 0) = u1 (x),

x ∈ RN ,

v(x, 0) = v0 (x), vt (x, 0) = v1 (x)x ∈ RN

(1.5)

with p, q > 1, ri , ki ∈ N ∪ {0}, ì ∈ N, ηi , ξi , γi ≥ 0 and 0 < βi ≤ αi ≤ 1 for i = 1, 2. The non-existence of global solutions in the case of a non fractional system of two (or more) equations with αi = 0 or 1 and βi = 0 or 1, is investigated in numerous studies of Aliev and colleagues: Aliev, Mammadzada, and Lichaei [8] considered the case βi = 1, γi = ηi = 1, ξi = 0, `1 = 1, `2 = 2, p = 27 and q = 52 ; Aliev and Kazymov [4] examined the case βi = 1, γi = ηi = 1, ξi = 0, ì ∈ N, and fi (u, v) ≥ Ci,1 |u|pi + Ci,2 |v|qi ; Aliev and Kazymov [2] considered the case βi = 1, γi = ηi = 1, ξi = 0, ì ∈ N, and f (v) ≥ C|v|p and g(u) ≥ C|u|q ; Aliev and Kazymov [3] dealt with a system of three equations that is similar to the case presented in [4]. Our work will complement these papers for the system of two equations in the cases γi , ηi , ξi > 0, ri , ki ∈ N ∪ {0}, ì ∈ N and extend it to the time-fractional case, using again the test function method. 2. Preliminaries For the convenience of the reader, we start by recalling some basic definitions and properties which will be useful throughout this paper.

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SEMI-LINEAR PSEUDO-HYPERBOLIC SYSTEMS

3

Definition 2.1. The left- and right-sided Riemann-Liouville integrals of order 0 < α < 1 for an integrable function are defined as Z t 1 α I0|t f (t) := (t − s)α−1 f (s)ds, (2.1) Γ(α) 0 Z T 1 α (s − t)α−1 f (s)ds, (2.2) It|T f (t) := Γ(α) t where Γ is the Euler gamma function. Definition 2.2. Let AC[0, T ] be the space of functions f which are absolutely continuous on [0, T ]. The left and right-handed Riemann-Liouville fractional derivatives of order n − 1 < γ < n for a function f ∈ AC n [0, T ] := {f : [0, T ] → R, Dn−1 f ∈ AC[0, T ]},

n∈N

is defined as (see [11]) γ n−γ D0|t f (t) := Dn (I0|t f )(t),

t > 0,

γ n−γ Dt|T f (t) := (−1)n Dn (It|T f )(t),

(2.3) (2.4)

where D is the usual time derivative. α α Furthermore, for every f, g ∈ C([0, T ]) such that D0|t f (t), D0|t g(t) exist and are continuous for all t ∈ [0, T ], 0 < α < 1, the formula of integration by parts can be given according to Love and Young [14] by Z T Z T α α g(t)(D0|t f )(t)dt = f (t)(Dt|T g)(t)dt. (2.5) 0

0

In addition, [19, Lemma 2.2] provides us with the formula Z T f (T ) 1 α [ − (t − s)−α f 0 (s)ds] Dt|T f (t) := Γ(1 − α) (T − t)α t

(2.6)

or α Dt|T f (t) :=

1 d Γ(1 − α) dt

Z

T

(t − s)−α f (s)ds.

(2.7)

t

3. Non-existence of global solutions of one equation In this section, we study the non-existence of global solutions for the timefractional semi-linear pseudo-hyperbolic equation (1.1) for certain initial data with f (u) = |u|p . Before we state our result. let us define the weak solution of problem (1.1)-(1.3). In this article, QT denotes the set QT := RN × (0, T ), 0 < T ≤ +∞. We set Z Z Z T Z Z Z ∞ f := f (x, t) dx dt, f := f (x, t) dx dt, QT RN 0 Q∞ RN 0 Z Z f := f (x, 0)dx. RN

RN

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Definition 3.1. The function u ∈ Lploc (Q∞ ) is a weak solution of problem (1.1)(1.3) on QT with initial data u0 (x), u1 (x) ∈ L1loc (RN ) if it satisfies Z

p

Z

Z

u1 (x)ϕ(x, 0) + η u1 (x)(−∆)k ϕ(x, 0) RN Z Z Z k = u0 (x)ϕt (x, 0) + η u0 (x)(−∆) ϕt (x, 0) + uϕtt RN RN QT Z Z Z β α ϕ ϕ+γ uDt|T +η u(−∆)k ϕtt − ξ u(−∆)r Dt|T QT Q QT Z T + u(−∆)` ϕ, |u| ϕ +

RN

QT

(3.1)

QT

for any test-function ϕ ∈ Cx2d t 2 (QT ) with d = max{`, k, r} such that ϕ is positive, β α ϕ ≡ 0 outside a compact K ⊂ Rn , ϕ(x, T ) = ϕt (x, T ) = 0 and Dt|T ϕ, Dt|T ϕ ∈ C(QT ). As for the result on the non-existence of a global solution, the constants η, ξ and γ will not play a role, and thus will be taken equal to one. Theorem 3.2. Assume that (1) r, k ∈ N ∪ {0}, ` ∈ N and 0 R< β ≤ α < 1; R (2) u0 , u1 , ∈ L1 (RN ) such that RN u0 (x)dx > 0, RN u1 (x)dx > 0 2` =: pc . (3) 1 < p ≤ 1 + N + 2`( β1 − 1) Then problem (1.1)–(1.3) does not admit any global in time nontrivial solution. Proof. The proof is by contraction. Let u be a global weak solution of problem (1.1)-(1.3) and ϕ be a non-negative function (satisfying the conditions of Definition 3.1) that will be specified later. Using -Young’s inequality ab ≤ ap + c()bpe,

p > 1, a ≥ 0, b ≥ 0, p + pe = pe p, > 0,

we can write Z

Z Z uϕtt ≤ |u|p ϕ + c() |ϕtt |peϕ−ep/p , QT Q Q Z Z T Z T k p u(−∆) ϕtt ≤ |u| ϕ + c() |(−∆)k1 ϕtt |peϕ−ep/p , QT Q Q Z Z T Z T r α p α u(−∆) Dt|T ϕ ≤ |u| ϕ + c() |(−∆)r Dt|T ϕ|peϕ−ep/p , QT Q Q Z Z T Z T β β p uDt|T ϕ ≤ |u| ϕ + c() |Dt|T ϕ|peϕ−ep/p , Q Q Q Z T Z T Z T ` p u(−∆) ϕ ≤ |u| ϕ + c() |(−∆)` ϕ|peϕ−ep/p . QT

QT

QT

(3.2)

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5

Using inequalities (3.2) in (3.1), we obtain the inequality Z Z Z |u|p ϕ + u1 (x)ϕ(x, 0) + u1 (x)(−∆)k ϕ(x, 0) N N QT R R Z Z − u0 (x)ϕt (x, 0) − u0 (x)(−∆)k ϕ(x, 0) N N R R Z Z (3.3) nZ p ≤ 1 |u| ϕ + C1 |ϕtt |peϕ−ep/p + |(−∆)k ϕtt |peϕ−ep/p QT QT QT Z Z Z o −p e β α |Dt|T ϕ|peϕ−ep/p + |(−∆)` ϕ|peϕ−ep/p . + |(−∆)r Dt|T ϕ|peϕ p + QT

QT

QT

Setting Z Z A1 = |ϕtt |peϕ−ep/p , A2 = |(−∆)k ϕtt |peϕ−ep/p , QT QT Z Z β α A3 = |(−∆)r Dt|T ϕ|peϕ−ep/p , A4 = |Dt|T ϕ|peϕ−ep/p , QT QT Z A5 = |(−∆)` ϕ|peϕ−ep/p , QT

and taking = 1/2, inequality (3.3) becomes Z Z Z p |u| ϕ + u1 (x)ϕ(x, 0) + u1 (x)(−∆)k ϕ(x, 0) QT RN RN Z Z − u0 (x)ϕt (x, 0) − u0 (x)(−∆)k ϕt (x, 0) RN

(3.4)

RN

≤ C{A1 + A2 + A3 + A4 + A5 }. At this stage, we set ϕ(x, t) = Ψν

t2 + |x|4ρ

, R > 0, ν 1, ρ > 0, R4 where Ψ ∈ Cc∞ (R+ ) is a decreasing function defined as ( 1 if r ≤ 1 Ψ(r) = 0 if r ≥ 2,

(3.5)

with 0 ≤ Ψ ≤ 1 and r|Ψ0 (r)| < C. Note that with this choice of ϕ, we have ϕt (x, t) = 2νtR−4 Ψν−1 ((t2 + |x|4ρ )/R4 )Ψ0 ((t2 + |x|4ρ )/R4 ), leading to ϕt (x, 0) = 0.

(3.6)

We also assume that ϕ satisfies Z β α ϕ−p/ep (|ϕtt |pe + |(−∆)k ϕtt |pe + |(−∆)r Dt|T ϕ|pe + |Dt|T ϕ|pe + |(−∆)` ϕ|pe) < ∞, QT

for that, we will choose ν 1. Therefore, the inequality (3.4) becomes Z Z Z |u|p ϕ + u1 (x)ϕ(x, 0) + u1 (x)(−∆)k ϕ(x, 0) QT

RN

≤ C{A1 + A2 + A3 + A4 + A5 }.

RN

(3.7)

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Let us pass to the scaled variables y = R−1/ρ x, τ = R−2 t and the function ϕ e given by ϕ(x, t) = ϕ(y, e τ ). In doing so, it follows that ϕt = R−2 ϕ eτ , 2

ϕtt = R−4 ϕ eτ τ , ∗

β ϕ = R−2β Dτβ|T ∗ ϕ, e Dt|T

(−∆)m ϕ = R

−2m ρ

∆ϕ, e

∗

where T = R T and T is a positive constant. Now, let us set Q = {(y, τ ), 0 ≤ τ 2 + y 4ρ ≤ 2}. Using these definitions, A1 , . . . , A5 can be rewritten as Z N A1 ≤ R−4ep+ ρ +2 |ϕ eτ τ |peϕ e−ep/p , Q Z −p e N +4)e p + +2 −( 2k ρ |(−∆)k ϕ A2 ≤ R ρ eτ τ |peϕ ep , Q Z N +2α)e p + +2 −( 2r ρ |(−∆)r Dτα|T ∗ ϕ| e peϕ−ep/p , A3 ≤ R ρ Q Z −p e −2β p e+ N +2 ρ A4 ≤ R |Dτβ|T ∗ ϕ| e peϕ ep , Q Z −2` N p e + +2 ρ A5 ≤ R ρ |(−∆)` ϕ| e peϕ e−ep/p .

(3.8)

Q

In a short form, we can write Ai = Ci Rθi

for i = 1, 2, . . . , 5,

where

N 2k N + 2, θ2 = − + 4 pe + + 2, ρ ρ ρ 2r N N θ3 = − + 2α pe + + 2, θ4 = −2β pe + + 2, ρ ρ ρ N −2` pe + + 2. θ5 = ρ ρ As β ≤ α < 1, we observe that θ1 = −4e p+

θ2 ≤ θ1 ≤ θ4 θi

θ4

(3.9)

and θ3 ≤ θ4 .

θ5

For R ≥ 1, we have R ≤ R + R for i = 1, 2, . . . , 5 and then inequality (3.7) becomes Z Z Z |u|p ϕ + u1 (x)ϕ(x, 0) + u1 (x)(−∆)k ϕ(x, 0) ≤ K(Rθ4 + Rθ5 ), (3.10) QT

RN

RN

where K is a positive constant. We have Z Z u1 (x)(−∆)k ϕ(x, 0) = u1 (x)(−∆)k ϕ(x, 0) RN QR Z N 2k =Rρ−ρ u1 (R1/ρ y)(−∆k )ϕ(y, e 0), Q

where QR = {(x, 0); R1/ρ ≤ |x| ≤ 21/(4ρ) R1/ρ },

Q = {(y, 0); 1 ≤ |y| ≤ 2}.

We assume that ϕ satisfies

(−∆k )ϕ(·, e 0) ∞ < ∞,

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for that, we will choose ν 1. It follows that Z Z Z N 2k 2k u1 (x)(−∆)k ϕ(x, 0) ≤ CR ρ − ρ |u1 (R1/ρ y)| ≤ CR− ρ RN

Q

7

|u1 (x)|,

QR

then, passing to the limit as R → +∞, we have Z u1 (x)(−∆)k ϕ(x, 0) → 0.

(3.11)

RN

Now, if θ4 < 0 and θ5 < 0, (i.e. max(θ4 , θ5 ) < 0), that means 2βρ , N + 2ρ(1 − β) 2` p < p2 (ρ) = 1 + N + 2(ρ − `)

p < p1 (ρ) = 1 +

which is equivalent to p < p(ρ) = min(p1 (ρ), p2 (ρ)). As p1 is a decreasing function and p2 is an increasing function, the maximum value of the function min(p1 , p2 ) will be at the point ρ = β` , when the two functions p1 and p2 are equal ` 2` ` p1 ( ) = p2 ( ) = 1 + =: pc , (i.e. θ4 = θ5 ). β β N + 2`( β1 − 1) Therefore, for θ4

we have R +R yield

θ5

1 < p < pc , (3.12) tends to zero when R → ∞ and the inequalities (3.10) and (3.11) Z

|u|p +

Z u1 (x) ≤ 0. RN

Q∞

As

Z u1 (x) > 0, RN

we get a contradiction. This proves the theorem in the case (3.12). For the border case where p = pc which corresponds to θ4 = θ5 = 0 and ρ = β` , let QT,R = {(x, t), R4 ≤ t2 + |x|4ρ ≤ 2R4 }, R if we use the H¨ older inequality in the estimate of QT uϕtt instead of the -Young inequality, we obtain Z Z Z 1/p Z 1/ep uϕtt = uϕtt ≤ |u|p ϕ ϕ−ep/p |ϕtt |pe QT

QT ,R

QT ,R

1/e p

≤ (A1 )

Z

p

|u| ϕ)

(

1/p

QT ,R

,

QT ,R

and similarly Z

Z

k

k

u(−∆) ϕtt ≤ (A2 )

u(−∆) ϕtt = QT

Z QT

α u(−∆)r Dt|T ϕ=

QT ,R

Z QT ,R

1/e p

Z

|u|p ϕ,

QT ,R

Z α u(−∆)r Dt|T ϕ ≤ (A3 )1/ep QT ,R

1/p |u|p ϕ ,

8


Z QT

Z

β ϕ= uDt|T

u(−∆)` ϕ =

Z QT ,R

Z

QT

u1 (x) ≤

1/p |u|p ϕ ,

QT ,R

Z

1/p |u|p ϕ .

u(−∆)` ϕ ≤ (A5 )1/ep

1/e p A1 +

RN

QT

Z β ϕ ≤ (A4 )1/ep uDt|T

QT ,R

Thus, we obtain Z Z p |u| ϕ +

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QT ,R

1/e p A2 +

1/e p A3 +

1 p e

A4 +

1/e p A5

Z

1/p |u|p ϕ

QT ,R

≤C

Z

1/p |u|p ϕ .

QT ,R

As

q

R QT

|u| < +∞, we have Z lim R→+∞

QT ,R

|u|q ϕ ≤ lim

R→+∞

Z

|u|q = 0.

QT ,R

R R Passing to the limit as R → +∞, we find that Q∞ |u|q + RN u1 (x) = 0, which R contradicts RN u1 > 0. This prove the theorem in the case p = pc . 4. A pseudo-hyperbolic system This section is concerned with the fractional time pseudo-hyperbolic system (1.4))-(1.5). Definition 4.1. The couple of functions (u, v), u ∈ Lqloc (Q∞ ) and v ∈ Lploc (Q∞ ) is a weak solution of (1.4)-(1.5) on QT with initial data u0 (x), u1 (x), v0 (x) and v1 (x) ∈ L1loc (RN ), if it satisfies Z Z Z |v|p ϕ + u1 (x)ϕ(x, 0) + η1 u1 (x)(−∆)k1 ϕ(x, 0) QT RN RN Z Z Z uϕtt = u0 (x)ϕt (x, 0) + η1 u0 (x)(−∆)k1 ϕt (x, 0) + RN RN QT Z Z Z (4.1) α1 + u(−∆)`1 ϕ + η1 u(−∆)k1 ϕtt − ξ1 u(−∆)r1 Dt|T ϕ QT QT QT Z β1 + γ1 uDt|T ϕ, QT

and Z

Z

q

Z

v1 (x)ϕ(x, 0) + η2 v1 (x)(−∆)k2 ϕ(x, 0) RN Z Z Z k2 = v0 (x)ϕt (x, 0) + η2 v0 (x)(−∆) ϕt (x, 0) + vϕtt RN RN QT Z Z Z α2 + v(−∆)`2 ϕ + η2 v(−∆)k2 ϕtt − ξ2 v(−∆)r1 Dt|T ϕ QT QT QT Z β2 + γ2 vDt|T ϕ, |u| ϕ +

RN

QT

(4.2)

QT

for any test-function ϕ ∈ Cx2`2t (QT ), ` = max{`1 , `2 } being positive, ϕ ≡ 0 outside β1 β2 α1 α2 a compact K ⊂ Rn , ϕ(x, T ) = ϕt (x, T ) = 0 and Dt|T ϕ, Dt|T ϕ, Dt|T ϕ, Dt|T ϕ ∈ C(QT ).

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Theorem 4.2. Assume that (1) ri , ki ∈ N ∪ {0}, ì ∈ N and 0 < βRi ≤ αi < 1, i = R 1, 2; R (2) u0 , u1 , Rv0 , v1 ∈ L1 (RN ) such that RN u0 (x) > 0, RN u1 (x) > 0, RN v0 (x) > 0 and RN v1 (x) > 0; (3) p > 1, q > 1, 2(pβ2 + β1 )ρ 2(qβ1 + β2 )ρ pq ≤ min 1 + ,1 + N + 2(1 − β1 )ρ N + 2(1 − β2 )ρ where ρ = min( β`11 , β`22 ). Then problem (1.4)-(1.5) does not admit any global non trivial solution. Proof. The proof is by contraction. Let (u, v) be a global weak solution of (1.4)(1.5) and ϕ be a non-negative function (satisfying the conditions of Definition 4.1). R Applying H¨ older inequality to QT uϕtt , we obtain Z 1/q Z Z 1/q 1/eq Z −q e ϕ q |ϕtt |qe |u|q ϕ uϕtt ≤ ≤ (A1 )1/eq |u|q ϕ QT

QT

QT

QT

and similarly Z

k1

Z

u(−∆) ϕtt ≤ (A2 ) QT

Z

QT α1 u(−∆)r1 Dt|T ϕ ≤ (A3 )

QT

1/q , |u|q ϕ

1/e q

Z

Z

β1 uDt|T ϕ ≤ (A4 )1/eq

1/q |u|q ϕ ,

QT

Z

QT

Z

1 q e

|u|q ϕ

1/q

(4.3) ,

QT

Z

u(−∆)`1 ϕ ≤ (A5 )1/eq

QT

1/q |u|q ϕ ,

QT

where Z

−q e

A1 = |ϕtt |qeϕ q dx dt, QT Z A2 = |(−∆)k1 ϕtt |qeϕ−eq/q dx dt, QT Z α1 A3 = |(−∆)r1 Dt|T ϕ|qeϕ−eq/q dx dt, QT Z β1 A4 = |Dt|T ϕ|qeϕ−eq/q dx dt, QT Z A5 = |(−∆)`1 ϕ|qeϕ−eq/q dx dt. QT

Now, let ϕ be the test function defined by the expression (3.5). Using the previous estimates (4.3) and the properties (3.5) and (3.6) of the function ϕ in equation (4.1), we obtain the inequality Z Z p |v| ϕ + u1 ϕ(x, 0) QT RN (4.4) Z 1/q h i 1/e q 1/e q 1/e q 1/e q 1/e q q ≤ |u| ϕ A1 + A2 + A3 + A4 + A5 . RN

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Similarly, for the equation (4.2), we have Z Z p |u| ϕ + v1 ϕ(x, 0) RN

QT

≤

Z

1/q 1 1 1/e q 1/e q 1/e q |v| ϕ B1 + B2 + B3qe + B4 + B5qe ,

(4.5)

q

RN

where Z −q e B1 = |ϕtt |qeϕ q dx dt, QT Z −q e B2 = |u(−∆)k2 ϕtt |qeϕ q dx dt, Q Z T −q e α2 B3 = ϕ|qeϕ q dx dt, |(−∆)r2 Dt|T QT Z −q e β2 B4 = |Dt|T ϕ|qeϕ q dx dt, Q Z T −q e |(−∆)`2 ϕ|qeϕ q dx dt. B5 = QT

Now, we estimate A1 , . . . , A5 and B1 , . . . , B5 in the same way as in Section 3, we obtain inequalities similar to those given in (3.7) and (3.8) Ai = C i R θ i ,

Bi = Di Rδi

for i = 1, 2, . . . , 5,

(4.6)

where θ1 = −4e q+ 2k1 + 4 qe + ρ 2r1 + 2α1 qe + θ3 = − ρ θ2 = −

θ4 = −2β1 qe + θ5 =

−2`1 qe + ρ

N ρ N ρ N ρ N ρ N ρ

N + 2, ρ 2k2 N =− + 4 pe + + 2, ρ ρ 2r2 N =− + 2α2 pe + + 2, ρ ρ N = −2β2 pe + + 2, ρ −2`2 N = pe + + 2. ρ ρ

+ 2,

δ1 = −4e p+

+ 2,

δ2

+ 2,

δ3

+ 2,

δ4

+ 2,

δ5

(4.7)

If we set I=

Z

1/p |v|p ϕ

Z and J = (

QT

|u|q ϕ)1/q ,

QT

inequalities (4.4) and (4.5) become Z IP + u1 ϕ(x, 0) ≤ J(C1 Rθ1 /eq + C2 Rθ2 /eq + C3 Rθ3 /eq + C4 Rθ4 /eq + C5 Rθ5 /eq ), RN

Jq +

(4.8)

Z v1 ϕ(x, 0) ≤ I(D1 R

δ1 q e

+ D2 Rδ2 /eq + D3 Rδ3 /eq + D4 Rδ4 /eq + D5 Rδ5 /eq ).

RN

(4.9)

We can observe that under the conditions on αi and βi , we have θ2 ≤ θ1 ≤ θ4 ,

θ3 ≤ θ4 ,

δ2 ≤ δ1 ≤ δ4 ,

δ3 ≤ δ4 .

EJDE-2016/20


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Hence, for R ≥ 1, we have Rθi ≤ Rθ4 + Rθ5 and Rδi ≤ Rδ4 + Rδ5 , and consequently, inequalities (4.8) and (4.9) can be rewritten as Z θ4 Ip + u1 ϕ(x, 0) ≤ CJ R qe + Rθ5 /eq , (4.10) RN Z δ5 δ4 JP + (4.11) v1 ϕ(x, 0) ≤ DI(R qe + R qe ), RN

where 5 X

C=

Ci ,

and D =

i=1

Since

R RN

u1 ϕ(x, 0) ≥ 0 and

5 X

Di .

i=1

R

v1 ϕ(x, 0) ≥ 0, inequalities (4.10) and (4.11) yield (4.12) I p ≤ CJ Rθ4 /eq + Rθ5 //eq , J P ≤ DI Rδ4 /eq + Rδ5 /eq . (4.13) RN

The constants C and D will be updated at each step of the calculation and will not play a role. This implies that I pq ≤ CI(Rδ4 /ep + Rδ5 /ep )(Rθ4 /eq + R

θ4 q e

θ5 q e

)q , J pq ≤ CJ(R

θ5 q e

)q , J pq−1 ≤ C(R

+R

θ5 q e

)(Rδ4 /ep + Rδ5 /ep )p ,

leading to I pq−1 ≤ C(Rδ4 /ep + Rδ5 /ep )(Rθ4 /eq + R

θ4 q e

+R

θ5 q e

)(Rδ4 /ep + Rδ5 /ep )p . (4.14)

Now, let S1 =

1 q max(δ4 , δ5 ) + max(θ4 , θ5 ), pe qe

S2 =

1 p max(θ4 , θ5 ) + max(δ4 , δ5 ). qe pe

If S1 < 0,

and S2 < 0,

(4.15)

θ4 q e

we have (Rδ4 /ep +Rδ5 /ep )(Rθ4 /eq +Rθ5 /eq )q → 0 and (R +Rθ5 /eq )(Rδ4 /ep +Rδ5 /ep )p → 0 as R → ∞. R Hence, by (4.14), both R I and J vanish as RR → ∞. This implies that J q = QT |u|q ϕ converges to Q∞ |u|q ϕ = 0 and I p = QT |v|p ϕ converges to R |v|p ϕ = 0. Consequently, u ≡ 0 and v ≡ 0. Q∞ As S1 < 0, then we have max(θ4 , θ5 ) < 0 or max(δ4 , δ5 ) 0. Now, we return to the condition (4.15) that lead to the contradiction. Inequalities S1 < 0 and S2 < 0 are equivalent to `1 `2 N pq − 1 S1 = −2 q min(β1 , ) + min(β2 , ) + ( + 2)