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Nouvelle série, tome 93 (107) (2013), 173–186 DOI: 10.2298/PIM1307173H

BLOW UP RESULTS FOR FRACTIONAL DIFFERENTIAL

EQUATIONS AND SYSTEMS Ali Hakem and Mohamed Berbiche

Communicated by Stevan Pilipović

Abstract. The aim of this research paper is to establish sufficient conditions for the nonexistence of global solutions for the following nonlinear fractional differential equation

Dα

0|tu+ (−∆)β/2|u|m−1u+a(x)· ∇|u|q−1u=h(x, t)|u|p, (t, x)Q, u(0, x) =u0(x), xRN

where (−∆)β/2, 0< β <2 is the fractional power of−∆, andDα

0|t, (0< α <1) denotes the time-derivative of arbitraryα(0; 1) in the sense of Caputo. The results are shown by the use of test function theory and extended to systems of the same type.

1. Introduction

In his article [3], Fujita considered the Cauchy problem (1.1) ut= ∆u+u1+˜p inQ=Rn×R+

u(0, x) =a(x) inRn where ˜p >0. Ifpc= 2n, he proved that:

1. If 06p˜6pc anda(x0)>0 for somex0, then any solution to (1.1) blows-up in a finite time.

2. If ˜p > pc, then there exist a solution onQ.

The critical case ˜p=pc was decided later by Hayakawa [6] forN= 1,2 and by Kobayashi, Sirao and Tanaka [9] forn>3.

2010Mathematics Subject Classification: 58J45, 26A33, 35B44.

Key words and phrases: blow-up; fractional derivatives; critical exponent.

This research was partly supported by MESRS-ALGERIA (CNEPRU B02120120054).

173

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In a more recent article, Guedda and Kirane [5] extended the previous results to the equations

ut= (−∆)β/2u+h(x, t)u1+p in Q=RN ×R+ u(0, x) =a(x) in RN

where h(x, t) =O(tσ|x|ρ) for|x| large.

Finally, Kirane and Qafsaoui [8] treated the more general equation ut+ (−∆)β/2(um) +a(x, t)· ∇uq =h(x, t)u1+˜p, in Q.

The technique we use has been introduced by Mitidieri and Pohozaev [10], [11], Pohozaev and Tesei [12], Pohozaev and Veron [14] and used by Hakem and Berb- iche [1].

Let us consider the following nonlinear fractional differential equation (1.2) Dα0|tu+ (−∆)β2(|u|m−1u) +a(x)· ∇(|u|q−1u) =h(x, t)|u|p

u(0, x) =u0(x), x∈RN,

whereDα0|tdenotes the time-derivative of an arbitrary orderα∈(0, α) in the sense of Caputo [14], (−∆)β/2,β ∈[1,2], is the (β2)-fractional power of the Laplacian−∆x

in the xvariable; a(x) := (a1(x), . . . , aN(x)) andh(x, t) are given functions,a(x)·

∇(|u|q−1u) is the scalar product ofa(x) and ∇(|u|q−1u) and the exponents p >1, q >1 andm>1 are positive constants. The nonlocal operator (−∆)β2 is defined by (−∆)β2v(x) =F−1 |ξ|βF(v)(ξ)

(x) for everyvD((−∆)β2) =Hβ(RN), where Hβ(RN) is the homogeneous Sobolev space of orderβ defined by

Hβ(RN) =

u∈ S; (−∆)β2u∈L2(RN) if β /∈N, Hβ(RN) =

u∈L2(RN); (−∆)β2u∈L2(RN) if β∈N,

where S is the space of Schwartz distributions;F denotes the Fourier transform and F−1 its inverse. The fractional Laplacian (−∆)β2 is related to Lévy flights in physics. Many observations and experiments related to Lévy flights (super- diffusion), e.g., collective slip diffusion on solid surfaces, quantum optics or Richard- son turbulent diffusion, have been recently performed. The symmetric β-stable processes (β ∈ (0,2)) are the basic characteristics for a class of jumping Lévy’s processes. Compared with the continuous Brownian motion (β = 2), symmetric β-stable processes have infinite jumps in an arbitrary time interval. The large jumps of these processes make their variances and expectations infinite according to β ∈(0,2) or β ∈(0,1], respectively. It is worth mentioning that when β = 32, the symmetric β-stable processes appear in the study of stellar dynamics. The time fractional derivative has been found to be very effective means to describe the anomalous attenuation behaviors. We here recall some definitions of fractional derivative.

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The left-handed derivative and the right-handed derivative in the Riemann–

Liouville sense for Ψ∈L1(0, T), 0< α <1 are defined as follows:

(Dα0|tΨ)(t) = 1 Γ(1−α)

d dt

Z t

0

Ψ(σ) (t−σ)αdσ, where the symbol Γ stands for the usual Euler gamma function, and

(Dαt|TΨ)(t) =− 1 Γ(1−α)

d dt

Z T

t

Ψ(σ) (σ−t)αdσ, respectively. The Caputo derivative

(Dα0|tΨ)(t) = 1 Γ(1−α)

Z t

0

Ψ(σ) (t−σ)αdσ, requires ΨL1(0, T).

Clearly we have

(Dα0|tg)(t) = 1 Γ(1−α)

g(0) tα +

Z t

0

g(σ) (t−σ)α

,

(Dαt|Tf)(t) = 1 Γ(1−α)

f(T) (T−t)α

Z T

t

f(σ) (σ−t)α

. (1.3)

Therefore the Caputo derivative is related to the Riemann–Liouville derivative by Dα0|tΨ(t) =Dα0|t[Ψ(t)−Ψ(0)].

We will use the formula of integration by parts [13, p. 46]

Z T

0 f(t)(Dα0|tg)(t)dt= Z T

0 g(t)(Dt|Tα f)(t)dt.

Solutions to problem (1.2) are meant in the following sense.

Definition1.1. A functionuLploc(QT), whereQT :=RN×(0, T), is a local weak solution to (1.2) defined on QT, ifuh1/pL1loc(QT, dx dt) such that

Z

QT

h(x, t)ξ|u|pdx dt+ Z

QT

u0Dt|Tα ξ dx dt (1.4)

= Z

QT

uDαt|Tξ dx dt+ Z

QT

|u|m−1u(−∆)β2ξ dx dt

N

X

i=1

Z

QT

|u|q−1uξ∂ai

∂xidx dt− Z

QT

|u|q−1ua· ∇ξ dx dt

for any test functionξCx,t2,1(QT), such thatξ(x, T) = 0.

The integrals in the definition are supposed to be convergent. If in the above definition,T = +∞the solution is called global.

To begin, we set some hypotheses. For the functionh, we require the condition (Hh) h(yR, τ Tβ/α)>ChRσTρβ/α, Ch>0,

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for some σ, ρ > 0 to be determined later, R, T large and τ > 0, y in a bounded domain. It can easily be seen that there is no conditions imposed onσ. The vector a(x) = (a1(x), . . . , aN(x)) is required to satisfy

(Ha) |ai(x)| ∼c|x|δi forxlarge andδi >2.

For later use, we defineδ= max(δi).

1.1. The Results. Now, we may state our first result.

Theorem 1.1. LetN >1,p >max(m, q)>1. The exponentρsatisfies (ρ+ 1)>maxnp

m,(1−α)p,p q

o.

Assume that (Hh)and(Ha)are satisfied and u0(x)satisfiesu0(x)>0. If p6min

1 + α(σ+β) +βρ

αN+β(1α),((αN+β) + (ασ+βρ))q ((δ−1)α+ (N α+β))

, then problem (1.2)admits no global weak solutions other than the trivial one.

Proof. The proof proceeds by contradiction. Suppose that uis a nontrivial solution which exists globally in time. That is exists in (0, T) for any arbitrary T >0. LetT and R be two positive real numbers such that 0 < T Rβ/α < T. For later use, let Φ be a smooth nonincreasing function such that

Φ(z) =

(1, ifz61, 0, ifz>2.

and 06Φ61. The test functionξis chosen so that Z

QT

(hξ)−m/(p−m)

(−∆)β/2ξ

p/(p−m)

dx dt <∞, Z

QT

(hξ)−1/(p−1)|Dt|Tα ξ|p/(p−1)dx dt <∞, Z

QT

h−q/(p−q)ξ

N

X

i=1

∂a

∂xi

p/(p−q)

dx dt <∞, Z

QT

(hξ)−q/(p−q)|a· ∇ξ|p/(p−q)dx dt <∞.

To estimate the right-hand side of (1.2) onQT R2/θ, we write Z

QT R2/θ

|u|m−1u(−∆)β/2ξdx dt= Z

QT R2/θ

|u|m−1u(hξ)m/p(hξ)−m/p(−∆)β/2ξdx dt.

Using theε-Young inequality

XY 6εXp+C(ε)Yp, p+p=pp, X, Y >0, we have the estimate

Z

QT R2/θ

|u|m−1u(−∆)β2ξdx dt

6ε Z

QT R2/θ

hξ|u|pdx dt+C(ε) Z

QT R2/θ

(hξ)p−mm|(−∆)β2ξ|ppmdx dt.

Similarly,

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Z

QT R2/θ

uDt|T Rα 2/θξdx dt= Z

QT R2/θ

u(hξ)1p(hξ)−1p Dt|T Rα 2/θξdx dt 6ε

Z

QT R2/θ

hξ|u|pdx dt+C(ε) Z

QT R2/θ

(hξ)p−11|Dαt|T R2/θξ|pp1dx dt.

Integrating by parts, we get Z

QT R2/θ

a· ∇(|u|q−1u)ξdx dt

=− Z

QT R2/θ

|u|q−1ua· ∇ξdx dt− Z

QT R2/θ

N

X

i=1

|u|q−1uξ∂a

∂xidx dt.

Now writing Z

QT R2/θ

|u|q−1

N

X

i=1

∂a

∂xi

dx dt= Z

QT R2/θ

|u|q−1u(hξ)qphpqξ(p−q)p

N

X

i=1

∂a

∂xi

dx dt,

and using theε-Young inequality, we get Z

QT R2/θ

|u|q−1

N

X

i=1

∂a

∂xi

dx dt

6ε Z

hξ|u|pdx dt+C(ε) Z

QT R2/θ

hp−qqξ

N

X

i=1

∂a

∂xi

p p−q

dx dt.

Similarly, we have Z

QT R2/θ

|u|q−1u(a· ∇ξ)dx dt 6ε

Z

QT R2/θ

hξ|u|pdx dt+C(ε) Z

QT R2/θ

(hξ)p−qq|a· ∇ξ|

p pq

dx dt.

Combining the above estimates with (1.4) and takingεsmall enough, we infer that

Z

QT R2/θ

u0Dt|T Rα 2/θξ dx dt+ Z

QT R2/θ

|u|pξh dx dt (1.5)

6C(ε) Z

QT R2/θ

(hξ)p−1−1

Dαt|T R2/θξ

p

p−1dx dt+ Z

QT R2/θ

hp−qq ξ

N

X

i=1

∂a

∂xi

p p−q

dx dt

+ Z

QT R2/θ

(hξ)p−qq|a· ∇ξ|

p pq

dx dt+ Z

QT R2/θ

(hξ)p−mm

(−∆)β2ξ

m pmdx dt

.

At this stage, we set

ξ(x, t) := Φ|x|2+tθ R2

,

where Randθ are positive real numbers to be determined latter.

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We note that ξ(x, T R2/θ) = 0 forTθ>2, then by (1.3) we have Z

QT R2/θ

u0Dαt|T R2/θξ dx dt>0.

Let us perform the change of variablesτ=t/R2/θ, y=x/R, and set Ω :=

(y, τ)∈RN×R+, |y|2+τθ<2 , µ(y, τ) :=|y|2+τθ. We have the estimates

Z

QT R2/θ

|Dαt|T R2/θξ|p/(p−1)(hξ)−1/(p−1)dx dt 6Rθ2αp/(p−1)+N+2θp−11 (σ+θ)

Z

|Dατ|TΦoµ|p−1p (Ch|y|στρΦoµ)p−11 dy dτ, Z

QT R2/θ

|(−∆)β2ξ|p−mp (hξ)p−mmdx dt 6Rp−βpm+N+2θpmm(σ+θ)

Z

|(−∆)β2Φoµ|ppm(Ch|y|στρΦoµ)p−mmdy dτ, Z

QT R2/θ

hpqq

N

X

i=1

∂a

∂xi

p p−q

ξdx dt

6R

−q

pq(σ+2θρ)+ppq(δ−2)+N+θ2Z

(Ch|y|στρ)

−q pq|

N

X

i=1

∂ai

∂yi

|ppqΦoµdy dτ, and

Z

QT R2/θ

(hξ)−q/(p−q)|a· ∇ξ|p/(p−q)dx dt 6Rp−qq(σ+θ2ρ)+(δ−1)ppq+N+2θ

Z

(Ch|y|στρΦoµ)p−qq|a· ∇ξ|ppqdτ dy.

Now, we chooseθ such that

βpθ+ (p−m)θN+ 2(p−m)m(θσ+ 2ρ)

6−2αp+ (p−1)θN+ 2(p−1)−(θσ+ 2ρ), then it is sufficient to takeθ= β . We then have the estimate

(1.6)

Z

QT Rβ/α

h|u|pξ dx dt6C(ε)(Rs1+Rs2+Rs3), where

(p−1)θs1=−2αp+ (p−1)θN+ 2(p−1)−(θσ+ 2ρ) (p−m)θs2=−βpθ+ (p−m)θN+ 2(p−m)m(θσ+ 2ρ)

(p−q)θs3= (δ−1)pθ+ (N θ+ 2)(p−q)q(θσ+ 2ρ)

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and C(ε) is a generic positive constant depending on ε. Now if we choose max(s1,s2, s3)<0, that is

p <min

1 + α(σ+β) +βρ

αN+β(1−α),((αN+β) + (ασ+βρ))q ((δ−1)α+ (N α+β))

and letR→ ∞in (1.6); we obtain Z

RN×R+

h|u|pdx dt60.

This implies thatu= 0 a.e., which is a contradiction. Ifp=pc(i.e., max(s1,s2, s3)

= 0) the critical case, we have from (1.6) (1.7)

Z

Rn×R+

h|u|pdx dt6C.

We modify the test functionξby introducing a new fixed constantS (0< S < R), such that

ξ(x, t) := Φ |x|2

R2 + tθ (SR)2

. We set

CR,S =n

(x, t)∈Rn×R+:R26|x|2+ tθ

S2 62R2o . See that because of the convergence of the integral in (1.7), then

(1.8) lim

R→∞

Z

CR,S

h|u|pξ dx dt= 0.

By using the Hölder inequality, we get Z

QT(SR)2

|u|q−1ua· ∇ξ dx dt= Z

CR,S

|u|q−1ua· ∇ξ dx dt

6 Z

CR,S

|u|phξ dx dt pqZ

CR,S

(hξ)pqq|a· ∇ξ|ppqdx dt p−qp

,

where we have used that the support ofa· ∇ξisCR,S. Taking into account of the scaled variables: t = (RS)2θτ, x=Ry ξ(x, t) =ξ(Ry,(RS)2θτ) =χ(y, τ) and the fact thatp=pc then instead of estimate (1.5), we get

(1.9) (1−3ε) Z

QT(SR)2

h|u|pξ dx dt

6 Z

CR,S

|u|phξ dx dt qpZ

CR,S

(hξ)p−qq |a· ∇ξ|p−qp dx dt ppq

+C(ε) L1Sp11(θ)−2θαpp1+2θ +L2Spmm(θ)+2θ +L3S

−q

pq(2θρ)+2θ , where

L1:=

Z

χp−11 Dt|Tα χ

p p−1dy dτ,

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L2:=

Z

χpmm

(−∆)β2χ

m pmdy dτ,

L3:=

Z

χ

n

X

i=1

∂a

∂yi

p p−q

dy dτ

.

Using (1.9), we obtain via (1.8), after passing to the limit asR→ ∞ (1.10)

Z

Rn×R+

h|u|pdx dt6C

Sp−11 (θ)−2θαp−1p +2θ +Sp−mm (θ)+2θ +S

q

p−q(2θρ)+2θ .

Finally, we realize that the left-hand side of (1.10) is independent of S, then by passing to the limit whenSgoes to infinity, we obtainu= 0, which is contradiction

and this completes the proof.

Remark 1.1. When the vector a = 0 and q = m = 1, we recover the case studied by Kirane-Tatar [6]. When a= 0,q=m= 1,σ=ρ= 0,α= 1 andβ= 2, the critical exponent coincides with the well known Fujita exponent [2].

2. System of Fractional Differential Equations

This section is devoted to the following system of reaction-diffusion equations (FDS) D0|tα (u−u0) + (−∆)β2(|u|m−1u) =h(t, x)|v|p+|g(t, x)||u|r in Q

D0|tδ (v−v0) + (−∆)γ2(|v|m−1v) =k(t, x)|u|q+|l(t, x)||v|s in Q subject to the initial conditions u(x,0) =u0(x)>0,v(x,0) =v0(x)>0,x∈RN, where 0< α, δ <1 and 06γ, β62.

The functions h, g, k, lare assumed to satisfy the conditions h(t, x)>C1tω1|x|d1, g(t, x)tω2|x|d2 when|x|large k(t, x)>C3tω3|x|d3, l(t, x)tω4|x|d4 when|x|large,

for t > 0, x≫ 1, ω1 > 0, ω2 > 0, ω3 > 0, ω4 > 0, d1 >0, d2 > 0, d3 >0 and d4 >0. We set λi =ωi+αβdi and ηi =ωi+γδdi for i= 1, . . . ,4. For the system (FDS) we have

Theorem 2.1. Letp, q >1, m, r,ands>1 and assume that (1) pq >max(m2, sm, sr, mr).

(2) 32+1)+1) >q

r >1, 14+1)+1) > ps >1, 34+1)+1) > msq2 >1.

If

(2.1) N 6max(θ1, θ2)

where θ1= min16j67θ1j2= min16j67θ2j

θ11=1+p2λ3+pq(δ−(1−1p)) +p2q(α−(1−1q)) ((p−1)qδγ +(q−1)pβ 2α) , θ12=mqη1+p2λ3+pq(δ−(1−mp)) +p2q(α−(1−1q))

((p−m)qδγ +(q−1)pβ 2α)

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θ13=sq(λ1+ 1)−pq(λ4+ 1) +p2λ3+p2q(α−(1−1q)) ((p−s)q+ (q−1)p2)αβ , θ14=qmη1+p23+pqm(δ−(1−1p)) +p2q(α−(1−mq))

(δ(p−1)mqγ +αp2(q−m)β ) , θ15=spλ3+sq21+ 1)−pq24+ 1) +pqs(α−(1−1q))

((q−1)sp+ (p−s)q2)αβ , θ16=rqη1p2q(η2+ 1) +rp23+ 1) +rpq(δ−1 +1p)

((p−1)rq+ (q−r)p2)γδ , θ17=qrs(λ1+ 1)−pqr(λ4+ 1) +p2r(λ3+ 1)−p2q(λ2+ 1)

((p−s)rq+ (q−r)p2)αβ . θ21=q2η1+pq(α−(1−1q)) +3+pq2(δ−(1−1p))

hp(q−1)αβ + (p−1)q2γδi , θ22=(η3+ 1)rp−(η2+ 1)pq+q2η1+pq2(δ−1 + 1p)

((q−r)p+ (p−1)q2)δγ , θ23=mpλ3+mq2η1+mpq(α−(1−1q)) +pq2(δ−(1−mp))

(mp(q−1)αβ +(p−m)qγ 2δ) , θ24=prm(η3+ 1)−mpq(η2+ 1) +mq2η1+pq2(δ−(1−mp))

((q−r)mp+ (p−m)q2)δγ , θ25=smq(λ1+ 1)−mpq(λ4+ 1) +p23+p2q(α−(1−mq)) ((p−s)mq+p2(q−m))αβ , θ26=msp(λ3+ 1) +sq2λ1pq24+ 1) +pqs(α−(1−qp))

((q−m)sp+ (p−s)q2)αβ , θ27=rsp(λ3+ 1)−spq(λ2+ 1) +sq21+ 1)−pq24+ 1)

((q−r)sp+ (p−s)q2)αβ ,

then the system (FDS)(with the initial data) does not admit nontrivial global weak solutions.

Proof. Here again the proof proceeds by contradiction. Let ξj(x, t) = Φt2+|x|j

R2

, j= 1,2 where R >0,µ1=β/α andµ2=γ/δ.

The weak formulation of solutions to (FDS) reads Z

QT R

h|v|pξ1+ Z

QT R

u0Dt|Tα ξ1

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= Z

QT R

uDαt|Tξ1+ Z

QT R

(|u|m−1u)(−∆)β2ξ1− Z

QT R

g|u|rξ1,

Z

QT R

2|u|q+ Z

QT R

v0Dt|δT Rξ2

= Z

QT R

vDδt|Tξ2+ Z

QT R

(|v|m−1v)(−∆)γ2ξ2− Z

QT R

2|v|s.

Using the Hölder inequality, we may write Z

QT R

uDt|Tα ξ16 Z

QT R

2|u|q 1qZ

QT R

(kξ2)q−11 |Dαt|Tξ1|q−1q q−1q

,

Z

QT R

(|u|m−1u)(−∆)β2ξ16 Z

QT R

2|u|q mqZ

QT R

|(−∆)β2ξ1|m−q−q (kξ2)m−qm m−qq

,

Z

QT R

g|u|rξ16 Z

QT R

2|u|q rq Z

QT R

(kξ2)qrr(gξ1)qqr

qr q

. Consequently

Z

QT R

h|v|pξ16 Z

QT R

2|u|q 1q

· A+ Z

QT R

2|u|q mq

· B+ Z

QT R

2|u|q rq

· C

where

A= Z

QT R

(kξ2)q−11 |Dαt|Tξ1|q−1q

q1 q

,

B= Z

QT R

|(−∆)β2ξ1|m−qq(kξ2)mmq m−qq

,

C= Z

QT R

(kξ2)qrr(gξ1)qqr

qr q

.

Similarly we obtain the estimates Z

QT R

vDt|Tδ ξ26 Z

QT R

|v|p(hξ1) 1pZ

QT R

(hξ1)p−11 |Dδt|Tξ2|p−1p p−1p

,

Z

QT R

(|v|m−1v)(−∆)γ2ξ26 Z

QT R

|v|p1

mpZ

QT R

(hξ1)mmp|(−∆)γ2ξ2|m−pp m−pp

,

Z

QT R

2|v|s6 Z

QT R

|v|p1

psZ

QT R

(hξ1)s−ps (lξ2)s−pp s−pp

.

So we get Z

QT R

2|u|q 6 Z

QT R

|v|p(hξ1) p1

·D+

Z

QT R

|v|p1

mp

·E+

Z

QT R

|v|p1

sp

·F,

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with

D:=

Z

QT R

(hξ1)p11|Dδt|Tξ2|pp1 pp1

,

E :=

Z

QT R

(hξ1)m−pm |(−∆)γ2ξ2|m−pp m−pp

,

F :=

Z

QT R

(hξ1)ssp(lξ2)spp spp

.

If we set

Y:=

Z

QT R

h|v|pξ1, Z :=

Z

QT R

2|u|q,

then we have

Yq 6Z · A+Zm· B+Zr· C, Zp6Y · D+Ym· E+Ys· F which yields

Ypq63p−1(Zp· Ap+Zpm· Bp+Zpr· Cp), (2.2)

Zpq63p−1(Yq· Dq+Yqm· Eq+Yqs· Fq).

(2.3)

We have used in (2.2) and (2.3) the inequality

(a+b+c)p63p−1(ap+bp+cp), p>1, a, b, c>0.

It then follows from (2.2), (2.3) that

Ypq6c(p, m, r) (Y · D+Ym· E+Ys· F)· Ap +(Ym· Dm+Ym2· Em+Ysm· Fm)· Bp +(Yr· Dr+Ymr· Er+Ysr· Fr)· Cp

,

where c(p, m, r) is a positive constant depending on p, m and r. Using ε-Young inequality, we get

Ypq6c(p, m, r, ε)

(DAp)pqpq1 + (EAp)pqpqm + (F Ap)pqpqs (2.4)

+ (DmBp)pqpqm + (EmBp)

pq pq−m2

+ (FmBp)pqpqsm + (DrCp)pqpqr + (ErCp)pqpqmr + (FrCp)pqpqsr

.

Now, using the scaled variables (y, τ) defined byt= andx=Rαβy, inA,B,F while in D,E,C we use the variables (y, τ) defined by t = and x =Rδγy, we obtain

(2.5) Ypq6c Rl1+Rl2+Rl3+Rl4+Rl5+Rl6+Rl7+Rl8+Rl9

(12)

where

(pq−1)l1:=N1+p2λ3+pq(δ−(1−1p)) +p2q(α−1 + 1q) ((p−1)qδγ +(q−1)pβ 2α) , (pq−m)l2:=Nmqη1+p2λ3+pq(δ−(1−mp)) +p2q(α−(1−1q))

((p−m)qδγ +(q−1)pβ 2α) , (pq−s)l3:=Nsq(λ1+ 1)−pq(λ4+ 1) +p2λ3+p2q(α−(1−1q))

((p−s)q+ (q−1)p2)αβ , (pq−m)l4:=Nqmη1+p23+pqm(δ−(1−1p)) +p2q(α−(1−mq))

(δ(p−1)mqγ +αp2(q−m)β ) , (pq−m2)l5:=Nm21+p23+pqm(δ−(1−mp)) +p2q(α−1 + mq)

(δm(p−m)qγ +p2(q−m)αβ ) , (pq−sm)l6:=Nsmq(λ1+ 1)−mpq(λ4+ 1) +p23+p2q(α−(1−mq))

((p−s)mq+p2(q−m))αβ , (pq−r)l7:=Nrqη1p2q(η2+ 1) +rp23+ 1) +rpq(δ−1 +1p)

((p−1)rq+ (q−r)p2)δγ , (pq−mr)l8:=Nmrqη1+ (η3+ 1)p2rp2q(η2+ 1) +rpq(δ−(1−mp))

((p−m)rq+ (q−r)p2)γδ , (pq−sr)l9:=Nqrs(λ1+ 1)−pqr(λ4+ 1) +p2r(λ3+ 1)−p2q(λ2+ 1)

((p−s)rq+ (q−r)p2)αβ . In the same way we find

Zpq6c(ε)

(ADq)pq−1pq + (BDq)pq−mpq + (CDq)pq−rpq (2.6)

+ (AmEq)pq−mpq + (BmEq)pq−mpq2 + (CmEq)pq−mrpq + (AsFq)pqpqs + (BsFq)pqpqms + (CsFq)pqpqrs

. Similarly, we have forZ

(2.7) Zpq6c Rj1+Rj2+Rj3+Rj4+Rj5+Rj6+Rj7+Rj8+Rj9 , where

(pq−1)j1:=Nq2η1+pq(α−(1−1q)) +3+pq2(δ−(1−1p)) p(q−1)αβ+ (p−1)q2γδ , (pq−m)j2:=N

mpλ3+q2η1+pq(α−(1−mq)) +pq2(δ−(1−1p)) (p(q−m)αβ +(p−1)qγ 2δ) , (pq−r)j3 :=N−(η3+ 1)rp−(η2+ 1)pq+q2η1+pq2(δ−1 +1p)

((q−r)p+ (p−1)q2)γδ ,

参照

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