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 Bulletin of the Transilvania University of Braşov • Vol xx(xx), No. x - 2018

 Series III: Mathematics, Informatics, Physics, xx-xx


  Medjahed DJILALI  and Ali HAKEM  


In the present paper, we prove nonexistence results for a nonlinear evolution equation

u_tt+(-Δ)^(α/2) (u)+u_t=f(t,x)|u|^p

  posed in (0,T)×R^N, where (-Δ)^(α/2),0<α≤2 is α/2- fractional power of -Δ.

  Our method of proof is based on suitable choices of the test functions in the weak

  formulation of the sought solutions. Then, we extend this result to the case

  of a 2×2 system of the same type.

 2010 Mathematics Subject Classification: 35A01, 35D30, 47J35

Key words: Damping term, fractional Laplacian, test function, weak solution.

1  Introduction

 In this article, we are concerned with the following problem:

(■(&u_tt+(-Δ)^(α/2) (u)+u_t=f(t,x)|u|^p,  (t,x)∈(0,T)×(R^N)@&@&u(0,x)=u_0 (x)≥0,   u_t (0,x)=u_1 (x)≥0, x∈R^N,)┤ (1)

 for some 0<T≤+∞, where (-Δ)^(α/2) with 0<α≤2 is the fractional power of the (-Δ), p>1 .

The integral representation of the fractional Laplacian in the N-dimensional space is

(-Δ)^(α/2) ψ(x)=-c_N (α)∫_(R^N)▒‍  (ψ(x+z)-ψ(x))/|z|^(N+α)  dz, ∀x∈R^N, (2)

where c_N (α)=Γ((N+α)/2)/(2π^(N/2+α) Γ(1-α/2)), and Γ denotes the gamma function.

Note that The fractional Laplacian ((-Δ)^(α/2)) [8, 12] with α∈(0;2]

is a pseudo-differential operator defined by:

(-Δ)^(α/2) u(x)=F^(-1) {|ζ|^α F(u)(ζ)}(x)  for  all  x∈R^N,

where F and F^(-1) are Fourier transform and its inverse, respectively.

Set Σ_T=(0,T)×(R^N).

Before beginning this work, let us point out that many authors were interested in studying the Cauchy problem for a nonlinear wave equation with damping term:

(■(&u_tt+u_t-Δu=|u|^p,  (t,x)∈(0,∞)×(R^N)@&@&u(0,x)=u_0 (x),   u_t (0,x)=u_1 (x), x∈R^N,)┤ (3)

Todorova-Yordanov [14] showed that, if p_c<p≤N/(N-2),

for n≥3 and p_c<p<∞, for N=1,2, where p_c=1+2/N, then (3) subjected to initial data u(0,x)=εu_0 (x),u_t (0,x)=εu_1 (x),ε>0,x∈R^N, admits a unique global solution, and they proved that if 1<p<1+2/N, then the solution u blows up in a finite time.

Qi. Zhang [15] studied the case 1<p<1+2/N, when u_i,i=0,1 is compactly supported and ∫▒‍ u_i (x)dx>0, he proved that global solution of (3) does not exist. Therefore, he showed that p=1+2/N belongs to the blow-up case.

Let us point that T. Ogawa and H. Takeda [9] showed that when 1<p<1+2/N and the support of data is not far away from the obstacle, then the weak solution of (3) does not exists globally, but if the supports of the initial data are sufficiently away from the boundary, they treated the problem as in the Cauchy problem.

Fino-Ibrahim and Wehbe [3] generalized the results of Ogawa-Takeda [9] by proving the blow-up of solutions of (3) under weaker assumptions on the initial data and they extended this results to the critical case p=1+2/N.

Observe that A. Hakem [7] treated the problem:

(■(&u_tt+g(t)u_t-Δu=|u|^p,  (t,x)∈(0,∞)×(R^N),@&@&u(0,x)=u_0 (x),   u_t (0,x)=u_1 (x), x∈R^N,)┤ (4)

 where g(t) is a function behaving like t^β, 0≤β<1. He obtained the non-existence of weak solution for the problem (4), when 1<p≤(N+2)/(N+2β) , then he extended this result to the case of a system :

(■(&u_tt+-Δu+g(t)u_t=|v|^p,  (t,x)∈(0,+∞)×R^[email protected]&v_tt+-Δv+f(t)v_t=|u|^q,  (t,x)∈(0,+∞)×R^[email protected]&u(0,x)=u_0 (x), u_t (0,x)=u_1 (x)@&v(0,x)=v_0 (x), v_t (0,x)=v_1 (x),)┤ (5)

 g(t) and f(t) are functions behaving like t^β and t^α, respectively, where 0≤β,α<1.

Hakem [7] showed that, if

N/2≤1/(pq-1) max[1-β+p(1-α),1-α+q(1-β)]-max(α,β),

then the problem (5) has only the trivial solution.

F. Sun and M. Wang [13] studied the same type of (5) in the case t^β=t^α=1, they showed that if max{(1+p)/(pq-1),(1+q)/(pq-1)}≥N/2 for N≥1 where p,q≥1 and satisfy pq>1, then every solution with initial data having positive average value does not exist globally.

Our purpose of this work is to generalize some of the above results, so in the first part of our research and with the suitable choice of the test function, we prove the non-existence of nontrivial global weak solution of (1), and in the second part we extend the results of A. Hakem’s [7] work to the fractional Laplacian [8, 12]. The same technique is used to prove the non-existence of solutions to the system:

(■(&u_tt+(-Δ)^(α/2) (u)+u_t=g(t,x)|v|^p,  (t,x)∈(0,+∞)×R^[email protected]&@&v_tt+(-Δ)^(β/2) (v)+v_t=f(t,x)|u|^q,  (t,x)∈(0,+∞)×R^N )┤ (6)

 subjected to the conditions

u(0,x)=u_0 (x)≥0, u_t (0,x)=u_1 (x)≥0

v(0,x)=v_0 (x)≥0, v_t (0,x)=v_1 (x)≥0.

2  Preliminaries

 The results of our research are based on the following definitions:

Definition 1 We say that u≥0 is a local weak solution to (1), defined in Σ_T,

0<T<+∞, if u is a locally integrable function such that u^p f∈L_loc^1 (Σ_T) and

■(∫_(Σ_T)▒‍ f|u|^p Ψ dx dt&+∫_(R^N)▒‍ u_0 (x)Ψ(0,x) dx+∫_(R^N)▒‍ u_1 (x)Ψ(0,x) dx-∫_(R^N)▒‍ u_0 (x)Ψ_t (0,x) [email protected]&=∫_(Σ_T)▒‍ uΨ_tt  dx dt-∫_(Σ_T)▒‍ uΨ_t  dx dt+∫_(Σ_T)▒‍ u(-Δ)^(α/2) Ψ dx dt,)

 (see ) is satisfied for any Ψ∈C_0^∞ (¯(Σ_T )) which Ψ(T,x)=Ψ_t (T,x)=0,  for large |x|.

 (see [2] p 5501).

Definition 2 We say that u≥0 is global weak solution to (1) if it is a local solution to (1) defined in Σ_T for any T>0.

Definition 3  Let Σ_T=(0,T)×R^N , 0<T<+∞.

We say that (u,v)∈(L_loc^1 (Σ_T))^2 is a local weak solution to problem (19) on Σ_T,

if (gv^p,fu^q)∈(L_loc^1 (Σ_T))^2, and it satisfies

■(∫_(Σ_T)▒‍ g|v|^p ζ dx dt&+∫_(R^N)▒‍ u_0 (x)ζ(0,x) dx+∫_(R^N)▒‍ u_1 (x)ζ(0,x) dx-∫_(R^N)▒‍ u_0 (x)ζ_t (0,x) [email protected]&=∫_(Σ_T)▒‍ uζ_tt  dx dt+∫_(Σ_T)▒‍ u_t ζ dx dt+∫_(Σ_T)▒‍ u(-Δ)^(α/2) ζ dx dt.) (7)


■(∫_(Σ_T)▒‍ f|u|^q ζ dx dt&+∫_(R^N)▒‍ v_0 (x)ζ(0,x) dx+∫_(R^N)▒‍ v_1 (x)ζ(0,x) dx-∫_(R^N)▒‍ v_0 (x)ζ_t (0,x) [email protected]&=∫_(Σ_T)▒‍ vζ_tt  dx dt+∫_(Σ_T)▒‍ v_t ζ dx dt+∫_(Q_T)▒‍ v(-Δ)^(β/2) ζ dx dt.) (8)

 for all test function ζ∈C_(t,x)^2,2 (Σ_T) such as ζ≥0 and ζ(T,x)=ζ_t (T,x)=0,

 (see [13] p 2890).

We notice that, in all steps of proof , C>0 is a real positive number which may change from line to line.

3  Main results

3.1  Nonexistence results for semi-linear hyperbolic equation

 We consider the following Cauchy problem :

(■(&u_tt+(-Δ)^(α/2) (u)+u_t=f(t,x)|u|^p,  (t,x)∈(0,T)×(R^N)@&@&u(0,x)=u_0 (x)≥0,   u_t (0,x)=u_1 (x)≥0, x∈R^N,)┤ (9)

 for some 0<T≤+∞, where (-Δ)^(α/2) with 0<α≤2 is the fractional power of the -Δ,

p>1, and the function f is positive and satisfy the condition:

For every compact Ω⊂R_+×R^N, there exists a real l>0 such that:

f(tR_ ,xR^(1/α))=O(R^l),where  R>0  and large, (t,x)∈Ω. (10)

Theorem 1  Assume that (u_0,u_1)∈(L^1 (R^N))^2 and the condition (10) is satisfied, if

p≤1+((1+l)α)/N, (11)

then the problem (9) has no nontrivial global weak solution.

Proof. Let Φ the test function such that

Φ(r)=(■(&0, if r≥2,@&1, if r≤1,)┤


0≤Φ≤1, |Φ^' |≤C/r, forall r>0.

Now multiplying the equation (9) by Ψ and integrating by parts on Σ_T=(0,T)×(R^N), we get

■(∫_(Σ_T)▒‍ f|u|^p Ψ dx dt&+∫_(R^N)▒‍ u_0 (x)Ψ(0,x) [email protected]&+∫_(R^N)▒‍ u_1 (x)Ψ(0,x) dx-∫_(R^N)▒‍ u_0 (x)Ψ_t (0,x) [email protected]&=∫_(Σ_T)▒‍ uΨ_tt  dx dt-∫_(Σ_T)▒‍ uΨ_t  dx dt+∫_(Σ_T)▒‍ u(-Δ)^(α/2) Ψ dx dt.) (12)


Ψ(t,x)=Φ((t^2+|x|^2α)/R^2 ), R>0.

with the fact that

Ψ_t (t,x)=2tR^(-2) Φ^' ((t^2+|x|^2α)/R^2 )

we have

Ψ_t (0,x)=0.

Thus the formula (12) will be on the shape

■(∫_(Σ_T)▒‍ f|u|^p Ψ dx dt&+∫_(R^N)▒‍ u_0 (x)Ψ(0,x) dx+∫_(R^N)▒‍ u_1 (x)Ψ(0,x) [email protected]&=∫_(Σ_T)▒‍ uΨ_tt  dx dt-∫_(Σ_T)▒‍ uΨ_t  dx dt+∫_(Σ_T)▒‍ u((-Δ)^(α/2) Ψ) dx dt.) (13)

 To estimate

∫_(Σ_T)▒‍ uΨ_tt  dx dt,

we observe that

∫_(Σ_T)▒‍ uΨ_tt  dx dt=∫_(Σ_T)▒‍ u(fΨ)^(1/p) Ψ_tt (fΨ)^((-1)/p)  dx dt,

we have olso

∫_(Σ_T)▒‍ uΨ_t  dx dt=∫_(Σ_T)▒‍ u(fΨ)^(1/p) Ψ_t (fΨ)^((-1)/p)  dx dt,


∫_(Σ_T)▒‍ u((-Δ)^(α/2) Ψ) dx dt=∫_(Σ_T)▒‍ u(fΨ)^(1/p) ((-Δ)^(α/2) Ψ)(fΨ)^((-1)/p)  dx dt.

An application of the ε-Young’s inequality

ab≤εa^p+C(ε) b^p ̃   where  a>0,b>0,ε>0,

  pp ̃=p+p ̃   and  C(ε)=(εp)^((-p ̃)/p) p ̃^(-1),

in the first integral of the right hand side of (13), we obtain

∫_(Σ_T)▒‍ uΨ_tt  dx dt≤ε∫_(Σ_T)▒‍ |u|^p fΨ dx dt+C(ε)∫_(Σ_T)▒‍ |Ψ_tt |^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt,

in the second integral of the right hand side of (13), we get

■(∫_(Σ_T)▒‍ uΨ_t  dx dt&≤ε∫_(Σ_T)▒‍ |u|^p fΨ dx dt+C(ε)∫_(Σ_T)▒‍(|Ψ_t | )^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx [email protected]&≤ε∫_(Σ_T)▒‍ |u|^p fΨ dx dt+C(ε)∫_(Σ_T)▒‍ |Ψ_t |^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt,)

 And in the third integral of the right hand side of (13), we have

|∫_(Σ_T)▒‍ u(-Δ)^(α/2) Ψ dx dt|≤ε∫_(Σ_T)▒‍ |u|^p fΨ dx dt+C(ε)∫_(Σ_T)▒‍ |(-Δ)^(α/2) (Ψ)|^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt.

Finally, we get

■(∫_(Σ_T)▒‍ |u|^p fΨ dx dt≤C∫_(Σ_T)▒‍ |Ψ_tt |^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt&+C∫_(Σ_T)▒‍ |Ψ_t |^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx [email protected]&+C∫_(Σ_T)▒‍ |(-Δ)^(α/2) (Ψ)|^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt.) (14)

 By the choice of Ψ, it is easy to show that

(■(&∫_(Σ_T)▒‍ |(Ψ_tt)|^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt<∞@&@&∫_(Σ_T)▒‍ |(Ψ_t)|^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt<∞@&@&∫_(Σ_T)▒‍ |(-Δ)^(α/2) Ψ|^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt<∞)┤

At this stage, we introduce the scaled variables : τ=tR^(-1), ζ=xR^((-1)/α) and use the fact that Ψ_t=R^(-1) Ψ_τ,Ψ_tt=R^(-2) Ψ_ττ,(-Δ)_x^(α/2) Ψ=R^(-1) (-Δ)_ζ^(α/2) Ψ, and set

Ω={(τ,ζ)∈R^+×R^N;1≤τ^2+|ζ|^2α≤2}, φ(τ,ζ)=τ^2+|ζ|^2α,

we arrive at

■(∫_(Σ_T)▒‍ |u|^p fΨ dx dt≤&CR^(θ_1 ) ∫_Ω▒‍ |(Ψ_ττ)(φ)|^(p/(p-1)) (fΨ)^((-1)/(p-1))  dζ dτ@&+CR^(θ_2 ) ∫_Ω▒‍ |(Ψ_τ)(φ)|^(p/(p-1)) (fΨ)^((-1)/(p-1))  dζ dτ@&+CR^(θ_3 ) ∫_Ω▒‍ |(-Δ)^(α/2) Ψ(φ)|^(p/(p-1)) (fΨ)^((-1)/(p-1))  dζ dτ.) (15)



One can easily observe that: θ_1<θ_2=θ_3, we infer that

■(∫_(Σ_T)▒‍ |u|^p fΨ dx dt&≤CR^θ [∫_Ω▒‍ |(Ψ_ττ)(φ)|^(p/(p-1)) (fΨ)^((-1)/(p-1))  dζ dτ+∫_Ω▒‍ |(Ψ_τ)(φ)|^(p/(p-1)) (fΨ)^((-1)/(p-1))  dζ dτ@&+∫_Ω▒‍ |(-Δ)^(α/2) Ψ(φ)|^(p/(p-1)) (fΨ)^((-1)/(p-1))  dζ dτ],) (16)

where R>0, large and θ:=θ_2=1/(p-1)[(N(p-1))/α-1-l].

We have two cases:  

    • If


then the right-hand side of (16) gos to 0 when R tends to infinity, while the left hand side converge to

∫_(R^+×R^N)▒‍ |u|^p f dx dt.

Therefore, if u exists then necessarily u≡0.

    • If


then we have

∫_(R^+×R^N)▒‍ |u|^p f dx dt<+∞. (17)

By using (13) we obtain

■(∫_(Σ_T)▒‍ f|u|^p Ψ dx dt&+∫_(R^N)▒‍ u_0 (x)Ψ(0,x) dx+∫_(R^N)▒‍ u_1 (x)Ψ(0,x) [email protected]&≤∫_(Σ_T)▒‍ u(fΨ)^(1/p) |Ψ_tt |(fΨ)^((-1)/p)  dx [email protected]&+∫_(Σ_T)▒‍ u(fΨ)^(1/p) |Ψ_t |(fΨ)^((-1)/p)  dx [email protected]&+∫_(Σ_T)▒‍ u(fΨ)^(1/p) |(-Δ)^(α/2) Ψ|(fΨ)^((-1)/p)  dx dt.) (18)

 Accordingly, using Hölder’s inequality in the right hand side of (18), yields

■(∫_(Σ_T)▒‍ |u|^p Ψ dx dt≤&(∫_(Σ_T)▒‍ u^p fΨ dx dt)^(1/p) (∫_(Σ_T)▒‍ |Ψ_tt |^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt)^((p-1)/p)@&+(∫_(Σ_T)▒‍ u^p fΨ dx dt)^(1/p) (∫_(Σ_T)▒‍(|Ψ_t | )^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt)^((p-1)/p)@&+(∫_(Σ_T)▒‍ u^p fΨ dx dt)^(1/p) (∫_(Σ_T)▒‍(|(-Δ)^(α/2) Ψ| )^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt)^((p-1)/p).)

we obtain

■(∫_(Σ_T)▒‍ f|u|^p Ψ dx dt≤&(∫_(Σ_T)▒‍ u^p fΨ dx dt)^(1/p) [(∫_(Σ_T)▒‍ |Ψ_tt |^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt)^((p-1)/p)@&+(∫_(Σ_T)▒‍(|Ψ_t | )^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt)^((p-1)/p)@&+(∫_(Σ_T)▒‍(|(-Δ)^(α/2) Ψ| )^(p/(p-1)) (fΨ)^((-1)/(p-1))  dx dt)^((p-1)/p)].)

Because (N(p-1))/α-1-l=0, we get from (17) that

■(∫_(Σ_T)▒‍ f|u|^p Ψ dx dt≤&(∫_(Ω_2)▒‍ u^p fΨ dx dt)^(1/p) [(∫_(Ω_1)▒‍ |Ψ_ττ (φ)|^(p/(p-1)) (fΨ(φ))^((-1)/(p-1))  dζ dτ)^((p-1)/p)@&+(∫_(Ω_1)▒‍(|Ψ_τ (φ)| )^(p/(p-1)) (fΨ(φ))^((-1)/(p-1))  dζ dτ)^((p-1)/p)@&+(∫_(Ω_1)▒‍(|(-Δ)^(α/2) Ψ(φ)| )^(p/(p-1)) (fΨ(φ))^((-1)/(p-1))  dζ dτ)^((p-1)/p)],)


Ω_1={(τ,ζ)∈R^+×R^N;  1≤τ^2+|ζ|^2α≤2},


Ω_2={(t,x)∈R^+×R^N;  R^2≤t^2+|x|^2α≤2R^2}.

Taking into account the fact that

∫_(R^+×R^N)▒‍ |u|^p f dx dt<+∞,

we obtain  

lim┬■(R→+∞) ∫_(Ω_2)▒‍ |u|^p fΨ dx dt=0,

hence, we conclude that

∫_(R^+×R^N)▒‍ |u|^p f dx dt=0.

Thus, u≡0. We deduce that no global positive solution is possible. This finishes the proof.   ∎

Remark 1 We can observe that in the case α=2,l=0, we retrieve the results obtained by Todorova-Yordanov [14] and Qi. Zhang [15].

3.2  Case of system of equations

 In this section we consider the problem

(■(&u_tt+(-Δ)^(α/2) (u)+u_t=g(t,x)|v|^p,  (t,x)∈(0,+∞)×R^[email protected]&@&v_tt+(-Δ)^(β/2) (v)+v_t=f(t,x)|u|^q,  (t,x)∈(0,+∞)×R^N )┤ (19)

 subjected to the conditions

u(0,x)=u_0 (x)≥0, u_t (0,x)=u_1 (x)≥0

v(0,x)=v_0 (x)≥0, v_t (0,x)=v_1 (x)≥0,

where p>1,q>1,0<α≤2,0<β≤2,f,g are positive functions.

Suppose that the functions f and g are satisfied what follows: For all compact

Ω∈R^+×R^N, exists two positive constants μ and ν such that:

■(&f(tR,xR^(1/σ))=O(R^μ), and g(tR,xR^(1/σ))=O(R^ν) where  R>0  andlarge,@&σ=max(α,β), (t,x)∈Ω.) (20)

Using the same reasoning above, one gets the following assertion.

Theorem 2 Let




 where p>1, q>1. Assume that the condition (20) is fulfilled and that


 then the problem (19) admits only the trivial solution as a global one.

Proof. We notice that, in all steps of proof , C>0 is a real positive number which may change from line to line.

Set  ζ(t,x)=Φ((t^2+|x|^2σ)/R^2 ), where R>0, Φ∈C_c^∞ (R^+) satisfies 0≤Φ≤1 and

Φ(r)=(■(1& if   r≤1,@0& if   r≥2.)┤

Multiplying the first equation of (19) by ζ and integrating by parts on Q_T=(0,T)×R^N, we get

■(∫_(Σ_T)▒‍ g|v|^p ζ dx dt&+∫_(R^N)▒‍ u_0 (x)ζ(0,x) dx+∫_(R^N)▒‍ u_1 (x)ζ(0,x) [email protected]&-∫_(R^N)▒‍ u_0 (x)ζ_t (0,x) dx=∫_(Σ_T)▒‍ uζ_tt  dx [email protected]&-∫_(Σ_T)▒‍ uζ_t  dx dt+∫_(Σ_T)▒‍ u(-Δ)^(α/2) ζ dx dt.) (21)

With the fact that   ζ_t (0,x)=0, we obtain

■(∫_(Σ_T)▒‍ g|v|^p ζ dx dt&+∫_(R^N)▒‍ u_0 (x)ζ(0,x) dx+∫_(R^N)▒‍ u_1 (x)ζ(0,x) [email protected]&=∫_(Σ_T)▒‍ uζ_tt  dx dt-∫_(Σ_T)▒‍ uζ_t  dx dt+∫_(Σ_T)▒‍ u(-Δ)^(α/2) ζ dx dt.) (22)


∫_(Σ_T)▒‍ g|v|^p ζ dx dt≤∫_(Σ_T)▒‍ |u||ζ_tt |  dx dt+∫_(Σ_T)▒‍ |u||ζ_t |  dx dt+∫_(Σ_T)▒‍ |u||(-Δ)^(α/2) ζ|  dx dt. (23)

 We have also

∫_(Σ_T)▒‍ f|u|^q ζ dx dt≤∫_(Σ_T)▒‍ |v||ζ_tt |  dx dt+∫_(Σ_T)▒‍ |v||ζ_t |  dx dt+∫_(Σ_T)▒‍ |v||(-Δ)^(β/2) ζ|  dx dt. (24)

 To estimate

∫_(Σ_T)▒‍ |u||ζ_tt |  dx dt,

we observe that it can be rewritten as

∫_(Σ_T)▒‍ |u||ζ_tt |  dx dt=∫_(Σ_T)▒‍ |u|(fζ)^(1/q) |ζ_tt |(fζ)^((-1)/q)  dx dt.

Using Hölder’s inequality, we obtain

∫_(Σ_T)▒‍ |u||ζ_tt |  dx dt≤(∫_(Σ_T)▒‍ |u|^q (fζ) dx dt)^(1/q) (∫_(Σ_T)▒‍ |ζ_tt |^(q/(q-1)) (fζ)^((-1)/(q-1))  dx dt)^((q-1)/q).

Proceeding as above, we have

∫_(Σ_T)▒‍ |u||ζ_t |  dx dt≤(∫_(Σ_T)▒‍ |u|^q (fζ) dx dt)^(1/q) (∫_(Σ_T)▒‍ |ζ_t |^(q/(q-1)) (hζ)^((-1)/(q-1))  dx dt)^((q-1)/q),


∫_(Σ_T)▒‍ |u||(-Δ)^(α/2) ζ|  dx dt≤(∫_(Σ_T)▒‍ |u|^q (fζ) dx dt)^(1/q) (∫_(Σ_T)▒‍ |(-Δ)^(α/2) ζ|^(q/(q-1)) (fζ)^((-1)/(q-1))  dx dt)^((q-1)/q).

Finally, we infer

∫_(Σ_T)▒‍ g|v|^p ζ dx dt≤(∫_(Σ_T)▒‍ |u|^q (fζ) dx dt)^(1/q) K_q, (25)


■(K_q=(∫_(Σ_T)▒‍ |ζ_tt |^(q/(q-1)) (fζ)^((-1)/q)  dx dt)^((q-1)/q)&+(∫_(Σ_T)▒‍ |ζ_t |^(q/(q-1)) (fζ)^((-1)/(q-1))  dx dt)^((q-1)/q)@&+(∫_(Σ_T)▒‍ |(-Δ)^(α/2) ζ|^(q/(q-1)) (fζ)^((-1)/(q-1))  dx dt)^((q-1)/q)@&=:K_(q,ζ_tt )+K_(q,ζ_t )+K_(q,(-Δ)_x^(β/2) ).)

 Arguing as above we have likewise

∫_(Σ_T)▒‍ f|u|^q ζ dx dt≤(∫_(Σ_T)▒‍ |v|^p (gζ) dx dt)^(1/p) L_p, (26)


■(L_p=(∫_(Σ_T)▒‍ |ζ_tt |^(p/(p-1)) (gζ)^((-1)/(p-1))  dx dt)^((p-1)/p)&+(∫_(Σ_T)▒‍ |ζ_t |^(p/(p-1)) (gζ)^((-1)/(p-1))  dx dt)^((p-1)/p)@&+(∫_(Σ_T)▒‍ |(-Δ)^(β/2) ζ|^(p/(p-1)) (gζ)^((-1)/(p-1))  dx dt)^((p-1)/p)@&=:L_(p,ζ_tt )+L_(p,ζ_t )+L_(p,(-Δ)_x^(β/2) ).)

By the choice of ζ, it is easy to show that (■(&K_(q,ζ_tt )<∞@&K_(q,ζ_t )<∞@&K_(q,(-Δ)_x^(β/2) )<∞)┤ and (■(&L_(p,ζ_tt )<∞@&L_(p,ζ_t )<∞@&L_(p,(-Δ)_x^(β/2) )<∞.)┤

By Substituting (26) in (25), it yields

(∫_(Σ_T)▒‍ g|v|^p ζ dx dt)^((pq-1)/pq)≤K_q L_p^(1/q). (27)

 Similarly, we get

(∫_(Σ_T)▒‍ f|u|^q ζ dx dt)^((pq-1)/pq)≤L_p K_q^(1/p). (28)

 Now we conseder the scale of variables

t=τR, x=yR^(1/σ),

 we have ζ_t=R^(-1) ζ_τ,ζ_tt=R^(-2) ζ_ττ,(-Δ)_x^(α/2) ζ=R^((-α)/σ) (-Δ)_y^(α/2) ζ,(-Δ)_x^(β/2) ζ=R^((-β)/σ) (-Δ)_y^(β/2) ζ, then

(∫_(Σ_T)▒‍ g|v|^p ζ dx dt)^((pq-1)/pq)≤C[R^(γ_1 )+R^(γ_2 )+R^(γ_3 )]×[R^(λ_1 )+R^(λ_2 )+R^(λ_3 ) ]^(1/q), (29)

 similarly, we have

(∫_(Σ_T)▒‍ f|u|^q ζ dx dt)^((pq-1)/pq)≤C[R^(λ_1 )+R^(λ_2 )+R^(λ_3 )]×[R^(γ_1 )+R^(γ_2 )+R^(γ_3 ) ]^(1/p), (30)

 where (■(&γ_1=(N/σ+1)(q-1)/q-2-μ/[email protected]&@&γ_2=(N/σ+1)(q-1)/q-1-μ/[email protected]&@&γ_3=(N/σ+1)(q-1)/q-α/σ-μ/q)┤ and (■(&λ_1=(N/σ+1)(p-1)/p-2-ν/[email protected]&@&λ_2=(N/σ+1)(p-1)/p-1-ν/[email protected]&@&λ_3=(N/σ+1)(p-1)/p-β/σ-ν/p)┤

we remark that γ_1<γ_2≤γ_3 and λ_1<λ_2≤λ_3,


(∫_(Σ_T)▒‍ g|v|^p ζ dx dt)^((pq-1)/pq)≤CR^(γ_3+λ_3/q) (31)


(∫_(Σ_T)▒‍ f|u|^q ζ dx dt)^((pq-1)/pq)≤CR^(λ_3+γ_3/p). (32)

 We conclude that  

    • If γ_3+λ_3/q<0, it yield


 Then the right hand side of (31) goes to 0, when R tends to infinity, while the left hand side  converge to

(∫_(Σ_T)▒‍ g|v|^p  dx dt)^((pq-1)/pq).

This implies that v≡0 .

Similarly, if λ_3+γ_3/p<0, it yield


by using also (32) to proceeding as above, we obtain u≡0.

    • If γ_3+λ_3/q=0, we get

(∫_(R^+×R^N)▒‍ g|v|^p  dx dt)<+∞.

Using again Hölder’s inequality we obtain

∫_(Σ_T)▒‍ f|u|^q ζ dx dt≤(∫_(B_R)▒‍ |v|^p (gζ) dx dt)^(1/p) L_p,


B_R={(t,x)∈R^+×R^N;  R^2≤t^2+|x|^2θ≤2R^2}.


∫_(R^+×R^N)▒‍ g|v|^p  dx dt<+∞,

we get

lim┬(R→+∞) ∫_(B_R)▒‍ |v|^p gζ dx dt=0,

hence, we infer that

∫_(R^+×R^N)▒‍ f|u|^q  dx dt=0,

This implies that u≡0 .

Similarly, if λ_3+γ_3/p=0, proceeding as above, we infer that v≡0.

which ends the proof.   ∎

Remark 2 We notice that in the case where α=β=2,μ=ν=0, we obtain the same result of A. Hakem, when g(t),f(t) behaves like t^β,t^α, and β=α=0, (see [7], section 5), and also we recover the case who studied by F. Sun and M. Wang [13].


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