Blow up of the solutions for the Petrovsky equation ...

Malaya Journal of Matematik, Vol. 6, No. 1, 85-90, 2018 https://doi.org/10.26637/MJM0601/0013

Blow up of the solutions for the Petrovsky equation with fractional damping terms Erhan Pis¸kin 1 * and Turgay Uysal2 Abstract In this paper, we prove a blow up result for solutions of the Petrovsky equation with fractional damping term with negative initial energy. Keywords Blow up, Petrovsky equation, Fractional damping term. 1,2 Department of Mathematics, Dicle University, 21280 Diyarbakır Turkey. *Corresponding author: 1 [email protected]; 2 [email protected] Article History: Received 24 October 2017; Accepted 19 December 2017

Contents 1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

2

Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

3

Blow up . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

c

2017 MJM.

to be weak damping and strong damping term, respectively. The fractional damping term plays a dissipative role, which is stronger than weak damping and weaker than strong damping [3]. Messaoudi [5] studied the local existence and blow up of the solution to the equation utt + ∆2 u + |ut |q−1 ut = |u| p−1 u.

1. Introduction In this paper, we investigate the following Petrovsky equation with fractional damping terms  p−1 1+α u, x ∈ Ω, t > 0,  utt + ∆2 u + ∂t u = |u| ∂ u(x,t) u(x,t) = ∂ ν = 0, x ∈ ∂ Ω, t > 0, (1.1)  u(x, 0) = u0 (x), ut (x, 0) = u1 (x), x ∈ Ω where Ω is a bounded domain with smooth boundary ∂ Ω in Rn , ν is the outer normal. The constants p > 1,and −1 < α < 1. The notation ∂t1+α stands for the Caputo’s fractional derivative of order 1 + α with respect to the time variable [6, 8]. It is defined as follows ( I −α dtd w(t) for − 1 < α < 0 1+α ∂t w(t) = 2 I 1−α dtd 2 w(t) for 0 < α < 1 where

Iβ , β Iβ

> 0 is fractional integral

d 1 w(t) = dt Γ(β )

Zt

(t − τ)β −1 w(τ)dτ.

(1.2)

Wu and Tsai [11] obtained global existence and blow up of the solution of the problem (1.2). Later, Chen and Zhou [2] studied blow up of the solution of the problem (1.2) for positive initial energy. Li et al. [4] studied global existence and blow up of the solution to the equation utt + ∆2 u − ∆u + |ut |q−1 ut = |u| p−1 u.

(1.3)

Recently, Pis¸kin and Polat [7] proved decay of solution of problem (1.3). Tatar [9] studied exponential growth of the solution to the wave equation with fractional damping utt − ∆u + ∂t1+α u = |u| p−1 u.

(1.4)

Also, he [1, 10] studied blow up of the solution to the equation (1.4). In this paper, we establish the blow up of the solution with negative initial energy by using the technique of [1].

2. Preliminaries

0

∂t1+α u

When −1 < α < 0, the term is said to be fractional damping. Also, α = −1 and α = 0 the term ∂t1+α u is said

In this section, we present some materials needed for our main results. Furthermore we will consider only the case

Blow up of the solutions for the Petrovsky equation with fractional damping terms — 86/90

−1 < α < 0. The classical energy functional associated to problem (1.1) is E(t) =

1 1 1 kut k2 + k∆uk2 − kuk p+1 2 2 p+1

Substituting (2.8), (2.2) and (2.3) into (2.7), we obtain

(2.1)

H 0 (t) =

In deriative of (2.1), we have 1 E (t) = − Γ(−α) 0

Zt

Z

ut Ω

−(α+1)

(t − τ)

+

Z

−σ ε 2 e−σ εt . uut dx Ω Z σε + ε− e−σ εt . |u| p+1 dx p+1 Ω

uτ (τ)dτdx.

0

Is obtained. Now, if we define our modified energy function as Eε (t) = E (t) − ε

Z i ε − µβ α Γ(−α) e−σ εt . ut2 dx 2 2 Ω Z σ 2 −σ εt + + 1 εe . |∆u| dx 2 Ω

hσε

e−σ εt + Γ(−α)

Z

(2.2)

uut dx

Zt

Z

0

Ω

−

where 0 < ε < 1 and to be determined later. (2.1) replace and a differation of (2.2) with respect to t yields Eε0 (t) = −

1 Γ(−α)

−ε

Z

Zt

Z

ut Ω

|ut |2 dx − ε

Z

|u| p+1 dx − ε

Ω

Zt

Z

u Ω

Ω

0

0

Ω

ε + Γ(−α)

Γ(−α)

+µ

(t − τ)−(α+1) uτ (τ)dτdx Z

0

For the fifth term on the right side of (2.9), using Young inequality, we get

(t − τ)−(α+1) uτ (τ)dτdx(2.3)

0

e−σ εt

Zt

Z

0

Z Zt

Z

= e G (t − τ) e−σ ετ u2τ dxdτ,

ut dx Ω

(2.5)

Ω

Z

≤ δ1 e−σ εt

and Z∞

G (t) = e

−β τ −(α+1)

e

τ

+ (2.6)

dτ

(t − τ)−(α+1) uτ (τ)dτdx

ut

Ω

βt

0

|∆u|2 dx

−σ εt

Ω


(t − τ)−(α+1) e−σ ετ u2τ dxdτ − µβ F(2.9) (t) .

Ω

F (t) =

u

Ω

Ω

Also, we define the following functions for use in our theorem, H (t) = − e−σ εt Eε (t) + µF (t) + d , (2.4)

Zt Z

Zt

Z εe−σ εt

Zt Z


ut Ω


0

ut2 dx 

1 −σ εt e 4δ1

2

Zt

Z

 (t − τ)−(α+1) uτ (τ)dτ  dx.

Ω

0

t

where σ =

p+1 2

and β , µ, d are positive consants. α+1 Writing − (α + 1) = − α+1 and using the Cauchy2 − 2 Schwarz inequality, we have

Lemma 2.1. If Eε (0) < 0 and p is sufcifiently large, then H (t) > 0 and H 0 (t) > 0. Proof. By taking a derivative of (2.4) and (2.5), we have H 0 (t) = σ εe−σ εt Eε (t) − e−σ εt Eε0 (t) − µF 0 (t) , (2.7)

e−σ εt .

Zt

Z

ut Ω

0

α

F (t) = β Γ (−α) e

−σ εt

Z

−σ εt

ut2 dx

≤ δ1 e

Ω

−

Zt Z 0

Ω


0

Z

ut2 dx +

Ω

(σ ε)α Γ (−α) 4δ1

(t − τ)−(α+1) e−σ ετ u2τ dxdτ + β F(2.8) (t) .

86

Z Zt Ω

0

e−σ ετ u2τ dτdx. (2.10) (t − τ)(α+1)


Adding and subtracting C1 H (t), we get

Smilarly, we have

H 0 (t)

Zt

Z


u Ω

≥ C1 H (t) +

0

2

 ≤ δ2

Z

1 4δ2

|u|2 dx +

Ω

e−σ εt .

Zt

Z

 (t − τ)−(α+1) uτ (τ)dτ  dx

Ω

0

0

≤ δ2 e−σ εt .

Z Ω

+

Z σ ε2 δ1 −µβ Γ(−α) − − e−σ εt ut2 dx 4δ3 Γ (−α) Ω Z C1 −σ εt σ +1 ε + e . |∆u|2 dx + 2 2 Ω Z εδ C p 2 1 − σ ε 2 δ3C p1 + e−σ εt . |∇u|2 dx Γ (−α) Ω Z C1 σε + ε− − e−σ εt . |u| p+1 dx p+1 p+1 Ω α


u Ω

Zt

Z

1 −σ εt e 4δ2

|u|2 dx 

2

Zt

Z

 (t − τ)−(α+1) e −σ2 εt uτ (τ)dτ  dx,

Ω

C1 σ ε ε + + 2 2 2

−C1 εe−σ εt .

Z

uut dx Ω

0

(σ ε)α (σ ε)α ε − + µ− 4δ1 4δ2

e−σ εt

Zt

Z

u Ω

≤ δ2 e

−σ εt

0

Z

Ω 0

e−σ ετ u2τ d τ dx (2.11) (t − τ)(α+1)

Used Sobolev-Poincare inequality. For the third term on the right side of (2.9), using the Young and Sobolev-Poincare inequalities, we get Z

uut dx Ω

≤ δ3

Z

|u|2 dx +

Ω

≤ δ3C∗

1 4δ3

Z

|ut |2 dx

Ω

1 |∇u| dx + 4δ Ω 3

Z

0

α

Ω

Z Z t

(t − τ)−(α+1) e−σ ετ u2τ dτdx

Z σ ε2 δ1 C1 ε −σ εt −µβ Γ(−α) − − − e ut2 dx 4δ3 Γ (−α) 4δ3 Ω Z σ C1 −σ εt + +1 ε + e . |∆u|2 dx 2 2 Ω Z C1 σε −σ εt + ε− − e . |u| p+1 dx p+1 p+1 Ω Z εδ2C p1 −σ εt 2 − σ ε δ3C p1 + +C1 εC p1 δ3 e . |∇u|2 dx Γ (−α) Ω t Z Z (σ ε)α (σ ε)α ε + µ− − (t − τ)−(α+1) e−σ ετ u2τ dτdx 4δ1 4δ2 Ω

|∇u|2 dx

(σ ε)α Γ (−α) + 4δ2

Ω

+ (C1 − β ) µF (t) +C1 d C1 σ ε ε + + ≥ C1 H (t) + 2 2 2


C∗

Z Zt

2

Z

0

2

+ (C1 − β ) µF (t) +C1 d

|ut | dx(2.12)

Ω

We choose C1 = obtain

By (2.9), (2.10), (2.11) and (2.12), it yields

p+1 2 ε,

δ1 = δ2 =

Γ(−α)ε 2

and δ3 = 12 , we

H 0 (t) Z σ ε2 δ1 σε ε α + − µ Γ(−α) − − e−σ εt . ut2 dx ≥ 2 2 4δ3 Γ (−α) Ω Z σ 2 −σ εt + + 1 εe . |∆u| dx 2 Ω Z εδ C p1 2 2 − σ ε δ3C p1 + e−σ εt . |∇u|2 dx Γ (−α) Ω Z σε + ε− e−σ εt . |u| p+1 dx p+1 Ω α α Z Zt (σ ε) (σ ε) ε + µ− − (t − τ)−(α+1) e−σ ετ u2τ dτdx 4δ1 4δ2 Ω

H 0 (t)

− µβ F (t) .

For the third term on the right side of (2), using the Sobolev-

p+1 p+2 2 εH (t) − ε C p1 e−σ εt . |∇u|2 dx 2 2 Ω Z p+1 + ε (1 − ε) − µβ α Γ(−α) e−σ εt . ut2 dx 2 Ω Z p + 3 −σ εt + εe . |∆u|2 dx 2 Ω Z Zt α α−1 (p + 1) ε + µ − α+1 (1 + ε) (t − τ)−(α+1) e−σ ετ u2τ dτdx 2 Γ(−α) Ω 0 p+1 p+1 + ε − β µF (t) + εd. 2 2 ≥

Z

0

87


3. Blow up

Poincare inequality, we get H 0 (t)

In this section, we state and prove the blow up result for negative initial energy.

p+1 p+2 2 εH (t) − ε C p1 e−σ εt C p2 |∆u|2 dx 2 2 Ω Z p+1 α −σ εt ε (1 − ε) − µβ Γ(−α) e . ut2 dx + 2 Ω Z p + 3 −σ εt + εe . |∆u|2 dx 2 Ω Z Zt α α−1 e−σ ετ u2τ (p + 1) ε (1 + ε) + µ − α+1 dτdx 2 Γ(−α) (t − τ)(α+1) Ω 0 p+1 p+1 ε − β µF (t) + εd + 2 2 p+1 εH (t) ≥ 2 Z p+1 + ε (1 − ε) − µβ α Γ(−α) e−σ εt . ut2 dx 2 Ω Z p + 3 − (p + 2) εC p + εe−σ εt . |∆u|2 dx 2 Ω Z Zt α α−1 (p + 1) ε e−σ ετ u2τ + µ − α+1 (1 + ε) dτdx 2 Γ(−α) (t − τ)(α+1) Ω 0 p+1 p+1 + ε − β µF (t) + εd 2 2 Z

≥

Theorem 3.1. Assume that −1 < α < 0, Z Ω

Then the solution u of the problem (1.1) blow up in finite time for sufficiently large values of p. Proof. Set Ψ (t) = H 1−γ (t) + ϕe−σ εt . where γ=

+

Ω

0

0

Ω

p+1 p + 3 −σ εt εH (t) + εe . 2 4 Z Zt

−γ

−σ εt

Ψ (t) = (1 − γ) H (t) H (t) − ϕσ εe . Z Z + ϕe−σ εt . ut2 dx + uutt dx

Ω

Z

|∆u|2 dx

Ω

e−σ ετ u2τ (2.13) dτdx. (t − τ)(α+1)

1 − Γ(−α)

Z

+ ϕe−σ εt

Z

Z

uut dx Ω

Ω

= (1 − γ) H −γ (t) H 0 (t) − ϕσ εe−σ εt . Z Z |u| p+1 dx + |∆u|2 dx + ϕe−σ εt

is positive. Consequently, we find

(p + 1)α 2α+1 ε 1−α Γ(−α)

p−1 2 (p + 1)

This, of course, will lead to a blow up result in finite time. Now, by taking a derivative of Ψ (t) and using (1.1), we have

it appears that the third coefficient is nonnegative. We can choose µ so that the second coefficient is nonnegative and α the forth coefficient greater than 2α+1(p+1) . Also, if p ε 1−α Γ(−α) H 0 (t) ≥

uut dx

and ϕ is a positive constant to be determined later. Our goal is to show that Ψ (t) satisfies a differential inequality of the form 1 Ψ 1−γ (t) ≤ KΨ0 (t) .

Next, we choose β = 1 and assuming that p+3 ε < ε1 = min 1, 2 (p + 2)C p

p+1 2 ε −β

Z Ω

0

sufficiently large

u1 u0 dx ≥ 0.

E (0) < 0 and

Z

uut dx Ω

Ω

Zt

u Ω

 (t − τ)−(α+1) uτ (τ)dτdx

0

ut2 dx.

Ω

By using the inequalities (2.12) and (2.11) with the constant δ4 , δ5 > 0 we obtain

If we select Eε (0) < −d, then H (0) > 0. Consequently of (2.13), that H (t) > 0 and H 0 (t) > 0.

Ψ0 (t) = (1 − γ) H −γ (t) H 0 (t) − ϕσ εδ4 e−σ εt .

Z

|u|2 dx

Ω

Now, we state the local existence theorem.

+ ϕe−σ εt .

Theorem 2.2. (Local Existence) [5]. Suppose that 1 < p < ∞, n = 1, 2, 2n 1 < p ≤ n−2 , n ≥ 3.

−

88

|u| p+1 dx

Ω Z ϕσ εe−σ εt

4δ4

+ ϕe−σ εt .

For every initial data (u0 , u1 ) ∈ H02 (Ω) × L2 (Ω) , there is T > 0 and a unique weak solution u (t) of (1.1) such that u ∈ C [0, T ) ; H02 (Ω) ∩C1 [0, T ) ; L2 (Ω) and ut ∈ L2 ((0, T ) × Ω) .

Z

ut2 dx + ϕe−σ εt .

Ω

Z

|∆u|2 dx −

Ω

ϕ (σ ε)α − 4δ5

Z Zt Ω

0

Z

ut2 dx

Ω Z e−σ εt

ϕδ5 Γ(−α)

|u|2 dx

Ω

(t − τ)−(α+1) e−σ ετ u2τ dτdx.


The last term of this inequality is derived from the (2.13), we see

Also, we take ϕ=

Ψ0 (t) ϕΓ(−α)ε ≥ (1 − γ) H −γ (t) − H 0 (t) 2δ5

to get

ϕ (p + 1) Γ(−α)ε 2 H (t) 4δ5 Z (p + 1) ε −σ εt +ϕ 1− e . ut2 dx 8δ4 Ω

(p + 1) ε ε ≥0 ϕ 1− − 8δ4 4δ6

+

−σ εt

Z

and

Z

The fifth coefficient is nonnegative as soon as ε and C p is chosen small enough, namely (p + 3) ε 2 p+1 γ 1− − εδ + LH (t) C p ≥ 0. 4 8LH γ (t) 2

|∆u|2 dx.

Ω

We pick δ5 = LΓ(−α)H γ (t) , we have h ϕε i −γ Ψ0 (t) ≥ (1 − γ) − H (t) H 0 (t) 2L ϕ (p + 1) ε 2 −γ + H (t) H (t) 4L Z (p + 1) ε −σ εt e ut2 dx +ϕ 1 − 8δ4 Ω

p−1 1 = . p + 1 4 (p + 1)

ϕ−

p+1

+ ϕe . |u| dx Ω p+1 (p + 3) Γ(−α)ε 2 δ5 − +ϕ 1− εδ4 + Cp 8δ5 2 Γ(−α) .e−σ εt

p+3 1 4 (p + 3) , δ4 = δ6 = and ε < ε3 = 4 (p + 1) 2 (p + 1) (p + 11) ,

Consequently, we have (3.1)

Ψ0 (t) ≥ H (t) +

1 2

Z

ut2 dx +

Ω

p−1 4 (p + 1)

Z

|u| p+1 dx. (3.4)

Ω

On the other hand, from the Ψ (t) , we have (3.2) 1 1−γ

γ 1−γ

"

1 1−γ

Z

1 1−γ

#

H (t) + ϕ uut dx Ψ (t) ≤ 2 . (3.5) +ϕe−σ εt |u| p+1 dx Ω Ω (p + 3) ε 2 p+1 γ +ϕ 1 − − εδ4 + LH (t) C p By the Cauchy-Schwarz and H¨older inequalities, we have 8LH γ (t) 2 Z Z Z γ 1 1 (3.3) Ψ 1−γ (t) ≤ 2 1−γ H (t) + ϕ 1−γ B .e−σ εt |∆u|2 dx |ut |2 dx + |u| p+1 dx . Z

Ω

Ω

Ω

Adding and substracting H (t) to the right hand side of (3.3), we have h If K is chosen large enough so that ϕε i −γ Ψ0 (t) ≥ (1 − γ) − H (t) H 0 (t) 2L γ 2 1−γ ≤ K, ϕ (p + 1) ε 2 −γ + H (t) + 1 H (t) 4L γ 1 K Z 2 1−γ ϕ 1−γ B ≤ , (p + 1) ε 1 ε 2 + − u2 dx + ϕ 1− 8δ4 2 4δ6 Ω t γ 1 p−1 Z 2 1−γ ϕ 1−γ B ≤ K, 1 p+1 −σ εt 4 (p + 1) + ϕ− e . |u| dx p+1 Ω That is K has to be selected so that 2 (p + 3) ε p+1 γ +ϕ 1 − − εδ4 + LH (t) C p γ 1 1 4 (p + 1) 1−γ 8LH γ (t) 2 1−γ max 1, 2ϕ 1−γ B, K ≥ 2 b B . Z p−1 2 −σ εt |∆u| dx + µF (t) + d. .e Ω Combining (3.4) and (3.6), we have By using the inequality 1

Ψ 1−γ (t) ≤ KΨ0 (t) ,

2L (1 − γ) ε < ε2 = ϕ

(3.6)

(3.7)

From (3.4) it is clear that Ψ0 (t) ≥ 0 . Hence, by the definition of Ψ (t) and the hypotheses on the initial data, we have

and ϕ (p + 1) ε 2 −γ H (t) ≥ 0. 4L

Ψ (t) ≥ Ψ (0) > ϕ

Z Ω

89

u1 u0 dx ≥ 0.


Thus Ψ (t) > 0. Integrating (3.7) over (0,t) , we find 1

γ

Ψ 1−γ (t) ≥ Ψ

−γ 1−γ

γ (0) − K(1−γ) t

Consequently, Ψ (t) blows up at some time −γ

T∗ ≤

K (1 − γ) Ψ 1−γ (0) . γ

References [1]

[2]

[3]

[4]

[5]

[6] [7]

[8] [9] [10] [11]

M.R. Alaimia, N.E.Tatar, Blow up for the wave equation with a fractional damping, J. Appl. Anal. 11(1) (2005) 133-144. W. Chen, Y. Zhou, Global nonexistence for a semilinear Petrovsky equation, Nonlinear Anal. 70 (2009) 32033208. S. Chen, R. Triggiani, Proof of extension of two conjectures on structural damping for elastic systems: the case 1/2 ≤ α ≤ 1, Pacific J. Math. 136 (1989) 15-55. G. Li, Y. Sun, W. Liu, Global existence and blow up of solutions for a strongly damped Petrovsky system with nonlinear damping, Appl. Anal. 91(3) (2012) 575-586. S. A. Messaoudi, Global existence and nonexistence in a system of Petrovsky, J. Math. Anal. Appl. 265 (2) (2002) 296-308. K.B. Oldham, J. Spanier, The Fractional Calculus, Academic Press, 1974. E. Pis¸kin, N. Polat, On the decay of solutions for a nonlinear Petrovsky equation, Math. Sci. Letters, 3(1) (2014) 43-47. I.Podlubny, Fractional Differential Equations, Math. Sci. Engrg. 198, Academic Press, San Diago, CA,1999. N.E. Tatar, A wave equation with fractional damping, Z. Anal. Anw. 22(3) (2003) 609-617. N.E. Tatar, A blow up result for a fractionally damped wave equation, NoDEA, 12 (2005) 215-226. S.T. Wu, L.Y. Tsai, On global solutions and blow-up of solutions for a nonlinearly damped Petrovsky system, Taiwanese J. Math. 13 (2A) (2009) 545-558. ????????? ISSN(P):2319 − 3786

Malaya Journal of Matematik ISSN(O):2321 − 5666

?????????

90