Global Existence and Nonexistence in a System of Petrovsky

Transcription

1 Journal of Matheatical Analysis and Applications 265, (2002) doi: /jaa , available online at on Global Existence and Nonexistence in a Syste of Petrovsky Sali A.Messaoudi Matheatical Sciences Departent, King Fahd University of Petroleu and Minerals, Dhahran 31261, Saudi Arabia E-ail: essaoud@kfup.edu.sa Subitted by Colin Rogers Received June 26, 2000 In this paper we consider the nonlinearly daped seilinear Petrovsky equation u tt + 2 u + au t u t 2 = buu p 2 in a bounded doain, where a b > 0.We prove the existence of a local weak solution and show that this solution blows up in finite tie if p>and the energy is negative.we also show that the solution is global if p Elsevier Science Key Words: nonlinear daping; nonlinear source; negative initial energy; local; global; blow up; finite tie. 1.INTRODUCTION In 4], Guesia considered the proble u tt x t+ 2 ux t+qxux t+gu t x t = 0 x t > 0 ux t = ν ux t =0 x t 0 (1.1) ux 0 =u 0 x u t x 0 =u 1 x x where is a bounded doain of n n 1, with a sooth boundary, and ν is the unit outer noral on.for g continuous, increasing, satisfying g0 =0, and q +, a bounded function, Guesia 4] proved a global existence and a regularity result.he also established, under suitable growth conditions on g, decay results for weak, as well as strong, solutions. Precisely, the author showed that the solution decays exponentially if g behaves like a linear function, whereas the decay is of a polynoial order X/02 $ Elsevier Science All rights reserved. 296

2 existence in a syste of petrovsky 297 otherwise.results siilar to the above syste, coupled with a seilinear wave equation, have been established by Guesia 5].Also the syste coposed of the equation (1.1), with 2 u t x t+gux t in the place of qxux t+gu t x t, has been treated by Aassila and Guesia 1], and an exponential decay theore, through the use of an iportant lea of Koornik 6], has been established. In this paper we are concerned with the proble u tt + 2 u + au t u t 2 = buu p 2 x t > 0 ux t = ν ux t =0 x t 0 (1.2) ux 0 =φx u t x 0 =ϕx x where a b > 0 and p > 2.This is a proble siilar to (1.1), which contains a nonlinear source ter copeting with the daping factor.we will establish an existence result and show that the solution continues to exist globally if p; however, it blows up in finite tie if <p.it is worth entioning that it is only for siplicity that q is taken to be zero, gu t x t = au t u t 2, and the source ter has a power for.the sae theores could be established for ore general functions. 2.LOCAL EXISTENCE In this section, we establish a local existence result for (1.2) under suitable conditions on and p.first we consider, for v given, the linear proble u tt + 2 u + au t u t 2 = bv p 2 v x t > 0 ux t = ν ux t =0 x t > 0 (2.1) ux 0 =φx u t x 0 =ϕx x where u is the sought solution. Lea 2.1. Assue that 2 <p n 4 2 <p 2n 2/n 4 n 5 (2.2) Then given any v in C0T C0 and φ ϕ in C 0, the proble (2.1) has a unique solution u satisfying u L 0T W u tt L 0T L 2 (2.3) u t L 0T H0 2 L 0T Here H 2 0 =w H2 w = ν w = 0 on and W =w H 4 H 2 0 w = νw = 0 on.

3 298 sali a. essaoudi This lea is a direct result of 7, Theore 3.1, Chap. 1] (see also 2] and 4, Theore 1.2]). Lea 2.2. Assue that (2.2) holds. Assue further that 2n/n 4 n 5 (2.4) Then given any φ in H0 2 ϕ in L2, and v in C0T H0 2, the proble (2.1) has a unique weak solution, u C0T H 2 0 u t C0T L 2 L 0T Moreover, we have 1 t u 2 t 2 +u2 x t dx + a u t x s dx ds 0 = 1 t ϕ 2 +φ 2 x dx + b v p 2 vu 2 t x s dx ds 0 t 0T (2.5) (2.6) Proof. We approxiate φ ϕ by sequences φ µ ϕ µ in C0, and v by a sequence v µ in C0T C0.We then consider the set of linear probles u µ tt + 2 u µ + au µ t u µ t 2 = bv µ p 2 v µ x t > 0 u µ x t = ν u µ x t =0 x t > 0 (2.7) u µ x 0 =φ µ x u µ t x 0 =ϕ µ x x Lea 2.1 guarantees the existence of a sequence of unique solutions u µ satisfying (2.3). Now we proceed to show that the sequence u µ u µ t is Cauchy in Y = w w C0T H 2 0 w t C0T L 2 L 0T For this ai, we set U = u µ u ν V = v µ v ν It is straightforward to see that U satisfies U tt + 2 U + au µ t u µ t 2 u ν t uν t 2 =bv µ p 2 v µ v ν p 2 v ν Ux t =0 x t > 0 (2.8) Ux 0 =U 0 x =φ µ x φ ν x U t x 0 =U 1 x =ϕ µ x ϕ ν x

4 existence in a syste of petrovsky 299 We ultiply Eq.(2.8) by U t and integrate over 0t to get 1 t Ut 2 +U 2 x t dx + a u µ t u µ t 2 u ν t 2 uν t 2 U t x s dx ds 0 = 1 t U U 0 2 x dx + b v µ p 2 v µ v ν p 2 v ν 0 U t x s dx ds (2.9) We then estiate the last ter in (2.9) as follows: v µ p 2 v µ v ν p 2 v ν U t x s dx ] CU t 2 V 2n/n 4 v µ p 2 np 2/2 +vν p 2 np 2/2 (2.10) The Sobolev ebedding and condition (2.2) give V 2n/n 2 CV 2 v µ p 2 np 2/2 +vν p 2 np 2/2 Cvµ p 2 2 +v ν p 2 2 where C is a constant depending on only.therefore (2.10) takes the for v µ p 2 v µ v ν p 2 v ν U t x s dx CU t 2 V 2 v µ p 2 2 +v ν p 2 2 Since u µ t u µ t 2 u ν t uν t 2 u µ t u ν t 0 then (2.9) yields 1 Ut 2 +U 2 x t dx U U 0 2 x dx + Ɣ t 0 U t s 2 V s 2 ds where Ɣ is a generic positive constant depending on C and the radius of the ball in C0T H0 2 containing vµ and v ν.young s inequality then gives ax 0 t T Ut 2 +U 2 x t dx Ɣ U1 2 +U 0 2 x dx + ƔT ax 0 t T V 2 t +V 2 x t dx Since φ µ is Cauchy in H 2 0 ϕµ is Cauchy in L 2, and v µ is Cauchy in C0T H 2 0, we conclude that uµ u µ t is Cauchy

5 300 sali a. essaoudi in C0T H0 2 C0T L2.To show that u t is Cauchy in L 0T, weuse t U t L 0T C u µ t u µ t 2 u ν t uν t 2 U t x s dx ds (2.11) which yields, by (2.9), 0 U t L 0T Ɣ + Ɣ U 2 1 +U 0 2 x dx T 0 U t s 2 V s 2 ds Therefore u µ t is Cauchy in L 0T and hence u µ is Cauchy in Y.We now show that the liit u is a weak solution of (2.1) in the sense of 7].That is for each θ in H0 2 we ust show that d dt +a u t x tθx dx + ux tθx dx u t u t 2 x tθx dx = b v p 2 vx tθx dx (2.12) for alost all t in 0T.To establish this, we ultiply Eq.(2.7) by θ and integrate over, so we obtain d dt +a u µ t x tθx dx + u µ x tθx dx u µ t u µ t 2 x tθx dx = b As µ, we see that u µ x tθx dx v µ p 2 v µ x tθx dx in C0T and u µ t u µ t 2 x tθx dx v µ p 2 v µ x tθx dx (2.13) ux tθx dx v p 2 vx tθx dx u t u t 2 x tθx dx in L 1 0T.We thus have u tx tθx dx = li uµ t x tθx dx is an absolutely continuous function on 0T, so (2.12) holds for alost all t in 0T.For the energy equality (2.6), we start fro the energy equality for u µ and proceed in the sae way to establish it for u.to prove uniqueness,

6 existence in a syste of petrovsky 301 we take v µ and v ν and let u µ and u ν be the corresponding solutions of (2.1). It is clear that U = u µ u ν satisfies 1 t Ut 2 +U 2 x t dx + a u µ t u µ t 2 u ν t 2 uν t 2 U t x s dx ds 0 t = b v µ p 2 v µ v ν p 2 v ν U t x s dx ds (2.14) 0 If v µ = v ν then (2.14) shows that U = 0, which iplies uniqueness.this copletes the proof. Reark 21. Note that the condition (2.4) on is needed so that uµ t u µ t 2 x tθx dx and u tu t 2 x tθx dx ake sense. Theore 2.3. Assue that (2.2) and (2.4) hold. Then given any φ in H0 2, and ϕ in L2, the proble (1.2) has a unique weak solution u Y, for T sall enough. Proof. For M>0 large and T>0, we define a class of functions ZM T which consists of all functions w in Y satisfying the initial conditions of (1.2) and 1 T ax wt 2 +w 2 x t dx + a w 0 t T 2 t x s dx ds M 2 (2.15) 0 ZM T is nonepty if M is large enough.this follows fro the trace theore (see 8]).We also define the ap f fro ZM T into Y by u = f v, where u is the unique solution of the linear proble (2.1). We would like to show, for M sufficiently large and T sufficiently sall, that f is a contraction fro ZM T into itself. By using the energy equality (2.5) we get t u 2 t +u2 x t dx + 2a u t x s dx ds 0 t u 2 1 +u 0 2 x dx + 2b v p 1 u t x s dx ds t 0T 0 t u 2 1 +u 0 2 x dx + 2b u t 2 v p 1 2 t 0T (2.16) consequently u 2 Y C 0 u 2 1 +u 0 2 x dx + CM p 1 T u Y where C is independant of M.By choosing M large enough and T sufficiently sall, (2.15) is satisfied; hence u ZM T.This shows that f aps ZM T into itself.

7 302 sali a. essaoudi Next we verify that f is a contraction.for this ai we set U = u ū and V = v v, where u = f v and ū = f v.it is straightforward to see that U satisfies U tt + 2 U + au t 2 u t aū t 2 ū t = bv p 2 v b v p 2 v Ux t =0 x t > 0 (2.17) Ux 0 =U t x 0 =0 x By ultiplying Eq.(2.17) by U t and integrating over 0t, we arrive at t Ut 2 +U 2 x t dx + u t 2 u t ū t 2 ū t U t x s dx ds 0 t C v p 2 v v p 2 vu t x s dx ds (2.18) 0 By using (2.2), (2.10), and (2.11), we obtain t Ut 2 +U 2 x t dx + U t x s dx ds 0 t Ɣ U t 2 V 2 v p v p 2 2 s ds 0 Thus we have U 2 Y CTMp 2 V 2 Y (2.19) By choosing T so sall that ƔTM p 2 < 1, (2.19) shows that f is a contraction.the contraction apping theore then guarantees the existence of a unique u satisfying u = f u.obviously it is a solution of (1.2).The uniqueness of this solution follows fro the energy inequality (2.18). The proof is copleted. 3.BLOW-UP RESULT In this section we show that the solution (2.5) blows up in finite tie if p>and E0 < 0, where Et = 1 2 u 2 t +u2 x t dx b ux t p dx (3.1) p Lea 3.1. Suppose that (2.2) holds. Then there exists a positive constant C>1, depending on only, such that for any u H0 2 and 2 s p. u s p C u2 2 +up p (3.2)

8 existence in a syste of petrovsky 303 Proof. If u p 1 then u s p u2 p Cu2 2 by Sobolev ebedding theores and the boundary conditions.if u p > 1 then u s p up p. Therefore (3.2) follows. We set Ht = Et (3.3) and use, throughout this section, C to denote a generic positive constant depending on only. As a result of (3.1) (3.3), we have Corollary 3.2. Let the assuptions of the lea hold. Then we have for any u H0 2 and 2 s p. u s p CHt+u t 2 2 +up p (3.4) Theore 3.3. further that Let the conditions of the Theore 2.3 be fulfilled. Assue Then the solution (2.5) blows up in finite tie. Proof. E0 < 0 (3.5) We ultiply Eq.(1.2) by u t and integrate over to get H t =a u t x t dx 0 for alost every t in 0T since Ht is absolutely continuous (see 2]); hence 0 <H0 Ht b p up p (3.6) for every t in 0T, by virtue of (3.1) and (3.3). We then define Lt =H 1 α t+ε uu t x t dx (3.7) for ε sall to be chosen later and { } p 2 p 0 <α in (3.8) 2p p 1 By taking a derivative of (3.7) and using Eq. (1.2) we obtain L t =1 αh α th t+ε u 2 t u2 x t dx + εb ux t p dx aε u t 2 u t ux t dx (3.9)

9 304sali a. essaoudi We then exploit Young s inequality, XY δr r Xr + δ q q Y q 1 XY 0 δ > 0 r + 1 q = 1 for r = and q = / 1 to estiate the last ter in (3.9) as u t 1 u dx δ u + 1 δ / 1 u t which yields, by substitution in (3.9), L t 1 αh α t 1 ] εδ / 1 H t +ε u 2 t u2 xtdx+ε pht+ p ] u 2 t 2 +u2 xtdx εa δ u δ>0 (3.10) Of course (3.10) reains valid even if δ is tie dependent, since the integral is taken over the x variable.therefore by taking δ so that δ / 1 = kh α t, for large k to be specified later, and substituting in (3.10) we arrive at L t 1 α 1 ] εk H α th t ( ) p + ε ε pht k1 p u 2 t x t dx + ε( 2 1) ux t 2 dx ] ahα 1 tu (3.11) By exploiting (3.6) and the inequality u C u p, we obtain H α 1 tu ( b p hence (3.11) yields L t 1 α 1 ( ) p + ε ε pht k1 a ) α 1 Cup +αp 1 ] εk H α th t p u 2 t x t dx + ε( 2 1) ux t 2 dx ( b p ) α 1 Cu +αp 1 p ] (3.12)

10 existence in a syste of petrovsky 305 We then use Corollary 3.2 and relation (3.8), for s = + αp 1 p, to deduce fro (3.12), L t 1 α 1 ] εk H α th t ( ) p p + ε u 2 t x t dx + ε( 2 1) u 2 x t dx + εpht C 1 k 1 Ht+u t 2 2 +up p (3.13) where C 1 = ab/p α 1 C/.By noting that Ht = b p up p 1 2 u t u2 2 and writing p =p + 2/2 +p 2/2, the estiate (3.13) gives L t 1 α 1 ] εk H α th t+ε p 2 u ( ( ) p + 2 p 2 + ε C 2 1 k )Ht+ 1 2p b C 1k 1 u p p ( ] p C 4 1 k )u 1 t 2 2 (3.14) At this point, we choose k large enough so that the coefficients of Ht u t 2 2, and up p in (3.14) are strictly positive; hence we get L t 1 α 1 ] εk H α th t + εγht+u t 2 2 +up p (3.15) where γ>0 is the iniu of these coefficients.once k is fixed (hence γ), we pick ε sall enough so that 1 α εk 1/ 0 and L0 =H 1 α 0+ε u 0 u 1 x dx > 0 Therefore (3.15) takes the for Consequently we have L t γεht+u t 2 2 +up p (3.16) Lt L0 > 0 t 0 Next we estiate the second ter in (3.7) as follows: uu t x t dx u 2u t 2 Cu p u t 2

11 306 sali a. essaoudi So we have 1/1 α uu t x t dx Cup 1/1 α u t 1/1 α 2 Again Young s inequality gives 1/1 α ] uu t x t dx C u µ/1 α p +u t θ/1 α 2 (3.17) for 1/µ + 1/θ = 1.We take θ = 21 α to get µ/1 α =2/1 2α p by condition (3.8). Therefore (3.17) becoes 1/1 α uu t x t dx Cu s p +u t 2 2 where s = 2/1 2α p.by using Corollary 3.2 we obtain 1/1 α uu t x t dx CHt+u p p +u t 2 2 t 0 (3.18) Consequently we have ( ) 1/1 α L 1/1 α t = H 1 α t+ε uu t x t dx 2 (Ht+ 1/1 α 1/1 α uu t x t dx CHt+u p p +u t 2 2 (3.19) We then cobine (3.16) and (3.19), to arrive at L t ƔL 1/1 α t (3.20) where Ɣ is a constant depending on C γ, and ε only (and hence is independent of the solution u).a siple integration of (3.20) over 0 t then yields L α/1 α 1 t L α/1 α 0 Ɣtα/1 α Therefore Lt blows up in a tie T 1 α (3.21) ƔαL0α/1 α Reark 21. By following the steps of the proof of Theore 3.3 closely, one can easily see that the blow-up result holds even for 1 <<p.therefore this ethod is a unified one for both linear and nonlinear daping cases. Reark 22. The estiate (3.21) shows that L0 is larger when the blow-up takes place ore quickly.

12 existence in a syste of petrovsky GLOBAL EXISTENCE In this section, we show that the solution (2.5) is global if p. Theore 4.1. Assue that (2.2) and (2.4) hold such that p. Then for any φ in H0 2 and ϕ in L2, the proble (1.2) has a unique weak solution u Y, for any T>0. Proof. Siilar to 3], we define the functional Ft = 1 2 u 2 t +u2 x t dx + b ux t p dx p By taking a derivative and using Eq.(1.2), we obtain F t = au t + 2b u t uux t p 2 dx By using Young s inequality, we get F t au t + δu t p p + C δu p p By noting that p, we easily see that F t au t + Cδu t p + C δu p p where C is a constant depending on only and C δ is a constant depending on δ.at this point we distinguish two cases: either u t > 1, so we choose δ sall enough so that au t + Cδu t p 0, and hence F t C δ u p p.or u t 1; in this case we have F t Cδ + C δ u p p. Therefore, in either case, we have A siple integration then yields F t c 1 + cft Ft F0+c 1 /ce ct The last estiate, together with the continuation principle, copletes the proof. ACKNOWLEDGMENT The author expresses his sincere thanks to King Fahd University of Petroleu and Minerals for its support.

13 308 sali a. essaoudi REFERENCES 1.M.Aassila and A.Guesia, Energy decay for a daped nonlinear hyperbolic equation, Appl. Math Lett. 12 (1999), V.Barbu, Analysis and Control of Nonlinearinfinite Diensional Systes, Acadeic Press, New York V.Georgiev and G.Todorova, Existence of solutions of the wave equation with nonlinear daping and source ters, J. Differential Equations 109, No.2 (1994), A.Guesia, Existence globale et stabilisation interne non linéaire d un systèe de Petrovsky, Bell. Belg. Math. Soc. 5 (1998), A.Guesia, Energy decay for a daped nonlinear coupled syste, J. Math. Anal. Appl. 239 (1999), V.Koornik, Exact Controllability and Stabilization.The Multiplier Method, Masson, Paris, J.L.Lions, Quelques ethodes de resolution des problèes aux liites nonlinéaires, Dunod Gautier-Villars, Paris, J.L.Lions and E.Magenes, Problèes aux liites nonhoogènes et applications, Vols. 1 and 2, Dunod, Paris, 1968.