International Journal of Analysis and Applications ISSN 2291-8639 Volume 3, Number 1 (2013), 1-13 http://www.etamaths.com EXPONENTIAL DECAY AND NUMERICAL SOLUTION FOR A TIMOSHENKO SYSTEM WITH DELAY TERM IN THE INTERNAL FEEDBACK C. A. RAPOSO1,∗, J. A. D. CHUQUIPOMA1, J. A. J. AVILA1, M. L. SANTOS2 Abstract. In this work we study the asymptotic behavior as t → ∞ of the solution for the Timoshenko system with delay term in the feedback. We use the semigroup theory for to prove the well-posedness of the system and for to establish the exponential stability. As far we know, there exist few results for problems with delay, where the asymptotic behavior is based on the Gearhart- Herbst-Pruss-Huang theorem to dissipative system. See [4],[5],[6]. Finally, we present numerical results of the solution of the system. 1. Introduction In this paper we consider the following Timoshenko system ρ1φtt(x,t) − K(φx + ψ)x(x,t) + µ1φt(x,t) + µ2φt(x,t − τ) = 0,(1) ρ2ψtt(x,t) − bψxx(x,t) + K(φx + ψ)(x,t) + µ3ψt(x,t) + µ4ψt(x,t − τ) = 0,(2) where φ is the transverse displacement of the beam, ψ is the rotation angle of the filament of the beam, (x,t) ∈ (0,L) × (0,∞), τ > 0 represents the time delay and ρ1,ρ2,b,K,µi, i = 1,2,3,4, are positive constants. This beam, of length L is subjected to the following boundary conditions φ(0, t) = φ(L,t) = ψ(0, t) = ψ(L,t) = 0, t > 0,(3) and initial conditions (φ0,φ1,ψ0,ψ1,f0,g0) belongs to a suitable functional space, defined for all x ∈ (0,L) by φ(x,0) = φ0(x), φt(x,0) = φ1(x), ψ(x,0) = ψ0(x), ψt(x,0) = ψ1(x),(4) and for (x,t) ∈ (0,L) × [0,τ], that implies past history with t − τ ≤ 0, by φt(x,t − τ) = f0(x,t − τ), ψt(x,t − τ) = g0(x,t − τ).(5) Note that f0(x,0) = φ1(x) and g0(x,0) = ψ1(x). In the study of the asymptotic behavior, we use the result due to Gearhart. See [4, 5, 6]. 2010 Mathematics Subject Classification. 35B40, 93D15. Key words and phrases. Timoshenko system; weak damping; exponential decay; delay. c⃝2013 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 1 2 RAPOSO, CHUQUIPOMA, SANTOS Theorem 1.1. Let S(t) = eA t be a C0−Semigroup of contractions on a Hilbert space. Then S(t) is exponentially stable if and only if ρ(A) ⊇ iβ,β ∈ R(6) and lim |β|→∞ ||(iβI − A)−1|| < ∞(7) hold. Certainly, this approach is very different from other works in the literature, espe- cially for problems with delay, where the exponential decay is made by the method of energy, see, for example [8, 9, 10], and references therein. The method of energy, in general, imposes a additional condition on the wave speeds, that is, Kρ2 = bρ1, see [1, 2, 3]. Here we do not use any additional condition for the coefficients of the system. Our work improves the result obtained in [7] in the sense that delays has been introduced in the control ( damping terms ). The delays µ2φt(x,t − τ), µ4ψt(x,t − τ) makes the problem different from that considered in the literature. It is well known that small delays in the controls might turn such well-behaving system into a wild one. In recent years, the PDEs with time delay effects have become an active area of research. The plan of this work is follows: in the next section, we introduce the Energy Space and prove that the full energy of the system decay. In the section 3, we introduce the semigroup representation for the system and prove that A the infin- itesimal generator of the semigroup is dissipative, and more, that A generates a eA t, C0-semigroup of contractions, that implies, prove the existence and regularity of solution. Finally in the section 4 by Theorem of Gearhart we prove that eA t is exponentially stably. 2. Energy Space For the Sobolev spaces we use the standard notation as in [11]. Let us proceed as [12]. We introduce the followings new dependents variables as in z(x,ρ,t) = φt(x,t − τρ), w(x,ρ,t) = ψt(x,t − τρ), ρ ∈ (0,1),(8) that satisfies for (x,ρ,t) ∈ (0,L) × (0,1) × (0,∞) τ zt(x,ρ,t) + zρ(x,ρ,t) = 0, τ wt(x,ρ,t) + wρ(x,ρ,t) = 0.(9) Therefore, problem (1)-(2) is equivalent to ρ1φtt(x,t) − K(φx + ψ)x(x,t) + µ1φt(x,t) + µ2z(x,1, t) = 0, ρ2ψtt(x,t) − bψxx(x,t) + K(φx + ψ)(x,t) + µ3ψt(x,t) + µ4w(x,1, t) = 0,(10) τ zt(x,ρ,t) + zρ(x,ρ,t) = 0, τ wt(x,ρ,t) + wρ(x,ρ,t) = 0, EXPONENTIAL DECAY AND NUMERICAL SOLUTION 3 The above system subjected to the following initial and boundary conditions φ(0, t) = φ(L,t) = ψ(0, t) = ψ(L,t) = 0, t > 0, z(x,0, t) = φt(x,t), w(x,0, t) = ψt(x,t), x ∈ (0,L), t > 0, φ(·,0) = φ0, φt(·,0) = φ1, x ∈ (0,L),(11) ψ(·,0) = ψ0, ψt(·,0) = ψ1, x ∈ (0,L), z(x,1,0) = f0(x,t − τ), w(x,1,0) = g0(x,t − τ), in (0,L) × (0,τ). Now, the energy space H is defined as H = {H10 × L 2 × H10 × L 2 × L2(0,1; L2) × L2(0,1; L2)}. For µ1 > µ2, µ3 > µ4 satisfying τµ2 < ξ < τ(2µ1 − µ2), τµ4 < η < τ(2µ3 − µ4)(12) respectively, we define the full energy of the system in the energy space as E(t) = 1 2 ∫ L 0 (ρ1|φt|2 + ρ2|ψt|2 + K|φx + ψ|2 + b|ψx|2)dx + ξ 2 ∫ L 0 ∫ 1 0 z2(x,ρ,t)dρdx + η 2 ∫ L 0 ∫ 1 0 w2(x,ρ,t)dρdx. Lemma 2.1. There exists a positive constant C such that for any regular solution (φ,ψ,z,w) of the problem (10)-(11) and for any t ≥ 0, we have d dt E(t) ≤ −C ∫ L 0 (|φt|2 + |ψt|2 + z2(x,1) + w2(x,1))dx.(13) Proof. 2.1. We multiplying (1) by φt, (2) by ψt, and using integration by part to get 1 2 d dt ∫ L 0 (ρ1|φt|2 + ρ2|ψt|2 + K|φx + ψ|2 + b|ψx|2)dx = − µ1 ∫ L 0 |φt|2 dx − µ2 ∫ L 0 φt z(1, t)dx − µ3 ∫ L 0 |ψt|2 dx − µ4 ∫ L 0 ψt w(1, t)dx and using the Energy E(t) of the system, we obatin d dt E(t) = −µ1 ∫ L 0 |φt|2 dx − µ2 ∫ L 0 φt z(1, t)dx − d dt { ξ 2 ∫ L 0 ∫ 1 0 z2(x,ρ,t)dρdx } −µ3 ∫ L 0 |ψt|2 dx − µ4 ∫ L 0 ψt w(1, t)dx − d dt { η 2 ∫ L 0 ∫ 1 0 w2(x,ρ,t)dρdx } 4 RAPOSO, CHUQUIPOMA, SANTOS using (14) and (15) d dt ξ 2 ∫ L 0 ∫ 1 0 z2(x,ρ,t)dρdx = − ξ τ ∫ L 0 ∫ 1 0 z(x,ρ,t)zρ(x,ρ,t)dρdx = − ξ 2τ ∫ L 0 ∫ 1 0 ∂ ∂ ρ z2(x,ρ,t)dρdx = ξ 2τ ∫ L 0 (z2(x,0) − z2(x,1))dx(14) d dt η 2 ∫ L 0 ∫ 1 0 w2(x,ρ,t)dρdx = − η τ ∫ L 0 ∫ 1 0 w(x,ρ,t)wρ(x,ρ,t)dρdx = − η 2τ ∫ L 0 ∫ 1 0 ∂ ∂ ρ w2(x,ρ,t)dρdx = η 2τ ∫ L 0 (w2(x,0) − w2(x,1))dx(15) we obtain d dt E(t) = −µ1 ∫ L 0 |φt|2 dx − µ2 ∫ L 0 φt z(1, t)dx(16) + ξ 2τ ∫ L 0 (z2(x,0) − z2(x,1))dx −µ3 ∫ L 0 |ψt|2 dx − µ4 ∫ L 0 ψt w(1, t)dx + η 2τ ∫ L 0 (w2(x,0) − w2(x,1))dx(17) Now, using Youngs’s inequality we can rewritten the last equation as d dt E(t) ≤ − ( µ1 − ξ 2τ − µ2 2 ) ∫ L 0 |φt|2 dx − ( ξ 2τ − µ2 2 ) ∫ L 0 z2(x,1)dx − ( µ3 − η 2τ − µ4 2 ) ∫ L 0 |ψt|2 dx − ( η 2τ − µ4 2 ) ∫ L 0 w2(x,1)dx from where our conclusion holds. 3. Existence of solution Let us introduce the semigroup representation. To this end, let U = (φ,φt,ψ,ψt,z,w) T and rewrite the problem (10)-(11) as Ut = AU U(0) = U0(18) EXPONENTIAL DECAY AND NUMERICAL SOLUTION 5 where the operator A is defined for U = (φ,u = φt,ψ,v = ψt,z,w)T by AU =   u K ρ1 (φx + ψ)x − µ1ρ1 u − µ2 ρ1 z(·,1) v b ρ2 ψxx − Kρ2 (φx + ψ) − µ3 ρ2 u − µ4 ρ2 w(·,1) − 1 τ zρ − 1 τ wρ   , with domain D(A) = {(φ,φt,ψ,ψt,z,w)T ∈ H : u = z(·,0, ·), v = w(·,0, ·), in(0,1)}, where for x ∈ (0,L) we denote H = H(0,L), L = L(0,L) and H = {(H2 ∩ H10) × H 1 0 × (H 2 ∩ H10) × H 1 0 × L 2(0,1; H1) × L2(0,1; H1)}. For U = (φi,ui,ψi,vi,zi,wi)T ∈ H, i = 1,2 and ξ, η as in (12) we defined the following inner product in the energy space as ⟨U1,U2⟩ = ∫ L 0 [ρ1u 1u2 + ρ2v 1v2 + K(φ1 + ψ1)(φ2 + ψ2) + bψ1ψ2)dx + ξ ∫ L 0 ∫ 1 0 z1(x,ρ)z2(x,ρ)dρdx + η ∫ L 0 ∫ 1 0 w1(x,ρ)w2(x,ρ)dρdx. For to prove the existence of solution we begin with the proof that the operator A is dissipative. Lemma 3.1. For U = (φ,u = φt,ψ,v = ψt,z,w) T ∈ D(A), we have ⟨AU,U⟩ ≤ 0. Proof. 3.1. ⟨AU,U⟩ = −µ1 ∫ L 0 |φt|2 dx − µ2 ∫ L 0 φt z(1, t)dx − ξ τ ∫ L 0 ∫ 1 0 z(x,ρ,t)zρ(x,ρ,t)dρdx −µ3 ∫ L 0 |ψt|2 dx − µ4 ∫ L 0 ψt w(1, t)dx − η τ ∫ L 0 ∫ 1 0 w(x,ρ,t)wρ(x,ρ,t)dρdx. Using (14) and (15) in the equation above we obtain ⟨AU,U⟩ = −µ1 ∫ L 0 |φt|2 dx − µ2 ∫ L 0 φt z(1, t)dx + ξ 2τ ∫ L 0 (z2(x,0) − z2(x,1))dx −µ3 ∫ L 0 |ψt|2 dx − µ4 ∫ L 0 ψt w(1, t)dx + η 2τ ∫ L 0 (w2(x,0) − w2(x,1))dx. Now using (16) and Lemma 3.1 we concludes ⟨AU,U⟩ = d dt E(t) ≤ −C ∫ L 0 (|u|2 + |v|2 + z2(x,1) + w2(x,1))dx.(19) In the next lemma, we will prove an important property of resolvent of the operator A. Lemma 3.2. 0 ∈ ρ(A). 6 RAPOSO, CHUQUIPOMA, SANTOS Proof. 3.2. For any F = (f1,f2,f3,f4,f5,f6) T ∈ H consider the equation AU = F. This implies u = f1,(20) K(φx + ψ)x − µ1u − µ2z(·,1) = ρ1f2,(21) v = f3,(22) bψxx − K(φx + ψ) − µ3v − µ4w(·,1) = ρ2f4,(23) −zρ = τf5,(24) −wρ = τf6,(25) We plug u = f1 obtained from (20) into (21) to get K(φx + ψ)x = µ1u + µ2z(·,1) + ρ1f2 ∈ L2(0,L). By Poincarè inequality we have K(φx + ψ) ∈ L2(0,L).(26) Now we plug v = f3 obtained from (26) and (22) into (23) to get bψxx = K(φx + ψ) + µ3v + µ4w(·,1) + ρ2f4 ∈ L2(0,L).(27) By the standard theory in the linear elliptic equations , we have a unique ψ ∈ H2 ∩ H10 satisfying (27). Then we plug ψ just obtained from solving (27) into (21) to get Kφxx = −Kψx + µ1u + µ2z(·,1) + ρ1f2 ∈ L2(0,L).(28) Applying the standard theory in the linear elliptic equations again yields a unique solvability of φ ∈ H2 ∩ H10 for (28). From (24) we have using Poincarè inequality, 1 Cp ∫ L 0 ∫ 1 0 |z|2 dρdx ≤ ∫ L 0 ∫ 1 0 |zρ|2 dρdx ∈ L2(0,1 : L2(0,L)), then z ∈ L2(0,1 : H1(0,L)). The same idea we use for w. Thus the unique solv- ability of AU = F follows. It is clear from the theory of the linear elliptic equation, see Chapter 1 of [13], that ||U||H ≤ C||F||H with C being a positive constant inde- pendent of U, and then 0 ∈ ρ(A). Now we will to prove that A generates a C0−Semigroup of contractions. Lemma 3.3. The operator A generates a C0−Semigroup of contractions on a Hilbert space H. Proof. 3.3. From Lemma 3.1 we have that A is dissipative operator, and from Lemma 3.2 follows that 0 ∈ ρ(A), them from Theorem 1.2.4, page 3 of [13], we concludes that A generates a C0−Semigroup of contractions on H. In this step, we prove that the problem (10)-(11) is well-posedness, and in this direction, we have the following result Theorem 3.4. If µ2 ≤ µ1 and µ4 ≤ µ3, then there exists a unique solution U ∈ C([0,∞),H) of the (10)-(11). Moreover if U0 ∈ D(A), then U ∈ C([0,∞),D(A))∩ C1([0,∞),H). EXPONENTIAL DECAY AND NUMERICAL SOLUTION 7 Proof. 3.5. From the classical semigroup theory, see for example [14], follows by Lemma 3.3 that U(t) = eA t U0 is the unique solution of the problem (10)-(11) in the conditions of theorem. The proof is complete. 4. Asymptotic behavior Now we are in position to present our principal result Theorem 4.1. The semigroup eA t is exponentially stably. Proof. 4.2. We now use Theorem 1.1 and we use a contradiction argument. We first prove (6). From Lemma 3.2 we have that 0 ∈ ρ(A) and follows from this fact and the contraction mapping theorem that for any real number β with |β| < ||A||−1, the operator iβI − A = A(iβA−1 − I) is invertible. Moreover, ||(iβI − A)−1|| is a continuous function of β in the interval (−||A||−1, ||A||−1). If sup{||(iβI − A)−1|| : |β| < ||A||−1} = M < ∞, then by the contraction map- ping theorem, the operator iβI − A = (iβ0I − A)(I + i(β − β0)(iβ0I − A)−1) with |β0| < ||A||−1 is invertible for |β − β0| < 1/M. It turns out that by choosing |β0| as close to ||A||−1 as we can, we conclude that {β : |β| < ||A||−1 + 1/M} ⊂ ρ(A) and ||(iβI − A)−1|| is continuous function of β in the interval (−||A||−1 − 1/M , ||A||−1 + 1/M). From argument above, it follows that if (6) is not true, then there is w ∈ R with ||A||−1 ≤ |w| < ∞ such that {iβ ; |β| < |w|} ⊂ ρ(A) and sup{||(iβ − A)−1|| : |β| < |w|} = ∞. It turns out that there exists a sequence βn ∈ R with βn → w, |βn| < |w| and a sequence of complex vector functions Un = (φn,un,ψn,vn,zn,wn)T satisfying Un ∈ D(A) with ||Un||H = 1 such that ||(iβn − A)Un|| → 0, as n → ∞, and then iβnφn − un → 0 in H10(29) iβnρ1u n − K(φnx + ψ n)x + µ1u n + µ2z n(·,1) → 0 in L2 iβnψn − vn → 0 in H10(30) iβnρ2v n − bψnxx + K(φ n x + ψ n) + µ3v n + µ4w n(·,1) → 0 in L2 iβnτzn − znρ → 0 in L 2(0,1; L2) iβnτwn − wnρ → 0 in L 2(0,1; L2) Making the inner product of (iβnI − A)Un with Un in H, taking the real part, and using (19) we have∫ L 0 (|un|2 + |vn|2 + zn(x,1)2 + wn(x,1)2)dx → 0, 8 RAPOSO, CHUQUIPOMA, SANTOS from where follows that un → 0(31) vn → 0(32) zn → 0(33) wn → 0(34) Using (31) into (29) we obtain φn → 0,(35) and using (32) into (30) we obtain φn → 0.(36) Now using (31),(32),(33),(34),(35),(36) we concludes that ||Un|| → 0 which is a contradiction with ||Un|| = 1 and the proof of (6) is complete. Finally we prove (7) by a contradiction argument again. Suppose that (7) is not true. Then there exists a sequence βn with |βn| → ∞ and a sequence of complex vector functions Un ∈ D(A) with unit norm in H such that ||(iβnI − A)Un|| → 0, as n → ∞. Again we have∫ L 0 (|un|2 + |vn|2 + zn(x,1)2 + wn(x,1)2)dx = −⟨AUn,Un⟩ → 0.(37) Making the inner product of (iβnI − A)Un with Un in H we obtain iβn||Un||2 − ⟨AUn,Un⟩ → 0. From (37) we get βn||Un||2 → 0.(38) As βn → ∞ and ||Un|| is limited, we concludes that (38) is true only if ||Un|| → 0 contradict ||Un|| = 1. The proof of theorem is complete. 5. Numerical Solution We will solve numerically the system of Timoshenko (1)-(5) in the one-dimension domain Ω of the length L, using high-order schemes. We used the Implicit Compact Finite Difference Method of fourth-order for discretization of spacial variable and the classic Finite Difference for discretization of temporal variable. 5.1. Discretization. In order to get the discretization of the problem (1)-(5), we define the following sets: Ωh = {xi : xi = ih, i = 0,1, ... ,I + 1; h = L/(I + 1)}, ⊤k = {tn : tn = nk, n = 0,1, ...,N; k = Ch}, †k = {tn : tn = nk, n = −M,−M + 1, ...,0; 0 < M < N} where Qkh = Ωh × ⊤ k and Dkh = Ωh × † k are the computational mesh, and mesh of delay, respectively. The width of mesh Dkh is τ = Mk. In Figure 1 we show a mesh model for the full-domain Qkh ∪ D k h. The points (xi, tn) are called nodes of EXPONENTIAL DECAY AND NUMERICAL SOLUTION 9 ( ,1)i ( , )i N (1, )n ( , )I n ( , 0)i ( 1, )n- ( 2, )I n+( 1, )I n+(0, )n ( , )i n ( , 1)i - ( , )i M- ( , )i M k h Q k h D Figure 1. Model mesh for the full-domain: Qkh ∪ D k h. the mesh and, usually denote by (i,n). The classification of nodes is as follows: interiors (circles), boundaries (stars), initials (squares) and ghosts (diamonds). Let χ = χ(x,t) be a function with second order partial derivatives. Henceforth consider the following notation χni ≡ χ(xi, tn). We define the following approxima- tion of the derivatives of χ, according to Taylor, (χt) n i ≈ 1 k δ−t χ n i , (χt) n i ≈ 1 2k δ0t χ n i , (χt) (n−M) i ≈ 1 k δ−t χ (n−M) i (χx) n i ≈ 1 2h δ0xχ n i , (χtt) n i ≈ 1 k2 δ2t χ n i , (χxx) n i ≈ 1 h2 [ δ2x 1 + 1 12 δ2x ] χni(39) where the finite difference operators are given by δ−t χ n i := χ n i − χ n−1 i , δ 0 t χ n i := χ n+1 i − χ n−1 i , δ−t χ (n−M) i := χ (n−M) i − χ (n−M)−1 i , δ 0 xχ n i := χ n i+1 − χ n i−1, δ2t χ n i := χ n+1 i − 2χ n i + χ n−1 i , δ 2 xχ n i := χ n i+1 − 2χ n i + χ n i−1, (40) [ 1 + 1 12 δ2x ] χni := 1 12 χni+1 + 5 6 χni + 1 12 χni−1 The discrete formulation of equations (1)-(5) is obtained using (39), ρ1 [ 1 + 1 12 δ2x ] δ2t φ n i − α1δ 2 xφ n i − α2 [ 1 + 1 12 δ2x ] δ0xψ n i + α3 [ 1 + 1 12 δ2x ] δ0t φ n i + α4 [ 1 + 1 12 δ2x ] δ−t φ (n−M) i = 0 in (xi, tn) ∈ Qkh(41) 10 RAPOSO, CHUQUIPOMA, SANTOS ρ2 [ 1 + 1 12 δ2x ] δ2t ψ n i − β1δ 2 xψ n i + β2 [ 1 + 1 12 δ2x ] δ0xφ n i + β0 [ 1 + 1 12 δ2x ] ψni + β3 [ 1 + 1 12 δ2x ] δ0t ψ n i + β4 [ 1 + 1 12 δ2x ] δ−t ψ (n−M) i = 0 in (xi, tn) ∈ Qkh(42) (43) φ0i = (φ0)i, ψ 0 i = (φ0)i, (φt) 0 i = (φ1)i, (ψt) 0 i = (ψ1)i in xi ∈ Ω̊h (44) φn0 = φ n I+1 = ψ n 0 = ψ n I+1 = 0 on tn ∈ ⊤ k (45) φ (n−M) i = (f0) (n−M) i , ψ (n−M) i = (g0) (n−M) i , in (xi, t(n−M)) ∈ D k h where, the parameters, are defined by α1 = Kk 2/h2, α2 = Kk 2/2h, α3 = µ1k/2, α4 = µ2k, β1 = bk 2/h2, β2 = α2, β0 = Kk 2, β3 = µ3k/2, β4 = µ4k Substituting (40) in (41)-(42), we have the following linear algebraic system: A1Φ n+1 = B1Φ n + C1Ψ n + D1Φ n−1 − E1δ−t Φ (n−M) + Υn1(46) A2Ψ n+1 = B2Ψ n + C2Φ n + D2Ψ n−1 − E2δ−t Ψ (n−M) + Υn2(47) where, Φn+1 = (φn+11 ,φ n+1 2 , ...,φ n+1 I ) T and Ψn+1 = (ψn+11 ,ψ n+1 2 , ...,ψ n+1 I ) T, n = 0,1, ...,N − 1, are unknown vectors, A1 = tridiag ( 1 12 (ρ1 + α3), 5 6 (ρ1 + α3), 1 12 (ρ1 + α3) ) , B1 = tridiag (1 6 (ρ1 + 6α1), 1 3 (5ρ1 − 6α1), 1 6 (ρ1 + 6α1) ) , C1 = pentadiag ( − 1 12 α2,− 5 6 α2,0, 5 6 α2, 1 12 α2 ) , D1 = tridiag ( 1 12 (−ρ1 + α3), 5 6 (−ρ1 + α3), 1 12 (−ρ1 + α3) ) , E1 = tridiag ( 1 12 α4, 5 6 α4, 1 12 α4 ) , A2 = tridiag ( 1 12 (ρ2 + β3), 5 6 (ρ2 + β3), 1 12 (ρ2 + β3) ) , B2 = tridiag ( 1 12 (2ρ2 + 12β1 − β0), 1 6 (10ρ2 − 12β1 − 5β0), 1 12 (2ρ2 + 12β1 − β0) ) , C2 = −C1, D2 = tridiag ( 1 12 (−ρ2 + β3), 5 6 (−ρ2 + β3), 1 12 (−ρ2 + β3) ) , E2 = tridiag ( 1 12 β4, 5 6 β4, 1 12 β4 ) , are matrices of order I×I. Υn1 and Υn2 are vectors of order I that load the boundary data. EXPONENTIAL DECAY AND NUMERICAL SOLUTION 11 5.2. Numerical test. In order to verify the asymptotic behavior of the solution of the Timoshenko system, we consider the following data: L = 2π, ρ1 = ρ2 = K = b = 1. Boundary condition: φ(0, t) = φ(2π,t) = ψ(0, t) = ψ(2π,t) = 0 Initial condition: φ0(x) = 0, ψ0(x) = 0, φ1(x) = sin(x), ψ1(x) = cos(x) Delay condition: f0(x,t − τ) = sin(x) cos(t − τ), g0(x,t − τ) = cos(x) cos(t − τ) Numerical data: I = 18, C = 0.3, τ = 10% of the width of the mesh Qkh, TOL = 4 × 10−5 (tolerance). Table 1 shows seven cases where Timoshenko system may behave differently with the presence the terms of delay and damping. Each of these cases are plotted in Figure 2-8. Note that the asymptotic behavior of the solution was calculated by taking the maximum value of the function φ, in x ∈ [0,2π], throughout time. In Figure 2, it is observed that there is no asymptotic behavior of the solution, in contrast to Figures 3-6, where the asymptotic behavior of the solution is increasingly more acute. Figure 7 represents the case without delay, the presence of damping is very evident, obtaining the asymptotic behavior of the solution immediately. Figure 8 represents the case without delay and damping, and as was expected, there is no convergence of the solution. In Figure 9 we show the graph of function φ(x,t), where x ∈ [0,2π], t ∈ [−2.97,29.76], µ1 = µ3 = 1, µ2 = µ4 = 0.8 and we choose only 300 iterations along time. With respect to rotation angle ψ we observe that it exhibits the same behavior that the function φ. Table 1. Table for different cases. Case Damping Delay Iterations in time Asymptotic be- havior 1 µ1 = µ3 = 1 µ2 = µ4 = 1 3000 diverges 2 µ1 = µ3 = 1 µ2 = µ4 = 0.9 3000 converges 3 µ1 = µ3 = 1 µ2 = µ4 = 0.8 3000 converges 4 µ1 = µ3 = 1 µ2 = µ4 = 0.7 3000 converges 5 µ1 = µ3 = 1 µ2 = µ4 = 0.6 3000 converges ... 6 µ1 = µ3 = 1 µ2 = µ4 = 0 159 converges 7 µ1 = µ1 = 0 µ2 = µ4 = 0 3000 diverges 12 RAPOSO, CHUQUIPOMA, SANTOS t p h i 500 1000 1500 2000 2500 3000 0.2 0.4 0.6 0.8 1 Figure 2. Case 1. t p h i 500 1000 1500 2000 2500 3000 0.2 0.4 0.6 0.8 Figure 3. Case 2. t p h i 500 1000 1500 2000 2500 3000 0.2 0.4 0.6 0.8 Figure 4. Case 3. t p h i 500 1000 1500 2000 2500 3000 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Figure 5. Case 4. t p h i 500 1000 1500 2000 2500 3000 0.1 0.2 0.3 0.4 0.5 0.6 Figure 6. Case 5. t p h i 50 100 150 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 Figure 7. Case 6. EXPONENTIAL DECAY AND NUMERICAL SOLUTION 13 t p h i 500 1000 1500 2000 2500 3000 0.2 0.4 0.6 0.8 1 Figure 8. Case 7. x 0 2 4 6 t 0 10 20 30 p h i -1 0 1 Figure 9. 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Zheng, Semigroups Associated with dissipative systems, Chapman, New Y & Hallo/CRC, New York, 1999. [14] A. Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations. Springer, New York, 1993. 1Federal University of São João del-Rei, 36.307-352, São João del-Rei - MG, Brazil 2Federal University of Pará, 36.307-352, Belém - PA, Brazil ∗Corresponding author