Wave type spectrum of the Gurtin-Pipkin equation of the second order

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1 Wave type spectrum of the Gurtin-Pipkin equation of the second order arxiv:2.283v2 [math.cv] 26 Apr 2 S. A. Ivanov Abstract We study the complex part of the spectrum of the the Gurtin- Pipkin integral-differential equation of the second order in time. We consider the model case when the kernel is a sum of exponentials a k exp( b k ) with a k = /k α, b k = k β. We show that there are two complex sequences of points of the spectrum asymptotically close to the spectrum points of the wave equation. Introduction. Notations. Main theorem. Introduction In several fields of physics such as heat transfer with finite propagation speed [2], systems with thermal memory [6], viscoelasticity problems [3], and acoustic waves in composite media [], the following integro-differential equation arises θ tt (x,t) = aθ xx t k(t s)θ xx ds x (,π), t >. () with the Dirichlet boundary condition and with the initial data θ(, x) = ξ(x), θ t (,x) = η(x). Regularity of this equation is studied in [5] and in [7]. 2 First, apply the Fourier method: we set ϕ n = sinnx and expand the π solution and the initial data in series in ϕ n θ(x,t) = θ n (t)ϕ n (x), ξ(x) = ξ n ϕ n (x), η(x) = η n ϕ n (x). State Marine Technical University of St. Petersburg. Research of Sergei Ivanov was supported in part by the Russia Foundation for Basic Research, grant a. sergei.ivanov@pobox.spbu.ru

2 For the components we obtain t θ n (t) = αn 2 θ n (t)+n 2 k(t s)θ n (s)ds, t >, θ n () = ξ n, θn () = η n. (2) We will denote the Laplace image by the capital characters. Applying the Laplace Transform to (2) we find or z 2 Θ n (z) zξ n η n = an 2 Θ n (z)+n 2 K(z)Θ n (z) Θ n (z) = zξ n +η n z 2 +an 2 n 2 K(z). The set Λ of all zeros of the denominators z 2 +an 2 n 2 K(z), n =,2,... is called the spectrum of the equation ()..2 The statement and the class of kernels In application [] the kernel k(t) is a sum of exponentials We consider the model case k(t) = a k e bkt. a =, a k = /k α, b k = k β, α >. The problem is to describe the zeros z n with Iz n of the functions z 2 /n 2 + K(z), n N, (3) where K is the Laplace transform of k(t) in our case, K(z) = We are interested in the case a k =, what gives a k z +b k. α,a+β >. a k b k <. (4) The conditions (4) means that k(t) has an integrable singularity at zero: k / L (, ), k L (, ). 2

3 .3 The main result Set r = α+β, < r. β Theorem For any n there exist two zeros z n ±, z n = z+ n (i) If r <, z + n = in+c r e i(r+)π/2 n r +O()+O(n 2r ), c r = of (3) such that π βsinπr. (ii)if r =, z ± n = ±in 2β logn+o(). 2 The proof of the main theorem We are going to find the zeros near the positive imaginary axis and we set Then we obtain the equation Consider the case τ n. z n = in+τ n n. τ n (τ n +2i) = K(z n ). (5) Remark 2 If we consider the case where the second factor in(5) is small: τ n +2i we obtain the complex conjugated roots. Indeed z n = in+ τ n n is also the solution to (3). Rewrite (5) as τ = g n (τ), g n (τ) = K(in+τn). (6) τ +2i Show that for large n we have a contraction map and we can applied the fixed point theorem. We need Lemma 3 If for some δ > argz < π δ, (7) then for z (i) r < K(z) = c r z r +O( z ). 3

4 (ii) r =. (iii) K(z) = β zk (z) log(+z) z +O(/ z ). { /ρ r, r < logρ ρ, r =. (8) The proof is in Sec. 3. Using this lemma, we find the asymptotic of τ n. Let τ < /2 (and z = in+nτ) g n (τ) = K (z)n τ +2i K(z) (τ +2i) 2 K n (z) τ +2ı + K(z) τ +2i 2 K (z)z + K(z) { /ρ r, r < logρ, r =. ρ as z. Therefore we can find R such that g n(τ) < ρ < for z > R and τ < /2. This implies that (6) has a zero τ n with τ n < /2. The iterations τ n () =, τ n () = g n (τn ) = g n() = i K(in),...,τ(k+) 2 n = g n (τ n (k+) ),..., converge and we may write τ n = τ n () +(τ(2) n τ() n )+ +(τ(k+) n Estimate τ (2) n τ () n. τ (2) n τ() n = g n(τ () zk (z) n ) g n(τ n () τ n (k) )+ = τ() n +O( τ(2) n τ() n ). ) max g n (τ) τ τ τ n () n () { /ρ 2r, r < log 2 ρ, r =.. ρ 2 Now Theorem follows from Lemma 3 and from z n + = in+nτ n = in in 2 K(in)+O(n τ(2) n τ n () ). 4

5 3 Asymptotic of K(z) 3. Integral instead series Find the error when we replace the series K(z) by the integral h(z) = dx x α (z +x β ). Lemma 4 Under (7) and x, b > z +x b 2 z 2 +x 2b. Proof. Set z = ρe iϕ, ρ = z z +x b 2 = (ρcosϕ+x b ) 2 +ρ 2 sin 2 ϕ = ρ 2 +2x b ρcosϕ+x 2b. Evidently and ρ 2 +2x b ρcosϕ+x 2b (ρ+x b ) 2 2(ρ 2 +x 2b ) ρ 2 +2x b ρcosϕ+x 2b ρ 2 2x b ρcosδ +x 2b (ρ 2 +x 2b ) (ρ 2 +x 2b )cosδ = ( cosδ)(ρ 2 +x 2b ). Lemma 5 K(z) h(z) /ρ. (9) Proof. Set w(x) = x α (z +x β ). k+ K(z) h(z) = [w(k) w(x)]dx k k+ k w(k) w(x) dx We have for x [k,k +] w(x) w(k) (x k)max w (x), w (x) = αx α z +(α+β)x α+β x 2α (z +x β ) 2 ρ+x β x a+ (ρ 2 +x 2β ), 5

6 what gives For α > this gives w k α+ (ρ+k β ) ρ K(z) h(z) = Const <. kα+ Remark 6 The restriction r in the theorem connected with the estimate (9). If ρ >, then the error /ρ of replacing the series by the integral is more than the integral, which h is O(/ρ r ). 3.2 Evaluating of the integral Lemma 7 (i) For r < h(z) = c r z r +O(/ z ) (ii) For r = h(z) = β log(z +). z Proof. (i) First, reduce one parameter. Set x β = t, x = t /β, dx = β t/β. Then h(z) = β dt t r (z +t). Replace this integral by the integral on [, ) with the error dt t r (z +t) z. Therefore h(z) = β ( ) dt t r (z +t) +O = ( ) z β J(z)+O. z 6

7 In J(z) we change the variable y = t/z. We obtain the integral by the ray R(z) = {ae iargz a }. J(z) = z r R(z) dy y r (+y). The integrand is an analytical function in the upper half plane and is o(/m) on the circle z = M. This implies that the integral does not depend on the ray. Now J(z) = z r dt t r (+t) = π sinπr z r, see [4, Probl (7)] or [8, Probl. 878] (ii) This is direct evaluating. Proof of Lemma 3. parts (i) and (ii) follows from Lemmas 3 and 5. Part (iii). zr z Lemma4 (z) ρ k α z +k β 2 k α (ρ 2 +k 2β ). Replace the series by integral: zk (z) ρ ρ Estimate the last integral setting x β = ρt J = 2 k α (ρ 2 +k 2β ) ρ + dx x α (ρ 2 +x 2β ) ρ/β α/β 2 ρ dx x α (ρ 2 +x 2β ). dt y r (+t 2 ). This gives J { /ρ r, r < logρ, r =. ρ Acknowledgements The author is very grateful to Prof. A. E. Eremenko for his constructive comments and consultations. The author is also grateful to Prof. V. V. Vlasov for the fruitful discussions on the topic. 7

8 References [] A. A. Gavrikov, S.A. Ivanov, D.Yu. Knyazkov, V.A. Samarain, A.S. Shamaev, V. V. Vlasov, Spectral properties of composite medaia, Contemporary Problems of Mathematic and Mechanic, v., 29, (Russian). [2] M. E. Gurtin, A. C. Pipkin A general theory of heat conduction with finite wave speeds. Archive for Rational Mechanics and Analysis 968; 32:3-26. [3] C. M. Dafermos, Asymptotic stability in viscoelasticity, Arch. Rational Mech. Anal., 37 (97), [4] M. A. Evgrafov, Collection of problems in the theory of analytic functions, Moscow, Nauka, 972 (Russian). [5] L. Pandolfi,The controllability of the Gurtin-Pipkin equation: a cosine operator approach. Appl. Math. Optim. 52 (25), no. 2, [6] F.M. Vegni Dissipativity of a consedved phase field systems with memory, Discrete and contiouns dynamical systems, Volume 9, Number 4, July 23. [7] V. V. Vlasov, J. Wu, Solvability and Spectral Analysis of Abstract Hyperbolic Equations with Delay. Funct. Differ. Equ. 6 (29), no. 4, [8] L. I Volkovyski, G. D. Lunz and I. G. Aramanovich, Collection of problems in the theory of functions of a complex variable, Moscow, Fizmatgiz, 96, (Russian) 8