Well-posedness of the linear heat equation with a second order memory term

where the source term f ∈ L2 ( (0,T ), L2(T) ) . In many problems arising in mathematical physics such as flow of fluid through fissured rocks, diffusion process of gas in a transparent tube, heat conduction in materials, and viscoelasticity, one may encounter memory effects that are relevant from a physical point of view and can be modeled by nonlocal terms. For instance, in [3], a modified Fourier’s law which is independent of the present value of the temperature gradient, but depends on its history, is introduced to correct the unphysical property of instantaneous propagation for the heat equation. By following [1, 5] (see, also, [2]) has been proposed a linearized theory for the heat transfer in isotropic media in which the heat flux depends both on the present value of the temperature gradient and its history being given by


Introduction and preliminaries
Let T := R/2πZ be the one-dimensional torus, let Q := (0, T )×T and M > 0. In this article we study the existence and uniqueness of the solution of the following heat equation involving a memory term: where the source term f ∈ L 2 (0, T ), L 2 (T) . In many problems arising in mathematical physics such as flow of fluid through fissured rocks, diffusion process of gas in a transparent tube, heat conduction in materials, and viscoelasticity, one may encounter memory effects that are relevant from a physical point of view and can be modeled by nonlocal terms. For instance, in [3], a modified Fourier's law which is independent of the present value of the temperature gradient, but depends on its history, is introduced to correct the unphysical property of instantaneous propagation for the heat equation. By following [1,5] (see, also, [2]) has been proposed a linearized theory for the heat transfer in isotropic media in which the heat flux depends both on the present value of the temperature gradient and its history being given by where k(s) is the heat conduction relaxation function and ϑ represents the temperature. A particular and simplified case of this modified Fourier's law is used for the equation with memory term (1).
Let us introduce the following space with its corresponding norm: We observe that (1) may be written as the following system which is equivalent to where f ∈ L 2 (0, T ); L 2 (T) and the operator A is given by I d being the identity operator. The domain of the operator A is given by D (A) = H 2 (T) × L 2 (T) and it is characterized by the following property Notice that in (3) the function z denotes t 0 u xx (s, x) ds and, consequently, z(0, ·) = 0. However, we can solve (3) for arbitrary value z(0, ·) = z 0 .

Spectral analysis
In this section we present the spectral analysis of the operator A. We have the following result.

Lemma 2.1
The eigenvalues of the operator A are given by the family where µ 0 = 0 and Moreover, each of the eigenvalues µ ± n , n ∈ N * , is double and has associated two eigenvectors The eigenvalue µ 0 has associated one eigenvector Φ + 0 = 1 0 and a generalized eigenvector If we are looking for u and z in the form u = n∈Z a n e inx , z = n∈Z b n e inx , from (9) it follows that                (n 2 − µ)a n + Mb n = 0 (n 0), n 2 a n − µb n = 0 (n 0), µb 0 = 0, Mb 0 = µa 0 .
The above system has a non-zero solution (a n , b n ) if µ verifies the following equation Solving (10) we obtain that µ are given by (7), for each n ∈ N * . Since we have z = −u xx µ we deduce the form (8) for the eigenvectors Φ ± n and Φ ± −n , for each n ∈ N * . If b 0 0 we obtain µ = M = 0. If b 0 = 0 we obtain µ = 0 is a double eigenvalue and the corresponding eigenvector and generalized eigenvector are given by The proof is complete. The following remark gives some useful properties of the eigenvalues given by (7).

Remark 2.1
We remark that, the eigenvalues (µ ± n ) n∈N * given by (7) can be written in the following form µ + n = n 2 + r n , µ − n = −r n (n ∈ N * ), where r n := 2M Moreover, the following estimates hold Also, notice that . We prove that, given an arbitrary element in L 2 (T) there exists a unique sequence a ± n n∈Z such that From the form (8) of Φ ± n , the form of Φ + 0 , Φ − 0 given by Lemma 2.1 and by relations (14) and (15) we obtain the following system By taking into account (13) it results that the above system has a unique solution a ± n given by , that is, there exist two positive constants K 1 and K 2 such that for all sequences of scalars a ± n n∈Z in the space ℓ 2 . We have that Taking into account the fact that γ ± |n| 2 ≤ 1, γ ± |n| 2 n 2 µ ± |n| 2 ≤ 1 and by using the inequalities we deduce that the second inequality (16) holds for K 2 = 4. To obtain the first inequality in (16), the following estimate is used, for any z 1 , z 2 ∈ C, α 1 , α 2 , β 1 , β 2 ∈ R and δ > 0 such that In the above inequality if we consider We remark that C 1 n , C 2 n > 0 for every δ n > 0 such that If we choose and we take into account the fact that From (19) it follows that there exists N 0 ∈ N * such that, for any |n| > N 0 , we have that From the above estimate it follows that and we deduce that (20) On the other hand, we have that From (17), (18), (20) and (21) we deduce that the left inequality (16) holds for and the proof is complete.

Remark 2.2 In Theorem 2.1 we show that
and Moreover, we have that F ∈ L 2 (0, T ); L 2 (T) 2 if and only if, for a.e. t ∈ (0, T ), there exists a unique sequence f ± n (t) and On the other hand, given f ∈ L 2 (0, T ); L 2 (T) such that we have that F = f 0 is given by (25), where the Fourier coefficients have the particular

Well posedness results
In this section we study the existence, uniqueness and regularity properties of the solution of (4). We shall introduce two concepts of solution: weak and classical.
Next, we study the solutions of the following system where A is the operator given by (5) and F ∈ L 2 (0, T ); L 2 (T) . The first result gives the simplest solutions of (29).
Proposition 3.1 Let n ∈ Z, a 0,± n ∈ C and f ± n ∈ L 2 (0, T ). If n 0, then the unique solution of (29) with F = f ± n (t)Φ ± n and u 0 z 0 = a 0,± n Φ ± n is given by On the other hand, if F = f + 0 (t)Φ + 0 and then the corresponding solution of (29) is given by Proof. Firstly, we consider the case n 0. We look for a solution of (29) of the form u z (t) = a ± n (t)Φ ± n . We deduce that Moreover, since Solving (32) with initial conditions (33) it follows that a ± n (t) = a 0,± n e −µ ± |n| t + t 0 e −µ ± |n| (t−s) f ± n (s) ds, and (30) holds.
On the other hand, if n = 0 we look for a solution of (29) of the form u z Since Solving (34) with initial conditions (35) we deduce that   For N ∈ N * , we consider and We have that where the Fourier coefficients f ± n (t) 0≤|n|≤N are given by (27).
The corresponding solution of the problem (29) with F = f N 0 and u 0 N z 0 N given by where the Fourier coefficients a ± n (t) 0≤|n|≤N are given by (37).
Taking into account the form of u N z N given by (43), it follows that the relation Taking into account the last three convergences and passing to the limit in (44), when N → ∞, we deduce that u z given by (36) verifies (28), for any φ, ψ ∈ C 1 [0, T ]; H 2 (T) and t ∈ [0, T ].