On initial inverse problem for nonlinear couple heat with Kirchhoff type

The main objective of the paper is to study the final model for the Kirchhoff-type parabolic system. Such type problems have many applications in physical and biological phenomena. Under some smoothness of the final Cauchy data, we prove that the problem has a unique mild solution. The main tool is Banach’s fixed point theorem. We also consider the non-well-posed problem in the Hadamard sense. Finally, we apply truncation method to regularize our problem. The paper is motivated by the work of Tuan, Nam, and Nhat [Comput. Math. Appl. 77(1):15–33, 2019].


Introduction
In this paper, we consider the following Kirchhoff-type problem for parabolic equation systems: x ∈ ∂ , t ∈ (0, T], (1.1) with the following terminal condition example, Almeida [16] studied the following system of two population densities u, v: where the death in species u is proportional to |u| p-2 u by the factor λ 1 , the death in species v is proportional to |v| p-2 v by the factor λ 2 , and f 1 , f 2 are the supplies of external sources. The author obtained the results on the existence, uniqueness, and long-time behavior of a smooth global solution of the system. Ferreira [17] also proved the well-posedness of the system of nonlocal reaction-diffusion equations with both homogeneous Dirichlet or Neumann boundary conditions. Since model (1.1) is a system having a gradient element, perhaps, the techniques for this problem are more complex. To the best of our knowledge, there is no result concerning problem (1.1)-(1.2). The current main applications of backward in time parabolic equations are hydrological inversion and image processing. The parabolic equation with terminal conditions plays an important role in physics and engineering, especially with thermal conductivity dependent on both time and space. We refer the reader to some interesting papers [18][19][20][21]. Our paper is motivated by the recent results of Baleanu et al. [21], Nam et al. [19], and Tuan et al. [20]. The techniques of this paper are based on the previous paper [22].
The main tool in the paper is the Fourier series technique in H s spaces, combined with Banach's fixed point theorem. The main and novel contributions of the paper are as follows: • The first contribution result is the proof of the existence and uniqueness of a solution of our backward problem. To this end, we had to increase the smoothness properties of the input Cauchy data. • The second result is showing that our backward problem is ill-posed in the Hadamard sense. Furthermore, we also regularize our inverse problem using the Fourier truncation method. We then obtain an error assessment between the regularized and exact solutions. This paper is organized as follows. In Sect. 2, we introduce some preliminaries and the mild solution of problem (1.1)-(1.2). Using the Banach fixed point theorem, we show that our problem has a unique mild solution. In Sect. 3, we prove the ill-posedness of our problem. Applying the Fourier truncation method, we give a stability estimate of logarithmic type between the regularized and exact solutions.

Some preliminaries and the mild solution of problem (1.1)-(1.2)
In this section, we introduce some properties of the eigenvalues of the operator -; see, for example, [6]. We have the equality e n (x) = -λ n e n (x), x ∈ ; e n (x) = 0, x ∈ ∂ , n ∈ N, where {λ n } ∞ n=1 is the set of eigenvalues of -satisfying and lim n→∞ λ n = ∞. Let us recall the following Hilbert scale space for ν > 0: associated with the norm Let L ∈ C 1 (R 2 ) be a function such that: • There exists two positive constants M 0 , M 1 such that • There exists a positive constant K l > 0 such that Due to conditions (2.5) and (2.6), we have the following lemma.  It is easy to see that system (3.2) allows us to get that the equalities and v n (t) = exp -λ n t 0 L ∇u(·, τ ) L 2 , ∇v(·, τ ) L 2 dτ v n (0). (3.4) Due to the terminal condition (1.2), we get that u n (T) = f , e n , v n (T) = g, e n .
By a simple calculation we get the Fourier coefficients of u and v: which allow us to get that L ∇u(·, s) L 2 , ∇v(·, s) L 2 ds g, e n e n (x). (3.7) for two constants B f , B g > 0. Then problem (1.1) has a mild solution on the space Proof To show the existence of a mild solution, we define the operator ) and show that Q has a fixed point in the space (L ∞ θ (0, T; H 1 ( ))) 2 . Here the operators Q 1 and Q 2 are defined as follows: (3.9) We will prove by induction that if (u 1 , v 1 ) ∈ (L ∞ (0, T; H 1 ( ))) 2 and (u 2 , v 2 ) ∈ (L ∞ (0, T; H 1 ( ))) 2 , then For m = 1, using the inequality (c + d) 2 ≤ 2c 2 + 2d 2 , we have Applying Lemma 2.1 and the inequality |e re q | ≤ |r -q| max(e r , e q ) for r, q ∈ R, we have (3.12) where in the above line, we applied the inequality ∇ψ L 2 ( ) ≤ C ψ H 1 ( ) . In a similar way, we get that Combining (3.11), (3.12), and (3.13), we deduce that (3.14) Let (3.10) hold for m = p. We will show that (3.10) holds for m = p + 1. Indeed, we have From two above observations we find that By the induction assumption on (3.10), from (3.15) it follows that (L ∞ (0,T;H 1 ( ))) 2 . (3.16) Hence (3.10) holds for any positive integer m. As a consequence, we derive that there exists a positive constant m 0 such that the term Using the Banach fixed point theorem, we conclude that Q m 0 has a fixed point (u * , v * ) on the space (L ∞ (0, T; H 1 ( ))) 2 . It is easy to get that (u * , v * ) is also a solution of the equation Q(u, v) = (u, v). Under the Cauchy terminal data (f m , g m ) as above, by Theorem (3.1) we get that problem (1.1) has a mild solution (u m , v m ) ∈ (L ∞ (0, T; H 1 ( ))) 2 , which is given by  Recalling the lower bound of L, we derive that for any 0 ≤ t ≤ T, By a similar argument we also obtain that

Fourier truncation method and error estimate
Let us define the regularized solution by the Fourier truncation method as 1) where N := N(δ) is a regularization parameter. Here the function (f δ , g δ ) ∈ L 2 ( ) × L 2 ( ) satisfies Then we have the estimate Then the error u N,δ (·, t)u(·, t) 2 Proof To show the existence of a mild solution, we define the operator R m δ (u, v)(t) = (R 1,δ (u, v)(t), R 2,δ (u, v)(t)) and show that R m δ has a fixed point in the space (L ∞ θ (0, T; H 1 ( ))) 2 . Here the operators R 1,δ and R 2,δ are defined as follows:  We will prove by induction that if (u 1 , v 1 ) ∈ (L ∞ (0, T; H 1 ( ))) 2 and (u 2 , v 2 ) ∈ (L ∞ (0, T; H 1 ( ))) 2 , then For m = 1, using the inequality (c + d) 2 ≤ 2c 2 + 2d 2 , we get that By applying Lemma 2.1 and the inequality |e re q | ≤ |r -q| max(e r , e q ) for r, q ∈ R, we have where in the last line, we used the inequality ∇ψ L 2 ( ) ≤ C ψ H 1 ( ) . By a similar argument we obtain that Combining (4.6), (4.7), and (4.8), we find that This implies (4.5). Assume that (4.5) holds for m = j. We will check that (4.5) holds for m = j + 1. Indeed, by similar arguments as before, we also get the following two bounds: From two above observation we find that Using the induction assumption of (3.10), from (3.15) it follows that (L ∞ (0,T;H 1 ( ))) 2 . (4.11) Hence, (3.10) holds for any positive integer m. As a consequence, we conclude that (4.5) holds for any m ∈ N. Since there exists a positive constant j 0 such that Using the Banach fixed point theorem (see [31][32][33]), we conclude that R j 0 δ has a fixed point (u + , v + ) on the space (L ∞ (0, T; H 1 ( ))) 2 . It is easy to get that (u * , v * ) is also a solution of the nonlinear equation From (3.6) and (4.1) we find that exp λ n T t L ∇u(·, s) L 2 , ∇v(·, s) L 2 ds f , e n e n (x). (4.12) By a simple calculation we find that First of all, let us look at the first term. Using Parseval's equality and noting that L(z 1 , z 2 ) ≤ M 1 for all (z 1 , z 2 ) ∈ R 2 , Error 1 is bounded by which allows us to derive that and from the inequality (c + d) 2 ≤ 2c 2 + 2d 2 , c, d ≥ 0, it follows that T t L ∇u N,δ (·, s) L 2 , ∇v N,δ (·, s) L 2 ds -