Convexification of a Complex Ginzburg–Landau Equation with Reversed Time Using a New Carleman Estimate

Abstract

In this paper, we propose a convexification method for the reversed-time problem of a complex Ginzburg–Landau equation by introducing a new Carleman estimate for the complex Ginzburg–Landau operator, which holds on an arbitrary time interval. We construct a weighted Tikhonov-like functional and prove its global strict convexity. We establish both the existence and uniqueness of the minimizer, along with the global convergence of the associated gradient projection method toward the exact solution. These results ensure global convergence, unlike conventional least-squares minimization techniques, which only offer local convergence.

1. Introduction

The Ginzburg–Landau equation originated in the work of Ginzburg and Landau [1] as a model for superconductivity. Subsequently, the theory found applications in broader contexts such as phase transitions [2] and pattern-forming systems [3]. In 1969, Newell and Whitehead [4] and independently Segel [5] derived real-valued Ginzburg–Landau equations as amplitude equations for long waves. The complex-valued counterpart was later obtained by Stewartson and Stuart [6]. A typical example is the classical cubic complex Ginzburg–Landau equation (CGL), which takes the form

$$\begin{array}{c}\hfill z_t-({\alpha }_1+i{a}_1)\Delta z=Fz-({\alpha }_2+i{a}_2){\left|z\right|}^{2}z,\hspace{1em}t>0,\hspace{1em}x\in G\subset {\mathbb{R}}^n,\end{array}$$

(1)

where $i=\sqrt{-1}$ is the imaginary unit, $z=z(x,t)$ is a complex amplitude, $F>0$ is the coefficient of a linear driving term, and $a_1,a_2\in \mathbb{R}$. The coefficients ${\alpha }_1$ and ${\alpha }_2$ are positive real numbers, which can be normalized to 1 by rescaling t and z. Many techniques are illustrated through the linearized CGL:

$$\begin{array}{c}\hfill z_t-({\alpha }_1+i{a}_1)\Delta z=Fz,\hspace{1em}t>0,\hspace{1em}x\in G.\end{array}$$

(2)

This equation also serves as the amplitude equation in pattern formation and describes the one-dimensional reaction diffusion of two chemical species [3]. Note that the linear CGL operator includes the heat operator as a special case and approaches the Schrödinger operator as ${\alpha }_1\to 0$. This connection makes the inverse problems for these equations mathematically interesting.

The well-posedness of the CGL and related equations has been extensively studied; see, e.g., [7,8,9,10,11,12,13,14,15,16,17,18,19] and references therein. Among these, some works established the existence and uniqueness of solutions for the linear CGL equation [8,10,18]. In particular, we focus on $C^1$-in-time solutions as in [13,14]. The controllability and observability of parabolic-type equations have also been widely investigated [20,21,22,23,24,25,26]. In particular, the controllability of the CGL was studied in [27,28], which established null-controllability results based on Carleman estimates and sectorial operator theory.

In some physical regimes, the nonlinear term in the CGL is either neglected or treated linearly, so that the equation becomes linear. In the present paper, we consider the following reversed-time problem for the linear CGL. Let $T>0$ and $\Omega \subset {\mathbb{R}}^n$ be a bounded domain with $C^2$ boundary $\partial \Omega$. Denote

$$Q=(0,T)\times \Omega ,\hspace{1em}\Sigma =(0,T)\times \partial \Omega .$$

Fix $a^{jk}(·)\in C^2(\overline{Q};\mathbb{R})$ satisfying

$$\begin{array}{c}\hfill a^{jk}(x,t)=a^{kj}(x,t),\hspace{1em}(x,t)\in \overline{Q},\hspace{1em}j,k=1,\dots ,n,\end{array}$$

(3)

and for some constant $s_0>0$,

$$\begin{array}{c}\hfill {\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\xi }_j\overline{{\xi }_k}\ge s_0{\left|\xi \right|}^2,\hspace{1em}(x,t,\xi )\in \overline{Q}\times {\mathbb{C}}^n.\end{array}$$

(4)

We define the complex Ginzburg–Landau operator as

$$\begin{array}{c}\hfill Gz\triangleq (1+ib)z_t+{\displaystyle \sum _{j,k=1}^n}{\left(a^{jk}{z}_j\right)}_k,\end{array}$$

(5)

where $b\in \mathbb{R}$ is a given constant. Consider the linear CGL

$$\begin{array}{c}\hfill Gz=q(x,t)z,\hspace{1em}(x,t)\in Q,\end{array}$$

(6)

where $q(·)\in L^{\infty }(0,T;L^{\infty }(\Omega ))$ is a potential, subject to the zero Dirichlet boundary condition

$$\begin{array}{c}\hfill {z|}_{\Sigma }=0\end{array}$$

(7)

and the terminal condition

$$\begin{array}{c}\hfill z(x,T)=g\left(x\right),\hspace{1em}x\in \Omega .\end{array}$$

(8)

The target problem involves finding a function $z\in H^2\left(Q\right)$ satisfying (6)–(8) under conditions (3)–(5). Given the potential q and the terminal value g, we seek to reconstruct the solution z for $t<T$. In contrast to the classical forward problem, the reversed-time problem considered here is ill-posed in the Hadamard sense, meaning that the solution lacks continuous dependence on the terminal data.

Conventional numerical methods for such problems typically rely on the minimization of least-squares Tikhonov-like functionals. However, these functionals often suffer from multiple local minima and ravines, which cause gradient-type descent methods to converge to arbitrary local minima unless the initial guess is sufficiently close to the exact solution. Such a precise initial guess is rarely available in practice, resulting in only local convergence. In contrast, a numerical method is called globally convergent if it is guaranteed, by a rigorous theorem, to produce a point in a sufficiently small neighborhood of the exact solution without any a priori knowledge of that neighborhood.

The concept of the convexification method was first introduced by Klibanov and his coauthors in 1995–1997 [29,30] to address the global convergence issue for inverse problems. The existence and uniqueness of the minimizer of the convexified functional were first established in [31]. Subsequently, the convexification methodology was extended to a range of problems, including the Helmholtz equation [32,33,34], hyperbolic equations [35,36,37], and electrical impedance tomography [38], and more recently for coefficient inverse problems of mean field games [39] and for 1D inverse problems with experimental data [40]. For parabolic equations with reversed-time, a convergent numerical approach was developed in [41]. However, the convexification method is a concept rather than a ready-to-use algorithm: each new problem requires its own version of the convexification method. In particular, there are few results on convexification for complex-valued equations. The main difficulty is the imaginary term $ib{z}_t$ in the operator $Gz$. For a real parabolic operator, the derivation of the Carleman estimate follows a standard procedure where the signs of the resulting terms are well-controlled. In the complex case, however, the presence of i introduces additional cross-terms with indefinite signs, which must be carefully estimated.

For convexification, we construct a weighted Tikhonov-like functional $I_{\alpha ,\lambda }\left(u\right)$ for the reversed-time problem of the CGL, where $\lambda \ge 1$ is the parameter in the Carleman weight function $\theta \left(t\right)=\exp \left({(t+1)}^{\lambda }\right)$. Given a convex bounded set $B\left(R\right)\subset H^k\left(Q\right)$ of arbitrary diameter $R>0$, one can choose $\lambda$ sufficiently large such that the functional is strictly convex on $\overline{B\left(R\right)}$, eliminating multiple local minima. Following the methodology in [31], we prove the existence and uniqueness of the minimizer of $I_{\alpha ,\lambda }$ on $\overline{B\left(R\right)}$ and the global convergence of the gradient projection method to that minimizer. Moreover, as the noise level in the terminal data tends to zero, the minimizers converge to the correct solution of the reversed-time problem on the time interval $(\tau ,T)$, which means the minimizers stabilize with respect to small noise. Since the diameter R of $B\left(R\right)$ is arbitrary, this amounts to global convergence.

The rest of this paper is organized as follows. Section 2 presents the main results of the convexification approach for the CGL reversed-time problem, namely the global strict convexity of the weighted Tikhonov-type functional, the existence and uniqueness of its minimizer, and the global convergence of the gradient projection method. Section 3 is devoted to the proof of a new Carleman estimate for the CGL operator, which is established through a four-step argument. Section 4 contains the proofs of Theorems 1–4. Finally, Section 5 provides a brief summary of the paper and discusses possible directions for future work. This is a purely theoretical work. Numerical experiments are not included and will be addressed in future research.

2. Statement of Main Results

We now state the main theoretical results of this work. We first introduce a transformation that converts the reversed-time problem into a form suitable for the convexification framework. Based on this transformation, we construct a weighted Tikhonov-like functional and establish its global strict convexity. We then demonstrate that the minimizer exists and is unique. We also establish the global convergence of the gradient projection method toward the exact solution. Finally, we provide accuracy estimates in the presence of noise in the terminal data.

Recall that by (8), the function $g\left(x\right)$ is defined as $g\left(x\right)=z(x,T)$. The following transformation is applied to the function $u(x,t)$:

$$\begin{array}{c}\hfill u(x,t)=z(x,t)-g\left(x\right),\hspace{1em}(x,t)\in Q.\end{array}$$

(9)

Then, (6)–(8) lead to

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}Gu=q(x,t)u+Bg\left(x\right)\hfill & \mathrm{in} Q,\hfill \\ u=0\hfill & \mathrm{on} \Sigma ,\hfill \\ u(x,T)=0\hfill & \mathrm{in} \Omega ,\hfill \end{array}\right.\end{array}$$

(10)

where

$$\begin{array}{c}\hfill Bg\left(x\right)=q(x,t)g\left(x\right)+{\displaystyle \sum _{j,k=1}^n}{\left(a^{jk}{g}_j\right)}_k,\hspace{1em}(x,t)\in Q.\end{array}$$

(11)

Since we need $Bg\left(x\right)$ to be bounded in $\overline{Q}$, we assume that

$$g\left(x\right)\in C^2\left(\overline{\Omega }\right).$$

(12)

Let k be an integer satisfying $k>\left[\right(n+1)/2]+2$, where $\left[\right(n+1)/2]$ is the largest integer not exceeding $(n+1)/2$. The Sobolev embedding theorem then ensures that $H^k\left(Q\right)\subset C^2\left(\overline{Q}\right)$ and

$$\begin{array}{c}\hfill {\parallel g\parallel }_{C^2\left(\overline{Q}\right)}\le E{\parallel g\parallel }_{H^k\left(Q\right)},\hspace{1em}\forall g\in H^k\left(Q\right),\end{array}$$

(13)

where the constant $E=E\left(Q\right)>0$ depends only on Q. Define the subspace ${\mathring{H}}^k\left(Q\right)\subset H^k\left(Q\right)$ as

$${\mathring{H}}^k\left(Q\right)=\{u\in H^k\left(Q\right):u{|}_{\Sigma }=0, u(x,T)=0\}.$$

Given an arbitrary number $R>0$, we introduce the bounded function set $B\left(R\right)$ as

$$\begin{array}{c}\hfill B\left(R\right)=\{u\in {\mathring{H}}^k\left(Q\right):\parallel u{\parallel }_{H^k\left(Q\right)}<R\}.\end{array}$$

(14)

Hence, (13) implies that

$$\overline{B\left(R\right)}\subset C^2\left(\overline{Q}\right),$$

(15)

$${\parallel u\parallel }_{C^2\left(\overline{Q}\right)}\le E_R,\hspace{1em}\forall u\in \overline{B\left(R\right)},$$

(16)

where $E_R=E_R(Q,R)>0$ depends only on Q and R. The closure $\overline{B\left(R\right)}$ is taken with respect to the $H^k\left(Q\right)$-norm.

Then, we introduce our Carleman weight function (CWF), which depends only on time variable $t,$ that is,

$$\begin{array}{c}\hfill \theta \left(t\right)=\exp \left({(t+1)}^{\lambda }\right),\hspace{1em}t\in (0,T),\end{array}$$

(17)

where $l={(t+1)}^{\lambda }$, $\lambda \ge 1$, is a large parameter. We seek an approximate solution of problem (10)–(11) in $\overline{B\left(R\right)}$. To this end, we choose $\tau \in (0,T)$ and minimize the following weighted Tikhonov-like functional:

$$\begin{array}{c}\hfill I_{\alpha ,\lambda }\left(u\right)={\theta }^2(-\tau ){\int }_Q{|Gu-q(x,t)u-Bg|}^2{\theta }^2\left(t\right) \mathrm{d}x\mathrm{d}t+\alpha {\parallel u\parallel }_{H^k\left(Q\right)}^2,\hspace{1em}\forall u\in \overline{B\left(R\right)},\end{array}$$

(18)

where $\alpha \in (0,1)$ is a regularization parameter and the multiplier ${\theta }^2(-\tau )$ balances the two terms in (18).

• Minimization problem. Minimize the functional $I_{\alpha ,\lambda }\left(u\right)$ on the set $\overline{B\left(R\right)}$.

We now state the main results concerning the functional $I_{\alpha ,\lambda }$. The following theorem establishes the global strict convexity of $I_{\alpha ,\lambda }$ on $\overline{B\left(R\right)}$, which guarantees the absence of local minima and is the key result of the convexification method. We outline the strategy for proving the strict convexity of $I_{\alpha ,\lambda }$. The proof relies on the Carleman estimate established in Section 3. The main idea is to use this estimate to show that for sufficiently large $\lambda$, the quadratic term determines the convexity of the functional, yielding the strict convexity inequality (21). Details are given in Section 4.

Theorem 1 

(The central theorem of this paper). Assume that conditions (3)–(5) and (11) hold. Then, for parameter $\alpha >0$, the functional $I_{\alpha ,\lambda }\left(u\right)$ has the Fréchet derivative $I_{\alpha ,\lambda }^{\prime }\left(u\right)\in H^k\left(Q\right)$ at every point $u\in {\mathring{H}}^k\left(Q\right)$. Let ${\lambda }_0>1$ be the constant of Theorem 5. Then, there exist constants

$${\lambda }_1={\lambda }_1\left({Q,\parallel g\parallel }_{C^2\left(Q\right)},R,\tau \right)\ge {\lambda }_0>1,$$

(19)

$$C_2=C_2\left({Q,\parallel g\parallel }_{C^2\left(Q\right)},R\right)>0$$

(20)

depending only on listed parameters such that if $\lambda \ge {\lambda }_1$, the functional $I_{\alpha ,\lambda }\left(u\right)$ is strictly convex on $\overline{B\left(R\right)}$ for all $\lambda \ge {\lambda }_1$, i.e., for all $u_1,u_2\in \overline{B\left(R\right)}$, $\lambda \ge {\lambda }_1$, and $\tau \in (0,T)$, the following inequality holds:

$$\begin{array}{cc}\hfill \hspace{1em}& I_{\alpha ,\lambda }\left(u_2\right)-I_{\alpha ,\lambda }\left(u_1\right)-I_{\alpha ,\lambda }^{\prime }\left(u_1\right)(u_2-u_1)\hfill \\ \hfill \ge & C_2\parallel u_2-u_1{\parallel }_{H^{1,0}\left(Q_{\tau }\right)}^2+{\displaystyle \frac{\alpha }{2}}{\parallel u_2-u_1\parallel }_{H^k\left(Q\right)}^2,\hspace{1em}\forall u_1,u_2\in \overline{B\left(R\right)}.\hfill \end{array}$$

(21)

Remark 1. 

The constant $C_2$ in (21) is independent of λ and α. The strict convexity of $I_{\alpha ,\lambda }$ on $\overline{B\left(R\right)}$ ensures that the minimization problem has a unique global minimizer, and it is also a key to prove the global convergences, which are established in the following theorems.

Regarding the existence and uniqueness of the minimizer, we establish Theorem 2 in the following.

Theorem 2. 

Let $R>0$ be an arbitrary number and $\alpha ,\lambda >0$ be constants. Assume that conditions (3)–(5) and (11) hold. Then, there exists a unique point of the global minimum $u_{\min }$ of the functional $I_{\alpha ,\lambda }\left(u\right)$ on the set $\overline{B\left(R\right)}$.

To actually compute the minimizer $u_{\min }$ whose existence is guaranteed by Theorem 2, we develop the gradient projection algorithm for minimizing $I_{\alpha ,\lambda }$ over the convex set $\overline{B\left(R\right)}$.

$P_{\overline{B}}:{\mathring{H}}^k\left(Q\right)\to \overline{B\left(R\right)}$ denotes the orthogonal projection from ${\mathring{H}}^k\left(Q\right)$ onto $\overline{B\left(R\right)}$. Choose an arbitrary initial point $u_0\in B\left(R\right)$ and a parameter $\gamma \in (0,1)$. The gradient projection iteration is then defined as

$$u_{n+1}=P_{\overline{B}}\left(u_n-\gamma I_{\alpha ,\lambda }^{\prime }\left(u_n\right)\right),\hspace{1em}n=0,1,2,\dots .$$

(22)

The following theorem establishes the global convergence of the gradient projection method to the unique minimizer.

Theorem 3 

(Convergence of the gradient projection method). Let ${\lambda }_1$ be the number of Theorem 1. Choose $\lambda \ge {\lambda }_1$. Let $u_{\min }\in \overline{B\left(R\right)}$ be the unique minimizer of the functional $I_{\alpha ,\lambda }\left(u\right)$ on $\overline{B\left(R\right)}$. Then, there exists a sufficiently small number ${\gamma }_0\in (0,1)$ such that for every $\gamma \in (0,{\gamma }_0)$, there exists $\theta =\theta \left(\gamma \right)\in (0,1)$ such that

$${\parallel u_{n+1}-u_{\min }\parallel }_{H^k\left(Q\right)}\le {\theta }^n{\parallel u_{\min }-u_0\parallel }_{H^k\left(Q\right)}.$$

(23)

Consider now the case of the data containing noise. We use again one of Tikhonov’s concepts of regularization. More precisely, using (13), we assume that there exist exact noiseless data $g^{*}\left(x\right)\in C^2\left(\overline{\Omega }\right)$ in (11), and there exists an exact solution $u^{*}\in B\left(R\right)$ of problem (10), where g is, respectively, replaced with $g^{*}$. Let $\delta \in (0,1)$ be the level of noise in the data $g\left(x\right)$, i.e.,

$$\begin{array}{c}\hfill \parallel g-g^{*}{\parallel }_{C^2\left(\overline{\Omega }\right)}<\delta .\end{array}$$

(24)

The following theorem provides the accuracy and convergence rate estimates for the minimizer and the gradient projection method in the presence of noise.

Theorem 4 

(Estimates of the accuracy and the convergence rate). Assume that there exists the exact solution $u^{*}\in B\left(R\right)$ of problem (10) and condition (24) holds. Let ${\lambda }_1>1$ be the number of Theorem 1. Select an arbitrary $\tau \in (0,T)$. For any $\delta \in (0,1)$, set $\lambda =\lambda \left(\delta \right)$ as in Lemma 3.1 of [41]. Let

$${\delta }_0={\delta }_0\left(\parallel g^{*}{\parallel }_{C^2\left(\overline{\Omega }\right)},Q,{\parallel q\parallel }_{L^{\infty }\left(Q\right)},R\right)>0$$

be sufficiently small such that $\lambda \left({\delta }_0\right)\ge {\lambda }_1$. Let $u_{\min }\in B\left(R\right)$ be the unique minimizer of $I_{\alpha ,\lambda }$ on $B\left(R\right)$. Let ${\gamma }_0$ be the number defined in Theorem 3. Let $\gamma \in (0,{\gamma }_0)$ and $\theta =\theta \left(\gamma \right)\in (0,1)$ be the numbers of Theorem 3. Then, the following estimates hold:

$$\begin{array}{c}\hfill \parallel u^{*}-u_{\min }{\parallel }_{H^{1,0}\left(Q_{\tau }\right)}\le C_3\exp \left[-{\displaystyle \frac{2}{2^c}}{\left(\ln \left({\delta }^{-1}\right)\right)}^c\right],\end{array}$$

(25)

$$\begin{array}{cc}\hfill \parallel u^{*}-u_{n+1}{\parallel }_{H^{1,0}\left(Q_{\tau }\right)}& \le C_3\exp \left[-{\displaystyle \frac{2}{2^c}}{\left(\ln \left({\delta }^{-1}\right)\right)}^c\right]+{\theta }^n{\parallel u_{\min }-u_0\parallel }_{H^k\left(Q\right)},\hfill \end{array}$$

(26)

where $c=c(\tau ,T)={\displaystyle \frac{\ln (\tau +1)}{\ln (T+1)}}\in (0,1)$, and $C_3$ is a positive constant.

Remark 2. 

The choice of $\alpha =\alpha \left(\delta \right)=2e^{-\lambda \left(\delta \right)T^2}$ in Theorem 4 is necessary for the theoretical convergence analysis as $\delta \to 0$. However, from a computational perspective, such an exponentially small α may lead to ill-conditioning of the discrete problem. In practical implementations, one may choose a slightly larger α and use a posteriori parameter selection strategy. A detailed numerical study of this issue is left for future work.

Remark 3. 

Unlike the local convergence of conventional least-squares minimization techniques, the convergences in Theorems 3 and 4 are global.

The proofs of Theorems 1–4 are given in Section 4 below.

3. Carleman Estimate

In this section, we establish a Carleman estimate for the complex Ginzburg–Landau operator $Gz=(1+ib)z_t+{\sum }_{j,k=1}^n{\left(a^{jk}{z}_j\right)}_k$. We choose a suitable time-dependent CWF, which allows us to obtain an estimate on an arbitrary time interval $(0,T)$. The result is stated in Theorem 5.

We first recall an identity from Corollary 2.1 of [42] (see Lemma 1 below). Based on this identity, by choosing a suitable CWF and performing some appropriate constructions, we obtain our Carleman estimate.

Lemma 1 

([42] Corollary 2.1). Let $b\in \mathbb{R}$ and $a^{jk}\in C^2(\overline{Q};\mathbb{R})$ satisfy $a^{jk}=a^{kj}$ for $j,k=1,\dots ,n$. Let $z,v\in C^2(\overline{Q};\mathbb{C})$ and $\Psi ,\mathcal{l}\in C^2(\overline{Q};\mathbb{R})$. Set $\theta =e^{\mathcal{l}}$, $v=\theta z$, and

$$\Psi =-2{\displaystyle \sum _{j,k=1}^n}{\left(a^{jk}{\mathcal{l}}_j\right)}_k.$$

Then,

$$\begin{array}{cc}\hfill \hspace{1em}& {\theta }^2{\left|Gz\right|}^2+M_t+\mathrm{div}V\hfill \\ \hfill \ge & |I_1{|}^2+{\displaystyle \sum _{j,k,j^{\prime },k^{\prime }=1}^n}\left[2{\left(a^{j^{\prime }k}{\mathcal{l}}_{j^{\prime }}\right)}_k{a}^{j{k}^{\prime }}-a_{k^{\prime }}^{jk}{a}^{j^{\prime }{k}^{\prime }}{\mathcal{l}}_{j^{\prime }}+{\displaystyle \frac{1}{2}}{a}_t^{jk}+a^{jk}{\left(a^{j^{\prime }{k}^{\prime }}{\mathcal{l}}_{j^{\prime }}\right)}_{k^{\prime }}\right]\hfill \\ \hfill \hspace{1em}& (v_k{\overline{v}}_j+{\overline{v}}_k{v}_j)+ib{\displaystyle \sum _{j,k=1}^n}\left(a_t^{jk}{\mathcal{l}}_j+2a^{jk}{\mathcal{l}}_{jt}\right)({\overline{v}}_{k}v-v_k\overline{v})\hfill \\ \hfill \hspace{1em}& -ib{\displaystyle \sum _{j,k=1}^n}{\left(a^{jk}{\mathcal{l}}_j\right)}_k(\overline{v}{v}_t-v{\overline{v}}_t)+B{\left|v\right|}^2,\hfill \end{array}$$

(27)

where

$$M=\left[\left(1+b^2\right){\mathcal{l}}_t-A\right]{\left|v\right|}^2+{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{v}_j{\overline{v}}_k+ib{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\mathcal{l}}_j({\overline{v}}_{k}v-v_k\overline{v}),$$

(28)

$$\begin{array}{cc}\hfill V^k& ={\displaystyle \sum _{j,j^{\prime },k^{\prime }=1}^n}\{-ib\left[a^{jk}{\mathcal{l}}_j({\overline{v}}_{t}v-\overline{v}{v}_t)+a^{jk}{\mathcal{l}}_t(v_j\overline{v}-{\overline{v}}_{j}v)\right]-a^{jk}(v_j{\overline{v}}_t+{\overline{v}}_j{v}_t)\hfill \\ \hfill \hspace{1em}& \hspace{1em}+(2a^{j{k}^{\prime }}{a}^{j^{\prime }k}-a^{jk}{a}^{j^{\prime }{k}^{\prime }}){\mathcal{l}}_j(v_{j^{\prime }}{\overline{v}}_{k^{\prime }}+{\overline{v}}_{j^{\prime }}{v}_{k^{\prime }})-\Psi a^{jk}(v_j\overline{v}+{\overline{v}}_{j}v)\hfill \\ \hfill \hspace{1em}& \hspace{1em}+a^{jk}(2A{\mathcal{l}}_j+{\Psi }_j-2{\mathcal{l}}_j{\mathcal{l}}_t){\left|v\right|}^2\},\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill B& =(1+b^2){\mathcal{l}}_{tt}-A_t-2\left[{\displaystyle \sum _{j,k=1}^n}{\left(a^{jk}{\mathcal{l}}_j{\mathcal{l}}_t\right)}_k+\Psi {\mathcal{l}}_t\right]+{\displaystyle \sum _{j,k=1}^n}{\left(a^{jk}{\Psi }_k\right)}_j\hfill \\ \hfill & +2\left|{\displaystyle \sum _{j,k=1}^n}{\left(a^{jk}{\mathcal{l}}_{j}A\right)}_k+A\Psi \right|,\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill A& ={\displaystyle \sum _{j,k=1}^n}{a}^{jk}{l}_j{l}_k-{\displaystyle \sum _{j,k=1}^n}{\left(a^{jk}{l}_j\right)}_k-\Psi .\hfill \end{array}$$

(31)

Theorem 5 

(Carleman estimate). Let $b\in \mathbb{R}$ and $a^{jk}\in C^2({\mathbb{R}}^n;\mathbb{R})$ satisfy $a^{jk}=a^{kj}$ for $j,k=1,\dots ,n$. Let $z,v\in C^2({\mathbb{R}}^{1+n};\mathbb{C})$ and $\Psi ,\mathcal{l}\in C^2({\mathbb{R}}^{1+n};\mathbb{R})$. Set $\theta =e^{\mathcal{l}}$ and $v=\theta z$. By setting $\Psi =-2{\sum }_{j,k=1}^n{\left(a^{jk}{\mathcal{l}}_j\right)}_k$, there exist numbers ${\lambda }_0,C_1$,

$$\begin{array}{cc}\hfill \hspace{1em}& {\lambda }_0={\lambda }_0(\Omega ,b)>0,\hfill \\ \hfill \hspace{1em}& C_1=C_1(Q,b,s_0)>0\hfill \end{array}$$

(32)

depending only on listed parameters such that the following Carleman estimate holds:

$$\begin{array}{cc}\hfill {\int }_Q{\left|Gz\right|}^2{\theta }^2 \mathrm{d}x\mathrm{d}t& \ge C_1\sqrt{\lambda }(\lambda -4){\int }_Q{\left|z\right|}^2{\theta }^2 \mathrm{d}x\mathrm{d}t+C_1\sqrt{\lambda }{\int }_Q{|\nabla z|}^2{\theta }^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \hspace{1em}& \hspace{1em}-C_1\exp \left(3{(T+1)}^{\lambda }\right){\parallel z\left(T\right)\parallel }_{H^1(\Omega )}^2,\hspace{1em}\forall \lambda ⩾{\lambda }_0.\hfill \end{array}$$

(33)

We are now in a position to prove Theorem 5.

Proof of Theorem 5. 

Differentiate $l={(t+1)}^{\lambda }$ with respect to x and t to obtain

$$\begin{array}{c}\hfill l_j=l_{jk}=0,\hspace{1em}j,k=1,\cdots ,n,\end{array}$$

(34)

$$\begin{array}{c}\hfill l_t=\lambda {(t+1)}^{\lambda -1},\end{array}$$

(35)

$$\begin{array}{c}\hfill l_{tt}=\lambda (\lambda -1){(t+1)}^{\lambda -2}.\end{array}$$

(36)

The proof is divided into four steps.

• Step 1. Firstly, by integrating (27) over Q, we have

  $$\begin{array}{cc}\hfill \hspace{1em}& {\int }_Q{\theta }^2{\left|Gz\right|}^2 \mathrm{d}x\mathrm{d}t+{\int }_Q{M}_t \mathrm{d}x\mathrm{d}t+{\int }_Q\mathrm{div}V \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \ge & {\int }_Q{\left|I_1\right|}^2 \mathrm{d}x\mathrm{d}t+{\int }_Q{\displaystyle \sum _{j,k,j^{\prime },k^{\prime }=1}^n}\left[2{\left(a^{j^{\prime }k}{\mathcal{l}}_{j^{\prime }}\right)}_k{a}^{j{k}^{\prime }}-a_{k^{\prime }}^{jk}{a}^{j^{\prime }{k}^{\prime }}{\mathcal{l}}_{j^{\prime }}+{\displaystyle \frac{1}{2}}{a}_t^{jk}+a^{jk}{\left(a^{j^{\prime }{k}^{\prime }}{\mathcal{l}}_{j^{\prime }}\right)}_{k^{\prime }}\right]\hfill \\ \hfill \hspace{1em}& (v_k{\overline{v}}_j+{\overline{v}}_k{v}_j) \mathrm{d}x\mathrm{d}t+{\int }_{Q}ib{\displaystyle \sum _{j,k=1}^n}\left(a_t^{jk}{\mathcal{l}}_j+2a^{jk}{\mathcal{l}}_{jt}\right)({\overline{v}}_{k}v-v_k\overline{v}) \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \hspace{1em}& -{\int }_{Q}ib{\displaystyle \sum _{j,k=1}^n}{\left(a^{jk}{\mathcal{l}}_j\right)}_k(\overline{v}{v}_t-v{\overline{v}}_t) \mathrm{d}x\mathrm{d}t+{\int }_{Q}B{\left|v\right|}^2 \mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

  (37)

  Then, we estimate the left-hand side of the inequality (37). From (28) and (31), we obtain

  $$\begin{array}{cc}\hfill \hspace{1em}& {\int }_Q{M}_t \mathrm{d}x\mathrm{d}t\hfill \\ \hfill =& {\int }_{\Omega }\{[(1+b^2)\lambda {(T+1)}^{\lambda -1}-A(x,T)]{\left|v(x,T)\right|}^2+{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{v}_j{\overline{v}}_k(x,T)\hfill \\ \hfill \hspace{1em}& +ib{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{l}_j({\overline{v}}_{k}v-v_k\overline{v})(x,T)-[(1+b^2)\lambda -A(x,0)]{\left|v(x,0)\right|}^2\hfill \\ \hfill \hspace{1em}& -{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{v}_j{\overline{v}}_k(x,0)-ib{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{l}_j({\overline{v}}_{k}v-v_k\overline{v})(x,0)\} \mathrm{d}x\hfill \\ \hfill =& {\int }_{\Omega }(1+b^2)\lambda {(T+1)}^{\lambda -1}{\left|v(x,T)\right|}^2 \mathrm{d}x+{\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{v}_j{\overline{v}}_k(x,T) \mathrm{d}x\hfill \\ \hfill \hspace{1em}& -{\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\overline{v}}_j{v}_k(x,0) \mathrm{d}x-{\int }_{\Omega }(1+b^2)\lambda {\left|v(x,0)\right|}^2 \mathrm{d}x.\hfill \end{array}$$

  (38)

  Let $n\left(x\right)$ be the outward normal vector to $\partial \Omega$ at the point $x\in \partial \Omega$. Using Gauss’s formula, (29), and (35), we have

  $$\begin{array}{cc}\hfill \hspace{1em}& {\int }_Q\mathrm{div}V \mathrm{d}x\mathrm{d}t\hfill \\ \hfill =& {\int }_0^T{\int }_{\partial \Omega }{\displaystyle \sum _{k=1}^n}{V}^k·n_k\left(x\right) \mathrm{d}S \mathrm{d}t\hfill \\ \hfill =& {\int }_0^T{\int }_{\partial \Omega }{\displaystyle \sum _{j,k=1}^n}\left[-ib{a}^{jk}\lambda {(t+1)}^{\lambda -1}(v_j\overline{v}-{\overline{v}}_{j}v)-a^{jk}(v_j{\overline{v}}_t+{\overline{v}}_j{v}_t)\right]n_k\left(x\right) \mathrm{d}S \mathrm{d}t.\hfill \end{array}$$

  (39)

  Recall that $v=\theta z=0$ on $\Sigma$. Obviously, we get

  $$\begin{array}{c}\hfill {\int }_Q\mathrm{div}V \mathrm{d}x\mathrm{d}t=0.\end{array}$$

  (40)

  We now analyze the terms on the right-hand side of (37). By (34), we have

  $$\begin{array}{cc}\hfill \hspace{1em}& {\int }_Q{\displaystyle \sum _{j,k,j^{\prime },k^{\prime }=1}^n}\left[2{\left(a^{j^{\prime }k}{\mathcal{l}}_{j^{\prime }}\right)}_k{a}^{j{k}^{\prime }}-a_{k^{\prime }}^{jk}{a}^{j^{\prime }{k}^{\prime }}{\mathcal{l}}_{j^{\prime }}+{\displaystyle \frac{1}{2}}{a}_t^{jk}+a^{jk}{\left(a^{j^{\prime }{k}^{\prime }}{\mathcal{l}}_{j^{\prime }}\right)}_{k^{\prime }}\right](v_k{\overline{v}}_j+{\overline{v}}_k{v}_j) \mathrm{d}x\mathrm{d}t\hfill \\ \hfill =& {\int }_Q{\displaystyle \sum _{j,k=1}^n}{\displaystyle \frac{1}{2}}{a}_t^{jk}(v_k{\overline{v}}_j+{\overline{v}}_k{v}_j) \mathrm{d}x\mathrm{d}t\hfill \end{array}$$

  (41)

  and

  $$\begin{array}{c}\hfill ib{\int }_Q{\displaystyle \sum _{j,k=1}^n}\left(a_t^{jk}{\mathcal{l}}_j+2a^{jk}{\mathcal{l}}_{jt}\right)({\overline{v}}_{k}v-v_k\overline{v}) \mathrm{d}x\mathrm{d}t=0.\end{array}$$

  (42)

  Similarly, we obtain

  $$\begin{array}{c}\hfill -ib{\int }_Q\left[{\displaystyle \sum _{j,k=1}^n}{\left(a^{jk}{\mathcal{l}}_j\right)}_k(\overline{v}{v}_t-v{\overline{v}}_t)+B{\left|v\right|}^2\right] \mathrm{d}x\mathrm{d}t=0.\end{array}$$

  (43)

  From (30) and (34)–(36), we obtain

  $$\begin{array}{c}\hfill {\int }_Q{B\left|v\right|}^2 \mathrm{d}x\mathrm{d}t={\int }_Q(1+b^2)\lambda (\lambda -1){(t+1)}^{\lambda -2}{\left|v\right|}^2 \mathrm{d}x\mathrm{d}t.\end{array}$$

  (44)

  Substituting (38) and (40)–(44) into (37), and using $a^{jk}\in C^2\left(Q\right)$, we have

  $$\begin{array}{cc}\hfill \hspace{1em}& {\int }_Q{\theta }^2{\left|Gz\right|}^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \ge & -{\int }_{\Omega }(1+b^2)\lambda {(T+1)}^{\lambda -1}{\left|v(x,T)\right|}^2 \mathrm{d}x-{\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{v}_j{\overline{v}}_k(x,T) \mathrm{d}x+{\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\overline{v}}_j{v}_k(x,0) \mathrm{d}x\hfill \\ \hfill \hspace{1em}& +{\int }_{\Omega }\lambda (1+b^2){\left|v(x,0)\right|}^2 \mathrm{d}x+{\int }_Q{\displaystyle \sum _{j,k=1}^n}{\displaystyle \frac{1}{2}}{a}_t^{jk}(v_k{\overline{v}}_j+{\overline{v}}_k{v}_j) \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \hspace{1em}& +{\int }_Q(1+b^2)\lambda (\lambda -1){(t+1)}^{\lambda -2}{\left|v\right|}^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \ge & -{\int }_{\Omega }(1+b^2)\lambda {(T+1)}^{\lambda -1}{\left|v(x,T)\right|}^2 \mathrm{d}x-{\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{v}_j{\overline{v}}_k(x,T) \mathrm{d}x+{\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\overline{v}}_j{v}_k(x,0) \mathrm{d}x\hfill \\ \hfill \hspace{1em}& +{\int }_{\Omega }\lambda (1+b^2)|v(x,0){|}^2 \mathrm{d}x+{\int }_Q(1+b^2)\lambda (\lambda -1){(t+1)}^{\lambda -2}{\left|v\right|}^2 \mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

  By $v=\theta z$, we have

  $$\begin{array}{cc}\hfill \hspace{1em}& {\int }_Q{\theta }^2{\left|Gz\right|}^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \ge & -{\int }_{\Omega }(1+b^2)\lambda {(T+1)}^{\lambda -1}{\theta }^2\left(T\right){\left|z(x,T)\right|}^2 \mathrm{d}x-{\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\theta }^2{z}_j{\overline{z}}_k(x,T) \mathrm{d}x\hfill \\ \hfill \hspace{1em}& +{\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\theta }^2{\overline{z}}_j{z}_k(x,0) \mathrm{d}x+{\int }_{\Omega }\lambda (1+b^2){\theta }^2\left(0\right){\left|z(x,0)\right|}^2 \mathrm{d}x\hfill \\ \hfill \hspace{1em}& +{\int }_Q(1+b^2)\lambda (\lambda -1){(t+1)}^{\lambda -2}{\theta }^2{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

  (45)

The estimate in (45) involves only ${\left|z\right|}^2$. We now derive an additional term involving ${|\nabla z|}^2$. Therefore, we will proceed with the next steps.

• Step 2. Construct the following two terms:

  $$\begin{array}{cc}\hfill \hspace{1em}& {\theta }^{2}Gz(-\overline{z}){(1+bi)}^{-1}\hfill \\ \hfill =& {\theta }^2\left[(1+ib)z_t+{\displaystyle \sum _{j,k=1}^n}{\left(a^{jk}{z}_j\right)}_k\right]{(1+bi)}^{-1}(-\overline{z})\hfill \\ \hfill =& -{\theta }^2{z}_t\overline{z}-{(1+bi)}^{-1}{\displaystyle \sum _{j,k=1}^n}{\left({\theta }^2{a}^{jk}{z}_j\overline{z}\right)}_k+{(1+bi)}^{-1}{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{z}_j{\overline{z}}_k{\theta }^2,\hfill \end{array}$$

  (46)

  and

  $$\begin{array}{cc}\hfill \hspace{1em}& {\theta }^2\overline{Gz}(-z){(1-bi)}^{-1}\hfill \\ \hfill =& -{\theta }^2{\overline{z}}_{t}z-{(1-bi)}^{-1}{\displaystyle \sum _{j,k=1}^n}{\left({\theta }^2{a}^{jk}{\overline{z}}_{j}z\right)}_k+{(1-bi)}^{-1}{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\overline{z}}_j{z}_k{\theta }^2.\hfill \end{array}$$

  (47)

  Summing up (46) with (47), and using $a^{jk}=a^{kj}$ and $v=\theta z=0$ on $\Sigma$, we have

  $$\begin{array}{cc}\hfill \hspace{1em}& {\theta }^{2}Gz(-\overline{z}){(1+bi)}^{-1}+{\theta }^2\overline{Gz}(-z){(1-bi)}^{-1}\hfill \\ \hfill =& -{\theta }^2{\left[{\left|z\right|}^2\right]}_t-{(1+bi)}^{-1}{\displaystyle \sum _{j,k=1}^n}{\left({\theta }^2{a}^{jk}{z}_j\overline{z}\right)}_k-{(1-bi)}^{-1}{\displaystyle \sum _{j,k=1}^n}{\left({\theta }^2{a}^{jk}{\overline{z}}_{j}z\right)}_k\hfill \\ \hfill \hspace{1em}& +{\displaystyle \frac{1}{1+b^2}}{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{z}_j{\overline{z}}_k{\theta }^2+{\displaystyle \frac{1}{1+b^2}}{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\overline{z}}_j{z}_k{\theta }^2\hfill \\ \hfill =& -{\theta }^2{\left[{\left|z\right|}^2\right]}_t+{\displaystyle \frac{1}{1+b^2}}{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{z}_j{\overline{z}}_k{\theta }^2+{\displaystyle \frac{1}{1+b^2}}{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\overline{z}}_j{z}_k{\theta }^2.\hfill \end{array}$$

  (48)

  Next, we estimate the left-hand side of the above inequality (48) as

  $$\begin{array}{cc}\hfill \hspace{1em}& {\theta }^{2}Gz(-\overline{z}){(1+bi)}^{-1}+{\theta }^2\overline{Gz}(-z){(1-bi)}^{-1}\hfill \\ \hfill =& -{\displaystyle \frac{1}{1+b^2}}\left(Gz\overline{z}+\overline{Gz}z\right){\theta }^2-{\displaystyle \frac{bi}{1+b^2}}(\overline{Gz}z-Gz\overline{z}){\theta }^2\hfill \\ \hfill \le & {\displaystyle \frac{2(1+|b\left|\right)}{1+b^2}}\left|Gz\right|\left|z\right|{\theta }^2.\hfill \end{array}$$

  (49)

  Then, we estimate the right-hand side of the above inequality (48) as

  $$\begin{array}{cc}\hfill {\int }_Q{\theta }^2{\left[{\left|z\right|}^2\right]}_t \mathrm{d}x\mathrm{d}t=& {\int }_{\Omega }{\theta }^2{\left(T\right)\left|z(x,T)\right|}^2 \mathrm{d}x-e^2{\int }_{\Omega }{\left|z(x,0)\right|}^2 \mathrm{d}x\hfill \\ \hfill \hspace{1em}& +2{\int }_Q{\theta }^2\lambda {(t+1)}^{\lambda -1}{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

  (50)

  Finally, integrating (48) over Q and using (49) and (50), we obtain

  $$\begin{array}{cc}\hfill \hspace{1em}& {\displaystyle \frac{2(1+|b\left|\right)}{1+b^2}}{\int }_Q\left|Gz\right|\left|z\right|{\theta }^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \ge & {\int }_{\Omega }{\theta }^2{\left(T\right)\left|z(x,T)\right|}^2 \mathrm{d}x-e^2{\int }_{\Omega }{\left|z(x,0)\right|}^2 \mathrm{d}x+2{\int }_Q{\theta }^2\lambda {(t+1)}^{\lambda -1}{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \hspace{1em}& +{\displaystyle \frac{2}{1+b^2}}{\int }_Q{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{z}_j{\overline{z}}_k{\theta }^2 \mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

  (51)

  Multiplying (51) by $\sqrt{\lambda }$ and summing up with (45), we obtain

  $$\begin{array}{cc}\hfill \hspace{1em}& {\int }_Q{\theta }^2{\left|Gz\right|}^2 \mathrm{d}x\mathrm{d}t+{\displaystyle \frac{2\sqrt{\lambda }(1+|b\left|\right)}{1+b^2}}{\int }_Q\left|Gz\right|\left|z\right|{\theta }^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \ge & -{\int }_{\Omega }(1+b^2)\lambda {(T+1)}^{\lambda -1}{\theta }^2\left(T\right){\left|z(x,T)\right|}^2 \mathrm{d}x-{\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\theta }^2{z}_j{\overline{z}}_k(x,T) \mathrm{d}x\hfill \\ \hfill \hspace{1em}& +{\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\theta }^2{\overline{z}}_j{z}_k(x,0) \mathrm{d}x+{\int }_{\Omega }\lambda (1+b^2){\theta }^2\left(0\right){\left|z(x,0)\right|}^2 \mathrm{d}x\hfill \\ \hfill \hspace{1em}& +{\int }_Q(1+b^2)\lambda (\lambda -1){(t+1)}^{\lambda -2}{\theta }^2{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t+{\displaystyle \frac{2\sqrt{\lambda }}{1+b^2}}{\int }_Q{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{z}_j{\overline{z}}_k{\theta }^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \hspace{1em}& -\sqrt{\lambda }{\theta }^2\left(T\right){\int }_{\Omega }{\left|z(x,T)\right|}^2 \mathrm{d}x+e^2\sqrt{\lambda }{\int }_{\Omega }{\left|z(x,0)\right|}^2 \mathrm{d}x+2\sqrt{\lambda }{\int }_Q{\theta }^2\lambda {(t+1)}^{\lambda -1}{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

  (52)

• Step 3. In this step, we estimate the right-hand side of (52). Using (4), when $\lambda$ is sufficiently large, we have

  $$\begin{array}{cc}\hfill \hspace{1em}& -{\int }_{\Omega }(1+b^2)\lambda {(T+1)}^{\lambda -1}{\theta }^2\left(T\right){\left|z(x,T)\right|}^2 \mathrm{d}x-{\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\theta }^2{z}_j{\overline{z}}_k(x,T) \mathrm{d}x\hfill \\ \hfill \hspace{1em}& -\sqrt{\lambda }{\theta }^2\left(T\right){\int }_{\Omega }{\left|z(x,T)\right|}^2 \mathrm{d}x\hfill \\ \hfill \ge & -(1+b^2)\lambda {(T+1)}^{\lambda -1}\exp \left(2{(T+1)}^{\lambda }\right){\int }_{\Omega }{\left|z(x,T)\right|}^2 \mathrm{d}x\hfill \\ \hfill \hspace{1em}& -C\exp \left(2{(T+1)}^{\lambda }\right){\int }_{\Omega }{|\nabla z(x,T)|}^2 \mathrm{d}x-\exp \left(2{(T+1)}^{\lambda }\right)\sqrt{\lambda }{\int }_{\Omega }{\left|z(x,T)\right|}^2 \mathrm{d}x\hfill \\ \hfill \ge & -(2+b^2)\exp \left(3{(T+1)}^{\lambda }\right){\int }_{\Omega }{\left|z(x,T)\right|}^2 \mathrm{d}x-C\exp \left(2{(T+1)}^{\lambda }\right){\int }_{\Omega }{|\nabla z(x,T)|}^2 \mathrm{d}x.\hfill \end{array}$$

  (53)

Then, (4) implies that

$$\begin{array}{c}\hfill {\int }_{\Omega }{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{\theta }^2{\overline{z}}_j{z}_k(x,0) \mathrm{d}x+{\int }_{\Omega }\lambda (1+b^2){\theta }^2\left(0\right){\left|z(x,0)\right|}^2 \mathrm{d}x+e^2\sqrt{\lambda }{\int }_{\Omega }{\left|z(x,0)\right|}^2 \mathrm{d}x\ge 0\end{array}$$

(54)

and

$$\begin{array}{c}\hfill {\displaystyle \frac{2}{1+b^2}}{\int }_Q{\displaystyle \sum _{j,k=1}^n}{a}^{jk}{z}_j{\overline{z}}_k{\theta }^2 \mathrm{d}x\mathrm{d}t\ge {\displaystyle \frac{2s_0\sqrt{\lambda }}{1+b^2}}{\int }_Q{|\nabla z|}^2{\theta }^2 \mathrm{d}x\mathrm{d}t.\end{array}$$

(55)

Then, from (52)–(55), we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& {\int }_Q{\theta }^2{\left|Gz\right|}^2 \mathrm{d}x\mathrm{d}t+{\displaystyle \frac{2\sqrt{\lambda }(1+|b\left|\right)}{1+b^2}}{\int }_Q\left|Gz\right|\left|z\right|{\theta }^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \ge & {\int }_Q(1+b^2)\lambda (\lambda -1){(t+1)}^{\lambda -2}{\theta }^2{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t+2\sqrt{\lambda }{\int }_Q{\theta }^2\lambda {(t+1)}^{\lambda -1}{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \hspace{1em}& +{\displaystyle \frac{2s_0\sqrt{\lambda }}{1+b^2}}{\int }_Q{|\nabla z|}^2{\theta }^2 \mathrm{d}x\mathrm{d}t-(2+b^2)\exp \left(3{(T+1)}^{\lambda }\right){\int }_{\Omega }{\left|z(x,T)\right|}^2 \mathrm{d}x\hfill \\ \hfill \hspace{1em}& -C\exp \left(2{(T+1)}^{\lambda }\right){\int }_{\Omega }{|\nabla z(x,T)|}^2 \mathrm{d}x.\hfill \end{array}$$

(56)

• Step 4. We now analyze (56). Using the triangle inequality, we have

  $$\begin{array}{cc}\hfill \hspace{1em}& {\int }_Q{\theta }^2{\left|Gz\right|}^2 \mathrm{d}x\mathrm{d}t+{\displaystyle \frac{2\sqrt{\lambda }(1+|b\left|\right)}{1+b^2}}{\int }_Q\left|Gz\right|\left|z\right|{\theta }^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \le & {\displaystyle \frac{b^2+\left|b\right|+2}{1+b^2}}{\int }_Q{\left|Gz\right|}^2{\theta }^2 \mathrm{d}x\mathrm{d}t+{\displaystyle \frac{\lambda (1+|b\left|\right)}{1+b^2}}{\int }_Q{\left|z\right|}^2{\theta }^2 \mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

  (57)

  Choosing $C_1={\displaystyle \frac{2+2\left|b\right|}{{(1+b^2)}^2}}>0$, with ${\lambda }_0=2C_1+3$, we obtain that for $\lambda \ge {\lambda }_0$ and $t\in (0,T)$,

  $$\begin{array}{cc}\hfill \hspace{1em}& {\int }_Q(1+b^2)\lambda (\lambda -1){(t+1)}^{\lambda -2}{\theta }^2{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t+2\sqrt{\lambda }{\int }_Q{\theta }^2\lambda {(t+1)}^{\lambda -1}{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \hspace{1em}& -{\displaystyle \frac{\lambda (1+|b\left|\right)}{1+b^2}}{\int }_Q{\theta }^2{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \ge & (1+b^2)\lambda (\lambda -3){\int }_Q{\theta }^2{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t-{\displaystyle \frac{\lambda (1+|b\left|\right)}{1+b^2}}{\int }_Q{\theta }^2{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \ge & {\displaystyle \frac{1+b^2}{2}}\lambda (\lambda -3){\int }_Q{\theta }^2{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t.\hfill \end{array}$$

  (58)

  Then, from (56)–(58), we obtain

  $$\begin{array}{cc}\hfill \hspace{1em}& {\displaystyle \frac{b^2+\left|b\right|+2}{1+b^2}}{\int }_Q{\theta }^2{\left|Gz\right|}^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \ge & {\displaystyle \frac{1+b^2}{2}}\lambda (\lambda -3){\int }_Q{\theta }^2{\left|z\right|}^2 \mathrm{d}x\mathrm{d}t+{\displaystyle \frac{2s_0\sqrt{\lambda }}{1+b^2}}{\int }_Q{|\nabla z|}^2{\theta }^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \hspace{1em}& -(2+b^2)\exp \left(3{(T+1)}^{\lambda }\right){\int }_{\Omega }{|z(x,T)|}^2 \mathrm{d}x-C\exp \left(2{(T+1)}^{\lambda }\right){\int }_{\Omega }{|\nabla z(x,T)|}^2 \mathrm{d}x,\hfill \end{array}$$

  (59)

  which implies (33).

This completes the proof of Theorem 5. □

Remark 4. 

By applying the chosen weight function $\theta \left(t\right)=\exp \left({(t+1)}^{\lambda }\right)$ to Corollary 2.1 of [42], we initially derive (45), which primarily involves only the ${\left|z\right|}^2$ term. We then develop additional terms (46)–(47). This process ultimately yields our improved Carleman estimate (33), which includes both the ${\left|z\right|}^2$ term and the gradient term ${|\nabla z|}^2$. This allows us to obtain the accuracy estimates in $H^{1,0}\left(Q_{\tau }\right)$. Moreover, the constants $C_1$ depends on b. As $b\to 0$, the estimate formally recovers the real parabolic case [41].

4. Proofs of the Main Theorems

In this section, we prove Theorems 1–4 stated in Section 2. The proof of Theorem 1 is based on the Carleman estimate established in Section 3, combined with the transformation and functional construction introduced in Section 2.

Proof of Theorem 1. 

Consider arbitrary functions $u_1,u_2\in B\left(R\right)$. We denote $h=u_2-u_1$. Then,

$$h\in B\left(2R\right).$$

(60)

Hence, by (16),

$$h\in C^2\left(\overline{Q}\right),\hspace{1em}\hspace{1em}{\parallel h\parallel }_{C^2\left(\overline{Q}\right)}\le E_{2R}.$$

(61)

Then,

$$\begin{array}{cc}\hfill \hspace{1em}& {|G(u_1+h)-q(x,t)(u_1+h)-Bg\left(x\right)|}^2-{|G{u}_1-q(x,t)u_1-Bg\left(x\right)|}^2\hfill \\ \hfill =& |G{u}_1+Gh-q(x,t)u_1{-q(x,t)h-Bg\left(x\right)|}^2-{|G{u}_1-q(x,t)u_1-Bg\left(x\right)|}^2\hfill \\ \hfill =& \mathrm{Lin}\left(h\right)(x,t)+\mathrm{Nonlin}\left(h\right)(x,t),\hfill \end{array}$$

(62)

where $\mathrm{Lin}\left(h\right)$ and $\mathrm{Nonlin}\left(h\right)$ refer to the linear and nonlinear parts of (62) in h, respectively. The linear part is given by

$$\begin{array}{cc}\hfill \hspace{1em}& \mathrm{Lin}\left(h\right)(x,t)\hfill \\ \hfill :=& [G{u}_1-q(x,t)u_1+Bg\left(x\right)][\overline{Gh}-\overline{q(x,t)h}]+[\overline{G{u}_1}-\overline{q(x,t)u_1}-\overline{Bg\left(x\right)}][Gh-q(x,t)h].\hfill \end{array}$$

(63)

We represent the term $\mathrm{Nonlin}\left(h\right)$ in (62) as

$$\mathrm{Nonlin}\left(h\right)(x,t):={|Gh-q(x,t)h|}^2.$$

(64)

Thus, (18) and (62)–(64) imply that

$$\begin{array}{cc}\hfill \hspace{1em}& I_{\alpha ,\lambda }(u_1+h)-I_{\alpha ,\lambda }\left(u_1\right)\hfill \\ \hfill =& {\theta }^2(-\tau ){\int }_Q\mathrm{Lin}\left(h\right){\theta }^2\left(t\right) \mathrm{d}x\mathrm{d}t+{\theta }^2(-\tau ){\int }_Q\mathrm{Nonlin}\left(h\right){\theta }^2\left(t\right) \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \hspace{1em}& +2\alpha {(u_1,h)}_k+\alpha {\parallel h\parallel }_{H^k\left(Q\right)}^2.\hfill \end{array}$$

(65)

where ${(·,·)}_k$ is the scalar product in $H^k\left(Q\right)$. Assuming temporarily that h is an arbitrary function of $H^k\left(Q\right)$, consider the expression $X\left(h\right)$:

$$X\left(h\right)={\theta }^2(-\tau ){\int }_Q\mathrm{Lin}\left(h\right){\theta }^2\left(t\right) \mathrm{d}x\mathrm{d}t+2\alpha {(u_1,h)}_k.$$

(66)

Clearly, $X:H^k\left(Q\right)\to \mathbb{R}$ is a bounded linear functional. By Riesz theorem, there exists a unique function $S\in H^k\left(Q\right)$ such that

$$X\left(h\right)={(S,h)}_k.$$

(67)

Also, (65)–(67) imply that

$$\underset{{\parallel h\parallel }_{H^k\left(Q\right)}\to 0}{\lim }{\displaystyle \frac{|I_{\alpha ,\lambda }(u_1+h)-I_{\alpha ,\lambda }\left(u_1\right)-{(S,h)}_k|}{{\parallel h\parallel }_{H^k\left(Q\right)}}}=0.$$

(68)

Hence, S is the Fréchet derivative $I_{\alpha ,\lambda }^{\prime }\left(u_1\right)\in H^k\left(Q\right)$ of the functional $I_{\alpha ,\lambda }$ at $u_1$,

$${(S,h)}_k=I_{\alpha ,\lambda }^{\prime }\left(u_1\right)\left(h\right),\hspace{1em}\forall h\in H^k\left(Q\right).$$

(69)

We now come back again to the case when $h\in B\left(2R\right)$. Using (65)–(69), we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& I_{\alpha ,\lambda }(u_1+h)-I_{\alpha ,\lambda }\left(u_1\right)-I_{\alpha ,\lambda }^{\prime }\left(u_1\right)\left(h\right)\hfill \\ \hfill =& {\theta }^2(-\tau ){\int }_Q\mathrm{Nonlin}\left(h\right){\theta }^2\left(t\right) \mathrm{d}x\mathrm{d}t+\alpha {\parallel h\parallel }_{H^k\left(Q\right)}^2.\hfill \end{array}$$

(70)

It follows from (33), (64), $h(x,T)=0$, and the Cauchy–Schwarz inequality that

$$\begin{array}{cc}\hfill \hspace{1em}& I_{\alpha ,\lambda }(u_1+h)-I_{\alpha ,\lambda }\left(u_1\right)-I_{\alpha ,\lambda }^{\prime }\left(u_1\right)\left(h\right)\hfill \\ \hfill \ge & {\displaystyle \frac{1}{2}}{\theta }^2(-\tau ){\int }_Q{\left|Gh\right|}^2{\theta }^2\left(t\right) \mathrm{d}x\mathrm{d}t-{\theta }^2(-\tau ){\int }_Q{\left|q(x,t)h\right|}^2{\theta }^2\left(t\right) \mathrm{d}x\mathrm{d}t+\alpha {\parallel h\parallel }_{H^k\left(Q\right)}^2\hfill \\ \hfill \ge & {\displaystyle \frac{1}{2}}{\theta }^2(-\tau )\lambda (\lambda -3){\int }_Q{\left|h\right|}^2{\theta }^2\left(t\right) \mathrm{d}x\mathrm{d}t+\widetilde{C_2}\sqrt{\lambda }{\theta }^2(-\tau ){\int }_Q{\theta }^2\left(t\right){|\nabla h|}^2 \mathrm{d}x\mathrm{d}t\hfill \\ \hfill \hspace{1em}& -{\theta }^2(-\tau ){\int }_Q{\left|q(x,t)h\right|}^2{\theta }^2\left(t\right) \mathrm{d}x\mathrm{d}t+{\displaystyle \frac{\alpha }{2}}{\parallel h\parallel }_{H^k\left(Q\right)}^2,\hfill \end{array}$$

(71)

Since $\tau \in (0,T)$, we obtain

$$\underset{t\in [\tau ,T]}{\min }\left\{\exp (-2{(\tau +1)}^{\lambda })\exp \left(2{(t+1)}^{\lambda }\right)\right\}=1.$$

Then, by recalling $q(·)\in L^{\infty }(0,T;L^{\infty }(\Omega ))$ and choosing ${\lambda }_1\ge {\lambda }_0>1$ based on the same parameters as those listed in (19) such that ${\lambda }_1\ge 2\sqrt{C_2}+3$, we have

$$\begin{array}{cc}\hfill \hspace{1em}& I_{\alpha ,\lambda }(u_1+h)-I_{\alpha ,\lambda }\left(u_1\right)-I_{\alpha ,\lambda }^{\prime }\left(u_1\right)\left(h\right)\hfill \\ \hfill \ge & {\displaystyle \frac{1}{4}}\lambda (\lambda -3){\int }_{Q_{\tau }}{\left|h\right|}^2 \mathrm{d}x\mathrm{d}t+C_2\sqrt{\lambda }{\int }_{Q_{\tau }}{|\nabla h|}^2 \mathrm{d}x\mathrm{d}t+{\displaystyle \frac{\alpha }{2}}{\parallel h\parallel }_{H^k\left(Q\right)}^2,\hfill \end{array}$$

(72)

where $C_2>0$ depends on Q and R. (72) implies (21).

This completes the proof of Theorem 1. □

We now prove Theorem 2.

Proof of Theorem 2. 

The strict convexity of the functional $I_{\alpha ,\lambda }^{\prime }\left(u\right)$ on $\overline{B\left(R\right)}$ follows from Theorem 1. Next, we prove the Lipschitz continuity of the Fréchet derivative $I_{\alpha ,\lambda }^{\prime }\left(u\right)$.

Using (65)–(69) and the Cauchy–Schwarz inequality, we obtain

$$|I_{\alpha ,\lambda }^{\prime }\left(u_1\right)\left(h\right)-I_{\alpha ,\lambda }^{\prime }\left(u_2\right)\left(h\right)|\le 2\alpha \parallel u_1-u_2{\parallel }_{H^k\left(Q\right)}{\parallel h\parallel }_{H^k\left(Q\right)}$$

(73)

for all $u_1,u_2\in \overline{B\left(R\right)}$ and $h\in {\mathring{H}}^k\left(Q\right)$. Thus, we have

$${\parallel I_{\alpha ,\lambda }^{\prime }\left(u_1\right)-I_{\alpha ,\lambda }^{\prime }\left(u_2\right)\parallel }_{H^k\left(Q\right)}\le 2\alpha {\parallel u_1-u_2\parallel }_{H^k\left(Q\right)}.$$

As soon as (21) is established, the existence and uniqueness of the global minimizer $u_{\min }$ of the functional $I_{\alpha ,\lambda }\left(u\right)$ on $\overline{B\left(R\right)}$ follows immediately from Lemma 2.1 and Theorem 2 of [31]. □

We now prove Theorem 3.

Proof of Theorem 3. 

Using Lemma 2.3 of [31] and Theorem 2, we obtain

$$u_{\min }=P_B\left(u_{\min }-\gamma I_{\alpha ,\lambda }^{\prime }\left(u_{\min }\right)\right).$$

(74)

Next, by (22) and (74), we obtain

$$\parallel u_{n+1}-u_{\min }{\parallel }_{H^k\left(Q\right)}\le \theta \left(\gamma \right){\parallel u_n-u_{\min }\parallel }_{H^k\left(Q\right)},\hspace{1em}n=0,1,2,\dots ,$$

(75)

which implies (23).

This completes the proof of Theorem 3. □

Following the lines of the proof of Theorem 5.4 in [41], we prove Theorem 4.

Proof of Theorem 4. 

From (18), and where $u^{*}$ is the exact solution of problem (10), we have

$$I_{\alpha ,\lambda }\left(u^{*}\right)={\theta }^2(-\tau ){\int }_Q|B{g}^{*}{-Bg|}^2{\theta }^2\left(t\right) \mathrm{d}x\mathrm{d}t+\alpha {\parallel u^{*}\parallel }_{H^k\left(Q\right)}^2.$$

By $u^{*}\in B\left(R\right)$ and (24), which implies $\parallel B{g}^{*}{-Bg\parallel }_{L_2\left(Q\right)}\le C\delta$. Hence,

$$I_{\alpha ,\lambda }\left(u^{*}\right)\le C_3\exp \left(2{(T+1)}^{\lambda }\right){\delta }^2+\alpha R^2.$$

(76)

Applying Theorem 2 and Lemma 2.1 of [31], we obtain

$$-I_{\alpha ,\lambda }^{\prime }\left(u_{\min }\right)(u^{*}-u_{\min })\le 0.$$

(77)

Combining (76) and (77), we have

$$\begin{array}{cc}\hfill \hspace{1em}& I_{\alpha ,\lambda }\left(u^{*}\right)-I_{\alpha ,\lambda }\left(u_{\min }\right)-I_{\alpha ,\lambda }^{\prime }\left(u_{\min }\right)(u^{*}-u_{\min })\le C_3\exp \left(2{(T+1)}^{\lambda }\right){\delta }^2+\alpha R^2.\hfill \end{array}$$

(78)

By (21), we obtain

$$\begin{array}{cc}\hfill \hspace{1em}& C_2\parallel u^{*}-u_{\min }{\parallel }_{H^{1,0}\left(Q_{\tau }\right)}^2+\alpha {\parallel u^{*}-u_{\min }\parallel }_{H^k\left(Q\right)}^2\hfill \\ \hfill \le & I_{\alpha ,\lambda }\left(u^{*}\right)-I_{\alpha ,\lambda }\left(u_{\min }\right)-I_{\alpha ,\lambda }^{\prime }\left(u_{\min }\right)(u^{*}-u_{\min }).\hfill \end{array}$$

(79)

Then, (78) and (79) imply that

$$C_2\parallel u^{*}-u_{\min }{\parallel }_{H^{1,0}\left(Q_{\tau }\right)}^2+\alpha {\parallel u^{*}-u_{\min }\parallel }_{H^k\left(Q\right)}^2\le C_3\exp \left(2{(T+1)}^{\lambda }\right){\delta }^2+\alpha R^2.$$

(80)

Choose $\lambda =\lambda \left(\delta \right)$ such that

$$\exp \left(2{(T+1)}^{\lambda \left(\delta \right)}\right)={\displaystyle \frac{1}{\delta }},$$

(81)

Then,

$$\lambda \left(\delta \right)=\ln \left[{\left(\ln \left({\delta }^{-1/2}\right)\right)}^{1/\ln (T+1)}\right].$$

For any $\tau \in (0,T)$, we have

$$\exp (-2{(\tau +1)}^{\lambda \left(\delta \right)})=\exp \left[-{\displaystyle \frac{2}{2^c}}{\left(\ln \left({\delta }^{-1}\right)\right)}^c\right],$$

(82)

where $c=c(\tau ,T)={\displaystyle \frac{\ln (\tau +1)}{\ln (T+1)}}\in (0,1)$. From (80) and (81), we obtain

$$\parallel u^{*}-u_{\min }{\parallel }_{H^{1,0}\left(Q_{\tau }\right)}^2\le C_3(\delta +\alpha R^2).$$

(83)

We choose $\alpha =\alpha \left(\delta \right)=2e^{-\lambda \left(\delta \right)T^2}$, which implies that $\alpha R^2=2R^2{e}^{-\lambda \left(\delta \right)T^2}\le C\delta$ for sufficiently small $\delta$. Conversely, from (81) and (82), we have $\delta \le \exp \left[-{\displaystyle \frac{2}{2c}}{\left(\ln \left({\delta }^{-1}\right)\right)}^c\right]$, and it follows that

$$\delta +\alpha R^2\le C_3\exp \left[-{\displaystyle \frac{2}{2c}}{\left(\ln \left({\delta }^{-1}\right)\right)}^c\right].$$

(84)

By (83) and (84), we have

$$\parallel u^{*}-u_{\min }{\parallel }_{H^{1,0}\left(Q_{\tau }\right)}^2\le C_3\exp \left[-{\displaystyle \frac{2}{2^c}}{\left(\ln \left({\delta }^{-1}\right)\right)}^c\right].$$

Finally, by combining (25) with (23), we obtain

$$\begin{array}{cc}\hfill \parallel u^{*}-u_{n+1}{\parallel }_{H^{1,0}\left(Q_{\tau }\right)}& \le \parallel u^{*}-u_{\min }{\parallel }_{H^{1,0}\left(Q_{\tau }\right)}+{\parallel u_{\min }-u_{n+1}\parallel }_{H^{1,0}\left(Q_{\tau }\right)}\hfill \\ \hfill \hspace{1em}& \le C_3\exp \left[-{\displaystyle \frac{2}{2^c}}{\left(\ln \left({\delta }^{-1}\right)\right)}^c\right]+{\theta }^n{\parallel u_{\min }-u_0\parallel }_{H^k\left(Q\right)}.\hfill \end{array}$$

(85)

This completes the proof of Theorem 4. □

5. Conclusions

In this paper, we developed a convexification method for the reversed-time problem of the linear CGL. We established a new Carleman estimate for the complex Ginzburg–Landau operator $Gz=(1+ib)z_t+{\sum }_{j,k}{\left(a^{jk}{z}_j\right)}_k$ on an arbitrary time interval. Based on this estimate, we constructed a weighted Tikhonov-like functional $I_{\alpha ,\lambda }$ and proved its global strict convexity on the $B\left(R\right)$ of arbitrary radius $R>0$. We proved the existence and uniqueness of the minimizer of $I_{\alpha ,\lambda }$ on $\overline{B\left(R\right)}$ and established the global convergence of the gradient projection method to this minimizer. We also provided stability estimates in the presence of noisy terminal data. This is a purely theoretical work; numerical experiments are not included. Possible future work includes a numerical implementation of the gradient projection method and an extension to the nonlinear CGL.