Global Weak Solutions to an Initial-Boundary Value Problem for a Three-phase Field Model of Solidification

Abstract

In this article, we study an initial-boundary value problem for a three-phase field model of nonisothermal solidification processes in the case of two possible crystallization states. The governing equations of the model are the three phase-field equations coupled with a nonlinear heat equation. Each equation of the model has strong nonlinearities involving the higher-order derivatives. We prove the existence of global-in-time weak solutions to our problem for one-dimensional case.

1 Introduction

Phase-field method is a very powerful computational tool in describing the evolution of microstructure without tracking the interface position. In phase-field method, an order parameter or field variable is used to describe the physical state of various regions or components in a system. Several phase-field models have been successfully developed for material processes, such as solidification, precipitate growth and coarsening, solid-state phase transformations, and crack propagation, etc.

The first phase-field model for solidification of pure material was proposed by Langer [1]. After that, important contributions on development of the phase-field models were made by Caginalp [2], Penrose and Fife [3], Wang and Sekerka [4], Wheeler, Boettinger and McFadden [5], Karma and Rappel [6].

In recent years, the phase-field methodology has been extended to describe the evolution of more than two phases by adopting multiple field variables. The main concept of multiphase-field system lies on the work of Steinbach et al. [7, 8]. Later, several multiphase systems were developed by Nestler et al. [9, 10], Folch and Plapp [11], Kim et al. [12], and Bollada et al. [13].

In the standard multiphase-field system for $N$-phases, each phase-field $u_{i}\in [0,1]$ is defined as a local volume fraction of a phase $i$ and related by the constraint $\sum \limits _{i}^Nu_{i}=1$.

In this article, we study a multiphase-field model based on the Steinbach et al. in [8]. In order to get our multiphase-field model, we consider the following energy functional

$$F[u_{i} ,u_{j} ,\theta ] = \int\limits_{V} {\left( {\sum _{{i,j(i < j)}}^{N} \frac{{\varepsilon _{{ij}}^{2} }}{2}|u_{j} \nabla u_{i} - u_{i} \nabla u_{j} |^{2} + \frac{{a_{{ij}} }}{2}u_{i}^{2} u_{j}^{2} + \frac{1}{2}\lambda \theta ^{2} } \right)dV,}$$

(1.1)

where $\varepsilon _{ij}^2=\varepsilon _{ji}^2>0$ and $a_{ij}=a_{ji}>0$ are the gradient energy coefficient and energy barrier height of the phases $i$ and $j$, respectively; $\lambda >0$ is the coupling constant. The dimensionless temperature $\theta$ is related to the internal energy $e$ by

$$\begin{aligned} \qquad \theta =e+\sum \limits _{i}^N l_{i}h(u_i), \end{aligned}$$

where $l_i (i=1,\cdots , N)$ are related to the latent heats associated to each kind of physical state. The interpolation functions $h(u_i)=u_{i}^2(3-2u_i)$ satisfy $h(u_i)=1$ at $u_i=1$ and $h(u_i)=0$ at $u_i=0$.

The dynamics of the phases $u_{i}$ are derived by the minimization of the free energy functional $F$,

$$\begin{aligned} \frac{\partial u_i}{\partial t}=-\sum \limits _{j(j\ne i)}^N\frac{1}{\tau _{ij}}\frac{\delta F(u_i,u_j,\theta )}{\delta u_i},\quad i=1,\cdots ,N. \end{aligned}$$

(1.2)

Here, $\tau _{ij}=\tau _{ji}>0$ is a relaxation time at an interface between $i$ and $j$ phases, and we assumed that in the triple point the transition between the phases occurs by the movement of the dual phase boundaries, which do not influence each other [8].

For a system of three-phases, we consider $u_1,u_2$, and $u_3$ are three phase-field variables that represent the solid fractions of the two possible different kinds of crystallization states and the liquid state, respectively, such that $\sum \limits _{i=1}^{3}u_{i}=1$ and $\frac{\partial u_i}{\partial u_j}=-1$ for $i\ne j$.

Thus, we have the following PDE system from the dynamic equation (1.2) for three-phases coupling with an energy equation,

$$\begin{aligned} \left. \begin{aligned}& \frac{\partial u_1}{\partial t}-\frac{\epsilon _{13}^2}{\tau _{13}}(u_3\Delta u_1-u_1\Delta u_3)-\frac{\epsilon _{12}^2}{\tau _{12}}(u_2\Delta u_1-u_1\Delta u_2)=-\frac{a_{13}}{\tau _{13}}u_1u_3\\&\qquad\times (u_3-u_1)-\frac{a_{12}}{\tau _{12}}u_1u_2(u_2-u_1)-\frac{6\lambda }{\tau _{13}}\theta \big (l_1(u_1-u_1^2)-l_3(u_3-u_3^2)\big )\\&\qquad- \frac{6\lambda }{\tau _{12}}\theta \big (l_1(u_1-u_1^2)-l_2(u_2-u_2^2)\big ), \\ &\frac{\partial u_2}{\partial t}-\frac{\epsilon _{23}^2}{\tau _{23}}(u_3\Delta u_2-u_2\Delta u_3)-\frac{\epsilon _{21}^2}{\tau _{21}}(u_1\Delta u_2-u_2\Delta u_1)=-\frac{a_{23}}{\tau _{23}} u_2 u_3\\&\qquad\times (u_3-u_2)-\frac{a_{21}}{\tau _{21}} u_2u_1(u_1-u_2)-\frac{6\lambda }{\tau _{23}}\theta \big (l_2(u_2-u_2^2)\\&\qquad-l_3(u_3-u_3^2)\big ) -\frac{6\lambda }{\tau _{21}}\theta \big (l_2(u_2-u_2^2)-l_1(u_1-u_1^2)\big ),\\& \frac{\partial u_3}{\partial t}-\frac{\epsilon _{31}^2}{\tau _{31}}(u_1 \Delta u_3-u_3\Delta u_1)-\frac{\epsilon _{32}^2}{\tau _{32}}(u_2\Delta u_3-u_3\Delta u_2)=-\frac{a_{31}}{\tau _{31}} u_3 u_1\\&\qquad\times (u_1-u_3)-\frac{a_{32}}{\tau _{32}} u_3 u_2(u_2-u_3) -\frac{6\lambda }{\tau _{31}}\theta \big (l_3(u_3-u_3^2)\\&\qquad-l_1(u_1-u_1^2)\big ) -\frac{6\lambda }{\tau _{32}}\theta \big (l_3(u_3-u_3^2)-l_2(u_2-u_2^2)\big ),\\ &\frac{\partial \theta }{\partial t}-b\Delta \theta =l_{1}6(u_1-u_1^2)\frac{\partial u_1}{\partial t}+l_{2}6(u_2-u_2^2)\frac{\partial u_2}{\partial t}+l_{3}6(u_3-u_3^2)\frac{\partial u_3}{\partial t}+g. \end{aligned} \right\} \end{aligned}$$

(1.3)

Here, the function $g$ is related to the density of heat sources or sinks and the given constant $b>0$ stands for thermal conductivity. We consider the system (1.3) equipped with the following initial and Dirichlet’s boundary conditions

$$\begin{aligned} \left. \begin{aligned} u_1(t,x)=u_2(t,x)=u_3(t,x)=\theta (t,x)=0,\;\;\qquad (t,x)\in [0,T]\times {\partial \Omega },\\ u_1=u_{10}(x),u_2=u_{20}(x),u_3=u_{30}(x),\theta =\theta _{0}(x),\qquad t=0,\;\; x\in \Omega , \end{aligned} \right\} \end{aligned}$$

(1.4)

where $\Omega \subset {\mathbb {R}}^n$ is an open bounded domain and $0<T<\infty$, and the initial conditions of phase-field variables satisfy $u_{10}+u_{20}+u_{30}=1$.

Our aforementioned mathematical model (1.3)–(1.4) plays a vital role in describing the complex growth phenomena during solidification or melting of certain metallic alloys in which two different kinds of crystallization are possible [7,8,9].

In this article, we will study the existence of global weak solutions for the initial-boundary value problem (1.3)–(1.4) in one-dimensional domain. The problem has very strong nonlinearities involving the higher-order derivatives. The mathematical analysis of such a model is much more difficult than any single phase-field model.

Analytical results for various phase-field models have been studied in Caginalp et al. [14, 15], Colli et al. [16,17,18], Hoffman and Jiang [19], Boldrini et al. [20, 21], and Alber and Zhu [22, 23]. Boldrini et al. [24] proved the existence of local solutions to a three-phase field model for one-dimensional case. Recently, Tang and Gao [25] investigated global weak solutions to a three-phase field model of solidification where they assumed the variable co-efficients of highest order derivative terms in each equation as positive constants. To our knowledge, there are a few theoretical results available for multi-phase systems.

We first introduce some notations and then formulate the main result.

Let us consider that all functions depend only on the variables $x_1$ and $t$. To simplify the notation, we write $x_1$ by $x$. Let $\Omega =(a,b)$ be a bounded open interval with $a<b$ and $Q_{T}=(0,T)\times \Omega$ for $0<T<\infty$. We denote the usual Sobolev spaces for $1\le p \le +\infty$, $k\in {\mathbb {N}}$ by

$$\begin{aligned} W^{k,p}(\Omega )=\{f\in L^p(\Omega ): D^{\alpha }f\in L^p(\Omega ),\,\vert \alpha \vert \le k\} \end{aligned}$$

endowed with the norm

$$\begin{aligned} \Vert f\Vert _{W^{k,p}(\Omega )}= {\left\{ \begin{array}{ll} \left( \sum _{\vert \alpha \vert \le k}\int _{\Omega }\vert D^{\alpha }f\vert ^pdx\right) ^{1/p} &{} (1\le p<\infty ) \\ \sum _{\vert \alpha \vert \le k}\text {ess sup}_{\Omega }\vert D^{\alpha }f\vert \, &{} ( p=\infty ). \end{array}\right. } \end{aligned}$$

For $p=2$, the Hilbert space $W^{k,2}(\Omega )=H^k(\Omega )$ is defined by

$$\begin{aligned} (f,g)_{H^k(\Omega )}=\sum \limits _{\vert \alpha \vert \le k}\int _{\Omega }D^{\alpha }fD^{\alpha }gdx. \end{aligned}$$

Let $B$ be a Banach space and $1\le p<\infty$, we consider the functional spaces by

$$\begin{aligned}&L^p(0,T;B)=\{f:(0,T)\rightarrow B: f\; \text {measurable}, \int _{0}^{T}\Vert f(s)\Vert _{B}^{p}ds<+\infty \},\;\;\\& \text {and}\;\;L^{\infty }(0,T;B)=\{f:(0,T)\rightarrow B: f\; \text {measurable}, \text {ess sup}_{{s} \in (0,T)}\Vert f(s)\Vert _{B}<+\infty \}. \end{aligned}$$

We will frequently use the following result from the Sobolev embedding that for some constant $C>0$,

$$\begin{aligned} \Vert f\Vert _{L^{\infty }(\Omega )}\le C\Vert f\Vert _{H_{0}^{1}(\Omega )},\quad \text {for any}\quad f\in H_{0}^{1}(\Omega ). \end{aligned}$$

Throughout this article, we denote $C$ as a generic positive constant depending on $\Omega$ and known quantities which varies from line to line.

Next, we state the main result of our article.

Theorem 1.1

Suppose that $\vert \frac{\varepsilon _{12}^2}{\tau _{12}}-\frac{\varepsilon _{13}^2}{\tau _{13}}\vert$, $\vert \frac{\varepsilon _{12}^2}{\tau _{12}}-\frac{\varepsilon _{23}^2}{\tau _{23}}\vert$ are sufficiently small, and $g\in L^2(Q_T)$. Then, for any $(u_{10},u_{20},u_{30})\in [H_0^1(\Omega )]^3$ with $u_{10}+u_{20}+u_{30}=1$ a.e. in $\Omega$, and $\theta _0\in H_0^1(\Omega )$, the initial-boundary value problem (1.3)–(1.4) possesses a global weak solution $(u_1,u_2,u_3,\theta )\in [C([0,T];H_{0}^1(\Omega ))]^4$.

The remaining parts of this work are devoted to the proof of Theorem 1.1. In Section 2, we formulate an auxiliary problem (2.1) by replacing $u_3=1-u_1-u_2$ in problem (1.3)–(1.4) and find local weak solutions of that auxiliary problem by using fixed-point argument. In Section 3, we derive uniform a priori estimates to deduce that auxiliary problem has global-in-time weak solutions. Finally, we complete the proof of Theorem 1.1.

2 An Auxiliary Problem

First, we need to consider an auxiliary problem to obtain the existence of solutions to problem (1.3)–(1.4). Owing to the condition $u_{10}+u_{20}+u_{30}=1$ and $u_{1t}+u_{2t}+u_{3t}=0$ in $Q_{T}$, we have $u_1+u_2+u_3=1$ in $Q_{T}$. So, by replacing $u_3=1-u_1-u_2$ in (1.3)–(1.4), we get the following equivalent problem:

$$\begin{aligned} \left. \begin{aligned} \frac{\partial u_1}{\partial t}-k_{1} u_{1xx}-(k_{3}-k_{1})(u_2u_{1xx}-u_1u_{2xx})=-\alpha _{1}u_1(1-u_1-u_2)\\ \times \,(1-2u_1-u_2)-\alpha _{3}u_1 u_2(u_2-u_1)-\theta \big (\beta _1(u_1-u_1^2)\\ \qquad \quad - \, \beta _2(u_2-u_2^2)-2\beta _{3}u_1u_2\big )\qquad \qquad \qquad \qquad \text {in}\quad Q_{T},\\ \frac{\partial u_2}{\partial t}-k_{2}u_{2xx}-(k_{3}-k_{2})(u_1u_{2xx}-u_2u_{1xx})=-\alpha _{2}u_2(1-u_1-u_2)\\ \times \,(1-u_1-2u_2)-\alpha _{3}u_2 u_1(u_1-u_2)-\theta \big (\gamma _1 (u_2-u_2^2)\\ \quad \qquad - \,\gamma _2(u_1-u_1^2)-2\gamma _3u_2u_1\big )\quad \qquad \qquad \qquad \quad \text {in}\quad Q_{T},\\ \frac{\partial \theta }{\partial t}-b \theta _{xx}=\big (6l_1(u_1-u_1^2)-6l_3(u_1+u_2-(u_1+u_2)^2)\big )\frac{\partial u_1}{\partial t}\qquad \;\\ \quad\qquad +\big (6l_2(u_2-u_2^2)-6l_3(u_1+u_2-(u_1+u_2)^2)\big )\frac{\partial u_2}{\partial t}+g\quad \text {in}\;\; Q_{T}, \\ u_1(t,x)=u_2(t,x)=\theta (t,x)=0\qquad \qquad \qquad (t,x)\in [0,T]\times {\partial \Omega },\\ u_1(0,x)=u_{10}(x),u_2(0,x)=u_{20}(x),\theta (0,x)=\theta _{0}(x)\quad \qquad x\in \Omega , \end{aligned} \right\} \end{aligned}$$

(2.1)

where $k_1=\frac{\varepsilon _{13}^2}{\tau _{13}},k_2=\frac{\varepsilon _{23}^2}{\tau _{23}}, k_3=\frac{\varepsilon _{12}^2}{\tau _{12}},\alpha _{1}=\frac{a_{13}}{\tau _{13}},\alpha _{2}=\frac{a_{23}}{\tau _{23}},\alpha _{3}=\frac{a_{12}}{\tau _{12}}$,

$\beta _1=\frac{6\lambda l_1}{\tau _{13}}+\frac{6\lambda l_1}{\tau _{12}}-\frac{6\lambda l_3}{\tau _{13}},\beta _{2}=\frac{6\lambda l_2}{\tau _{12}}, \beta _{3}=\frac{6\lambda l_3}{\tau _{13}}, \gamma _1=\frac{6\lambda l_2}{\tau _{21}}+\frac{6\lambda l_2}{\tau _{23}}-\frac{6\lambda l_3}{\tau _{23}}$,

$\gamma _2=\frac{6\lambda l_1}{\tau _{21}}$, and $\gamma _3=\frac{6\lambda l_3}{\tau _{23}}$.

It is convenient to rewrite that

$$\begin{aligned}&-(k_{3}-k_{1})(u_2 u_{1xx}-u_1u_{2xx})=-(k_{3}-k_{1})(u_2u_{1x}-u_1u_{2x})_x,\\&-(k_{3}-k_{2})(u_1u_{2xx}-u_2u_{1xx})=-(k_{3}-k_{2})(u_1u_{2x}-u_2u_{1x})_x.\end{aligned}$$

Let us define a weak solution for the auxiliary problem (2.1) as follows.

Definition 2.1

Let $u_{10},u_{20}\in H_0^1(\Omega ),$ and $\theta _0\in H_0^1(\Omega )$. A function $(u_1,u_2,\theta )$ with

$$\begin{aligned}& u_1,u_2,\theta \in L^{\infty }(0,T; \; H_0^1(\Omega ))\cap L^{2}(0,T;\; H^2(\Omega )), \\ & \qquad \qquad u_{1t},u_{2t},\theta _t\in L^2(0,T;\;L^2(\Omega ))\end{aligned}$$

is a weak solution to the auxiliary problem (2.1) for all $\phi \in C_{0}^{\infty }\left( (-\infty ,T)\times \Omega \right)$ such that

$$\begin{aligned}&(u_1,\phi _{t})_{Q_{T}}-k_{1}( u_{1x}, \phi _x)_{Q_{T}} -(k_{3}-k_{1})\left( u_2 u_{1x}-u_1 u_{2x},\phi _x\right) _{Q_{T}}\;\\&\quad\;-\alpha _{1}\big (u_1(1-u_1-u_2)(1-2u_1-u_2),\phi \big )_{Q_{T}} -\alpha _{3}\big (u_1 u_2(u_2-u_1),\phi \big )_{Q_{T}}\\&\quad -\big (\theta (\beta _{1}(u_1-u_1^2)-\beta _{2}(u_2-u_2^2)-2\beta _{3}u_1u_2,\phi \big )_{Q_{T}}+(u_{10},\phi (0))_{\Omega }=0,\,\\&(u_2,\phi _{t})_{Q_{T}}-k_{2}(u_{2x},\phi _x)_{Q_{T}} -(k_{3}-k_{2})\left( u_1 u_{2x}-u_2 u_{1x},\phi _x\right) _{Q_{T}}\qquad \qquad \qquad \\&\quad \;-\alpha _{2}\big (u_2(1-u_1-u_2)(1-u_1-2u_2),\phi \big )_{Q_{T}} -\alpha _{3}\big (u_2u_1(u_1-u_2),\phi \big )_{Q_{T}}\\&\quad -\big (\theta (\gamma _{1}(u_2-u_2^2)-\gamma _{2}(u_1-u_1^2)-2\gamma _{3}u_2u_1,\phi \big )_{Q_{T}}+(u_{20},\phi (0))_{\Omega }=0,\\&(\theta ,\phi _{t})_{Q_{T}}-b( \theta _x, \phi _x)_{Q_{T}}+\big (6l_1(u_1-u_1^2)-6l_3(u_1+u_2-(u_1+u_2)^2)u_{1t},\phi \big )_{Q_{T}}\qquad \;\\&\quad \;+\big (6l_2(u_2-u_2^2)-6l_3(u_1+u_2-(u_1+u_2)^2)u_{2t},\phi \big )_{Q_{T}}\\&\quad -(g,\phi )_{Q_{T}} +\left( \theta _{0},\phi (0)\right) _{\Omega }=0.\qquad \qquad \qquad \end{aligned}$$

Next, we state the result on the existence of global weak solutions concerning with the auxiliary problem (2.1). Then, by defining $u_3=1-u_1-u_2$, we obtain a solution $(u_1,u_2,u_3,\theta )$ to the original problem (1.3)–(1.4).

Proposition 2.2

Assume that $(u_{10},u_{20},\theta _{0})\in [H_0^1(\Omega )]^3$ and $g\in L^2(Q_T)$. If $\vert k_3-k_1\vert$ and $\vert k_3-k_2\vert$ are small enough, then the auxiliary problem (2.1) admits at least one global weak solution $(u_1,u_2,\theta )$ in the sense of Definition 2.1.

The proof of Proposition 2.2 consists of a couple of steps. First, we linearize the problem (2.1) to obtain a unique local solution. Then, we use the method of continuation of local solutions to obtain global solutions after deriving some uniform-in-time a priori estimates.

2.1 Existence of Local Solutions

Now let us define a nonlinear operator:

$$\mathcal {F} : (\mu _1,\mu _2,\vartheta )\rightarrow (u_1,u_2,\theta),$$

where $(u_1,u_2,\theta )$ is the solution of the following auxiliary linear problem

$$\begin{aligned} \left. \begin{aligned}& u_{1t}-k_{1} u_{1xx}-(k_{3}-k_{1})(\mu _2 u_{1x}-\mu _1\nu _{2x})_x=-\alpha _{1}\mu _1(1-\mu _1-\mu _2)\\ &\qquad \qquad \times (1-2\mu _1-\mu _2)-\alpha _{3}\mu _1 \mu _2(\mu _2-\mu _1)-\vartheta \big (\beta _{1}(\mu _1-\mu _1^2)\\ & \qquad \qquad-\beta _{2}(\mu _2-\mu _2^2)-2\beta _{3}\mu _1\mu _1\big )\qquad \qquad \qquad \qquad \qquad\text {in}\quad Q_{t^\star },\\& u_{2t}-k_{2} u_{2xx}-(k_{3}-k_{2})(\mu _1 u_{2x}-\mu _2 \mu _{1x})_x=-\alpha _{2}\mu _2(1-\mu _1-\mu _2)\\&\qquad \qquad \times (1-\mu _1-2\mu _2)-\alpha _{3} \mu _2\mu _1(\mu _1-\mu _2)-\vartheta \big (\gamma _{1}(\mu _2-\mu _2^2)\\ &\qquad \qquad -\gamma _{2}(\mu _1-\mu _1^2)-2\gamma _{3}\mu _2\mu _1\big )\qquad \qquad \qquad \qquad \qquad \text {in}\quad Q_{t^\star },\\ &\theta _{t}-b \theta _{xx}=\big (6l_1(u_1-u_1^2)-6l_3(u_1+u_2-(u_1+u_2)^2)\big )u_{1t}\qquad \;\;\\& \;\;\qquad +\big (6l_2(u_2-u_2^2)-6l_3(u_1+u_2-(u_1+u_2)^2)\big )u_{2t}+g\quad \text {in}\;\; Q_{t^\star },\\& u_1(t,x)=u_2(t,x)=\theta (t,x)=0\qquad \qquad \qquad\;\; (t,x)\in [0,t^\star ]\times {\partial \Omega },\\& u_1(0,x)=u_{10}(x),u_2(0,x)=u_{20}(x),\theta (0,x)=\theta _{0}(x),\;\quad \qquad x\in \Omega . \end{aligned} \right\} \end{aligned}$$

(2.2)

Here $Q_{t^\star }=(0,t)\times \Omega$ with $0<t<t^{\star }$.

For this linearized initial-boundary value problem (2.2), we study the existence of local solutions by using the Banach fixed-point theorem [24].

Lemma 2.3

Assume that $(u_{10},u_{20},\theta _{0})\in [H_0^1(\Omega )]^3$ and $g\in L^2(Q_T)$. If $\vert k_3-k_1\vert ,\vert k_3-k_2\vert,$ and $t^{\star }>0$ are small enough, then there exists a unique local solution $(u_1,u_2,\theta )\in [C([0,t^\star ]; H_{0}^1(\Omega ))]^3$ to problem (2.2).

Proof

First, we consider the following Banach spaces for $i=1,2,3$:

$$\begin{aligned} E_i=\{f: f\in L^{\infty }(0,t^\star ;H_{0}^1(\Omega ))\cap L^{2}(0,t^\star ;H^2(\Omega )),f_{t}\in L^{2}(0,t^\star ;L^2(\Omega ))\}, \end{aligned}$$

with norm

$$\begin{aligned} \Vert f\Vert _{E_{i}}=\text {max}\{\Vert f\Vert _{L^{\infty }(0,t^\star ;H_{0}^1(\Omega ))}, \Vert f\Vert _{L^{2}(0,t^\star ;H^2(\Omega ))}, \Vert f_{t}\Vert _{L^{2}(0,t^\star ;L^2(\Omega ))}\}. \end{aligned}$$

Next, we will apply the Banach fixed-point theorem on the following closed ball for $M>0$:

$$\begin{aligned} E:=\{(u_1,u_2,\theta )\in E_1\times E_2\times E_3: \Vert u_1\Vert _{E_1}\le M,\Vert u_2\Vert _{E_2}\le M,\Vert \theta \Vert _{E_3}\le M\} \end{aligned}$$

with norm $\Vert (u_1,u_2,\theta )\Vert _E=\text {max}\{\Vert u_1\Vert _{E_1},\Vert u_2\Vert _{E_2},\Vert \theta \Vert _{E_3}\}$.

Let us consider $\varepsilon _0>0$ so small that $\vert k_3-k_1\vert <\varepsilon _{0}$ and $\vert k_3-k_2\vert <\varepsilon _0$ then

$$\begin{aligned} 2\vert k_3-k_1\vert CM<k_1 \qquad \text {and} \qquad 2\vert k_3-k_2\vert CM<k_2. \end{aligned}$$

Observe that, if $\mu _1,\mu _2\in L^{\infty }(0,t^\star ;H_0^1(\Omega ))$, then owing to the Sobolev embedding $H_0^1(\Omega )\hookrightarrow L^\infty (\Omega )$, we have

$$\begin{aligned}&k_1+(k_3-k_1)\mu _2\ge k_1-\vert k_3-k_1\vert \Vert \mu _2\Vert _{L^{\infty }(Q_{t^\star })} \ge k_1-\vert k_3-k_1\vert CM \ge \frac{k_1}{2}>0, \end{aligned}$$

(2.3)

$$\begin{aligned}&k_2+(k_3-k_2)\mu _1\ge k_1-\vert k_3-k_1\vert \Vert \mu _1\Vert _{L^{\infty }(Q_{t^\star })}\ge k_3-\vert k_3-k_2\vert CM \ge \frac{k_2}{2}>0. \end{aligned}$$

(2.4)

Later, the inequalities (2.3) and (2.4) will be found very useful to deal with some divergence related terms to the left-hand side of first and second equations of our auxiliary linear problem (2.2).

1. Since $(\mu _1,\mu _2,\vartheta )\in E$, then owing to the Sobolev embedding $H_0^1(\Omega )\hookrightarrow L^\infty (\Omega )$, the right-hand side of each equation of (2.2) belongs to $L^{2}(Q_{t^\star })$. Thus for any suitable choice of $M$, one can apply the theory of linear parabolic equation (see e.g. Evans [26]) such that $(u_1,u_2,\theta ) \in E$, i.e., the operator $\mathcal {F}$ maps from $E$ to $E$.

2. Next, we need to prove that $\mathcal {F}$ is a contraction, i.e.,

$$\begin{aligned} \Vert \mathcal {F}(\mu _1,\mu _2,\vartheta ) - \mathcal {F}(\mu _1^{\prime },\mu _2^{\prime },\vartheta ^{\prime })\Vert _E\le \lambda \Vert (\mu _1,\mu _2,\vartheta )-(\mu _1^{\prime },\mu _2^{\prime },\vartheta ^{\prime })\Vert _E \end{aligned}$$

for some $0 \le \lambda <1$ and all $(\mu _1,\mu _2,\vartheta ),(\mu _1^{\prime },\mu _2^{\prime },\vartheta ^{\prime })\in E$.

Now, for any $(\mu _1,\mu _2,\vartheta ),(\mu _1^{\prime },\mu _2^{\prime },\vartheta ^{\prime })\in E$, we have by considering $({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }}) =(\mu _1,\mu _2,\vartheta )-(\mu _1^{\prime },\mu _2^{\prime },\vartheta ^{\prime })$ and $({\tilde{u}}_1,{\tilde{u}}_2,\tilde{\theta })=(u_1,u_2,\theta )-(u_1^{\prime },u_2^{\prime },\theta ^{\prime })$ such that $({\tilde{u}}_1,{\tilde{u}}_2,\tilde{\theta })=$ $\mathcal {F}$ $({\tilde{\mu }}_1,{\tilde{\mu }}_2, \tilde{\vartheta })$ satisfies

$$\begin{aligned} \left. \begin{aligned}& {\tilde{u}}_{1t}-k_{1}{\tilde{u}}_{1xx}-(k_{3}-k_{1})(\mu _2 {\tilde{u}}_{1x})_x=-(k_{3}-k_{1})({\tilde{\mu }}_2 u_{1x}^{\prime })_x\qquad \qquad \\ &\qquad -(k_{3}-k_{1})({\tilde{\mu }}_1 \mu _{2x}+\mu _1^{\prime }{\tilde{\mu }}_{2x})_x+A_1{\tilde{\mu }}_1+A_2{\tilde{\mu }}_2+A_3 {\tilde{\vartheta }}\; \;\;\text {in}\;\; Q_{t^\star },\\& {\tilde{u}}_{2t}-k_{2}{\tilde{u}}_{2xx}-(k_{3}-k_{2})(\mu _1 {\tilde{u}}_{2x})_x=-(k_{3}-k_{2})({\tilde{\mu }}_1 u_{2x}^{\prime })_x\qquad \qquad \\& \qquad -(k_{3}-k_{2})({\tilde{\mu }}_2 \mu _{1x}+\mu _2^{\prime }{\tilde{\mu }}_{1x})_x+B_1{\tilde{\mu }}_1+B_2{\tilde{\mu }}_2+B_3{\tilde{\vartheta }}\;\; \text {in}\;\; Q_{t^\star },\\& {{\tilde{\theta }}}_{t}-b{\tilde{\theta }}_{xx}=D_1{\tilde{u}}_{1t}+D_2{\tilde{u}}_{2t}+(D_3u_{1t}^{\prime }+D_4u_{2t}^{\prime }){\tilde{u}}_1\qquad \qquad \qquad \quad \\ &\qquad +(D_5u_{1t}^{\prime }+D_6u_{2t}^{\prime }){\tilde{u}}_2,\qquad \qquad\qquad \qquad \qquad \qquad \quad \text {in}\quad Q_{t^\star },\\& {\tilde{u}}_1(t,x)={\tilde{u}}_2(t,x)= {\tilde{\theta }}(t,x)=0,\qquad \qquad \quad\; (t,x)\in [0,t^\star ]\times {\partial \Omega },\\ &{\tilde{u}}_1(0,x)=0,{\tilde{u}}_2(0,x)=0,\tilde{\theta }(0,x)=0,\qquad \qquad \qquad\qquad \quad x\in \Omega , \end{aligned} \right\} \end{aligned}$$

(2.5)

where

$$\begin{aligned}&A_1=-\alpha _{3}(\mu _{2}^2-(\mu _1+\mu _{1}^{\prime })\mu _{2})-\alpha _{1}(1-3(\mu _1+\mu _{1}^{\prime })+2(\mu _1^{2}+\mu _1\mu _{1}^{\prime }+(\mu _{1}^{\prime })^{2})-2\mu _{2}^{\prime }\\&\qquad +3(\mu _1+\mu _{1}^{\prime })\mu _{2}^{\prime }+\mu _{2}^{2})-\beta _{1}(\vartheta ^{\prime }-\vartheta (\mu _1+\mu _{1}^{\prime }))+2\beta _{3}\mu _{2}\vartheta ^{\prime },\\&A_2=-\alpha _{3}(\mu _{1}^{\prime }(\mu _2+\mu _2^{\prime })-(\mu _{1}^{\prime })^2)-\alpha _{1}(-2\mu _1+3\mu _{1}^2+\mu _{1}^{\prime }(\mu _2+\mu _2^{\prime })-\beta _{2}(\vartheta (\mu _2+\mu _2^{\prime })-\vartheta ^{\prime })\\&\qquad +2\beta _{3}\mu _1^{\prime }\vartheta ^{\prime },\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\&A_3=-\beta _{1}(\mu _1-(\mu _{1}^{\prime })^2)+\beta _{2}(\mu _{2}-(\mu _{2}^{\prime })^2)+2\beta _{3}\mu _{1}\mu _2,\qquad \qquad \qquad \qquad \qquad \qquad \\&B_1=-\alpha _{3}(\mu _2^{\prime }(\mu _1+\mu _{1}^{\prime })-(\mu _2^{\prime })^2)-\alpha _{2}(-2\mu _2+3\mu _{2}^2+\mu _2^{\prime }(\mu _1+\mu _1^{\prime })-\beta _{2}(\vartheta (\mu _1+\mu _1^{\prime })-\vartheta ^{\prime })\\&\qquad +2\gamma _{3}\mu _2^{\prime }\vartheta ^{\prime },\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\&B_2=-\alpha _{3}(\mu _{1}^2-(\mu _2+\mu _2^{\prime })\mu _1)-\alpha _{2}(1-3(\mu _2+\mu _2^{\prime })+2(\mu _2^2+\mu _2\mu _2^{\prime }+(\mu _2^{\prime })^2)-2\mu _1^{\prime }\\&\qquad +3(\mu _2+\mu _2^{\prime })\mu _1^{\prime }+\mu _{1}^2)-\gamma _{1}(\vartheta ^{\prime }-\vartheta (\mu _2+\mu _2^{\prime }))+2\gamma _{3}\mu _1\vartheta ^{\prime },\\&B_3=-\gamma _{1}(\mu _2-(\mu _{2}^{\prime })^2)+\gamma _{2}(\mu _{1}-(\mu _{1}^{\prime })^2)-2\gamma _{3}\mu _{2}\mu _1,\qquad \qquad \qquad \qquad \qquad \qquad \quad \\&D_1=6l_1(u_1-u_1^2)-6l_3(u_1+u_2)(1-u_1-u_2),\qquad \qquad \qquad \qquad \qquad \qquad \qquad \\&D_2=6l_2(u_2-u_2^2)-6l_3(u_1+u_2)(1-u_1-u_2),\qquad \qquad \qquad \qquad \qquad \qquad \qquad \\&D_3=6l_1(1-u_1-u_1^{\prime })+6l_3(1-u_1-u_1^{\prime }-2u_2),\qquad \qquad \qquad \qquad \qquad \qquad \quad \;\;\\&D_4=6l_3(1-u_1-u_1^{\prime }-2u_2),\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \;\;\;\\&D_5=6l_3(1-v_1-v_2-2u_1^{\prime }),\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\&D_6=6l_2(1-u_2-u_2^{\prime }-2u_1^{\prime })+6l_3(1-u_2-u_2^{\prime }-2u_1^{\prime })\big ).\qquad \qquad \qquad \qquad \qquad \end{aligned}$$

Since $(\mu _1,\mu _2, \vartheta ),(\mu _1^{\prime },\mu _2^{\prime }, \vartheta ^{\prime }) \in {L^{\infty }(0,t^\star ;H_0^1(\Omega ))}$, then by $H_0^1(\Omega )\hookrightarrow L^\infty (\Omega )$, we have $A_i,B_i\in L^{\infty }(Q_{t^\star })$, for $i=1,2,3$ and $D_{i} \in L^{\infty }(Q_{t^\star })$ for $i=1,\cdots ,6$.

Now, by multiplying the first equation of (2.5) by $\tilde{u}_1$, integrating in $\Omega \times (0,t)$ with $0< t < t^\star$, using Hölder’s inequality, Sobolev embedding, and Young’s inequality, we obtain

$$\begin{aligned}&\frac{1}{2}\Vert \tilde{u}_{1}(t)\Vert _{L^2(\Omega )}^2+\int _{0}^{t}\int _{\Omega}\big (k_1+(k_3-k_1)\mu _{2}\big )\vert {\tilde{u}}_{1x}\vert ^2d(x,s) \nonumber \\&\quad \le \frac{1}{2}\Vert \tilde{u}_{10}\Vert _{L^2(\Omega )}^2+C_{\delta }\Big ( \Vert {\tilde{\mu }}_2\Vert _{L^{2}\left( 0,t^\star ;H_0^1(\Omega )\right) }^2+\Vert {\tilde{\mu }}_1\Vert _{L^{2}\left( 0,t^\star ;H_0^1(\Omega )\right) }^2\Big )\nonumber \\&\quad +C_{\varepsilon }\Big (\int _{0}^{t}\Vert {\tilde{\mu }}_1(s)\Vert _{L^2(\Omega )}^2ds +\int _{0}^{t}\Vert {\tilde{\mu }}_2(s)\Vert _{L^2(\Omega )}^2ds +\int _{0}^{t}\Vert {{\tilde{\vartheta }}}(s)\Vert _{L^2(\Omega )}^2ds \Big )\nonumber \\&\quad +\frac{\delta }{8}\int _{0}^{t}\Vert \tilde{u}_{1x}(s)\Vert _{L^2(\Omega )}^2ds +\frac{\varepsilon }{6}\int _{0}^{t}\Vert \tilde{u}_1(s)\Vert _{L^2(\Omega )}^2ds .\qquad \qquad \qquad \qquad \quad \;\; \end{aligned}$$

(2.6)

By using inequality (2.3), we have $k_1+(k_3-k_1)\mu _{2}\ge k_{1}/2$ for the second term on the left-hand side of inequality (2.6). Therefore, by choosing any appropriate constant $\delta >0$, using the definition of norm in $E$, and applying Gronwall’s lemma for all $0\le t\le t^\star$, we get

$$\begin{aligned}&\Vert {\tilde{u}}_1\Vert _{L^{\infty }(0,t^\star ;L^2(\Omega ))}\le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E, \end{aligned}$$

(2.7)

$$\begin{aligned}&\text {and}\quad \Vert {\tilde{u}}_1\Vert _{L^{2}(0,t^\star ;H_0^1(\Omega ))}\le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E. \end{aligned}$$

(2.8)

By multiplying the first equation of (2.5) by ${\tilde{u}}_{1t}$, integrating in $\Omega \times (0,t)$ with $0< t < t^\star$, using Hölder’s inequality, Sobolev embedding, Young’s inequality, and proceeding in a similar way, we deduce that

$$\begin{aligned} \int _{0}^{t}\Vert {\tilde{u}}_{1t}(s)\Vert _{L^2(\Omega )}^2ds+\frac{k_{1}}{2}\Vert \tilde{u}_{1x}(t)\Vert _{L^{2}(\Omega )}^2 \le Ct^\star \Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E^2. \end{aligned}$$

(2.9)

The above inequality implies that

$$\begin{aligned} \Vert {{\tilde{u}}}_{1t}\Vert _{L^{2}\left( 0,t^\star ;L^2(\Omega )\right) }\le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E. \end{aligned}$$

(2.10)

Observe that the first equation of problem (2.5) can be written as

$$\begin{aligned}&{{\tilde{u}}}_{1t}-\big (k_{1}+(k_{3}-k_{1})\mu _{2}\big ){\tilde{u}}_{1xx}=(k_{3}-k_{1}){\mu }_{2x}{\tilde{u}}_{1x}\nonumber \\&\quad -(k_{3}-k_{1})\big ({\tilde{\mu }}_2 u_{1xx}^{\prime }+{\tilde{\mu }}_{2x}u_{1x}^{\prime }+\tilde{\mu }_1\mu _{2xx} +{\tilde{\mu }}_{1x}\mu _{2x}\nonumber \\&\quad +\mu _{1}^{\prime } {\tilde{\nu }}_{xx} +\mu _{1x}^{\prime }{\tilde{\mu }}_{2x}\big )+A_1{\tilde{\mu }}_1+A_2{\tilde{\mu }}_2+A_3{\tilde{\vartheta }}.\qquad \qquad \end{aligned}$$

(2.11)

Next, by multiplying the equation (2.11) by $-\tilde{u}_{1xx}$, integrating in $\Omega \times (0,t)$ with $0< t <t^\star$, using Hölder’s inequality, Sobolev embedding, and Young’s inequality, we have

$$\begin{aligned}&\frac{1}{2}\Vert \tilde{u}_{1x}(t)\Vert _{L^2(\Omega )}^2+\int _{0}^{t}\int _{\Omega }\big (k_{1}+(k_{3}-k_{1})\mu _{2}\big )\vert \tilde{u}_{1xx}\vert ^2d(x,s) \nonumber \\ &\quad \le \frac{1}{2}\Vert \tilde{u}_{10x}\Vert _{L^2(\Omega )}^2+C_{\delta }\big (\Vert \mu _2\Vert _{L^{2}\left( 0,t^\star ;H^2(\Omega )\right) }^2\Vert {\tilde{u}}_1\Vert _{L^{\infty }\left( 0,t^\star ;H_0^1(\Omega )\right) }^2\nonumber \\ &\qquad +\Vert {\tilde{\mu }}_2\Vert _{L^{\infty }\left( 0,t^\star ;H_0^1(\Omega )\right) }^2\Vert u_1^{\prime }\Vert _{L^{2}\left( 0,t^\star ;H^2(\Omega )\right) }^2+\Vert {\tilde{\mu }}_2\Vert _{L^{2}\left( 0,t^\star ;H^2(\Omega )\right) }^2\nonumber \\ &\qquad \times \Vert u_{1}^{\prime }\Vert _{L^{\infty }\left( 0,t^\star ;H_0^1(\Omega )\right) }^2 +\Vert {\tilde{\mu }}_1\Vert _{L^{\infty }\left( 0,t^\star ;H_0^1(\Omega )\right) }^2\Vert \mu _2\Vert _{L^{2}\left( 0,t^\star ;H^2(\Omega )\right) }^2\nonumber \\ &\qquad +\Vert \mu _1^{\prime }\Vert _{L^{\infty }\left( 0,t^\star ;H_0^1(\Omega )\right) }^2 \Vert {\tilde{\mu }}_2\Vert _{L^{2}\left( 0,t^\star ;H^2(\Omega )\right) }^2\big )+\frac{\delta }{12}\int _{0}^{t}\Vert {{\tilde{u}}}_{1xx}(s)\Vert _{L^2(\Omega )}^2ds \nonumber \\ &\qquad +C_{\varepsilon }\big (\int _{0}^{t}\Vert {\tilde{\mu }}_1(s)\Vert _{L^2(\Omega )}^2ds+\int _{0}^{t}\Vert {\tilde{\mu }}_2(s)\Vert _{L^2(\Omega )}^2ds +\int _{0}^{t}\Vert {{\tilde{\vartheta(s) }}}\Vert _{L^2(\Omega )}^2ds \big )\nonumber \\ &\qquad +\frac{\varepsilon }{6}\int _{0}^{t}\Vert \tilde{u}_{1xx}(s)\Vert _{L^2(\Omega )}^2ds .\end{aligned}$$

(2.12)

Similarly, by using the inequality (2.3), we have $k_1+(k_3-k_1)\mu _{2}\ge k_{1}/2$ for the second term on the left-hand side of inequality (2.12). Therefore, by choosing any appropriate constants $\delta , \varepsilon >0$, using the definition of norm in $E$, utilizing (2.8), and employing standard elliptic estimate, we deduce that

$$\begin{aligned}&\Vert {\tilde{u}}_1\Vert _{L^{\infty }(0,t^\star ;H_0^1(\Omega ))}\le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E,\\& \text {and}\quad \Vert {\tilde{u}}_1\Vert _{L^{2}(0,t^\star ;H^2(\Omega ))}\le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E. \end{aligned}$$

Thus, we conclude that

$$\begin{aligned} \Vert {\tilde{u}}_1\Vert _{E_1} \le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E. \end{aligned}$$

(2.13)

Proceeding in the same way with the second equation of problem (2.5) and utilizing inequality (2.4), we can also derive that

$$\begin{aligned}&\Vert \tilde{u}_2\Vert _{L^{\infty }\left( 0,t^\star ;H_0^1(\Omega )\right) }\le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E,\\&\Vert {\tilde{u}}_2\Vert _{L^2\left( 0,t^\star ;H^2(\Omega )\right) }\le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E,\\&\text {and}\quad \Vert {\tilde{u}}_{2t}\Vert _{L^2(0,t^\star ;L^2(\Omega ))}\le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E. \end{aligned}$$

Thus,

$$\begin{aligned} \Vert {\tilde{u}}_2\Vert _{E_2} \le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E. \end{aligned}$$

(2.14)

Next, we need to find the necessary estimates for ${\tilde{\theta }}$. So, we will deal with the third equation of problem (2.5).

Therefore, by multiplying the third equation of problem (2.5) by ${{\tilde{\theta }}}$, integrating in $\Omega \times (0,t)$ with $0< t<t^\star$, and using Hölder’s and Young’s inequalities, we get

$$\begin{aligned}&\frac{1}{2}\Vert {{\tilde{\theta }}}(t)\Vert _{L^2(\Omega )}^2+b\int _{0}^{t}\Vert {\tilde{\theta }}_x(s)\Vert _{L^2(\Omega )}^2ds \nonumber \\&\quad \le \frac{1}{2}\Vert {\tilde{\theta }}_{0}\Vert _{L^2(\Omega )}^2+C_\delta \big (\Vert D_1\Vert _{L^\infty (Q_{t^\star })}^2\Vert {\tilde{u}}_{1t}\Vert _{L^2(Q_{t^\star })}^2+\Vert D_2\Vert _{L^{\infty }(Q_{t^\star })}^2\Vert {\tilde{u}}_{2t}\Vert _{L^2(Q_{t^\star })}^2\big )\nonumber \\&\quad +C_\varepsilon \big (\Vert D_3{\tilde{u}}_1\Vert _{L^{\infty }(Q_{t^\star })}^2\Vert u_{1t}^{\prime }\Vert _{L^2(Q_{t^\star })}^2+\Vert D_4{\tilde{u}}_1\Vert _{L^{\infty }(Q_{t^\star })}^2\Vert u_{2t}^{\prime }\Vert _{L^2(Q_{t^\star })}^2\nonumber \\&\quad +\Vert D_5{\tilde{u}}_2\Vert _{L^{\infty }(Q_{t^\star })}^2\Vert u_{1t}^{\prime }\Vert _{L^2(Q_{t^\star })}^2+\Vert D_6{\tilde{u}}_2\Vert _{L^{\infty }(Q_{t^\star })}^2\Vert u_{2t}^{\prime }\Vert _{L^2(Q_{t^\star })}^2\big )\nonumber \\&\quad +\frac{\delta }{4}\int _{0}^{t}\Vert {{\tilde{\theta }}}(s )\Vert _{L^2(\Omega )}^2ds +\frac{\varepsilon }{8}\int _{0}^{t}\Vert {{\tilde{\theta }}}(s )\Vert _{L^2(\Omega )}^2ds .\qquad \qquad \qquad \qquad \qquad \end{aligned}$$

(2.15)

So by choosing appropriate constants $\delta , \varepsilon >0$, using the definition of norm in $E$, utilizing (2.13)-(2.14), and applying Gronwall’s lemma, we deduce that

$$\begin{aligned} \Vert {{\tilde{\theta }}}\Vert _{L^{\infty }\left( 0,t^\star ;L^2(\Omega )\right) } \le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E\quad a.e.,\quad 0\le t\le t^\star . \end{aligned}$$

Similarly, by multiplying the third equation of problem (2.5) successively by $-{{\tilde{\theta }}}_{xx}$ and ${\tilde{\theta }}_t$, integrating in $\Omega \times (0,t)$ with $0< t<t^\star$, and proceeding as like the last one, we conclude that

$$\begin{aligned}&\Vert {\tilde{\theta }}\Vert _{L^{\infty }\left( 0,t^\star ;H_0^1(\Omega )\right) }\le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E,\\& \Vert {\tilde{\theta }}\Vert _{L^2\left( 0,t^\star ;H^2(\Omega )\right) }\le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E,\\& \text {and}\quad \Vert {\tilde{\theta }}_t\Vert _{L^2(0,t^\star ;L^2(\Omega ))}\le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E.\qquad \qquad \qquad \end{aligned}$$

Therefore,

$$\begin{aligned} \Vert {\tilde{\theta }}\Vert _{E_3} \le (Ct^\star )^{1/2}\Vert ({\tilde{\mu }}_1,{\tilde{\mu }}_2,{\tilde{\vartheta }})\Vert _E. \end{aligned}$$

(2.16)

Thus from inequalities (2.13), (2.14), and (2.16), we have that $\mathcal {F}$ $:E\rightarrow E$ is a contraction provided $t^{\star }>0$ so small that $(Ct^\star )^{1/2}=\lambda <1$. Thus, Lemma 3.1 is proved. $\square$

3 Uniform a Priori Estimates

In this section, we establish a priori estimates for solutions $(u_1,u_2,\theta )$ of the problem (2.1) for arbitrary $T>0$.

Lemma 3.1

There exits a constant $C$ independent of $t^{\star }$ such that, for any $T>0,$

$$\begin{aligned}&\int _{\Omega }\Big (\sum \limits _{i,j(i<j)}^3\frac{\varepsilon _{ij}^2}{2}\vert u_j u_{ix}-u_i u_{jx}\vert ^2+\frac{a_{ij}}{2}u_{i}^2u_{j}^2+\frac{1}{2}\lambda \theta ^2\Big )dx\big (t\big ) +\int _{Q_T}\Big (\sum \limits _{j(j\ne 1)}^3\frac{1}{\tau _{1j}}\\&\quad \times \Big (\frac{\delta F(u_1,u_j,\theta )}{\delta u_1}\Big )^2 +\sum \limits _{j(j\ne 2)}^3\frac{1}{\tau _{2j}}\Big (\frac{\delta F(u_2,u_j,\theta )}{\delta u_2}\Big )^2 +\sum \limits _{j(j\ne 3)}^3\frac{1}{\tau _{3j}}\Big (\frac{\delta F(u_3,u_j,\theta )}{\delta u_3}\Big )^2\\&\quad +\sum \limits _{i(i\ne j,k)}^3\Big (\frac{1}{\tau _{ij}}+\frac{1}{\tau _{ik}}\Big )\frac{\delta F(u_i,u_j,\theta )}{\delta u_i}\frac{\delta F(u_i,u_k,\theta )}{\delta u_i}+\lambda b\vert \theta _x\vert ^2\Big )d(x,s)\le C. \end{aligned}$$

Proof

By differentiating the free energy functional $F(u_i,u_j,\theta )$ in (1.1) with respect to $t$, we obtain

$$\begin{aligned} \frac{dF}{dt}=\int _{\Omega }\left( \frac{\delta F}{\delta u_1}\frac{\partial u_1}{\partial t}+\frac{\delta F}{\delta u_2}\frac{\partial u_2}{\partial t}+\frac{\delta F}{\delta u_3}\frac{\partial u_3}{\partial t}+\frac{\delta F}{\delta e}\frac{\partial e}{\partial t}\right) dx. \end{aligned}$$

Considering the dual phase interactions at the interfaces and applying the dynamic equation (1.2), it follows that

$$\begin{aligned}&\frac{dF}{dt}=-\int _{\Omega }\Big (\Big (\frac{\delta F(u_1,u_3,\theta )}{\delta u_1}+\frac{\delta F(u_1,u_2,\theta )}{\delta u_1}\Big )\sum \limits _{j(j\ne 1)}^3\frac{1}{\tau _{1j}}\frac{\delta F(u_1,u_j,\theta )}{\delta u_1}\nonumber \\&\quad +\Big (\frac{\delta F(u_2,u_3,\theta )}{\delta u_2}+\frac{\delta F(u_2,u_1,\theta )}{\delta u_2}\Big )\sum \limits _{j(j\ne 2)}^3\frac{1}{\tau _{2j}}\frac{\delta F(u_2,u_j,\theta )}{\delta u_2}\qquad \nonumber \\&\quad +\Big (\frac{\delta F(u_3,u_1,\theta )}{\delta u_3}+\frac{\delta F(u_3,u_2,\theta )}{\delta u_3}\Big )\sum \limits _{j(j\ne 3)}^3\frac{1}{\tau _{3j}}\frac{\delta F(u_3,u_j,\theta )}{\delta u_3}+\lambda b\vert \theta _x\vert ^2\Big )dx. \end{aligned}$$

(3.1)

By integrating (3.1) in time $t\in (0,T)$, we obtain that

$$\begin{aligned}&F(t)+\int _{Q_T}\Big (\sum \limits _{j(j\ne 1)}^3\frac{1}{\tau _{1j}}\Big (\frac{\delta F(u_1,u_j,\theta )}{\delta u_1}\Big )^2+\sum \limits _{j(j\ne 2)}^3\frac{1}{\tau _{2j}}\Big (\frac{\delta F(u_2,u_j,\theta )}{\delta u_2}\Big )^2\nonumber \\&\quad +\sum \limits _{j(j\ne 3)}^3\frac{1}{\tau _{3j}}\Big (\frac{\delta F(u_3,u_j,\theta )}{\delta u_3}\Big )^2+\sum \limits _{i(i\ne j,k)}^3\Big (\frac{1}{\tau _{ij}}+\frac{1}{\tau _{ik}}\Big )\frac{\delta F(u_i,u_j,\theta )}{\delta u_i}\frac{\delta F(u_i,u_k,\theta )}{\delta u_i}\nonumber \\&\quad +\lambda b\vert \theta _x\vert ^2\Big )d(x,s)= F(0).\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \end{aligned}$$

(3.2)

Thus, by using the definition of $F(u_i,u_j,\theta )$ and initial conditions (1.4)$_2$ into Eq. (3.2), we can prove Lemma 3.1. $\square$

Lemma 3.2

There exits a constant $C$ independent of $t^{\star }$ such that, for any $T>0,$

$$\begin{aligned}&\Vert (u_1,u_2)\Vert _{L^\infty (0,T;H_0^1(\Omega ))\cap L^2\left( 0,T;H^2(\Omega )\right) }\le C;\qquad \qquad \qquad \qquad \qquad \end{aligned}$$

(3.3)

$$\begin{aligned}&\Vert \theta \Vert _{L^\infty \left( 0,T;L^2(\Omega )\right) \cap L^2(0,T;H_0^1(\Omega )) }\le C;\qquad \qquad \qquad \qquad \qquad \qquad \quad \end{aligned}$$

(3.4)

Proof

From Lemma 3.1, we observe that $a.e.$ $t>0,$

$$\begin{aligned}&\int _{\Omega }\Big (\frac{\varepsilon _{12}^2}{2}\vert u_2 u_{1x}-u_1 u_{2x}\vert ^2+\frac{\varepsilon _{13}^2}{2}\vert u_3 u_{1x}-u_1 u_{3x}\vert ^2+\frac{\varepsilon _{23}^2}{2}\vert u_3 u_{2x}-u_2 u_{3x}\vert ^2\nonumber \\&\quad +\frac{a_{12}}{2}u_1^2u_2^2+\frac{a_{13}}{2}u_1^2u_3^2+\frac{a_{23}}{2}u_2^2u_3^2+\frac{1}{2}\lambda \theta ^2\Big )dx\big (t\big )<C. \end{aligned}$$

(3.5)

Thus, we deduce from (3.5) that

$$\begin{aligned}&\Vert u_2 u_{1x}-u_1u_{2x}\Vert _{L^\infty \left( 0,T;L^2(\Omega )\right) }\le C, \end{aligned}$$

(3.6)

$$\begin{aligned}&\Vert u_3 u_{1x}-u_1 u_{3x}\Vert _{L^\infty \left( 0,T;L^2(\Omega )\right) }\le C, \end{aligned}$$

(3.7)

and

$$\begin{aligned} \Vert u_3 u_{2x}-u_2 u_{3x}\Vert _{L^\infty \left( 0,T;L^2(\Omega )\right) }\le C. \end{aligned}$$

(3.8)

Then, by inserting $u_3=1-u_1-u_2$ in estimates (3.7)-(3.8), one yields

$$\begin{aligned} \qquad \qquad \qquad \left. \begin{aligned} \Vert u_{1x}+(u_1u_{2x}-u_2u_{1x})\Vert _{L^\infty \left( 0,T;L^2(\Omega )\right) }\le C,\\ \Vert u_{2x}+(u_2 u_{1x}-u_1 u_{2x})\Vert _{L^\infty \left( 0,T;L^2(\Omega )\right) }\le C. \end{aligned} \right\} \qquad \qquad \quad \; \end{aligned}$$

(3.9)

Therefore, we apply Minkowski’s inequality for (3.6) and (3.9), and then use Poincaré’s inequality to obtain

$$\begin{aligned} \Vert (u_1,u_2)\Vert _{L^\infty (0,T;H_0^1(\Omega ))}\le C. \end{aligned}$$

(3.10)

From inequality (3.5), we also infer that

$$\begin{aligned} \Vert \theta \Vert _{L^\infty (0,T;L^2(\Omega ))}\le C. \end{aligned}$$

(3.11)

Let us recall Lemma 3.1, which also implies that

$$\begin{aligned}&\int _{Q_T}\Big (\sum \limits _{j(j\ne 1)}^3\frac{1}{\tau _{1j}}\Big (\frac{\delta F(u_1,u_j,\theta )}{\delta u_1}\Big )^2+\sum \limits _{j(j\ne 2)}^3\frac{1}{\tau _{2j}}\Big (\frac{\delta F(u_2,u_j,\theta )}{\delta u_2}\Big )^2\nonumber \\&\quad +\sum \limits _{j(j\ne 3)}^3\frac{1}{\tau _{3j}}\Big (\frac{\delta F(u_3,u_j,\theta )}{\delta u_3}\Big )^2+\lambda b\vert \theta _x\vert ^2\Big )d(x,s)\le C.\qquad \end{aligned}$$

(3.12)

It is obvious that this inequality provides $\theta _x$ is bounded in $L^2(Q_T)$. Thus, by using Poincaré’s inequality, we have

$$\begin{aligned} \Vert \theta \Vert _{L^2(0,T;H_0^1(\Omega ))}\le C. \end{aligned}$$

(3.13)

Hence, by combining (3.11) together with (3.13), we arrive at (3.4).

Observe that the first three terms on the left-hand side of inequality (3.12) have the same structure. So, we will deal with only first term and others can be dealt with the same way.

Now, the first term of inequality (3.12) implies that

$$\begin{aligned} \frac{1}{\tau _{13}}\int _{Q_T}\Big (\frac{\delta F(u_1,u_3,\theta )}{\delta u_1}\Big )^2d(s,x)+\frac{1}{\tau _{12}}\int _{Q_T}\Big (\frac{\delta F(u_1,u_2,\theta )}{\delta u_1}\Big )^2d(x,s)\le C. \end{aligned}$$

The last inequality means that

$$\begin{aligned}&\frac{1}{\tau _{13}}\int _{Q_T}\Big (\varepsilon _{13}^2(u_{1xx}+ u_1u_{2xx}-u_2u_{1xx})-a_{13}u_1(1-u_1-u_2)(1-2u_1-u_2)\nonumber \\&\quad -\theta \big (6\lambda l_1(u_1-u_1^2)-6\lambda l_3(u_1+u_2)(1-u_1-u_2)\big )\Big )^2d(x,s)\le C, \end{aligned}$$

(3.14)

and

$$\begin{aligned}&\frac{1}{\tau _{12}}\int _{Q_T}\Big (\varepsilon _{12}^2(u_2 u_{1xx}-u_1u_{2xx})-a_{12}u_1u_2(u_2-u_1)-\theta \big (6\lambda l_1(u_1-u_1^2)\nonumber \\&\quad -6\lambda l_2(u_2-u_2^2)\big )\Big )^2d(x,s)\le C.\qquad \qquad \qquad \end{aligned}$$

(3.15)

Let us write inequality (3.14) in the following form

$$\begin{aligned}\int_{Q_T}(I+J)^2d(x,s)\leq C,\end{aligned}$$

(3.16)

where

$$\begin{aligned}&I:=\varepsilon _{13}^2(u_{1xx}+ u_1u_{2xx}-u_2u_{1xx}),\\&J:=-a_{13}u_1(1-u_1-u_2)(1-2u_1-u_2)-\theta \big (6\lambda l_1(u_1-u_1^2)\\&\quad -6\lambda l_3(u_1+u_2)(1-u_1-u_2)\big ).\qquad \qquad \qquad \qquad \qquad \end{aligned}$$

To estimate $J$, we use Hölder’s inequality, Sobolev embedding, and estimates $u_1,u_2\in L^{\infty }(0,T;H_0^1(\Omega ))$ and $\theta \in L^2(0,T;H_0^1(\Omega ))$ such that

$$\begin{aligned} \int _{Q_T}J^2d(x,s)\leq C. \end{aligned}$$

Let $\eta , \sigma >0$ be small constants such that

$$\begin{aligned} (I+J)^2+\eta J^2\ge \sigma I^2. \end{aligned}$$

(3.17)

Next, we integrate inequality (3.17) over $\Omega \times (0,T)$ and use the bound of $J$ to obtain

$$\begin{aligned} \sigma \int _{Q_T}I^2d(x,s)\le \int _{Q_T}\big ((I+J)^2+\eta J^2\big )d(x,s)\le C. \end{aligned}$$

Thus, it follows after replacing $I$, that

$$\begin{aligned} \Vert u_{1xx}+ u_1u_{2xx}-u_2u_{1xx}\Vert _{L^2(Q_T)}\le C. \end{aligned}$$

(3.18)

Similarly, we obtain from (3.15) that

$$\begin{aligned} \Vert u_2 u_{1xx}-u_1u_{2xx}\Vert _{L^2(Q_T)}\le C. \end{aligned}$$

(3.19)

Hence, by using Minkowski’s inequality for (3.18) and (3.19), we get that

$$\begin{aligned} \Vert u_{1xx}\Vert _{L^2\left( 0,T;L^2(\Omega )\right) }\le C. \end{aligned}$$

(3.20)

By using (3.20) with elliptic estimate, one yields

$$\begin{aligned} \Vert u_1\Vert _{L^2\left( 0,T;H^2(\Omega )\right) }\le C. \end{aligned}$$

(3.21)

Proceeding in the previous way, one can also obtain from the second term of left-hand side of (3.12) that

$$\begin{aligned} \qquad \Vert u_2\Vert _{L^2\left( 0,T;H^2(\Omega )\right) }\le C.\qquad \qquad \end{aligned}$$

(3.22)

Thus, combining (3.10) with (3.21) and (3.22), estimate (3.3) holds. Hence Lemma 3.2 is proved. $\square$

Lemma 3.3

There exits a constant $C$ independent of $t^{\star }$ such that, for any $T>0,$

$$\begin{aligned}&\int _0^{T}\big (\Vert u_{1t}\Vert _{L^2(\Omega )}^2+\Vert u_{2t}\Vert _{L^2(\Omega )}^2\big )ds+\frac{k_{1}}{2}\Vert u_{1x}(t)\Vert _{L^2(\Omega )}^2+\frac{k_{2}}{2}\Vert u_{2x}(t)\Vert _{L^2(\Omega )}^2<C; \end{aligned}$$

(3.23)

$$\begin{aligned}&\Vert (u_{1t},u_{2t})\Vert _{L^2\left( 0,T;L^2(\Omega )\right) }\le C;\qquad \qquad \qquad \qquad \qquad \qquad \end{aligned}$$

(3.24)

$$\begin{aligned}&\Vert \theta \Vert _{L^2\left( 0,T;H^2(\Omega )\right) }+ \Vert \theta _{t}\Vert _{L^2\left( 0,T;L^2(\Omega )\right) }\le C;\qquad \qquad \quad \qquad \end{aligned}$$

(3.25)

Proof

By multiplying the first and second equations of problem (2.1) by $u_{1t}$ and $u_{2t}$, respectively, integrating in $\Omega$, using Hölder’s inequality, Sobolev embedding, Young’s inequality, and adding the resulting inequalities, we have

$$\begin{aligned}&\Vert u_{1t}\Vert _{L^2(\Omega )}^2+\Vert u_{2t}\Vert _{L^2(\Omega )}^2+\frac{k_{1}}{2}\frac{d}{dt}\Vert u_{1x}\Vert _{L^2(\Omega )}^2+\frac{k_{2}}{2}\frac{d}{dt}\Vert u_{2x}\Vert _{L^2(\Omega )}^2\nonumber \\&\quad \le c_{1}\big (\Vert u_2\Vert _{H_0^1(\Omega )}^2\Vert u_{1xx}\Vert _{L^2(\Omega )}^2+\Vert u_1\Vert _{H_0^1(\Omega )}^2\Vert u_{2xx}\Vert _{L^2(\Omega )}^2\big )+c_{2}\big ((1+\Vert u_1\Vert _{H_0^1(\Omega )}\nonumber \\&\qquad +\Vert u_2\Vert _{H_0^1(\Omega )})(1+2\Vert u_1\Vert _{H_0^1(\Omega )}+\Vert u_2\Vert _{H_0^1(\Omega )})\big )^2\Vert u_1\Vert _{H_0^1(\Omega )}^2+c_{3}\big (\Vert (u_1,u_2)\Vert _{H_0^1(\Omega )}\nonumber \\&\qquad \times (\Vert u_1\Vert _{H_0^1(\Omega )}+\Vert u_2\Vert _{H_0^1(\Omega )}\big )^2+c_{4}\big ((1+\Vert u_1\Vert _{H_0^1(\Omega )}+\Vert u_2\Vert _{H_0^1(\Omega )})(1+\Vert u_1\Vert _{H_0^1(\Omega )}\nonumber \\&\qquad +2\Vert u_2\Vert _{H_0^1(\Omega )})\big )^2\Vert u_2\Vert _{H_0^1(\Omega )}^2+c_{5}\big (\Vert (u_1,u_2)\Vert _{H_0^1(\Omega )}(\Vert u_1\Vert _{H_0^1(\Omega )}+\Vert u_2\Vert _{H_0^1(\Omega )}\big )^2\nonumber \\&\qquad +c_6\Vert \theta \Vert _{L^2(\Omega )}^2\big (\Vert u_1\Vert _{H_0^1(\Omega )}^2+\Vert u_1^2\Vert _{H_0^1(\Omega )}^2+\Vert u_2\Vert _{H_0^1(\Omega )}^2+\Vert u_2^2\Vert _{H_0^1(\Omega )}^2\nonumber \\&\qquad +\Vert (u_1,u_2)\Vert _{H_0^1(\Omega )}^2\big ) +\delta \Vert u_{1t}\Vert _{L^2(\Omega )}^2+\varepsilon \Vert u_{2t}\Vert _{L^2(\Omega )}^2, \end{aligned}$$

(3.26)

where the positive constants $c_{i},$ for $i=1,\cdots ,6,$ depend on the known parameters.

Now, let us integrate inequality (3.26) over $0< t< T$, and use the facts that $u_1,u_2\in {L^{\infty }(0,T;H_0^1(\Omega ))}$ and $\theta \in {L^{2}(0,T;L^2(\Omega ))}$ such that

$$\begin{aligned}&\int _0^{T}\big (\Vert u_{1t}(s)\Vert _{L^2(\Omega )}^2+\Vert u_{2t}(s)\Vert _{L^2(\Omega )}^2\big )ds+\frac{k_{1}}{2}\Vert u_{1x}(t)\Vert _{L^2(\Omega )}^2+\frac{k_{2}}{2}\Vert u_{2x}(t)\Vert _{L^2(\Omega )}^2\nonumber \\&\quad \le C+\delta \int _0^{T}\Vert u_{1t}(s)\Vert _{L^2(\Omega )}^2ds+\varepsilon \int _0^{T}\Vert u_{2t}(s)\Vert _{L^2(\Omega )}^2ds.\qquad \end{aligned}$$

(3.27)

By choosing sufficiently small $\delta ,\varepsilon >0$, we arrive at (3.23). Observe that inequality (3.27) also implies (3.24).

Similarly, by multiplying the third equation of problem (2.1), successively, by $-\theta _{xx}$ and $\theta _t$, integrating over $\Omega \times (0,t)$ with $0< t< T$, using $g\in {L^{2}(Q_T)}$, utilizing estimates of $u_1,$ $u_2$ from (3.3) and corresponding time derivatives $u_{1t},$ $u_{2t}$ from (3.24), we can prove easily (3.25). Thus, the proof of Lemma 3.3 is completed. $\square$

Proof

Since Lemma 2.3 provides the existence of local weak solution $(u_1,u_2,\theta )$ to problem (2.1), therefore, Lemmas 3.2-3.3 for uniform a priori estimates allow us to extend the local weak solution $(u_1,u_2,\theta )$ to $[0,T]$ for arbitrary $T>0$. This means that $(u_1,u_2,\theta )$ is a global weak solution of (2.1). Thus, the proof is established. $\square$

Finally, let us use Proposition 2.2 to prove our main result.

Proof of Theorem 1.1

By Proposition 2.2, and using the result of continuity of the embedding $\{L^2(0,T; H^2(\Omega ))\cap H^1(0,T; L^2(\Omega ))\}\hookrightarrow C([0,T];H_{0}^1(\Omega ))$(see e.g. Evans [26]), we obtain that $(u_1,u_2,\theta )\in [C([0,T];H_{0}^1(\Omega ))]^3$. Recalling $u_3=1-u_1-u_2$ implies that $(u_1,u_2,u_3,\theta )\in [C([0,T];H_{0}^1(\Omega ))]^4$ is a global weak solution of (1.3)–(1.4) such that $u_1+u_2+u_3=1$. Hence, the proof is finished. $\square$