Global Existence and Decay Estimates of Solutions of a Parabolic–Elliptic–Parabolic System for Ion Transport Networks

Abstract

This paper deals with a parabolic-elliptic-parabolic system arising from ion transport networks originally proposed by Albi et al. (Anal Appl 14(1):185–206, 2016). Our investigation demonstrates that the local existence of the weak solution can always be achieved. Moreover, by establishing the uniform a-priori decay estimates of solution, the local solution can be extended to a global one provided that the initial data is suitably small. In addition, the time-decay rate of the weak solution is obtained. To our best knowledge, this seems to be the first analytical work on this system indeed.

Appendix

In this concluding section, we provide the rigorous proofs of Lemmas 2.4 and 3.1.

Proof of Lemma 2.4

When $\gamma =1$, by applying the theory of the linear parabolic equation, the solution $\mathbf{m}\in C\big ([0,T],L^p(\Omega )\big )$ follows directly. Moreover, by applying [22, Theorem 3.7.1] to the Dirichlet problem associated with the linear heat equation with a potential

$$\begin{aligned} \mathbf{m}_t-\Delta \mathbf{m}+\mathbf{A}(t,x)\mathbf{m}=0\quad \mathrm {with}\quad \mathbf{A}:=I-\nabla \psi \otimes \nabla \psi , \end{aligned}$$

we conclude from $\nabla \psi \in L^\infty \big ([0,T], H^2(\Omega )\big )$ and Sobolev’s embedding that $\mathbf{m}\in C\big ((0,T],L^\infty (\Omega )\big )$. From this, $\mathbf{m}\in C^{\alpha ,\frac{\alpha }{2}}(\Omega _T)$ by employing [22, Theorem 3.10.1].

Next, we consider the case of $\gamma >1$. We begin with $p=\infty$. A well-known construction, on the basis of the contraction mapping principle and Lemma 2.1, asserts the existence of solution of (2.8). This solution satisfies

$$\begin{aligned} \Vert \mathbf{m}(t)\Vert _{L^r(\Omega )}\le Ce^{kt}\Vert \mathbf{m}_0\Vert _{L^r(\Omega )},\ \ \Vert \mathbf{m}(t)\Vert _{L^\infty (\Omega )}\le Ce^{kt}t^{-\frac{n}{2r}}\Vert \mathbf{m}_0\Vert _{L^r(\Omega )} \end{aligned}$$

(5.1)

for any $t\in (0,T]$ and $r\in [1, \infty ]$, where k depends on $\Vert \mathbf{m}\Vert _{L^\infty ((0,T),L^{\frac{p}{2(\gamma -1)}}(\Omega ))}$ and $\Vert \nabla \psi \Vert _{L^\infty ((0,T),L^\sigma (\Omega ))}$ with some $\sigma >n$, see Theorem A1 in [3]. Moreover, we see from [22, Theorem 3.10.1] that $\mathbf{m}\in C^{\theta ,\frac{\theta }{2}}(\Omega _T)$ for some $\theta \in (0,1)$, due to $\psi \in L^\infty \big ([0,T];H^1_0(\Omega )\cap H^2(\Omega )\big )$.

It remains to show that the same conclusion holds for $p\in [2,\infty )$ and $p>n(\gamma -1)$ when $\gamma >1$. In this case, we can establish the existence of the solution of (2.8) by employing the contraction mapping principle in a somewhat unusual space (this idea is due to [3, 35]). Let $T\in (0,1)$ to be specified below. For simplicity, we set

$$\begin{aligned} X:=L^\infty \big ((0,T),L^p(\Omega )\big )\cap L^\infty _{\text {loc}}\big ((0,T),L^{p(2\gamma -1)}(\Omega )\big ), \end{aligned}$$

and denote the closed bounded convex subset $X_R$ of X by

$$\begin{aligned} X_R:=\big \{f\in X: \Vert f\Vert _{L^p(\Omega )}\le R+1 \quad \text {and}\quad t^\alpha \Vert f\Vert _{L^\beta (\Omega )}\le R+1\quad \text {for}\ t\in (0,T)\big \}, \end{aligned}$$

where $\beta :=p(2\gamma -1)$ and $\alpha :=\frac{n(\gamma -1)}{p(2\gamma -1)}<\frac{1}{2\gamma -1}<1$, due to $\gamma >1$. We equip $X_R$ with the distance

$$\begin{aligned} d(f,g)=\sup _{t\in (0,T)}t^\alpha \Vert f(t)-g(t)\Vert _{L^{\beta }(\Omega )}, \end{aligned}$$

so that $(X_R,d)$ is a nonempty complete metric space.

We first fix an $R\ge \max \{C_1,C_4\}\Vert \mathbf{m}_0\Vert _{L^p(\Omega )}$, where $C_1$ and $C_4$ are determined by (5.3) and (5.4) below. Then for any given $\mathbf{m}\in X_R$, we set

$$\begin{aligned} \Psi (\mathbf{m})(t):=e^{\Delta t}\mathbf{m}_0+\int _0^te^{\Delta (t-s)}\big ((\mathbf{m}\cdot \nabla \psi )\nabla \psi -|\mathbf{m}|^{2(\gamma -1)}\mathbf{m}\big )(s)ds, \end{aligned}$$

where $e^{\Delta t}$ satisfies the standard smoothing effect: let $1\le a\le b\le \infty$, one has

$$\begin{aligned} \Vert e^{\Delta t}f\Vert _{L^b(\Omega )}\le C t^{-\frac{n}{2}\big (\frac{1}{a}-\frac{1}{b}\big )}\Vert f\Vert _{L^a(\Omega )}, \end{aligned}$$

(5.2)

where we only use the special case of (2.1), due to $t\le T<1$. Since $\psi \in L^\infty \big ([0,T];H^1_0(\Omega )\cap H^2(\Omega )\big )$, we have $\nabla \psi \in L^\infty \big ([0,T], L^r(\Omega )\big )$ by Sobolev’s embedding, where $r\in [1,\infty ]$ if $n=1$, $r\in [1,\infty )$ if $n=2$ and $r\in [1,6]$ if $n=3$. Utilizing (5.2) we can obtain that, if $p\ge 2$,

$$\begin{aligned} \Vert \Psi (\mathbf{m})(t)\Vert _{L^p(\Omega )}&\le C\Vert \mathbf{m}_0\Vert _{L^p(\Omega )} +C\int _0^t(t-s)^{-\frac{n}{6}}\Vert (\mathbf{m}\cdot \nabla \psi ) \nabla \psi \Vert _{L^\frac{3p}{3+p}(\Omega )}\\&\quad +\big \Vert |\mathbf{m}|^{2(\gamma -1)} \mathbf{m}(s)\big \Vert _{L^p(\Omega )}ds\\ \le&C\Vert \mathbf{m}_0\Vert _{L^p(\Omega )}+C\Vert \nabla \psi \Vert _{L^\infty \left( [0,T], L^6(\Omega )\right) }^2\int _0^t(t-s)^{-\frac{n}{6}} \Vert \mathbf{m}(s)\Vert _{L^p(\Omega )}ds\\&\quad +C\int _0^t\big \Vert \mathbf{m}(s)\big \Vert _{L^\beta (\Omega )}^{2\gamma -1}ds\\&\le C\Vert \mathbf{m}_0\Vert _{L^p(\Omega )}+C\Vert \nabla \psi \Vert _{L^\infty \left( [0,T], L^6(\Omega )\right) }^2(R+1)T^{1-\frac{n}{6}}\\&\quad +C\big (t^\alpha \Vert \mathbf{m}\Vert _{L^\beta (\Omega )}\big )^{2\gamma -1} \int _0^ts^{-\alpha (2\gamma -1)}ds, \end{aligned}$$

which implies that there exist three positive constants $C_1, C_2$ and $C_3$ such that

$$\begin{aligned} \Vert \Psi (\mathbf{m})(t)\Vert _{L^p(\Omega )}&\le C_1\Vert \mathbf{m}_0\Vert _{L^p(\Omega )} +C_2\Vert \nabla \psi \Vert _{L^\infty \left( [0,T], L^6(\Omega )\right) }^2(R+1)T^{1-\frac{n}{6}}\nonumber \\&\quad +C_3\big (R+1\big )^{2\gamma -1} T^{1-\alpha (2\gamma -1)}. \end{aligned}$$

(5.3)

Similarly, we have

$$\begin{aligned} t^\alpha \Vert \Psi (\mathbf{m})(t)\Vert _{L^\beta (\Omega )}&\le Ct^\alpha t^{-\frac{n}{2}(\frac{1}{p}-\frac{1}{\beta })} \Vert \mathbf{m}_0\Vert _{L^p(\Omega )}\\&\quad +Ct^\alpha \int _0^t(t-s)^{-\frac{n}{6}} \Vert (\mathbf{m}\cdot \nabla \psi )\nabla \psi \Vert _{L^{\frac{3\beta }{3+\beta }}(\Omega )}ds\\&\quad +Ct^\alpha \int _0^t(t-s)^{-\alpha }\big \Vert |\mathbf{m}|^{2(\gamma -1)} \mathbf{m}\big \Vert _{L^p(\Omega )}ds\\&\le C\Vert \mathbf{m}_0\Vert _{L^p(\Omega )}+Ct^\alpha \int _0^t\big ((t-s)^{-\frac{n}{6}} \Vert \nabla \psi \Vert _{L^6(\Omega )}^2\Vert \mathbf{m}\Vert _{L^\beta (\Omega )}\\&\quad +(t-s)^{-\alpha }\Vert \mathbf{m}\Vert _{L^\beta (\Omega )}^{2\gamma -1}\big )ds\\&\le C\Vert \mathbf{m}_0\Vert _{L^p(\Omega )}+C\Vert \nabla \psi \Vert _{L^\infty \left( [0,T], L^6(\Omega )\right) }^2(R+1)t^\alpha \int _0^t(t-s)^{-\frac{n}{6}}s^{-\alpha }ds\\&\quad +C\big (R+1\big )^{2\gamma -1}t^\alpha \int _0^t(t-s)^{-\alpha }s^{-\alpha (2\gamma -1)}ds, \end{aligned}$$

where we use the fact $\beta \ge 2$ due to $p\ge 2$ and $\gamma >1$. Hence, there exist three positive constants $C_4, C_5$ and $C_6$ such that

$$\begin{aligned} t^\alpha \Vert \Psi (\mathbf{m})(t)\Vert _{L^\beta (\Omega )}&\le C_4\Vert \mathbf{m}_0\Vert _{L^p(\Omega )}+C_5\Vert \nabla \psi \Vert _{L^\infty \left( [0,T], L^6(\Omega )\right) }^2(R+1)T^{1-\frac{n}{6}}\nonumber \\&\quad +C_6\big (R+1\big )^{2\gamma -1}T^{1-\alpha (2\gamma -1)}. \end{aligned}$$

(5.4)

By taking T suitable small, we get that

$$\begin{aligned} C_2\Vert \nabla \psi \Vert _{L^\infty \left( [0,T], L^6(\Omega )\right) }^2(R+1)T^{1-\frac{n}{6}} +C_3\big (R+1\big )^{2\gamma -1}T^{1-\alpha (2\gamma -1)}<1,\\ C_5\Vert \nabla \psi \Vert _{L^\infty \left( [0,T], L^6(\Omega )\right) }^2(R+1)T^{1-\frac{n}{6}} +C_6\big (R+1\big )^{2\gamma -1}T^{1-\alpha (2\gamma -1)}<1. \end{aligned}$$

Hence, we infer from (5.3) and (5.4) that $\Psi$ maps $X_R$ into itself when T is suitable small.

We now turn to the proof of the contractivity of the mapping $\Psi$. For any given $\mathbf{m}^1, \mathbf{m}^2\in X_R$, we let $\delta \mathbf{m}:=\mathbf{m}^1-\mathbf{m}^2$ and $\delta \Psi (\mathbf{m}):=\Psi (\mathbf{m}^1)-\Psi (\mathbf{m}^2)$. The key is to estimate $t^\alpha \Vert \delta \Psi (\mathbf{m})(t)\Vert _{L^\beta (\Omega )}$. In fact, we have

$$\begin{aligned}&t^\alpha \Vert \delta \Psi (\mathbf{m})(t)\Vert _{L^\beta (\Omega )}\\&\quad \le Ct^\alpha \int _0^t(t-s)^{-\frac{n}{6}}\Vert (\delta \mathbf{m}\cdot \nabla \psi ) \nabla \psi \Vert _{L^{\frac{3\beta }{3+\beta }}(\Omega )}ds\\&\qquad +\,Ct^\alpha \int _0^t(t-s)^{-\alpha }\big \Vert \big (|\mathbf{m}^1|^{2(\gamma -1)} \mathbf{m}^1-|\mathbf{m}^2|^{2(\gamma -1)}\mathbf{m}^2\big )(s)\big \Vert _{L^p(\Omega )}ds, \end{aligned}$$

where we have used the fact $\Vert \delta \mathbf{m}_0\Vert _{L^p(\Omega )}=0$. Meanwhile, noticing the elementary inequality

$$\begin{aligned} \big ||x|^\kappa x-|y|^\kappa y\big |\le C\big (|x|+|y|\big )^\kappa |x-y| \le C(|x|^\kappa +|y|^\kappa )|x-y| \end{aligned}$$

(5.5)

for all $\kappa \ge 0$ and $x, y\in {\mathbb {R}}^n$, we have

$$\begin{aligned} t^\alpha \Vert \delta \Psi (\mathbf{m})(t)\Vert _{L^\beta (\Omega )}\le & {} Ct^\alpha \int _0^t(t-s)^{-\frac{n}{6}}\Vert \delta \mathbf{m}\Vert _{L^\beta (\Omega )} \Vert \nabla \psi \Vert _{L^6(\Omega )}^2ds\\&+\,Ct^\alpha \int _0^t(t-s)^{-\alpha }\big \Vert \big (|\mathbf{m}^1|^{2(\gamma -1)} +|\mathbf{m}^2|^{2(\gamma -1)}\big )\delta \mathbf{m}\big \Vert _{L^p(\Omega )}ds\\\le & {} C\Vert \nabla \psi \Vert _{L^\infty \left( [0,T], L^6(\Omega )\right) }^2t^\alpha d(\mathbf{m}^1,\mathbf{m}^2)\int _0^t(t-s)^{-\frac{n}{6}}s^{-\alpha }ds\\&+\,Ct^\alpha \int _0^t(t-s)^{-\alpha } \big (\Vert \mathbf{m}^1\Vert _{L^\beta (\Omega )}^{2(\gamma -1)} +\Vert \mathbf{m}^2\Vert _{L^\beta (\Omega )}^{2(\gamma -1)}\big )\Vert \delta \mathbf{m}\Vert _{L^\beta (\Omega )}ds\\\le & {} C\Vert \nabla \psi \Vert _{L^\infty \left( [0,T], L^6(\Omega )\right) }^2T^{1-\frac{n}{6}}d(\mathbf{m}^1,\mathbf{m}^2)\\&+\,Ct^\alpha \big (R+1\big )^{2(\gamma -1)}\int _0^t(t-s)^{-\alpha } s^{-2(\gamma -1)\alpha }\Vert \delta \mathbf{m}\Vert _{L^\beta (\Omega )}ds, \end{aligned}$$

which implies that there exists a positive constant $C_7$ such that

$$\begin{aligned} t^\alpha \Vert \delta \Psi (\mathbf{m})(t)\Vert _{L^\beta (\Omega )}\le & {} C_7\Big (\Vert \nabla \psi \Vert _{L^\infty \left( [0,T], L^6(\Omega )\right) }^2T^{1-\frac{n}{6}}\\&+\big (R+1\big )^{2(\gamma -1)} T^{1-\alpha (2\gamma -1)}\Big )d(\mathbf{m}^1,\mathbf{m}^2). \end{aligned}$$

From this, we can assert that, if T (depending on R) is small enough such that

$$\begin{aligned} C_7\Big (\Vert \nabla \psi \Vert _{L^\infty \left( [0,T], L^6(\Omega )\right) }^2T^{1-\frac{n}{6}}+\big (R+1\big )^{2(\gamma -1)} T^{1-\alpha (2\gamma -1)}\Big )<1, \end{aligned}$$

then $\Psi$ is a strict contraction mapping from $X_R$ to $X_R$. Thus $\Psi$ has a unique fixed point $\mathbf{m}$ in $X_R$. Moreover, $\mathbf{m}$ is a solution of (2.8).

We claim that there exists $\theta \in (0,1)$ such that $\mathbf{m}\in C^{\theta ,\frac{\theta }{2}}(\Omega _T)$. To this end, it suffices to verify that $\mathbf{m}\in C\big ([0,T], L^p(\Omega )\big )\cap L^\infty _{\text {loc}}\big ((0,T), L^\infty (\Omega )\big )$ by applying [22, Theorem 3.10.1] to the Dirichlet problem associated with the linear heat equation with a potential

$$\begin{aligned} \mathbf{m}_t-\Delta \mathbf{m}+\mathbf{A}_\gamma (t,x)\mathbf{m}=0\quad \mathrm {with}\quad \mathbf{A}_\gamma :=|\mathbf{m}|^{2(\gamma -1)}I-\nabla p\otimes \nabla p. \end{aligned}$$

In fact, due to $\mathbf{m}\in L^\infty _{\text {loc}}((0,T),L^\beta (\Omega ))$, we can apply Theorem A1 in [3] with $r=\beta$ and $\sigma =\frac{\beta }{2(\gamma -1)}$ on the interval $(\epsilon ,T-\epsilon )$ for any $\epsilon \in (0,\frac{T}{2})$, which also implies

$$\begin{aligned} \mathbf{m}\in C\big ([0,T], L^p(\Omega )\big )\cap L^\infty _{\text {loc}}\big ((0,T), L^\infty (\Omega )\big ). \end{aligned}$$

This completes the proof of Lemma 2.4. $\square$

Proof of Lemma 3.1

Here we will omit the superscript $\varepsilon$ for notational convenience. Without loss of generality, we just consider the case $\Vert S\Vert _{C^\theta (\Omega )}\ne 0$, since the case $\Vert S\Vert _{C^\theta (\Omega )}=0$ can be similarly dealt with. Let $T\in (0,1)$ to be specified below. For simplicity, we also set

$$\begin{aligned} E:=C\big ([0,T]; L^2(\Omega )\big ). \end{aligned}$$

For any $R>0$, we will denote the closed bounded convex subset $E_R$ of E by

$$\begin{aligned} E_R:=\big \{g\in E: \Vert g(\cdot ,t)\Vert _{L^2(\Omega )}\le R+1 \quad \text {for all}\ t\in [0,T]\big \}. \end{aligned}$$

We first fix an $R\ge C_1\Vert \varrho _0\Vert _{L^2(\Omega )}$, where $C_1$ is given in (5.10). Then for any given ${\bar{\varrho }}\in E_R$, we see from the standard elliptic theory [10] that the boundary value problem

$$\begin{aligned} \left\{ \begin{aligned}&-\Delta \psi ={\bar{\varrho }}, \quad x\in \Omega ,\\&\psi (x,t)=0, \quad x\in \partial \Omega \end{aligned} \right. \end{aligned}$$

(5.6)

possesses a unique solution ${\bar{\psi }}=\psi [{\bar{\varrho }}](\cdot , t)\in H^1_0(\Omega )\cap H^2(\Omega )$ satisfying

$$\begin{aligned} \Vert {\bar{\psi }}\Vert _{L^\infty ([0,T]; H^2(\Omega ))}\le C\Vert {\bar{\varrho }}\Vert _{L^\infty ([0,T];L^2(\Omega ))}. \end{aligned}$$

(5.7)

Invoking this, we can get from Lemma 2.4 that, if $\mathbf{m}_0\in L^\infty (\Omega )$, then the initial-boundary value problem

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t\mathbf{m}-\Delta \mathbf{m}-\big (\mathbf{m}\cdot \nabla \psi [{\bar{\varrho }}\big )\nabla \psi [{\bar{\varrho }}] +|\mathbf{m}|^{2(\gamma -1)}\mathbf{m}=0, \&x\in \Omega ,\\&\mathbf{m}(x,t)=\mathbf{0}, \qquad&x\in \partial \Omega ,\\&\mathbf{m}(x,0)=\mathbf{m}_0(x), \qquad&x\in \Omega \end{aligned} \right. \end{aligned}$$

(5.8)

admits a unique solution ${\bar{\mathbf{m}}}{=}\mathbf{m}[\psi [{\bar{\varrho }}]]$ belonging to $C([0,T];L^\infty (\Omega ))\cap C^{\theta ,\frac{\theta }{2}}(\Omega _T)$.

Thus, for any ${\bar{\varrho }}\in E_R$ we can introduce a mapping $\Psi$ by setting $\Psi ({\bar{\varrho }})=\varrho$, where $\varrho$ is a solution of the initial-boundary value problem

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t\varrho =\text {div}\left( (I+{\bar{\mathbf{m}}}\otimes {\bar{\mathbf{m}}})(\nabla \varrho +\varrho \nabla {\bar{\psi }})\right) +S\varrho , \quad&x\in \Omega ,\\&\varrho (x,t)=0, \qquad \quad&x\in \partial \Omega ,\\&\varrho (x,0)=\varrho _0(x), \qquad&x\in \Omega . \end{aligned} \right. \end{aligned}$$

(5.9)

Notice that the local existence of solution to (5.9) can be showed by utilizing the linear parabolic theory, where ${\bar{\psi }}=\psi [{\bar{\varrho }}](x,t)$ is determined by (5.6) and ${\bar{\mathbf{m}}}=\mathbf{m}[\psi [{\bar{\varrho }}]]$ is determined by (5.8). Moreover, $\varrho$ satisfies

$$\begin{aligned} \Vert \varrho \Vert _{C([0,T]\times L^2(\Omega ))}+\Vert \varrho \Vert _{L^2([0,T];H^1(\Omega ))}\le C_1\Vert \rho _0\Vert _{L^2(\Omega )}, \end{aligned}$$

(5.10)

see Theorem 2.1 and Theorem 4.1 in Chapter 3 of [22] for details. Moreover, we see from [22, Theorem 3.7.1] that there exists a positive constant $C=C(R)$ such that

$$\begin{aligned} \Vert \varrho \Vert _{L^\infty ([0,T]\times \Omega )}\le C\Vert j_\varepsilon *\varrho _0\Vert _{L^\infty ([0,T]\times \Omega )}. \end{aligned}$$

(5.11)

Invoking (5.10), it is clear that $\Psi$ maps $E_R$ into itself when T is suitable small.

Next, we shall certify that $\Psi$ is contractive and thus has a fixed point in $E_R$ provided that T is small enough. To this end, for any given ${\bar{\varrho }}^i\in E_R$, let ${\bar{\psi }}^i:=\psi [{\bar{\varrho }}^i]$ and ${\bar{\mathbf{m}}}^i:=\mathbf{m}[{\bar{\psi }}^i]$ be the solution of system (5.6) and (5.8) with ${\bar{\mathbf{m}}}={\bar{\mathbf{m}}}^i$ and ${\bar{\psi }}={\bar{\psi }}^i$, $i=1,2$, respectively. For simplicity, we also set

$$\begin{aligned} \delta {\bar{\mathbf{m}}}:= & {} {\bar{\mathbf{m}}}^1-{\bar{\mathbf{m}}}^2, \qquad \delta {\bar{\psi }}:={\bar{\psi }}^1-{\bar{\psi }}^2, \qquad \delta \varrho =\varrho ^1-\varrho ^2 \qquad \mathrm {and}\qquad \\ \delta {\bar{\varrho }}:= & {} {\bar{\varrho }}^1-{\bar{\varrho }}^2, \end{aligned}$$

where $\varrho ^i:=\Psi ({\bar{\varrho }}^i)$, $i=1,2$. The key of the proof is to estimate $\Vert \delta \varrho (t)\Vert _{L^2(\Omega )}$. To achieve this, we notice that $\delta \varrho$ satisfies the initial-boundary value problem

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t\delta \varrho =\Delta \delta \varrho +\text {div}\big (({\bar{\mathbf{m}}}^1\otimes {\bar{\mathbf{m}}}^1) \nabla \varrho ^1-({\bar{\mathbf{m}}}^2\otimes {\bar{\mathbf{m}}}^2)\nabla \varrho ^2\big )\\&+\text {div}\big (\varrho ^1\nabla {\bar{\psi }}^1-\varrho ^2\nabla {\bar{\psi }}^2\big ) +\text {div}\big (({\bar{\mathbf{m}}}^1\otimes {\bar{\mathbf{m}}}^1) \nabla {\bar{\psi }}^1\varrho ^1-({\bar{\mathbf{m}}}^2\otimes {\bar{\mathbf{m}}}^2) \nabla {\bar{\psi }}^2\varrho ^2\big )+\delta \varrho S,&x\in \Omega , \\&\delta \varrho =0, \quad&x\in \partial \Omega ,\\&\delta \varrho (x,0)=0, \quad&x\in \Omega . \end{aligned} \right. \nonumber \\ \end{aligned}$$

(5.12)

First of all, multiplication of (5.12) by $\delta \varrho$ and integration by parts give us

$$\begin{aligned} \frac{1}{2}\frac{d}{dt}\int _{\Omega }|\delta \varrho |^2dx +\int _\Omega |\nabla \delta \varrho |^2dx=&-\int _\Omega \big (({\bar{\mathbf{m}}}^1\otimes {\bar{\mathbf{m}}}^1)\nabla \varrho ^1-({\bar{\mathbf{m}}}^2 \otimes {\bar{\mathbf{m}}}^2)\nabla \varrho ^2\big ) \cdot \nabla \delta \varrho dx \nonumber \\&- \int _\Omega \big (\varrho ^1\nabla {\bar{\psi }}^1-\varrho ^2\nabla {\bar{\psi }}^2\big ) \cdot \nabla \delta \varrho dx \nonumber \\&-\int _\Omega \big (({\bar{\mathbf{m}}}^1\otimes {\bar{\mathbf{m}}}^1) \nabla {\bar{\psi }}^1\varrho ^1-({\bar{\mathbf{m}}}^2\otimes {\bar{\mathbf{m}}}^2) \nabla {\bar{\psi }}^2\varrho ^2\big )\cdot \nabla \delta \varrho dx\nonumber \\&+\int _\Omega |\delta \varrho |^2Sdx \nonumber \\ =:&\;\Pi _1+\Pi _2+\Pi _3+\Pi _4. \end{aligned}$$

(5.13)

For $\Pi _1$, a careful calculation shows that

$$\begin{aligned} \Pi _1 =&-\int _\Omega \big [({\bar{\mathbf{m}}}^1\otimes {\bar{\mathbf{m}}}^1) \nabla \delta \varrho +\big (\delta {\bar{\mathbf{m}}}\otimes {\bar{\mathbf{m}}}^1\big ) \nabla \varrho ^2+\big ({\bar{\mathbf{m}}}^2\otimes (\delta {\bar{\mathbf{m}}})\big ) \nabla \varrho ^2\big ]\cdot \nabla \delta \varrho dx\nonumber \\ \le&-\int _\Omega \big ({\bar{\mathbf{m}}}^1\cdot \nabla \delta \varrho \big )^2dx+\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )} \Vert \nabla \varrho ^2\Vert _{L^2(\Omega )}\Vert \nabla \delta \varrho \Vert _{L^2(\Omega )}\nonumber \\&\big (\Vert {\bar{\mathbf{m}}}^1\Vert _{L^\infty (\Omega )} +\Vert {\bar{\mathbf{m}}}^2\Vert _{L^\infty (\Omega )}\big )\nonumber \\ \le&-\Vert {\bar{\mathbf{m}}}^1\cdot \nabla \delta \varrho \Vert _{L^2(\Omega )}^2 +\frac{1}{8}\Vert \nabla \delta \varrho \Vert _{L^2(\Omega )}^2 +C\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}^2\Vert \nabla \varrho ^2 \Vert _{L^2(\Omega )}^2\nonumber \\&\big (\Vert {\bar{\mathbf{m}}}^1\Vert _{L^\infty (\Omega )}^2 +\Vert {\bar{\mathbf{m}}}^2\Vert _{L^\infty (\Omega )}^2\big ). \end{aligned}$$

(5.14)

Similarly, $\Pi _2$ can be controlled as follows:

$$\begin{aligned} \Pi _2&=-\int _\Omega \delta \varrho \nabla {\bar{\psi }}^1\cdot \nabla \delta \varrho dx-\int _\Omega \varrho ^2\nabla \delta {\bar{\psi }}\cdot \nabla \delta \varrho dx\nonumber \\&=-\frac{1}{2}\int _\Omega {\bar{\rho }}^1(\delta \varrho )^2dx -\int _\Omega \varrho ^2\nabla \delta {\bar{\psi }}\cdot \nabla \delta \varrho dx\nonumber \\&\le -\frac{1}{2}\int _\Omega {\bar{\rho }}^1(\delta \varrho )^2dx +\Vert \rho ^2\Vert _{L^3(\Omega )}\Vert \nabla \delta {\bar{\psi }}\Vert _{L^6(\Omega )} \Vert \nabla \delta \varrho \Vert _{L^2(\Omega )}. \end{aligned}$$

Moreover, employing Sobolev’s embedding $H^1(\Omega )\hookrightarrow L^6(\Omega )$ and Poincaré’s inequality, we have

$$\begin{aligned} \Pi _2\le -\frac{1}{2}\int _\Omega {\bar{\rho }}^1(\delta \varrho )^2dx +C\Vert \rho ^2\Vert _{L^3(\Omega )}\Vert \delta {\bar{\psi }}\Vert _{H^2(\Omega )} \Vert \nabla \delta \varrho \Vert _{L^2(\Omega )}. \end{aligned}$$

(5.15)

Here we need to estimate the factor $\Vert \delta {\bar{\psi }}\Vert _{H^2(\Omega )}$. Note that ${\bar{\psi }}^i\in H^2(\Omega )\cap H^1_0(\Omega )$ is a solution of (5.6) and $\delta {\bar{\psi }}$ is a solution to the boundary-value problem

$$\begin{aligned} \left\{ \begin{array}{ll} -\Delta \delta \psi =\delta {\bar{\varrho }}, \quad \quad &{} x\in \Omega ,\\ \delta \psi =0, \quad \quad &{} x\in \partial \Omega . \end{array} \right. \end{aligned}$$

Hence, it demonstrates that $\delta {\bar{\psi }}\in H^2(\Omega )\cap H^1_0(\Omega )$ and

$$\begin{aligned} \Vert \delta {\bar{\psi }}\Vert _{H^2(\Omega )}\le C\Vert \delta {\bar{\varrho }}\Vert _{L^2(\Omega )}, \end{aligned}$$

(5.16)

see Theorem 6.3.4 in [9]. Substitute (5.16) into (5.15), to find:

$$\begin{aligned} \Pi _2&\le -\frac{1}{2}\int _\Omega {\bar{\rho }}^1(\delta \varrho )^2dx +C\Vert \rho ^2\Vert _{L^3(\Omega )}\Vert \delta {\bar{\varrho }}\Vert _{L^2(\Omega )} \Vert \nabla \delta \varrho \Vert _{L^2(\Omega )}\nonumber \\&\le \frac{1}{2}\Vert {\bar{\rho }}^1\Vert _{L^2(\Omega )}\Vert \delta \varrho \Vert _{L^4(\Omega )}^2+ \frac{1}{8}\Vert \nabla \delta \rho \Vert _{L^2(\Omega )}^2 +C\Vert \rho ^2\Vert _{L^3(\Omega )}^2\Vert \delta {\bar{\varrho }}\Vert _{L^2(\Omega )}^2. \end{aligned}$$

(5.17)

Next, by employing (5.7) and (5.16) we can estimate the term $\Pi _3$ as follows:

$$\begin{aligned} \Pi _3&=-\int _\Omega \big ((\delta {\bar{\mathbf{m}}}\otimes {\bar{\mathbf{m}}}^1) \nabla {\bar{\psi }}^2\varrho ^2+({\bar{\mathbf{m}}}^2 \otimes \delta {\bar{\mathbf{m}}})\nabla {\bar{\psi }}^2\varrho ^2\big )\cdot \nabla \delta \varrho dx\nonumber \\&\quad -\int _\Omega ({\bar{\mathbf{m}}}^1\otimes {\bar{\mathbf{m}}}^1) \big (\nabla \delta {\bar{\psi }}\varrho ^1+\nabla {\bar{\psi }}^2\delta \varrho \big ) \cdot \nabla \delta \varrho dx\nonumber \\&\le \Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}\Vert \nabla {\bar{\psi }}^2 \Vert _{L^6(\Omega )}\Vert \varrho ^2\Vert _{L^3(\Omega )} \Vert \nabla \delta \varrho \Vert _{L^2(\Omega )}\big (\Vert {\bar{\mathbf{m}}}^1\Vert _{L^\infty (\Omega )} +\Vert {\bar{\mathbf{m}}}^2\Vert _{L^\infty (\Omega )}\big ) \nonumber \\&\quad +\Vert {\bar{\mathbf{m}}}^1\Vert _{L^\infty (\Omega )} \Vert {\bar{\mathbf{m}}}^1\cdot \nabla \delta \varrho \Vert _{L^2(\Omega )} \big (\Vert \nabla \delta {\bar{\psi }}\Vert _{L^6(\Omega )}\Vert \varrho ^1\Vert _{L^3(\Omega )} +\Vert \nabla {\bar{\psi }}^2\Vert _{L^6(\Omega )}\Vert \delta \rho \Vert _{L^3(\Omega )}\big ) \nonumber \\&\le C\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}\Vert {\bar{\varrho }}^2 \Vert _{L^2(\Omega )}\Vert \varrho ^2\Vert _{L^3(\Omega )} \Vert \nabla \delta \varrho \Vert _{L^2(\Omega )}\big (\Vert {\bar{\mathbf{m}}}^1 \Vert _{L^\infty (\Omega )}+\Vert {\bar{\mathbf{m}}}^2\Vert _{L^\infty (\Omega )}\big ) \nonumber \\&\quad +C\Vert {\bar{\mathbf{m}}}^1\Vert _{L^\infty (\Omega )}\Vert {\bar{\mathbf{m}}}^1\cdot \nabla \delta \varrho \Vert _{L^2(\Omega )}\big (\Vert \delta {\bar{\varrho }} \Vert _{L^2(\Omega )}\Vert \varrho ^1\Vert _{L^3(\Omega )}+\Vert {\bar{\varrho }}^2 \Vert _{L^2(\Omega )}\Vert \delta \rho \Vert _{L^3(\Omega )}\big ) \nonumber \\&\le \frac{1}{8}\Vert \nabla \delta \varrho \Vert _{L^2(\Omega )}^2 +C\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}^2\Vert {\bar{\varrho }}^2 \Vert _{L^2(\Omega )}^2\Vert \varrho ^2\Vert _{L^3(\Omega )}\big (\Vert {\bar{\mathbf{m}}}^1 \Vert _{L^\infty (\Omega )}^2+\Vert {\bar{\mathbf{m}}}^2\Vert _{L^\infty (\Omega )}^2\big ) \nonumber \\&\quad +\frac{1}{8}\Vert {\bar{\mathbf{m}}}^1\cdot \nabla \delta \varrho \Vert _{L^2(\Omega )}^2 +C\Vert {\bar{\mathbf{m}}}^1\Vert _{L^\infty (\Omega )}^2\big (\Vert \delta {\bar{\varrho }} \Vert _{L^2(\Omega )}^2\Vert \varrho ^1\Vert _{L^3(\Omega )}^2+\Vert {\bar{\varrho }}^2 \Vert _{L^2(\Omega )}^2\Vert \delta \rho \Vert _{L^3(\Omega )}^2\big ). \end{aligned}$$

(5.18)

Substituting (5.14), (5.17) and (5.18) into (5.13), we obtain that there exists a positive constant $C_2=C_2(R)$ such that

$$\begin{aligned}&\frac{d}{dt}\Vert \delta \varrho \Vert _{L^2(\Omega )}^2 +\Vert \nabla \delta \varrho \Vert _{L^2(\Omega )}^2\nonumber \\&\le C_2\Big (\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}^2 \Vert \nabla \varrho ^2\Vert _{L^2(\Omega )}^2 +\Vert \delta \varrho \Vert _{L^4(\Omega )}^2{+}\big (\Vert \rho ^1\Vert _{L^3(\Omega )}^2 +\Vert \rho ^2\Vert _{L^3(\Omega )}^2\big )\Vert \delta {\bar{\varrho }}\Vert _{L^2(\Omega )}^2 \nonumber \\&\quad +\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}^2\Vert \varrho ^2\Vert _{L^2(\Omega )} +\Vert \delta \rho \Vert _{L^3(\Omega )}^2\big (1+\Vert S\Vert _{L^3(\Omega )}\big )\Big ). \end{aligned}$$

(5.19)

It follows from Sobolev’s embedding and Poincaré’s inequality that

$$\begin{aligned} \Vert \delta \rho \Vert _{L^3(\Omega )}\le&\Vert \delta \rho \Vert _{L^2(\Omega )}^{\frac{1}{2}} \Vert \delta \varrho \Vert _{L^6(\Omega )}^{\frac{1}{2}}\le C\Vert \delta \rho \Vert _{L^2(\Omega )}^{\frac{1}{2}}\Vert \delta \rho \Vert _{H^1(\Omega )}^{\frac{1}{2}}\le C\Vert \delta \rho \Vert _{L^2(\Omega )}^{\frac{1}{2}} \Vert \nabla \delta \rho \Vert _{L^2(\Omega )}^{\frac{1}{2}};\\ \Vert \delta \rho \Vert _{L^4(\Omega )}\le&\Vert \delta \rho \Vert _{L^2(\Omega )}^{\frac{1}{4}} \Vert \delta \varrho \Vert _{L^6(\Omega )}^{\frac{3}{4}}\le C\Vert \delta \rho \Vert _{L^2(\Omega )}^{\frac{1}{4}}\Vert \delta \rho \Vert _{H^1(\Omega )}^{\frac{3}{4}} \le C\Vert \delta \rho \Vert _{L^2(\Omega )}^{\frac{1}{4}}\Vert \nabla \delta \rho \Vert _{L^2(\Omega )}^{\frac{3}{4}}. \end{aligned}$$

Invoking this, we get from (5.19) and Young’s inequality that there is a positive constant C such that

$$\begin{aligned}&\frac{d}{dt}\Vert \delta \varrho \Vert _{L^2(\Omega )}^2+\Vert \nabla \delta \varrho \Vert _{L^2(\Omega )}^2\\&\quad \le C\Big (\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}^2 \Vert \nabla \varrho ^2\Vert _{L^2(\Omega )}^2\\&\qquad +\Vert \delta \rho \Vert _{L^2(\Omega )}^{\frac{1}{2}}\Vert \nabla \delta \rho \Vert _{L^2(\Omega )}^{\frac{3}{2}} +\big (\Vert \rho ^1\Vert _{L^3(\Omega )}^2+\Vert \rho ^2\Vert _{L^3(\Omega )}^2\big ) \Vert \delta {\bar{\varrho }}\Vert _{L^2(\Omega )}^2\\&\qquad +\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}^2\Vert \varrho ^2\Vert _{L^2(\Omega )} +\Vert \delta \rho \Vert _{L^2(\Omega )}\Vert \nabla \delta \rho \Vert _{L^2(\Omega )} \big (1+\Vert S\Vert _{L^3(\Omega )}\big )\Big )\\&\quad \le \frac{1}{2}\Vert \nabla \delta \rho \Vert _{L^2(\Omega )}^2+ C\Big (\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}^2 \Vert \nabla \varrho ^2\Vert _{L^2(\Omega )}^2+\big (\Vert \rho ^1\Vert _{L^3(\Omega )}^2 +\Vert \rho ^2\Vert _{L^3(\Omega )}^2\big )\Vert \delta {\bar{\varrho }}\Vert _{L^2(\Omega )}^2\\&\qquad +\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}^2\Vert \varrho ^2\Vert _{L^2(\Omega )} +\Vert \delta \rho \Vert _{L^2(\Omega )}^2\big (1+\Vert S\Vert _{L^3(\Omega )}^2\big )\Big ). \end{aligned}$$

By employing Hölder’s inequality and (5.11), it arrives at

$$\begin{aligned} \frac{d}{dt}\Vert \delta \varrho \Vert _{L^2(\Omega )}^2&\le C\Big (\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}^2\Vert \nabla \varrho ^2\Vert _{L^2(\Omega )}^2 +\big (\Vert \rho ^1\Vert _{L^3(\Omega )}^2+\Vert \rho ^2\Vert _{L^3(\Omega )}^2\big ) \Vert \delta {\bar{\varrho }}\Vert _{L^2(\Omega )}^2\nonumber \\&\quad +\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}^2\Vert \varrho ^2\Vert _{L^2(\Omega )} +\Vert \delta \rho \Vert _{L^2(\Omega )}^2\big (1+\Vert S\Vert _{L^3(\Omega )}^2\big )\Big )\nonumber \\&\le C\Big (\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}^2 \big (1+\Vert \nabla \varrho ^2\Vert _{L^2(\Omega )}^2\big ) +\Vert \delta {\bar{\varrho }}\Vert _{L^2(\Omega )}^2\nonumber \\&\quad +\Vert \delta \rho \Vert _{L^2(\Omega )}^2\big (1+\Vert S\Vert _{L^3(\Omega )}^2\big )\Big ). \end{aligned}$$

(5.20)

Here we need to estimate the factor $\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}$.

On the other hand, $\delta {\bar{\mathbf{m}}}$ satisfies the initial-boundary value problem

$$\begin{aligned} \left\{ \begin{aligned}&\partial _t\delta {\bar{\mathbf{m}}}-\Delta \delta {\bar{\mathbf{m}}} +\big (|{\bar{\mathbf{m}}}^1|^{2(\gamma -1)}{\bar{\mathbf{m}}}^1-|{\bar{\mathbf{m}}}^2|^{2(\gamma -1)} {\bar{\mathbf{m}}}^2\big )\\&=\big (({\bar{\mathbf{m}}}^1\cdot \nabla {\bar{\psi }}^1)\nabla {\bar{\psi }}^1 -({\bar{\mathbf{m}}}^2\cdot \nabla {\bar{\psi }}^2)\nabla {\bar{\psi }}^2\big ),\&(x,t)\in \Omega _T,\\&\delta {\bar{\mathbf{m}}}(x,t)=\mathbf{0},\&(x,t)\in \Sigma _T, \\&\delta {\bar{\mathbf{m}}}(x,0)=\mathbf{0},\&x\in \Omega , \end{aligned} \right. \end{aligned}$$

(5.21)

where ${\bar{\psi }}^i\in L^\infty \big ([0,T];H^1_0(\Omega )\cap H^2(\Omega )\big )$ is a solution of (5.6) and ${\bar{\mathbf{m}}}^i\in L^\infty ([0,T]\times \Omega )\cap C^{\theta ,\frac{\theta }{2}}(\Omega _T)$ is a solution of (5.8). Similar to the proof of Lemma 2.4, due to ${\bar{\mathbf{m}}}^i\in L^\infty ([0,T]\times \Omega )$, we obtain from (5.2), (5.5) and (5.7) that

$$\begin{aligned} \Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}&\le C \int _0^t(t-s)^{-\frac{n}{6}}\Vert ({\bar{\mathbf{m}}}^1\cdot \nabla {\bar{\psi }}^1)\nabla {\bar{\psi }}^1 -({\bar{\mathbf{m}}}^2\cdot \nabla {\bar{\psi }}^2) \nabla {\bar{\psi }}^2\Vert _{L^3(\Omega )}ds\\&\quad +C\int _0^t\big \Vert |{\bar{\mathbf{m}}}^1|^{2(\gamma -1)} {\bar{\mathbf{m}}}^1-|{\bar{\mathbf{m}}}^2|^{2(\gamma -1)}{\bar{\mathbf{m}}}^2 \big \Vert _{L^\infty (\Omega )}ds\\&\le C \int _0^t(t-s)^{-\frac{n}{6}}\Vert (\delta {\bar{\mathbf{m}}} \cdot \nabla {\bar{\psi }}^1)\nabla {\bar{\psi }}^1\\&\quad +({\bar{\mathbf{m}}}^2\cdot \nabla \delta {\bar{\psi }})\nabla {\bar{\psi }}^1 +({\bar{\mathbf{m}}}^2\cdot \nabla {\bar{\psi }}^2) \nabla \delta {\bar{\psi }}\Vert _{L^3(\Omega )}ds\\&\quad +C\int _0^t\big (\Vert {\bar{\mathbf{m}}}^1\Vert _{L^\infty (\Omega )}^{2(\gamma -1)} +\Vert {\bar{\mathbf{m}}}^2\Vert _{L^\infty (\Omega )}^{2(\gamma -1)}\big ) \Vert \delta {\bar{\mathbf{m}}}\big \Vert _{L^\infty (\Omega )}ds \\&\le C\int _0^t(t-s)^{-\frac{n}{6}}\Vert \delta {\bar{\mathbf{m}}} \Vert _{L^\infty (\Omega )}\Vert \nabla {\bar{\psi }}^1\Vert _{L^6(\Omega )}^2ds\\&\quad +C\int _0^t(t-s)^{-\frac{n}{6}}\Vert {\bar{\mathbf{m}}}^2 \Vert _{L^\infty (\Omega )}\Vert \nabla \delta {\bar{\psi }}\Vert _{L^6(\Omega )} \big (\Vert \nabla {\bar{\psi }}^1\Vert _{L^6(\Omega )}+\Vert \nabla {\bar{\psi }}^2\Vert _{L^6(\Omega )}\big ) ds\\&\quad +C\int _0^t\big (\Vert {\bar{\mathbf{m}}}^1\Vert _{L^\infty (\Omega )}^{2(\gamma -1)} +\Vert {\bar{\mathbf{m}}}^2\Vert _{L^\infty (\Omega )}^{2(\gamma -1)}\big ) \Vert \delta {\bar{\mathbf{m}}}\big \Vert _{L^\infty (\Omega )}ds \\&\le C\int _0^t(t-s)^{-\frac{n}{6}}\Vert \delta {\bar{\mathbf{m}}} \Vert _{L^\infty (\Omega )}\Vert {\bar{\psi }}^1\Vert _{H^2(\Omega )}^2ds \\&\quad +C\int _0^t(t-s)^{-\frac{n}{6}}\Vert {\bar{\mathbf{m}}}^2\Vert _{L^\infty (\Omega )} \Vert \delta {\bar{\psi }}\Vert _{H^2(\Omega )}\big (\Vert {\bar{\psi }}^1\Vert _{H^2(\Omega )} +\Vert {\bar{\psi }}^2\Vert _{H^2(\Omega )}\big ) ds \\&\quad +C\int _0^t\big (\Vert {\bar{\mathbf{m}}}^1\Vert _{L^\infty (\Omega )}^{2(\gamma -1)} +\Vert {\bar{\mathbf{m}}}^2\Vert _{L^\infty (\Omega )}^{2(\gamma -1)}\big ) \Vert \delta {\bar{\mathbf{m}}}\big \Vert _{L^\infty (\Omega )}ds \\&\le C\int _0^t(t-s)^{-\frac{n}{6}}\Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}ds+ CT^{1-\frac{n}{6}}\Vert \delta {\bar{\psi }}\Vert _{L^\infty ([0,T];H^2(\Omega ))}\\&\quad +C\int _0^t\Vert \delta {\bar{\mathbf{m}}}\big \Vert _{L^\infty (\Omega )}ds. \end{aligned}$$

We can pick T suitable small such that $(t-s)^{-\frac{n}{6}}\ge t^{-\frac{n}{6}}\ge T^{-\frac{n}{6}}>1$. Hence there exist two positive constant $C_4$ and $C_5$ such that

$$\begin{aligned} \Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )} \le C_4T^{1-\frac{n}{6}}\Vert \delta {\bar{\psi }}\Vert _{L^\infty ([0,T];H^2(\Omega ))} +C_5\int _0^t(t-s)^{-\frac{n}{6}}\Vert \delta {\bar{\mathbf{m}}}\big \Vert _{L^\infty (\Omega )}ds. \end{aligned}$$

(5.22)

Thus, a singular Gronwall’s inequality, see Lemma 2.3, gives

$$\begin{aligned} \Vert \delta {\bar{\mathbf{m}}}\Vert _{L^\infty (\Omega )}\le C_4T^{1-\frac{n}{6}}\Vert \delta {\bar{\psi }}\Vert _{L^\infty ([0,T];H^2(\Omega ))} \big (1+C_5C_6T^{1-\frac{n}{6}}e^{(1+\epsilon )\mu T}\big ):=M_0 \end{aligned}$$

(5.23)

for some $\epsilon >0$ and $C_6>0$, where $\mu =\big (C_5\Gamma (1-\frac{n}{6})\big )^{\frac{6}{6-n}}$.

Then returning to (5.20), we conclude from (5.10) and (5.23) that

$$\begin{aligned} \Vert \delta \varrho \Vert _{L^2(\Omega )}^2&\le CM_0^2\int _0^t\Big (1+\Vert \nabla \varrho ^2\Vert _{L^2(\Omega )}^2\Big )ds\nonumber \\&\quad +C\int _0^t\Big (\big (1+\Vert S\Vert _{L^3(\Omega )}^2\big ) \Vert \delta \varrho \Vert _{L^2(\Omega )}^2+\Vert \delta {\bar{\varrho }}\Vert _{L^2(\Omega )}^2\Big )ds \nonumber \\&\le CM_0^2(1+T)+C\Vert \delta {\bar{\varrho }}\Vert _{L^\infty ([0,T]; L^2(\Omega ))}^2T\nonumber \\&\quad +C\big (1+\Vert S\Vert _{L^3(\Omega )}^2\big )\int _0^t \Vert \delta \varrho \Vert _{L^2(\Omega )}^2ds. \end{aligned}$$

(5.24)

Thus, Gronwall’s inequality gives that there are two positive constants $C_7$ and $C_8$ such that

$$\begin{aligned} \Vert \delta \varrho \Vert _{L^2(\Omega )}^2\le C_7 \Big (M_0^2(1+T)+\Vert \delta {\bar{\varrho }}\Vert _{L^\infty ([0,T];L^2(\Omega ))}^2T\Big ) e^{C_8\big (1+\Vert S\Vert _{L^3(\Omega )}^2\big )T}. \end{aligned}$$

(5.25)

Moreover, we can infer from (5.16), (5.23) and (5.25) that there exists a positive constant $C_9$ such that

$$\begin{aligned} \Vert \delta \varrho \Vert _{L^2(\Omega )}^2&\le C_9\Big [T^{2-\frac{n}{3}}\big (1+T^{1-\frac{n}{6}}e^{(1+\epsilon )\mu T}\big )^2(1+T)+T\Big ]\nonumber \\&\qquad e^{C_8\big (1+\Vert S\Vert _{L^3(\Omega )}^2\big )T} \Vert \delta {\bar{\varrho }}\Vert _{L^\infty ([0,T];L^2(\Omega ))}^2. \end{aligned}$$

(5.26)

We can pick T suitable small such that

$$\begin{aligned} \Big [T^{2-\frac{n}{3}}\big (1+C_8T^{1-\frac{n}{6}}e^{(1+\epsilon )\mu T}\big )^2(1+T)+T\Big ]e^{C\big (1+\Vert S\Vert _{L^3(\Omega )}^2\big )T}<\frac{1}{C_9}, \end{aligned}$$

and, thus, get from (5.26) that there exists $\kappa \in (0,1)$ such that

$$\begin{aligned} \Vert \delta \varrho \Vert _{L^2(\Omega )}\le \kappa \Vert \delta {\bar{\varrho }}\Vert _{L^\infty ([0,T];L^2(\Omega ))}, \end{aligned}$$

which implies that $\Psi$ is a contraction mapping from $E_R$ to $E_R$ and thus has a fixed point $\varrho$ in $E_R$. Hence, $(\varrho ,\psi ,\mathbf{m})$ is a solution to the initial-boundary value problem (1.1)–(1.2) on [0, T].

To show the higher regularity, we notice from elliptic equation $-\Delta \psi =\varrho$ with homogeneous Dirichlet boundary condition that $\psi \in L^\infty ((0,T);W^{2,p}(\Omega ))$ for any $p>1$ by using the inequality (5.11) and the $L^p$ regularity theory [10, 23]. Next, by applying [22, Theorem 3.10.1] to the Dirichlet problem associated with the linear parabolic equation

$$\begin{aligned} \varrho _t=\partial _i(a_{ij}\varrho _{x_i}+b_i\varrho )+S\varrho , \end{aligned}$$

we get that there exists $\theta _1\in (0,1)$ such that $\varrho \in C^{\theta _1,\frac{\theta _1}{2}}(\Omega _T)$, due to $\delta _{ij}+m_im_j=a_{ij}\in C^{\theta ,\frac{\theta }{2}}(\Omega _T)$ and $a_{ij}\partial _i\psi =b_i\in L^\infty (\Omega _T)$. Hence, using the Schauder theory [10, 23] for the elliptic equation we get $\psi \in C^{2+\theta _1,\frac{\theta _1}{2}}(\Omega _T)$, which also implies from Lemma 2.4 that $\mathbf{m}\in C^{2+\theta _2,1+\frac{\theta _2}{2}}(\Omega _T)$ for some $\theta _2\in (0,1)$ by utilizing the standard Schauder theory for the parabolic equation. In the sequel, by employing the standard Schauder theory for parabolic equation again, we infer that $\varrho \in C^{2+\theta _3,1+\frac{\theta _3}{2}}(\Omega _T)$ for some $\theta _3\in (0,1)$, due to $\mathbf{m}\in C^{2+\theta _2,1+\frac{\theta _2}{2}}(\Omega _T)$ and $\psi \in C^{2+\theta _1,\frac{\theta _1}{2}}(\Omega _T)$, provided that $S\in C^\alpha (\Omega )$ with $\alpha \in (0,1)$, which leads to $\psi \in C^{2+\theta _3,1+\frac{\theta _3}{2}}(\Omega _T)$. This implies the assertions in Lemma 3.1. $\square$