Vanishing heat conductivity limit for the 2D Cahn-Hilliard-Boussinesq system

Abstract

This article studies the vanishing heat conductivity limit for the 2D Cahn-Hilliard-boussinesq system in a bounded domain with non-slip boundary condition. The result has been proved globally in time.

2010 MSC: 35Q30; 76D03; 76D05; 76D07.

1 Introduction

Let Ω ⊆ ℝ² be a bounded, simply connected domain with smooth boundary ∂Ω, and n is the unit outward normal vector to ∂Ω. We consider the following Cahn-Hilliard-Boussinesq system in Ω × (0, ∞) [1]:

$${\partial }_{t}u+\left(u\cdot \nabla \right)u+\nabla \pi -\Delta u=\mu \nabla \varphi +\theta e_2,$$

(1.1)

$$\mathsf{\text{div}}u=0,$$

(1.2)

$${\partial }_t\theta +u\cdot \nabla \theta =\epsilon \Delta \theta ,$$

(1.3)

$${\partial }_t\varphi +u\cdot \nabla \varphi =\Delta \mu ,$$

(1.4)

$$-\Delta \varphi +f^{\prime }\left(\varphi \right)=\mu ,$$

(1.5)

$$u=0,\theta =0,\frac{\partial \varphi }{\partial n}=\frac{\partial \mu }{\partial n}=0on\partial \Omega \times \left(0,\infty \right),$$

(1.6)

$$\left(u,\theta ,\varphi \right)\left(x,0\right)=\left(u_0,{\theta }_0,{\varphi }_0\right)\left(x\right),x\in \Omega ,$$

(1.7)

where u, π, θ and ϕ denote unknown velocity field, pressure scalar, temperature of the fluid and the order parameter, respectively. ε > 0 is the heat conductivity coefficient and e₂ : = (0, 1) ^t . μ is a chemical potential and $f\left(\varphi \right):=\frac{1}{4}{\left({\varphi }^2-1\right)}^2$ is the double well potential.

When ϕ = 0, (1.1), (1.2) and (1.3) is the well-known Boussinesq system. In [2] Zhou and Fan proved a regularity criterion $\omega \dot{=}curlu\in L^1\left(0,T;{Ḃ}_{\infty ,\infty }^0\right)$ for the 3D Boussinesq system with partial viscosity. Later, in [3] Zhou and Fan studied the Cauchy problem of certain Boussinesq-α equations in n dimensions with n = 2 or 3. We establish regularity for the solution under $\nabla u\in L^1\left(0,T;{Ḃ}_{\infty ,\infty }^0\right)$. Here ${Ḃ}_{\infty ,\infty }^0$ denotes the homogeneous Besov space. Chae [4] studied the vanishing viscosity limit ε → 0 when Ω = ℝ². The aim of this article is to prove a similar result. We will prove that

Theorem 1.1. Let $\left(u_0,{\theta }_0\right)\in H_0^1\cap H^2$, ϕ₀ ∈ H⁴, div u₀ = 0 in Ω and $\frac{\partial {\varphi }_0}{\partial n}=\frac{\partial {\mu }_0}{\partial n}=0$ on∂Ω. Then, there exists a positive constant C independent of ε such that

$$\begin{array}{c}\parallel u_{\epsilon }{\parallel }_{L^{\infty }\left(0,T;H^2\right)}\le C,\parallel {\theta }_{\epsilon }{\parallel }_{L^{\infty }\left(0,T;H^2\right)}\le C,\\ \parallel {\varphi }_{\epsilon }{\parallel }_{L^{\infty }\left(0,T;H^4\right)}\le C,\parallel {\partial }_t\left(u_{ϵ},{\theta }_{\epsilon },{\varphi }_{ϵ}\right){\parallel }_{L^2\left(0,T;L^2\right)}\le C,\end{array}$$

(1.8)

for any T > 0, which implies

$$\left(u_{\epsilon },{\theta }_{\epsilon },{\varphi }_{\epsilon }\right)\to \left(u,\theta ,\varphi \right)stronglyin{L}^2\left(0,T;H^1\right)when\epsilon \to 0.$$

(1.9)

Here, (u, θ, ϕ) is the solution of the problem (1.1)-(1.7) with ε = 0.

2 Proof of Theorem 1.1

Since (1.9) follows easily from (1.8) by the Aubin-Lions compactness principle, we only need to prove the a priori estimates (1.8). From now on, we will drop the subscript ε and throughout this section C will be a constant independent of ε.

First, by the maximum principle, it follows from (1.2), (1.3), and (1.6) that

$$\parallel \theta {\parallel }_{L^{\infty }\left(0,T;L^{\infty }\right)}\le \parallel {\theta }_0{\parallel }_{L^{\infty }}\le C.$$

(2.1)

Testing (1.3) by θ, using (1.2) and (1.6), we see that

$$\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int {\theta }^2\mathsf{\text{d}}x+\epsilon \int \mid \nabla \theta {\mid }^2\mathsf{\text{d}}x=0,$$

whence

$$\sqrt{\epsilon }\parallel \theta {\parallel }_{L^2\left(0,T;H^1\right)}\le C.$$

(2.2)

Testing (1.1) and (1.4) by u and μ, respectively, using (1.2), (1.6), (2.1), and summing up the result, we find that

$$\begin{array}{c}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int \frac{1}{2}{u}^2+\frac{1}{2}\mid \nabla \varphi {\mid }^2+f\left(\varphi \right)\mathsf{\text{d}}x+\int \mid \nabla u{\mid }^2+\mid \nabla \mu {\mid }^2\mathsf{\text{d}}x\\ =\int \theta e_{2}u\mathsf{\text{d}}x\le \parallel \theta {\parallel }_{L^2}\parallel u{\parallel }_{L^2}\le C\parallel u{\parallel }_{L^2},\end{array}$$

which gives

$$\parallel \varphi {\parallel }_{L^{\infty }\left(0,T;H^1\right)}\le C,$$

(2.3)

$$\parallel u{\parallel }_{L^{\infty }\left(0,T;L^2\right)}+\parallel u{\parallel }_{L^2\left(0,T;H^1\right)}\le C,$$

(2.4)

$$\parallel \nabla \mu {\parallel }_{L^2\left(0,T;L^2\right)}\le C.$$

(2.5)

Testing (1.4) by ϕ, using (1.2), (1.5) and (1.6), we infer that

$$\begin{array}{c}\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int {\varphi }^2\mathsf{\text{d}}x+\int \mid \Delta \varphi {\mid }^2\mathsf{\text{d}}x=\int \left({\varphi }^3-\varphi \right)\Delta \varphi \mathsf{\text{d}}x\\ =-3\int {\varphi }^2\mid \nabla \varphi {\mid }^2\mathsf{\text{d}}x-\int \varphi \Delta \varphi \mathsf{\text{d}}x\le -\int \varphi \Delta \varphi \mathsf{\text{d}}x\\ \le \frac{1}{2}\int \mid \Delta \varphi {\mid }^2\mathsf{\text{d}}x+\frac{1}{2}\int {\varphi }^2\mathsf{\text{d}}x,\end{array}$$

which leads to

$$\parallel \varphi {\parallel }_{L^2\left(0,T;H^2\right)}\le C.$$

(2.6)

We will use the following Gagliardo-Nirenberg inequality:

$$\parallel \varphi {\parallel }_{L^{\infty }}^2\le C\parallel \varphi {\parallel }_{L^6}\parallel \varphi {\parallel }_{H^2}.$$

(2.7)

It follows from (2.6), (2.7), (2.5), (2.3) and (1.5) that

$$\begin{array}{c}{\int }_0^T\int \mid \nabla \Delta \varphi {\mid }^2\mathsf{\text{d}}x\mathsf{\text{d}}t\\ ={\int }_0^T\int \mid \nabla \left(f^{\prime }\left(\varphi \right)-\mu \right){\mid }^2\mathsf{\text{d}}x\mathsf{\text{d}}t\\ \le C{\int }_0^T\int \mid \nabla \mu {\mid }^2\mathsf{\text{d}}x\mathsf{\text{d}}t+C{\int }_0^T\int \mid \nabla \left({\varphi }^3-\varphi \right){\mid }^2\mathsf{\text{d}}x\mathsf{\text{d}}t\\ \le C+C{\int }_0^T\int {\varphi }^4\mid \nabla \varphi {\mid }^2\mathsf{\text{d}}x\mathsf{\text{d}}t\\ \le C+C\parallel \nabla \varphi {\parallel }_{L^{\infty }\left(0,T;L^2\right)}^2{\int }_0^T\parallel \varphi {\parallel }_{L^{\infty }}^4\mathsf{\text{d}}t\\ \le C+C{\int }_0^T\parallel \varphi {\parallel }_{L^6}^2\parallel \varphi {\parallel }_{H^2}^2\mathsf{\text{d}}t\\ \le C+C\parallel \varphi {\parallel }_{L^{\infty }\left(0,T;H^1\right)}^2\underset{0}{\overset{T}{\int }}\parallel \varphi {\parallel }_{H^2}^2\mathsf{\text{d}}t\le C,\end{array}$$

(2.8)

which yields

$$\parallel \varphi {\parallel }_{L^2\left(0,T;H^3\right)}\le C,$$

(2.9)

$$\parallel \varphi {\parallel }_{L^4\left(0,T;L^{\infty }\right)}\le C,$$

(2.10)

$$\parallel \nabla \varphi {\parallel }_{L^2\left(0,T;L^{\infty }\right)}\le C.$$

(2.11)

Testing (1.4) by Δ²ϕ, using (1.5), (2.4), (2.3), (2.10) and (2.11), we derive

$$\begin{array}{c}\frac{1}{2}\frac{d}{\mathsf{\text{d}}t}\int \mid \Delta \varphi {\mid }^2\mathsf{\text{d}}x+\int \mid {\Delta }^2\varphi {\mid }^2\mathsf{\text{d}}x\\ =-\int u\cdot \nabla \varphi \cdot {\Delta }^2\varphi \mathsf{\text{d}}x+\int \Delta \left({\varphi }^3-\varphi \right)\cdot {\Delta }^2\varphi \mathsf{\text{d}}x\\ \le \parallel u{\parallel }_{L^2}\parallel \nabla \varphi {\parallel }_{L^{\infty }}\parallel {\Delta }^2\varphi {\parallel }_{L^2}+\parallel \Delta \left({\varphi }^3-\varphi \right){\parallel }_{L^2}\parallel {\Delta }^2\varphi {\parallel }_{L^2}\\ \le C\parallel \nabla \varphi {\parallel }_{L^{\infty }}\parallel {\Delta }^2\varphi {\parallel }_{L^2}\\ +C\left(\parallel \varphi {\parallel }_{L^{\infty }}^2\parallel \Delta \varphi {\parallel }_{L^2}+\parallel \varphi {\parallel }_{L^{\infty }}\parallel \nabla \varphi {\parallel }_{L^{\infty }}\parallel \nabla \varphi {\parallel }_{L^2}+\parallel \Delta \varphi {\parallel }_{L^2}\right)\parallel {\Delta }^2\varphi {\parallel }_{L^2}\\ \le C\parallel \nabla \varphi {\parallel }_{L^{\infty }}\parallel {\Delta }^2\varphi {\parallel }_{L^2}\\ +C\left(\parallel \varphi {\parallel }_{L^{\infty }}^2\parallel \Delta \varphi {\parallel }_{L^2}+\parallel \varphi {\parallel }_{H^2}\parallel \nabla \varphi {\parallel }_{L^{\infty }}+\parallel \Delta \varphi {\parallel }_{L^2}\right)\parallel {\Delta }^2\varphi {\parallel }_{L^2}\\ \le \frac{1}{2}\parallel {\Delta }^2\varphi {\parallel }_{L^2}^2+C\parallel \nabla \varphi {\parallel }_{L^{\infty }}^2+C\parallel \varphi {\parallel }_{L^{\infty }}^4\parallel \Delta \varphi {\parallel }_{L^2}^2\\ +C\parallel \nabla \varphi {\parallel }_{L^{\infty }}^2\parallel \varphi {\parallel }_{H^2}^2+C\parallel \Delta \varphi {\parallel }_{L^2}^2,\end{array}$$

which implies

$$\parallel \varphi {\parallel }_{L^{\infty }\left(0,T;H^2\right)}+\parallel \varphi {\parallel }_{L^2\left(0,T;H^4\right)}\le C.$$

(2.12)

Testing (1.1) by -Δu + ∇π, using (1.2), (1.6), (2.12), (2.1) and (2.4), we reach

$$\begin{array}{c}\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int \mid \nabla u{\mid }^2\mathsf{\text{d}}x+\int {\left(-\Delta u+\nabla \pi \right)}^2\mathsf{\text{d}}x\\ =\int \left(\mu \nabla \varphi +\theta e_2-u\cdot \nabla u\right)\left(-\Delta u+\nabla \pi \right)\mathsf{\text{d}}x\\ \le \left(\parallel \mu {\parallel }_{L^2}\parallel \nabla \varphi {\parallel }_{L^{\infty }}+\parallel \theta {\parallel }_{L^2}+\parallel u{\parallel }_{L^4}\parallel \nabla u{\parallel }_{L^4}\right)\parallel -\Delta u+\nabla \pi {\parallel }_{L^2}\\ \le C\left(\parallel \nabla \varphi {\parallel }_{L^{\infty }}+1+\parallel u{\parallel }_{L^2}^{1∕2}\parallel \nabla u{\parallel }_{L^2}^{1∕2}\cdot \parallel \nabla u{\parallel }_{L^2}^{1∕2}\parallel \Delta u{\parallel }_{L^2}^{1∕2}\right)\parallel -\Delta u+\nabla \pi {\parallel }_{L^2}\\ \le C\parallel \nabla \varphi {\parallel }_{L^{\infty }}^2+C+C\parallel \nabla u{\parallel }_{L^2}^4+\frac{1}{2}\parallel -\Delta u+\nabla \pi {\parallel }_{L^2}^2,\end{array}$$

which yields

$$\parallel u{\parallel }_{L^{\infty }\left(0,T;H^1\right)}+\parallel u{\parallel }_{L^2\left(0,T;H^2\right)}\le C.$$

(2.13)

Here, we have used the Gagliardo-Nirenberg inequalities:

$$\begin{array}{c}\parallel u{\parallel }_{L^4}^2\le C\parallel u{\parallel }_{L^2}\parallel \nabla u{\parallel }_{L^2},\\ \parallel \nabla u{\parallel }_{L^4}^2\le C\parallel \nabla u{\parallel }_{L^2}\parallel u{\parallel }_{H^2},\end{array}$$

and the H²-theory of the Stokes system:

$$\parallel u{\parallel }_{H^2}+\parallel \pi {\parallel }_{H^1}\le C\parallel -\Delta u+\nabla \pi {\parallel }_{L^2}.$$

(2.14)

Similarly to (2.13), we have

$$\parallel {\partial }_{t}u{\parallel }_{L^2\left(0,T;L^2\right)}\le C.$$

(2.15)

(1.1), (1.2), (1.6) and (1.7) can be rewritten as

$$\left\{\begin{array}{c}{\partial }_{t}u-\Delta u+\nabla \pi =g:=\mu \nabla \varphi +\theta e_2-u\cdot \nabla u,in\Omega \times \left(0,\infty \right),\\ u=0,on\partial \Omega \times \left(0,\infty \right),\\ u\mathsf{\text{(}}x\mathsf{\text{, 0)}}\mathsf{\text{ = }}{u}_0\mathsf{\text{(}}x\mathsf{\text{)}}\mathsf{\text{.}}\end{array}\right.$$

Using (2.12), (2.1), (2.13), and the regularity theory of Stokes system, we have

$$\begin{array}{c}\parallel {\partial }_{t}u{\parallel }_{L^2\left(0,T;L^p\right)}+\parallel u{\parallel }_{L^2\left(0,T;W^{2,p}\right)}\le C\parallel g{\parallel }_{L^2\left(0,T;L^p\right)}\\ \le C\parallel \mu {\parallel }_{L^2\left(0,T;L^{\infty }\right)}\parallel \nabla \varphi {\parallel }_{L^{\infty }\left(0,T;L^p\right)}+C\parallel \theta {\parallel }_{L^{\infty }\left(0,T;L^{\infty }\right)}\\ +C\parallel u{\parallel }_{L^{\infty }\left(0,T;L^{2p}\right)}\parallel \nabla u{\parallel }_{L^2\left(0,T;L^{2p}\right)}\le C,\end{array}$$

(2.16)

for any 2 < p < ∞.

(2.16) gives

$$\parallel \nabla u{\parallel }_{L^2\left(0,T;L^{\infty }\right)}\le C.$$

(2.17)

It follows from (1.3) and (1.6) that

$$\Delta \theta =0on\partial \Omega \times \left(0,\infty \right).$$

(2.18)

Applying Δ to (1.3), testing by Δθ, using (1.2), (1.6), (2.16), (2.17) and (2.18), we obtain

$$\begin{array}{c}\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int \mid \Delta \theta {\mid }^2\mathsf{\text{d}}x+\epsilon \int \mid \nabla \Delta \theta {\mid }^2\mathsf{\text{d}}x\\ =-\int \left(\Delta \left(u\cdot \nabla \theta \right)-u\nabla \Delta \theta \right)\Delta \theta \mathsf{\text{d}}x\\ \le C\left(\parallel \Delta u{\parallel }_{L^4}\parallel \nabla \theta {\parallel }_{L^4}+\parallel \nabla u{\parallel }_{L^{\infty }}\parallel \Delta \theta {\parallel }_{L^2}\right)\parallel \Delta \theta {\parallel }_{L^2}\\ \le C\left(\parallel \Delta u{\parallel }_{L^4}+\parallel \nabla u{\parallel }_{L^{\infty }}\right)\parallel \Delta \theta {\parallel }_{L^2}^2,\end{array}$$

which implies

$$\parallel \theta {\parallel }_{L^{\infty }\left(0,T;H^2\right)}+\sqrt{\epsilon }\parallel \theta {\parallel }_{L^2\left(0,T;H^3\right)}\le C.$$

(2.19)

It follows from (1.3), (1.6), (2.19) and (2.13) that

$$\parallel {\partial }_t\theta {\parallel }_{L^{\infty }\left(0,T;L^2\right)}\le C.$$

(2.20)

Taking ∂ _t to (1.4) and (1.5), testing by ∂ _t ϕ, using (1.2), (1.6), (2.12), and (2.15), we have

$$\begin{array}{c}\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int \mid {\partial }_t\varphi {\mid }^2\mathsf{\text{d}}x+\int \mid \Delta {\partial }_t\varphi {\mid }^2\mathsf{\text{d}}x\\ =-\int {\partial }_{t}u\cdot \nabla \varphi \cdot {\partial }_t\varphi \mathsf{\text{d}}x+\int \Delta \left(3{\varphi }^2{\partial }_t\varphi -{\partial }_t\varphi \right)\cdot {\partial }_t\varphi \mathsf{\text{d}}x\\ =-\int {\partial }_{t}u\cdot \nabla \varphi \cdot {\partial }_t\varphi \mathsf{\text{d}}x+\int \left(3{\varphi }^2{\partial }_t\varphi -{\partial }_t\varphi \right)\Delta {\partial }_t\varphi \mathsf{\text{d}}x\\ \le \parallel {\partial }_{t}u{\parallel }_{L^2}\parallel \nabla \varphi {\parallel }_{L^{\infty }}\parallel {\partial }_t\varphi {\parallel }_{L^2}+\left(\parallel 3\varphi {\parallel }_{L^{\infty }}^2+1\right)\parallel {\partial }_t\varphi {\parallel }_{L^2}\parallel \Delta {\partial }_t\varphi {\parallel }_{L^2}\\ \le \parallel {\partial }_{t}u{\parallel }_{L^2}\parallel \nabla \varphi {\parallel }_{L^{\infty }}\parallel {\partial }_t\varphi {\parallel }_{L^2}+\frac{1}{2}\parallel \Delta {\partial }_t\varphi {\parallel }_{L^2}^2+C\parallel {\partial }_t\varphi {\parallel }_{L^2}^2,\end{array}$$

which gives

$$\parallel {\partial }_t\varphi {\parallel }_{L^{\infty }\left(0,T;L^2\right)}+\parallel {\partial }_t\varphi {\parallel }_{L^2\left(0,T;H^2\right)}\le C.$$

(2.21)

By the regularity theory of elliptic equation, it follows from (1.4), (1.5), (1.6), (2.21), (2.13) and (2.12) that

$$\begin{array}{cc}\hfill \parallel \varphi {\parallel }_{L^{\infty }\left(0,T;H^4\right)}& \le C\parallel \Delta \varphi {\parallel }_{L^{\infty }\left(0,T;H^2\right)}\le C\parallel \mu -f^{\prime }\left(\varphi \right){\parallel }_{L^{\infty }\left(0,T;H^2\right)}\hfill \\ \le C\parallel \mu {\parallel }_{L^{\infty }\left(0,T;H^2\right)}+C\parallel f^{\prime }\left(\varphi \right){\parallel }_{L^{\infty }\left(0,T;H^2\right)}\hfill \\ \le C\parallel \Delta \mu {\parallel }_{L^{\infty }\left(0,T;L^2\right)}+C\parallel f^{\prime }\left(\varphi \right){\parallel }_{L^{\infty }\left(0,T;H^2\right)}\hfill \\ \le C\parallel {\partial }_t\varphi +u\cdot \nabla \varphi {\parallel }_{L^{\infty }\left(0,T;L^2\right)}+C\parallel f^{\prime }\left(\varphi \right){\parallel }_{L^{\infty }\left(0,T;H^2\right)}\hfill \\ \le C\parallel {\partial }_t\varphi {\parallel }_{L^{\infty }\left(0,T;L^2\right)}+C\parallel u{\parallel }_{L^{\infty }\left(0,T;L^4\right)}\parallel \nabla \varphi {\parallel }_{L^{\infty }\left(0,T;L^4\right)}\hfill \\ +C\parallel f^{\prime }\left(\varphi \right){\parallel }_{L^{\infty }\left(0,T;H^2\right)}\le C.\hfill \end{array}$$

(2.22)

Taking ∂ _t to (1.1), testing by ∂ _t u, using (1.2), (1.6), (2.17), (2.22), (2.21) and (1.5), we conclude that

$$\begin{array}{c}\frac{1}{2}\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\int \mid {\partial }_{t}u{\mid }^2\mathsf{\text{d}}x+\int \mid \nabla {\partial }_{t}u{\mid }^2\mathsf{\text{d}}x\\ =-\int {\partial }_{t}u\cdot \nabla u\cdot {\partial }_{t}u\mathsf{\text{d}}x+\int \left({\partial }_t\mu \cdot \nabla \varphi +\mu \cdot \nabla {\partial }_t\varphi +{\partial }_t\theta e_2\right){\partial }_{t}u\mathsf{\text{d}}x\\ \le \parallel \nabla u{\parallel }_{L^{\infty }}\parallel {\partial }_{t}u{\parallel }_{L^2}^2+\left(\parallel {\partial }_{t}u{\parallel }_{L^2}\parallel \nabla \varphi {\parallel }_{L^{\infty }}+\parallel \mu {\parallel }_{L^{\infty }}\parallel \nabla {\partial }_t\varphi {\parallel }_{L^2}+\parallel {\partial }_t\theta {\parallel }_{L^2}\right)\parallel {\partial }_{t}u{\parallel }_{L^2}\\ \le \parallel \nabla u{\parallel }_{L^{\infty }}\parallel {\partial }_{t}u{\parallel }_{L^2}^2+C\left(\parallel \Delta {\partial }_t\varphi {\parallel }_{L^2}+\parallel {\partial }_t\left({\varphi }^3-\varphi \right){\parallel }_{L^2}+\parallel \nabla {\partial }_t\varphi {\parallel }_{L^2}+1\right)\parallel {\partial }_{t}u{\parallel }_{L^2},\end{array}$$

which implies

$$\parallel {\partial }_{t}u{\parallel }_{L^{\infty }\left(0,T;L^2\right)}+\parallel {\partial }_{t}u{\parallel }_{L^2\left(0,T;H^1\right)}\le C.$$

(2.23)

Using (2.23), (2.22), (2.1), (2.13), (1.1), (1.2), (1.6) and the H²-theory of the Stokes system, we arrive at

$$\parallel u{\parallel }_{L^{\infty }\left(0,T;H^2\right)}\le C.$$

This completes the proof.