Accurate runoff forecasting plays a vital role in water resource management. Therefore, various forecasting models have been proposed in the literature. Among them, the decomposition-based models have proved their superiority in runoff series forecasting. However, most of the models simulate each decomposition sub-signals separately without considering the potential correlation information. A neoteric hybrid runoff forecasting model based on variational mode decomposition (VMD), convolution neural networks (CNN), and long short-term memory (LSTM) called VMD-CNN-LSTM, is proposed to improve the runoff forecasting performance further. The two-dimensional matrix containing both the time delay and correlation information among sub-signals decomposing by VMD is firstly applied to the CNN. The feature of the input matrix is then extracted by CNN and delivered to LSTM with more potential information. The experiment performed on monthly runoff data investigated from Huaxian and Xianyang hydrological stations at Wei River, China, demonstrates the VMD-superiority of CNN-LSTM to the baseline models, and robustness and stability of the forecasting of the VMD-CNN-LSTM for different leading times.
Citation: Xin Jing, Jungang Luo, Shangyao Zhang, Na Wei. Runoff forecasting model based on variational mode decomposition and artificial neural networks[J]. Mathematical Biosciences and Engineering, 2022, 19(2): 1633-1648. doi: 10.3934/mbe.2022076
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Accurate runoff forecasting plays a vital role in water resource management. Therefore, various forecasting models have been proposed in the literature. Among them, the decomposition-based models have proved their superiority in runoff series forecasting. However, most of the models simulate each decomposition sub-signals separately without considering the potential correlation information. A neoteric hybrid runoff forecasting model based on variational mode decomposition (VMD), convolution neural networks (CNN), and long short-term memory (LSTM) called VMD-CNN-LSTM, is proposed to improve the runoff forecasting performance further. The two-dimensional matrix containing both the time delay and correlation information among sub-signals decomposing by VMD is firstly applied to the CNN. The feature of the input matrix is then extracted by CNN and delivered to LSTM with more potential information. The experiment performed on monthly runoff data investigated from Huaxian and Xianyang hydrological stations at Wei River, China, demonstrates the VMD-superiority of CNN-LSTM to the baseline models, and robustness and stability of the forecasting of the VMD-CNN-LSTM for different leading times.
In this paper, we study the coupled chemotaxis-fluid models with the initial-bounary conditions
{nt+u⋅∇n=Δn−∇⋅(n∇c)+γn−μn2,in Q≡(0,T)×Ω,ct+u⋅∇c=Δc−c+n+f,in Q,ut+u⋅∇u=Δu−∇π+n∇φ,in Q,∇⋅u=0,in Q,∂n∂ν=∂c∂ν=0,u=0,on (0,T)×∂Ω,n(x,0)=n0(x),c(x,0)=c0(x),u(x,0)=u0(x),in Ω, | (1.1) |
where
In order to understand the development of system (1.1), let us mention some previous contributions in this direction. Jin [11] dealed with the time periodic problem of (1.1) in spatial dimension
Espejo and Suzuki [6] discussed the chemotaxis-fluid model
nt+u⋅∇n=Δn−∇⋅(n∇c)+n(γ−μn), | (1.2) |
ct+u⋅∇c=Δc−c+n, | (1.3) |
ut=Δu−∇π+n∇φ, | (1.4) |
∇⋅u=0, | (1.5) |
∂n∂ν=∂c∂ν=0,u=0. | (1.6) |
They proved the global existence of weak solution. Tao and Winkler [17] proved the existence of global classical solution and the uniform boundedness. Tao and Winkler [18] also obtained the global classical solution and uniform boundedness under the condition of
The optimal control problems governed by the coupled partial differential equations is important. Colli et al. [4] studied the distributed control problem for a phase-field system of conserved type with a possibly singular potential. Liu and Zhang [14] considered the optimal control of a new mechanochemical model with state constraint. Chen et al. [3] studied the distributed optimal control problem for the coupled Allen-Cahn/Cahn-Hilliard equations. Recently, Guillén-González et al. [9] studied a bilinear optimal control problem for the chemo-repulsion model with the linear production term. The existence, uniqueness and regularity of strong solutions of this model are deduced. They also derived the first-order optimality conditions by using a Lagrange multipliers theorem. Frigeri et al. [8] studied an optimal control problem for two-dimensional nonlocal Cahn-Hilliard-Navier-Stokes systems with degenerate mobility and singular potential. Some other results can be found in [2,5,13,15,19].
In this paper, we discuss the optimal control problem for (1.1). We adjust the external source
In this section, we will construct the existence and some priori estimates of the linearized problem for the chemotaxis-Navier-Stokes system in a bounded domain
In the following lemmas we will state the Gagliardo-Nirenberg interpolation inequality [7].
Lemma 2.1. Let
1p−lN=a(1q−kN)+(1−a)1r. | (2.1) |
Then, for any
‖Dlu‖Lp⩽c1‖Dku‖aLq‖u‖1−aLr+c2‖u‖Lr |
with the following exception: If
The following log-interpolation inequality has been proved by [1].
Lemma 2.2. Let
‖u‖3L3(Ω)≤δ‖u‖2H1(Ω)‖(u+1)log(u+1)‖L1(Ω)+p(δ−1)‖u‖L1(Ω), |
where
We first consider the existence of solutions to the linear problem of system (1.1). Assume functions
{ut−Δu+ˆu⋅∇u=−∇π+ˆn∇φ,in Q,∇⋅u=0,in Q,u=0,on ∂Ω,u(x,0)=u0(x),in Ω. | (2.2) |
By using fixed point method, the existence of solutions can be easily obtained. Therefore, we ignore the process of proof and just give the regularity estimate.
Lemma 2.3. Let
Proof. Multiplying the first equation of (2.2) by
12ddt∫Ωu2dx+∫Ω|∇u|2dx+∫Ωu2dx=∫Ωˆn∇φ⋅udx+∫Ωu2dx≤‖ˆn‖L2‖u‖L2+‖u‖2L2≤C(‖ˆn‖2L2+‖u‖2L2). |
By Gronwall's inequality, we have
‖u‖2L2+∫T0‖u‖2H1dτ≤C(∫T0‖ˆn‖2L2dτ+‖u0‖2L2). |
Operating the Helmholtz projection operator
ut+Au+P(ˆu⋅∇u)=P(ˆn∇φ), |
where
12ddt∫Ω|∇u|2dx+∫Ω|Δu|2dx+∫Ω|∇u|2dx=∫ΩP(ˆu∇u)Δudx−∫ΩP(ˆn∇φ)Δudx+∫Ω|∇u|2dx. |
For the terms on the right, we have
∫ΩP(ˆu∇u)Δudx−∫ΩP(ˆn∇φ)Δudx+∫Ω|∇u|2dx≤‖ˆu‖L4‖∇u‖L4‖Δu‖L2+‖ˆn‖L2‖Δu‖L2+‖∇u‖2L2≤‖ˆu‖L4‖∇u‖1/2L2‖Δu‖3/2L2+‖ˆu‖L4‖∇u‖L2‖Δu‖L2+‖ˆn‖L2‖Δu‖L2+‖∇u‖2L2≤12‖Δu‖2L2+C(‖ˆu‖4L4+‖ˆu‖2L4+1)‖∇u‖2L2+‖ˆn‖2L2. |
Therefore, we get
ddt‖∇u‖2L2+‖∇u‖2H1≤C(‖ˆu‖4L4+‖ˆu‖2L4+1)‖∇u‖2L2+C‖ˆn‖2L2+C. |
By Gronwall's inequality, we derive
‖∇u‖2L2+∫T0‖∇u‖2H1dτ≤C. |
Multiplying the first equation of (2.2) by
∫T0∫Ω|ut|2dxdt≤C. |
Summing up, we complete the proof.
For the above solution
{ct−Δc+u⋅∇c+c=ˆn++f,in Q,∂c∂ν=0,on (0,T)×∂Ω,c(x,0)=c0(x),in Ω. | (2.3) |
Along with fixed point method, the existence of solutions can be easily obtained. Thus we omit the proof and only give the regularity estimate.
Lemma 2.4. Let
Proof. Multiplying the first equation of (2.3) by
12ddt∫Ωc2dx+∫Ω|∇c|2dx+∫Ωc2dx≤‖ˆn‖L2‖c‖L2+‖f‖L2‖c‖L2. |
Therefore, we have
‖c‖2L2+‖c‖2H1≤C(‖c0‖2L2+∫t0(‖ˆn‖2L2+‖f‖2L2)dτ). |
Multiplying the first equation of (2.3) by
12ddt∫Ω|∇c|2dx+∫Ω|Δc|2dx+∫Ω|∇c|2dx=∫Ωu∇cΔcdx−∫ΩΔcˆndx−∫ΩΔcfdx. |
Using the Young inequality and the Hölder inequality, we obtain
∫Ωu∇cΔcdx−∫ΩΔcˆndx−∫ΩΔcfdx≤‖u‖L4‖∇c‖L4‖Δc‖L2+‖ˆn‖L2‖Δc‖L2+‖f‖L2‖Δc‖L2≤C‖u‖H1(‖∇c‖12L2‖Δc‖12L2+‖∇c‖L2)‖Δc‖L2+‖ˆn‖L2‖Δc‖L2+‖f‖L2‖Δc‖L2=C‖u‖H1‖∇c‖12L2‖Δc‖32L2+C‖∇c‖L2‖Δc‖L2+‖ˆn‖L2‖Δc‖L2+‖f‖L2‖Δc‖L2≤12‖Δc‖2L2+C‖u‖4H1‖∇c‖2L2+C(‖ˆn‖2L2+‖f‖2L2). |
Combining this and above inequalities, we conclude
ddt‖∇c‖2L2+‖∇c‖2H1≤C‖u‖4H1‖∇c‖2L2+C(‖ˆn‖2L2+‖f‖2L2). |
We therefore verify that
‖∇c‖2L2+∫t0‖∇c‖2H1≤C(∫t0‖ˆn‖2L2dτ+∫t0‖f‖2L2dτ). |
Applying
12ddt∫Ω|Δc|2dx+∫Ω|∇Δc|2dx+∫Ω|Δc|2dx=∫Ω∇(u∇c)∇Δcdx−∫Ω∇ˆn+∇Δcdx−∫Ω∇f∇Δcdx. |
For the terms on the right, we obtain
∫Ω∇(u∇c)∇Δcdx−∫Ω∇ˆn+∇Δcdx−∫Ω∇f∇Δcdx≤‖∇Δc‖L2(‖u‖L4‖Δc‖L4+‖∇u‖L4‖∇c‖L4)+‖∇ˆn‖L2‖∇Δc‖L2+‖∇f‖L2‖∇Δc‖L2≤‖∇Δc‖L2(‖u‖L4‖Δc‖12L2‖∇Δc‖12L2+‖u‖L4‖Δc‖L2+‖∇u‖12L2‖Δu‖12L2‖∇c‖12L2‖Δc‖12L2+‖∇u‖L2‖∇c‖12L2‖Δc‖12L2+‖∇u‖12L2‖Δu‖12L2‖∇c‖L2+‖∇u‖L2‖∇c‖L2)+‖∇ˆn‖L2‖∇Δc‖L2+‖∇f‖L2‖∇Δc‖L2≤12‖∇Δc‖2L2+C(1+‖Δc‖2L2+‖Δu‖2L2+‖∇ˆn‖2L2+‖∇f‖2L2). |
Straightforward calculations yield
‖Δc‖2L2+∫t0‖Δc‖2H1dτ≤C(1+∫t0‖ˆn‖2H1dτ+∫t0‖f‖2H1dτ). |
Multiplying the first equation of (2.3) by
∫T0∫Ω|ct|2dxdt≤C, |
and thereby precisely arrive at the conclusion.
With above solutions
{nt−Δn+u⋅∇n+n=−∇⋅(n∇c)+(1+γ)ˆn+−μˆn+n,in Q,∂n∂ν|∂Ω=0,n(x,0)=n0(x),in Ω. | (2.4) |
By a similar argument as the above two problems, the existence of solutions can be easily obtained. Therefore, we only give the regularity estimate.
Lemma 2.5. Suppose
Proof. Firstly, we verify the nonnegativity of
ddt∫A(t)ndx−∫∂A(t)∂n∂νds+∫A(t)ndx=(1+γ)∫A(t)ˆn+dx−μ∫A(t)ˆn+ndx. |
Since
∫A(t)ndxdτ+∫t0∫A(t)ndxdτ=0. |
Then, we get
Next, multiplying the first equation of (2.4) by
12ddt∫Ωn2dx+∫Ω(n2+|∇n|2)dx+μ∫Ωˆn+n2dx=∫Ωn∇c∇ndx+(1+γ)∫Ωnˆn+dx≤‖n‖L4‖∇c‖L4‖∇n‖L2+(1+γ)‖ˆn‖L2‖n‖L2≤C(‖n‖12L2‖∇n‖12L2+‖n‖L2)‖c‖H2‖∇n‖L2+(1+γ)‖ˆn‖L2‖n‖L2≤C(‖n‖2L2‖c‖4H2+‖n‖2L2‖c‖2H2+‖ˆn‖L2)+12‖n‖2H1. |
So, we derive that
‖n‖2L2+∫T0‖n‖2H1dt≤C(1+∫T0‖ˆn‖2L2dt). |
Multiplying the first equation of (2.4) by
12ddt∫Ω|∇n|2dx+∫Ω|Δn|2dx+∫Ω|∇n|2dx=∫Ωu∇nΔndx+∫Ω(∇⋅(n∇c)Δn−(1+γ)ˆn+Δn+μˆn+nΔn)dx≤‖u‖L4‖∇n‖L4‖Δn‖L2+‖n‖L4‖Δc‖L4‖Δn‖L2+‖∇n‖L4‖∇c‖L4‖Δn‖L2+(1+γ)‖ˆn‖L2‖Δn‖L2+μ‖n‖L4‖ˆn‖L4‖Δn‖L2≤C‖u‖H1(‖∇n‖12L2‖Δn‖12L2+‖∇n‖L2)‖Δn‖L2+‖n‖L4(‖Δc‖12L2‖∇Δc‖12L2+‖Δc‖L2)‖Δn‖L2+μ‖n‖L4‖ˆn‖L4‖Δn‖L2+(‖∇n‖12L2‖Δn‖12L2+‖∇n‖L2)‖∇c‖H1‖Δn‖L2+(1+γ)‖ˆn‖L2‖Δn‖L2≤12‖Δn‖2L2+C(‖∇n‖2L2+‖n‖4L4+‖Δc‖4L2+‖∇Δc‖2L2+‖ˆn‖2L2+‖ˆn‖4L4)≤12‖Δn‖2L2+C(1+‖∇n‖2L2+‖n‖4L2+‖n‖2L2‖∇n‖2L2+‖∇Δc‖2L2+‖ˆn‖2L2+‖ˆn‖4L4). |
Straightforward calculations yield
‖∇n‖2L2+∫T0∫Ω(|Δn|2+|∇n|2+ˆn+|∇n|2)dxdt≤C. |
Multiplying the first equation of (2.4) by
∫T0∫Ω|nt|2dxdt≤C. |
The proof is complete.
Introduce the spaces
Xu=L4(0,T;L4(Ω)),Xn=L4(0,T;L4(Ω))∩L2(0,T;H1(Ω)),Yu=L∞(0,T;H1(Ω))∩L2(0,T;H2(Ω)),Yn=L∞(0,T;H1(Ω))∩L2(0,T;H2(Ω)). |
Define a map
F:Xu×Xn→Xu×Xn,F(ˆu,ˆn)=(u,n), |
where the
{nt−Δn+u⋅∇n+n=−∇⋅(n∇c)+(1+γ)ˆn+−μˆn+n,in (0,T)×Ω≡Q,ct−Δc+u⋅∇c+c=ˆn++f,in (0,T)×Ω≡Q,ut−Δu+ˆu⋅∇u=−∇π+ˆn∇φ,in (0,T)×Ω≡Q,∇⋅u=0,in (0,T)×Ω≡Q,∂n∂ν=∂c∂ν=0,u=0,on (0,T)×∂Ω,n(x,0)=n0(x),c(x,0)=c0(x),u(x,0)=u0(x),in Ω. |
Next, we use fixed point method to prove the local existence of solutions of the problem (1.1).
Lemma 2.6. The map
Proof. Let
From Lemma 2.6,
{nt−Δn+u⋅∇n+n=−∇⋅(n∇c)+α(1+γ)n−μn2,in Q,ct−Δc+u⋅∇c+c=n+αf,in Q,ut−Δu+u⋅∇u=−∇π+αn∇φ,in Q,∇⋅u=0,in Q,∂n∂ν=∂c∂ν=0,u=0,on (0,T)×∂Ω,n(x,0)=n0(x),c(x,0)=c0(x),u(x,0)=u0(x),in Ω. | (3.1) |
In order to prove the existence of solution, we first give some a priori estimates.
Lemma 3.1. Let
‖n‖L1+∫t0(‖n‖L1+‖n‖L2)dτ≤C, | (3.2) |
‖∇u‖2L2+∫t0‖∇u‖2H1dτ≤C, | (3.3) |
‖∇c‖2L2+∫t0‖∇c‖2H1dτ≤C. | (3.4) |
Proof. With Lemma 2.5 in hand, we get
ddt∫Ωndx+∫Ωndx+μ∫Ωn2dx=α(1+γ)∫Ωndx≤μ2∫Ωn2dx+C. |
Solving this differential inequality, we obtain that
‖n‖L1+∫t0(‖n‖L1+‖n‖L2)dτ≤C. |
Multiplying the third equation of (3.1) by
12ddt∫Ωu2dx+∫Ω|∇u|2dx+∫Ωu2dx=α∫Ωn∇φ⋅udx+∫Ωu2dx≤‖n‖L2‖u‖L2+‖u‖2L2≤C(‖n‖2L2+‖u‖2L2). |
Therefore, we see that
‖u‖2L2+∫t0‖u‖H1dτ≤C. |
By the Gagliardo-Nirenberg interpolation inequality, we deduce that
∫t0‖u‖4L4dτ≤C∫t0(‖u‖2L2‖∇u‖2L2d+‖u‖2L2)τ≤‖u‖2L2∫t0‖∇u‖2L2dτ+∫t0‖u‖2L2dτ≤C. |
Multiplying the third equation of (3.1) by
ddt‖∇u‖2L2+‖∇u‖2H1≤C(‖u‖4L4+‖u‖2L4+1)‖∇u‖2L2+C‖n‖2L2+C. |
Thus, we know
‖∇u‖2L2+∫t0‖∇u‖2H1dτ≤C. |
Multiplying the second equation of (3.1) by
12ddt∫Ωc2dx+∫Ω|∇c|2dx+∫Ωc2dx≤‖n‖L2‖c‖L2+α‖f‖L2‖c‖L2. |
Then, we have
‖c‖L2+∫t0‖c‖H1dτ≤C. |
Multiplying the second equation of (3.1) by
ddt‖∇c‖2L2+‖∇c‖2H1≤C‖u‖4H1‖∇c‖2L2+C(‖n‖2L2+‖f‖2L2). |
Further, we have
‖∇c‖2L2+∫t0‖∇c‖2H1dτ≤C. |
The proof is complete.
Lemma 3.2. Let
‖(n+1)ln(n+1)‖L1+‖∇c‖2L2+‖∇c‖2H1≤C. | (3.5) |
Proof. We rewrite the first equation of (3.1) as
ddt(n+1)+u⋅∇(n+1)−Δ(n+1)=−∇⋅((n+1)⋅∇c)+Δc+α(1+γ)n−μn2. |
Multiplying the above equation by
ddt∫Ω(n+1)ln(n+1)dx+4∫Ω|∇√n+1|2dx≤∫Ω∇(n+1)⋅∇cdx+∫ΩΔcln(n+1)dx+α(1+γ)∫Ωnln(n+1)dx=I1+I2+I3. |
For
I1=−∫ΩnΔcdx≤‖n‖L2‖Δc‖L2≤δ‖Δc‖2L2+C‖n‖2L2. |
For the term
I2=∫ΩΔcln(n+1)dx≤δ‖Δc‖2L2+C‖ln(n+1)‖2L2≤δ‖Δc‖2L2+C∫Ω(n+1)ln(n+1)dx. |
For the rest term
I3=α(1+γ)∫Ωnln(n+1)dx≤(1+γ)∫Ω(n+1)ln(n+1)dx. |
Combining
ddt∫Ω(n+1)ln(n+1)dx+4∫Ω|∇√n+1|2dx≤δ‖Δc‖2L2+C∫Ω(n+1)ln(n+1)dx+C‖n‖2L2. | (3.6) |
Multiplying the second equation of (3.1) by
12ddt∫Ω|∇c|2dx+∫Ω|Δc|2dx+∫Ω|∇c|2dx=∫Ωu∇cΔcdx−∫ΩΔcndx−α∫ΩΔcfdx. |
Straightforward calculations yield
ddt‖∇c‖2L2+‖∇c‖2H1≤C‖∇c‖2L2+C(‖n‖2L2+‖f‖2L2). | (3.7) |
Combing (3.6) and (3.7), it follows that
ddt∫Ω(n+1)ln(n+1)dx+ddt‖∇c‖2L2+(1−δ)‖∇c‖2H1+4∫Ω|∇√n+1|2dx≤C∫Ω(n+1)ln(n+1)dx+C(‖f‖2L2+‖n‖2L2). |
Taking
‖(n+1)ln(n+1)‖L1+‖∇c‖2L2+‖∇c‖2H1≤C. |
The proof is complete.
Lemma 3.3. Assume
‖n‖2L2+‖Δc‖2L2+∫t0‖n‖H1dτ+∫t0‖Δc‖H1dτ≤C. | (3.8) |
Proof. Taking the
12ddt∫Ωn2dx+∫Ω(n2+|∇n|2)dx+μ∫Ωn3dx=∫Ωn∇c∇ndx+α(1+γ)∫Ωn2dx=−12∫Ωn2Δcdx+α(1+γ)∫Ωn2dx. |
Here, we note that
|∫Ωn2Δcdx|≤‖n‖2L3‖Δc‖L3≤C‖n‖2L3(‖∇Δc‖23L2‖∇c‖13L2+‖∇c‖L2)≤C‖n‖2L3(‖∇Δc‖23L2+1). |
From Lemma 2.2 and (3.2), it follows that
−χ2∫Ωn2Δcdx≤C(δ‖n‖2H1‖(n+1)log(n+1)‖L1+p(δ−1)‖n‖L1)23(‖∇Δc‖23L2+1)≤C(δ‖n‖2H1+p(δ−1))23(‖∇Δc‖23L2+1)≤C(δ‖n‖43H1‖∇Δc‖23L2+δ‖n‖43H1+p23(δ−1)‖∇Δc‖23L2+p23(δ−1))≤δ‖∇Δc‖2L2+Cδ12‖n‖2H1+C−1/2δp(δ−1). |
As an immediate consequence
ddt‖n‖2L2+‖n‖2H1≤δ‖∇Δc‖2L2+Cδ12‖n‖2H1+C‖n‖2L2. | (3.9) |
Applying
12ddt∫Ω|Δc|2dx+∫Ω|∇Δc|2dx+∫Ω|Δc|2dx=∫Ω∇(u∇c)∇Δcdx−∫Ω∇n∇Δcdx−∫Ω∇f∇Δcdx=I4+I5. |
For
I4=∫Ω∇(u∇c)∇Δcdx≤‖∇Δc‖L2(‖u‖L4‖Δc‖L4+‖∇u‖L4‖∇c‖L4)≤‖∇Δc‖L2(‖u‖L4‖Δc‖12L2‖∇Δc‖12L2+‖u‖L4‖Δc‖L2+‖∇u‖12L2‖Δu‖12L2‖∇c‖12L2‖Δc‖12L2+‖∇u‖L2‖∇c‖12L2‖Δc‖12L2+‖∇u‖12L2‖Δu‖12L2‖∇c‖L2+‖∇u‖L2‖∇c‖L2)≤14‖∇Δc‖2L2+C(1+‖Δc‖2L2+‖Δu‖2L2). |
For the term
\begin{align*} I_5 = &- \int_{\Omega} \nabla n \nabla \Delta c d x- \int_{\Omega} \nabla f \nabla \Delta c d x \\ \leq&C(\|\nabla n\|^2_{L^2}+\|\nabla f\|^2_{L^2})+\frac{1}{4}\|\nabla \Delta c\|_{L^{2}}^{2}. \end{align*} |
Along with
\begin{align} &\frac{d}{d t}\|\Delta c\|^2_{L^2}+\|\nabla\Delta c\|^2_{L^2}+ \|\Delta c\|^2_{L^2} \\ \leq & C(1+\|\Delta c\|_{L^{2}}^{2}+\|\Delta u\|_{L^{2}}^{2}+\|\nabla n\|^2_{L^2}+\|\nabla f\|^2_{L^2}) . \end{align} | (3.10) |
Combining (3.9) and (3.10), it follows that
\begin{align*} &\frac{d}{d t}(\|n\|^2_{L^2}+\|\Delta c\|^2_{L^2})+ \|\Delta c\|^2_{L^2}+(1-C\delta^\frac{1}{2})\| n\|^2_{H^1}+(1-\delta)\|\nabla\Delta c\|^2_{L^2} \\ \leq&C(1+\|\Delta c\|_{L^{2}}^{2}+\|\Delta u\|_{L^{2}}^{2}+\|\nabla n\|^2_{L^2}+\|\nabla f\|^2_{L^2}) . \end{align*} |
By choosing
\begin{align*} \|n\|^2_{L^2}+\|\Delta c\|^2_{L^2}+\int_{0}^t\|n\|_{H^1}d\tau +\int_{0}^t\|\Delta c\|_{H^1}d\tau\leq C. \end{align*} |
The proof is complete.
Lemma 3.4. Assume
\begin{align} \|\nabla n\|^2_{L^2} +\int_{0}^t\|n\|^2_{H^2} d\tau \leq C. \end{align} | (3.11) |
Proof. Taking the
\begin{align*} &\frac{1}{2} \frac{d}{d t} \int_{\Omega}|\nabla n|^{2} d x+\int_{\Omega}|\Delta n|^{2} d x+\int_{\Omega}|\nabla n|^{2} d x \\ = &\int_{\Omega}u\nabla n \Delta n d x+\int_{\Omega} \nabla \cdot(n\nabla c)\Delta n d x+(1+\gamma)\int_{\Omega}|\nabla n|^2 d x+\mu\int_{\Omega}n^2\Delta n d x \\ = &I_6+I_7+I_8. \end{align*} |
For the term
\begin{align*} I_6 = &\int_{\Omega}u\nabla n \Delta n d x = -\frac{1}{2}\int_{\Omega}\nabla u(\nabla n)^2 d x\leq \|\nabla u\|_{L^2}\|\nabla n\|^2_{L^4} \\ \leq& \|\nabla u\|_{L^2}(\|\nabla n\|^{\frac{1}{2}}_{L^2}\|\Delta n\|^{\frac{1}{2}}_{L^2}+\|\nabla n\|_{L^2})^2 \\ \leq& \delta\|\Delta n\|^{2}_{L^2}+C\|\nabla n\|_{L^2}^2. \end{align*} |
For the term
\begin{align*} I_7 = &\int_{\Omega} \nabla \cdot(n\nabla c)\Delta n d x \\ = &\int_{\Omega}(\nabla n\nabla c+n\Delta c)\Delta n d x \\ \leq&\|\Delta n\|_{L^{2}}\left(\|\nabla n\|_{L^{3}}\|\nabla c\|_{L^{6}}+\|n\|_{\mathcal{C}}\|\Delta c\|_{L^{2}}\right) \nonumber \\ \leq& C\|\Delta n\|_{L^{2}}\left(\|\nabla n\|_{H^{\frac{1}{3}}}\|\nabla c\|_{H^{1}}+\|n\|_{H^{\frac{4}{3}}}\|\Delta c\|_{L^{2}}\right) \nonumber \\ \leq& C \|n\|_{H^{2}}\|n\|_{H^{\frac{4}{3}}}\|c\|_{H^{2}} \leq C\|n\|_{H^{2}}^{\frac{5}{3}}\|n\|_{L^{2}}^{\frac{1}{3}}\|c\|_{H^{2}} \nonumber \\ \leq& \delta\| n\|_{H^{2}}^{2}+C(\delta) \|n\|_{L^{2}}^{2}\|c\|_{H^{2}}^{6}\leq \delta\| n\|_{H^{2}}^{2}+C. \end{align*} |
For the term
\begin{align*} I_8 = &(1+\gamma)\int_{\Omega}|\nabla n|^2 d x+\mu\int_{\Omega}n^2\Delta n d x \\ = &(1+\gamma)\int_{\Omega}|\nabla n|^2 d x-2\mu\int_{\Omega}|\nabla n|^2 n d x \\ \leq&(1+\gamma)\|\nabla n\|^2_{L^2}. \end{align*} |
Combine the estimates about
\begin{align*} \frac{d}{d t}\|\nabla n\|^2_{L^2}+(1-4\delta)\|n\|^2_{H^2}\leq C\|\nabla n\|^2_{L^2}+C. \end{align*} |
By taking
\begin{align*} \|\nabla n\|^2_{L^2} +\int_{0}^t\|n\|^2_{H^2} d\tau \leq C. \end{align*} |
Therefore, this proof is complete.
Lemma 3.5. The operator
Proof. Let
\begin{align*} \mathcal{F}(\hat{n}_m,\hat{u}_{m})\rightarrow (\hat{n},\hat{u}) \text{ weakly in } Y_u\times Y_n \text{ and strongly in } X_{u} \times X_{n}. \end{align*} |
Let
Theorem 3.1. Let
\begin{align} \|n\|&_{L^{\infty}(0,T;H^1(\Omega))}+\|n\|_{L^2(0,T;H^2(\Omega))}+\|n_t\|_{L^2(0,T;L^2(\Omega))}+\|c\|_{L^{\infty}(0,T;H^2(\Omega))} \\ &+\|c\|_{L^{2}(0,T;H^3(\Omega))}+\|c_t\|_{L^2(0,T;L^2(\Omega))} +\|u\|_{L^{\infty}(0,T;H^1(\Omega))} \\ &+\|u\|_{L^2(0,T;H^2(\Omega))}+\|u_t\|_{L^2(0,T;L^2(\Omega))}\leq C. \end{align} | (3.12) |
Proof. From Lemmas 3.1, 3.3 and 3.4, it is easy to verify the existence of solution and (3.11). Therefore, we will prove the uniqueness of the solution in the following part. For convenience, we set
\begin{align} n_{t}-\Delta n+u_1 \cdot \nabla n+u\nabla n_2 = - \nabla \cdot(n_1 \nabla c) \\ -\nabla(n\nabla c_2)+\gamma n -\mu n(n_1+n_2), &&\text {in } (0, T) \times \Omega \equiv Q, \end{align} | (3.13) |
\begin{align} &c_{t}-\Delta c+u_1 \cdot \nabla c+u\nabla c_2+c = n, &&\text{in } (0, T) \times \Omega \equiv Q, \end{align} | (3.14) |
\begin{align} &u_{t}-\Delta u+u_1 \cdot \nabla u+u\cdot \nabla u_2 = n \nabla \varphi, &&\text{in } (0, T) \times \Omega \equiv Q, \end{align} | (3.15) |
\begin{align} &\nabla \cdot u = 0, &&\text{in } (0, T) \times \Omega \equiv Q, \end{align} | (3.16) |
\begin{align} &\frac{\partial n}{\partial \nu} = \frac{\partial c}{\partial \nu} = 0,\quad u = 0, &&\text{on } (0, T) \times \partial\Omega, \end{align} | (3.17) |
\begin{align} &n_0(x) = c_0(x) = u_0(x) = 0, &&\text{in }\Omega. \end{align} | (3.18) |
Taking the
\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega}n^2 d x+\int_{\Omega}|\nabla n|^2 d x+\int_{\Omega} n^2 d x \\ \leq&-\int_{\Omega}u \nabla n_2 n d x+\int_{\Omega}n_1\nabla c \nabla n d x +\int_{\Omega} n\nabla c_2 \nabla n d x+(1+\gamma) \int_{\Omega} n^2 d x \\ = &I_9+I_{10}+I_{11}+I_{12}. \end{align*} |
For the term
\begin{align*} I_9 = &-\int_{\Omega}u \nabla n_2 n d x \leq \|\nabla n_2\|_{L^2}\|u\|_{L^4}\|n\|_{L^4} \\ \leq &C\|\nabla n_2\|_{L^2}\|u\|_{H^1}(\|n\|^{\frac{1}{2}}_{L^2}\|\nabla n\|^\frac{1}{2}_{L^2}+\|n\|_{L^2}) \\ \leq &\frac{\delta}{3} \|\nabla n\|^2_{L^2}+C\|n\|^2_{L^2}. \end{align*} |
For the term
\begin{align*} I_{10} = &\int_{\Omega}n_1\nabla c \nabla n d x \leq \|\nabla n\|_{L^2}\|n_1\|_{L^4}\|\nabla c\|_{L^4} \\ \leq&C\|\nabla n\|_{L^2}\|n_1\|_{H^1}\|\nabla c\|_{H^1} \\ \leq&\frac{\delta}{3}\|\nabla n\|^2_{L^2}+C. \end{align*} |
For the term
\begin{align*} I_{11} = &\int_{\Omega} n\nabla c_2 \nabla n d x \leq\|\nabla n\|_{L^2}\|\nabla c_2\|_{L^4}\|n\|_{L^4} \\ \leq& \|\nabla n\|_{L^2}\|\nabla c_2\|_{H^1}\|n\|_{H^1} \\ \leq&\frac{\delta}{3}\|\nabla n\|^2_{L^2}+C. \end{align*} |
With the use of estimates
\begin{align} \frac{d}{d t}\|n\|^2_{L^2}+\|n\|_{H^1} \leq \delta\|\nabla n\|^2_{L^2}+ C\|n\|^2_{L^2}+C. \end{align} | (3.19) |
Taking the
\begin{align*} &\frac{1}{2}\frac{d}{ d t}\int_{\Omega}c^2 d x+\int_{\Omega}|\nabla c|^2 d x+ \int_{\Omega}c^2 d x \nonumber \\ = &-\int_{\Omega}u_1\nabla c c d x-\int_{\Omega}u\nabla c_2 c d x+ \int_{\Omega}n c d x \nonumber \\ \leq &\|c\|^2_{L^4}\|\nabla u_1\|_{L^2}+\|u\|_{L^2}\|\nabla c_2\|_{L^4}\|c\|_{L^4}+\|n\|_{L^2}\|c\|_{L^2} \nonumber \\ \leq& C(\|c\|^{\frac{1}{2}}_{L^2}\|\nabla c\|^{\frac{1}{2}}_{L^2}+\|c\|_{L^2})^2\|\nabla u_1\|_{L^2}+(\|c\|^{\frac{1}{2}}_{L^2}\|\nabla c\|^{\frac{1}{2}}_{L^2}+\|c\|_{L^2})\|u\|_{L^2}\|\nabla c_2\|_{H^1} \nonumber \\ &+\|n\|_{L^2}\|c\|_{L^2} \nonumber \\ \leq&\delta\|\nabla c\|^2_{L^2}+C\|c\|^2_{L^2}. \end{align*} |
Then, we get
\begin{align} \frac{d}{d t}\|c\|^2_{L^2}+\|c\|_{H^1}\leq \delta \|\nabla c\|^2_{L^2}+C\|c\|^2_{L^2}. \end{align} | (3.20) |
Taking the
\begin{align*} \frac{1}{2} \int_{\Omega}u^2 d x+ \int_{\Omega}|\nabla u|^2 d x = \int_{\Omega} n \nabla \varphi u d x . \end{align*} |
Straightforward calculations yield
\begin{align} \frac{d}{d t}\|u\|^2_{L^2}+\|u\|_{H^1}\leq C (\|u\|^2_{L^2}+\|n\|^2_{L^2}). \end{align} | (3.21) |
Then, a combination of (3.19), (3.20) and (3.21) yields
\begin{align*} &\frac{d}{d t}(\|n\|^2_{L^2}+\|c\|^2_{L^2}+\|u\|^2_{L^2})+(\|n\|_{H^1}+\|c\|_{H^1}+\|u\|_{H^1}) \\ \leq&\delta(\|\nabla n\|^2_{L^2}+\|\nabla c\|^2_{L^2}+\|\nabla u\|^2_{L^2})+(\|n\|^2_{L^2}+\|c\|^2_{L^2}+\|u\|^2_{L^2})+C. \end{align*} |
By choosing
\begin{align*} \frac{d}{d t}(\|n\|^2_{L^2}+\|c\|^2_{L^2}+\|u\|^2_{L^2}) \leq C(\|n\|^2_{L^2}+\|c\|^2_{L^2}+\|u\|^2_{L^2})+C. \end{align*} |
Applying Gronwall's lemma to the resulting differential inequality, we finally obtain the uniqueness of the solution.
In this section, we will prove the existence of the optimal solution of control problem. The method we use for treating this problem was inspired by some ideas of Guillén-González et al [9]. Assume
Minimize the cost functional
\begin{align} J(n,c,u,f) = &\frac{\beta_1}{2}\|n-n_d\|_{L^2(Q_d)}^2+\frac{\beta_2}{2}\|c-c_d\|_{L^2(Q_d)}^2 +\frac{\beta_3}{2}\|u-u_d\|_{L^2(Q_d)}^2 \\ &+\frac{\beta_4}{2}\|n(T)-n_{\Omega}\|_{L^2(\Omega_d)}^2+\frac{\beta_5}{2}\|c(T)-c_{\Omega}\|_{L^2(\Omega_d)}^2 \\ &+\frac{\beta_6}{2}\|u(T)-u_{\Omega}\|_{L^2(\Omega_d)}^2+\frac{\beta_7}{2}\|f(x,t)\|_{L^2(Q_c)}^2, \end{align} | (4.1) |
subject to the system (1.1). Moreover, the nonnegative constants
\begin{align*} &n_d\in L^2(Q_d), c_d\in L^2(Q_d), u_d\in L^2(Q_d), \\ &n_{\Omega}\in L^2(\Omega_c), c_{\Omega}\in L^2(\Omega_c), u_{\Omega}\in L^2(\Omega_c), \;f\in \mathcal{U}. \end{align*} |
The set of admissible solutions of optimal control problem (4.1) is defined by
\begin{align*} \mathcal{S}_{a d} = \{s = (n,c,u,f) \in \mathcal{H}: s \text { is a strong solution of }(1.1)\}. \end{align*} |
The function space
\begin{align*} \mathcal{H} = Y_n\times Y_c \times Y_u\times\mathcal{U}, \end{align*} |
where
Now, we prove the existence of a global optimal control for problem (1.1).
Theorem 4.1. Suppose
Proof. Along with Theorem 3.1, we conduct that
\begin{align} \lim\limits_{m\rightarrow +\infty}J(n_m,c_m,u_m,f_m) = \inf\limits_{(n,c,u,f)\in \mathcal{S}_{ad}}J(n,c,u,f). \end{align} | (4.2) |
According to the definition of
\begin{align} \left\{\begin{aligned} &n_{mt}+u_{m} \cdot \nabla n_{m} = \Delta n_{m}-\nabla \cdot(n_{m} \cdot \nabla c_{m})+\gamma n_{m} -\mu n_{m}^{2}, &&\text {in } Q, \\ &c_{mt}+u_{m} \cdot \nabla c_{m} = \Delta c_{m}-c_{m}+n_{m}+f_{m}, &&\text{in } Q, \\ &u_{mt}+u_{m} \cdot \nabla u_{m} = \Delta u_{m}-\nabla \pi+n_{m} \nabla \varphi, &&\text{in } Q, \\ &\nabla \cdot u_{m} = 0, &&\text{in } Q, \\ &\frac{\partial n_{m}}{\partial \nu}\Big|_{\partial\Omega} = \frac{\partial c_{m}}{\partial \nu}\Big|_{\partial\Omega} = 0,\quad u_{m}|_{\partial\Omega} = 0, \\ &n_{m}(0) = n_0, c_{m}(0) = c_0, u_{m}(0) = u_0,&&\text{in }\Omega. \end{aligned} \right. \end{align} | (4.3) |
Observing that
\begin{align*} &n_{m} \rightarrow \bar{n}, \text { weakly in } L^{2}\left(0, T ; H^{2}(\Omega)\right) \text { and weakly* } \text { in } L^{\infty}\left(0, T ; H^{1}(\Omega)\right), \\ &c_{m} \rightarrow \bar{c}, \text { weakly in } L^{2}\left(0, T ; H^{3}(\Omega)\right) \text { and weakly* } \text { in } L^{\infty}\left(0, T ; H^{2}(\Omega)\right), \\ &u_{m} \rightarrow \bar{u}, \text { weakly in } L^{2}\left(0, T ; H^{2}(\Omega)\right) \text { and weakly* } \text { in } L^{\infty}\left(0, T ; H^{1}(\Omega)\right), \\ &f_{m} \rightarrow \bar{f}, \text { weakly in } L^2(0,T;H^1(\Omega_c)), \text { and } \tilde{f} \in \mathcal{U}. \end{align*} |
According to the Aubin-Lions lemma [16] and the compact embedding theorems, we obtain
\begin{align*} &n_{m} \rightarrow \bar{n}, \quad \text { strongly in } C\left([0, T] ; L^{2}(\Omega)\right) \cap L^{2}\left(0, T ; H^{1}(\Omega)\right), \\ &c_{m} \rightarrow \bar{c}, \quad \text{ strongly in } C\left([0, T] ; H^{1}(\Omega)\right) \cap L^{2}\left(0, T ; H^{2}(\Omega)\right), \\ &u_{m} \rightarrow \bar{u}, \quad \text { strongly in } C\left([0, T] ; L^{2}(\Omega)\right) \cap L^{2}\left(0, T ; H^{1}(\Omega)\right). \end{align*} |
Since
\begin{align*} \nabla \cdot\left(n_{m} \nabla c_{m}\right)\rightarrow \chi, \text{ weakly in } L^2(0,T;L^2(\Omega)). \end{align*} |
Recalling that
\begin{align*} n_{m} \nabla c_{m}\rightarrow \bar{n}\nabla\bar{c}, \text{ weakly in } L^{\infty}(0,T; L^2(\Omega)). \end{align*} |
Therefore, we get that
\begin{align*} \lim\limits_{m\rightarrow +\infty}J(n_m,c_m,u_m,f_m) = \inf\limits_{(u,c,u,f)\in \mathcal{S}_{ad}}J(u,c,u,f)\leq J(\bar{n},\bar{c},\bar{u},\bar{f}). \end{align*} |
On the other hand, we deduce from the weak lower semicontinuity of the cost functional
\begin{align*} J(\bar{n},\bar{c},\bar{u},\bar{f})\leq \liminf\limits_{m\rightarrow +\infty} J(n_m,c_m,u_m,f_m). \end{align*} |
Therefore, this implies that
In order to derive the first-order necessary optimality conditions for a local optimal solution of problem (4.1). To this end, we will use a result on existence of Lagrange multipliers in Banach spaces ([20]). First, we discuss the following problem
\begin{align} \min J(s) \text{ subject to } s\in \mathcal{S} = \{s\in \mathcal{H}:G(s)\in \mathcal{N}\}, \end{align} | (5.1) |
where
\begin{align*} A^+ = \{\rho\in X^{\prime}:\langle \rho,a\rangle_{X^{\prime}}\geq 0, \forall a \in A\}. \end{align*} |
We consider the following Banach spaces
\begin{align*} &X = V_n\times V_c\times V_u\times L^2(0,T;H^1(\Omega_c)), \\ &Y = L^2(Q)\times L^2(0,T;H^1(\Omega))\times L^2(Q)\times H^1(\Omega)\times H^2(\Omega) \times H^1(\Omega), \end{align*} |
where
\begin{align*} &V_n = \{n \in Y_n:\frac{\partial n}{\partial \nu}\text{ on } (0,T)\times \partial \Omega \}, \\ &V_c = \{n \in Y_c:\frac{\partial c}{\partial \nu}\text{ on } (0,T)\times \partial \Omega \}, \\ &V_u = \{n \in Y_u: u = 0 \text{ on } (0,T)\times \partial \Omega \text{ and } \nabla \cdot u = 0 \text{ in } (0,T)\times \Omega\} \end{align*} |
and the operator
\begin{align*} &G_1: X \rightarrow L^2(Q),&& G_2: X \rightarrow L^2(0,T;H^1(\Omega)),&&&G_3: X \rightarrow L^2(Q), \\ &G_4: X \rightarrow H^1(\Omega),&&G_5: X \rightarrow H^2(\Omega),&&&G_6: X \rightarrow H^1(\Omega), \end{align*} |
which are defined at each point
\begin{align} \left\{\begin{aligned} &G_1 = n_{t}+u \cdot \nabla n-\Delta n+\nabla \cdot(n \cdot \nabla c)-\gamma n+\mu n^{2}, \\ &G_2 = c_{t}+u \cdot \nabla c-\Delta c+c-n-f, \\ &G_3 = u_{t}+u \cdot \nabla u-\Delta u+\nabla \pi-n \nabla \varphi, \\ &G_4 = n(0)-n_0, \\ &G_5 = c(0)-c_0, \\ &G_6 = u(0)-u_0. \end{aligned} \right. \end{align} | (5.2) |
The function spaces are given as follows
\begin{align*} \mathcal{H} = V_n\times V_c \times V_u \times \mathcal{U}. \end{align*} |
We see that
\begin{align} \min J(s) \text{ subject to } s\in \mathcal{S}_{ad} = \{s\in \mathcal{H}:G(s) = 0\}. \end{align} | (5.3) |
Taking the differentiability of
Lemma 5.1. The functional
\begin{align} J^{\prime}(\bar{s})[r] = &\beta_1\int_{0}^T\int_{\Omega_d}(\bar{n}-n_d)\tilde{n}d x d t+\beta_2\int_{0}^T\int_{\Omega_d}(\bar{c}-c_d)\tilde{c}d x d t \\ &+\beta_3\int_{0}^T\int_{\Omega_d}(\bar{u}-u_d)\tilde{u}(T)d x d t +\beta_4\int_{\Omega_d} (\bar{n}(T)-n_{\Omega})\tilde{n}(T)d x \\ &+\beta_5\int_{\Omega_d} (\bar{c}(T)-c_{\Omega})\tilde{c}d x+\beta_6\int_{\Omega_d} (\bar{u}(T)-u_{\Omega})\tilde{u}(T)d x \\ &+\beta_7\int_{0}^T\int_{\Omega_d}\bar{f}\tilde{f} d x d t. \end{align} | (5.4) |
Lemma 5.2. The operator
\begin{align*} G^{\prime}(\bar{s})[r] = (G_1^{\prime}(\bar{s})[r],G_2^{\prime}(\bar{s})[r],G_3^{\prime}(\bar{s})[r],G_4^{\prime}(\bar{s})[r],G_5^{\prime}(\bar{s})[r],G_6^{\prime}(\bar{s})[r]) \end{align*} |
defined by
\begin{align*} \left\{\begin{aligned} &G_1^{\prime}(\bar{s})[r] = \tilde{n}_{t}-\Delta \tilde{n}+\bar{u} \cdot \nabla \tilde{n}+\tilde{u}\nabla \bar{n}+\nabla \cdot(\bar{n} \nabla \tilde{c}) \\ &+\nabla(\tilde{n}\nabla \bar{c})-\gamma \tilde{n} +2\mu \tilde{n}\bar{n}, &&\mathit{\text{in}}\;Q , \\ &G_2^{\prime}(\bar{s})[r] = \tilde{c}_{t}-\Delta \tilde{c}+\bar{u} \cdot \nabla \tilde{c}+\tilde{u}\cdot\nabla \bar{c}+\tilde{c}- \tilde{n}-\tilde{f},&&\mathit{\text{in}}\;Q , \\ &G_3^{\prime}(\bar{s})[r] = \tilde{u}_{t}-\Delta \tilde{u}+\bar{u} \cdot \nabla \tilde{u}+\tilde{u}\cdot \nabla \bar{u}-\tilde{n} \nabla \varphi,&&\mathit{\text{in}}\;Q , \\ &\nabla\cdot \tilde{u} = 0,&&\mathit{\text{in}}\;Q , \\ &\frac{\partial \tilde{n}}{\partial \nu} = \frac{\partial \tilde{c}}{\partial \nu} = 0, \tilde{u} = 0, &&\mathit{\text{on}}\;(0,T)\times\partial\Omega, \\ &\tilde{n}(0) = \tilde{n}_0, \tilde{c}(0) = \tilde{c}_0,\tilde{u}(0) = \tilde{u}_0, &&\mathit{\text{in}}\;\Omega. \end{aligned} \right. \end{align*} |
Lemma 5.3. Let
Proof. For any fixed
\begin{align} \left\{\begin{aligned} &\tilde{n}_{t}-\Delta \tilde{n}+\bar{u} \cdot \nabla \tilde{n}+\tilde{u}\nabla \bar{n}+\nabla \cdot(\bar{n} \nabla \tilde{c}) \\ &+\nabla(\tilde{n}\nabla \bar{c})-\gamma \tilde{n} +2\mu \tilde{n}\bar{n} = g_n, &&\text{in }Q , \\ &\tilde{c}_{t}-\Delta \tilde{c}+\bar{u} \cdot \nabla \tilde{c}+\tilde{u}\cdot\nabla \bar{c}+\tilde{c}- \tilde{n} = g_c,&&\text{in }Q , \\ &\tilde{u}_{t}-\Delta \tilde{u}+\bar{u} \cdot \nabla \tilde{u}+\tilde{u}\cdot \nabla \bar{u}-\tilde{n} \nabla \varphi = g_u,&&\text{in }Q , \\ &\nabla\cdot \tilde{u} = 0,&&\text{in }Q , \\ &\frac{\partial \tilde{n}}{\partial \nu} = \frac{\partial \tilde{c}}{\partial \nu} = 0, \tilde{u} = 0, &&\text{on }(0,T)\times\partial\Omega, \\ &\tilde{n}(0) = \tilde{n}_0, \tilde{c}(0) = \tilde{c}_0,\tilde{u}(0) = \tilde{u}_0, &&\text{in }\Omega. \end{aligned} \right. \end{align} | (5.5) |
Next, we use Leray-Schauder's fixed point method to prove the existence of solutions of the problem (5.5), the operator
\begin{align} \left\{\begin{aligned} &\tilde{n}_{t}-\Delta \tilde{n}+\bar{u} \cdot \nabla \tilde{n}+\tilde{u}\nabla \bar{n}+\nabla \cdot(\bar{n} \nabla \tilde{c}) \\ &+\nabla(\tilde{n}\nabla \bar{c})-\gamma \tilde{n} +2\mu \dot{n}\bar{n} = g_n, &&\text{in }Q , \\ &\tilde{c}_{t}-\Delta \tilde{c}+\bar{u} \cdot \nabla \tilde{c}+\tilde{u}\cdot\nabla \bar{c}+\tilde{c}- \dot{n} = g_c,&&\text{in }Q , \\ &\tilde{u}_{t}-\Delta \tilde{u}+\bar{u} \cdot \nabla \tilde{u}+\dot{u}\cdot \nabla \bar{u}-\dot{n} \nabla \varphi = g_u,&&\text{in }Q. \end{aligned} \right. \end{align} | (5.6) |
The system (5.6) is complemented by the corresponding Neumann boundary and initial conditions. Similar to the proof of Lemmas 2.3, 2.4, 2.5 and 2.6, we conduct that operator
Similar to the proof of Theorem 3.1,
\begin{align} \left\{\begin{aligned} &\tilde{n}_{t}-\Delta \tilde{n}+\tilde{n} = -\bar{u} \cdot \nabla \tilde{n}-\tilde{u}\cdot\nabla \bar{n}-\nabla \cdot(\bar{n} \nabla \tilde{c}) \\ &-\nabla(\tilde{n}\nabla \bar{c})+\alpha(\gamma+1) \tilde{n} -2\mu \tilde{n}\bar{n}+\alpha g_n, &&\text{in }Q , \\ &\tilde{c}_{t}-\Delta \tilde{c}+\tilde{c} = -\bar{u} \cdot \nabla \tilde{c}-\tilde{u}\cdot\nabla \bar{c}+ \alpha \tilde{n}+\alpha g_c,&&\text{in }Q , \\ &\tilde{u}_{t}-\Delta \tilde{u} = -\bar{u} \cdot \nabla \tilde{u}-\tilde{u}\cdot \nabla \bar{u}+\alpha\tilde{n} \nabla \varphi+\alpha g_u,&&\text{in }Q, \end{aligned} \right. \end{align} | (5.7) |
complemented by the corresponding Neumann boundary and initial conditions.
Taking the
\begin{align*} \frac{1}{2}\int_{\Omega} \tilde{u}^2 d x+\int_{\Omega} |\nabla \tilde{u}|^2 d x = \alpha\int_{\Omega} \tilde{n}\nabla \varphi \tilde{u} d x +\alpha \int_{\Omega} \tilde{u} g_u d x. \end{align*} |
By the Poincaré inequality and Young's inequality, we have
\begin{align} \frac{d}{d t}\|\tilde{u}\|^2_{L^2}+\|\tilde{u}\|^2_{H^1}\leq C(\|\tilde{n}\|^2_{L^2}+\|g_u\|^2_{L^2})+C\|\tilde{u}\|^2_{L^2}. \end{align} | (5.8) |
Taking the
\begin{align*} &\frac{1}{2}\int_{\Omega} \tilde{c}^2 d x+\int_{\Omega} |\nabla \tilde{c}|^2 d x+\int_{\Omega} \tilde{c}^2 d x \\ = &\int_{\Omega} \tilde{u}\nabla \bar{c} \tilde{c} d x+ \alpha\int_{\Omega} \tilde{n} \tilde{c} d x +\alpha\int_{\Omega} g_c \tilde{c} d x. \end{align*} |
With the Poincaré inequality and Young's inequality in hand, we see that
\begin{align} \frac{d}{d t}\|\tilde{c}\|^2_{L^2}+\|\tilde{c}\|^2_{H^1}\leq C(\|\tilde{n}\|^2_{L^2}+\|g_c\|^2_{L^2})+C\|\tilde{c}\|^2_{L^2}. \end{align} | (5.9) |
Taking the
\begin{align*} &\frac{1}{2}\int_{\Omega} |\nabla \tilde{c}|^2 d x+\int_{\Omega} |\Delta \tilde{c}|^2 d x+\int_{\Omega} |\nabla \tilde{c}|^2 d x \\ = &\int_{\Omega} \tilde{u}\nabla \bar{c} \Delta \tilde{c} d x+\int_{\Omega} \bar{u}\nabla \tilde{c} \Delta \tilde{c} d x- \alpha\int_{\Omega} \tilde{n} \Delta \tilde{c} d x -\alpha\int_{\Omega} g_c \Delta \tilde{c} d x \\ = &J_1+J_2+J_3. \end{align*} |
For the term
\begin{align*} J_1 = &\int_{\Omega} \tilde{u}\nabla \bar{c} \Delta \tilde{c} d x \leq \|\Delta \tilde{c}\|_{L^2}\|\nabla \bar{c}\|_{L^4}\|\tilde{u}\|_{L^4} \\ \leq &\frac{1}{6}\|\Delta \tilde{c}\|^2_{L^2}+C\|\nabla \bar{c}\|^2_{H^1}\|\tilde{u}\|^2_{H^1}. \end{align*} |
For the term
\begin{align*} J_2 = & \int_{\Omega} \bar{u}\nabla \tilde{c} \Delta \tilde{c} d x = - \frac{1}{2} \int_{\Omega} \nabla \bar{u}|\nabla \tilde{c}|^2 d x \\ \leq&\|\nabla\bar{u}\|_{L^2}\|\nabla \tilde{c}\|^2_{L^4} \\ \leq& \|\nabla\bar{u}\|_{L^2}(\|\nabla \tilde{c}\|^{\frac{1}{2}}_{L^2}\|\Delta \tilde{c}\|^{\frac{1}{2}}_{L^2}+\|\nabla \tilde{c}\|_{L^2}) \\ \leq&\frac{1}{6}\|\Delta \tilde{c}\|^2_{L^2}+C\|\nabla \tilde{c}\|^2_{L^2}. \end{align*} |
For the term
\begin{align*} J_3 = &- \alpha\int_{\Omega} \tilde{n} \Delta \tilde{c} d x -\alpha\int_{\Omega} g_c \Delta \tilde{c} d x \\ \leq&\frac{1}{6}\|\Delta \tilde{c}\|^2_{L^2}+C(\|\tilde{n}\|^2_{L^2}+\|g_c\|^2_{L^2}). \end{align*} |
Therefore, combining
\begin{align} \frac{d}{d t}\|\nabla \tilde{c}\|^2_{L^2}+\|\nabla \tilde{c}\|^2_{H^1} \leq C\|\nabla \tilde{c}\|^2_{L^2}+ C(\|\tilde{n}\|^2_{L^2}+\|g_c\|^2_{L^2}). \end{align} | (5.10) |
Taking the
\begin{align*} &\frac{d}{d t}\int_{\Omega} \tilde{n}^2 d x+\int_{\Omega} |\nabla\tilde{n}|^2 d x+ \int_{\Omega} \tilde{n}^2 d x \\ = &-\int_{\Omega}\tilde{u}\nabla\bar{n}\tilde{n}d x+\int_{\Omega}\nabla \tilde{n}\bar{n}\nabla\tilde{c} d x+\int_{\Omega}\nabla \tilde{n}\tilde{n}\nabla\bar{c} d x+\alpha (\gamma +1)\int_{\Omega} \tilde{n}^2 d x \\ &+2\mu \int_{\Omega}\bar{n}\tilde{n}^2 d x+\alpha \int_{\Omega}\tilde{n} g_n d x \\ = &J_4+J_5+J_6+J_7. \end{align*} |
For the term
\begin{align*} J_4 = &-\int_{\Omega}\tilde{u}\nabla\bar{n}\tilde{n}d x\leq \|\tilde{u}\|_{L^4}\|\nabla\bar{n}\|_{L^2}\|\tilde{n}\|_{L^4} \\ \leq& C (\|\nabla\tilde{u}\|^{{\frac{1}{2}}}_{L^2}\|\tilde{u}\|^{\frac{1}{2}}_{L^2}+\|\tilde{u}\|_{L^2}) \|\nabla\bar{n}\|_{L^2}\|\tilde{n}\|_{H^1} \\ \leq &\delta \|\tilde{n}\|^2_{H^1}+C\|\nabla\tilde{u}\|_{L^2}\|\tilde{u}\|_{L^2}+C\|\tilde{u}\|^2_{L^2} \\ \leq &\delta \|\tilde{n}\|^2_{H^1}+\delta \|\nabla\tilde{u}\|^2_{L^2}+C\|\tilde{u}\|^2_{L^2}. \end{align*} |
For the term
\begin{align*} J_5 = &\int_{\Omega}\nabla \tilde{n}\bar{n}\nabla\tilde{c} d x \leq \|\nabla\tilde{n}\|_{L^2} \|\bar{n}\|_{L^4}\|\nabla\tilde{c}\|_{L^4} \\ \leq &\|\nabla\tilde{n}\|_{L^2}\|\bar{n}\|_{H^1}(\|\nabla\tilde{c}\|^{\frac{1}{2}}_{L^2}\|\Delta\tilde{c}\|^{\frac{1}{2}}_{L^2}+\|\nabla\tilde{c}\|_{L^2}) \\ \leq&\delta \|\nabla\tilde{n}\|^2_{L^2}+\|\nabla\tilde{c}\|_{L^2}\|\Delta\tilde{c}\|_{L^2}+C\|\nabla\tilde{c}\|^2_{L^2} \\ \leq& \delta \|\nabla\tilde{n}\|^2_{L^2}+\delta \|\Delta\tilde{c}\|_{L^2} +C\|\nabla\tilde{c}\|^2_{L^2}. \end{align*} |
For the term
\begin{align*} J_6 = &\int_{\Omega}\nabla \tilde{n}\tilde{n}\nabla\bar{c} d x\leq \|\tilde{n}\|^2_{L^4}\|\Delta \bar{c}\|_{L^2} \\ \leq&(\|\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\nabla \tilde{n}\|^{\frac{1}{2}}_{L^2}+\| \tilde{n}\|_{L^2})\|\Delta \bar{c}\|_{L^2} \\ \leq&\delta \|\nabla \tilde{n}\|^2_{L^2}+C\|\tilde{n}\|^2_{L^2}+C. \end{align*} |
For the term
\begin{align*} J_7 = &\alpha (\gamma +1)\int_{\Omega} \tilde{n}^2 d x+2\mu \int_{\Omega}\bar{n}\tilde{n}^2 d x+\alpha \int_{\Omega}\tilde{n} g_n d x \\ \leq &(\gamma +1)\|\tilde{n}\|^2_{L^2}+\|g_n\|_{L^2}\|\tilde{n}\|_{L^2}+\|\bar{n}\|_{L^2}\|\tilde{n}\|^2_{L^4} \\ \leq& (\gamma +1)\|\tilde{n}\|^2_{L^2}+\|g_n\|_{L^2}\|\tilde{n}\|_{L^2}+\|\bar{n}\|_{L^2}(\|\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\nabla\tilde{n}\|^{\frac{1}{2}}_{L^2}+\|\tilde{n}\|_{L^2} ) \\ \leq& \delta\|\nabla\tilde{n}\|_{L^2}+ C\|\tilde{n}\|^2_{L^2}+C\|g_n\|^2_{L^2}. \end{align*} |
Therefore, by choosing
\begin{align} &\frac{d}{d t}\|\tilde{n}\|^2_{L^2}+\|\tilde{n}\|^2_{H^1} \\ \leq& C(\|\tilde{n}\|^2_{L^2}+\|\nabla\tilde{c}\|^2_{L^2}+\|\tilde{u}\|^2_{L^2}) +\delta \|\Delta\tilde{c}\|_{L^2} +\delta \|\nabla\tilde{u}\|^2_{L^2}+C\|g_n\|^2_{L^2}. \end{align} | (5.11) |
By choosing
\begin{align*} &\frac{d}{d t}(\|\tilde{n}\|^2_{L^2}+\|\tilde{c}\|^2_{H^1}+\|\tilde{u}\|^2_{L^2})+\|\tilde{n}\|^2_{H^1}+\|\tilde{c}\|^2_{H^2}+\|\tilde{u}\|^2_{H^1} \\ \leq& C(\|g_n\|^2_{L^2}+\|g_c\|^2_{L^2}+\|g_u\|^2_{L^2})+C(\|\tilde{n}\|^2_{L^2}+\|\tilde{c}\|^2_{H^1}+\|\tilde{u}\|^2_{L^2}). \end{align*} |
Applying Gronwall's lemma to the resulting differential inequality, we obatin
\begin{align} \|\tilde{n}\|^2_{L^2}+\|\tilde{c}\|^2_{H^1}+\|\tilde{u}\|^2_{L^2} +\int_0^t\|\tilde{n}\|^2_{H^1}d\tau+\int_0^t\|\tilde{c}\|^2_{H^2}d\tau+\int_0^t\|\tilde{u}\|^2_{H^1}d\tau\leq C. \end{align} | (5.12) |
Taking the
\begin{align*} &\frac{1}{2} \frac{d}{d t}\int_{\Omega} |\nabla\tilde{u}|^2 d x+ \int_{\Omega} |\Delta\tilde{u}|^2 d x \\ = &\int_{\Omega}\bar{u}\cdot\nabla \tilde{u} \Delta\tilde{u} d x+\int_{\Omega}\tilde{u}\cdot\nabla\bar{u}\Delta \tilde{u} d x-\alpha \int_{\Omega} \tilde{n} \nabla \varphi \Delta \tilde{u} d x-\alpha \int_{\Omega}g_u \Delta \tilde{u} d x \\ = &J_8+J_9+J_{10}. \end{align*} |
With the use of the Gagliardo-Nirenberg interpolation inequality, we derive
\begin{align*} J_8 = &\int_{\Omega}\bar{u}\cdot\nabla \tilde{u} \Delta\tilde{u} d x \leq \|\bar{u}\|_{L^4} \|\nabla\tilde{u}\|_{L^4}\|\Delta\tilde{u}\|_{L^2} \\ \leq&\|\bar{u}\|_{H^1}(\|\nabla\tilde{u}\|^{\frac{1}{2}}_{L^2}\|\Delta\tilde{u}\|^{\frac{1}{2}}_{L^2}+\|\nabla\tilde{u}\|_{L^2})\|\Delta\tilde{u}\|_{L^2} \\ \leq& \delta\|\Delta \tilde{u}\|^2_{L^2}+C\|\nabla\tilde{u}\|^2_{L^2} \end{align*} |
and
\begin{align*} J_9 = &\int_{\Omega}\tilde{u}\cdot\nabla\bar{u}\Delta \tilde{u} d x\leq \|\Delta \tilde{u}\|_{L^2}\|\nabla\bar{u}\|_{L^4}\|\tilde{u}\|_{L^4} \\ \leq& C\|\Delta \tilde{u}\|_{L^2}\|\nabla\bar{u}\|_{H^1}(\|\nabla\tilde{u}\|^{\frac{1}{2}}_{L^2}\|\tilde{u}\|^{\frac{1}{2}}_{L^2}+\|\tilde{u}\|_{L^2}) \\ \leq& \delta\|\Delta \tilde{u}\|^2_{L^2}+C\|\nabla\tilde{u}\|^2_{L^2}. \end{align*} |
For the term
\begin{align*} J_{10} = &\alpha \int_{\Omega} \tilde{n} \nabla \varphi \Delta \tilde{u} d x-\alpha \int_{\Omega}g_u \Delta \tilde{u} d x \\ \leq&\delta\|\Delta \tilde{u}\|^2_{L^2}+C(\|\tilde{n}\|^2_{L^2}+\|g_u\|^2_{L^2}) . \end{align*} |
By choosing
\begin{align} \frac{d}{d t}\|\nabla\tilde{u}\|^2_{L^2}+\|\Delta \tilde{u}\|^2_{L^2}\leq C\|\nabla\tilde{u}\|^2_{L^2}+C\|g_u\|^2_{L^2}. \end{align} | (5.13) |
Applying
\begin{align*} &\frac{1}{2}\frac{d}{ d t}\int_{\Omega} |\Delta c|^2 d x + \int_{\Omega}|\nabla\Delta c|^2 d x+\int_{\Omega} |\Delta c|^2 d x \\ = &-\int_{\Omega}\nabla (\bar{u}\nabla\tilde{c})\nabla\Delta\tilde{c}d x-\int_{\Omega}\nabla(\tilde{u}\nabla\bar{c})\nabla\Delta\tilde{c} d x+\alpha\int_{\Omega}\nabla \tilde{n}\nabla\Delta\tilde{c} d x \\ &+\alpha \int_{\Omega}\nabla g_c\nabla\Delta\tilde{c} d x \\ = &J_{11}+J_{12}+J_{13}. \end{align*} |
For the first term
\begin{align*} J_{11} = &-\int_{\Omega}\nabla (\bar{u}\nabla\tilde{c})\nabla\Delta\tilde{c}d x = - \int_{\Omega}\nabla \bar{u}\nabla\tilde{c} \nabla\Delta\tilde{c}d x-\int_{\Omega}\bar{u}\Delta\tilde{c} \nabla\Delta\tilde{c}d x \\ \leq& \|\nabla\Delta\tilde{c}\|_{L^2}\|\nabla \bar{u}\|_{L^4}\|\nabla\tilde{c}\|_{L^4}+\|\nabla\Delta\tilde{c}\|_{L^2}\|\bar{u}\|_{L^4}\|\Delta\tilde{c}\|_{L^4} \\ \leq &\|\nabla\Delta\tilde{c}\|_{L^2} (\|\nabla \bar{u}\|^{\frac{1}{2}}_{L^2}\|\Delta \bar{u}\|^{\frac{1}{2}}_{L^2}+\|\nabla \bar{u}\|_{L^2})(\|\nabla \bar{c}\|^{\frac{1}{2}}_{L^2}\|\Delta \bar{c}\|^{\frac{1}{2}}_{L^2}+\|\nabla \bar{c}\|_{L^2}) \\ &+\|\nabla\Delta\tilde{c}\|_{L^2}\|\bar{u}\|_{H^1}(\|\nabla\Delta\tilde{c}\|^\frac{1}{2}_{L^2}\|\Delta\tilde{c}\|^\frac{1}{2}_{L^2}+\|\Delta\tilde{c}\|_{L^2}) \\ \leq&\delta\|\nabla\Delta\tilde{c}\|^2_{L^2}+C\|\Delta \bar{u}\|^2_{L^2}+C\|\Delta \tilde{c}\|^2_{L^2}. \end{align*} |
Similarly, for the term
\begin{align*} J_{12} = &-\int_{\Omega}\nabla(\tilde{u}\nabla\bar{c})\nabla\Delta\tilde{c} d x = -\int_{\Omega}\nabla\tilde{u}\nabla\bar{c} \nabla\Delta\tilde{c} d x-\int_{\Omega}\tilde{u}\Delta\bar{c}\nabla\Delta\tilde{c} d x \\ \leq&\|\nabla\Delta\tilde{c}\|_{L^2}\|\nabla \tilde{u}\|_{L^4}\|\nabla \bar{c}\|_{L^4}+\|\tilde{u}\|_{L^4}\|\Delta\bar{c}\|_{L^4}\|\nabla\Delta\tilde{c}\|_{L^2} \\ \leq &C\|\nabla\Delta\tilde{c}\|_{L^2}(\|\nabla \tilde{u}\|^{\frac{1}{2}}_{L^2}\|\Delta \tilde{u}\|^{\frac{1}{2}}_{L^2}+\|\nabla \tilde{u}\|_{L^2})\|\nabla \bar{c}\|_{H^1} \\ &+(\|\tilde{u}\|^{\frac{1}{2}}_{L^2}\|\nabla\tilde{u}\|^{\frac{1}{2}}_{L^2}+\|\tilde{u}\|_{L^2})(\|\Delta\bar{c}\|^{\frac{1}{2}}_{L^2}\|\nabla\Delta\bar{c}\|^{\frac{1}{2}}_{L^2}+\|\Delta\bar{c}\|_{L^2})\|\nabla\Delta\tilde{c}\|_{L^2} \\ \leq&\delta\|\nabla\Delta\tilde{c}\|^2_{L^2}+\delta \|\Delta \tilde{u}\|^2_{L^2}+C \|\nabla\Delta\bar{c}\|^2_{L^2}+C\|\nabla\tilde{u}\|^2_{L^2}. \end{align*} |
For the rest term
\begin{align*} J_{13} = &\alpha\int_{\Omega}\nabla \tilde{n}\nabla\Delta\tilde{c} d x +\alpha \int_{\Omega}\nabla g_c\nabla\Delta\tilde{c} d x \\ \leq &\delta\|\nabla\Delta\tilde{c}\|^2_{L^2}+C(\|\nabla \tilde{n}\|^2_{L^2}+\|\nabla g_c\|^2_{L^2}). \end{align*} |
By choosing
\begin{align} &\frac{d}{d t}\|\Delta\tilde{c}\|^2_{L^2}+\|\Delta\tilde{c}\|^2_{H^1} \\ \leq& C(\|\nabla \tilde{n}\|^2_{L^2}+\|\Delta \tilde{c}\|^2_{L^2}+\|\nabla\tilde{u}\|^2_{L^2})+ C \|\Delta \bar{u}\|^2_{L^2}+\delta \|\Delta \tilde{u}\|^2_{L^2} \\ &+C \|\nabla\Delta\bar{c}\|^2_{L^2}+C\|\nabla g_c\|^2_{L^2}. \end{align} | (5.14) |
From (5.13) and (5.14), along with
\begin{align*} &\frac{d}{d t}(\|\nabla\tilde{u}\|^2_{L^2}+\|\Delta\tilde{c}\|^2_{L^2})+\|\Delta \tilde{u}\|^2_{L^2}+\|\Delta\tilde{c}\|^2_{H^1} \\ \leq& C(\|\nabla\tilde{u}\|^2_{L^2}+\|\Delta\tilde{c}\|^2_{L^2})+(\|\nabla \tilde{n}\|^2_{L^2}+\|\Delta \bar{u}\|^2_{L^2}+ \|\nabla\Delta\bar{c}\|^2_{L^2}+\|\nabla g_c\|^2_{L^2}) +C\|g_u\|^2_{L^2}. \end{align*} |
Applying Gronwall's lemma to the resulting differential inequality, we know
\begin{align*} \|\nabla\tilde{u}\|^2_{L^2}+\|\Delta\tilde{c}\|^2_{L^2}+\int_{0}^t \|\Delta \tilde{u}\|^2_{L^2} d\tau+\int_{0}^t \|\Delta\tilde{c}\|^2_{H^1} d\tau\leq C. \end{align*} |
Taking the
\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega} |\nabla \tilde{n}|^2 d x +\int_{\Omega} |\Delta \tilde{n}|^2 d x+\int_{\Omega} |\nabla \tilde{n}|^2 d x \\ = &-\int_{\Omega}\bar{u}\cdot \nabla\tilde{n} \Delta \tilde{n}d x-\int_{\Omega}\tilde{u}\cdot \nabla\bar{n}\Delta \tilde{n}d x-\int_{\Omega} \nabla(\tilde{n}\nabla \bar{c})\Delta\tilde{n} d x-\int_{\Omega} \nabla(\bar{n}\nabla \tilde{c})\Delta\tilde{n} d x \\ &-\alpha(1+\gamma)\int_{\Omega}\tilde{n}\Delta\tilde{n} d x+2\mu\int_{\Omega}\tilde{n}\bar{n} \Delta\tilde{n}d x-\alpha \int_{\Omega}g_n\Delta\tilde{n} d x \\ = &J_{14}+J_{15}+J_{16}+J_{17}+J_{18}. \end{align*} |
With the Gagliardo-Nirenberg interpolation inequality in hand, we can estimate
\begin{align*} J_{14} = &-\int_{\Omega}\bar{u}\cdot \nabla\tilde{n} \Delta \tilde{n}d x\leq \|\bar{u}\|_{L^4}\|\nabla\tilde{n}\|_{L^4}\|\Delta\tilde{n}\|_{L^2} \\ \leq &C\|\bar{u}\|_{H^1}(\|\nabla\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\Delta\tilde{n}\|^{\frac{1}{2}}_{L^2}+\|\nabla\tilde{n}\|_{L^2})\|\Delta\tilde{n}\|_{L^2} \\ \leq &\delta\|\Delta\tilde{n}\|^2_{L^2}+C\|\nabla\tilde{n}\|^{2}_{L^2}. \end{align*} |
Similar to above estimates, we see
\begin{align*} J_{15} = &-\int_{\Omega}\tilde{u}\cdot \nabla\bar{n}\Delta \tilde{n}d x\leq \|\tilde{u}\|_{L^4}\|\nabla \bar{n}\|_{L^4}\|\Delta\tilde{n}\|_{L^2} \\ \leq&C\|\tilde{u}\|_{H^1}\|\nabla \bar{n}\|_{H^1}\|\Delta\tilde{n}\|_{L^2} \\ \leq& \delta \|\Delta\tilde{n}\|_{L^2}+C\|\nabla \bar{n}\|^2_{H^1}. \end{align*} |
Similarly, we derive
\begin{align*} J_{16} = &-\int_{\Omega} \nabla(\tilde{n}\nabla \bar{c})\Delta\tilde{n} d x = -\int_{\Omega}\nabla\tilde{n}\nabla \bar{c}\Delta\tilde{n} d x-\int_{\Omega}\tilde{n}\Delta \bar{c}\Delta\tilde{n} d x \\ \leq &\|\nabla\tilde{n}\|_{L^4}\|\nabla\bar{c}\|_{L^4}\|\Delta\tilde{n}\|_{L^2}+\|\tilde{n}\|_{L^4}\|\Delta\bar{c}\|_{L^4}\|\Delta\tilde{n}\|_{L^2} \\ \leq&(\|\nabla\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\Delta\tilde{n}\|^{\frac{1}{2}}_{L^2} +\|\nabla\tilde{n}\|_{L^2})\|\nabla\bar{c}\|_{H^1}\|\Delta\tilde{n}\|_{L^2} \\ &+(\|\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\nabla\tilde{n}\|^{\frac{1}{2}}_{L^2}+\|\tilde{n}\|_{L^2})(\|\Delta\bar{c}\|^{\frac{1}{2}}_{L^2}\|\nabla\Delta\bar{c}\|^{\frac{1}{2}}_{L^2}+\|\Delta\bar{c}\|_{L^2})\|\Delta\tilde{n}\|_{L^2} \\ \leq&\delta\|\Delta\tilde{n}\|^2_{L^2}+C\|\nabla\tilde{n}\|^{2}_{L^2}+C\|\nabla\Delta\bar{c}\|^{2}_{L^2}+C \end{align*} |
and
\begin{align*} J_{17} = &-\int_{\Omega} \nabla(\bar{n}\nabla \tilde{c})\Delta\tilde{n} d x = -\int_{\Omega} \nabla\bar{n}\nabla \tilde{c}\Delta\tilde{n} d x-\int_{\Omega} \nabla\bar{n}\Delta \tilde{c}\Delta\tilde{n} d x \\ \leq&\|\nabla\bar{n}\|_{L^4}\|\nabla\tilde{c}\|_{L^4}\|\Delta\tilde{n}\|_{L^2}+\|\bar{n}\|_{L^4}\|\Delta\tilde{c}\|_{L^4}\|\Delta\tilde{n}\|_{L^2} \\ \leq &(\|\nabla\bar{n}\|^{\frac{1}{2}}_{L^2}\|\Delta\bar{n}\|^{\frac{1}{2}}_{L^2}+\|\nabla\bar{n}\|_{L^2})\|\nabla \tilde{c}\|_{H^1}\|\Delta\tilde{n}\|_{L^2} \\ &+\|\bar{n}\|_{H^1}(\|\Delta\tilde{c}\|^{\frac{1}{2}}_{L^2}\|\nabla\Delta\tilde{c}\|^{\frac{1}{2}}_{L^2}+\|\Delta\tilde{c}\|_{L^2})\|\Delta\tilde{n}\|_{L^2} \\ \leq &\delta\|\Delta\tilde{n}\|^2_{L^2}+C\|\nabla\Delta\tilde{c}\|^{2}_{L^2}+C. \end{align*} |
For the rest terms, we know
\begin{align*} J_{18} = &-\alpha(1+\gamma)\int_{\Omega}\tilde{n}\Delta\tilde{n} d x+2\mu\int_{\Omega}\tilde{n}\bar{n} \Delta\tilde{n}d x-\alpha \int_{\Omega}g_n\Delta\tilde{n} d x \\ \leq&(1+\gamma)\|\tilde{n}\|_{L^2}\|\Delta\tilde{n}\|_{L^2}+2\mu\|\tilde{n}\|_{L^4}\|\bar{n}\|_{L^4}\|\Delta\tilde{n}\|_{L^2}+\|g_n\|_{L^2}\|\Delta\tilde{n}\|_{L^2} \\ \leq& (1+\gamma)\|\tilde{n}\|_{L^2}\|\Delta\tilde{n}\|_{L^2}+C(\|\tilde{n}\|^{\frac{1}{2}}_{L^2}\|\nabla\tilde{n}\|^{\frac{1}{2}}_{L^2}+\|\tilde{n}\|_{L^2})\|\bar{n}\|_{H^1}\|\Delta\tilde{n}\|_{L^2} \\ &+\|g_n\|_{L^2}\|\Delta\tilde{n}\|_{L^2} \\ \leq &\delta\|\Delta\tilde{n}\|^2_{L^2}+C\|\nabla\tilde{n}\|^{2}_{L^2}+C\|g_n\|^2_{L^2}. \end{align*} |
Therefore, Taking
\begin{align*} &\frac{d}{d t}\|\nabla\tilde{n}\|^{2}_{L^2}+ \|\nabla\tilde{n}\|^{2}_{H^1} \\ \leq& C(\|\nabla\tilde{n}\|^{2}_{L^2} +\|\nabla \bar{n}\|^2_{H^1}+ \|\nabla\Delta\bar{c}\|^{2}_{L^2}+\|\nabla\Delta\tilde{c}\|^{2}_{L^2}+\|g_n\|^2_{L^2})+C. \end{align*} |
Applying Gronwall's lemma to the resulting differential inequality, we know
\begin{align*} \|\nabla\tilde{n}\|^{2}_{L^2}+\int_0^t \|\nabla\tilde{n}\|^{2}_{H^1}d\tau \leq C. \end{align*} |
Therefore, from Leray-Schauder theorem, we derive the existence of solution for (5.5). Along with the regularity of
Theorem 5.1. Assume that
\begin{align} &\beta_1\int_{0}^T\int_{\Omega_d} (\bar{n}-n_d)\tilde{n}d x d t + \beta_2\int_{0}^T\int_{\Omega_d} (\bar{c}-c_d)\tilde{c}d x d t + \beta_3\int_{0}^T\int_{\Omega_d} (\bar{u}-u_d)\tilde{u}d x d t \\ &+\beta_4\int_{\Omega_d} (\bar{n}(T)-n_{\Omega})\tilde{n}(T)d x +\beta_5\int_{\Omega_d} (\bar{c}(T)-c_{\Omega})\tilde{c}(T)d x \\ &-\int_{0}^T\int_{\Omega}(\tilde{n}_{t}-\Delta \tilde{n}+\bar{u} \cdot \nabla \tilde{n}+\tilde{u}\cdot\nabla \bar{n}+\nabla \cdot(\bar{n} \nabla \tilde{c}) +\nabla(\tilde{n}\nabla \bar{c})-\gamma \tilde{n} +2\mu \tilde{n}\bar{n})\lambda d x d t \\ &-\int_{0}^T\int_{\Omega}\left(\tilde{c}_{t}-\Delta \tilde{c}+\bar{u} \cdot \nabla \tilde{c}+\tilde{u}\cdot\nabla \bar{c}+\tilde{c}- \tilde{n}\right)\eta d x d t+\beta_7\int_{0}^T\int_{\Omega_d} \tilde{f}\bar{f}d x d t \\ &-\int_{0}^T\int_{\Omega}\left(\tilde{u}_{t}-\Delta \tilde{u}+\bar{u} \cdot \nabla \tilde{u}+\tilde{u}\cdot \nabla \bar{u}-\tilde{n} \nabla \varphi \right) \rho d x d t +\int_{\Omega}\tilde{n}(0)\xi d x+\int_{\Omega}\tilde{c}(0)\varphi d x \\ &+\int_{\Omega}\tilde{u}(0)\omega d x+\beta_6\int_{\Omega_d} (\bar{u}(T)-u_{\Omega})\tilde{u}(T)d x+\int_{0}^T\int_{\Omega}\tilde{f}\eta d x d t \geq 0, \end{align} | (5.15) |
where
Proof. With the Lemma 5.3 in hand, we get that
\begin{align*} &J^{\prime}(\bar{s})[r]-\langle G_1^{\prime}(\bar{s})[r],\lambda \rangle-\langle G_2^{\prime}(\bar{s})[r],\eta \rangle-\langle G_3^{\prime}(\bar{s})[r],\rho \rangle-\langle G_4^{\prime}(\bar{s})[r],\xi \rangle \\ &-\langle G_5^{\prime}(\bar{s})[r],\varphi \rangle -\langle G_6^{\prime}(\bar{s})[r],\omega \rangle \geq 0, \end{align*} |
for all
Corollary 5.1. Assume that
\begin{align} &\int_{0}^T\int_{\Omega}(\tilde{n}_{t}-\Delta \tilde{n}+\bar{u} \cdot \nabla \tilde{n} +\nabla(\tilde{n}\nabla \bar{c})-\gamma \tilde{n} +2\mu \tilde{n}\bar{n})\lambda d x d t -\int_{0}^T\int_{\Omega}\tilde{n}\eta d x d t \\ &-\int_{0}^T\int_{\Omega}\tilde{n} \nabla \varphi \rho d x d t = \beta_1\int_{0}^T\int_{\Omega_d}(\bar{n}-n_d)\tilde{n}d x d t, \end{align} | (5.16) |
\begin{align} &\int_{0}^T\int_{\Omega}\left(\tilde{c}_{t}-\Delta \tilde{c}+\bar{u} \cdot \nabla \tilde{c}+\tilde{c}\right)\eta d x d t+\int_{0}^T\int_{\Omega} \nabla \cdot(\bar{n} \nabla \tilde{c})\lambda d x d t \\ = &\beta_2\int_{0}^T\int_{\Omega_d} (\bar{c}-c_d)\tilde{c}d x d t, \end{align} | (5.17) |
\begin{align} &\int_{0}^T\int_{\Omega}\left(\tilde{u}_{t}-\Delta \tilde{u}+\bar{u} \cdot \nabla \tilde{u}+\tilde{u}\cdot \nabla \bar{u} \right) \rho d x d t + \int_{0}^T\int_{\Omega}\tilde{u}\nabla \bar{n} \lambda d x d t \\ &+\int_{0}^T\int_{\Omega} \tilde{u}\cdot\nabla \bar{c}\eta d x d t = \beta_3\int_{0}^T\int_{\Omega_d} (\bar{u}-u_d)\tilde{u}d x d t, \end{align} | (5.18) |
which corresponds to the linear system
\begin{align} \left\{\begin{aligned} &-\lambda_t-\Delta \lambda +\bar{u}\cdot\nabla \lambda -\nabla \lambda\nabla \bar{c}-\gamma \lambda +2\mu\lambda\bar{n}-\eta - \nabla\varphi\rho \\ & = \beta_1(\bar{n}-n_d), \\ &-\eta_t-\Delta \eta+\bar{u}\cdot\nabla\eta+\eta+\nabla(\bar{n}\nabla\lambda) = \beta_2 (\bar{c}-c_d), \\ &-\rho_t-\Delta \rho+(\bar{u}\cdot\nabla)\rho+(\rho\cdot\nabla^{T})\bar{u}+\lambda \nabla \bar{n}+\eta \nabla \bar{c} = \beta_3(\bar{u}-u_d), \end{aligned} \right. \end{align} | (5.19) |
subject to the following boundary and final conditions
\begin{align*} \left\{\begin{aligned} &\nabla\cdot \rho = 0, &&\mathit{\text{in}}\; Q, \\ &\frac{\partial \lambda }{\partial \nu} = \frac{\partial \eta }{\partial \nu}, \rho = 0,&& \mathit{\text{on}}\; (0,T)\times \partial\Omega, \\ &\lambda(T) = \beta_4(\bar{n}(T)-n_{\Omega}),\eta(T) = \beta_5 (\bar{c}(T)-c_{\Omega}), \\ &\rho(T) = \beta_5(\bar{c}(T)-c_{\Omega}),&& \mathit{\text{in}}\; \Omega, \end{aligned} \right. \end{align*} |
and the following identities hold
\begin{align} \int_{0}^T\int_{\Omega_d}(\beta_7\bar{f}+\eta)(f-\bar{f}) d x d t \geq 0, \;\forall f \in\mathcal{U}. \end{align} | (5.20) |
Proof. By taking
\begin{align*} \beta_7\int_{0}^{T}\tilde{f}\bar{f} d x d t+\int_{0}^{T}\tilde{f}\eta d x d t\geq 0, \quad \forall \tilde{f} \in \mathcal{C}(\bar{f}). \end{align*} |
By choosing
Theorem 5.2. Under the assumptions of Theorem 5.1, system (5.19) has a unique weak solution such that
\begin{align*} \|\lambda\|^2_{H^1}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2}+\int_0^t\|\lambda\|^2_{H^2}d \tau +\int_0^t\|\eta\|^2_{H^1}d \tau+\int_0^t\|\rho\|^2_{H^1} d \tau \leq C. \end{align*} |
Proof. For convenience, we set
\begin{align} \left\{\begin{aligned} &\lambda_t-\Delta \lambda +\bar{u}\cdot\nabla \lambda -\nabla \lambda\nabla \bar{c}-\gamma \lambda +2\mu\lambda\bar{n}-\eta - \nabla\varphi\rho \\ & = \beta_1(\bar{n}-n_d),&&\text{ in } Q, \\ &\eta_t-\Delta \eta+\bar{u}\cdot\nabla\eta+\eta+\nabla(\bar{n}\nabla\lambda) = \beta_2 (\bar{c}-c_d),&&\text{ in } Q, \\ &\rho_t-\Delta \rho+(\bar{u}\cdot\nabla)\rho+(\rho\cdot\nabla^{T})\bar{u}+\lambda \nabla \bar{n}+\eta \nabla \bar{c} = \beta_3(\bar{u}-u_d),&&\text{ in } Q, \end{aligned} \right. \end{align} | (5.21) |
subject to the following boundary and final conditions
\begin{align*} \left\{\begin{aligned} &\nabla\cdot \rho = 0, &&\text{ in } Q, \\ &\frac{\partial \lambda }{\partial \nu} = \frac{\partial \eta }{\partial \nu}, \rho = 0,&& \text{ on } (0,T)\times\partial \Omega, \\ &\lambda(0) = \beta_4(\bar{n}(T)-n_{\Omega}),\eta(0) = \beta_5 (\bar{c}(T)-c_{\Omega}), \\ &\rho(0) = \beta_5(\bar{c}(T)-c_{\Omega}),&& \text{ in } \Omega. \end{aligned} \right. \end{align*} |
Following an analogous reasoning as in the proof of Lemma 5.3, we omit the process and just give a number of a priori estimates as follows.
Taking the
\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega}\lambda^2 d x+ \int_{\Omega}|\nabla\lambda|^2 d x +2\mu\int_{\Omega}\lambda^2\bar{n} d x \\ = &\int_{\Omega} \nabla \lambda \nabla \bar{c}d x+\gamma\int_{\Omega}\lambda^2 d x +\int_{\Omega}\lambda \eta d x+\int_{\Omega}\lambda \nabla \varphi \rho d x+ \beta_1\int_{\Omega}(\bar{n}-n_d) \lambda d x \\ \leq &\|\nabla\lambda\|_{L^2}\|\nabla\bar{c}\|_{L^2}+\gamma\|\lambda\|^2_{L^2}+\|\lambda\|_{L^2}(\|\eta\|_{L^2}+\|\rho\|_{L^2})+\beta_1\|\bar{n}-n_d\|_{L^2}\|\lambda\|_{L^2} \\ \leq& \frac{1}{2}\|\nabla\lambda\|^2_{L^2}+C(\|\lambda\|^2_{L^2}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2})+C\|\bar{n}-n_d\|^2_{L^2}. \end{align*} |
Then, we have
\begin{align} \frac{d}{d t}\|\lambda\|^2_{L^2}+\|\lambda\|^2_{H^1}\leq C(\|\lambda\|^2_{L^2}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2})+C\|\bar{n}-n_d\|^2_{L^2}. \end{align} | (5.22) |
Taking the
\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega}|\nabla\lambda|^2 d x + \int_{\Omega}|\Delta\lambda|^2 d x \\ = &\int_{\Omega}\bar{u}\cdot \nabla \lambda \Delta\lambda d x -\int_{\Omega}\nabla \lambda \nabla \bar{c}\Delta \lambda d x-\gamma \int_{\Omega}\lambda \Delta \lambda d x +2\mu \int_{\Omega}\lambda \bar{n}\Delta \lambda d x \\ &- \int_{\Omega}\eta \Delta \lambda d x-\int_{\Omega}\nabla \varphi \rho\Delta \lambda d x +\beta_1\int_{\Omega}(\bar{n}-n_d)\Delta\lambda d x \\ \leq&\|\bar{u}\|_{L^4}\|\nabla \lambda\|_{L^4}\|\Delta \lambda \|_{L^2} +\|\nabla \lambda\|_{L^4}\|\nabla \bar{c}\|_{L^4}\|\Delta \lambda \|_{L^2} +\gamma\|\nabla \lambda\|^2_{L^2} \\ &+\| \lambda\|_{L^4}\|\bar{n}\|_{L^4}\|\Delta \lambda \|_{L^2}+\|\eta\|_{L^2}\|\Delta \lambda \|_{L^2}+ \| \rho\|_{L^2} \|\Delta \lambda \|_{L^2} \\ &+\beta_1\|\Delta \lambda \|_{L^2}\|\bar{n}-n_d\|^2_{L^2} \\ \leq& \|\bar{u}\|_{H^1}(\|\nabla \lambda\|^{\frac{1}{2}}_{L^2}\|\Delta \lambda\|^{\frac{1}{2}}_{L^2}+\|\nabla \lambda\|_{L^2})\|\Delta \lambda\|_{L^2}+\gamma\|\nabla \lambda\|^2_{L^2} \\ &+(\|\nabla \lambda\|^{\frac{1}{2}}_{L^2}\|\Delta \lambda\|^{\frac{1}{2}}_{L^2}+\|\nabla \lambda\|_{L^2})\|\nabla\bar{c}\|_{H^1}\|\Delta \lambda\|_{L^2}+\|\eta\|_{L^2}\|\Delta \lambda \|_{L^2} \\ &+\| \rho\|_{L^2} \|\Delta \lambda \|_{L^2}+\beta_1\|\Delta \lambda \|_{L^2}\|\bar{n}-n_d\|^2_{L^2} \\ \leq&\frac{1}{2}\|\Delta \lambda\|^{2}_{L^2}+C(\|\nabla \lambda\|^{2}_{L^2}+\|\eta\|^2_{L^2}+\| \rho\|^2_{L^2} ). \end{align*} |
Thus, we get
\begin{align} \frac{d}{d t}\|\nabla\lambda\|^2_{L^2}+\|\nabla\lambda\|^2_{H^1}\leq C(\|\nabla\lambda\|^2_{L^2}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2})+C\|\bar{n}-n_d\|^2_{L^2}. \end{align} | (5.23) |
Taking the
\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega}\eta^2 d x+ \int_{\Omega}|\nabla\eta|^2 d x+ \int_{\Omega}\eta^2 d x \\ = & \int_{\Omega}\bar{n}\nabla\lambda \nabla \eta d x+\beta_2 \int_{\Omega}\eta (\bar{c}-c_d) d x \\ \leq& \|\bar{n}\|_{L^{4}}\|\nabla \lambda\|_{L^4}\|\nabla \eta\|_{L^2}+\beta_2\|\eta \|_{L^2}\|\bar{c}-c_d\|_{L^2} \\ \leq&\|\bar{n}\|_{H^1}(\|\nabla \lambda\|^{\frac{1}{2}}_{L^2}\|\Delta \lambda\|^{\frac{1}{2}}_{L^2}+\|\nabla \lambda\|_{L^2})\|\nabla \eta\|_{L^2}+\beta_2\|\eta \|_{L^2}\|\bar{c}-c_d\|_{L^2} \\ \leq &\frac{1}{2}\|\nabla \eta\|^2_{L^2}+\delta\|\Delta \lambda\|^2_{L^2}+C\|\nabla \lambda\|_{L^2}+C\|\eta \|^2_{L^2}+C\|\bar{c}-c_d\|^2_{L^2}. \end{align*} |
As an immediate consequence, we obtain
\begin{align} \frac{d}{d t}\|\eta \|^2_{L^2}+\|\eta \|^2_{H^1} \leq \delta\|\Delta \lambda\|^2_{L^2}+C\|\nabla \lambda\|_{L^2}+C\|\eta \|^2_{L^2}+C\|\bar{c}-c_d\|^2_{L^2}. \end{align} | (5.24) |
Taking the
\begin{align*} &\frac{1}{2}\frac{d}{d t}\int_{\Omega}\rho^2 d x+ \int_{\Omega}|\nabla\rho|^2 d x \\ = &-\int_{\Omega} (\rho\cdot\nabla^{T})\bar{u} \rho d x-\lambda\int_{\Omega}\nabla \bar{n} \rho d x-\int_{\Omega}\eta \nabla \bar{c} \rho d x+\beta_3\int_{\Omega}(\bar{u}-u_d) \rho d x \\ \leq &\|\rho\|_{L^2}\|\nabla \bar{u}\|_{L^4}\|\rho\|_{L^4}+\lambda \|\nabla \bar{n}\|_{L^2}\|\rho\|_{L^2}+\|\eta \|_{L^2}\|\nabla \bar{c}\|_{L^4}\|\rho\|_{L^4} \\ &+\beta_3\|\rho\|_{L^2}\|\bar{u}-u_d\|_{L^2} \\ \leq& \|\rho\|_{L^2}\|\nabla \bar{u}\|_{H^1}(\|\rho\|^{\frac{1}{2}}_{L^2}\|\nabla\rho\|^{\frac{1}{2}}_{L^2}+\|\rho\|_{L^2})+\lambda \|\nabla \bar{n}\|_{L^2}\|\rho\|_{L^2} \\ &+\|\eta \|_{L^2}\|\nabla \bar{c}\|_{H^1}(\|\rho\|^{\frac{1}{2}}_{L^2}\|\nabla\rho\|^{\frac{1}{2}}_{L^2}+\|\rho\|_{L^2})+\beta_3\|\rho\|_{L^2}\|\bar{u}-u_d\|_{L^2} \\ \leq&\frac{1}{2}\|\nabla\rho\|^{2}_{L^2}+C\|\rho\|^2_{L^2}(\|\nabla \bar{u}\|^2_{H^1}+1)+C\|\eta \|^2_{L^2}+C\|\bar{u}-u_d\|^2_{L^2}. \end{align*} |
Therefore, we see that
\begin{align} \frac{d}{d t}\|\rho\|^{2}_{L^2}+\|\rho\|^{2}_{H^1}\leq C\|\rho\|^2_{L^2}(\|\nabla \bar{u}\|^2_{H^1}+1)+C\|\eta \|^2_{L^2}+C\|\bar{u}-u_d\|^2_{L^2}. \end{align} | (5.25) |
Combining (5.22)-(5.25) and taking
\begin{align*} &\frac{d}{ d t}(\|\lambda\|^2_{H^1}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2})+ \|\lambda\|^2_{H^2}+ \|\eta\|^2_{H^1} +\|\rho\|^2_{H^1} \\ \leq& C(\|\nabla \bar{u}\|^2_{H^1}+1)( \|\lambda\|^2_{H^1}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2})+C\|\bar{n}-n_d\|^2_{L^2} \\ &+C\|\bar{c}-c_d\|^2_{L^2}+C\|\bar{u}-u_d\|^2_{L^2}. \end{align*} |
Applying Gronwall's lemma to the resulting differential inequality, we know
\begin{align*} \|\lambda\|^2_{H^1}+\|\eta\|^2_{L^2}+\|\rho\|^2_{L^2}+\int_0^t\|\lambda\|^2_{H^2}d \tau +\int_0^t\|\eta\|^2_{H^1}d \tau+\int_0^t\|\rho\|^2_{H^1} d \tau \leq C. \end{align*} |
The proof is complete.
The authors would like to express their deep thanks to the referee's valuable suggestions for the revision and improvement of the manuscript.
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