# Existence of Strong Solutions for Nonlinear Systems of PDEs Arising in Convective Flow

Authors: Khaled Bouazzaoui; Mohammed Aiboudi; Sameh Elsayed Ahmed
Journal: International Journal of Differential Equations
 (2022)

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## Abstract

In this paper, we will study the existence of strong solutions for a nonlinear system of partial differential equations arising in convective flow, modeling a phenomenon of mixed convection created by a heated and diving plate in a porous medium saturated with a fluid. The main tools are Schäfer’s fixed-point theorem, the Fredholm alternative, and some theorems on second-order elliptic operators.


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## Body

## 1. Introduction

In recent years, many authors have studied the case of the semi-infinite vertical plane plate immersed in a porous medium saturated with a fluid. The following problem is derived from this phenomenon:(1)∂2Ψ∂x2+∂2Ψ∂y2=k∂yT,λ∂2T∂x2+∂2T∂y2=∂xT∂yΨ−∂yT∂xΨ,with mixed boundary conditions(2)∂xΨx,0=−ωxm−1/2,Tx,0=Tωx=T∞+Axm,∂yΨx,∞=0,Tx,∞=T∞,where x,y is the rectangular Cartesian coordinates system. The constants k and λ depend on the density, the viscosity, and the thermal expansion coefficient of the fluid. They also depend on the permeability and the thermal diffusivity of the saturated porous medium.In the framework of boundary layer approximations, by introducing similarity variables, we can transform the system of partial differential equations into a system of ordinary differential equations of the third order with appropriate boundary values (see [1], p.3). These two-point boundary value problems can be studied by using a shooting method (see for example [2, 3]). For the auxiliary dynamical system, we refer the reader to [4]. For an integral equation, we refer the reader to [5]. For nonstandard analysis techniques, we refer the reader to [6]. For numerical method, we refer the reader to [7]. The second natural way of dealing with this problem, which is the framework of this paper, is straightly related to the coupled partial differential equations (see [1, 8, 9]).The aim of this paper is to generalize the existence of strong solutions of the problem introduced in [1, 8, 9]. More precisely, we will give new results about the existence of strong solutions, in particular, without assuming that ΨH2 is small (as in [8], Theorem 4). We choose the data (the function K) in the larger space inspired by other works, and we will give conditions which make each solution in C∞Ω¯2.Let us introduce now the problem in which we are interested. LetΩ be a bounded domain of ℝ2 whose boundary Γ=∂Ω is sufficiently smooth and divided into two parts Γ1andΓ2 such that measΓ1≠0 and(3)Γ¯1∪Γ¯2=Γ,Γ1∩Γ2=∅.OnΩ, we consider the boundary value problem defined by(4)ΔΨ=K⋅∇Ta.e.Ω,(5)λΔT=∇T⋅∇Ψ⊥,a.e.Ω,with boundary conditions(6)Ψ=0onΓ1,∂Ψ∂ν=0onΓ2,(7)T=Tωon∂Ω,where λ is a strictly positive constant, ν is the unit outward normal vector on ∂Ω, and ∇Ψ⊥=∂yΨ,−∂xΨ.


## 2. Preliminaries

In this section, we will introduce the notations, definitions, and preliminary facts which are used throughout the paper.


### 2.1. Sobolev Spaces and Notations

For a positive integerk and for 1≤p≤∞, the Sobolev space is defined as follows:(8)Wk,pΩ=u∈LpΩ,∀α∈ℕ2withα≤k:Dαu∈LpΩ,where Dαu is the α-derivative of u in the sense of distributions. For p=2, we denote by HkΩ the space Wk,2Ω. The boundary space is defined as(9)Bk−1/p,p∂Ω=the space of boundary valuesφof functionsu∈Wk,pΩ,1<p<∞, equipped with the norm(10)φBk−1/p,p=infuWk,p:u|∂Ω=φ.Throughout the paper, we use the following notations:(i)
Wk,2Ω=HkΩ, Bk−1/2,2∂Ω=Hk−1/2∂Ω, and C is Poincaré’s constant of Ω(ii)
DΩ,Γi=u∈DΩ¯:⊃pu∩Γi=∅∪pu(iii)
W01,pΩ,Γi=DΩ,Γi¯W1,pΩ,W01,pΩ=W01,pΩ,Γ0, and W01,2Ω=H01Ω(iv)
⋅p,1≤p≤∞, is the norm of LpΩ (or the norm of LpΩ×LpΩ)(v)
⋅ is the norm of LE,F, i.e., the space of linear operators from E into F (normed spaces)


### 2.2. Variational Formulation

Let us assume(11)Tω∈H3/2∂Ω.Then, from the existence and uniqueness theorem for the Dirichlet problem for strong solutions (see [10], Theorem 5.13), there exists a unique function Θ∈H2Ω satisfying(12)ΔΘ=0a.e.Ω,Θ=Tωon∂Ω.Furthermore,Θ∈L∞Ω and ∇Θ∈L4Ω2. For the coefficient K=k1,k2, it is supposed that(13)K∈W24div,Ω,where(14)Wqpdiv,Ω=f∈LpΩ,ℝ2:divK∈LqΩ.Let us notice thatDΩ¯,ℝ2 is dense in W24div,Ω (see [11], Proposition 3.57). Let us set H=T−Θ. Then, Ψ,T is the solution of problem (5)–(7) if and only if Ψ,H is the solution of the problem(15)ΔΨ=K⋅∇H+Θa.e.Ω,(16)λΔH=∇H⋅∇Ψ⊥+∇Θ⋅∇Ψ⊥a.e.Ω,with boundary conditions(17)Ψ=0onΓ1,∂Ψ∂ν=0onΓ2,(18)H=0on∂Ω.Let us assume thatΨ,H∈H2Ω2 is a solution of problem (15)–(18). Multiplying equation (15) by a function u∈H01Ω,Γ1 and equation (16) by a function v∈H01Ω, integrating on Ω, and using Green’s formula, we yield(19)∇Ψ,∇u=−K⋅∇H+Θ,u,∀u∈H01Ω,Γ1,λ∇H,∇v=−v∇H,∇Ψ⊥−v∇Θ,∇Ψ⊥,∀v∈H01Ω,where(20)f,g=∫Ωfg.


## 2.1. Sobolev Spaces and Notations

For a positive integerk and for 1≤p≤∞, the Sobolev space is defined as follows:(8)Wk,pΩ=u∈LpΩ,∀α∈ℕ2withα≤k:Dαu∈LpΩ,where Dαu is the α-derivative of u in the sense of distributions. For p=2, we denote by HkΩ the space Wk,2Ω. The boundary space is defined as(9)Bk−1/p,p∂Ω=the space of boundary valuesφof functionsu∈Wk,pΩ,1<p<∞, equipped with the norm(10)φBk−1/p,p=infuWk,p:u|∂Ω=φ.Throughout the paper, we use the following notations:(i)
Wk,2Ω=HkΩ, Bk−1/2,2∂Ω=Hk−1/2∂Ω, and C is Poincaré’s constant of Ω(ii)
DΩ,Γi=u∈DΩ¯:⊃pu∩Γi=∅∪pu(iii)
W01,pΩ,Γi=DΩ,Γi¯W1,pΩ,W01,pΩ=W01,pΩ,Γ0, and W01,2Ω=H01Ω(iv)
⋅p,1≤p≤∞, is the norm of LpΩ (or the norm of LpΩ×LpΩ)(v)
⋅ is the norm of LE,F, i.e., the space of linear operators from E into F (normed spaces)


## 2.2. Variational Formulation

Let us assume(11)Tω∈H3/2∂Ω.Then, from the existence and uniqueness theorem for the Dirichlet problem for strong solutions (see [10], Theorem 5.13), there exists a unique function Θ∈H2Ω satisfying(12)ΔΘ=0a.e.Ω,Θ=Tωon∂Ω.Furthermore,Θ∈L∞Ω and ∇Θ∈L4Ω2. For the coefficient K=k1,k2, it is supposed that(13)K∈W24div,Ω,where(14)Wqpdiv,Ω=f∈LpΩ,ℝ2:divK∈LqΩ.Let us notice thatDΩ¯,ℝ2 is dense in W24div,Ω (see [11], Proposition 3.57). Let us set H=T−Θ. Then, Ψ,T is the solution of problem (5)–(7) if and only if Ψ,H is the solution of the problem(15)ΔΨ=K⋅∇H+Θa.e.Ω,(16)λΔH=∇H⋅∇Ψ⊥+∇Θ⋅∇Ψ⊥a.e.Ω,with boundary conditions(17)Ψ=0onΓ1,∂Ψ∂ν=0onΓ2,(18)H=0on∂Ω.Let us assume thatΨ,H∈H2Ω2 is a solution of problem (15)–(18). Multiplying equation (15) by a function u∈H01Ω,Γ1 and equation (16) by a function v∈H01Ω, integrating on Ω, and using Green’s formula, we yield(19)∇Ψ,∇u=−K⋅∇H+Θ,u,∀u∈H01Ω,Γ1,λ∇H,∇v=−v∇H,∇Ψ⊥−v∇Θ,∇Ψ⊥,∀v∈H01Ω,where(20)f,g=∫Ωfg.


## 3. Main Results

We start with the following lemma which plays a key role in our main results.Lemma 1.
LetΨ∈W1,4Ω. The function H≡0 is the unique solution of problem(21)λΔH−∇H⋅∇Ψ⊥=0a.eΩ,H=0on∂Ω,in the space H2Ω∩H01Ω.Proof.
It is clear thatH≡0 is the solution of problem (21). Then, if H∈H2Ω∩H01Ω satisfies λΔH−∇H⋅∇Ψ⊥=0, multiplying the members of this equation by H, we get(22)λHΔH=H∇H⋅∇Ψ⊥,a.e.  onΩ.
By integration onΩ, we get(23)λ∫ΩHΔH=−λ∇H22=∫ΩH∇H⋅∇Ψ⊥=∇H⋅∇Ψ⊥,H.
On the other hand, we have(24)∇H⋅∇Ψ⊥=divH∇Ψ⊥−Hdiv∇Ψ⊥=divH∇Ψ⊥,because div∇Ψ⊥=0, in the sense of distributions. Since ∇H⋅∇Ψ⊥∈L2Ω, divH∇Ψ⊥∈L2Ω, and hence, by the density of DΩ in H01Ω (when Ω is at least a Lipschitz open set), we have(25)−λ∇H22=divH⋅∇Ψ⊥,HH−1Ω,H01Ω=−H⋅∇Ψ⊥,∇H=−∇H⋅∇Ψ⊥,H=λ∇H22,that is, ∇H2=0. Thus, H=0.Remark 1.
From the proof of the previous lemma, we observe that for allH∈H2Ω∩H01Ω and Ψ∈W1,4Ω, we have(26)H,∇H⋅∇Ψ⊥=0.
Thus, by the density ofDΩ in H01Ω and by the embedding H1Ω↬L4Ω, we can easily show that for all H∈H01Ω and Ψ∈W1,4Ω, we have(27)H,∇H⋅∇Ψ⊥=0.Now, we will prove the following proposition which plays an important role in proving our main results.Proposition 1.
Let us assume that∂Ω is of class C1,1 and Ψ∈W1,4Ω. Then, for all f∈L2Ω, the equation(28)λΔH−∇H⋅∇Ψ⊥=fa.e.Ωhas exactly one solution H in the space H2Ω∩H01Ω.Proof.
Let us consider(29)TΨ:H2Ω∩H01Ω⟶L2Ω,defined by(30)TΨH=∇H⋅∇Ψ⊥,T:H2Ω∩H01Ω⟶L2Ω,where the operator is defined by(31)TH=λΔH.
Then, it is clear that for allH∈H2Ω∩H01Ω, we have(32)λΔH−∇H⋅∇Ψ⊥=f⟺TH−TΨH=f.
From the existence and uniqueness theorems for the Dirichlet problem for strong solutions (see [12], Theorem 9.15, Lemma 9.17, problem 9.8) combined with the open mapping theorem (see [13], Corollary 2.7), the operator T is invertible. On the other hand, we can write TΨ=φ2∘φ1 where(33)φ1:H2Ω∩H01Ω⟶L4Ω2,defined by(34)φ1H=i¯c∘jH,with j as the operator from H2Ω∩H01Ω into H1Ω2 defined by jH=∇H and i¯c=ic,ic, where ic:H1Ω↬L4Ω, and(35)φ2:L4Ω2⟶L2Ω,defined by(36)φ2ξ=ξ⋅∇Ψ⊥.
Then,TΨ is compact. Indeed, φ1 is continuous because icandj are continuous, and the operator ic is compact (Rellich–Kondrachov theorem); it is clear that φ2 is continuous. Now, it is easy to show that for all H∈H2Ω∩H01Ω, (32) holds, if and only if(37)H−T−1∘TΨH=T−1fholds. By Lemma (21), H≡0 is the unique solution of equation(38)TH−TΨH=0,in the space H2Ω∩H01Ω. Hence, from the Fredholm alternative, equation (28) has exactly one solution H∈H2Ω∩H01Ω. Besides(39)HH2≤I−T−1∘TΨ−1⋅T−1⋅f2,as needed.Lemma 2.
Under the assumptions of the previous proposition, the mappingL from W1,4Ω into H01Ω is defined by LΨ=H where(40)λΔH−∇H⋅∇Ψ⊥=∇Θ⋅∇Ψ⊥a.eΩ,H=0on∂Ω.Then,L=i∘I−T−1∘TΨ−1withi:H2Ω↬H1Ω is continuous. Moreover, there exists R>0 such that for all Ψ∈W1,4Ω, if LΨ=H,(41)H+Θ≤R.Proof.
LetΨ∈W1,4Ω and let H∈H2Ω∩H01Ω be the unique solution of problem (40). Let Ψnn be a sequence of elements in DΩ¯ converging to Ψ in W1,4Ω. Then, there exists a sequence Hnn of elements in H2Ω∩H01Ω satisfying(42)λΔHn−∇Hn⋅∇Ψn⊥=∇Θ⋅∇Ψn⊥,Hn=0on∂Ω.
Then, for alln∈ℕ,Hn+Θ is the solution of the problem(43)λΔW−∇W⋅∇Ψn⊥=0,W=Tωon∂Ω.
By the weak maximum principle theorem (see [12], Theorem 8.1), for all n∈ℕ, we have(44)−sup∂ΩTω≤Hn+Θ≤sup∂ΩTω,and by variational formulation (system (19)), we have(45)λ∇H,∇v=−v∇H,∇Ψ⊥−v∇Θ,∇Ψ⊥,∀v∈H01Ω,λ∇Hn,∇v=−v∇Hn,∇Ψn⊥−v∇Θ,∇Ψn⊥,∀v∈H01Ω.
Replacingv by H−Hn and using Remark 1, we give(46)λ∇Hn−H22=−H−Hn∇H,∇Ψ⊥+H−Hn∇Hn,∇Ψn⊥+H−Hn⋅∇Θ,∇Ψn−Ψ⊥=Hn∇H,∇Ψ⊥+H∇Hn,∇Ψn⊥+H−Hn⋅∇Θ,∇Ψn−Ψ⊥.
On the other hand, forW=H−Hn, we have(47)W⋅∇Θ,∇Ψ⊥=∫ΩWdivΘ⋅∇Ψ⊥−W⋅Θ⋅div∇Ψ⊥=−∫ΩΘ⋅∇W⋅∇Ψ⊥,=−Θ∇W,∇Ψ⊥.
Similarly,H∇Hn,∇Ψn⊥=−Hn∇H,∇Ψn⊥. We replace the previous inequality in inequality (46), and we use Remark 1, to get(48)λ∇H−Hn22=Hn∇H,∇Ψ⊥−Hn∇H,∇Ψn⊥+Θ∇H−Hn,∇Ψ−Ψn⊥,=Hn∇H,∇Ψ−Ψn⊥+Θ∇H−Hn,∇Ψ−Ψn⊥=Hn∇H−Hn,∇Ψ−Ψn⊥+Θ∇H−Hn,∇Ψ−Ψn⊥=Hn+Θ∇H−Hn,∇Ψ−Ψn⊥,≤Hn+Θ∞∇Hn−H2∇Ψn−Ψ2.
Thus,(49)∇H−Hn2≤Rλ∇Ψn−Ψ2⟶0asn⟶∞,with(50)R=sup∂ΩTω.
Then, there exists a subsequenceHnkk of the sequence Hnn converging almost everywhere to H. That is,(51)H+Θ≤R.
From inequality (49), we get, for all Ψ1,Ψ2∈W1,4Ω,(52)∇H1−H22≤Rλ∇Ψ1−Ψ22≤C″∇Ψ1−Ψ24,where C″>0 depends only on Ω,R, and λ, as needed.Remarks 2.
(1)
The main advantage of Proposition1 (particularly, Fredholm’s alternative) is the generalization of the results ([8], Theorem 1 and Theorem 4, and [1], Theorem 4), and more precisely, it leads us to show that there exist strong solutions to problem (15)–(18) without assuming that ΨH2 is bounded by certain constants and without assuming that K⋅∇Θ2 is quite small.(2)
The previous lemma is similar to Lemma 3.1 in [8], but in this paper, we have seen a new technique for its proof (it is proved by a weak maximum principle).The following proposition is useful.Proposition 2.
Let us assume, in addition to assumptions (11)–(13), that ∂Ω is of class C1,1,Γ1=∂Ω, and K∈DΩ¯2. Then, problem (15)–(18) have at least one solution Ψ,H in the space W2,4Ω2.Proof.
For allH∈H01Ω, we have K⋅∇H+Θ∈L2Ω. Then, from the existence and uniqueness theorem for the Dirichlet problem for strong solutions ([12], Theorem 9.15), the problem(53)ΔΨ=K⋅∇H+Θa.e.Ω,Ψ=0on∂Ω,has exactly one solution Ψ∈H2Ω. Besides, there exists C¯>0 such that(54)ΨH2≤C¯K⋅∇H+Θ2,(see [12], Lemma 9.17). Let us denote by Λ the application from H01Ω into H2Ω defined by ΛH=Ψ, and let us consider the application J:H01Ω⟶H01Ω defined by(55)J=L∘i∘Λ,with L given in Lemma 2 and i:H2Ω↬W1,4Ω. Then, we have(56)JH=L∘i∘ΛH,∀H∈H01Ω,that is, the set of the solutions of problem (15)–(18) is(57)ΛH,Hsuch thatH∈H2Ω∩H01ΩandJH=H.
Then,J is continuous and compact. Indeed, LandΛ are continuous because for Ψi=ΛHi,i=1,2, we have(58)ΛΨ1−ΛΨ2H2≤C¯K∞∇H1−H22,and the continuity of L has been proved in Lemma 2. By the Rellich–Kondrachov theorem, we can easily show that i is compact. Now, let H be in the set(59)A=H∈H01Ω:H=εJHfor someε∈0,1.
Then, there existsε∈0,1 such that(60)ΔΨ=εK⋅∇H+K⋅∇Θ,a.eΩ,λΔH−∇H⋅∇Ψ⊥=∇Θ⋅∇Ψ⊥,a.eΩ,Ψ=0on∂Ω,H=0on∂Ω.
By the variational formulation, the previous problem is equivalent to(61)∇Ψ,∇u=−K⋅∇εH+Θ,u,∀u∈H01Ω,λ∇H,∇v=−v∇H,∇Ψ⊥−v∇Θ,∇Ψ⊥,∀v∈H01Ω.
From the first equation in system (61), we have(62)∇Ψ22=−εK⋅∇H,Ψ−K⋅∇Θ,Ψ=εHdivK,Ψ−divHK,Ψ−K⋅∇Θ,Ψ=εHdivK,Ψ+HK,∇Ψ−K⋅∇Θ,Ψ≤R′divK2Ψ2+R′K2∇Ψ2+K4∇Θ4Ψ2,with R′=R′Θ,Ω. Hence,(63)∇Ψ2≤CR′divK2+R′K2+CK4∇Θ4.
Now, from the second equation in system (61), since H∇H,∇Ψ⊥=0 (see Remark 1) and H∇Θ,∇Ψ⊥=−Θ∇H,∇Ψ⊥ (the same method as the one used to get inequality (47)), we have(64)∇H2≤Θ∞λ∇Ψ2.
From inequalities (63) and (64), we deduce that HH1<∞, i.e., A is bounded. From Schäfer’s fixed-point theorem (see [14], Theorem 11.1, p.59), we deduce that J has at least one fixed point H∈H2Ω∩H01Ω. That is, ΛH,H is the strong solution of problem (15)–(18). Furthermore, K⋅∇H+Θ∈L4Ω. Thus, Ψ∈W2,4Ω, i.e., ∇Ψ∈L∞Ω2. Hence, H∈W2,4Ω, as needed.Remark 3.
From the previous proof, it is observed that under the assumptions(65)K∈L∞Ω×L∞Ω,divK∈L∞Ω,the previous result (the previous proposition) is still valid, which is the generalization of ([8], Theorem 1 and Theorem 4). In addition, Proposition 1 helps us to generalize the previous result in the case where K∈W24div,Ω.Now, we are going to generalize the previous result in the case whereK∈W24div,Ω. To this end, we start with the following remark.Remark 4.
LetΨ,H∈W2,4Ω∩W01,4Ω2 be a solution of the problem(66)ΔΨ=K⋅∇H+K⋅∇Θ,a.e.Ω,λΔH−∇H⋅∇Ψ⊥=∇Θ⋅∇Ψ⊥,a.e.Ω,Ψ=0on∂Ω,H=0on∂Ω.
Then, from the first equation with boundary conditions, for allu∈H01Ω, we have(67)∫Ω∇Ψ⋅∇u=−K⋅∇H+Θ,u=−∫ΩK⋅∇H⋅u−∫ΩK⋅∇Θ⋅u=∫ΩHdivK−divHK⋅u−∫ΩK⋅∇Θ⋅u=∫ΩHdivK⋅u−∫ΩdivHK⋅u−∫ΩK⋅∇Θ⋅u∫ΩHdivK⋅u+∫ΩHK⋅∇u−∫ΩK⋅∇Θ⋅u=HdivK,u+HK,∇u−K⋅∇Θ,u.
Now, from the second equation in system (66) with boundary conditions, for all v∈H01Ω, we have(68)λ∇H,∇v=−v∇H,∇Ψ⊥−v∇Θ,∇Ψ⊥.
On the other hand, we have(69)H∇v,∇Ψ⊥=−v∇H,∇Ψ⊥,the same method as the one used to get inequality (47). Then, we have(70)λ∇H,∇v=H∇v,∇Ψ⊥−v∇Θ,∇Ψ⊥.
Thereby, ifΨ,H∈W2,4Ω∩W01,4Ω2 is a solution of problem (66), then for all u,v∈H01Ω, we have(71)∇Ψ,∇u=K⋅∇H+Θ,u=HdivK,u+HK,∇u−K⋅∇Θ,u,(72)λ∇H,∇v=H∇v,∇Ψ⊥−v∇Θ,∇Ψ⊥.Now, we will prove the following result.Theorem 1.
Let us assume, in addition to assumptions (11)–(13), that ∂Ω is of class C1,1 and Γ1=∂Ω. Then, problem (15)–(18) have at least one solution Ψ,H in the space W2,4Ω∩W01,4Ω2.Proof.
LetKnn⊆DΩ¯ be a sequence converging to K in W24div,Ω. Then, from Proposition 2, there exists Ψn,Hnn⊆W2,4Ω∩W01,4Ω2 which is bounded in H1Ω2 and satisfying(73)ΔΨn=Kn⋅∇Hn+Kn⋅∇Θ,a.eΩ,λΔHn−∇Hn⋅∇Ψn⊥=∇Θ⋅∇Ψn⊥,a.eΩ,Ψn=0on∂Ω,Hn=0on∂Ω.
By the compactness of the embeddingH1Ω↬LpΩ,p≥2, there exists a subsequence Ψnk,Hnkk of the sequence Ψn,Hnn converging weakly in H1Ω2 and strongly in LpΩ,p≥2, to an element denoted Ψ∗,H∗, and we have H∗∈L∞Ω. Now, we shall complete the proof in three steps.Step 1.
Ψnk,Hnkk converges in H1 to Ψ∗,H∗. Indeed, from Remark 4, for all u∈DΩ, we have(74)∇Ψnk+p−Ψnk,∇u=Hnk+pdivKnk+p−HnkdivKnk,u+Hnk+pKnk+p−HnkKnk,u−Knk+p−Knk∇Θ,u=Hnk+pdivKnk+p−Knk,u+divKnkHnk+p−Hnk,u+Hnk+pKnk+p−Knk,∇u+KnkHnk+p−Hnk,∇u−Knk+p−Knk∇Θ,u.
Then, foru=Ψnk+p−Ψnk, we have(75)∇Ψnk+p−Ψnk22≤Hnk+p∞divKnk+p−Knk2Ψnk+p−Ψnk2+divKk2Hnk+p−Hnk4Ψnk+p−Ψk4+Hnk+p∞Knk+p−Knk2∇Ψnk+p−Ψnk2+Knk4Hnk+p−Hnk4∇Ψnk+p−Ψnk2+Knk+p−Knk4∇Θ4∇Ψnk+p−Ψnk2.
By Lemma2 and the fact that Knn is convergent in W24div,Ω, we get that there are strictly positive constants C1,C2, and C3, such that(76)∇Ψnk+p−Ψnk2≤C1divKnk+p−Knk2+C2Knk+p−Knk4+C3Hnk+p−Hnk4.
That is,Ψnkk converges strongly to Ψ∗. On the other hand, for all v∈H01Ω, we have(77)λ∇Hnk,∇v=Hnk∇v,∇Ψnk⊥−v∇Θ,∇Ψnk⊥,
By the same process as the one used to obtain inequality (52), we have(78)∇Hnk+p−Hnk2≤Θ∞λ∇Ψnk+p−Ψnk2,that is, Hnkk converges strongly to H∗.Step 2.
Ψ∗,H∗ satisfies systems (71) and (72). Indeed, for all u∈DΩ, we have(79)Knk⋅∇Hnk+Θ,u=Knk⋅∇Hnk−H∗,u+Knk⋅∇H∗+Θ,u.
For allu∈DΩ, we have(80)Knk∇Hnk−H∗,u≤Knk4∇Hnk−H∗2u4⟶0,Knk⋅∇H∗+Θ,u−K⋅∇H∗+Θ,u≤Knk−K4∇H∗+Θ2u4⟶0.
Taking the limit ask⟶∞ in (71) when Ψ,H,K=Ψnk,Hnk,Knk gives(81)∇Ψ∗,∇u=K⋅∇H∗+Θ,u,∀u∈DΩ.
Now, for allv∈H01Ω, we have(82)v∇Hnk,∇Ψnk⊥=−Hnk⋅∇v,∇Ψnk⊥=−Hnk⋅∇v,∇Ψnk−Ψ∗⊥−Hnk⋅∇v,∇Ψ∗⊥⟶−H∗⋅∇v,∇Ψ∗⊥,because(83)Hnk⋅∇v,∇Ψnk−∇Ψ∗⊥≤R′∇v2∇Ψnk−Ψ∗⊥2⟶0.By the dominated convergence theorem, we have(84)Hnk⋅∇v,∇Ψ∗⊥⟶H∗∇v,∇Ψ∗⊥.
Obviously,(85)∇Hnk−H∗,∇v,v∇Θ,∇Ψnk−Ψ∗⊥⟶0.
Taking the limit in (77) as k⟶∞ yields(86)λ∇H∗,∇v=H∗∇v,∇Ψ∗⊥−v∇Θ,∇Ψ∗⊥.Step 3.
Ψ∗,H∗∈W2,4Ω2. Indeed(87)∇Ψ∗,∇u=−K⋅∇H∗+Θ,u,∀u∈DΩ.
SinceK∈L4Ω2, we have K⋅∇H∗+Θ∈L4/3Ω. Then, from the existence and uniqueness theorems for the Dirichlet problem for strong solutions and since the application(88)Ψ,u∈H01Ω2⟶∇Ψ,∇uis continuous and coercive, we deduce that Ψ∗∈W2,4/3Ω. Thus, ∇Ψ∗∈W1,4/3Ω2 in which W1,4/3Ω↬L4Ω (see [11], Theorem 2.72), that is, Ψ∗∈W1,4Ω2. Then, from Proposition 1, there exists a unique H∈H2Ω∩H01Ω satisfying(89)λ∇H,∇v−H∇v,∇Ψ∗⊥=v∇Θ,∇Ψ∗⊥,∀v∈H01Ω,λΔH−∇H∇Ψ∗⊥=∇Θ∇Ψ∗⊥.
On the other hand, from Remark1, the bilinear form(90)H,v∈H01Ω2⟶∇H,∇v−H∇v,∇Ψ∗⊥is continuous and coercive. Thus, from the Lax–Milgram theorem, we deduce that H=H∗. Hence, K⋅∇H∗∈L2Ω, that is, K⋅∇H∗+Θ∈L2Ω, and thereby, from the existence and uniqueness theorems for the Dirichlet problem for strong solutions ([12], Theorem 9.15, problem 9.8) and from the Lax–Milgram theorem, Ψ∗ is the unique function in H2Ω∩H01Ω obeying(91)∇Ψ∗,∇u=−K⋅∇H∗+Θ,u,∀u∈H01Ω,ΔΨ∗=K⋅∇H∗+Θ.
By the same method as the one used in the previous proposition, we get(92)Ψ,H∈W2,4Ω2.
This completes the proof.Remark 5.
Notice that in Step3 of the previous proof, Proposition 1 is necessary. Also, we need Lemma 1 and Remark 1, particularly, H∇H,∇Ψ⊥=0 for H∈H01Ω and Ψ∈W1,4Ω which is a different result from ([8], Lemma 2.1) and it is proved by the generalized Green formula.In addition to the previous results, we will study the existence of solutions inCnΩ,n≥3.Theorem 2.
Let us assume, in addition to the hypotheses of the previous theorem, that∂Ω is of class Cn, Tω∈Hn−1/2∂Ω, and K∈Wn−2,4Ω with n≥2. Then, problem (15)–(18) have at least one solution Ψ,H in the space HnΩ×HnΩ.Proof.
Let us assumen≥3, and let Ψ,H∈W2,4Ω2 be a solution of problem (15)–(18). Then, we have(93)∇Ψ,∇H∈H1Ω∩L∞Ω2.
On the other hand, from (12) and by the regularity theorem of elliptic problem ([12], Theorem 8.13), Θ is in HnΩ. It then follows that(94)∂xiΘ∈Hn−1Ω∩L∞Ω.
SinceK∈Wn−2,4Ω2,n≥3, we have(95)K⋅∇Θ∈H1Ω,K⋅∇H∈H1Ω.
From the regularity theorem of elliptic problem ([12], Theorem 8.13), we deduce that Ψ is in H3Ω. Hence,(96)∇Θ⋅∇Ψ⊥,∇H⋅∇Ψ⊥∈H1Ω,that is, H∈H3Ω. Continuing the above process, we obtain, after n−2 steps, Ψ,H∈HnΩ2.Let us denote byγ0 the trace operator on ∂Ω. Then, we have the following result.Corollary 1.
Let us assume, in addition to the hypotheses of the previous theorem, that∂Ω is of class C∞ and K∈C∞Ω¯2 and there exists φ∈C∞Ω¯ such that γ0φ=Tω. Then, problem (15)–(18) have at least one solution Ψ,H in the space C∞Ω¯×C∞Ω¯.Remark 6.
IfΓ1andΓ2 are closed and open subsets and Ω is sufficiently smooth (being able to apply Poincaré’s inequality), then by applying Theorem 2.24 given in ([15], p. 132) or Theorem 5.8 given in ([16], p. 146), we can prove that the previous results of the existence of strong solutions are still valid in the general case where Γ2≠∅.


## 4. Conclusion

In this paper, we study a problem given by two strongly coupled partial differential equations in a two-dimensional bounded domain, modeling a phenomenon of convective flow created by a heated and diving plate in a porous medium saturated with a fluid. We have established the existence of strong solutions without assuming thatΨH2 is bounded by a certain constant (nor K∇Θ2 is small enough) and by choosing K in W24div,Ω. Besides, we have added a new result (which does not exist in the literature) which is the existence of solutions of class Cn,n≥3 (particularly, solutions of class C∞).



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