# Global Analysis of a Blood Flow Model with Artificial Boundaries

Authors: S. C. Oukouomi Noutchie
Journal: ISRN Applied Mathematics
 (2013)

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## Abstract

A theoretical model for blood flow in ramifying arteries was introduced and studied numerically (Quarteroni and Veneziani, 2003). A special experimental condition was considered on the artificial boundaries. In this paper, the aim is to analyze the well-posedness of this model, with the focus on the stilted boundary conditions. We use Brouwer’s fixed point theorem to show the existence of a solution to the stationary problem. For the evolutionary version, we use some energy estimates and Galerkin’s method to prove global existence, uniqueness, and stability of a weak solution.


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## Body

## 1. Introduction

Evolution systems with artificial boundaries are difficult to analyze as a result of the complex dynamics at the boundaries and very few papers have attempted to capture these processes from an analytical point of view. This paper explores this question regarding the flow of blood in a portion of a large artery and addresses the analysis of the Navier-Stokes problem, having provided boundary conditions which can be considered as a generalization of the mean pressure drop problem investigated in [1–3], as they arise in bioengineering applications. Our purpose is to consider both the stationary case and the nonstationary case. In this regard, we will prove the existence of a weak solution for the stationary case based on Brouwer’s fixed point theorem and afterwards, we will establish a well-posedness analysis for the nonstationary case based on a suitable energy estimate that we are going to derive as well as a well-known compactness argument.


### 1.1. Basic Notations

In this subsection, we summarize some notations that will occur throughout the paper. Vectors and tensors are denoted by bold-face lettersx=(x1,x2,x3): location of fluid particle,u: velocity field of the flow,p: pressure,∇p: the pressure gradient,I: identity tensor,T: symmetric Cauchy stress tensor,μ: dynamic viscosity,ρ: fluid mass density,ℝ: the set of real numbers,|·|: the absolute value of ℝ and correspondingly, the norm ofℝ3,Ω: a bounded domain inℝ3,Γ: the boundary ofΩ,∇u: the gradient ofu,∇·u: the divergence ofu,Δu: the Laplacian ofu.The spacesC(Ω),Ck(Ω),C0k(Ω), andC0∞(Ω)and their vector-valued analoguesC(Ω),Ck(Ω),C0k(Ω), and C0∞(Ω) are defined as usual, the superscript indicating continuous derivatives to a certain order and the subscript zero indicating functions with compact support.The spaceLP(Ω), the Hölder space Ck,γ(Ω), and the Sobolev space Wpk(Ω) and their vector-valued analogues LP(Ω), Ck,γ(Ω), and Wpk(Ω) are also defined as usual. In particular, H1(Ω)=W21(Ω).For functions depending on space and time, for a given spaceV of space-dependent functions, we define (for some T>0)
(1)L2(0,T;V)={∥v(t)∥V2∫0T‍v:(0,T)⟶V∣vismeasurableand∫0T‍∥v(t)∥V2dt<∞}
with norm ∥v∥L2(0,T;V)=(∫0T‍∥v(t)∥V2dt)1/2.When considering functions which depend only on time, we define the spaceL∞(0,T)={z:(0,T)→ℝ3∣esssupt∈(0,T)|z(t)|<∞} endowed with the norm
(2)∥z∥L∞(0,T)=esssupt∈(0,T)|z(t)|.


### 1.2. Preliminaries

In this subsection, we recall some assumptions used in [2, 4], and we are going to make use of them throughout our analysis. In what follows, Ω is a bounded domain of ℝ3 with boundary Γ sufficiently smooth. Γ consists of the artery wall denoted by Γwall and some artificial sections. The velocity is required to be zero on the artery wall. To account for homogeneous Dirichlet boundary conditions on the artery wall, we define
(3)X≡{φ∈H1(Ω):φ|Γwall=0}.Note that Poincaré’s inequality(4)∥φ∥≤CΩ∥∇φ∥
holds for φ∈X [2, 4]. The artificial sections consist of the upstream section on the side of the heart and the downstream sections on the side of the peripheral vessels. Rather than giving serious thought to the artificial sections boundary conditions, in seeking a variational formulation, the test space is left free on these portions of the boundary. Accordingly, we introduce
(5)V≡{φ∈H1(Ω):φ|Γwall=0,∇·φ=0}
as the test space. To prove an existence theorem for a Navier-Stokes problem, either steady or nonsteady, it is convenient to construct the solution as a limit of Galerkin approximations in terms of the eigenfunctions of the corresponding steady Stokes problem. This use of the Stokes eigenfunctions originated with Prodi and was further developed by Heywood [5]. To define the corresponding Stokes operator, we introduce V⋆ as the completion of V with respect to the norm of L2(Ω). Then for every f∈V⋆, there exists exactly one v∈V satisfying
(6)(∇v,∇φ)=(f,φ),∀φ∈V.
Moreover, for each v∈V, there exist at most one f∈V⋆ such that (6) holds. In this way, a one-to-one correspondence can be defined between elements f of V⋆ and functions v belonging to an allowing suitable subspace of V that we denote by D(Δ~). The Stokes operator
(7)Δ~:D(Δ~)⟶V⋆
is defined setting
(8)-Δ~v=f
so that (6) is satisfied.The inverse operatorΔ~-1 is completely continuous and self-adjoint. Therefore it possesses a sequence of eigenfunctions {ak}k=1∞, which are complete and orthogonal in bothVandV⋆.In what follows,ci (i=1,2,…) will denote generic constants, not necessarily the same at different places. The inequalities
(9)supΩ|v|≤c1∥∇v∥1/2∥Δ~v∥1/2,(10)∥∇v∥≤c2∥Δ~v∥
are satisfied for everyv∈D(Δ~), provided thatΩis a bounded domain (see [4] page 178).Next, we provide a few theorems [1, 3, 6] used throughout our analysis.Theorem 1 (Brouwer’s fixed point theorem).
LetXbe a finite dimensional linear space endowed with a norm. Assume
(11)u:B(0,1)⟶B(0,1)
is continuous, whereB(0,1)denotes the closed-unit ball inX. Thenuhas a fixed point; that is, there exists a pointx∈B(0,1)with
(12)u(x)=x.Theorem 2 (Rellich-Kondrachov compactness theorem).
AssumeUis a bounded open subset ofℝn, with aC1boundary. Suppose1≤p<n. Then
(13)Wp1(U)⊂⊂Lq(U)
for each1≤q<p⋆, where1/p⋆=(1/p)-(1/n).As a result of that, we have(14)∥u∥Lq(U)≤C∥u∥Wp1(U).Theorem 3 (trace inequality for solenoidal functions).
Letube a solenoidal function defined onU. Then the following inequality holds:
(15)∫∂U‍(u·n)2ds≤c∥u∥2.Theorem 4 (weak compactness).
LetX be a reflexive Banach space and suppose that the sequence {uk}k=1∞⊂X is bounded. Then there exists a subsequence {ukj}j=1∞⊂{uk}k=0∞ and u∈X such that
(16)ukj⇀u.In other words, bounded sequences in a reflexive Banach space are weakly precompact. In particular, a bounded sequence in a Hilbert space contains a weakly convergent subsequence.


## 1.1. Basic Notations

In this subsection, we summarize some notations that will occur throughout the paper. Vectors and tensors are denoted by bold-face lettersx=(x1,x2,x3): location of fluid particle,u: velocity field of the flow,p: pressure,∇p: the pressure gradient,I: identity tensor,T: symmetric Cauchy stress tensor,μ: dynamic viscosity,ρ: fluid mass density,ℝ: the set of real numbers,|·|: the absolute value of ℝ and correspondingly, the norm ofℝ3,Ω: a bounded domain inℝ3,Γ: the boundary ofΩ,∇u: the gradient ofu,∇·u: the divergence ofu,Δu: the Laplacian ofu.The spacesC(Ω),Ck(Ω),C0k(Ω), andC0∞(Ω)and their vector-valued analoguesC(Ω),Ck(Ω),C0k(Ω), and C0∞(Ω) are defined as usual, the superscript indicating continuous derivatives to a certain order and the subscript zero indicating functions with compact support.The spaceLP(Ω), the Hölder space Ck,γ(Ω), and the Sobolev space Wpk(Ω) and their vector-valued analogues LP(Ω), Ck,γ(Ω), and Wpk(Ω) are also defined as usual. In particular, H1(Ω)=W21(Ω).For functions depending on space and time, for a given spaceV of space-dependent functions, we define (for some T>0)
(1)L2(0,T;V)={∥v(t)∥V2∫0T‍v:(0,T)⟶V∣vismeasurableand∫0T‍∥v(t)∥V2dt<∞}
with norm ∥v∥L2(0,T;V)=(∫0T‍∥v(t)∥V2dt)1/2.When considering functions which depend only on time, we define the spaceL∞(0,T)={z:(0,T)→ℝ3∣esssupt∈(0,T)|z(t)|<∞} endowed with the norm
(2)∥z∥L∞(0,T)=esssupt∈(0,T)|z(t)|.


## 1.2. Preliminaries

In this subsection, we recall some assumptions used in [2, 4], and we are going to make use of them throughout our analysis. In what follows, Ω is a bounded domain of ℝ3 with boundary Γ sufficiently smooth. Γ consists of the artery wall denoted by Γwall and some artificial sections. The velocity is required to be zero on the artery wall. To account for homogeneous Dirichlet boundary conditions on the artery wall, we define
(3)X≡{φ∈H1(Ω):φ|Γwall=0}.Note that Poincaré’s inequality(4)∥φ∥≤CΩ∥∇φ∥
holds for φ∈X [2, 4]. The artificial sections consist of the upstream section on the side of the heart and the downstream sections on the side of the peripheral vessels. Rather than giving serious thought to the artificial sections boundary conditions, in seeking a variational formulation, the test space is left free on these portions of the boundary. Accordingly, we introduce
(5)V≡{φ∈H1(Ω):φ|Γwall=0,∇·φ=0}
as the test space. To prove an existence theorem for a Navier-Stokes problem, either steady or nonsteady, it is convenient to construct the solution as a limit of Galerkin approximations in terms of the eigenfunctions of the corresponding steady Stokes problem. This use of the Stokes eigenfunctions originated with Prodi and was further developed by Heywood [5]. To define the corresponding Stokes operator, we introduce V⋆ as the completion of V with respect to the norm of L2(Ω). Then for every f∈V⋆, there exists exactly one v∈V satisfying
(6)(∇v,∇φ)=(f,φ),∀φ∈V.
Moreover, for each v∈V, there exist at most one f∈V⋆ such that (6) holds. In this way, a one-to-one correspondence can be defined between elements f of V⋆ and functions v belonging to an allowing suitable subspace of V that we denote by D(Δ~). The Stokes operator
(7)Δ~:D(Δ~)⟶V⋆
is defined setting
(8)-Δ~v=f
so that (6) is satisfied.The inverse operatorΔ~-1 is completely continuous and self-adjoint. Therefore it possesses a sequence of eigenfunctions {ak}k=1∞, which are complete and orthogonal in bothVandV⋆.In what follows,ci (i=1,2,…) will denote generic constants, not necessarily the same at different places. The inequalities
(9)supΩ|v|≤c1∥∇v∥1/2∥Δ~v∥1/2,(10)∥∇v∥≤c2∥Δ~v∥
are satisfied for everyv∈D(Δ~), provided thatΩis a bounded domain (see [4] page 178).Next, we provide a few theorems [1, 3, 6] used throughout our analysis.Theorem 1 (Brouwer’s fixed point theorem).
LetXbe a finite dimensional linear space endowed with a norm. Assume
(11)u:B(0,1)⟶B(0,1)
is continuous, whereB(0,1)denotes the closed-unit ball inX. Thenuhas a fixed point; that is, there exists a pointx∈B(0,1)with
(12)u(x)=x.Theorem 2 (Rellich-Kondrachov compactness theorem).
AssumeUis a bounded open subset ofℝn, with aC1boundary. Suppose1≤p<n. Then
(13)Wp1(U)⊂⊂Lq(U)
for each1≤q<p⋆, where1/p⋆=(1/p)-(1/n).As a result of that, we have(14)∥u∥Lq(U)≤C∥u∥Wp1(U).Theorem 3 (trace inequality for solenoidal functions).
Letube a solenoidal function defined onU. Then the following inequality holds:
(15)∫∂U‍(u·n)2ds≤c∥u∥2.Theorem 4 (weak compactness).
LetX be a reflexive Banach space and suppose that the sequence {uk}k=1∞⊂X is bounded. Then there exists a subsequence {ukj}j=1∞⊂{uk}k=0∞ and u∈X such that
(16)ukj⇀u.In other words, bounded sequences in a reflexive Banach space are weakly precompact. In particular, a bounded sequence in a Hilbert space contains a weakly convergent subsequence.


## 2. Formulation of the Problem

LetΩ⊂ℝ3be the artery portion where we aim at providing a detailed flow analysis. For each x∈Ω and at any time t>0, we denote by u(x,t) and p(x,t) the blood velocity and pressure, respectively. In larger vessels, it is reasonable to assume that blood has a constant viscosity, because the vessel diameters are large compared with the individual cell diameters and because shear rates are high enough for viscosity to be independent of them. Hence, in these vessels the non-Newtonian behavior becomes insignificant and blood can be considered to be a Newtonian fluid ([4] and references therein). In what follows, we assume that the vessels are large enough; blood density and blood viscosity are assumed to be constant. Under these assumptions, blood flow can be described by the Navier-Stokes equations
(17)ρ∂u∂t+ρ(u·∇)u+∇·T=0,∇·u=0,000000000000000000000000x∈Ω,t>0,
where T=T=pI-μ(∇u+(∇u)T). The equations are expressions of balance of linear momentum and incompressibility. We are neglecting the presence of any external forces. Then (18) is obtained by substitution of the divergence-free constraint into the expression for the stress in (17) as follows
(18)ρ∂u∂t+ρ(u·∇)u-μΔu+∇p=0,∇·u=0,00000000000000000000000000000x∈Ω,t>0.In fact,(19)∇·T=∇·(pI-μ(∇u+(∇u)T))=∇p-μΔuprovidedthat∇·u=0.
For the sake of simplicity, we normalizeρto 1.


### 2.1. Initial and Boundary Conditions

The initial condition requires the specification of the flow velocity at the initial time; for example, fort0=0;
(20)u(x,0)=u0(x),x∈Ω,
where the given initial velocity fieldu0is divergence-free. The system (18) has to be provided with boundary conditions. In this respect, we split the boundaryΓinto different parts. In the present work, we are assuming that the wall is rigid so that no-slip boundary condition
(21)u(x,t)=0,x∈Γwall,t>0,
holds. The other parts ofΓare the artificial boundaries which bound the computational domain. For the sake of clarity, we distinguish the upstream section on the side of the heart denoted byΓ1and the downstream sections on the side of the peripheral vessels denoted byΓ2andΓ3. As introduced in [4], the following boundary conditions are provided:
(22)pn-μ(∇u)n-Ki(u·n)n=Pin0000fort>0,x∈Γi(i=1,2,3),
whereKiis a suitable nonnegative constant,Pi=Pi(t)is the prescribed mean pressure on each artificial sectionΓi, andnrepresents the outward normal unit vector on every part of the vessel boundary. These conditions can be considered as a generalization of the mean pressure drop problem investigated in [2] in the sense that whenKi=0 (i=1,2,3), we recover the usual Neumann or natural conditions associated with (18). The physical justification for the case in whichKi≠0is provided in ([4] Figure 4.1).The mathematical formulation of the problem is therefore described by the system of differential equations:(23)∂u∂t+(u·∇)u-μΔu+∇p=0,x∈Ω,t>0,∇·u=0,x∈Ω,t>0,u(x,0)=u0(x),x∈Ω,u(x,t)=0,x∈Γwall,t>0,pn-μ(∇u)n-Ki(u·n)n=Pin,fort>0,x∈Γi.Next we define the following bilinear and trilinear forms:a:H1(Ω)×H1(Ω)→ℝ such  that
(24)a(v(1),v(2))=μ(∇v(1),∇v(2))+∑i=13‍Ki∫Γi‍(v(1)·n)n·v(2)ds,b:H1(Ω)×H1(Ω)×H1(Ω)→ℝ such  that
(25)b(v(1),v(2),v(3))=((v(1)·∇)v(2),v(3)).Following [5] we set
(26)c(t,v)=-∑i=13‍Pi(t)∫Γin·vds,foreachv∈H1(Ω).
Furthermore, if we set
(27)P~=supt≥0∑i=13‍|Pi|,K=max1≤i≤3Ki
and consider the sequence{ak}k=1∞of eigenfunctions ofΔ~-1, the following estimates hold for every v∈span{ak,k=1,2,…}:
(28)|∑i=13‍Pi∫Γi‍Δ~v·nds|≤c3P~∥Δ~v∥,
wherec3depends on the trace inequality (15) and
(29)|∑i=13‍Ki∫Γi‍(v·n)n·Δ~vds|≤c4K∥∇v∥∥Δ~v∥,
wherec4depends on the trace inequality (15) and Poincaré’s inequality [3].The former is a consequence of the trace inequality for solenoidal functions [2] and the proof of (29) is given in [4].


## 2.1. Initial and Boundary Conditions

The initial condition requires the specification of the flow velocity at the initial time; for example, fort0=0;
(20)u(x,0)=u0(x),x∈Ω,
where the given initial velocity fieldu0is divergence-free. The system (18) has to be provided with boundary conditions. In this respect, we split the boundaryΓinto different parts. In the present work, we are assuming that the wall is rigid so that no-slip boundary condition
(21)u(x,t)=0,x∈Γwall,t>0,
holds. The other parts ofΓare the artificial boundaries which bound the computational domain. For the sake of clarity, we distinguish the upstream section on the side of the heart denoted byΓ1and the downstream sections on the side of the peripheral vessels denoted byΓ2andΓ3. As introduced in [4], the following boundary conditions are provided:
(22)pn-μ(∇u)n-Ki(u·n)n=Pin0000fort>0,x∈Γi(i=1,2,3),
whereKiis a suitable nonnegative constant,Pi=Pi(t)is the prescribed mean pressure on each artificial sectionΓi, andnrepresents the outward normal unit vector on every part of the vessel boundary. These conditions can be considered as a generalization of the mean pressure drop problem investigated in [2] in the sense that whenKi=0 (i=1,2,3), we recover the usual Neumann or natural conditions associated with (18). The physical justification for the case in whichKi≠0is provided in ([4] Figure 4.1).The mathematical formulation of the problem is therefore described by the system of differential equations:(23)∂u∂t+(u·∇)u-μΔu+∇p=0,x∈Ω,t>0,∇·u=0,x∈Ω,t>0,u(x,0)=u0(x),x∈Ω,u(x,t)=0,x∈Γwall,t>0,pn-μ(∇u)n-Ki(u·n)n=Pin,fort>0,x∈Γi.Next we define the following bilinear and trilinear forms:a:H1(Ω)×H1(Ω)→ℝ such  that
(24)a(v(1),v(2))=μ(∇v(1),∇v(2))+∑i=13‍Ki∫Γi‍(v(1)·n)n·v(2)ds,b:H1(Ω)×H1(Ω)×H1(Ω)→ℝ such  that
(25)b(v(1),v(2),v(3))=((v(1)·∇)v(2),v(3)).Following [5] we set
(26)c(t,v)=-∑i=13‍Pi(t)∫Γin·vds,foreachv∈H1(Ω).
Furthermore, if we set
(27)P~=supt≥0∑i=13‍|Pi|,K=max1≤i≤3Ki
and consider the sequence{ak}k=1∞of eigenfunctions ofΔ~-1, the following estimates hold for every v∈span{ak,k=1,2,…}:
(28)|∑i=13‍Pi∫Γi‍Δ~v·nds|≤c3P~∥Δ~v∥,
wherec3depends on the trace inequality (15) and
(29)|∑i=13‍Ki∫Γi‍(v·n)n·Δ~vds|≤c4K∥∇v∥∥Δ~v∥,
wherec4depends on the trace inequality (15) and Poincaré’s inequality [3].The former is a consequence of the trace inequality for solenoidal functions [2] and the proof of (29) is given in [4].


## 3. The Stationary Problem

Our purpose is to establish the existence of a weak solution to the stationary version of problem (23). We are going to consider the inner product withφ∈Vin order to formulate the weak formulation.


### 3.1. Weak Formulation

Assume thatuis a solution of the boundary value problem described above andφis a smooth solenoidal vector-valued function defined onΩ. The functionusatisfies the following identities:
(30)((u·∇)u-μΔu+∇p,φ)=0,x∈Ω,(pn-μ(∇u)n-Ki(u·n)n,φ)=(Pin,φ),000000000000000000000x∈Γi,i=1,2,3.
We notice thatφ|Γwall=0sinceφ∈V.We calculate(31)(∇p,φ)=-∫‍Ωp(∇·φ)dx+∫‍Γp(φ·n)ds=∑‍i=13∫‍Γi(Pin+μ(∇u)n+Ki(u·n)n)·φds,(32)(Δu,φ)=-∫Ω‍∑i,j=13‍∂φi∂xj∂ui∂xjdx+∫Γ‍∑i=13‍φi∑j=13‍∂ui∂xjnjds=-(∇φ,∇u)+∑‍i=13∫‍Γi(n·∇)u·φds
to obtain
(33)((u·∇)u,φ)+μ(∇u,∇φ)=-∑‍i=13∫‍ΓiKi(u·n)n·φds-∑‍i=13Pi∫‍Γin·φds.
It is clear that (33) does not depend on the pressurep; therefore, the weak formulation of the problem reads as follows.Given a divergence free velocity fieldu0∈V and nonnegative constants Pi and Ki (i=1,2,3), find u∈Vsuch that
(34)((u·∇)u,φ)+μ(∇u,∇φ)=-∑‍i=13∫‍ΓiKi(u·n)n·φds-∑‍i=13Pi∫‍Γin·φds
holds for all φ∈V, whereV, defined in (5), is the space of test functions.


### 3.2. Galerkin Approximations

The inverse operatorΔ~-1of the Stokes operatorΔ~is self-adjoint and possesses a sequence of eigenfunctions{ak}which are orthogonal inV. Following [6] we fix a positive integern. Galerkin approximations
(35)un=∑k=1n‍dnkak
are defined as solutions of the finite system of equations (k=1,…,n) as follows:
(36)((un·∇)un,ak)+μ(∇un,∇ak)=-∑‍i=13∫‍ΓiKi(un·n)n·akds-∑‍i=13Pi∫‍Γin·akds.
This is a system of linear equations for constant unknowns dnk (k=1,…,n). The identity (37) forunis obtained by multiplying (36) through bydnkand summing overk=1,…,n:
(37)((un·∇)un,un)+μ(∇un,∇un)=-∑‍i=13∫‍ΓiKi(un·n)n·unds-∑‍i=13Pi∫‍Γin·unds.
This implies that
(38)μ∥∇un∥2≤|((un·∇)un,un)|+|∑‍i=13∫‍ΓiKi(un·n)n·unds|+|∑‍i=13Pi∫‍Γin·unds|.
Together with (4), we make use of (28) to obtain
(39)|∑i=13‍Pi(t)∫Γi‍n·(un)ds|≤c3P~∥un∥≤c5P~∥∇un∥.
Likewise, from (29), we have
(40)|∑i=13‍Ki∫Γi‍(un·n)n·unds|≤c4K∥∇un∥∥un∥≤c6K∥∇un∥2.
Furthermore, making use of Hölder’s inequality ([6], page 623), one obtains
(41)((un·∇)un,un)≤∥un∥L6∥∇un∥∥un∥L3≤c7∥∇un∥3,
where c7 depends on Rellich-Kondrachov compactness inequality [1], Poincaré’s inequality [3], and Sobolev inequality ([6], page 270). We make use of these inequalities in (38) and we obtain
(42)μ∥∇un∥2≤c7∥∇un∥3+c5P~∥∇un∥+c6K∥∇un∥2.
Set ξ:=μ-c6K. It follows that
(43)ξ∥∇un∥≤c7∥∇un∥2+c5P~,
where P~=∑i=13‍|Pi|, K=max1≤i≤3Ki.Theorem 5.
Assume(44)K≤μc6,P~≤ξ24c5c7.
For each integer n=1,2,…, there exists a function un of the form (35) satisfying (36) such that
(45)∥∇un∥≤ξ2c7[1-(1-4c5c7P~ξ2)].Proof.
Owing to Poincaré’s inequality (4), ∥∇φ∥ is a norm equivalent to ∥φ∥1 for all φ∈V; therefore, (45) defines a closed ball in span{a1,…,an}. To prove the solvability of the finite-dimensional problem (36), we follow Fujita in using Brouwer’s fixed point theorem (see Theorem 1.1 in [7]), applying it to the continuous mapping v→w defined by the linear problem (k=1,…,n):
(46)((v·∇)w,ak)+μ(∇w,∇ak)=-∑‍i=13∫‍ΓiKi(w·n)n·akds-∑‍i=13Pi∫‍Γin·akds.
Equation (46) is a system of n linear equations. These linear equations are uniquely solvable if v lies in the ball defined by (45), because then w=0 is the only solution of the corresponding homogeneous equation (Ki=Pi=0, 1≤i≤3). In fact, if v satisfies (45) and w satisfies (46) with Ki=Pi=0, we have
(47)ξ∥∇w∥2≤c7∥∇v∥∥∇w∥2≤c7ξ2c7[1-(1-4c5c7P~ξ2)]∥∇w∥2≤ξ2∥∇w∥2.
Together with Poincaré’s inequality (4), this imply that w=0. To see that the mapping v→w takes the ball defined by (45) into itself, suppose that v satisfies (45). Then, similarly to (43), we obtain that
(48)ξ∥∇w∥≤c7∥∇v∥∥∇w∥+c5P~,
and therefore,
(49)∥∇w∥≤c5P~ξ-c7∥∇v∥≤c5P~ξ-(ξ/2)[1-(1-(4c5c7P~/ξ2))]≤ξ2c7[1-(1-4c5c7P~ξ2)].
Thus, (46) defines a continuous mapping v→w of the closed ball
(50){φ∈span{a1,…,an}:∥∇φ∥≤ξ2c7[1-(1-4c5c7P~ξ2)]}
into itself. The map has at least one fixed point, and any such fixed point is a solution of (36). un is chosen to be any one of these fixed points. Hence, Brouwer’s fixed point has been applied and has given the existence of Galerkin approximations
(51)un=∑k=1n‍dnkak
satisfying
(52)∥∇un∥≤ξ2c7[1-(1-4c5c7P~ξ2)].


### 3.3. Existence of a Weak Solution

Theorem 6.
Assume(53)K≤μc6,P~≤ξ24c5c7.
There exists a weak solution to the stationary version of problem (23).Proof.
By Poincaré’s inequality [3], the fact that the sequence {∥∇un∥}n=1∞ is bounded implies that the sequence {un}n=1∞ is bounded in V. As a result, Theorem 4 yields the existence of a subsequence {unq}q=1∞⊂{un}n=1∞ such that
(54)unq⇀uweaklyinV.
Next we show that the weak limit u is in fact a weak solution. In this respect, we are going to show that
(55)((u·∇)u,φ)+μ(∇u,∇φ)=-∑i=13∫ΓiKi(u·n)n·φds-∑‍i=13Pi∫Γin·φds,
for each φ∈V.
Fix an integerk, (k=1,2,…). From (36), we have the identity
(56)((un·∇)un,ak)+μ(∇un,∇ak)=-∑‍i=13∫‍ΓiKi(un·n)n·akds-∑‍i=13Pi∫‍Γin·akds.
We recall (54), to find upon passing to weak limits that
(57)((u·∇)u,ak)+μ(∇u,∇ak)=-∑‍i=13∫‍ΓiKi(u·n)n·akds-∑‍i=13Pi∫‍Γin·akds.
Since {ak}k=1∞ is a basis of V, it follows that
(58)((u·∇)u,φ)+μ(∇u,∇φ)=-∑‍i=13∫‍ΓiKi(u·n)n·φds-∑‍i=13Pi∫‍Γin·φds,
for each φ∈V.Note that, in taking this limit, there is no difficulty with the nonlinear term. In fact, we have that(59)(unq·∇)unq-(u·∇)u=(unq·∇)unq+(unq·∇)u-(unq·∇)u-(u·∇)u=(unq·∇)(unq-u)+((unq-u)·∇)u.
Also, ak∈H2(Ω) is an eigenfunction of the inverse of the stokes operator Δ~ and H2(Ω)⊂C(Ω¯). This implies that ak∈C(Ω¯) [6]. It follows that
(60)((unq·∇)unq-(u·∇)u,ak)=((unq·∇)(unq-u),ak)+(((unq-u)·∇)u,ak)⟶0
(see [5], page 651).


## 3.1. Weak Formulation

Assume thatuis a solution of the boundary value problem described above andφis a smooth solenoidal vector-valued function defined onΩ. The functionusatisfies the following identities:
(30)((u·∇)u-μΔu+∇p,φ)=0,x∈Ω,(pn-μ(∇u)n-Ki(u·n)n,φ)=(Pin,φ),000000000000000000000x∈Γi,i=1,2,3.
We notice thatφ|Γwall=0sinceφ∈V.We calculate(31)(∇p,φ)=-∫‍Ωp(∇·φ)dx+∫‍Γp(φ·n)ds=∑‍i=13∫‍Γi(Pin+μ(∇u)n+Ki(u·n)n)·φds,(32)(Δu,φ)=-∫Ω‍∑i,j=13‍∂φi∂xj∂ui∂xjdx+∫Γ‍∑i=13‍φi∑j=13‍∂ui∂xjnjds=-(∇φ,∇u)+∑‍i=13∫‍Γi(n·∇)u·φds
to obtain
(33)((u·∇)u,φ)+μ(∇u,∇φ)=-∑‍i=13∫‍ΓiKi(u·n)n·φds-∑‍i=13Pi∫‍Γin·φds.
It is clear that (33) does not depend on the pressurep; therefore, the weak formulation of the problem reads as follows.Given a divergence free velocity fieldu0∈V and nonnegative constants Pi and Ki (i=1,2,3), find u∈Vsuch that
(34)((u·∇)u,φ)+μ(∇u,∇φ)=-∑‍i=13∫‍ΓiKi(u·n)n·φds-∑‍i=13Pi∫‍Γin·φds
holds for all φ∈V, whereV, defined in (5), is the space of test functions.


## 3.2. Galerkin Approximations

The inverse operatorΔ~-1of the Stokes operatorΔ~is self-adjoint and possesses a sequence of eigenfunctions{ak}which are orthogonal inV. Following [6] we fix a positive integern. Galerkin approximations
(35)un=∑k=1n‍dnkak
are defined as solutions of the finite system of equations (k=1,…,n) as follows:
(36)((un·∇)un,ak)+μ(∇un,∇ak)=-∑‍i=13∫‍ΓiKi(un·n)n·akds-∑‍i=13Pi∫‍Γin·akds.
This is a system of linear equations for constant unknowns dnk (k=1,…,n). The identity (37) forunis obtained by multiplying (36) through bydnkand summing overk=1,…,n:
(37)((un·∇)un,un)+μ(∇un,∇un)=-∑‍i=13∫‍ΓiKi(un·n)n·unds-∑‍i=13Pi∫‍Γin·unds.
This implies that
(38)μ∥∇un∥2≤|((un·∇)un,un)|+|∑‍i=13∫‍ΓiKi(un·n)n·unds|+|∑‍i=13Pi∫‍Γin·unds|.
Together with (4), we make use of (28) to obtain
(39)|∑i=13‍Pi(t)∫Γi‍n·(un)ds|≤c3P~∥un∥≤c5P~∥∇un∥.
Likewise, from (29), we have
(40)|∑i=13‍Ki∫Γi‍(un·n)n·unds|≤c4K∥∇un∥∥un∥≤c6K∥∇un∥2.
Furthermore, making use of Hölder’s inequality ([6], page 623), one obtains
(41)((un·∇)un,un)≤∥un∥L6∥∇un∥∥un∥L3≤c7∥∇un∥3,
where c7 depends on Rellich-Kondrachov compactness inequality [1], Poincaré’s inequality [3], and Sobolev inequality ([6], page 270). We make use of these inequalities in (38) and we obtain
(42)μ∥∇un∥2≤c7∥∇un∥3+c5P~∥∇un∥+c6K∥∇un∥2.
Set ξ:=μ-c6K. It follows that
(43)ξ∥∇un∥≤c7∥∇un∥2+c5P~,
where P~=∑i=13‍|Pi|, K=max1≤i≤3Ki.Theorem 5.
Assume(44)K≤μc6,P~≤ξ24c5c7.
For each integer n=1,2,…, there exists a function un of the form (35) satisfying (36) such that
(45)∥∇un∥≤ξ2c7[1-(1-4c5c7P~ξ2)].Proof.
Owing to Poincaré’s inequality (4), ∥∇φ∥ is a norm equivalent to ∥φ∥1 for all φ∈V; therefore, (45) defines a closed ball in span{a1,…,an}. To prove the solvability of the finite-dimensional problem (36), we follow Fujita in using Brouwer’s fixed point theorem (see Theorem 1.1 in [7]), applying it to the continuous mapping v→w defined by the linear problem (k=1,…,n):
(46)((v·∇)w,ak)+μ(∇w,∇ak)=-∑‍i=13∫‍ΓiKi(w·n)n·akds-∑‍i=13Pi∫‍Γin·akds.
Equation (46) is a system of n linear equations. These linear equations are uniquely solvable if v lies in the ball defined by (45), because then w=0 is the only solution of the corresponding homogeneous equation (Ki=Pi=0, 1≤i≤3). In fact, if v satisfies (45) and w satisfies (46) with Ki=Pi=0, we have
(47)ξ∥∇w∥2≤c7∥∇v∥∥∇w∥2≤c7ξ2c7[1-(1-4c5c7P~ξ2)]∥∇w∥2≤ξ2∥∇w∥2.
Together with Poincaré’s inequality (4), this imply that w=0. To see that the mapping v→w takes the ball defined by (45) into itself, suppose that v satisfies (45). Then, similarly to (43), we obtain that
(48)ξ∥∇w∥≤c7∥∇v∥∥∇w∥+c5P~,
and therefore,
(49)∥∇w∥≤c5P~ξ-c7∥∇v∥≤c5P~ξ-(ξ/2)[1-(1-(4c5c7P~/ξ2))]≤ξ2c7[1-(1-4c5c7P~ξ2)].
Thus, (46) defines a continuous mapping v→w of the closed ball
(50){φ∈span{a1,…,an}:∥∇φ∥≤ξ2c7[1-(1-4c5c7P~ξ2)]}
into itself. The map has at least one fixed point, and any such fixed point is a solution of (36). un is chosen to be any one of these fixed points. Hence, Brouwer’s fixed point has been applied and has given the existence of Galerkin approximations
(51)un=∑k=1n‍dnkak
satisfying
(52)∥∇un∥≤ξ2c7[1-(1-4c5c7P~ξ2)].


## 3.3. Existence of a Weak Solution

Theorem 6.
Assume(53)K≤μc6,P~≤ξ24c5c7.
There exists a weak solution to the stationary version of problem (23).Proof.
By Poincaré’s inequality [3], the fact that the sequence {∥∇un∥}n=1∞ is bounded implies that the sequence {un}n=1∞ is bounded in V. As a result, Theorem 4 yields the existence of a subsequence {unq}q=1∞⊂{un}n=1∞ such that
(54)unq⇀uweaklyinV.
Next we show that the weak limit u is in fact a weak solution. In this respect, we are going to show that
(55)((u·∇)u,φ)+μ(∇u,∇φ)=-∑i=13∫ΓiKi(u·n)n·φds-∑‍i=13Pi∫Γin·φds,
for each φ∈V.
Fix an integerk, (k=1,2,…). From (36), we have the identity
(56)((un·∇)un,ak)+μ(∇un,∇ak)=-∑‍i=13∫‍ΓiKi(un·n)n·akds-∑‍i=13Pi∫‍Γin·akds.
We recall (54), to find upon passing to weak limits that
(57)((u·∇)u,ak)+μ(∇u,∇ak)=-∑‍i=13∫‍ΓiKi(u·n)n·akds-∑‍i=13Pi∫‍Γin·akds.
Since {ak}k=1∞ is a basis of V, it follows that
(58)((u·∇)u,φ)+μ(∇u,∇φ)=-∑‍i=13∫‍ΓiKi(u·n)n·φds-∑‍i=13Pi∫‍Γin·φds,
for each φ∈V.Note that, in taking this limit, there is no difficulty with the nonlinear term. In fact, we have that(59)(unq·∇)unq-(u·∇)u=(unq·∇)unq+(unq·∇)u-(unq·∇)u-(u·∇)u=(unq·∇)(unq-u)+((unq-u)·∇)u.
Also, ak∈H2(Ω) is an eigenfunction of the inverse of the stokes operator Δ~ and H2(Ω)⊂C(Ω¯). This implies that ak∈C(Ω¯) [6]. It follows that
(60)((unq·∇)unq-(u·∇)u,ak)=((unq·∇)(unq-u),ak)+(((unq-u)·∇)u,ak)⟶0
(see [5], page 651).


## 4. The Nonstationary Problem

The mathematical analysis of the Navier-Stokes problem is based on its weak formulation.


### 4.1. Weak Formulation

Assume thatu is a solution to problem (23) and φis a smooth solenoidal vector-valued function defined on Ω. u satisfies the following identities:
(61)(∂u∂t+(u·∇)u-μΔu+∇p,φ)=0,x∈Ω,t>0,(u(0),φ)=(u0,φ),x∈Ω,(pn-μ(∇u)n-Ki(u·n)n,φ)=(Pin,φ),000000000000000000000x∈Γi,i=1,2,3.
We calculate
(62)(∇p,φ)=∑‍i=13∫‍Γi(Pin)·φds+μ∑‍i=13∫‍Γi(n·∇)u·φds+∑‍i=13∫‍ΓiKi(u·n)n·φds,(Δu,φ)=-(∇φ,∇u)+∑‍i=13∫‍Γi(n·∇)u·φds,
to obtain
(63)(∂u∂t,φ)+b(u,u,φ)+a(u,φ)=c(t,φ),
where a, b, and c are defined by (24), (25), and (26), respectively. We notice that (63) does not depend on the pressure p. Therefore, the weak formulation of the problem reads as follows.Given a divergence free velocity fieldu0∈V, Pi∈L∞(0,T), and a nonnegative constant Ki for i=1,2,3, find u∈L2(0,T;V) such that for all t≥0,
(64)(∂u∂t,φ)+b(u,u,φ)+a(u,φ)=c(t,φ),(u(0),φ)=(u0,φ),∀φ∈V,
where V, defined in (5), is the space of test functions.


### 4.2. Galerkin Approximations

We are going to follow the same approach as in [6]. We will construct the weak solution of the initial boundary value problem by first solving for a finite dimensional approximation. We recall that the inverse operator Δ~-1 of the Stokes operator Δ~ is self-adjoint and possesses a sequence of eigenfunctions {ak} which are orthogonal in V and orthonormal in L2(Ω). Fix a positive integer n and write
(65)un(t):=∑k=1n‍dnk(t)ak,
where we intend to select the coefficients dnk(t) (0≤t≤T, k=1,…,n) to satisfy
(66)dnk(0)=(u0,ak),(67)(∂un∂t,ak)+a(un,ak)+b(un,un,ak)=c(t,ak).Theorem 7.
For each integern=1,2,…, there exists a unique function un of the form (65) satisfying (66)-(67).Proof.
Assuming thatun is given by (65), we observe using the fact that {ak}k=1∞ is an orthonormal basis of L2(Ω) that
(68)(∂un∂t,ak)=(dnk)′(t),
where (dnk)′(t) is the derivative of dnk(t) with respect to t.
Furthermore, from (24) and the fact that a is a bilinear form, we have that
(69)a(un,ak)=∑i=1n‍Aikdni(t),
where Aik=a(ai,ak).
From (25) and recalling that b is a trilinear form, we have that
(70)b(un,un,ak)=∑i,j=1n‍Bi,jkdni(t)dnj(t),
where Bi,jk=b(ai,aj,ak).
Moreover, we set(71)fk(t):=c(t,ak).
It follows that (67) becomes the system of ODEs
(72)dnk′(t)+∑i=1n‍Aikdni(t)+∑i,j=1n‍Bi,jkdni(t)dnj(t)=fk(t),(k=1,…,n),
subject to the initial condition (66). According to standard existence theory for ordinary differential equations, there exists a unique function dn(t)=(dn1(t),…,dnn(t)) satisfying (66) and (72) for a.e. 0≤t≤T. And then un defined by (65) solves (67) for a.e. 0≤t≤T.


### 4.3. Energy Estimate

Our plan hereafter is to letn→∞. Before that, we will need some estimates, uniform in n.Theorem 8.
For each approximate solutionun, the following energy estimates hold:
(73)ddt∥∇un∥2+(μ-2Kc2c4)∥Δ~un∥2≤16c14μ3∥∇un∥6+2c32P~2μ.Proof.
We denote byλk the eigenvalue associated with the eigenfunction ak. Multiplying (63) by -λkdnk and summing over k=1,…,n, one obtains
(74)(∂un∂t,-Δ~un)+b(un,un,-Δ~un)+a(un,-Δ~un)=c(t,-Δ~un).
Now,
(75)(∂un∂t,-Δ~un)=(∇(∂un∂t),∇un)=12ddt∥∇un∥2,
and also
(76)a(un,-Δ~un)=μ(∇un,∇(-Δ~un))+∑i=13‍Ki∫Γi‍(un·n)n·(-Δ~un)ds=μ∥Δ~un∥2-∑i=13‍Ki∫Γi‍(un·n)n·(Δ~un)ds;b(un,un,-Δ~un)=-((un·∇)un,Δ~un),
Moreover,
(77)c(t,-Δ~un)=∑i=13‍Pi(t)∫Γi‍n·Δ~unds.
It follows that
(78)12ddt∥∇un∥2+μ∥Δ~un∥2=((un·∇)un,Δ~un)+∑i=13‍Pi(t)∫Γi‍n·(Δ~un)ds+∑i=13‍Ki∫Γi‍(un·n)n·Δ~unds.
We now make use of (28) to obtain
(79)|∑i=13‍Pi(t)∫Γi‍n·(Δ~un)ds|≤c3P~∥Δ~un∥.
Likewise, from (29), we have
(80)|∑i=13‍Ki∫Γi‍(un·n)n·Δ~unds|≤c4K∥∇un∥∥Δ~un∥.
Furthermore,
(81)((un·∇)un,Δ~un)≤supΩ|un|∥∇un∥∥Δ~un∥.
We make use of these inequalities in (78) to find that
(82)12ddt∥∇un∥2+μ∥Δ~un∥2≤supΩ|un|∥∇un∥∥Δ~un∥+c3P~∥Δ~un∥+c4K∥∇un∥∥Δ~un∥.
Now from (9),
(83)supΩ|un|≤c1∥∇un∥1/2∥Δ~un∥1/2,
which implies that
(84)supΩ|un|∥∇un∥∥Δ~un∥≤c1∥∇un∥3/2∥Δ~un∥3/2,
and so, inequality (82) now reads
(85)12ddt∥∇un∥2+μ∥Δ~un∥2≤c1∥∇un∥3/2∥Δ~un∥3/2+c3P~∥Δ~un∥+c4K∥∇un∥∥Δ~un∥.
We make use of Cauchy’s inequality with epsilon to estimate the right-hand side of (85). We have
(86)c1∥∇un∥3/2∥Δ~un∥3/2=(c1∥∇un∥3/2∥Δ~un∥1/2)∥Δ~un∥≤14ϵ1c12∥∇un∥3∥Δ~un∥+ϵ1∥Δ~un∥2≤8c14μ3∥∇un∥6+μ4∥Δ~un∥2,c3P~∥Δ~un∥≤c32P~2μ+μ4∥Δ~un∥2(setϵ=1μ).
Also inequality (10) implies that
(87)∥∇un∥≤c2∥Δ~un∥,
so that
(88)c4K∥∇un∥∥Δ~un∥≤Kc2c4∥Δ~un∥2.
Making use of theseinequalities on (85), we obtain that
(89)12ddt∥∇un∥2+μ∥Δ~un∥2≤8c14μ3∥∇un∥6+μ4∥Δ~un∥2+c32P~2μ+μ4∥Δ~un∥2+Kc2c4∥Δ~un∥2,
which leads us to the required energy estimates.


### 4.4. Local Existence of a Weak Solution

In this subsection, we use Galerkin method to build up a local weak solution of the initial/boundary-value problem. We have already constructed the Galerkin approximations sequence{un}n=1∞ in Section 4.2. Our goal now is to extract from this sequence a subsequence that converges to the weak solution. In this respect, we are going to show that this sequence is bounded and thereafter, we will make use of a compactness result.Lemma 9.
LetY be a nonnegative function satisfying the inequality
(90)Y′≤ηY3+ζ
and let Y(0)=M0 be a strictly positive real number; then there exists a time interval (0,T⋆) where Y(t) is bounded by a positive constant M and M depends only on M0, η and ζ.Proof.
We have(91)Y′≤η(Y+τ)3,whereτ=(ζη)1/3.
We set Z=Y+τ to obtain
(92)Z′Z3≤η.
Integrate over (0,t), with 0≤t≤T:
(93)Z(t)-2≥Z(0)-2-2ηT.
We select T⋆ in (0,T) such that Z(0)-2-2ηT⋆>0; that is,
(94)T⋆<(Z(0))-22η=(M0+(ζ/η)1/3)-22η.
Then
(95)Z(t)-2-Z(0)-2≥-2ηt≥-2ηT⋆,t∈(0,T⋆),
so that
(96)Z(t)≤(Z(0)-2-2ηT⋆)-1/2,t∈(0,T⋆).
Considering that Z=Y+τ and τ=(ζ/η)1/3, we obtain
(97)Y(t)≤((M0+(ζη)1/3)-2-2ηT⋆)-1/2-(ζη)1/3,t∈(0,T⋆).
In order to obtain ((M0+(ζ/η)1/3)-2-2ηT⋆)-1/2-(ζ/η)1/3>0, we must have
(98)T⋆>(M0+(ζ/η)1/3)-2-(ζ/η)-2/32η.
Hence, we choose T⋆ such that
(99)(M0+(ζ/η)1/3)-2-(ζ/η)-2/32η<T⋆<(M0+(ζ/η)1/3)-22η.
It follows that Y(t) is bounded by a positive constant
(100)M=((M0+(ζη)1/3)-2-2ηT⋆)-1/2-(ζη)1/3
for all t such that 0≤t≤T⋆ and that M depends only on M0, η, and ζ.Theorem 10.
LetK be such that
(101)κ=μ-2Kc2c4>0;
there is a time interval (0,T⋆) on which a weak solution of (23) exists.Proof.
Considering (101), the energy estimates (73) now read:
(102)ddt∥∇un∥2+κ∥Δ~un∥2≤16c14μ3∥∇un∥6+2c32P~2μ.
Since κ>0, (102) implies that
(103)ddt∥∇un∥2≤16c14μ3∥∇un∥6+2c32P~2μ.
Defining
(104)Yn(t)=∥∇un∥2,
we see that inequality (103) has the following form:
(105)Yn′≤ηYn3+ζ,
where η>0 and ζ>0 are some constants depending on the data.
Note that for each integern, the function Yn(t)=∥∇un∥2 verifies the requirements of Lemma 9 with Yn(0)=∥∇un∥2 since the initial condition is the same for all functions un, n=0,1,2,…. Consequently, the sequence {∥∇un∥2}n=1∞ is bounded by a real number M which depends only on the initial data. We make use of Poincaré’s inequality (4) to find that for each integer n,
(106)∥un∥12=∥un∥2+∥∇un∥2≤(1+CΩ2)M,∥un∥L2(0,T⋆;V)2=∫0T⋆‍∥un∥12dt≤(1+CΩ2)MT⋆.
Therefore, the sequence {∥un∥L2(0,T⋆;V)}n=1∞ is bounded.
Furthermore, the spaceL2(0,T⋆;V) being a Hilbert space, Theorem 4 can be applied to the sequence {∥un∥L2(0,T⋆;V)}n=1∞. In fact, the sequence {un}n=1∞ is bounded in L2(0,T⋆;V). Consequently, there exists a subsequence {unq}q=1∞⊂{un}n=1∞ such that
(107)unq⇀uweaklyinL2(0,T⋆;V).
Next, we show that the weak limit u is in fact a weak solution. In this respect, we are going to show first that
(108)(∂u∂t,φ)+b(u,u,φ)+a(u,φ)=c(t,φ),∀φ∈V,
and, thereafter,
(109)(u(0),φ)=(u0,φ),∀φ∈V.(1)
Fix an integerN and choose a function w∈C(0,T⋆;V) having the form
(110)w(t)=∑k=1N‍dk(t)ak,
where {dk}k=1N are given functions and {ak} is the basis of V. We choose n≥N, multiply (67) by dk(t), sum k=1,…,N, and then integrate with respect to t to find that
(111)∫0T⋆‍[(∂unq∂t,w)+a(unq,w)+b(unq,unq,w)]dt=∫0T⋆‍c(t,w)dt.
We recall (107) to find upon passing to weak limits (see remark 3.1 and [5]) that
(112)∫0T⋆‍[(∂u∂t,w)+a(u,w)+b(u,u,w)]dt=∫0T⋆‍c(t,w)dt.
Equality (112) then holds for all functions w∈L2(0,T;V), as functions of the form (110) are dense in this space [5]. Hence, in particular,
(113)(∂u∂t,φ)+b(u,u,φ)+a(u,φ)=c(t,φ)
for each φ∈V and a.e. t≥0.(2)
In order to prove that(u(0),φ)=(u0,φ) for every φ∈V, we first note from (112) that
(114)∫0T⋆‍[-(u,∂w∂t)+a(u,w)+b(u,u,w)]dt=∫0T⋆‍c(t,w)dt+(u(0),w(0)),
for each w∈H1(0,T⋆;V) with w(T⋆)=0. Similarly, from (111) we deduce that
(115)∫0T⋆‍[-(unq,∂w∂t)+a(unq,w)+b(unq,unq,w)]dt=∫0T⋆‍c(t,w)dt+(unq(0),w(0)).
We once again employ (107) to find that
(116)∫0T⋆‍[-(u,∂w∂t)+a(u,w)+b(u,u,w)]dt=∫0T⋆‍c(t,w)dt+(u0,w(0)),
since unq(0)→u0. As w(0) is arbitrary, comparing (114) and (116), we conclude that (u(0),φ)=(u0,φ) for each φ∈V. Therefore, u is a local weak solution of (23).


### 4.5. Global Existence of a Weak Solution for Small Data

In this section, we make use of Galerkin’s method to establish the global existence of a weak solution. We are going to show that under certain circumstances, the weak solutionu is defined at any time t. In this regard, we are going to consider the energy estimate and the following lemma.Lemma 11.
LetY be a nonnegative, absolutely continuous function satisfying inequality
(117)Y′+λY≤ηY3+ζ,
and let a real number M be such that 0<M<(λ/2η)1/2, if Y(0)≤M, and ζ≤λM/2, then Y(t) is bounded by M for all t>0.Proof.
This result is proven by contradiction [8]. Suppose that there exists a t such that Y(t)>M, and define
(118)t⋆=inf{t∈ℝ+,Y(t)>M};
we have Y(t⋆)=M and Y′(t⋆)>0. (1)
We first show thatY(t⋆)=M. Set I={t∈ℝ+,Y(t)>M} and choose any ϵ>0; we have
(119)t⋆-ϵ<t⋆,t⋆=inf(I),
and so t⋆-ϵ does not belong to the set I. It follows that Y(t⋆-ϵ)≤M, and this is true for each ϵ>0; hence
(120)Y(t⋆)≤M.
On the other hand, for each natural number n, we have t⋆+(1/n)>t⋆. We make use of the fact that t⋆=inf(I) to see that there exists ϕn∈I such that
(121)t⋆≤ϕn≤t⋆+1n,
where ϕn is not necessarily unique. We choose one value that ϕn may take and we denote it by tn. This defines a real sequence {tn}n=1∞. For each natural number n, since tn∈I, we have that Y(tn)>M. The fact that t⋆≤tn≤t⋆+(1/n) implies that the sequence {tn}n=1∞ converges to t⋆. We make use of the continuity of Y to see that the sequence
(122){Y(tn)}n=1∞convergestoY(t⋆).
Since tn∈I for each n, we have Y(tn)>M, and then Y(tn)→Y(t⋆) implies that
(123)Y(t⋆)≥M.
It follows that Y(t⋆)≤M and Y(t⋆)≥M; therefore,
(124)Y(t⋆)=M.(2)
Next we show thatY′(t⋆)>0. Suppose that Y′(t⋆)≤0; then there exists a nonnegative natural number n such that Y decreases on the interval (t⋆,t⋆+(1/n)). Also we have that
(125)t⋆≤tn≤t⋆+1n;
this implies that
(126)Y(t⋆)≥Y(tn).
But from part (1), we have Y(t⋆)=M and Y(tn)>M; because tn∈I, it follows that
(127)Y(t⋆)<Y(tn).
Hence, Y(t⋆)≥Y(tn) and Y(t⋆)<Y(tn), which is impossible. Therefore,
(128)Y′(t⋆)>0.
Finally that we show that Y′(t⋆)≤0 and this contradicts the result obtained in (2). From (117), we have
(129)Y′(t⋆)≤-λY(t⋆)+ηY3(t⋆)+ζ=-λM+ηM(M2)+ζ(becauseY(t⋆)=M)≤-λM+ηM(λ2η)+ζ≤0,because  of  the  hypothesis(ζ≤λM2).
Therefore, Y(t)≤M for all t∈ℝ+.Theorem 12.
Assume that the initial and boundary data are sufficiently small, precisely(130)∥∇u0∥≤(κμ364c2c14)1/4,P~<μ4c1c3(κ3μ4c23)1/4.
Then, for all T≥0, there exists a weak solution u∈L2(0,T;V) to problem (23), and it satisfies the inequality
(131)∥∇u∥≤(κμ364c2c14)1/4.Proof.
We make use of the estimates (10), (73), and (101) in order to obtain
(132)ddt∥∇un∥2+κc2∥∇un∥2≤16c14μ3∥∇un∥6+2c32P~2μ.
Set
(133)Yn=∥∇un∥2.
The energy estimate (132) takes the form
(134)Yn′+λYn≤ηYn3+ζ,
where
(135)λ=κc2,η=16c14μ3,ζ=2c32P~2μ.
Let
(136)M=(κμ364c2c14)1/2;
we have
(137)0<M<(λ2η)1/2.
Also
(138)Yn(0)=∥∇u0∥2≤Mforeachintegern,ζ=2c32P~2μ≤κ2c2(κμ364c2c14)1/2=λM2.
We make use of Lemma 11 to find that Yn=∥∇un∥2 is bounded by M for all t>0 and the bound does not depend on n. Poincaré’s inequality (4) implies that
(139)∥un∥12=∥un∥2+∥∇un∥2≤CΩ2∥∇un∥2+∥∇un∥2≤(1+CΩ2)M.
Let T>0 be any positive real number,
(140)∥un∥L2(0,T;V)2=∫0T‍∥un∥12dt≤(1+CΩ2)MT.
Thus, for each nonnegative real number T, the sequence {∥un∥L2(0,T;V)}n=1∞ is bounded. We make use of Lemma 9 to extract a subsequence {unq}q=1∞ that converges to an element u of L2(0,T;V), and we use the same steps as we did for the local existence to show that u is a weak solution of problem (23).
Furthermore, the fact that the sequence{∇un}n=1∞ is bounded by M implies that the subsequence {∇unq}q=1∞ is also bounded by M.
Therefore, since∇u is the limit of this subsequence, it follows that
(141)∥∇u∥≤(κμ364c2c14)1/4.


## 4.1. Weak Formulation

Assume thatu is a solution to problem (23) and φis a smooth solenoidal vector-valued function defined on Ω. u satisfies the following identities:
(61)(∂u∂t+(u·∇)u-μΔu+∇p,φ)=0,x∈Ω,t>0,(u(0),φ)=(u0,φ),x∈Ω,(pn-μ(∇u)n-Ki(u·n)n,φ)=(Pin,φ),000000000000000000000x∈Γi,i=1,2,3.
We calculate
(62)(∇p,φ)=∑‍i=13∫‍Γi(Pin)·φds+μ∑‍i=13∫‍Γi(n·∇)u·φds+∑‍i=13∫‍ΓiKi(u·n)n·φds,(Δu,φ)=-(∇φ,∇u)+∑‍i=13∫‍Γi(n·∇)u·φds,
to obtain
(63)(∂u∂t,φ)+b(u,u,φ)+a(u,φ)=c(t,φ),
where a, b, and c are defined by (24), (25), and (26), respectively. We notice that (63) does not depend on the pressure p. Therefore, the weak formulation of the problem reads as follows.Given a divergence free velocity fieldu0∈V, Pi∈L∞(0,T), and a nonnegative constant Ki for i=1,2,3, find u∈L2(0,T;V) such that for all t≥0,
(64)(∂u∂t,φ)+b(u,u,φ)+a(u,φ)=c(t,φ),(u(0),φ)=(u0,φ),∀φ∈V,
where V, defined in (5), is the space of test functions.


## 4.2. Galerkin Approximations

We are going to follow the same approach as in [6]. We will construct the weak solution of the initial boundary value problem by first solving for a finite dimensional approximation. We recall that the inverse operator Δ~-1 of the Stokes operator Δ~ is self-adjoint and possesses a sequence of eigenfunctions {ak} which are orthogonal in V and orthonormal in L2(Ω). Fix a positive integer n and write
(65)un(t):=∑k=1n‍dnk(t)ak,
where we intend to select the coefficients dnk(t) (0≤t≤T, k=1,…,n) to satisfy
(66)dnk(0)=(u0,ak),(67)(∂un∂t,ak)+a(un,ak)+b(un,un,ak)=c(t,ak).Theorem 7.
For each integern=1,2,…, there exists a unique function un of the form (65) satisfying (66)-(67).Proof.
Assuming thatun is given by (65), we observe using the fact that {ak}k=1∞ is an orthonormal basis of L2(Ω) that
(68)(∂un∂t,ak)=(dnk)′(t),
where (dnk)′(t) is the derivative of dnk(t) with respect to t.
Furthermore, from (24) and the fact that a is a bilinear form, we have that
(69)a(un,ak)=∑i=1n‍Aikdni(t),
where Aik=a(ai,ak).
From (25) and recalling that b is a trilinear form, we have that
(70)b(un,un,ak)=∑i,j=1n‍Bi,jkdni(t)dnj(t),
where Bi,jk=b(ai,aj,ak).
Moreover, we set(71)fk(t):=c(t,ak).
It follows that (67) becomes the system of ODEs
(72)dnk′(t)+∑i=1n‍Aikdni(t)+∑i,j=1n‍Bi,jkdni(t)dnj(t)=fk(t),(k=1,…,n),
subject to the initial condition (66). According to standard existence theory for ordinary differential equations, there exists a unique function dn(t)=(dn1(t),…,dnn(t)) satisfying (66) and (72) for a.e. 0≤t≤T. And then un defined by (65) solves (67) for a.e. 0≤t≤T.


## 4.3. Energy Estimate

Our plan hereafter is to letn→∞. Before that, we will need some estimates, uniform in n.Theorem 8.
For each approximate solutionun, the following energy estimates hold:
(73)ddt∥∇un∥2+(μ-2Kc2c4)∥Δ~un∥2≤16c14μ3∥∇un∥6+2c32P~2μ.Proof.
We denote byλk the eigenvalue associated with the eigenfunction ak. Multiplying (63) by -λkdnk and summing over k=1,…,n, one obtains
(74)(∂un∂t,-Δ~un)+b(un,un,-Δ~un)+a(un,-Δ~un)=c(t,-Δ~un).
Now,
(75)(∂un∂t,-Δ~un)=(∇(∂un∂t),∇un)=12ddt∥∇un∥2,
and also
(76)a(un,-Δ~un)=μ(∇un,∇(-Δ~un))+∑i=13‍Ki∫Γi‍(un·n)n·(-Δ~un)ds=μ∥Δ~un∥2-∑i=13‍Ki∫Γi‍(un·n)n·(Δ~un)ds;b(un,un,-Δ~un)=-((un·∇)un,Δ~un),
Moreover,
(77)c(t,-Δ~un)=∑i=13‍Pi(t)∫Γi‍n·Δ~unds.
It follows that
(78)12ddt∥∇un∥2+μ∥Δ~un∥2=((un·∇)un,Δ~un)+∑i=13‍Pi(t)∫Γi‍n·(Δ~un)ds+∑i=13‍Ki∫Γi‍(un·n)n·Δ~unds.
We now make use of (28) to obtain
(79)|∑i=13‍Pi(t)∫Γi‍n·(Δ~un)ds|≤c3P~∥Δ~un∥.
Likewise, from (29), we have
(80)|∑i=13‍Ki∫Γi‍(un·n)n·Δ~unds|≤c4K∥∇un∥∥Δ~un∥.
Furthermore,
(81)((un·∇)un,Δ~un)≤supΩ|un|∥∇un∥∥Δ~un∥.
We make use of these inequalities in (78) to find that
(82)12ddt∥∇un∥2+μ∥Δ~un∥2≤supΩ|un|∥∇un∥∥Δ~un∥+c3P~∥Δ~un∥+c4K∥∇un∥∥Δ~un∥.
Now from (9),
(83)supΩ|un|≤c1∥∇un∥1/2∥Δ~un∥1/2,
which implies that
(84)supΩ|un|∥∇un∥∥Δ~un∥≤c1∥∇un∥3/2∥Δ~un∥3/2,
and so, inequality (82) now reads
(85)12ddt∥∇un∥2+μ∥Δ~un∥2≤c1∥∇un∥3/2∥Δ~un∥3/2+c3P~∥Δ~un∥+c4K∥∇un∥∥Δ~un∥.
We make use of Cauchy’s inequality with epsilon to estimate the right-hand side of (85). We have
(86)c1∥∇un∥3/2∥Δ~un∥3/2=(c1∥∇un∥3/2∥Δ~un∥1/2)∥Δ~un∥≤14ϵ1c12∥∇un∥3∥Δ~un∥+ϵ1∥Δ~un∥2≤8c14μ3∥∇un∥6+μ4∥Δ~un∥2,c3P~∥Δ~un∥≤c32P~2μ+μ4∥Δ~un∥2(setϵ=1μ).
Also inequality (10) implies that
(87)∥∇un∥≤c2∥Δ~un∥,
so that
(88)c4K∥∇un∥∥Δ~un∥≤Kc2c4∥Δ~un∥2.
Making use of theseinequalities on (85), we obtain that
(89)12ddt∥∇un∥2+μ∥Δ~un∥2≤8c14μ3∥∇un∥6+μ4∥Δ~un∥2+c32P~2μ+μ4∥Δ~un∥2+Kc2c4∥Δ~un∥2,
which leads us to the required energy estimates.


## 4.4. Local Existence of a Weak Solution

In this subsection, we use Galerkin method to build up a local weak solution of the initial/boundary-value problem. We have already constructed the Galerkin approximations sequence{un}n=1∞ in Section 4.2. Our goal now is to extract from this sequence a subsequence that converges to the weak solution. In this respect, we are going to show that this sequence is bounded and thereafter, we will make use of a compactness result.Lemma 9.
LetY be a nonnegative function satisfying the inequality
(90)Y′≤ηY3+ζ
and let Y(0)=M0 be a strictly positive real number; then there exists a time interval (0,T⋆) where Y(t) is bounded by a positive constant M and M depends only on M0, η and ζ.Proof.
We have(91)Y′≤η(Y+τ)3,whereτ=(ζη)1/3.
We set Z=Y+τ to obtain
(92)Z′Z3≤η.
Integrate over (0,t), with 0≤t≤T:
(93)Z(t)-2≥Z(0)-2-2ηT.
We select T⋆ in (0,T) such that Z(0)-2-2ηT⋆>0; that is,
(94)T⋆<(Z(0))-22η=(M0+(ζ/η)1/3)-22η.
Then
(95)Z(t)-2-Z(0)-2≥-2ηt≥-2ηT⋆,t∈(0,T⋆),
so that
(96)Z(t)≤(Z(0)-2-2ηT⋆)-1/2,t∈(0,T⋆).
Considering that Z=Y+τ and τ=(ζ/η)1/3, we obtain
(97)Y(t)≤((M0+(ζη)1/3)-2-2ηT⋆)-1/2-(ζη)1/3,t∈(0,T⋆).
In order to obtain ((M0+(ζ/η)1/3)-2-2ηT⋆)-1/2-(ζ/η)1/3>0, we must have
(98)T⋆>(M0+(ζ/η)1/3)-2-(ζ/η)-2/32η.
Hence, we choose T⋆ such that
(99)(M0+(ζ/η)1/3)-2-(ζ/η)-2/32η<T⋆<(M0+(ζ/η)1/3)-22η.
It follows that Y(t) is bounded by a positive constant
(100)M=((M0+(ζη)1/3)-2-2ηT⋆)-1/2-(ζη)1/3
for all t such that 0≤t≤T⋆ and that M depends only on M0, η, and ζ.Theorem 10.
LetK be such that
(101)κ=μ-2Kc2c4>0;
there is a time interval (0,T⋆) on which a weak solution of (23) exists.Proof.
Considering (101), the energy estimates (73) now read:
(102)ddt∥∇un∥2+κ∥Δ~un∥2≤16c14μ3∥∇un∥6+2c32P~2μ.
Since κ>0, (102) implies that
(103)ddt∥∇un∥2≤16c14μ3∥∇un∥6+2c32P~2μ.
Defining
(104)Yn(t)=∥∇un∥2,
we see that inequality (103) has the following form:
(105)Yn′≤ηYn3+ζ,
where η>0 and ζ>0 are some constants depending on the data.
Note that for each integern, the function Yn(t)=∥∇un∥2 verifies the requirements of Lemma 9 with Yn(0)=∥∇un∥2 since the initial condition is the same for all functions un, n=0,1,2,…. Consequently, the sequence {∥∇un∥2}n=1∞ is bounded by a real number M which depends only on the initial data. We make use of Poincaré’s inequality (4) to find that for each integer n,
(106)∥un∥12=∥un∥2+∥∇un∥2≤(1+CΩ2)M,∥un∥L2(0,T⋆;V)2=∫0T⋆‍∥un∥12dt≤(1+CΩ2)MT⋆.
Therefore, the sequence {∥un∥L2(0,T⋆;V)}n=1∞ is bounded.
Furthermore, the spaceL2(0,T⋆;V) being a Hilbert space, Theorem 4 can be applied to the sequence {∥un∥L2(0,T⋆;V)}n=1∞. In fact, the sequence {un}n=1∞ is bounded in L2(0,T⋆;V). Consequently, there exists a subsequence {unq}q=1∞⊂{un}n=1∞ such that
(107)unq⇀uweaklyinL2(0,T⋆;V).
Next, we show that the weak limit u is in fact a weak solution. In this respect, we are going to show first that
(108)(∂u∂t,φ)+b(u,u,φ)+a(u,φ)=c(t,φ),∀φ∈V,
and, thereafter,
(109)(u(0),φ)=(u0,φ),∀φ∈V.(1)
Fix an integerN and choose a function w∈C(0,T⋆;V) having the form
(110)w(t)=∑k=1N‍dk(t)ak,
where {dk}k=1N are given functions and {ak} is the basis of V. We choose n≥N, multiply (67) by dk(t), sum k=1,…,N, and then integrate with respect to t to find that
(111)∫0T⋆‍[(∂unq∂t,w)+a(unq,w)+b(unq,unq,w)]dt=∫0T⋆‍c(t,w)dt.
We recall (107) to find upon passing to weak limits (see remark 3.1 and [5]) that
(112)∫0T⋆‍[(∂u∂t,w)+a(u,w)+b(u,u,w)]dt=∫0T⋆‍c(t,w)dt.
Equality (112) then holds for all functions w∈L2(0,T;V), as functions of the form (110) are dense in this space [5]. Hence, in particular,
(113)(∂u∂t,φ)+b(u,u,φ)+a(u,φ)=c(t,φ)
for each φ∈V and a.e. t≥0.(2)
In order to prove that(u(0),φ)=(u0,φ) for every φ∈V, we first note from (112) that
(114)∫0T⋆‍[-(u,∂w∂t)+a(u,w)+b(u,u,w)]dt=∫0T⋆‍c(t,w)dt+(u(0),w(0)),
for each w∈H1(0,T⋆;V) with w(T⋆)=0. Similarly, from (111) we deduce that
(115)∫0T⋆‍[-(unq,∂w∂t)+a(unq,w)+b(unq,unq,w)]dt=∫0T⋆‍c(t,w)dt+(unq(0),w(0)).
We once again employ (107) to find that
(116)∫0T⋆‍[-(u,∂w∂t)+a(u,w)+b(u,u,w)]dt=∫0T⋆‍c(t,w)dt+(u0,w(0)),
since unq(0)→u0. As w(0) is arbitrary, comparing (114) and (116), we conclude that (u(0),φ)=(u0,φ) for each φ∈V. Therefore, u is a local weak solution of (23).


## 4.5. Global Existence of a Weak Solution for Small Data

In this section, we make use of Galerkin’s method to establish the global existence of a weak solution. We are going to show that under certain circumstances, the weak solutionu is defined at any time t. In this regard, we are going to consider the energy estimate and the following lemma.Lemma 11.
LetY be a nonnegative, absolutely continuous function satisfying inequality
(117)Y′+λY≤ηY3+ζ,
and let a real number M be such that 0<M<(λ/2η)1/2, if Y(0)≤M, and ζ≤λM/2, then Y(t) is bounded by M for all t>0.Proof.
This result is proven by contradiction [8]. Suppose that there exists a t such that Y(t)>M, and define
(118)t⋆=inf{t∈ℝ+,Y(t)>M};
we have Y(t⋆)=M and Y′(t⋆)>0. (1)
We first show thatY(t⋆)=M. Set I={t∈ℝ+,Y(t)>M} and choose any ϵ>0; we have
(119)t⋆-ϵ<t⋆,t⋆=inf(I),
and so t⋆-ϵ does not belong to the set I. It follows that Y(t⋆-ϵ)≤M, and this is true for each ϵ>0; hence
(120)Y(t⋆)≤M.
On the other hand, for each natural number n, we have t⋆+(1/n)>t⋆. We make use of the fact that t⋆=inf(I) to see that there exists ϕn∈I such that
(121)t⋆≤ϕn≤t⋆+1n,
where ϕn is not necessarily unique. We choose one value that ϕn may take and we denote it by tn. This defines a real sequence {tn}n=1∞. For each natural number n, since tn∈I, we have that Y(tn)>M. The fact that t⋆≤tn≤t⋆+(1/n) implies that the sequence {tn}n=1∞ converges to t⋆. We make use of the continuity of Y to see that the sequence
(122){Y(tn)}n=1∞convergestoY(t⋆).
Since tn∈I for each n, we have Y(tn)>M, and then Y(tn)→Y(t⋆) implies that
(123)Y(t⋆)≥M.
It follows that Y(t⋆)≤M and Y(t⋆)≥M; therefore,
(124)Y(t⋆)=M.(2)
Next we show thatY′(t⋆)>0. Suppose that Y′(t⋆)≤0; then there exists a nonnegative natural number n such that Y decreases on the interval (t⋆,t⋆+(1/n)). Also we have that
(125)t⋆≤tn≤t⋆+1n;
this implies that
(126)Y(t⋆)≥Y(tn).
But from part (1), we have Y(t⋆)=M and Y(tn)>M; because tn∈I, it follows that
(127)Y(t⋆)<Y(tn).
Hence, Y(t⋆)≥Y(tn) and Y(t⋆)<Y(tn), which is impossible. Therefore,
(128)Y′(t⋆)>0.
Finally that we show that Y′(t⋆)≤0 and this contradicts the result obtained in (2). From (117), we have
(129)Y′(t⋆)≤-λY(t⋆)+ηY3(t⋆)+ζ=-λM+ηM(M2)+ζ(becauseY(t⋆)=M)≤-λM+ηM(λ2η)+ζ≤0,because  of  the  hypothesis(ζ≤λM2).
Therefore, Y(t)≤M for all t∈ℝ+.Theorem 12.
Assume that the initial and boundary data are sufficiently small, precisely(130)∥∇u0∥≤(κμ364c2c14)1/4,P~<μ4c1c3(κ3μ4c23)1/4.
Then, for all T≥0, there exists a weak solution u∈L2(0,T;V) to problem (23), and it satisfies the inequality
(131)∥∇u∥≤(κμ364c2c14)1/4.Proof.
We make use of the estimates (10), (73), and (101) in order to obtain
(132)ddt∥∇un∥2+κc2∥∇un∥2≤16c14μ3∥∇un∥6+2c32P~2μ.
Set
(133)Yn=∥∇un∥2.
The energy estimate (132) takes the form
(134)Yn′+λYn≤ηYn3+ζ,
where
(135)λ=κc2,η=16c14μ3,ζ=2c32P~2μ.
Let
(136)M=(κμ364c2c14)1/2;
we have
(137)0<M<(λ2η)1/2.
Also
(138)Yn(0)=∥∇u0∥2≤Mforeachintegern,ζ=2c32P~2μ≤κ2c2(κμ364c2c14)1/2=λM2.
We make use of Lemma 11 to find that Yn=∥∇un∥2 is bounded by M for all t>0 and the bound does not depend on n. Poincaré’s inequality (4) implies that
(139)∥un∥12=∥un∥2+∥∇un∥2≤CΩ2∥∇un∥2+∥∇un∥2≤(1+CΩ2)M.
Let T>0 be any positive real number,
(140)∥un∥L2(0,T;V)2=∫0T‍∥un∥12dt≤(1+CΩ2)MT.
Thus, for each nonnegative real number T, the sequence {∥un∥L2(0,T;V)}n=1∞ is bounded. We make use of Lemma 9 to extract a subsequence {unq}q=1∞ that converges to an element u of L2(0,T;V), and we use the same steps as we did for the local existence to show that u is a weak solution of problem (23).
Furthermore, the fact that the sequence{∇un}n=1∞ is bounded by M implies that the subsequence {∇unq}q=1∞ is also bounded by M.
Therefore, since∇u is the limit of this subsequence, it follows that
(141)∥∇u∥≤(κμ364c2c14)1/4.


## 5. Uniqueness of the Solution

Theorem 13.
There exists a time intervalT1 where the solution of problem (23) is unique.Proof.
Assume that there exist two solutionsu(1) and u(2) associated with the same data. Set w=u(1)-u(2). Consider (63) with φ=w. By substraction, we obtain
(142)(∂w∂t,w)+a(w,w)=b(u(2),u(2),w)-b(u(1),u(1),w).
We recall that a is a bilinear form, b is a trilinear form, and c is a linear form described by (24), (25), and (26), respectively. We evaluate every term of (142):
(143)(∂w∂t,w)=12ddt∥w∥2,a(w,w)=μ∥∇w∥2+∑i=13‍Ki∫Γ‍i|w·n|2ds,
and also,
(144)b(u(2),u(2),w)-b(u(1),u(1),w)=b(u(2),u(2),w)-b(u(2),u(2)+w,w)-b(w,u(2)+w,w)=-b(w,w,w)-b(u(2),w,w)-b(w,u(2),w).
It follows that
(145)12ddt∥w∥2+μ∥∇w∥2+∑i=13‍Ki∫Γ‍i|w·n|2ds=-b(w,w,w)-b(u(2),w,w)-b(w,u(2),w).
We now make use of some classical estimates to evaluate the rights hand side of (145) as follow:
(146)b(w,w,w)=((w·∇)w,w)=-∫Ω‍∑i=13‍(∑j=13‍∂(wiwj)∂xj)widx+∫Γ‍(w·w)(w·n)ds=-b(w,w,w)+∫Γ‍(w·w)(w·n)ds,because∇·w=0.
This implies that
(147)b(w,w,w)=12∫Γ‍(w·w)(w·n)ds≤12(∫Γ‍(w·w)2ds)1/2(∫Γ‍(w·n)2ds)1/2≤c5∥w∥L42∥w∥(15)≤c6∥w∥∥∇w∥2(14),(4).
Also,
(148)|b(u(2),w,w)|≤supΩ|u(2)|∥∇w∥∥w∥≤μ2∥∇w∥2+12μ(supΩ|u(2)|)2∥w∥2,|b(w,u(2),w)|≤supΩ|∇u(2)|∥w∥2.
It follows that
(149)|b(u(2),w,w)+b(w,u(2),w)|≤μ2∥∇w∥2+12μ(supΩ|u(2)|)2∥w∥2+supΩ|∇u(2)|∥w∥2.
We now make use of these estimates in (145) to find that
(150)12ddt∥w∥2+μ∥∇w∥2+∑i=13‍Ki∫Γ‍i|w·n|2ds≤μ2∥∇w∥2+12μ(supΩ|u(2)|)2∥w∥2+supΩ|∇u(2)|∥w∥2+c6∥w∥∥∇w∥2.
Thus,
(151)ddt∥w∥2+(μ-2c6∥w∥)∥∇w∥2≤(1μ(supΩ|u(2)|)2+2(supΩ|∇u(2)|))∥w∥2.
Since w(0,x)=0, there exists T1>0 such that ∥w∥≤μ/2c6 with 0<T1<T⋆. On the interval (0,T1), we have
(152)ddt∥w∥2≤ϑ(t)∥w∥2,
where
(153)ϑ(t)=(1μ(supΩ|u(2)|)2+2(supΩ|∇u(2)|))
is a real function depending on t.
The uniqueness theorem follows from Gronwall’s inequality [6], as w(0,x)=0.


## 6. Stability of the Solution

In this section, we are going to prove that there exists a time interval of continuous dependence on the data. We start with the boundary conditions and afterward we move on to the initial condition.


### 6.1. Boundary Conditions Stability

Theorem 14.
There exists a time interval where the solution of problem (23) depends continuously on the prescribed data as the real functions Pi (i=1,2,3) are varied.Proof.
Denote byu(1) the solution associated with the data Pi(1) (i=1,2,3) and, correspondingly, by u(2) the solution associated with Pi(2) (i=1,2,3).
Set(154)w=u(1)-u(2),P¯=maxi=1,2,3|Pi(1)-Pi(2)|.
We make use of (63) for u(1) and u(2) with φ=w. By substraction, we obtain
(155)(∂w∂t,w)+a(w,w)=b(u(2),u(2),w)-b(u(1),u(1),w)+c1(t,w)-c2(t,w),
where
(156)cj(t,v)=-∑i=13‍Pi(j)(t)∫Γi‍n·vds0foreachv∈H1(Ω),j=1,2.
This gives us that
(157)12ddt∥w∥2+μ∥∇w∥2+∑i=13‍Ki∫Γ‍i|w·n|2ds=-b(w,w,w)-b(u(2),w,w)-b(w,u(2),w)+∑i=13‍(Pi(2)-Pi(1))∫Γ‍iw·nds.
Also,
(158)∑i=13‍(Pi(2)-Pi(1))∫Γ‍iw·nds≤∑i=13‍P¯∫Γ‍i|w·n|ds≤c7P¯2+∑i=13‍Ki∫Γ‍i|w·n|2ds.
It follows that
(159)ddt∥w∥2+(μ-2c6∥w∥)∥∇w∥2≤(1μ(supΩ|u(2)|)2+2(supΩ|∇u(2)|))∥w∥2+2c7P¯2.
Since w(0,x)=0, there exists T1>0 such that ∥w∥≤μ/2c6 with 0<T1<T⋆. On the interval (0,T1), we have that
(160)ddt∥w∥2≤c8∥w∥2+2c7P¯2,
where
(161)c8=sup(0,T1)[(1μ(supΩ|u(2)|)2+2(supΩ|∇u(2)|))].
From Gronwall’s inequality [6], we deduce that in the interval (0,T1)(162)∥w∥≤cmax0<t≤T1P¯,whereP¯=maxi=1,2,3|Pi(1)-Pi(2)|.
Hence, a small change in the given data produces a correspondingly small change in the solution.
Therefore, the solution of problem (23) is stable in the interval (0,T1) as the real functions Pi (i=1,2,3) are varied.


### 6.2. Initial Condition Stability

Theorem 15.
There exists a time interval where the solution of problem (23) depends continuously on the prescribed data as the initial condition u0(x) is varied.Proof.
Denote byu(1) the solution associated with the initial data u0(1) and, correspondingly, by u(2) the solution associated with the initial data u0(2).
Setw=u(1)-u(2).
Consider (63) with φ=w. By substraction, we obtain
(163)(∂w∂t,w)+a(w,w)=b(u(2),u(2),w)-b(u(1),u(1),w),
and this gives us
(164)12ddt∥w∥2+μ∥∇w∥2+∑i=13‍Ki∫Γ‍i|w·n|2ds=-b(w,w,w)-b(u(2),w,w)-b(w,u(2),w).
We estimate the right-hand side of this equality like we did in the case of uniqueness and we obtain
(165)12ddt∥w∥2+μ∥∇w∥2+∑i=13‍Ki∫Γ‍i|w·n|2ds≤μ2∥∇w∥2+12μ((supΩ|u(2)|)2∥w∥2+supΩ|∇u(2)|∥w∥2+c6∥w∥∥∇w∥2.
Thus,
(166)ddt∥w∥2+(μ-2c6∥w∥)∥∇w∥2≤(1μ(supΩ|u(2)|)2+2(supΩ|∇u(2)|))∥w∥2.
Therefore, on any interval during which ∥w∥≤μ/4c6, making use of Poincaré’s inequality (4) gives us
(167)ddt∥w∥2≤-c7∥w∥2,
where
(168)c7=supJ(μ2CΩ2-1μ(supΩ|u(2)|)2-2(supΩ|∇u(2)|)),
and J is an interval of time during which ∥w∥≤μ/4c6.
From Gronwall’s inequality [6], we deduce that in the interval J(169)∥w∥≤cmaxJw0,wherew0=u0(1)-u0(2).
Hence, a small change in the given data produces a correspondingly small change in the solution.


## 6.1. Boundary Conditions Stability

Theorem 14.
There exists a time interval where the solution of problem (23) depends continuously on the prescribed data as the real functions Pi (i=1,2,3) are varied.Proof.
Denote byu(1) the solution associated with the data Pi(1) (i=1,2,3) and, correspondingly, by u(2) the solution associated with Pi(2) (i=1,2,3).
Set(154)w=u(1)-u(2),P¯=maxi=1,2,3|Pi(1)-Pi(2)|.
We make use of (63) for u(1) and u(2) with φ=w. By substraction, we obtain
(155)(∂w∂t,w)+a(w,w)=b(u(2),u(2),w)-b(u(1),u(1),w)+c1(t,w)-c2(t,w),
where
(156)cj(t,v)=-∑i=13‍Pi(j)(t)∫Γi‍n·vds0foreachv∈H1(Ω),j=1,2.
This gives us that
(157)12ddt∥w∥2+μ∥∇w∥2+∑i=13‍Ki∫Γ‍i|w·n|2ds=-b(w,w,w)-b(u(2),w,w)-b(w,u(2),w)+∑i=13‍(Pi(2)-Pi(1))∫Γ‍iw·nds.
Also,
(158)∑i=13‍(Pi(2)-Pi(1))∫Γ‍iw·nds≤∑i=13‍P¯∫Γ‍i|w·n|ds≤c7P¯2+∑i=13‍Ki∫Γ‍i|w·n|2ds.
It follows that
(159)ddt∥w∥2+(μ-2c6∥w∥)∥∇w∥2≤(1μ(supΩ|u(2)|)2+2(supΩ|∇u(2)|))∥w∥2+2c7P¯2.
Since w(0,x)=0, there exists T1>0 such that ∥w∥≤μ/2c6 with 0<T1<T⋆. On the interval (0,T1), we have that
(160)ddt∥w∥2≤c8∥w∥2+2c7P¯2,
where
(161)c8=sup(0,T1)[(1μ(supΩ|u(2)|)2+2(supΩ|∇u(2)|))].
From Gronwall’s inequality [6], we deduce that in the interval (0,T1)(162)∥w∥≤cmax0<t≤T1P¯,whereP¯=maxi=1,2,3|Pi(1)-Pi(2)|.
Hence, a small change in the given data produces a correspondingly small change in the solution.
Therefore, the solution of problem (23) is stable in the interval (0,T1) as the real functions Pi (i=1,2,3) are varied.


## 6.2. Initial Condition Stability

Theorem 15.
There exists a time interval where the solution of problem (23) depends continuously on the prescribed data as the initial condition u0(x) is varied.Proof.
Denote byu(1) the solution associated with the initial data u0(1) and, correspondingly, by u(2) the solution associated with the initial data u0(2).
Setw=u(1)-u(2).
Consider (63) with φ=w. By substraction, we obtain
(163)(∂w∂t,w)+a(w,w)=b(u(2),u(2),w)-b(u(1),u(1),w),
and this gives us
(164)12ddt∥w∥2+μ∥∇w∥2+∑i=13‍Ki∫Γ‍i|w·n|2ds=-b(w,w,w)-b(u(2),w,w)-b(w,u(2),w).
We estimate the right-hand side of this equality like we did in the case of uniqueness and we obtain
(165)12ddt∥w∥2+μ∥∇w∥2+∑i=13‍Ki∫Γ‍i|w·n|2ds≤μ2∥∇w∥2+12μ((supΩ|u(2)|)2∥w∥2+supΩ|∇u(2)|∥w∥2+c6∥w∥∥∇w∥2.
Thus,
(166)ddt∥w∥2+(μ-2c6∥w∥)∥∇w∥2≤(1μ(supΩ|u(2)|)2+2(supΩ|∇u(2)|))∥w∥2.
Therefore, on any interval during which ∥w∥≤μ/4c6, making use of Poincaré’s inequality (4) gives us
(167)ddt∥w∥2≤-c7∥w∥2,
where
(168)c7=supJ(μ2CΩ2-1μ(supΩ|u(2)|)2-2(supΩ|∇u(2)|)),
and J is an interval of time during which ∥w∥≤μ/4c6.
From Gronwall’s inequality [6], we deduce that in the interval J(169)∥w∥≤cmaxJw0,wherew0=u0(1)-u0(2).
Hence, a small change in the given data produces a correspondingly small change in the solution.



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*Source: 750185-2013-09-19.xml*