Abstract
In this paper, we are concerned with the nonsteady Boussinesq system under mixed boundary conditions. The boundary conditions for fluid may include Tresca slip, leak and one-sided leak conditions, velocity, static (or total) pressure, rotation, and stress (or total stress) together, and the boundary conditions for temperature may include Dirichlet, Neumann, and Robin conditions together. Relying on the relations among strain, rotation, normal derivative of velocity, and shape of the boundary surface, we get variational formulation. The formulations consist of a variational inequality for velocity due to the boundary conditions of friction type and a variational equation for temperature. For the case of boundary conditions including the static pressure and stress, we prove that if the data of the problem are small enough and compatibility conditions at the initial instance are satisfied, then there exists a unique solution on the given interval. For the case of boundary conditions including the total pressure and total stress, we prove the existence of a solution without restriction on the data and parameters of the problem.
1. Introduction
In this paper, we are concerned with the Boussinesq equation for heat convectionunder mixed boundary conditions. Here, , and are, respectively, velocity, pressure, and temperature, and is the parameter for buoyancy effect, is the body force, is the heat source, is the viscosity, and is the thermal conductivity. The strain tensor is the one with the components . System (1) is a special case ofwhich is a mathematical model for nonsteady motion of heat-conducting incompressible Newtonian fluid. Here, is the parameter for dissipation of energy due to expansion, is a positive real number, and for two matrices and , and . The term represents the dissipation of energy due to viscosity (the Joule effect). Owing to the dissipation of energy due to viscosity , study of (2) is usually more difficult than the Boussinesq system.
For the papers concerned with (2), we refer to Introduction of [1]. Here, we more mention [2–5] concerned with (2), where . In [2], the problem under nonhomogeneous Dirichlet boundary conditions for velocity and temperature in the time-dependent domain was studied, and existence of a local-in-time solution or existence of the solution on the given interval for small data was proved. In [3], existence of a strong solution and periodic solution for the 2D problem was studied under the boundary conditions and domain as above. In [4], under homogeneous Dirichlet boundary conditions for velocity and temperature, existence of a strong solution and periodic solution were studied when data of the problem are small enough. Łukaszewicz and Krzyżanowski [5] dealt with the initial boundary value problem on a time-dependent domain with the homogeneous Dirichlet boundary condition for velocity and temperature, and they proved the existence and uniqueness of local weak solutions and the existence of a global weak solution for small initial data.
Several papers are concerned with (1). In [6, 7], the existence and uniqueness (for 2D) of a solution to the problem were studied under the homogeneous Dirichlet boundary condition for velocity and mixture of nonhomogeneous Dirichlet and Neumann boundary conditions for temperature. In [8], for the problem with nonhomogeneous Dirichlet boundary conditions for velocity and temperature, the existence of the time periodic solution was proved (see [9]). In [10–13], problem (1) on the time-dependent domain was studied under the nonhomogeneous Dirichlet boundary condition for velocity and temperature. In [14, 15], the problem on exterior domains with the homogeneous Dirichlet boundary condition for velocity and nonhomogeneous Dirichlet boundary condition for temperature was studied. In [16], problem (1) was studied under the mixture of the nonhomogeneous Dirichlet boundary condition and the stress boundary condition for fluid and the mixture of nonhomogeneous Dirichlet, Neumann, and Robin boundary conditions for temperature. They proved the existence of a unique local-in-time solution under a compatibility condition at the initial instance (see (27) and (31) of [16]). In [17], problem (1) in the cylindrical pipe with inflow and outflow was studied under slip boundary conditions for velocity and the Neumann conditions for temperature. In that, it was proved that there exists a solution on the given interval when norms of derivatives in the direction along the cylinder of the initial velocity, initial temperature, and the external force are small enough. In [18], the existence of a solution to problem (1) on the time-dependent domain was studied under the mixture of the Dirichlet condition of velocity, total pressure, and rotation boundary conditions for fluid and the mixture of Dirichlet, Neumann, and Robin boundary conditions for temperature.
On the contrary, for movement of fluid (), different kinds of boundary conditions are used, and in practice, we deal with the mixture of some kinds of boundary conditions. On some portions of the boundary, we can use boundary conditions with stress or rotation, whereas when there is flux through a portion of the boundary, we can deal with the static pressure or the total pressure (Bernoulli’s pressure) boundary conditions. There are many literature studies for the Navier–Stokes problem with mixed boundary conditions (see Introduction of [19, 20] and references therein). Recently, Navier–Stokes system with mixed boundary conditions including friction-type conditions was studied (cf. [20, 21]).
In [1], problem (2) is studied under mixed boundary conditions, and the boundary conditions for fluid may include Tresca slip, leak and one-sided leak conditions, velocity, total pressure, rotation, and total stress together, and the conditions for temperature may include Dirichlet, Neumann, and Robin conditions together. From the result of [1], we can get results for (1) with the boundary conditions as in [1]; however, the result demands that the parameter for buoyancy effect is small enough in accordance with the data of the problem, and the solution includes “defect measure” as in [22]. Also, for (2) and (1), the problem with a mixed boundary condition including the static pressure (not total pressure) and stress (not total stress) together is not yet considered.
When one of static pressure, stress, or the outflow boundary condition is given on a portion of the boundary, for the initial boundary value problems of the Navier–Stokes equations, the existence of a unique local-in-time solution and a unique solution on a given interval for small data of the problem are proved. From the mathematical point of view, the main difficulty of such problems results from the fact that in a priori estimation of solution, the term arising from the nonlinear term is not canceled (cf. Preface in [23]).
In the present paper, we are first concerned with heat convection equation (1) under mixed boundary conditions including the static pressure and stress. The boundary conditions for fluid may include conditions of friction type (Tresca slip, threshold leak, and one-sided leak conditions), velocity, static pressure, rotation, and stress together, and the conditions for temperature may include Dirichlet, Neumann, and Robin conditions together. Due to the boundary conditions of friction type, it is difficult to follow the methods in [16, 20]. The main difficulty of this problem is from the estimate of approximate solutions, and due to simultaneous velocity and temperature, the estimate is more difficult than the case of the Navier–Stokes equations. Also, in this paper, we prove the existence of a solution to (1) with the boundary conditions as in [1] without restriction on the parameter for buoyancy effect .
This paper consists of 5 sections. In the last part of Section 1, we give notations.
In Section 2, the problems to study and assumptions for future are stated. According to the boundary conditions for fluid, Problems I and II are distinguished. Problem I includes the static pressure and the stress conditions, whereas Problem II includes the total pressure and the total stress boundary conditions. Assumption for Problem I is stronger than the one for Problem II.
In Section 3, we first get a variational formulation for Problem I which consists of six formulae with six unknown functions, that is, using velocity, tangent stress on slip surface, normal stress on the leak surface, normal stresses on one-sided leak surfaces, and temperature together as unknown functions (Problem I-VE). Then, we get a new variational formulation for Problem I consisting of one variational inequality for velocity and a variational equation for temperature (Problem I-VI).
The variational formulation for Problem II is obtained in the same way as in [1], and smoothness of the solution with respect to is weaker than the one in Problem I. In the end of Section 3, the main results of this paper are stated (Theorems 1 and 2). The main result for Problem I asserts that if the data of the problem are small enough and compatibility conditions at the initial time (conditions 4 and 6 of Theorem 1) hold, then there exists a unique smooth solution. The main result for Problem II asserts the existence of a solution without restriction on the parameter for buoyancy unlike [1].
Section 4 is devoted to the proof of Theorem 1. To this end, first in Section 4.1, we consider an approximate problem, where the variational inequality for velocity is replaced by an equation with the gradient of the Moreau regularization of the functional due to the boundary conditions of friction type. Developing the method for the proof of Theorem 4.4 of [21], we get existence and estimations of approximate solutions for small data under the compatibility conditions at initial time. In Section 4.2, we complete the proof of the existence and uniqueness of a solution.
Section 5 is devoted to the proof of Theorem 2. The existence of solutions to an approximate problem and relative compactness of the set of solutions are studied. Then, passing to limit, we get the conclusion.
Throughout this paper, we will use the following notation. Let be a connected bounded open subset of . ,, for , and , where are connected open subsets of , and for and for others. When is a Banach space, and is a dual space of . Let be Sobolev spaces; , and so . An inner product and norm in the spaces and are denoted, respectively, by and , and means the duality pairing between a Sobolev space and its dual one. Also, is an inner product in or , and means the duality pairing between and or between and . The inner product and norms in , respectively, are denoted by and . Sometimes, the inner product between and in is denoted by .
Let and be, respectively, outward normal and tangent unit vectors at in . When , if with , then we denote by on . For convergence in spaces, , , and mean, respectively, strong, weak, and weak convergence. Derivative of with respect to is denoted by . We also assume that .
2. Problems and Assumptions
For temperature, we are concerned with the boundary conditions
Stress tensor is the one with components , and stress vector on the boundary surface is . The value of the normal stress vector on the boundary surface is . And . Total stress tensor is the one with components , and the total stress vector on the boundary surface is . The value of the total normal stress vector on the boundary surface is . And .
For Problem I, we assume that and are independent of . Thus, Problem I is the one with the boundary conditionsand Problem II is the one with the boundary conditionswhere , , , , , and (components of matrix ) are given functions or vectors of functions of on . For convenience in what follows, the problems with boundary conditions (5) and (6) are called, respectively, the case of static pressure and the case of total pressure.
Let
We assume that , , , and that , and , , , and for a.e. of the portions of boundary. Also, we use the following assumption.
Assumption 1. (for the case of static pressure). We assume the following:(1), , and(2)If , where is 10 or 11, is nonempty, then at least one of is nonempty, and there exists diffeomorphism in between and .(3)For the functions of (1),(4)For the functions of (4) and (5),
Assumption 2. (for the case of total pressure). We assume (1) and (2) of Assumption 1 and the following:
(3′) For the functions of (1),(4′) For the functions of (4) and (6),
Remark 1. On , only outflow (inflow) is possible, and so (2) of Assumption 1 is used to guarantee . In Theorems 3.3 and 3.5 of [24], for the proof of equivalence of variational formulations to variational inequalities, this assumption was used via Lemma 3.2 of [24]. In this paper, this assumption is also necessary to guarantee equivalence between Problems I-VE and I-VI in Remark 4.
3. Variational Formulations and Main Results
Since and , by the Korn inequality and Poincaré inequality, we use
3.1. Variational Formulations: The Case of Static Pressure
By Theorems 2.1 and 2.2 of [19], for and , we havewhere is the shape operator of the boundary surface (cf. (A.1) in [19]), are expressions of in a local coordinate system on , and .
For and , we havewhere was used. By (4), for and , we have
Taking into account (8) and , for , , and , we have
By (14)–(17), we can see that smooth solutions of problems (1), (4), and (5) satisfy
Define , and by
Define and by
Remark 2. Under (4) of Assumption 1, the duality product of (19) has a meaning (cf. Remark 3.1 in [24]). By (9) and (10),Then, taking into accountand (18), we introduce the following variational formulation for problems (1), (4), and (5).
Problem (I-VE). Find , , and a.e. such that , , andwhere is the subspace of consisting of functions such that .
Remark 3. Under Assumption 1, if a solution is smooth enough, (, , ), then Problem I-VE is equivalent to problems (1), (4), and (5) in the following sense.
By Theorem 3.4 of [24], at a.e., there exists satisfying the first equation of (1), and satisfies boundary condition (5). As given in Section 1, ch. 2 of [25], it is proved that satisfies the third equation of (1) and boundary condition (4).
We will find another variational formulation consisting of a variational inequality and a variational equation, which is equivalent to Problem I-VE if the solution is smooth enough (cf. Remark 4).
For fixed , let us consider the problemSubtracting the first formula of (21) with from the first formula of (21), we getDefine the functionals , respectively, bySince if , then , , , and , in what follows, for convenience, we use the notationDefine a functional byNote that since . Then, the functional is proper (cf. Definition A.1 of [21]), convex, lower semicontinuous, andDefine a functional byIn the same way as Problem I in [24], from (25), we getDefine operators , , and , respectively, byIf is a solution to (24), then we can see that the solution satisfies (cf. (3 20) of [21])Therefore, we have the following variational formulation for problems (1), (4), and (5).
Problem (I-VI). Find such that
Remark 4. If the solutions to Problem I-VI are smooth as much as and , then the first one of (34) is equivalent to(Remark, pp. 114 in [26]). In (31), putting by Theorem 3.5 of [24], we can see existence of for a.e. such that is a solution to Problem I-VE.
3.2. Variational Formulations: The Case of Total Pressure
Taking into account, by (14)–(17) with depending on , we can see that smooth solutions of problems (1), (4), and (6) satisfy the following:
Define , and by
Define and by
Then, taking into accountand (36), we introduce the following variational formulation for problems (1), (4), and (6).
Problem (II-VE). Find , , and , in a.e., , such that , , andwhere is the subspace of consisting of functions such that .
Define operators and , respectively, byLet functional be defined by (25)–(28). Then, in the same way as Problem I-VI of [21], we find a variational inequality for velocity. Then, we get another variational formulation consisting of a variational inequality for velocity and a variational equation for temperature, which is equivalent to Problem II-VE if the solution is smooth enough.
Problem (II-VI). Find such that
3.3. Main Results
The main results of this paper are as follows.
Theorem 1 (the case of static pressure). Let Assumption 1 be satisfied. Suppose that(1)The norms of , in the spaces they belong to are small enough(2) and (3)(4) (compatibility condition at initial time for velocity)(5) and are small enough(6) (compatibility condition at initial time for temperature)(7) and are small enoughThen, there exists a solution to (34) such that
The solution satisfying and for a constant small enough is unique.
Theorem 2 (the case of total pressure). Let Assumption 2 be satisfied; and . Then, there exists a solution to (43).
4. Proof of Theorem 1
4.1. Existence and Estimation of Solutions to an Approximate Problem
We first consider a problem approximating (34).
For every , define a functional bywhich is called the Moreau regularization of . When is the subdifferential of in the Hilbert space , let and (the Yosida approximation of ) for all . Then, the functional is convex, continuous, Fréchet differentiable, and for all . Moreover,(cf. Theorem 2.9 in [27]). The operator is Lipschitz continuous with the constant (cf. Proposition 2.3 in [27]) and monotone (cf. Lemma 4.10 of ch. III in [25]).
By the fact that are in and 4 of Assumption 1, there exists a constant such that
Thus, there exists such that(cf. .5.1.10 of [28]). Thus,where the operators are the ones in (32).
Let and be, respectively, bases of the space and . Without loss of generality, we assume that and as in [26]. We find a solution and to the problemwhich gives us a system for and . The solutions to (52) depend on , but for convenience of notation, here and in what follows, we use subindex instead of subindex . For , there exist absolute continuous functions and on . Since , , , and is Lipschitz continuous, and are in fact absolute continuous. If and are bounded and are integrable, then and are prolonged over . Under smallness of the data of the problem and the compatibility condition of the data at the initial instant, we will find estimates for and in the following, by which we obtain (111) and see that .
Multiplying the first and the second equation of (52), respectively, by and and adding for , we get
We will find a priori estimates for
Since is convex, continuous, and Fréchet differentiable, we haveand so by ,
Also,
By virtue of (50), (51), (56), and (57), we have from the first equation of (53)where and are, respectively, the ones in (49) and (51), and so
Here and in what follows are the constants independent of the data of problem which are denoted by with the exceptions of .
Setting in the first equation of (52) and multiplying the resulting equation by and adding for , we get
By condition (2) of Theorem 1, for any , we have , which by (47), it implies that and . Then, from (60), we havewhich is valid by the compatibility condition at the initial time for velocity (condition (4)) and the conditions for . On the contrary, taking into account (50), (51), and (56), we have from the first equation of (53)and sowhere is such that . Since by (17), we get from the second equation of (53)
From (64), we have
Setting in the second equation of (52) and multiplying the resulting equation by and adding for , we getwhere was used. From (67), we havewhich is valid by the compatibility condition at the initial time for temperature (condition (6)).
On the contrary, taking into account , from the second equation of (53), we havewhere is such that , and so
Taking into account (66), we have from (63)
Differentiating the first equality of (52) with respect to , we have that
Multiplying (72) by and adding for , we have
Calculating , we havewhere is the one in (51). Also, by the Hölder inequality and the Young inequality, we have
Taking into account (50), (74), and (75) and the fact that , which is owing to monotonicity of r from (cf. [26], pp. 116), from (73), we havethat is,
Differentiating the second equality of (52) with respect to , we have that
Multiplying (78) by and adding for , we have that
On the contrary, we get
Taking into account (see (17) and (80)), we have from (79)
We have from (81)
Adding (59), (77), (65), and (82), we have
Integrating (83), we havewhere
By the condition of theorem, let be so small that
Ifare valid, then there exists such thaton . Therefore, taking into account (86), by the Gronwall inequality, we haveon all intervals of satisfying (90).
Using the estimate, we will obtain a quadratic inequality satisfied by .
Put
Note that depends only on the data of the problem.
Then, when satisfies (87), we can see from (91) thaton , where (90) holds. Let the data of the problem be so small that
By (70) and (93), for the small data of the problem, we haveon , which implies
Therefore, for such small data of the problem that (94) is valid, ifthen owing to (96), step by step, we have
From the above, we see that, for the small data of the problem satisfying (87)–(89) and (94),is valid on the interval where the first inequality of (90) is valid.
Put
By (66), (93), and (63), for the small data satisfying (87)–(89) and (94), we have a quadratic inequality for , which is the one we want,on the intervals where the first inequality of (90) is satisfied.
By the conditions of the theorem, we can assume that the data of the problem are so small that (87)–(89) and (94) are valid, and satisfies the following inequality:
Now, let us prove that ifthen for any ,
Since , on an interval ,
Let us first prove that if the first inequality of (90) is valid on an interval , then more strongeris valid. Putting in (101) (which is valid on the interval where the first inequality of (90) holds when (87)–(89) and (94) are valid), we get
By virtue of (102), there exist two real roots of :and on the intervals and , (107) holds. Thus, by continuity of with respect to from , we have that , that is,
Thus,which shows (106). Thus, by step by step, we see that the fist inequality of (90) is valid on , and (104) is valid.
If (103) is valid, then so is (88). Therefore, for the small data satisfying (87), (89), (94), (102), and (103), we also have (99) on . By (104) and (99), we have
Then, by (91) and (83), we have
By (111),and so by (46) and (47),
4.2. Existence and Uniqueness of a Solution
Let us prove the existence of a solution. Owing to (111) and (112), we can extract subsequences, which are denoted with the subindex as before, such thatwhen and .
Putting , where and is the positive integer, multiply the first equation of (52) by and add for . Then, multiply the first equation of (52) by and add for . Substituting the resulting equations, we have
Since is convex, continuous, and Fréchet differentiable, we have
Taking into account (117), we have from (116)
Since and (see (47)), we have from (118)
By (114), in as and , and by lower semicontinuity of ,
In the routine way, we can prove thatas and .
Since , by (120) and (121), we have from (119)
Since , by (115), it is obvious that .
, and the set is dense in , and so (122) is valid for all .
By (115), we can get from the second equation of (52) that
Easily, we see that
Also,
Since , by (115), . Thus,
It is easy to prove that
Therefore, from (123), we have
Since , by (115), it is obvious that . Therefore, we proved the existence of a solution.
Let us prove uniqueness of a solution. Let be the two solutions to Problem I-VI satisfying inequality (111) instead of approximate solutions. Then, taking into account (44) (Remark 4), from (35), we havewhich imply
By virtue of (50) and (51), we havewhere is the one in (51). By (111),and so we have
Also, fromwe have
Taking into account (see (17)), by (80), we haveand so
By (111),
Therefore, adding (133) and (137), we get
We have from (139)which implies and for all .
5. Proof of Theorem 2
5.1. Existence and Estimation of a Solution to an Approximate Problem
We first consider a problem approximating (43). For every , let a functional be defined by (45).
Let and be, respectively, bases of the space and . Without loss of generality, we assume that and as in [26]. We find a solution and to the problemwhich gives us a system for and . The solutions to (141) depend on , but for convenience of notation, here and in what follows, we use subindex . For , there exist absolute continuous functions and on . If and are bounded and are integrable, then and are prolonged over . We will find estimates (157) in the following, by which we see that .
Multiplying the first and the second equation of (141), respectively, by and and adding for , we get
Let us estimate terms on the left-hand side above. It is easy to see
By the fact that are in and Assumption 2, there exists a constant such that
Thus,(cf. Theorem 1.5.1.10 of [28]). Obviously,
Since is convex, continuous and Fréchet differentiable, we have
Thus,
Also, by the Hölder inequality, we havewhere .
Also, we have
By (17), we have
Takingby (143)–(154), we have from (142)
Applying the Gronwall inequality, we have from (156)
Note that in (157) depends on and (via ) but independent of and .
By (46) and (29) and the third inequality of (157), we havewith independent of . Multiplying the first equation of (141) by , summing for , and taking into account (147), we havewhere the operators are the ones in (42). By (145) and (144), we havewhere is the one in (145). Taking into account (160) and the fact that , we have from (159)
Let us integrate every term of (161) first with respect to from to and then with respect to from 0 to , where when .
By the third inequality of (157),
Since ,and so by (157), we have
Also, by (157), we have
In the same way, we get
Note that constants , are independent of . By virtue of (162)–(168), uniformly with respect to ,and the set is relatively compact in (see Theorem 5 of [29]). Also, we have
By (170), from the second equation of (141), we have
Hence, by (157), we know thatwhere is independent of . Thus, the set is relatively compact in (see Corollary 5 of [29]).
5.2. Existence of a Solution
We can extract subsequences, which are denoted as before, such thatas and .
On the contrary, putting , where and is a positive integer, multiply the first equation of (141) by and add for . Then, multiply the first equation of (141) by and add for . Substituting the resulting equations, we have
Sincetaking into account (147), we have from (174)
Since and (cf. (47)), we have from (176)
By (173) and Corollary Appendix B.2 of [1], taking a subsequence if necessary, we have
Owing to (173),
Thus, taking a subsequence if necessary, we have (see Lemma Appendix B.1 of [1])
Therefore, taking into account(see Corollary Appendix B.3 of [1]), we have
By (158) and (173), in as and . Since the functional is lower weak semicontinuous, we have
In a rather routine way, we can prove thatas . For convenience of readers, we give a proof in the following:
By (165), the Hölder inequality with exponents , and (173), we have
By (165),which means that the mappingis continuous and linear on , that is, there exists such that
Thus, by (173),
Let us prove that
By (173), we have
By (193) and (194), we get (192).
Therefore, by (183)–(185) and (192), from (177), we have
Since (cf. (165)) and , . Therefore, by density of the set in , (195) is valid for all .
Thus, the first formula of (43) is valid.
Putting , where , multiply the second equation of (141) by and add for . Thus, we have
By Corollary Appendix B.2 of [1], we have
Since , we have
By (198), we havewhich implies
Therefore, taking into account (197) and (200), from (196), we have
Since the set is dense in , from (201), we have the second equation of (43).
Data Availability
No data were used to support this study.
Conflicts of Interest
The author declares that there are no conflicts of interest.
Acknowledgments
The author was supported by the Scientific and Technical Development Plan Fund SCDP-5.