On the Convex and Convex-Concave Solutions of Opposing Mixed Convection Boundary Layer Flow in a Porous Medium

Abstract

In this paper, we are concerned with the solution of the third-order nonlinear differential equation , satisfying the boundary conditions , , and , as , where and The problem arises in the study of the opposing mixed convection approximation in a porous medium. We prove the existence, nonexistence, and the sign of convex and convex-concave solutions of the problem above according to the mixed convection parameter and the temperature parameter .

1. Introduction

Owing to their numerous applications in industrial manufacturing processes, the convection phenomena about heated or cooled surfaces embedded in fluid-saturated porous media have attracted considerable attention during the last few decades. In this paper, our interest focuses on the analysis of the boundary value problems where . This problem derives from the study of mixed convection boundary layer near a semi-infinite vertical plate embedded in a saturated porous medium, with a prescribed power law of the distance from the leading edge for the temperature. The parameter is a temperature power-law profile and is the mixed convection parameter, namely, , with the Rayleigh number and the Péclet number. The interested reader can consult references [1, 2] for more details on the physical derivation and the numerical treatments.

Mathematical results about the problem with can be found in [3–7]. The case where , , and was treated by Aïboudi and al. in [3], and the results obtained generalize the ones of [6]. In [4], Brighi and Hoernel established some results about the existence and uniqueness of convex and concave solution of where and . These results can be recovered from [8], where the general equation is studied.

In [5], some theoretical results can be found about the problem with , , and . In particular, the authors prove that there exist and , such that(i) has no convex solution for any and each .(ii) has a convex solution for each and each .

In [7] one can find an interesting new result about the existence of convex solutions of where under some conditions. In [5, 7], the method used by the authors allows them to prove the existence of a convex solution for the case and seems difficult to generalize for .

The problem with is the well known Blasius problem. For a broad view, see [9]. See also [10].

Great interest is given to analytical studies of similarity solutions because of their applications in different fields, for example, in magnetohydrodynamic (see [11–13]) or in boundary layer flows (see [8, 14]).

The main goal of this paper is to study the question of existence and nonexistence of the solutions of with and . We will focus our attention on convex and convex-concave solutions of the equationAs usual, to get a convex or convex-concave solution of , we use the shooting technique which consists of finding the values of a parameter for which the solution of (1) satisfying the initial conditions , , and exists on and is such that as . We denote by the solution of the following initial value problem and by the right maximal interval of existence:

2. On Blasius Equation

In this section, we recall some basic properties of the subsolutions and -subsolutions of the Blasius equation. Let be an interval and be a function.

Definition 1. We say that is a subsolution of the Blasius equation if is of class and if on .

Definition 2 (let ). We say that is an -subsolution of the Blasius equation if is of class and if on .

Proposition 3 (let ). There does not exist nonpositive concave subsolution of the Blasius equation on the interval .

Proof. See [8], Proposition 2.11.

Proposition 4 (let and ). There does not exist any -subsolution of the Blasius equation on the interval

Proof. See [8], Proposition 2.18.

3. Preliminary Results

Proposition 5. Let be a solution of (1) on some maximal interval .(1)If is any antiderivative of on , then (2)Assume that and that as If moreover is of constant sign at infinity, then as (3)If and if as , then or (4)If , then and are unbounded near (5)If there exists a point satisfying and , where or , then, for all , we have .

Proof. The first item follows immediately from (1). For the proof of items -, see [8], Proposition 3.1 with

Lemma 6. Let and be a solution of (1) on some maximal interval . If there exists such that then and as . Moreover, on .

Proof. See [3], Lemma 9.

4. The Boundary Value Problem in the Convex and Convex-Concave Case with

In the following, we take and with and . We are interested here in convex and convex-concave solutions of the boundary value problem . As mentioned in the introduction, we will use the shooting method to find these solutions. Define the following sets:

Remark 7. It is easy to prove that and are disjoint nonempty open subsets of and that there exist such that , , and (see Appendix A of [8] with and .

Lemma 8 (let ). Then, is a convex solution of the boundary value problem if and only if .

Proof. See Appendix A of [8] with and .

Lemma 9 (let ). If , then . Moreover, is convex-concave, decreasing and as .

Proof. If then there exists such that and . From Proposition 5, items (1) and (3), we have and for all , and as . Thus, is convex-concave solution on and as .
Let us assume that ; then there exists such that and are negative on and we obtain on . Hence, is a nonpositive concave subsolution of the Blasius equation on . This contradicts the Proposition 3 and thus .

Remark 10. From Proposition 5, items (1), (3), and (5), if , then there are only three possibilities for the solution of the initial value problem : (1) is convex and as (with ).(2)There exists a point such that and .(3) is a convex solution of .

The next proposition shows that case (1) cannot hold.

Proposition 11 (let ). There does not exist , such that is convex on its right maximal interval of existence and as .

Proof. Assume that is convex on its right maximal interval of existence and as . Then there exists such that, for all , and Consequently, is a -subsolution of the Blasius equation on . Therefore from Proposition 4 we have .
Furthermore, there exists such that and , and then and for all . Thus,for all . Next, integrating (5) on for , we obtain and using Proposition 5, item (4), yields a contradiction as .

5. The Case

Lemma 12 (let and ). If and if there exists such that and , then .

Proof. Let and assume that there exists such that and .
Let us consider the function ). Since on , then is nondecreasing on and hence Thus,

For the rest of this section we will set .

Proposition 13 (let ). If either or and , then the boundary value problem has no convex solution.

Proof. Suppose that and that is a convex solution of the boundary value problem . Then, there exists such that .
Let . From (1), we obtain on . Therefore, is decreasing on and hence . It follows thatwhich implies that . This is a contradiction. The same contradiction is obtained where and .

Theorem 14. Let and with and and . (1)The boundary value problem has at least one convex solution.(2)If either or and , then the boundary value problem has no convex solution and has infinitely many convex-concave solutions.

Proof. The first result follows from Remark 7 and Lemma 8. The second result follows from Remark 7, Remark 10, Proposition 11, Proposition 13, and Lemma 6.

6. The Case

Let with and . We assume and consider the solution of the initial value problem on the right maximal interval of existence .

Let us set .

Lemma 15 (let . let ). If and if there exists such that is the first point where , then and .

Proof. Let be such that on and . Suppose that on . Then, necessarily, we have on . Moreover, since is increasing and , we also have on .
Let . From (1), we have . Consequently, on and since , it follows that on . Integrating from to gives Thus which is a contradiction.
Therefore, there exists such that on and . From Proposition 5, items (1) and (5), we have either or . The second case cannot happen. Assume, for the sake of contradiction, that . Then and on , so that we have . From Lemma 6, we obtain that , as , and on . It follows that is a positive convex-concave solution of the boundary value problem on , which contradicts the existence of . Consequently, we have . This implies that on and that . By virtue of Lemma 9, we see that remains negative after . The proof is complete.

Lemma 16 (let ). If there exists such that and , then .

Proof. Assume that there exists such that , , and . Then, there would exist such that .
Let . From (1), we have on . Therefore, is nondecreasing on . Since , we get . But, this and Proposition 5, item (1), imply that remains negative on , a contradiction. Hence .

Lemma 17. If and , then there exists such that if then is a convex-concave solution of .

Proof (let ). From Remark 10 and Proposition 11, we see that either is a convex solution of or there exists such that and . Now, as we have seen in the proof of Lemma 15, in the second case, is a convex-concave solution of .
Let be such that is a convex solution of . Therefore, we have on and, from Lemma 15, we have . It follows that on . Integrating between and , and using the fact that , we obtain Integrating once again we getLet us set . We have for all . It means that has no positive roots. Thus cannot be too large, because, on the contrary, its discriminant and would be positive, and hence the polynomial would have two positive roots, a contradiction.
Therefore, there exists such that is convex-concave solution of the problem for . This completes the proof.

Theorem 18. Let , with , , and . (1)The boundary value problem has at least one convex solution. If in addition , then any convex solution of is positive.(2)If , then the boundary value problem has infinitely many positive convex-concave solutions.

Proof. The first part of (1) follows from Remark 7 and Lemma 8. The second part follows from Lemma 15, because if there was a point such that on and then , a contradiction. The second result follows from Remark 7, Remark 10, Proposition 11, Lemma 16, and Lemma 6.

7. Conclusion

In this work, in particular in Theorems 14 and 18, we have presented some new and important results about the boundary value problems and , which we summarize below. The parameters and satisfy and . The constants and are defined in Sections 5 and 6.(1)For (a)The boundary value problem has at least one convex solution.(b)If either or and , then the boundary value problem has no convex solution and has infinitely many convex-concave solutions.(2)For (a)If , then the boundary value problem has at least one positive convex solution.(b)If , then the boundary value problem has infinitely many positive convex-concave solutions.

Numerical simulations prompt us to formulate the following conjecture.

Conjecture 19. Let , with , , and . The boundary value problem has no convex solution.

To finish, we give the following proposition concerning the case .

Proposition 20 (let ). If , then the boundary value problem has no convex solution.

Proof. Assume that is a convex solution of the boundary value problem . Then, there exists , such that on , , and . Consider again the function We have on . Thus, is a decreasing function and hence . It follows that which implies that , which is a contradiction.

Data Availability

The data used to support the findings of this study are available from the corresponding author upon request.

Conflicts of Interest

The authors declare that they have no conflicts of interest.