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Research Article | Open Access

Volume 2018 |Article ID 4340204 | https://doi.org/10.1155/2018/4340204

M. Aïboudi, K. Boudjema Djeffal, B. Brighi, "On the Convex and Convex-Concave Solutions of Opposing Mixed Convection Boundary Layer Flow in a Porous Medium", Abstract and Applied Analysis, vol. 2018, Article ID 4340204, 5 pages, 2018. https://doi.org/10.1155/2018/4340204

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

Accepted11 Oct 2018
Published01 Nov 2018

#### 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 . The case where , , and was treated by Aïboudi and al. in , and the results obtained generalize the ones of . In , 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 , where the general equation is studied.

In , 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  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 . See also .

Great interest is given to analytical studies of similarity solutions because of their applications in different fields, for example, in magnetohydrodynamic (see ) 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 , Proposition 2.11.

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

Proof. See , 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 , 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 , 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  with and .

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

Proof. See Appendix A of  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.

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