SOME RESULTS ABOUT A BOUNDARY VALUE PROBLEM ON MIXED CONVECTION

The purpose of this paper is to study the autonomous third order non linear differential equation f ′′′ + ff ′′ + g(f ′) = 0 on [0,+∞[ with g(x) = βx(x − 1) and β > 1, subject to the boundary conditions f(0) = a ∈ R, f ′(0) = b < 0 and f ′(t) → λ ∈ {0, 1} as t → +∞. This problem arises when looking for similarity solutions to problems of boundary-layer theory in some contexts of fluids mechanics, as mixed convection in porous medium or flow adjacent to a stretching wall. Our goal, here is to investigate by a direct approach this boundary value problem as completely as possible, say study existence or non-existence and uniqueness solutions and the sign of this solutions according to the value of the real parameter β.


Introduction
In fluid mechanics, the problems are usually governed by systems of partial differential equations. In modeling of boundary layer, this is sometimes possible, and in some cases, the system of partial differential equations reduces to a systems involving a third order differential equation of the form f + f f + g(f ) = 0, (1.1) where the function g : R → R is assumed to be locally Lipschitz.
If g(x) = 0, the equation is the Blasius equation (1907) see [6], [15]. The case g(x) = β(x 2 − 1) was first given by Falkner and Skan (1931) see [13]. The case g(x) = βx 2 , this case occurs in the study of free convection (1966) see [3], [5], [7], [9], [12]. And for g(x) = βx(x − 1) is the mixed convection (2003) see [1], [2], [4], [8], [10], [11], [14], [16]. In this paper is to investigate this last case with β > 1. We consider the equation And we associate to equation (1.2) the boundary value problem: where λ ∈ {0, 1} and β > 1. This problem arises in 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 β is a temperature power-law profile and b is the mixed convection parameter, namely b = Ra P e − 1, with R a the Rayleigh number and P e the Péclet number. The case where a ≥ 0, b ≥ 0, β > 0 and λ ∈ {0, 1} was treated by Aïboudi and al.see [1], and for a ∈ R, b ≤ 0, 0 < β < 1 see [2], and the results obtained generalize the ones of [11]. In [8], Brighi and Hoernel established some results about the existence and uniqueness of convex and concave solution of (P β;a,b,1 ) where −2 < β < 0 and b > 0. These results can be recovered from [10], where the general equation f + f f + g(f ) = 0 is studied. In [16], some theoretical results can be found about the problem (P β;0,b,1 ) with −2 < β < 0, and b < 0. In [14] and [16], the method used by the authors allows them to prove the existence of a convex solution for the case a = 0 and seems difficult to generalize for a = 0. The problem (P β;a,b,λ ) with β = 0 is the well known Blasius problem. In the following, we note by f c a solution of the problem to the initial values below and by [0, T c ) the right maximal interval of its existence: To solve the boundary value problem (P β;a,b,λ ) we will use the shooting method, which consists of finding the values of a reel parameter c for which the solution of (1.2) satisfying the initial conditions.

On Blasius Equation
In this section, we recall some results about subsolutions and supersolutions of the Blasius equation.
Recall that the so-called Blasius equation is the third order ordinary differential equation f + f f = 0 i.e Eq.(1.1) with g = 0. Let us notice that, for any τ ∈ R, the function h τ : t → 3 t−τ is a solution of Blasius equation on each (−∞, τ ) and (τ, +∞). Let I ⊂ R be an interval and f : I −→ R be a function.   Proof. See [10], Proposition 2.5. 1. If F is any anti-derivative of f on I, then (f e F ) = −βf (f − 1)e F .

4.
If T + < +∞, then f and f are unbounded near T + .
Proof. The first item follows immediately from equation (1.2). For the proof of items 2-5, see [3], and item Proof. This statements follows immediately from equation (1.2).

The boundary value problem (P β;a,b,λ )
Let the boundary value problem (P β;a,b,λ ), we are interested here in a concave, convex and convex-concave solutions of a problem (P β;a,b,λ ) and there sign. We used shooting method to find these solutions, this method consists of finding the values of a parameter c ∈ R for which the solution of (P β;a,b,c ) satisfying the initial conditions f (0) = a, f (0) = b and f (0) = c, exists up to infinity and is such that f (t) → λ as t → +∞.
Define the following sets: Remark 4.1. It is easy to prove that C 0 , C 1 , C 2 and C 3 are disjoint nonempty open subsets of R, [10] with g(x) = βx(x − 1) and β > 0).
The next proposition shows that the case (1) cannot hold.
Proof. Assume that f c is convex on its right maximal interval of existence [0, T c ) and f c (t) → +∞ as t → T c .
There exist t 0 ∈ [0, T 0 ) , which the function H 2 is decreasing for t > t 0 , this is a contradiction as t →T c . Proof. Let f c is convex on its right maximal interval of existence [0, T c ), suppose there exist t 1 > t 0 such that which yields a contradiction.
Proof. Let f c is convex on maximal interval of existence [0, T c ), such that f c (t) → 1 as t → T c , then there Proof  Proof. Let f c is solution on maximal interval of existence [0, T c ), if c ∈ C 1 , then T c = +∞, the function H 2 is creasing on [0, T c ) , it follows that 3c 2 + βb 2 (2b − 3) < 0, we obtain c < −b β(3−2b)

3
, and the solution f c is negative because a < 0 and f c < 0. Proof. If c ∈ C 3 , then f c → −∞, and T c < +∞, other results same proof that lemma 5.1.

Moreover, the function H 3 is decreasing on
, we obtain a contradiction.
Proof. If c ∈ C 2 , there exist t 0 ∈ [0, T c ) such that f c (t 0 ) = 0 and f c (t 0 ) > 0, there exist t 1 > t 0 such that , this is a contradiction.
Proof. The first result follows from remark 4.1 and lemma 4.2, the second result follows from proposition 5.1 and the third result follows from proposition 4.1.

The a > 0 case
Let a, b ∈ R with b < 0 and a > 0. We assume β > 1, and f c be a solution of the initial value problem (P β;a,b,c ) on the right maximal interval of existence [0, T c ), c > 0.
Before that, and in order to complete the study, let us divide the sets C 2 and C 3 into the following two subsets: Proof. Let c ∈ C 2.2 and s c be as in the definition of C 2.2 , the function H 3 is creasing on [0, s c ), we have: which implies that f c (s c ) < b 2 +(β−2b)a 2 β . From the proposition 4.2, the conclusion follows from that , for Lemma 6.2. If c ∈ C 1 ∪ C 2.1 and b > − 1 2 a 2 . Then T c = +∞ and there exist c * > 0 such that c > c * .
Proof. This follows immediately from remark 6.4, lemma 6.2 and proposition 6.3. (1) The boundary value problem (P β;a,b,0 ) has as least one convex solution on [0, +∞) if in addition we have b > − 1 2 a 2 it will be no negative convex solution. (2) The boundary value problem (P β;a,b,−∞ ) has infinity convex-concave solutions on the maximal interval of existence [0, T c ) with T c < +∞, if in addition we have b > − 1 2 a 2 the convex part of these solutions will be no negative.
Proof. The first result follows from remark 4.1, lemma 4.2 and proposition 6.2 , the second result follows from proposition 3.1, proposition 4.2 and remark 6.1, and the third result follows from proposition 4.1.

Conflicts of Interest:
The author(s) declare that there are no conflicts of interest regarding the publication of this paper.