ON THE SOLUTIONS OF FALKNER-SKAN EQUATION

We consider the differential equation f ′′′ +ff ′′ +β ( f ′2 − 1 ) = 0, with β > 0. In order to prove the existence of solutions satisfying the boundary conditions f (0) = a ≥ 0, f ′ (0) = b ≥ 0 and f ′ (+∞) = −1 or 1 for 0 < β ≤ 1 2 . We use shooting technique and consider the initial conditions f (0) = a, f ′ (0) = b and f ′′ (0) = c. We prove that there exists an infinitely many solutions such that f ′ (+∞) = 1.


introduction
In 1931 the Falkner-Skan equation is introduced for studying the boundary layer flow past a semi infinite wedge, it is defined by The solution of this equation have been studied by numerous authors as, for example, D. R. Hartree (1937), H. Weyl (1942), W. A. Coppel (1960), P. Hartman (1964) and G.C. Yang (2003Yang ( ,2004. The more general equation of (1.1) is where g : R −→ R is some function. The solutions obtained are called similarity solutions.
The most famous example is perhaps the Blasius equation (1908), which corresponds to g (x) = 0 and arises in the study of laminar boundary layer on a flat plate.
More recently, the equation (1.2) with g (x) = βx 2 and g (x) = βx (x − 1) has been considered. These cases occur, for example, in the study of free convection and of mixed convection boundary layer flows over a vertical surface embedded in a porous medium.
Most of the time, associated with the equation (1.1) is the boundary value problem: To obtain a solution of (P β; a, b, λ ) amounts to find a value of c such that T c = +∞ and We must assume that β λ 2 − 1 = 0 to have solutions, in our case of Falkner-Skan equation, the only In the following, we will study the existence of concave, convex, concave-convex and convex-concave solutions to the boundary value problem (P β; a, b, −1 ) and (P β; a, b, 1 ) for 0 < β ≤ 1 2 , a ≥ 0 and b ≥ 0.

Preliminary results
Let f be a solution of the equation (1.1) on some interval I, we consider the function H f : This function is obtained by integrating the equation (1.1), in fact, if f is a solution of (1.1) then We give lemmas that will be useful later. Proof. See [4], proposition 3.1 item 3.  (2) If there exists s < r ∈ I such that f (s) ≥ 0 and f 2 − 1 < 0 on ]s, r[, then f (t) > 0 for all t ∈ ]s, r] .
Proof. Let F denote any primitive function of f . From (1.1) we deduce the relation All the assertions 1-4 follow easily from this relation and from previous lemma. Let us verify the first and the third of these assertions. For the first one, since ψ = f exp F is decreasing on [s, r], we obtain For the third one, since ψ < 0 on ]s, r[ and ψ (r) = 0, ψ (t) ≥ 0 on [s, r], then ψ (r) ≥ 0.
Proof. See [4], proposition 3.1 item 4 and 5. Proof. Let L = L f be the function defined on I by Easily, using (1.1), we obtain that and, since f ≥ 0 on I this implies that L is decreasing. Hence It follows that f is bounded on I and thanks to lemma (2.3) that T + = +∞.
Proof. Assume for contradiction that f > 0 on I. Then f (t) ≥ 0, f (t) ≥ 1 for all t ∈ I. Then, we have It follows that 0 < f (t) ≤ c for all t ∈ I and hence, by lemma (2.3), we have T + = +∞.
Next, let s > τ and = β f (s) 2 − 1 . One has > 0 and, comming back to (2.3), we obtain We claim that δ = T + , assume for contradiction that δ < T + . From lemma (2.2), item 2, we get that f (δ) > 0, which implies, by definition of δ, that f (δ) = 1. Therefore, since the function H f defined by  To this end, let us partition R into the four sets C 0 , C 1 , C 2 and C 3 defined as follows.
Let C 0 = ]0, + ∞[ and, according to the notations used in [4], let us set This is obvious that C 0 , C 1 , C 2 and C 3 are disjoint sets and that their union is the whole line of real numbers. Thanks to lemma 2.3 and 2.4 if c ∈ C 1 then T + = +∞ and f c (t) −→ 1 as t −→ +∞. In fact, C 1 is the set of values of c for which f c is a concave solution of (P β; a, b, 1 ). Since β > 0, the study done in [4] (specially in section 5.2) says, on the one hand, that C 3 = ∅ (which can easily be deduced from lemma 2, item 1) and, on the other hand, that either C 1 = ∅ and C 2 = ]−∞, 0] , or there exist c * ≤ 0 such that In addition, from the lemma 5.16 in [4], if β ∈ 0, 1 2 then we are in the second case and c * ≤ −a (b − 1). In order to complete the study, let us divide the set C 2 into the following two subsets And let us give properties of each of them that hold for all β ∈ 0, 1 2 .  Proof. Let c ∈ C 2.1 . By proposition 4.2, we have c < 0.
Assume first that f c < 0 on ]0, +∞[. Then f c is decreasing, and thus f c has a finite limit λ at infinity.
Moreover, by definition of the set C 2.1 we get and by lemma 2.4, we finally get that λ = −1.
Assume now that f c vanishes on ]0, +∞[, let t 0 be the first point where f c vanishes. Thanks to lemma 2.2 item 3, we have 0 < f c (t 0 ) < 1, and the conclusion follows from lemma 2.7 (λ = 1).

Remark 4.3.
If c ∈ C 2.1 then either f c is a concave solution of (P β; a, b, −1 ) or f c is a concave-convex solution of (P β; a, b, 1 ). Proof. See [3] in the case of the mixed convection equation. and f c < 0. By the change of variable, as done in [4], section 4, we can define a function v : From (3.1), we deduce that v b 2 = a and v b 2 = 1 2c . Moreover, since f c is bounded, it is so for v. Assume that there exists c 1 > c 2 such that f c1 (t) → 0 and f c2 (t) → 0 as t → +∞, and denote by v 1 and v 2 the functions associated to f c1 and f c2 by (4.1). If we set w = v 1 − v 2 then w b 2 = 0 and w b 2 < 0.
We claim that w < 0 on 0, b 2 . For contradiction, assume there exists x ∈ 0, b 2 such that w < 0 on ]0, x[ and w (x) = 0. Hence we have w (x) ≤ 0 and w (x) > 0. But thanks to (4.1), we have and a contradiction.

Now, let us set
In the other hand, thanks to (4.2); we have Therefore, we have the last equality following from the fact that w (y) tends to finite limit as y → 0. Since w < 0, we finally obtain W b 2 < 0 and a contradiction.  Theorem 4.1. Let β ∈ 0, 1 2 , a ≥ 0 and b ≥ 1. There exists c * < 0 such that: (1) f c is not defined on the whole interval [0, + ∞[ if c < c * ; (2) f c * is a concave solution of (P β; a, b, −1 ); (3) f c is a solution of (P β; a, b, 1 ) for all c ∈ ]c * , + ∞[ (1) f c is a convex-concave solution of (P β; a, b, 1 ) for all c ∈ ]0, + ∞[ ; (2) f c is a concave solution of (P β; a, b, 1 ) for all c ∈ [c * , 0] ; (3) f c is a concave-convex solution of (P β; a, b, 1 ) for all c ∈ ]c * , c * [ ; 5. The case β ∈ 0, 1 2 and −1 < b < 1 Let β ∈ 0, 1 2 , a ≥ 0 and −1 < b < 1. In this situation, it is easy to see that R can be partitioned into the four sets C 0,1 , C 0,2 , C 1 and C 2 where The arguments used in the previous section, can be applied here.
First, since g (x) = β x 2 − 1 < 0 for x ∈ ]−1, b] where b ∈ ]−1, 0[, the function g is nonincreasing on ]−1, b], it follows from theorem 5.5 of [4] that there exists a unique c * such that f c * is a concave solution of (P β; a, b, −1 ). Moreover, we have c * < 0. As in the previous section, this implies that C 0,2 = ]−∞, c * [. Next, in the same way as in the proof of proposition 3.1, we can proof that c * = sup C 1 is finite, and hence that C 1 = [0, c * ] and C 2 = ]c * , + ∞[. Moreover, from lemma 2.7, we have c * ≥ a (1 − b). On the other hand, it follows from lemma 2.6 that, if c ∈ C 2 , then f c vanishes at a first point where f c > 1.
All this, combined with an appropriate use of lemmas 2.7 and 2.8 allows to state the following theorem.