International Journal of Analysis and Applications Volume 18, Number 3 (2020), 409-420 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-18-2020-409 ON THE SOLUTIONS OF FALKNER-SKAN EQUATION LABBAOUI FATMA1,2,∗, AND AIBOUDI MOHAMMED1,2, 1Département de Mathématique, Faculté des Sciences Exactes et Appliquées, Université Oran 1 Ahmed Ben Bella, Oran, Algérie 2Laboratoire de recherche d’Analyse Mathématique et Application L.A.M.A, Oran, Algerie ∗Corresponding author: fatimazahra107@yahoo.fr Abstract. 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. 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 f ′′′ + ff ′′ + β ( f ′2 − 1 ) = 0 (1.1) 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 (2003,2004). The more general equation of (1.1) is f ′′′ + ff ′′ + g ( f ′ ) = 0 (1.2) where g : R −→ R is some function. The solutions obtained are called similarity solutions. Received February 12th, 2020; accepted March 9th, 2020; published May 1st, 2020. 2010 Mathematics Subject Classification. 76D10, 34B15. Key words and phrases. boundary layer; shooting technique; convex solution; concave solution; convex-concave solution. ©2020 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 409 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-18-2020-409 Int. J. Anal. Appl. 18 (3) (2020) 410 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) = βx2 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:  f ′′′ + ff ′′ + β ( f ′2 − 1 ) = 0 f (0) = a f ′ (0) = b f ′ (t) −→ λ as t −→ +∞ (Pβ;a,b,λ) To solve the boundary value problem (Pβ; a, b, λ) we will use the shooting technique. To this end, let fc denote the solution of the initial value problem (Qβ; a, b, c) consisting in the equation (1.1) together with the initial conditions fc (0) = a, f ′ c (0) = b and f ′′ c (0) = c, and let [0, Tc[ be the right maximal interval of existence of fc. To obtain a solution of (Pβ; a, b, λ) amounts to find a value of c such that Tc = +∞ and f ′ c (t) −→ λ as t −→ +∞. We must assume that β ( λ2 − 1 ) = 0 to have solutions, in our case of Falkner-Skan equation, the only relevant conditions are f ′ c (t) −→−1 or f ′ c (t) −→ 1 as t −→ +∞. In the following, we will study the existence of concave, convex, concave-convex and convex-concave solutions to the boundary value problem (Pβ; a, b, −1) and (Pβ; a, b, 1) for 0 < β ≤ 12 , a ≥ 0 and b ≥ 0. 2. Preliminary results Let f be a solution of the equation (1.1) on some interval I, we consider the function Hf : I −→ R defined by Hf = f ′′ + f ( f ′ − 1 ) . (2.1) This function is obtained by integrating the equation (1.1), in fact, if f is a solution of (1.1) then H ′ f = ( f ′ − 1 )( (1 −β) f ′ −β ) . We give lemmas that will be useful later. Lemma 2.1. Let f be a solution of (1.1) on some maximal interval I. If there exists t0 ∈ I such that f ′ (t0) ∈{−1, 1} and f ′′ (t0) = 0, then I = R and f ′′ (t) = 0 for all t ∈ R. Proof. See [4], proposition 3.1 item 3. � Lemma 2.2. Let β > 0 and f be a solution of equation (1.1) on some interval I, such that f ′ is not constant. Int. J. Anal. Appl. 18 (3) (2020) 411 (1) 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] . (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] . (3) If there exists s < r ∈ I such that f ′′ < 0 on ]s, r[ and f ′′ (r) = 0, then ( f ′2 (r) − 1 ) < 0. (4) If there exists s < r ∈ I such that f ′′ > 0 on ]s, r[ and f ′′ (r) = 0, then ( f ′2 (r) − 1 ) > 0. Proof. Let F denote any primitive function of f. From (1.1) we deduce the relation ( f ′′ exp F )′ = −β ( f ′2 − 1 ) exp F. 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 t ≥ s =⇒ ψ (t) ≤ ψ (s) =⇒ f ′′ (t) exp F (t) ≤ f ′′ (s) exp F (s) =⇒ f ′′ (t) ≤ f ′′ (s) exp (F (s) −F (t)) =⇒ f ′′ (t) ≤ 0, ∀t ∈ ]s, r] . For the third one, since ψ < 0 on ]s, r[ and ψ (r) = 0, ψ ′ (t) ≥ 0 on [s, r], then ψ ′ (r) ≥ 0. ψ ′ (r) = −β ( f ′2 (r) − 1 ) exp F ≥ 0. This and lemma 2.1 imply that ( f ′2 (r) − 1 ) < 0. � Lemma 2.3. Let f be a solution of (1.1) on some maximal interval ]T−, T+[ . If T+ is finite, then f ′ and f ′′ are unbounded in any neighborhood of T+. Proof. See [4], proposition 3.1 item 6. � Lemma 2.4. Let β 6= 0. If f is a solution of (1.1)on some interval ]τ, + ∞[ such that f ′ (t) −→ λ as t −→ +∞, then λ ∈{−1, 1}. Moreover, if f is of constant sign at infinity, then f ′′ (t) −→ 0 as t −→ +∞. Proof. See [4], proposition 3.1 item 4 and 5. � Lemma 2.5. Let β > 0 and f be a solution of equation (1.1) on some right maximal interval I = [τ, + ∞[. If f ≥ 0 and f ′ ≥ 0 on I, then T+ = +∞ and f ′ is bounded on I. Int. J. Anal. Appl. 18 (3) (2020) 412 Proof. Let L = Lf be the function defined on I by L (t) = 3f′′ 2 (t) + βf′ (t) ( 2f′ 2 (t) − 6 ) . (2.2) Easily, using (1.1), we obtain that L′ (t) = −6f (t) f′′ 2 (t) ∀t ∈ I, and, since f ≥ 0 on I this implies that L is decreasing. Hence ∀t ∈ I = [τ, T+[ : t > τ =⇒ L (t) ≤ L (τ) βf′ (t) ( 2f′ 2 (t) − 6 ) ≤ 3f′′ 2 (t) + βf′ (t) ( 2f′ 2 (t) − 6 ) ≤ L (τ) , ∀t ∈ I. It follows that f ′ is bounded on I and thanks to lemma (2.3) that T+ = +∞. � Lemma 2.6. Let β > 0 and f be a solution of equation (1.1) on some right maximal interval I = [τ, T+[ . If f (τ) ≥ 0, f ′ (τ) ≥ 1 and f ′′ (τ) > 0, then there exists t0 ∈ ]τ, T+[ such that f ′′ > 0 on [τ, t0[ and f ′′ (t0) = 0. Proof. Assume for contradiction that f ′′ > 0 on I. Then f (t) ≥ 0, f ′ (t) ≥ 1 for all t ∈ I. Then, we have f ′′′ = −ff ′′ −β ( f ′2 − 1 ) ≤ 0. (2.3) 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 f ′′′ ≤−� on [s, + ∞[ . After integrating, we get ∀t ≥ s, f ′′ (t) −f ′′ (s) ≤−� (t−s) , and a contradiction with the fact that f ′′ (t) > 0. Consequently, there exists t0 ∈ ]τ, T+[such that f ′′ > 0 on [τ, t0[and f ′′ (t0) = 0. � Lemma 2.7. Let β ∈ ] 0, 1 2 [ and f be a solution of equation (1.1) on some right maximal interval I = ]T−, T+[ . If there exists t0 ∈ I such that β1−β < f ′ (t0) < 1 and 0 ≤ f ′′ (t0) ≤ f (t0) ( 1 −f ′ (t0) ) , then T+ = +∞ and f ′ (t) −→ 1 as t −→ +∞. Moreover f ′′ > 0 on [t0, + ∞[ . Int. J. Anal. Appl. 18 (3) (2020) 413 Proof. Let δ = sup X (t0), where X (t0) = { t ∈ ]t0, T+[ : f ′ (t0) < f ′ < 1 and f ′′ > 0 on ]t0, t[ } . The set X (t0) is not empty. This is clear if f ′′ (t0) > 0, and if f ′′ (t0) = 0 it follows from the fact that f ′′′ (t0) = −β ( f ′2 (t0) − 1 ) > 0. 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 Hf defined by (2.1) is nonincreasing on [t0, δ] , we obtain δ ≥ t0 =⇒ Hf (δ) ≤ Hf (t0) =⇒ f ′′ (δ) ≤ f ′′ (t0) −f (t0) ( 1 −f ′ (t0) ) =⇒ f ′′ (δ) < 0, a contradiction. Thus, we have δ = T+. From lemma 2.3, it follows that T+ = +∞. Since f ′′ > 0 on [t0, + ∞[ , by virtue of lemma 2.4, we get that f ′ (t) −→ 1 as t −→ +∞. � Lemma 2.8. Let β ∈ ] 0, 1 2 ] and f be a solution of equation (1.1) on some right maximal interval I = ]T−, T+[ . If there exists t0 ∈ I such that f ′ (t0) > 1 and f (t0) ( 1 −f ′ (t0) ) ≤ f ′′ (t0) ≤ 0. Then T+ = +∞ and f ′ (t) −→ 1 as t −→ +∞. Moreover f ′′ < 0 on [t0, + ∞[ . Proof. If we set ω = sup Y (t0), where Y (t0) = { t ∈ ]t0, T+[ : 1 < f ′ < f ′ (t0) and f ′′ < 0 on ]t0, t[ } . The conclusion will follow by proceeding in the same way as the previous proof. � 3. Description of our approach when b ≥ 1 Let β > 0, a ≥ 0 and b ≥ 1. As said in the introduction, the method we will use to obtain solutions of the boundary value problems (Pβ; a, b, −1) and (Pβ; a, b, 1) is the shooting technique. Specifically, for c ∈ R, let us denote by fc the solution of equation (1.1) satisfying the initial conditions fc (0) = a, f ′ c (0) = b and f ′′ c (0) = c, (3.1) and let [0, Tc[ be the right maximal interval of existence of fc. Hence, finding a solution of one of the problems (Pβ; a, b, −1) and (Pβ; a, b, 1) amounts to finding a value of c such that T+ = +∞ and f ′ c (t) −→−1 or 1 as t −→ +∞. Int. J. Anal. Appl. 18 (3) (2020) 414 To this end, let us partition R into the four sets C0, C1, C2 and C3 defined as follows. Let C0 = ]0, + ∞[ and, according to the notations used in [4], let us set C1 = { c ≤ 0; 1 ≤ f ′ c ≤ b and f ′′ c ≤ 0 on [0, Tc[ } , C2 =   c ≤ 0; ∃ tc ∈ [0, Tc[ , ∃ �c > 0 such that f ′ c > 1 on ]0, tc[ , f ′ c < 1 on ]tc, tc + �c[ and f ′′ c ≤ 0 on [0, tc + �c[   , C3 =   c ≤ 0; ∃ rc ∈ [0, Tc[ , ∃ ηc > 0 such that f ′′ c < 0 on ]0, rc[ f ′′ c > 0 on ]rc, rc + ηc[ and f ′ c > 1 on ]0, rc + ηc[   . This is obvious that C0, C1, C2 and C3 are disjoint sets and that their union is the whole line of real numbers. Thanks to lemma 2.3 and 2.4 if c ∈ C1 then T+ = +∞ and f ′ c (t) −→ 1 as t −→ +∞. In fact, C1 is the set of values of c for which fc is a concave solution of (Pβ; a, b, 1). Since β > 0, the study done in [4] (specially in section 5.2) says, on the one hand, that C3 = ∅ (which can easily be deduced from lemma 2, item 1) and, on the other hand, that either C1 = ∅ and C2 = ]−∞, 0] , or there exist c∗ ≤ 0 such that C1 = [c ∗, 0] and C2 = ]−∞, c∗[. 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 C2 into the following two subsets C2,1 = { c ∈ C2 : f ′ c > −1 on [0, Tc[ } , C2,2 = { c ∈ C2 : ∃ sc ∈ ]0, Tc[ such that f ′ c > −1 on [0, sc[ and f ′ c (sc) = −1 } . And let us give properties of each of them that hold for all β ∈ ] 0, 1 2 ] . Lemma 3.1. If c ∈ R such that f ′ c > 0 on [0, Tc[, then Tc = +∞ and f ′ c is bounded. Moreover, if c ≤ 0, then f ′ c ≤ max { b, √ 3 } on [0, + ∞[. Proof. Let c ∈ R is such that f ′ c > 0 on [0, Tc[, then fc ≥ a ≥ 0 on [0, Tc[ and thanks to lemma 2.5, it follows that Tc = +∞ and f ′ c is bounded. It remains to show that f ′ c ≤ max { b, √ 3 } in the case where c ≤ 0. As in [4], let us define the function Lc on [0, + ∞[ by Lc (t) = 3f ′′2 c (t) + βf ′ c (t) ( 2f′ 2 c (t) − 6 ) , and since fc ≥ 0, it implies that Lc is decreasing. Int. J. Anal. Appl. 18 (3) (2020) 415 If f ′′ c ≤ 0 on ]0, + ∞[, then f ′ c ≤ b. Otherwise, there exists t0 such that f ′′ c < 0 on ]0, t0[ and f ′′ c (t0) = 0. By lemma 2.2 item 3, it follows that f ′ c < 1, and thus Lc (t0) < 0. Then, Lc < 0 on ]t0, + ∞[ which implies that f ′ c ≤ √ 3 on ]t0, + ∞[. Since f ′ c ≤ b on ]0, t0[, the proof is complete. � Proposition 3.1. Let c∗ = inf (C1 ∪C2,1). Then c∗ is finite. Proof. let c ∈ C1 ∪C2,1. By definition of C1 and C2,1, and thanks to lemma 2.3, we have Tc = +∞ and 0 < f ′ c < d on [0, + ∞[ where d = max { b, √ 3 } . Since ( f ′′ c + fcf ′ c )′ = f ′′′ c + fcf ′′ c + f ′2 c = −β ( f ′2 c − 1 ) + f ′2 c = −βf ′2 c + β + f ′2 c ≤ β + d 2. By integrating, we then have ∀t ≥ 0, f ′′ c (t) + fc (t) f ′ c (t) ≤ c + ab + ( β + d2 ) t. Integrating once again, for all t ≥ 0, we get 0 < f ′ c (t) ≤ f ′ c (t) + 1 2 f2c (t) ≤ b + 1 2 a2 + (c + ab) + 1 2 ( β + d2 ) t2. Which implies that c ≥−ab− √ (2b + a2) (β + d2). � Remark 3.1. As we have seen above, if C1 6= ∅, then C1 = [c∗, 0] and thus C2,1 ⊂ [c∗,c∗[. 4. The case β ∈ ] 0, 1 2 ] and b > 1 In this section we assume that β ∈ ] 0, 1 2 ] , a > 0 and b > 1. Proposition 4.1. If c > 0, then Tc = +∞ and f′c (t) → 1 as t → +∞. Proof. From lemma 2.6, there exists t0 ∈ ]0,Tc[ such that f′′c > 0 on [0, t0[ and f ′′ c (t0) = 0. Since fc (t0) > 0 and f ′ c (t0) > b > 1. Thus fc (t0) ( 1 −f ′ c (t0) ) 6 f ′′ c (t0) = 0, the conclusion follows from lemma 2.8. � Remark 4.1. Thanks to the previous proposition, we see that fc is a convex-concave solution of (Pβ; a, b, 1) for all c > 0. Proposition 4.2. There exists c∗≤−a (b− 1) such that C1 = [c∗, 0]. Int. J. Anal. Appl. 18 (3) (2020) 416 Proof. If b = 1 then C1 = {0}. If b > 1, as we already said in the previous section, this result is proven in [4] (see corollary 5.13 and lemma 5.16), let us recall briefly the main arguments which where used to get it. On the one hand, from lemma 2.8 with t0 = 0 (or lemma 5.16 of [4]), it follows that [−a (b− 1) , 0] ⊂ C1. On the other hand, lemma 5.12 of [4] implies that C2 is an interval of the type ]−∞,c∗[. This complete the proof since C1 = ]−∞, 0] �C2. � Remark 4.2. From the previous proposition, we have that 0 /∈ C2.2. Proposition 4.3. If c ∈ C2.1, then Tc = +∞ and f′c has a finite limit at infinity, equal either to −1 or to 1. Proof. Let c ∈ C2.1. 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 C2.1 we get ∃ tc ∈ [0, +∞[ such that f ′ c (tc) = 1 and by lemma 2.4, we finally get that λ = −1. Assume now that f ′′ c vanishes on ]0, +∞[, let t0 be the first point where f ′′ c vanishes. Thanks to lemma 2.2 item 3, we have 0 < f ′ c (t0) < 1, and the conclusion follows from lemma 2.7 (λ = 1). � Remark 4.3. If c ∈ C2.1 then either fc is a concave solution of (Pβ; a, b, −1) or fc is a concave-convex solution of (Pβ; a, b, 1). Proposition 4.4. Let c be a point of the boundary of C2.2. Then c ∈ C2.1 and f ′ c (t) −→−1 as t −→ +∞. Proof. See [3] in the case of the mixed convection equation. � Proposition 4.5. There exists at most one c such that f ′ c (t) −→−1 as t −→ +∞. Proof. From proposition 4.2 and 4.3 , we see that if c is such that f ′ c (t) −→ −1 as t −→ +∞, then c < 0 and f ′′ c < 0. By the change of variable, as done in [4], section 4, we can define a function v : ] 0, b2 ] → R such that ∀ t ≥ 0, v ( f′ 2 c (t) ) = fc (t) . (4.1) By setting y = f′ 2 c (t), we get fc (t) = v (y) , f ′ c (t) = √ y, f′′c (t) = 1 2v′ (y) and f′′′c (t) = − v′′ (y) √ y 2v′ 3 (y) , Int. J. Anal. Appl. 18 (3) (2020) 417 and using (1.1) we obtain ∀ y ∈ ] 0, b2 ] , v′′ (y) = v (y) v′ 2 (y) √ y + 2β (y − 1) √ y v′ 3 (y) 3 . (4.2) From (3.1), we deduce that v ( b2 ) = a and v ′ ( b2 ) = 1 2c . Moreover, since fc is bounded, it is so for v. Assume that there exists c1 > c2 such that f ′ c1 (t) → 0 and f ′ c2 (t) → 0 as t → +∞, and denote by v1 and v2 the functions associated to fc1 and fc2 by (4.1). If we set w = v1 −v2 then w ( b2 ) = 0 and w ′ ( b2 ) < 0. We claim that w ′ < 0 on ] 0, b2 ] . For contradiction, assume there exists x ∈ ] 0, b2 [ 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 w′′ (x) = w (x) √ x v′ 2 1 (x) , and a contradiction. Now, let us set Vi = 1 v′i for i = 1, 2 and W = V1 −V2. Then W ( b2 ) = 2 (c1 − c2) > 0 and W (y) → 0 as y → 0. In the other hand, thanks to (4.2); we have ∀ y ∈ ] 0, b2 ] , W ′ (y) = − w (y) √ y − 2β y − 1 √ y w′ (y) . Therefore, we have W ( b2 ) = ∫ b2 0 W ′ (y) dy = − ∫ b2 0 ( w (y) √ y + 2β y − 1 √ y w ′ (y) ) dy = −2 [ √ yw (y)] b2 0 + 2 ∫ b2 0 ( (1 −β) √ y + 2β √ y ) w ′ (y) dy = 2 ∫ b2 0 ( (1 −β) √ y + 2β √ y ) w ′ (y) dy, (4.3) the last equality following from the fact that w (y) tends to finite limit as y → 0. Since w ′ < 0, we finally obtain W ( b2 ) < 0 and a contradiction. � Remark 4.4. The change of variable (4.1) is particularly efficient to obtain some uniqueness results. In [4], it is used for the general equation (1.2) (cf. Section 4, lemma 5.4 and lemma 5.17). The case we examined in the previous proposition is part of lemma 5.17 of [4] with λ = −1. Corollary 4.1. One has C2.2 = ]−∞, c∗[ and C2.1 = [c∗, c∗[. Proof. From remark 4.2, proposition 4.4 and 4.5, we see that C2.2 is open, contains ]−∞, c∗[ . Therefore, since c∗ = inf (C1 ∪C2,1), we necessarily have C2.2 = ]−∞, c∗[ and C2.1 = [c∗, c∗[. � Int. J. Anal. Appl. 18 (3) (2020) 418 To finish this section, let us express the results of proposition 4.1, proposition 4.2 and corollary 4.1 in terms of the boundary problems (Pβ; a, b, −1) and (Pβ; a, b, 1). Theorem 4.1. Let β ∈ ] 0, 1 2 ] , a ≥ 0 and b ≥ 1. There exists c∗ < 0 such that: (1) fc is not defined on the whole interval [0, + ∞[ if c < c∗; (2) fc∗ is a concave solution of (Pβ; a, b, −1); (3) fc is a solution of (Pβ; a, b, 1) for all c ∈ ]c∗, + ∞[ Moreover, there exists c∗ ∈ ]c∗, −a (b− 1)] ; (1) fc is a convex-concave solution of (Pβ; a, b, 1) for all c ∈ ]0, + ∞[ ; (2) fc is a concave solution of (Pβ; a, b, 1) for all c ∈ [c∗, 0] ; (3) fc is a concave-convex solution of (Pβ; 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 C ′ 0,1 = { c < 0 : f ′ c > −1 on [0, Tc[ } , C ′ 0,2 = { c < 0 : ∃ sc ∈ ]0, Tc[ such that f ′ c > −1 on [0, sc[ and f ′ c (sc) = −1 } , C ′ 1 = { c ≥ 0; b ≤ f ′ c ≤ 1 and f ′′ c ≥ 0 on [0, Tc[ } , C ′ 2 =   c ≥ 0; ∃ tc ∈ [0, Tc[ , ∃ �c > 0 such that f ′ c < 1 on ]0, tc[ , f ′ c > 1 on ]tc, tc + �c[ and f ′′ c > 0 on ]0, tc + �c[   . The arguments used in the previous section, can be applied here. First, since g (x) = β ( x2 − 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 fc∗ is a concave solution of (Pβ; a, b, −1). Moreover, we have c∗ < 0. As in the previous section, this implies that C ′ 0,2 = ]−∞, c∗[. Hence C ′ 0,1 = [c∗, 0[, and if c ∈ ]c∗, 0[, then f ′′ c vanishes at a first point where f ′ c < 1. 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. Int. J. Anal. Appl. 18 (3) (2020) 419 Theorem 5.1. Let β ∈ ] 0, 1 2 ] and −1 < b < 1. There exist c∗ < 0 and c∗ ≥ a (1 − b) such that: (1) fc is not defined on the whole interval [0, + ∞[ if c < c∗; (2) fc∗ is a concave solution of (Pβ; a, b, −1) if b ∈ ]−1, 0[ ; (3) fc is a concave-convex solution of (Pβ; a, b, 1) for all c ∈ ]c∗, 0[ ; (4) fc is a convex solution of (Pβ; a, b, 1) for all c ∈ [0, c∗] ; (5) fc is a convex-concave solution of (Pβ; a, b, 1) for all c ∈ ]c∗, + ∞[ . 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The case ] 0,12] and b1 5. The case ] 0, 12] and -1