International Journal of Analysis and Applications Volume 18, Number 3 (2020), 356-365 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-18-2020-356 SOME RESULTS ABOUT A BOUNDARY VALUE PROBLEM ON MIXED CONVECTION M. BOULEKBACHE1,∗, M. AÏBOUDI1 AND K. BOUDJEMA DJEFFAL2 1Département de Mathématiques, Faculté des Sciences Exactes et Appliquées. Université Oran 1 Ahmed Ben Bella, Laboratoire de recherche d’Analyse Mathématique et Applications"L.A.M.A", Oran, Algérie 2Département de Mathématiques, Faculté des Sciences Exactes et Informatique. Université de Hassiba Benbouali, Chlef, Algérie ∗Corresponding author: medbouke@gmail.com Abstract. 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. 1. 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 Received November 3rd, 2019; accepted November 27th, 2019; published May 1st, 2020. 2010 Mathematics Subject Classification. 34B15, 34C11, 76D10. Key words and phrases. third order non linear differential equation; boundary value problem; mixed convection; concave solution; convex 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. 356 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-18-2020-356 Int. J. Anal. Appl. 18 (3) (2020) 357 equations reduces to a systems involving a third order differential equation of the form f′′′ + ff′′ + 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(x2 − 1) was first given by Falkner and Skan (1931) see [13]. The case g(x) = βx2, 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 f′′′ + ff′′ + βf′(f′ −1) = 0 (1.2) And we associate to equation (1.2) the boundary value problem:  f′′′ + ff′′ + βf′(f′ −1) = 0 f(0) = a, a ∈ R f′(0) = b < 0 f′(t) −→ λ as t −→ +∞ (Pβ;a,b,λ) 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 Pe − 1, with Ra the Rayleigh number and Pe 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 (Pβ;a,b,1) where −2 < β < 0 and b > 0. These results can be recovered from [10], where the general equation f′′′ + ff′′ + g(f′) = 0 is studied. In [16], some theoretical results can be found about the problem (Pβ;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 6= 0. The problem (Pβ;a,b,λ) with β = 0 is the well known Blasius problem. In the following, we note by fc a solution of the problem to the initial values below and by [0,Tc) the right maximal interval of its existence:  f′′′ + ff′′ + βf′(f′ −1) = 0 f(0) = a f′(0) = b f′′(0) = c (Pβ;a,b,c) Int. J. Anal. Appl. 18 (3) (2020) 358 To solve the boundary value problem (Pβ;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. 2. 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′′′ + ff′′ = 0 i.e Eq.(1.1) with g = 0. Let us notice that, for any τ ∈ R, the function hτ : t 7→ 3t−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. Definition 2.1. We say that f is a subsolution (resp. a supersolution) of the Blasius equation f′′′+ff′′ = 0 if f is of class C3 and if f′′′ + ff′′ ≤ 0 on I (resp.f′′′ + ff′′ ≥ 0 on I). Proposition 2.1. Let t0 ∈ R. There does not exist no positive concave subsolution of the Blasius equation on the interval [t0,+∞). Proof. See [10], Proposition 2.11. � Proposition 2.2. Let t0 ∈ R. There does not exist no positive convex supersolution of the Blasius equation on the interval [t0,+∞). Proof. See [10], Proposition 2.5. � 3. Preliminary Results Proposition 3.1. Let f be a solution of the equation (1.2) on some maximal interval I = (T−,T+) and β > 1. 1. If F is any anti-derivative of f on I, then (f′′eF )′ = −βf′(f′ −1)eF . 2. Assume that T+ = +∞ and that f′(t) −→ λ ∈ R as t → +∞. If moreover f is of constant sign at infinity, then f′′(t) −→ 0 as t → +∞. 3. If T+ = +∞ and if f′(t) −→ λ ∈ R as t → +∞, then λ = 0 or λ = 1. 4. If T+ < +∞, then f′′ and f′ are unbounded near T+. 5. If there exists a point t0 ∈ I satisfying f′′(t0) = 0 and f′(t0) = µ, where µ = 0 or 1 then for all t ∈ I, we have f(t) = µ(t− t0) + f(t0). 6. If f′(t) → 0 as t → +∞, then f(t) does not tend to −∞ or +∞ as t → +∞. Proof. The first item follows immediately from equation (1.2). For the proof of items 2-5, see [3], and item 6 see [1]. � Int. J. Anal. Appl. 18 (3) (2020) 359 Proposition 3.2. Let us suppose that f be a solution of equation (1.2) on the maximal interval I = (T−,T+) (1) Let H1 = f′′ + f(f′ −1) then H′1 = (1−β)f′(f′ −1), for all t ∈ I; (2) Let H2 = 3f′′2 + βf′2(2f′ −3) then H′2 = −6ff′′2, for all t ∈ I ; (3) Let H3 = 2ff′′ −f′2 + (2f′ −β)f2 then H′3 = 2(2−β)ff′2, for all t ∈ I; (4) Let H4 = f′′ + ff′ then H′4 = (1−β)f′2 + βf′, for all t ∈ I; (5) Let H5 = f′ + 12f 2 then H′5 = H4 = f ′′ + ff′, for all t ∈ I. Proof. This statements follows immediately from equation (1.2). � 4. The boundary value problem (Pβ;a,b,λ) Let the boundary value problem (Pβ;a,b,λ), we are interested here in a concave, convex and convex-concave solutions of a problem (Pβ;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 (Pβ;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: C0 = {c ≤ 0 : f′′c ≤ 0 on [0,Tc )}, C1 = {c > 0 : f′c ≤ 0 and f ′′ c ≥ 0 on [0,Tc)}, C2 = {c > 0 : ∃tc ∈ [0,Tc),∃εc > 0 s.t f′c < 0 on (0, tc), f′c > 0 on (tc, tc + εc) and f ′′ c > 0 on (0, tc + εc)}, C3 = {c > 0 : ∃sc ∈ [0,Tc),∃εc > 0 s.t f′′c > 0 on (0,sc), f′′c < 0 on (sc,sc + εc) and f ′ c < 0 on (0,sc + εc)}. Int. J. Anal. Appl. 18 (3) (2020) 360 Remark 4.1. It is easy to prove that C0, C1, C2 and C3 are disjoint nonempty open subsets of R, C0 = ]−∞,0] and C1 ∪C2 ∪C3 =]0,+∞[ (see Appendix A of [10] with g(x) = βx(x−1) and β > 0). Lemma 4.1. Let β > 0. If c ∈ C0, then Tc < +∞. Moreover, fc is concave solution, decreasing and f′c(t)→−∞ as t→Tc. Proof. If c ∈ C0, the result follows from proposition 3.1 item 1, we have f′′c (t) < 0 and f′′c (t) < 0 for all t ∈ [0,+∞), then fc is a no positive concave subsolution of the Blasius equation on [0,+∞) if a < 0, and on [t0,+∞) such that fc(t0) = 0 if a > 0, with f′c(t)→ −∞ as t→Tc. If we assume that Tc = +∞, This leads to a contradiction with proposition 3.1, then Tc < +∞. � Lemma 4.2. Let β > 0. Then fc is a convex solution of the boundary value problem (Pβ,a,b,0) if and only if c ∈ C1. Proof. See Appendix A of [10] with g(x) = βx(x−1) and β > 0. � Lemma 4.3. Let β > 0. If c ∈ C3, then Tc < +∞. Moreover, fc is convex-concave, decreasing and f′c(t) →−∞ as t → Tc. Proof. See [2], lemma 5.3. � Remark 4.2. From proposition 3.1 items 1,3 and 5, if c ∈ C2, then there are only three possibilities for the solution of the initial value problem (Pβ;a,b,c): (1) fc is convex and f′c(t) → +∞ as t → Tc (with Tc ≤ +∞); (2) there exists a point t0 ∈ [0,Tc) such that f′′c (t0) = 0 and f′c(t0) > 1; (3) fc is a convex solution of (Pβ;a,b,1). The next proposition shows that the case (1) cannot hold. Proposition 4.1. Let β > 0. There does not exit c ≥ 0, such that fc is convex and f′c(t) → +∞ as t → Tc on its right maximal interval of existence [0,Tc). Proof. Assume that fc is convex on its right maximal interval of existence [0,Tc) and f′c(t) → +∞ as t → Tc. There exist t0 ∈ [0,T0), which the function H2 is decreasing for t > t0, this is a contradiction as t →Tc. � Proposition 4.2. Let β > 1. If there exist t0 ∈ [0,Tc) such that f′c(t0) = 0 and f′′c (t0) < 0, then for all t > t0, f′′c (t) < 0 and f ′ c(t) 6= 0. Proof. Let fc is convex on its right maximal interval of existence [0,Tc), suppose there exist t1 > t0 such that f′c(t1) = 0, hence the function H4 is decreasing on [t0, t1], therefore H4(t0) > H4(t1), we have f ′′ c (t0) > f ′′ c (t1), which yields a contradiction. � Int. J. Anal. Appl. 18 (3) (2020) 361 5. The a < 0 case Proposition 5.1. Let β > 1, the boundary value problem (Pβ;a,b,1) has no convex solution. Proof. Let fc is convex on maximal interval of existence [0,Tc), such that f′c(t) → 1 as t → Tc, then there exist t0 ∈ [0,Tc), such that f′c(t0) = 0, the function H1 is creasing for all t > t0, therefore H1(t) > H1(t0) for t > t0, we have f′′c (t) −f′′c (t0)>−fc(t) (f′c(t)−1) > 0, we obtain a contradiction for t large enough because f′′c (t) −→ 0 and fc(t) > 0. � Proposition 5.2. The boundary value problem (Pβ;a,b,0) has no negative convex-concave solution. Proof. Let fc is convex-concave on maximal interval of existence [0,Tc), such that f′c(t) → 0 as t → Tc, then there exist t0 ∈ [0,Tc) such that f′c(t0) = 0, the function H2 is creasing for all t > t0, we have 3f′′2c (t0) < H2(t) for all t > t0, H2(t) → 0 as t → +∞, a contradiction. � Remark 5.1. If the boundary value problem (Pβ;a,b,0) has a convex-concave solution, then this solution changes the sign. Lemma 5.1. If c ∈ C1, then there exist c∗ such that 0 < c < c∗, Tc = +∞, and the solution fc is negative on [0,+∞). Proof. Let fc is solution on maximal interval of existence [0,Tc), if c ∈ C1, then Tc = +∞, the function H2 is creasing on [0,Tc), it follows that 3c2 + βb2(2b− 3) < 0, we obtain c < −b √ β(3−2b) 3 , and the solution fc is negative because a < 0 and f′c < 0. � Lemma 5.2. If c ∈ C3, then there exist c∗ such that 0 < c < c∗, Tc < +∞ and the solution fc is negative on [0,Tc). Proof. If c ∈ C3, then f′c →−∞, and Tc < +∞, other results same proof that lemma 5.1. � Remark 5.2. It follows from lemma 5.1 and lemma 5.2, there exist c∗ > 0 such that c > c∗, C2 6= ∅ and here the solution fc is convex-concave. Lemma 5.3. Let 1 < β < 2, if c ∈ C2 and fc is a no positive solution on maximal interval of existence [0,Tc), then for all t ∈ [0,Tc) we have fc(t) ≤ max { a, b√ β } , Tc < +∞ and f′c(t) →−∞ as t → Tc. Proof. Let c ∈ C2 and fc is a no positive solution on maximal interval of existence [0,Tc). From the proposition 3.1, 4.2 and 5.2, there exist t0 ∈ [0,Tc) such that f′c(t0) = 0. Moreover, the function H3 is decreasing on [0,Tc), we have H3(0) > H3(t0), it follows that, −b2 > 2ac − b2 + (2b − β)a2 > 2fc(t0)f′′c (t0) − βf2c (t0) > −βf2c (t0), we get fc(t0) < b√ β for all t ∈ [0,Tc), the conclusion follows from that, for all t ∈ [0,Tc ) , if a < b√β we have fc(t) ≤ fc(t0) and, if a > b√ β we have fc(t) ≤ a with Tc < +∞, and f′c →−∞ as t → Tc. � Int. J. Anal. Appl. 18 (3) (2020) 362 Lemma 5.4. If c ∈ C2, and if there exist t1 ∈[0,Tc) such that f′′c (t1) = 0 and fc(t1) < 0, then f′c(t1) > 3 2 . Proof. If c ∈ C2, there exist t0 ∈ [0,Tc) such that f′c(t0) = 0, fc(t0) < 0, and there exist t1 > t0 such that f′′c (t1) = 0, we suppose fc(t1) < 0 and f ′ c(t1) < 3 2 , the function H2 is creasing on [0, t1), we have 3f′′c (t0) < βf ′ c 2(t1)(2f ′ c(t1)−3), we obtain a contradiction. � Remark 5.3. Thanks to the previous lemma, if we have f′c(t1) < 3 2 and fc is convex-concave solution on maximal interval of existence [0,Tc), then fc changes the sign. Lemma 5.5. For 1 < β < 2 and b < −1, if c ∈ C2 and if there exist t0 ∈ [0,Tc) such that fc(t0) = 0, then f′c(t0) > 1. Proof. Let fc is convex-concave solution on maximal interval of existence [0,Tc), 1 < β < 2 and b < −1, if there exist t0 ∈ [0,Tc) such that fc(t0) = 0, the function H3 is decreasing on [0, t0), we have H3(0) > H3(t0), therefore −b2 > −f′2c (t0), and we obtain f′c(t0) > −b > 1. � Lemma 5.6. If c ∈ C2 and there exist t1 ∈ [0,Tc) such that f′′c (t1) = 0 and if fc(t1) < 0, then f′c(t1) > −β 1−β . Proof. If c ∈ C2, there exist t0 ∈ [0,Tc) such that f′c(t0) = 0 and f′′c (t0) > 0, there exist t1 > t0 such that f′′c (t0) = 0, we suppose f ′ c(t1) < −β 1−β , the function H4 is creasing on [t0, t1], we have f ′′ c (t0) < fc(t1)f ′ c(t1), this is a contradiction. � Theorem 5.1. Let β > 1, a < 0 and b < 0. (1) The boundary value problem (Pβ;a,b,0) has as least one negative convex solution on [0,+∞). (2) The boundary value problem (Pβ;a,b,1) has no convex solution on [0,+∞). (3) The boundary value problem (Pβ;a,b,+∞) has no convex solution on [0,Tc). 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. � 6. The a > 0 case Let a,b ∈ R with b < 0 and a > 0. We assume β > 1, and fc be a solution of the initial value problem (Pβ;a,b,c) on the right maximal interval of existence [0,Tc), c > 0. Before that, and in order to complete the study, let us divide the sets C2 and C3 into the following two Int. J. Anal. Appl. 18 (3) (2020) 363 subsets: C2.1 = {c ∈ C2; f′c > 0 on [tc,Tc )}, C2.2 = {c ∈ C2; ∃sc > tc s.t f′c > 0 on [tc,sc) and f′c(sc) = 0}, C3.1 = {c ∈ C3; fc(sc) < 0}, C3.2 = {c ∈ C3; fc(sc) > 0}. Proposition 6.1. If c ∈ C1 ∪C2 ∪C3.1, then c > −ab Proof. If c ∈ C1, Tc = +∞, f′c(t) → 0 as t → +∞, the function H4 is decreasing on [0,+∞), we have c + ab > 0, if c ∈ C2 ∪ C3.1, there exist t0 ∈ [0,Tc) such that f′c(t0) = 0 or fc(t0) = 0, we have c + ab ≥ f′′c (t0) > 0. � Remark 6.1. If c ≤−ab then c ∈ C3.2 and Tc < +∞. Thus C3.2 6= ∅ and the convex part of the solution fc is positive. Proposition 6.2. If c ∈ C1 ∪C2.1 and b > −12a 2, then Tc = +∞ and the solution fc is positive. Proof. Let fc solution of the initial value problem (Pβ;a,b,c) on the right maximal interval of existence [0,Tc), c > 0, if c ∈ C1 ∪ C2.1, thanks to propositions 3.1 and 4.1 it follows that Tc = +∞, no we suppose there exist t0 ∈ [0,Tc) such that fc(t0) = 0, the function H4 is decreasing for all t > 0, we have H4(t0) = f′′c (t0), therefore H5 is creasing on [0, t0), we obtain b + 12a 2 < f′c(t0) < 0, this is a contradiction. � Remark 6.2. If c ∈ C2.2 and b > −12a 2, the solution fc is positive on [0, t0), t0 is the point such that t0 > sc with fc(t0) = 0 and sc be as in definition of C2.2. Lemma 6.1. Let β > 1 and −1 2 a2 < b < 0. If fc be solution of the initial value problem (Pβ;a,b,c), on the right maximal interval of existence [0,Tc) and if there exist t0 ∈ [0,Tc) such that fc(t0) = 0 and f′c(t0) < 0 then f′′c (t0) < 0. Proof. For contradiction, let us that t0 ∈ [0,Tc) with fc(t0) = 0 and f′c(t0) < 0, the function H4 is decreasing on [0, t0) and H4(t0)=f′′c (t0) > 0 then for all t∈ [0, t0), H4 > 0, and H5 is creasing on [0, t0), we have b + 1 2 a2 < f′c(t0) < 0, this is a contradiction. � Proposition 6.3. Let 1 < β < 2, b > −1 2 a2 and c ∈ C2.2. For all t ∈ [0,Tc ) , one has fc(t) < √ b2+(β−2b)a2 β . Proof. Let c ∈ C2.2 and sc be as in the definition of C2.2, the function H3 is creasing on [0,sc), we have: −b2 + (2b−β)a2 < 2ac− b2 + (2b−β)a2 < 2fc(sc)f′′c (sc)−βf2c (sc) < −βf2c (sc), Int. J. Anal. Appl. 18 (3) (2020) 364 which implies that fc(sc) < √ b2+(β−2b)a2 β . From the proposition 4.2, the conclusion follows from that , for all t ∈ [0,Tc ) , we have fc(t) ≤ fc(sc). � Lemma 6.2. If c ∈ C1 ∪C2.1 and b > −12a 2. Then Tc = +∞ and there exist c∗ > 0 such that c > c∗. Proof. Let c ∈ C1 ∪ C2.1, and b > −12a 2. By the definition of C1 and C2.1, thanks to proposition 6.2, we have Tc = +∞ and f′c is bounded. Otherwise the function H2 is decreasing for t > 0, we obtain 3c2 + βb2(2b−3) > 0, which implies that c > −b √ β(3−2b) 3 . � Remark 6.3. There exist c∗ > 0, if c < c∗, then there exist t0 ∈ [0,Tc) such that fc(t0) = 0, f′c(t0) < 0 and f′′c (t0) < 0 say that c ∈ C2.2 ∪C3.2, since if c ∈ C2.1 then Tc = +∞. Let us divide the set C2.1 into the following two subsets: C2.1.1 = {c ∈ C2.1; f′c(t) → 0 as t → +∞}, C2.1.2 = {c ∈ C2.1; f′c(t) → 1 as t → +∞}. Proposition 6.4. Let 1 < β < 2, if c ∈ C1 ∪C3 ∪C2.2 ∪C2.1.1. Then there exist c∗ > 0 such that c < c∗. Proof. Let fc solution of the initial value problem (Pβ;a,b,c) on the right maximal interval of existence [0,Tc), either there exist t0 ∈ [0,Tc) such that fc(t0) = 0 or f′c(t0) = 0 if Tc < +∞, and if Tc = +∞, we have f′c(t) → 0 as t → +∞, from proposition 3.1 item 6, it follows that the function H3 is creasing on [0, t0) or [0,+∞), we get 2ac− b2 + (2b−β)a2 < 0, which implies that c < b 2+(β−2b)a2 2a . � Remark 6.4. From proposition 6.4 there exist c∗ > 0, such that for c ≥ c∗, then c ∈ C2.1.2. Thus C2.1.2 6= ∅. Corollary 6.1. If 1 < β < 2, a > 0, b < 0 and b > −1 2 a2, then the problem (Pβ;a,b,1) has as least one positive convex or positive convex-concave solution on [0,+∞). Proof. This follows immediately from remark 6.4, lemma 6.2 and proposition 6.3. � Theorem 6.1. Let β > 1, a > 0 and b < 0. (1) The boundary value problem (Pβ;a,b,0) has as least one convex solution on [0,+∞) if in addition we have b > −1 2 a2 it will be no negative convex solution. (2) The boundary value problem (Pβ;a,b,−∞) has infinity convex-concave solutions on the maximal interval of existence [0,Tc) with Tc < +∞, if in addition we have b > −12a 2 the convex part of these solutions will be no negative. (3) The boundary value problem (Pβ;a,b,+∞) has no convex solution on [0,Tc). Int. J. Anal. Appl. 18 (3) (2020) 365 Proof. 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Eur. J. Mech. B, Fluids, 43 (2014), 148-153. 1. Introduction 2. On Blasius Equation 3. Preliminary Results 4. The boundary value problem (P;a,b,) 5. The a<0 case 6. The a>0 case References