Existence and nonexistence of solutions on opposing mixed convection problems in boundary layer theory

European Journal of Mechanics B/Fluids 43 (2014) 148–153

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Existence and nonexistence of solutions on opposing mixed convection problems in boundary layer theory✩ G.C. Yang ∗ , L. Zhang, L.F. Dang College of Applied Mathematics, Chengdu University of Information Technology, Chengdu, Sichuan 610225, PR China

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abstract

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Article history: Received 8 September 2012 Received in revised form 28 June 2013 Accepted 6 August 2013 Available online 14 August 2013

We introduce an integral equation to study the opposing mixed convection problems in boundary layer theory. This equation is of singularities and two integrands take negative values. By means of some special analytical techniques, we prove the existence and the nonexistence of positive solutions of this equation and utilize it to treat analytically the mixed convection parameter ε < −1 and the temperature parameter λ > 0 involved in the problems mentioned above. Previous results only treated the case λ = 0 or ε ≥ −1. © 2013 Elsevier Masson SAS. All rights reserved.

Keywords: Opposing mixed convection Similarity solutions Existence and nonexistence Improper integrals

and −1 < ε <

1. Introduction In this research, we are concerned with the existence and the nonexistence of analytical solutions of opposing mixed convection for the following third-order nonlinear differential equation f ′′′ (η) + (1 + λ)f (η)f ′′ (η) + 2λ(1 − f ′ (η))f ′ (η) = 0 on [0, ∞)

(1.1)

subject to the boundary conditions f (0) = 0,

f ′ (0) = 1 + ε

and

f ′ (∞) = 1,

(1.2)

which has been used to describe the plane mixed convection boundary layer flow near a semi-infinite vertical plate embedded in a saturated porous media, with a prescribed power law of the distance from the leading edge for the temperature, where λ is the parameter of the temperature profile power-law and ε is the mixed convection parameter, namely, ε = RPa with Ra the Rayleigh nume

ber and Pe the P e´ clet number. For ε = 0, (1.1)–(1.2) corresponds to the forced convection, for ε > 0, it corresponds to aiding the mixed convection and for ε < 0, it corresponds to the opposing mixed convection. For more details on the physical derivation and the numerical treatments of (1.1)–(1.2), see [1,2]. Regarding the study of (1.1)–(1.2), Guedda [3] studied the existence of infinitely many solutions of (1.1)–(1.2) for −1 < λ < 0

✩ Project supported by the National Natural Science Foundation of China (Grant

No. 11171046). ∗ Corresponding author. Tel.: +86 2885966353. E-mail addresses: [email protected], [email protected] (G.C. Yang), [email protected] (L. Zhang), [email protected] (L.F. Dang). 0997-7546/$ – see front matter © 2013 Elsevier Masson SAS. All rights reserved. http://dx.doi.org/10.1016/j.euromechflu.2013.08.005

1 2

and the nonexistence of nonnegative solutions

for λ ≤ −1 and ε ≥ 12 ; Brighi and Hoernel [4] proved the existence and the uniqueness of convex and concave solutions of (1.1)–(1.2) for λ > 0, −1 < ε < 0 and ε > 0. For λ = 0, it is well known that (1.1)–(1.2) is the Blasius equation, Hussaini and Lakin [5] showed that there exists εc < −1 such that (1.1)–(1.2) has a solution for . ε ≥ εc and no solution for ε < εc . Numerical result showed εc = −1.3545. For further results on the Blasius equation, one may refer to [6,7]. However, to our knowledge, for λ ̸= 0 and ε < −1, there exists little analytic study on the existence and the nonexistence of analytical solutions of (1.1)–(1.2). In this research, our interest is focused on this case. We shall establish the relation between (1.1)–(1.2) and an integral equation (see (2.3)) and prove results on the existence and the nonexistence of positive solutions of this equation, which derive our desired conclusions as follows. Theorem 1.1. There exist ε∗ ∈ (−1.807, −1.806) and ε ∗ ∈ (−1.193, −1.192) such that (i) (1.1)–(1.2) has no convex solutions for any λ > 0 and each

ε ≤ ε∗ . (ii) (1.1)–(1.2) has a convex solution for each λ > 0 and each ε ∈ [ε ∗ , −1).

It is well known that analytical and numerical studies of similarity solutions are both of considerable practical importance in many fields and can provide an important standard of comparison without introducing the complication of non-similar solutions, much attention is always focused on them. One may refer to, for example, the review and extension of similarity solutions [8,9] and some

G.C. Yang et al. / European Journal of Mechanics B/Fluids 43 (2014) 148–153

recent studies such as boundary layer flows [10–13], magnetohydrodynamic (MHD) [12,14,15], etc. and the references therein.

149

t ∈ [0, 1] and noticing that 1





µ

s

β z (µ)

dµds

2. Relation between (1.1)–(1.2) and an integral equation

t

Let f (η) ∈ C (R ), f (η) is said to be a convex solution if f (η) satisfies (1.1)–(1.2) and f ′′ (η) > 0 for η ∈ R+ , where R+ = [0, ∞). Throughout this paper, we assume λ > 0 and ε < −1. For the sake of convenience, let

   1  1 µ µ ds dµ + ds dµ = z (µ) µ z (µ) t t β  1 s(1 − s) = (1 − t )Bz (t ) + ds, z (s) t

+

3

3λ Notation. β = 1 + ε, ϕ(λ) = 11++λ . We change (1.1) and (1.2) into (2.1) and (2.2), which can be found in the literature (see, e.g., Remark 1.2 in [10]).

 t 

1

we see that z (t ) satisfies (2.3).

(2.6)



Since (2.3) contains the improper integrals Az (t ) and Bz (t ), we first investigate some properties of solutions of (2.3). Let z ∈ Q be a solution of (2.3); then

Proposition 2.1. The problem (1.1)–(1.2) is equivalent to the following boundary value problem

z (1) = 0

F ′′′ + FF ′′ + (ϕ(λ) − 1)(1 − F ′ )F ′ = 0 for η ∈ R+

In fact, if z (1) > 0, then z (t ) > 0 for t ∈ [β, 1]. This implies that two integrands in (2.3) are continuous and then z (1) = 0, a contradiction. Since Az (t ) and (1 − t )Bz (t ) are well-defined on [β, 1] and Az (1) = 0, we immediately get that limt →1− (1 − t )Bz (t ) = 0. Proposition 2.3(ii) will show that limt →1− (1 − t )Bz (t ) is an indeterminate form of type 0 × ∞. The following proposition shows the equivalence between (2.3) and a first-order differential equation with suitable boundary condition.

(2.1)

subject to the boundary conditions F (0) = 0,

F ′ (0) = β

where F (η) =

√

and

F ′ (∞) = 1,

(2.2)

λf ( √1η+λ ).

1+

Let

Γ = {F ∈ C 3 (R+ ) : F ′′ (η) > 0 on R+ } Q = {z ∈ C [β, 1] : z (t ) > 0 for t ∈ [β, 1)}, where C [β, 1] denotes the continuous functions space with the maximum norm ∥z ∥ = max{|z (t )| : t ∈ [β, 1]}. Now, we establish the relation between (2.1)–(2.2) and an integral equation. Lemma 2.1. If F ∈ Γ satisfies (2.1)–(2.2), then z (t ) = ϕ(λ)Az (t ) + (1 − t )Bz (t ) for t ∈ [β, 1]

(2.3)

has a solution z ∈ Q , where Az (t ) =

1

s(1 − s) z (s)

t

ds

and

Bz (t ) =

t

 β

s z (s)

dη dt

=

1 F ′′ (η)

=

1 z (t )

,

F (η) =

η

ds.

F ′ (σ )dσ =

t



0

s

β z (s)

ds

by setting s = F ′ (σ ) and F ′′′ (η) = z ′ (t ) ddtη = z (t )z ′ (t ). Substituting F (η), F ′ (η), F ′′ (η) and F ′′′ (η) into (2.1) implies z ′ (t ) =

(1 − ϕ(λ))t (1 − t ) − Bz (t ) for t ∈ [β, 1) z (t )

(2.4)

(1 − ϕ(λ))t (1 − t ) − Bz (t ) for β ≤ t < 1. z (t )

(2.7)

Proof. Assume that z ∈ Q is a solution of (2.3). Differentiating (2.3) with t and combining z (1) = 0, we know that (2.7) holds. Conversely, integrating (2.7) from t to 1 and using the same arguments used in the proof of Lemma 2.1 and (2.6), we know that z is a solution of (2.3). 

Proof. Assume that z ∈ Q is a solution of (2.3). (i) If there exists t ∈ [0, 1] such that z (t ) < γ (t ), then t ∈ (0, 1) since z (0) = γ (0) and z (1) = 0 = γ (1). Let ψ(t ) = z (t ) − γ (t ) and ξ ∈ (0, 1) satisfying

ψ(ξ ) = min{ψ(t ) : t ∈ [0, 1]} < 0. Then ψ ′ (ξ ) = 0, ψ ′′ (ξ ) ≥ 0 by the second derivative test and z ′ (ξ ) = γ ′ (ξ ) = −z (0) and then z ′′ (ξ ) = ψ ′′ (ξ ) ≥ 0. Differentiating (2.7) with t, we have

(1 − ϕ(λ))(1 − 2t ) − t (ϕ(λ) − 1)(1 − t )tz ′ (t ) + z (t ) z 2 (t ) for 0 ≤ t < 1.

z ′′ (t ) =

Since z (ξ ) < γ (ξ ) = z (0)(1 − ξ ), we see

(1 − ξ )ξ z ′ (ξ ) ξ (1 − ξ )z (0) ξ =− <− . z 2 (ξ ) z (ξ ) z (ξ ) z (ξ )

and z (1) = 0. Since

(1 − ϕ(λ))t (1 − t ) − Bz (t ) z (t )    t (1 − ϕ(λ))t (1 − t ) s = − ds − Bz (0), z (t ) 0 z (s)

z ′ (t ) =

Proposition 2.3. Let z ∈ Q be a solution of (2.3); then (i) z(t ) ≥ z (0)(1 − t ) := γ (t ) for t ∈ [0, 1]. 1 (ii) β z (1s) ds = ∞.

Proof. Assume that F ∈ Γ satisfies (2.1)–(2.2). As the same as the proof of Proposition 3.1 (4) in [10], we have F ′′ (∞) = 0. Let t = F ′ (η) be a dependent variable, z (t ) = F ′′ (η). Since F ′ is strictly increasing on [0, ∞) by F ′′ (η) > 0 on R+ , then β ≤ t < 1, dη z (t ) > 0 on [β, 1), z (1) = F ′′ (∞) = 0 and z ∈ Q . By 1 = F ′′ (η) dt , we see



lim (1 − t )Bz (t ) = 0.

t →1−

Proposition 2.2. Let z ∈ Q ; then z is a solution of (2.3) if and only if z (1) = 0 and

and



and

λ This, together with ϕ(λ) − 1 = 12+λ > 0, implies

(1 − ϕ(λ))(1 − 2ξ ) − ξ (ϕ(λ) − 1)ξ − z (ξ ) z (ξ ) (1 − ϕ(λ))(1 − ξ ) − ξ = < 0. z (ξ )

z ′′ (ξ ) < (2.5)

integrating (2.4) from t to 1, using the fact that both terms in the bracket of the right hand side in (2.5) have the same sign for

This is a contradiction. Hence, (i) holds.

150

G.C. Yang et al. / European Journal of Mechanics B/Fluids 43 (2014) 148–153

(ii) By (2.3) and (i), we see for t ∈ [0, 1) z (t ) ≤ ϕ(λ)

s(1 − s)

1



z ( s)

t

≤ ϕ(λ)



≤ ϕ(λ)



1 t

=

z (0)



ds + (1 − t )

3. Proof of Theorem 1.1 s

z ( s) t

In this section, we introduce two functions to determine ε∗ and

ds

s

γ (s) 0 γ (s)  t 1−s 1 ds + (1 − t ) ds γ (s) 0 γ (s)

t

1

ds + (1 − t )

0

s(1 − s)

1

t



ε ∗ in Theorem 1.1. Let

ds

τ (β) = 3 − 3β 2 + 2β 3 , and

σ (β) = 1 − 24β 2 + 16β 3 ,

(ϕ(λ) − ln(1 − t ))(1 − t ).

(ϕ(λ)− ln(1 − t )); then = and we obtain  1  1  ∞ 1 1 du ds ≥ ds = z (0) = ∞. z ( s ) u ( s )( 1 − s ) u 0 0 0 1 Hence β z (1s) ds = ∞, i.e., (ii) holds.  Let u(t ) =

du dt

1 z (0)

1 z (0)(1−t )

Lemma 2.2. If z ∈ Q is a solution of (2.3), then (2.1)–(2.2) has a solution F (η) ∈ Γ . Proof. Assume that z ∈ Q is a solution of (2.3). Let

η = g (t ) =

t



1 z (s)

β

ds for β ≤ t < 1.

(2.8)

Then g (t ) is strictly increasing on [β, 1) and g (β) = 0. Moreover, by Proposition 2.3(ii), we have g (1 − 0) =

1



z (s)

β

ds = +∞.

Let t = h(η) be the inverse function to η = g (t ), we define the function F (η) =

η



  807 2431443 < 0 and τ − =− 1000 5 × 108   403 242423 τ − = > 0, 500

F ′ (η)

1 z (s)

β

ds for 0 ≤ η < +∞.

(2.9)

807 then there exists a unique β∗ ∈ (− 1000 , − 403 ) = (−0.807, 500 −0.806) such that τ (β∗ ) = 0 and τ (β) ≤ 0 for β ≤ β∗ . By σ ′ (β) = −48β + 48β 2 > 0 for β ∈ (−∞, 0) and

  193 562557 σ − =− <0 1000 62500000   3941 24 = > 0, σ −

F ′′ (η) = z (F ′ (η)) = z (t )

for 0 ≤ η < +∞.

0



1



Noticing that Bz (t ) =

β

z (s)

η

ds =

s(1 − s) z (0)

β

z (s)

1

 ds ≤

s(1 − s)

γ (s)

0

z (β) = ϕ(λ)

= ϕ(λ)

F ′ (σ ) dσ = F (η)

1



F ′′′ (η) = −(ϕ(λ) − 1)(1 − F ′ (η))F ′ (η) − F ′′ (η)F (η) for 0 ≤ η < +∞. 

By Lemmas 2.1 and 2.2, we see that (2.1)–(2.2) has a solution F ∈ Γ if and only if (2.3) has a solution z ∈ Q . Hence, we obtain the following by Proposition 2.1. Theorem 2.1. For λ > 0, (1.1)–(1.2) has a convex solution f if and only if (2.3) has a solution z ∈ Q .

s(1 − s) z ( s)

β



0

by setting s = F ′ (σ ) and utilizing t = F ′ (η), (2.7), (2.10), (2.11), we have

This is the desired result.

0



ds <

1

ds =

6z (0)

ds =

2z (0)

(2β 3 − 3β 2 )

1

.

By (2.11)



s(1 − s)

0

F ′′′ (η) = z ′ (F ′ (η))F ′′ (η) = z ′ (F ′ (η))z (F ′ (η)) = z ′ (t )z (t )

s

1953125

and (2.10)

Then F (η) > 0 for 0 ≤ η < +∞. Differentiating (2.10) with respect to η, we have

F ′ (η)

s(1 − s) z (s)

′′

for 0 ≤ t < 1.

and

193 24 we know that there exists a unique β ∗ ∈ (− 1000 , − 125 ) = ∗ (−0.193, −0.192) such that σ (β ) = 0 and σ (β) ≥ σ (β ∗ ) = 0 for β ∗ ≤ β < 0. 

β

Differentiating (2.9) with respect to η, we have



62500000

Proof. Assume that (2.3) has a solution z ∈ Q for some λ > 0 and some β ≤ β∗ . By (2.7) and ϕ(λ) − 1 > 0, we see z ′ (t ) > 0 in [β, 0] and 0 < z (t ) < z (0) in [β, 0). Since s(1 − s) < 0 on [β, 0), s(1 − s) ≥ 0 on [0, 1] and Proposition 2.3(i), we see

Clearly, F ′ (η) = h(η), F (0) = 0, F ′ (0) = β and F ′ (∞) = 1. From (2.8), we have



Proof. Since τ ′ (β) = −6β + 6β 2 > 0 for β ∈ (−∞, 0) and

Lemma 3.1. Eq. (2.3) has no solution in Q for any λ > 0 and β ≤ β∗ .

h(s) ds for 0 ≤ η < +∞.

0

η = g (F ′ (η)) =

β ∈ (−∞, 0].

Proposition 3.1. There exist a unique β∗ ∈ (−0.807, −0.806) and a unique β ∗ ∈ (−0.193, −0.192) such that τ (β) ≤ 0 for β ≤ β∗ and σ (β) ≥ 0 for β ∗ ≤ β < 0.

125

1

β ∈ (−∞, 0]

0

s(1 − s) z (s)

β

we obtain z (β) < ction.  Notation. cβ =

ds 1

 ds +

√ √

4 3

z (s)

0

ϕ(λ)(3−3β 2 +2β 3 ) 6z (0)

1+ σ (β)

s(1 − s)

=

 ds ,

ϕ(λ)τ (β) 6z (0)

≤ 0, a contradi-

√ ϕ(λ).

To prove that (2.3) has a solution in Q for each λ > 0 and each β ∗ ≤ β < 0, we consider the following integral equation of the form z (t ) = ϕ(λ)A(Sz )(t ) + (1 − t )B(Sz )(t )

for β ≤ t ≤ 1,

(3.1)

where A, B defined in (2.3) and S : C [β, 1] → C [β, 1] defined by Sz (t ) = max{z (t ), c (t )}

G.C. Yang et al. / European Journal of Mechanics B/Fluids 43 (2014) 148–153

and Sz (t ) ≥ c (t ), the Lebesgue dominated convergence theorem with the dominated function F (s) = c1 for s ∈ [β, 1] implies that

and



c (t ) =

cβ (1 − t ) cβ

if t ∈ [0, 1], if t ∈ [β, 0).

β

∥Tzn − Tz ∥ → 0 and T is continuous. By

Let

d(Tz (t ))

K = {z ∈ C [β, 1] : ∥z ∥ ≤ R}, where R = (ϕ(λ) + 1)

3β 2 −2β 3 +3 6cβ

dt

.

Lemma 3.2. For λ > 0 and β ∗ ≤ β < 0, (2.3) has a solution in K . Proof. The proof is divided into two steps. Step 1. (3.1) has a solution in K . Let z ∈ C [β, 1] and Tz (t ) =



ϕ(λ)A(Sz )(t ) + (1 − t )B(Sz )(t ) 0

if t ∈ [β, 1), if t = 1.

We prove that T maps K into K . Noticing that Sz (t ) ≥ c (t ) for t ∈ [β, 1], we have A(Sz )(t ) =

1



s(1 − s) Sz (s)

t

1

 ds ≤ t

1

ds =

cβ

1−t cβ

t 0

t



s Sz (s)

1

ds ≤

cβ (1 − s)

0

ds = −

ln(1 − t ) cβ

|s|(1 − s) ds + Sz (s)



1

|s|(1 − s) ds + Sz (s)



1

|s|(1 − s) ds + c (s)



1

 t

≤ ϕ(λ)



≤ ϕ(λ)



t

for t ∈ (0, 1).

β

|s|(1 − t ) ds Sz (s)

t

|s|(1 − s) ds Sz (s)

1

|s|(1 − s) ds = R. c (s)

β

β

Hence T maps K into K . Next, we show that T is continuous and compact. Let zn ∈ K , z ∈ K such that limn→+∞ ∥zn − z ∥ = 0. Since 1 − t ≤ 1 − s for β ≤ s ≤ t ≤ 1, we have

|Tzn (t ) − Tz (t )| ≤ ϕ(λ) |A(Szn )(t ) − A(Sz )(t )| + (1 − t ) |B(Szn )(t ) − B(Sz )(t )|   1  s(1 − s) s(1 − s)    ≤ ϕ(λ)  Sz (s) − Sz (s)  ds n t   t  s(1 − t ) s(1 − t )   ds  + −  Sz (s) Sz (s)  n β   1  s(1 − s) s(1 − s)    ≤ ϕ(λ)  Sz (s) − Sz (s)  ds n β   t  s(1 − s) s(1 − s)    +  Sz (s) − Sz (s)  ds n β   1  s(1 − s) s(1 − s)    ≤ ϕ(λ)  Sz (s) − Sz (s)  ds n β   1  s(1 − s) s(1 − s)    ds + −  Sz (s) Sz (s)  n β   1  s(1 − s) s(1 − s)   ds.  = (ϕ(λ) + 1) −  Sz (s) Sz (s)  β

n

Since lim

n→+∞

s(1 − s) Szn (s)

=

(1 − s)s on [β, 1) Sz (s)

t

 β

s Sz (s)

   t  d(Tz (t ))   ≤ |1 − ϕ(λ)||t |(1 − t ) +   dt  Sz (t ) β  t |1 − ϕ(λ)||t |(1 − t ) ≤ + c (t ) β := F (t ) on [β, 1).

for t < 1,

ds

|s| ds Sz (s) |s| ds c (s)

Noticing that 1



t

 β

β

1



|s| dsdt = c (s)

1



β

s

1



t

β

(1 − ϕ(λ))t (1 − t ) − Sz (t )

|s| dtds = c (s)



1

β

(1 − s)|s| ds, c ( s)

we have

Since limt →1− (1 − t ) ln(1 − t ) = 0, we have limt →1− Tz (t ) = 0 and then T maps K into C [β, 1]. By

|Tz (t )| ≤ ϕ(λ)

=

we have

for t ∈ (0, 1)

and



151

β

F (t )dt ≤ |1 − ϕ(λ)|

= ϕ(λ)

1

 β

1

 β

(1 − s)|s| ds + c ( s)

1

 β

(1 − s)|s| ds c ( s)

(1 − s)|s| ϕ(λ) ds = R, c (s) ϕ(λ) + 1

i.e., F ∈ L[β, 1]. This, together with the absolutely continuity of the Lebesgue integral, implies that T (K ) = {Tz (t ) : z ∈ K } is equicontinuous and T (K ) is relatively compact. Hence T : K → K is compact. Since K is a bounded closed convex, it follows from the Schauder fixed point theorem [16] that there exists z ∈ K such that (3.1) holds. Step 2. The solution z of (3.1) obtained in Step 1, is a solution of (2.3) in Q . It is clear that we need only to prove Sz (t ) = z (t ), i.e., z (t ) ≥ c (t ) on [β, 1]. For this, we prove the following fact: (P ) Let ν ∈ [0, 1) satisfy z (ν) ≥ cβ ; then z (t ) ≥ c (t ) on [ν, 1]. In fact, if there exists t ∈ (ν, 1) such that z (t ) < c (t ), by z (ν) ≥ cβ ≥ c (ν) and z (1) = c (1), we may choose a, b ∈ (ν, 1) with a < b such that z (a) = c (a),

z (b) = c (b)

and z (t ) < c (t )

on (a, b).

Let g (t ) = z (t ) − c (t ) and ξ ∈ (a, b) such that g (ξ ) = min{g (t ) : t ∈ [a, b]}; then g ′′ (ξ ) ≥ 0 by the second derivative test and then z ′′ (ξ ) = g ′′ (ξ ) ≥ 0. On the other hand, by Sz (t ) = c (t ) on (a, b) and (3.1), we have z ′ (t ) =

(1 − ϕ(λ))t (1 − t ) − Sz (t )

t

 β

s Sz (s)

ds

for β < t < 1. (3.2)

From this, we have z ′ (t ) =

(1 − ϕ(λ))t cβ

t

 − β

s Sz (s)

ds

for a < t < b

and z ′′ (t ) =

1 − ϕ(λ) cβ

−

t cβ (1 − t )

<0

for a < t < b.

This shows z ′′ (ξ ) < 0, a contradiction. Let t˜ ∈ [β, 1] such that z (t˜) = max{z (t ) : t ∈ [β, 1]} := M. We prove

√ cβ ≤ M ≤

3ϕ(λ) 3

.

(3.3)

152

G.C. Yang et al. / European Journal of Mechanics B/Fluids 43 (2014) 148–153

In fact, if M < cβ , then Sz (t ) ≤ cβ on [0, 1] since cβ ≥ c (t ) on [β, 1]. By (3.1) and Sz (t ) ≥ c (t ) = cβ on [β, 0], we have cβ = c (0) ≥ z (0) = ϕ(λ)A(Sz )(0) + B(Sz )(0)

= ϕ(λ)

s(1 − s)

1



Sz (s)

0

≥ ϕ(λ)



≥ ϕ(λ)



Sz (s)

0

cβ

0

=

ϕ(λ) 6cβ

+

ds +

Sz (s) 0

 ds +

s(1 − s) cβ

β

 t s (1 − ϕ(λ))t (1 − t ) − ds z (t )  t z (s)  t (1 − ϕ(λ))t (1 − t ) 1 ≥ − s ds z (t ) z (t ) t

z ′ (t ) ≥

ds

s(1 − s)

β

s(1 − s)

1

0



s Sz (s)

β

s(1 − s)

1

0

 ds +

Hence, we have by (3.4) and the decrease in z

for  t ≤ t < 1 and

ds

z (t )z ′ (t ) ≥ (1 − ϕ(λ))t (1 − t ) − ds



z ( t) ≤ 2 2

6cβ √

2 3

σ (β) − 1 8

≥−

1

(ϕ(λ) − 1)t (1 − t )dt +

1 8

1

(ϕ(λ) − 1)t (1 − t )dt +

≤2 =



t



s ds dt  t 1





t



s ds dt 0

0

on [β ∗ , 0]

1

  t

 t



.

Since σ (β) ≥ 0 on [β ∗ , 0] and ϕ(λ) ≥ 1, then 2β 3 − 3β 2 =

for  t ≤ t < 1.

s ds

Integrating this inequality and using z (1) = 0, we obtain

2β 3 − 3β 2

2ϕ(λ)+2(2β 3 −3β 2 )

t

 t

√ and cβ ≥



0

ϕ(λ) 3 √

3ϕ(λ)

and M = z ( t) ≤ . Hence, the right hand side of (3.3) holds. 3 Finally, we prove z (t ) ≥ c (t ) on [β, 1]. By (3.1) and (3.3), we have

and 2ϕ(λ) + 2(2β 3 − 3β 2 ) > 2ϕ(λ) − we know cβ > we have cβ ≤

√

ϕ(λ) √ .

√2 3 ϕ(λ) √ , 2 3

1 4

> ϕ(λ),

However, by σ (β) ≤ σ (0) = 1 on [β , 0], ∗

a contradiction. Hence, the left hand side of

(3.3) holds. Since 1 − ϕ(λ) < 0, by (3.2), we have that z ′ (t ) > 0 on [β, 0] and z (t ) is increasing [β, 0]. Hence t˜ ∈ [0, 1] and  t < 1 since z (1) = 0. To prove the right hand side of (3.3), we show that z (t ) is decreasing on [ t , 1). In fact, if there exist t1 , t2 ∈ ( t , 1) with t1 < t2 such that z (t1 ) < z (t2 ). Since z (1) = 0, we may choose ξ ∈ ( t , t2 ) and τ ∈ (ξ , 1) such that z (ξ ) = min{z (t ) : t ∈ [ t , t2 ]},

z (τ ) = max{z (t ) : t ∈ [ξ , 1]}.

Then z ′ (ξ ) = 0, z ′′ (ξ ) ≥ 0, z ′ (τ ) = 0 and z ′′ (τ ) ≤ 0. By (3.2) and (P )(z ( t ) ≥ cβ , Sz (t ) = z (t ) for t ∈ [ t , 1]), we have z ′ (t ) =

(1 − ϕ(λ))t (1 − t ) − z (t )



t

s

β Sz (s)

ds

for  t ≤t <1

(3.4)

and

(1 − ϕ(λ))(1 − 2t ) − t (ϕ(λ) − 1)(1 − t )tz ′ (t ) + z (t ) z 2 (t ) for  t ≤ t < 1.

z ′′ (t ) =

ω(ξ )

Hence z ′′ (ξ ) = z (ξ ) and z ′′ (τ ) = ϕ(λ))(1 − 2t ) − t. Since

ω′ (t ) = 2ϕ(λ) − 3,

ω(τ ) , z (τ )

where ω(t ) = (1 −

ω(0) = 1 − ϕ(λ) < 0

and ω(ξ ) ≥ 0 by z (ξ ) ≥ 0, we know ω′ (t ) = 2ϕ(λ) − 3 > 0 on [0, 1] and ω(t ) is strictly increasing on [0, 1]. This shows ω(τ ) > ω(ξ ) ≥ 0, which contradicts ω(τ ) = z (τ )z ′′ (τ ) ≤ 0. Hence z (t ) is decreasing on [ t , 1]. By (3.4) and z ′ ( t ) = 0, we see ′′



 t

s

β Sz (s)

ds =

z (β) = ϕ(λ)

(1 − ϕ(λ))(1 −  t ) t ≤ 0. z ( t)

= ϕ(λ)

1



s(1 − s) Sz (s)

β



1

≥ ϕ(λ)

s(1 − s) Sz (s)

0



1

s(1 − s) √

0

ds

3ϕ(λ) 3 3

0

 ds +

Sz (s)

β 0

 ds + β

s(1 − s)

s(1 − s) cβ

 ds

 ds

√ ϕ(λ) 2β − 3β 2 = √ + ϕ(λ) = cβ 2 3

6cβ

and z (t ) ≥ c (t ) = cβ on [β, 0] since z (t ) is increasing on [β, 0] and z (t ) ≥ c (t ) on [0, 1] by (P ). This completes the proof.  Proof of Theorem 1.1. Let ε∗ = β∗ − 1 and ε ∗ = β ∗ − 1; then ε∗ ∈ (−1.807, −1.806) and ε ∗ ∈ (−1.193, −1.192). (i) If (1.1)–(1.2) has a convex solution f ∈ C 3 (R+ ) for some λ > 0 and some ε ≤ ε∗ , by Theorem 2.1, (2.3) has a solution z in Q for λ > 0 and β = 1 + ε ≤ 1 + ε∗ = β∗ , which contradicts Lemma 3.1. (ii) Let λ > 0 and ε ∈ [ε ∗ , −1); then 0 > β = 1 + ε ≥ 1 + ε ∗ ≥ β ∗ , by Lemma 3.2, (2.3) has a solution z in Q for λ > 0 and 0 > β ≥ β ∗ and, by Theorem 2.1, (1.1)–(1.2) has a convex solution f ∈ C 3 (R+ ). The proof is completed. Remark 3.1. Theorem 1.1 extends the existence of convex solution of (1.1)–(1.2) [4] to ε < −1. Also, to our knowledge, there exists little study on the nonexistence of the convex solution of (1.1)–(1.2) for λ ̸= 0 and ε < −1; hence the nonexistence result in Theorem 1.1 is new. Remark 3.2. In [7], we studied the forced convection (λ = 0) with ε < −1. But, the methods could not be used to this research since z therein is concave down on [0, 1]. At the same time, we encounter the difficulties that two integrands in (2.3) are of singularities and take negative values and the usual fixed point theorems on cone maps that map cone into cone (see, for example, [16]) are not used in the study of the existence of positive solutions of (2.3). Previous methods [17–19] are only used to treat boundary layer problems for the case of ε ≥ −1.

G.C. Yang et al. / European Journal of Mechanics B/Fluids 43 (2014) 148–153

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