# On the Convex and Convex-Concave Solutions of Opposing Mixed Convection Boundary Layer Flow in a Porous Medium

Authors: M. Aïboudi; K. Boudjema Djeffal; B. Brighi
Journal: Abstract and Applied Analysis
 (2018)

---

## Abstract

In this paper, we are concerned with the solution of the third-order nonlinear differential equationf
″
′
+
f
f
″
+
β
f
′
(
f
′
-
1
)
=
0, satisfying the boundary conditions f
(
0
)
=
a
∈
R, f
′
(
0
)
=
b
<
0, and f
′
(
t
)
→
λ, as t
→
+
∞, where λ
∈
{
0,1
} and 0
<
β
<
1
. The problem arises in the study of the opposing mixed convection approximation in a porous medium. We prove the existence, nonexistence, and the sign of convex and convex-concave solutions of the problem above according to the mixed convection parameter b
<
0 and the temperature parameter 0
<
β
<
1.


---

## Body

## 1. Introduction

Owing to their numerous applications in industrial manufacturing processes, the convection phenomena about heated or cooled surfaces embedded in fluid-saturated porous media have attracted considerable attention during the last few decades. In this paper, our interest focuses on the analysis of the boundary value problemsP
λ
a
,
b
P
λ
a
,
b
f
″
′
+
f
f
″
+
β
f
′
f
′
-
1
=
0
f
0
=
a
,
a
∈
R
f
′
0
=
b
<
0
f
′
t
→
λ
a
s
t
→
+
∞where λ
∈
{
0,1
}. This problem derives from 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
=
R
a
/
P
e
-
1, with R
a the Rayleigh number and P
e the Péclet number. The interested reader can consult references [1, 2] for more details on the physical derivation and the numerical treatments.Mathematical results about the problemP
λ
a
,
b with λ
=
1 can be found in [3–7]. The case where a
≥
0, b
≥
0, β
>
0 and λ
∈
{
0,1
} was treated by Aïboudi and al. in [3], and the results obtained generalize the ones of [6]. In [4], Brighi and Hoernel established some results about the existence and uniqueness of convex and concave solution of P
1
(
a
,
b
) where -
2
<
β
<
0 and b
>
0. These results can be recovered from [8], where the general equation f
″
′
+
f
f
″
+
g
(
f
′
)
=
0 is studied.In [5], some theoretical results can be found about the problem P
1
(
0
,
b
) with -
2
<
β
<
0, b
=
1
+
ε, and ε
<
-
1. In particular, the authors prove that there exist ε
∗
∈
(
-
1.807
,
-
1.806
) and ε
∗
∈
(
-
1.193
,
-
1.192
), such that(i)
P
1
(
0
,
b
) has no convex solution for any β
<
0 and each ε
≤
ε
∗.
(ii)
P
1
(
0
,
b
) has a convex solution for each β
<
0 and each ε
∈
[
ε
∗
,
-
1
).In [7] one can find an interesting new result about the existence of convex solutions of P
1
(
0
,
b
) where 0
<
β
<
1 under some conditions. In [5, 7], 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 problemP
λ
a
,
b with β
=
0 is the well known Blasius problem. For a broad view, see [9]. See also [10].Great interest is given to analytical studies of similarity solutions because of their applications in different fields, for example, in magnetohydrodynamic (see [11–13]) or in boundary layer flows (see [8, 14]).The main goal of this paper is to study the question of existence and nonexistence of the solutions ofP
λ
a
,
b with 0
<
β
<
1 and λ
∈
{
0,1
}. We will focus our attention on convex and convex-concave solutions of the equation(1)
f
″
′
+
f
f
″
+
β
f
′
f
′
-
1
=
0
.As usual, to get a convex or convex-concave solution of P
λ
a
,
b, we use the shooting technique which consists of finding the values of a parameter c
≥
0 for which the solution of (1) satisfying the initial conditions f
(
0
)
=
a, f
′
(
0
)
=
b, and f
″
(
0
)
=
c exists on [
0
,
+
∞
) and is such that f
′
(
t
)
→
λ as t
→
+
∞. We denote by f
c the solution of the following initial value problem and by [
0
,
T
c
) the right maximal interval of existence: P
a
,
b
,
c
f
″
′
+
f
f
″
+
β
f
′
f
′
-
1
=
0
f
0
=
a
f
′
0
=
b
<
0
f
″
0
=
c
≥
0


## 2. On Blasius Equation

In this section, we recall some basic properties of the subsolutions andε-subsolutions of the Blasius equation. Let I
⊂
R be an interval and f
:
I
→
R be a function.Definition 1.
We say thatf is a subsolution of the Blasius equation f
″
′
+
f
f
″
=
0 if f is of class C
3 and if f
′
″
+
f
f
″
≤
0 on I.Definition 2 (letε
>
0).
We say thatf is an ε-subsolution of the Blasius equation f
″
′
+
f
f
″
=
0 if f is of class C
3 and if f
″
′
+
f
f
″
≤
-
ε on I.Proposition 3 (lett
0
∈
R).
There does not exist nonpositive concave subsolution of the Blasius equation on the interval[
t
0
,
+
∞
).Proof.
See [8], Proposition 2.11.Proposition 4 (letε
>
0 and t
0
∈
R).
There does not exist anyε-subsolution of the Blasius equation on the interval [
t
0
,
+
∞
)
.Proof.
See [8], Proposition 2.18.


## 3. Preliminary Results

Proposition 5.
Letf be a solution of (1) on some maximal interval I
=
(
T
-
,
T
+
).(1)
IfF is any antiderivative of f on I, then (
f
″
e
F
)
′
=
-
β
f
′
(
f
′
-
1
)
e
F
.
(2)
Assume thatT
+
=
+
∞ and that f
′
(
t
)
→
λ
∈
R as t
→
+
∞
. If moreover f is of constant sign at infinity, then f
″
(
t
)
→
0 as t
→
+
∞
.
(3)
IfT
+
=
+
∞ and if f
′
(
t
)
→
λ
∈
R as t
→
+
∞, then λ
=
0 or λ
=
1
.
(4)
IfT
+
<
+
∞, then f
″ and f
′ are unbounded near T
+
.
(5)
If there exists a pointt
0
∈
I satisfying f
″
(
t
0
)
=
0 and f
′
(
t
0
)
=
μ, where μ
=
0 or 1, then, for all t
∈
I, we have f
(
t
)
=
μ
(
t
-
t
0
)
+
f
(
t
0
).Proof.
The first item follows immediately from (1). For the proof of items (
2
)-(
5
), see [8], Proposition 3.1 with g
(
x
)
=
β
x
(
x
-
1
)
.Lemma 6.
Letβ
∈
(
0,1
] and f be a solution of (1) on some maximal interval I
=
(
T
-
,
T
+
). If there exists t
0
∈
I such that (2)
f
′
t
0
>
1
and
f
t
0
1
-
f
′
t
0
≤
f
″
t
0
≤
0
,then T
+
=
+
∞ and f
′
(
t
)
→
1 as t
→
+
∞. Moreover, f
″
<
0 on [
t
0
,
+
∞
).Proof.
See [3], Lemma 9.


## 4. The Boundary Value Problem in the Convex and Convex-Concave Case with0
<
β
<
1

In the following, we takea
,
b
∈
R and λ
∈
{
0,1
} with b
<
0 and 0
<
β
<
1. We are interested here in convex and convex-concave solutions of the boundary value problem P
λ
a
,
b. As mentioned in the introduction, we will use the shooting method to find these solutions. Define the following sets:(3)
C
1
=
c
≥
0
:
f
c
′
≤
0
and
f
c
″
≥
0
on
0
,
T
c
,
C
2
=
c
≥
0
:
∃
t
c
∈
0
,
T
c
,
∃
ε
c
>
0
s.t
f
c
′
<
0
on
0
,
t
c
,
f
c
′
>
0
on
t
c
,
t
c
+
ε
c
and
f
c
″
>
0
on
0
,
t
c
+
ε
c
,
C
3
=
c
≥
0
:
∃
s
c
∈
0
,
T
c
,
∃
ε
c
>
0
s.t
f
c
″
>
0
on
0
,
s
c
,
f
c
″
<
0
on
s
c
,
s
c
+
ε
c
and
f
c
′
<
0
on
0
,
s
c
+
ε
c
.Remark 7.
It is easy to prove thatC
2 and C
3 are disjoint nonempty open subsets of [
0
,
+
∞
) and that there exist c
0
>
c
∗
>
0 such that C
2
=
(
c
0
,
+
∞
), C
3
=
[
0
,
c
∗
), and C
1
∪
C
2
∪
C
3
=
[
0
,
+
∞
) (see Appendix A of [8] with g
(
x
)
=
β
x
(
x
-
1
) and β
>
0
).Lemma 8 (letβ
>
0).
Then,f
c is a convex solution of the boundary value problem P
0
(
a
,
b
) if and only if c
∈
C
1.Proof.
See Appendix A of [8] with g
(
x
)
=
β
x
(
x
-
1
) and β
>
0.Lemma 9 (letβ
>
0).
Ifc
∈
C
3, then T
c
<
+
∞. Moreover, f
c is convex-concave, decreasing and f
c
′
t
→
-
∞ as t
→
T
c.Proof.
Ifc
∈
C
3 then there exists s
c
∈
[
0
,
T
c
) such that f
c
′
(
s
c
)
<
0 and f
c
″
(
s
c
)
=
0. From Proposition 5, items (1) and (3), we have f
c
″
(
t
)
<
0 and f
c
′
(
t
)
<
0 for all t
∈
s
c
,
T
c, and f
c
′
(
T
)
→
-
∞ as t
→
T
c. Thus, f
c is convex-concave solution on [
0
,
T
c
) and f
c
′
(
t
)
→
-
∞ as t
→
T
c.
Let us assume thatT
c
=
+
∞; then there exists t
0
∈
s
c
,
+
∞ such that f
c
′ and f
c are negative on t
∈
t
0
,
+
∞ and we obtain f
c
″
′
+
f
c
f
c
″
=
-
β
f
c
′
(
f
c
′
-
1
)
<
0 on (
t
0
,
+
∞
). Hence, f
c is a nonpositive concave subsolution of the Blasius equation on (
t
0
,
+
∞
). This contradicts the Proposition 3 and thus T
c
<
+
∞.Remark 10.
From Proposition5, items (1), (3), and (5), if c
∈
C
2, then there are only three possibilities for the solution of the initial value problem P
a
,
b
,
c: (1)
f
c is convex and f
c
′
(
t
)
→
+
∞ as t
→
T
c (with T
c
≤
+
∞).
(2)
There exists a pointt
0
∈
[
0
,
T
c
) such that f
c
″
(
t
0
)
=
0 and f
c
′
(
t
0
)
>
1.
(3)
f
c is a convex solution of P
1
(
a
,
b
).The next proposition shows that case (1) cannot hold.Proposition 11 (letβ
>
0).
There does not existc
≥
0, such that f
c is convex on its right maximal interval of existence [
0
,
T
c
) and f
c
′
(
t
)
→
+
∞ as t
→
T
c.Proof.
Assume thatf
c is convex on its right maximal interval of existence [
0
,
T
c
) and f
c
′
(
t
)
→
+
∞ as t
→
T
c. Then there exists t
0
∈
[
0
,
T
c
) such that, for all t
∈
[
t
0
,
T
c
), f
c
′
(
t
)
>
1 and (4)
f
c
″
′
t
+
f
c
t
f
c
″
t
=
-
β
f
c
′
t
f
c
′
t
-
1
<
-
β
f
c
′
t
0
f
c
′
t
0
-
1
=
-
ε
.Consequently, f
c is a ε-subsolution of the Blasius equation on [
t
0
,
T
c
). Therefore from Proposition 4 we have T
c
<
+
∞.
Furthermore, there existst
1
∈
[
t
0
,
T
c
) such that f
c
(
t
1
)
=
α
>
0 and f
c
′
(
t
1
)
>
1, and then f
c
″
′
t
+
f
c
t
f
c
″
(
t
)
<
0 and f
c
(
t
)
>
f
c
(
t
1
)
=
α for all t
∈
[
t
1
,
T
c
). Thus,(5)
f
c
″
′
t
<
-
α
f
c
″
tfor all t
∈
[
t
1
,
T
c
). Next, integrating (5) on [
t
1
,
t
] for t
1
<
t
<
T
c, we obtain f
c
″
(
t
)
-
f
c
″
(
t
1
)
<
-
α
(
f
c
′
(
t
)
-
f
c
′
(
t
1
)
) and using Proposition 5, item (4), yields a contradiction as t
→
T
c.


## 5. Thea
≤
0 Case

Lemma 12 (let0
<
β
<
1 and a
≤
0).
Ifc
≥
0 and if there exists t
0
∈
[
0
,
T
c
) such that f
c
″
(
t
0
)
=
0 and f
c
′
(
t
0
)
>
1, then f
c
(
t
0
)
>
0.Proof.
Letc
≥
0 and assume that there exists t
0
∈
[
0
,
T
c
) such that f
c
″
(
t
0
)
=
0 and f
c
′
(
t
0
)
>
1.
Let us consider the functionH
c
=
f
c
″
+
f
c
(
f
c
′
-
β). Since H
c
′
=
(
1
-
β
)
f
c
′
2
≥
0 on [
0
,
T
c
), then H
c is nondecreasing on [
0
,
T
c
) and hence (6)
0
≤
H
c
0
=
c
+
a
b
-
β
<
H
c
t
0
=
f
c
t
0
f
c
′
t
0
-
β
.Thus, f
c
(
t
0
)
>
0
.For the rest of this section we will seta
∗
=
-
1
-
b
2
/
β
-
2
b.Proposition 13 (let0
<
β
<
1).
If eitherb
≤
-
1 or b
∈
(
-
1,0
] and a
≤
a
∗, then the boundary value problem P
1
(
a
,
b
) has no convex solution.Proof.
Suppose thatb
≤
-
1 and that f
c is a convex solution of the boundary value problem P
1
(
a
,
b
). Then, there exists t
∗
>
0 such that f
c
(
t
∗
)
=
0.
LetK
c
=
2
f
c
f
c
″
-
f
c
′
2
+
f
c
2
(
2
f
c
′
-
β
). From (1), we obtain K
c
′
=
2
(
2
-
β
)
f
c
f
c
′
2
<
0 on (
0
,
t
∗
). Therefore, K
c is decreasing on (
0
,
t
∗
) and hence K
c
(
0
)
>
K
c
(
t
∗
). It follows that(7)
f
c
′
2
t
∗
>
-
2
a
c
+
b
2
+
a
2
β
-
2
b
≥
b
2
,which implies that f
c
′
(
t
∗
)
>
1. This is a contradiction. The same contradiction is obtained where b
∈
(
-
1,0
] and a
≤
a
∗.Theorem 14.
Let0
<
β
<
1 and a
,
b
∈
R with b
<
0 and a
≤
0 and 0
<
β
<
1. (1)
The boundary value problemP
0
(
a
,
b
) has at least one convex solution.
(2)
If eitherb
≤
-
1 or b
∈
(
-
1,0
] and a
≤
a
∗, then the boundary value problem P
1
(
a
,
b
) has no convex solution and has infinitely many convex-concave solutions.Proof.
The first result follows from Remark7 and Lemma 8. The second result follows from Remark 7, Remark 10, Proposition 11, Proposition 13, and Lemma 6.


## 6. Thea
>
0 Case

Leta
,
b
∈
R with b
<
0 and a
>
0. We assume 0
<
β
<
1 and consider the solution f
c of the initial value problem P
(
a
,
b
,
c
) on the right maximal interval of existence [
0
,
T
c
).Let us setb
∗
=
max
⁡
-
1
/
2
a
2
,
-
β
/
1
-
β.Lemma 15 (let0
<
β
<
1. let c
≥
0).
Ifb
∈
(
b
∗
,
0
) and if there exists t
∗
∈
(
0
,
T
c
) such that t
∗ is the first point where f
c
(
t
∗
)
=
0, then f
c
′
(
t
∗
)
<
0 and f
c
″
(
t
∗
)
<
0.Proof.
Lett
∗
∈
(
0
,
T
c
) be such that f
c
>
0 on [
0
,
t
∗
) and f
c
(
t
∗
)
=
0. Suppose that f
c
″
>
0 on [
0
,
t
∗
). Then, necessarily, we have f
c
′
<
0 on [
0
,
t
∗
). Moreover, since f
c
′ is increasing and b
>
b
∗, we also have f
c
′
>
-
β
/
1
-
β on [
0
,
t
∗
).
LetE
c
=
f
c
″
+
f
c
f
c
′. From (1), we have E
c
′
=
(
1
-
β
)
f
c
′
2
+
β
f
c
′. Consequently, E
c
′
<
0 on [
0
,
t
∗
) and since E
c
(
t
∗
)
=
f
c
″
(
t
∗
)
≥
0, it follows that E
c
>
0 on [
0
,
t
∗
). Integrating from 0 to t
∗ gives (8)
0
<
∫
0
t
∗
E
c
t
d
t
=
f
c
′
t
∗
-
b
-
1
2
a
2
.Thus f
c
′
(
t
∗
)
>
b
+
1
/
2
a
2
≥
0 which is a contradiction.
Therefore, there existst
0
∈
[
0
,
t
∗
) such that f
c
″
>
0 on (
0
,
t
0
) and f
c
″
(
t
0
)
=
0. From Proposition 5, items (1) and (5), we have either f
c
′
(
t
0
)
<
0 or f
c
′
(
t
0
)
>
1. The second case cannot happen. Assume, for the sake of contradiction, that f
c
′
(
t
0
)
>
1. Then f
c
″
≥
0 and f
c
>
0 on [
0
,
t
0
], so that we have f
c
(
t
0
)
(
1
-
f
c
′
(
t
0
)
)
≤
f
c
″
(
t
0
)
≤
0. From Lemma 6, we obtain that T
c
=
+
∞, f
c
′
(
t
)
→
1 as t
→
+
∞, and f
c
″
<
0 on [
t
0
,
+
∞
). It follows that f
c is a positive convex-concave solution of the boundary value problem P
1
(
a
,
b
) on [
0
,
+
∞
), which contradicts the existence of t
∗. Consequently, we have f
c
′
(
t
0
)
<
0. This implies that f
c
′
<
0 on [
0
,
t
0
] and that c
∈
C
3. By virtue of Lemma 9, we see that f
c
″ remains negative after t
0. The proof is complete.Lemma 16 (let0
<
β
<
1).
If there existst
0
∈
[
0
,
T
c
) such that f
c
″
(
t
0
)
=
0 and f
c
′
(
t
0
)
>
1, then f
c
(
t
0
)
>
0.Proof.
Assume that there existst
0
∈
[
0
,
T
c
) such that f
c
″
(
t
0
)
=
0, f
c
′
(
t
0
)
>
1, and f
c
(
t
0
)
<
0. Then, there would exist t
1
<
t
0 such that f
c
(
t
1
)
=
0.
LetH
c
=
f
c
″
+
f
c
(
f
c
′
-
β
). From (1), we have H
c
′
=
(
1
-
β
)
f
c
′
2
≥
0 on [
0
,
T
c
). Therefore, H
c is nondecreasing on [
0
,
T
c
). Since H
c
(
t
0
)
=
f
c
(
t
0
)
(
f
c
′
(
t
0
)
-
β
)
<
0, we get f
c
″
(
t
1
)
=
H
c
(
t
1
)
<
0. But, this and Proposition 5, item (1), imply that f
c
″ remains negative on (
t
1
,
T
c
), a contradiction. Hence f
c
(
t
0
)
>
0.Lemma 17.
If0
<
β
<
1 and b
∈
(
b
∗
,
0
), then there exists c
0
∈
C
2 such that if c
≥
c
0 then f
c is a convex-concave solution of P
1
(
a
,
b
).Proof (letc
∈
C
2).
From Remark10 and Proposition 11, we see that either f
c is a convex solution of P
1
(
a
,
b
) or there exists t
0
∈
[
0
,
T
c
) such that f
c
″
(
t
0
)
=
0 and f
c
′
(
t
0
)
>
1. Now, as we have seen in the proof of Lemma 15, in the second case, f
c is a convex-concave solution of P
1
(
a
,
b
).
Letc
∈
C
2 be such that f
c is a convex solution of P
1
(
a
,
b
). Therefore, we have b
<
f
c
′
<
1 on [
0
,
+
∞
) and, from Lemma 15, we have f
c
>
0. It follows that (9)
f
c
″
+
f
c
f
c
′
-
1
′
=
1
-
β
f
c
′
f
c
′
-
1
≥
-
1
4
1
-
βon [
0
,
+
∞
). Integrating between 0 and t
≥
0, and using the fact that f
c
>
0, we obtain (10)
f
c
″
t
≥
-
1
4
1
-
β
t
+
a
b
-
1
+
c
-
f
c
t
f
c
′
t
-
1
≥
-
1
4
1
-
β
t
+
a
b
-
1
+
c
.Integrating once again we get(11)
∀
t
≥
0
,
(12)
1
>
f
c
′
t
≥
-
1
8
1
-
β
t
2
+
a
b
-
1
+
c
t
+
b
.Let us set P
c
(
t
)
=
-
1
/
8
1
-
β
t
2
+
(
a
(
b
-
1
)
+
c
)
t
+
b
-
1. We have P
c
(
t
)
<
0 for all t
≥
0. It means that P
c has no positive roots. Thus c cannot be too large, because, on the contrary, its discriminant Δ
=
(
a
(
b
-
1
)
+
c
)
2
+
1
/
2
(
1
-
β
)
(
b
-
1
) and a
(
b
-
1
)
+
c would be positive, and hence the polynomial P
c would have two positive roots, a contradiction.
Therefore, there existsc
0
>
0 such that f
c is convex-concave solution of the problem P
1
(
a
,
b
) for c
≥
c
0. This completes the proof.Theorem 18.
Leta
,
b
∈
R, with b
<
0, a
>
0, and 0
<
β
<
1. (1)
The boundary value problemP
0
(
a
,
b
) has at least one convex solution. If in addition b
∈
(
b
∗
,
0
), then any convex solution of P
0
(
a
,
b
) is positive.
(2)
Ifb
∈
(
b
∗
,
0
), then the boundary value problem P
1
(
a
,
b
) has infinitely many positive convex-concave solutions.Proof.
The first part of (1) follows from Remark7 and Lemma 8. The second part follows from Lemma 15, because if there was a point t
∗ such that f
c
>
0 on [
0
,
t
∗
) and f
c
(
t
∗
)
=
0 then f
c
″
(
t
∗
)
<
0, a contradiction. The second result follows from Remark 7, Remark 10, Proposition 11, Lemma 16, and Lemma 6.


## 7. Conclusion

In this work, in particular in Theorems14 and 18, we have presented some new and important results about the boundary value problems P
0
(
a
,
b
) and P
1
(
a
,
b
), which we summarize below. The parameters β and b satisfy 0
<
β
<
1 and b
<
0. The constants a
∗ and b
∗ are defined in Sections 5 and 6.(1)
Fora
≤
0
:
(a)
The boundary value problemP
0
(
a
,
b
) has at least one convex solution.
(b)
If eitherb
≤
-
1 or b
∈
(
-
1,0
] and a
≤
a
∗, then the boundary value problem P
1
(
a
,
b
) has no convex solution and has infinitely many convex-concave solutions.
(2)
Fora
>
0
:
(a)
Ifb
∈
(
b
∗
,
0
), then the boundary value problem P
0
(
a
,
b
) has at least one positive convex solution.
(b)
Ifb
∈
(
b
∗
,
0
), then the boundary value problem P
1
(
a
,
b
) has infinitely many positive convex-concave solutions.Numerical simulations prompt us to formulate the following conjecture.Conjecture 19.
Leta
,
b
∈
R, with b
≤
-
1, a
>
0, and 0
<
β
<
1. The boundary value problem P
1
(
a
,
b
) has no convex solution.To finish, we give the following proposition concerning the casea
=
0.Proposition 20 (letβ
<
2).
Ifb
≤
-
1, then the boundary value problem P
1
(
0
,
b
) has no convex solution.Proof.
Assume thatf
c is a convex solution of the boundary value problem P
1
(
0
,
b
). Then, there exists t
∗
≥
0, such that f
c
<
0 on (
0
,
t
∗
), f
c
(
t
∗
)
=
0, and f
c
′
(
t
∗
)
>
0. Consider again the function (13)
K
c
=
2
f
c
f
c
″
-
f
c
′
2
+
f
c
2
2
f
c
′
-
β
.We have K
c
′
=
2
(
2
-
β
)
f
c
f
c
′
2
<
0 on (
0
,
t
∗
). Thus, K
c is a decreasing function and hence K
c
(
0
)
>
K
c
(
t
∗
). It follows that f
c
′
2
(
t
∗
)
>
b
2 which implies that f
c
′
(
t
∗
)
>
1, which is a contradiction.



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*Source: 4340204-2018-11-01.xml*