Volume 2010, Article ID 960365,6pages doi:10.1155/2010/960365
Research Article
An Upper Bound on the Critical Value
β
∗
Involved
in the Blasius Problem
G. C. Yang
1, 21School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, China 2College of Mathematics, Chengdu University of Information Technology, Chengdu 610225, China
Correspondence should be addressed to G. C. Yang,[email protected]
Received 21 February 2010; Revised 29 April 2010; Accepted 6 May 2010 Academic Editor: Michel C. Chipot
Copyrightq2010 G. C. Yang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Utilizing the Schauder fixed point theorem to study existence on positive solutions of an integral equation, we obtain an upper bound of the critical valueβ∗ involved in the Blasius problem, in particular,β∗ <−18733/105−0.18733. Previous results only presented a lower boundβ∗ ≥ −1/2 and numerical investigationsβ∗. −0.3541.
1. Introduction
The following third-order nonlinear differential equation arising in the boundary-layer problems
fηfηfη0 on0,∞ 1.1
subject to the boundary conditions
f0 0, f0 β, f∞ 1, 1.2
called the Blasius problem1, has been used to describe the steady two-dimensional flow of a slightly viscous incompressible fluid past a flat plate, whereηis the similarity boundary-layer ordinate,fηis the similarity stream function, andfηandfηare the velocity and the shear stress functions, respectively.
Problem1.1-1.2also arises in the study of the mixed convection in porous media
Rayleighnumber and Pe theP´eclet number. The case of β < 0 corresponds to a flat plate
moving at steady speed opposite to that of a uniform mainstream3.
The boundary value problem1.1-1.2has been widely studied analytically. Weyl4 proved that1.1-1.2has one and only one solution forβ 0; Coppel5studied the case ofβ >0; the cases of 0< β <16andβ >17were also investigated, respectively. Also, see
8. Blasius problem is a special case of the Falkner-Skan equation, forβ0; we may refer to
9–13for some recent results on the Falkner-Skan equation.
Very recently, Brighi et al.14summarized historical study on the Blasius problem and analyzed the caseβ <0 in details, in which the shape and the number of solutions were determined. We may refer to14and the references therein for more recent results.
However, up to today, we know only that there exists a critical valueβ∗ ∈ −1/2,0 such that1.1-1.2has at least a solution forβ≥β∗, no solution forβ < β∗15. Numerical results showed thatβ∗. −0.354115.
An open question is what is exactlyβ∗? To our knowledge, there is little study on it. In this paper, we will study the open question mentioned above by studying the existence on positive solutions of an integral equation and present an upper bound ofβ∗, in particular,β∗<−18733/105−0.18733.
2. An Upper Bound of
β
∗By the basic fact in14, we know easily that iff is a solution of1.1-1.2, thenf > 0 for η∈0,∞. In this case, the most powerful method is the so-called Crocco transformationsee
14,15, which consists of choosingtfas independent variable and expressingzfas a function oft. Differentiatingzf fthe variabletis omitted for simplicity, we obtain zfff−ff; hencezf −f. Differentiating once again, we obtainzff−f. Then1.1-1.2becomes the Crocco equation14
d2z dt2 −
t
z, β≤t <1 2.1
with the boundary conditions
zβ0, z1 0. 2.2
Integrating2.1fromβtot, we have
zt −
t
β s
zsds on
β,1. 2.3
Integrating this equality from t to 1, we obtain the following integral equation that is equivalent to2.1-2.2:
zt
1
t
s1−s
zs ds 1−t
t
β s
zsds fort∈
Letgβ 1/3−81−ββ2 forβ ∈ −1/2,0, then gβ −82β−3β2 > 0 forβ ∈ −1/2,0. By direct computation
g
−1 5
<0, g
−18733 105
>0. 2.5
Hence there existsβ∈−1/5,−18733/105such thatgβ 0 andgβ>0 forβ∈β,0.
We shall prove that1.1-1.2has at least a solution forβ∈β, 0.
Letβ ∈β, 0andCβ,1be the Banach space of continuous functions onβ,1with the normzmax{|zt|:t∈β,1}andS:Cβ,1 → Cβ,1withSzt max{zt, ct}, wherect cβ1−tfort∈β,1and
cβ
√
3/3− gβ
41−β .
2.6
Clearly,Szt≥ctforz∈Cβ,1and 0< cβ≤
√ 3/12.
Notation. One has
Azt
1
t
s1−s
Szs ds, Bzt
t
β s
Szsds forβ≤t <1. 2.7
We consider the following integral equation of the form
zt Azt 1−tBzt forβ≤t <1. 2.8
Lemma 2.1. The integral equation2.8has a solutionz∈Cβ,1.
Proof. LetC{z∈Cβ,1:z ≤2M}withMβ11−s|s|/csds. We define an operator TonCby setting
Tzt
⎧ ⎨ ⎩
Azt 1−tBzt if t∈β,1,
0 if t1.
2.9
Since
Azt
1
t
s1−s Szs ds≤
1
t s cβ ds
1−t2
2cβ fort∈0,1, t
0 s Szs ds≤
t
0
1
cβ1−s ds−
ln1−t cβ
fort∈0,1,
lim
t→1−1−tln1−t 0,
we know that limt→1−Tzt 0 and thenTmapsCintoCβ,1. We show thatTis continuous
and compact fromCintoC.
Letzn∈C,z∈C, and limn→∞zn−z0. Since 1−t≤1−sforβ≤s≤t≤1, we have
|Tznt−Tzt| ≤ |Aznt−Azt| 1−t|Bznt−Bzt|
≤
1
β
sSz1n−ss−s1−s Szs
ds
1
β
sSz1n−ss−s1−s Szs
1−t 1−s
ds
≤2
1
β
sSz1n−ss−s1−s Szs
ds.
2.11
Since
lim
n→∞
s1−s Szns
1−ss
Szs fors∈
β,1 2.12
and Szt ≥ ct, the Lebesgue dominated convergence theorem, the dominated function Fs 1/cβfors∈β,1implies thatTzn−Tz → 0, that is,T is continuous.
BydTzt/dt−βts/Szsds, we have
dTzdtt≤
t
β
|s| Szsds≤
t
β
|s|
csds forβ≤t <1. 2.13
Noticing that
1
β t
β
|s|
csds dt
1
β 1
s
|s|
csdt ds
1
β
1−s|s|
cs dsM <∞, 2.14
we haveβ1|dTzs/ds|ds≤M. This, together with the absolute continuity of the Lebesgue integral, implies thatTC {Tzt:z∈C}is equicontinuous.
On the other hand,
|Tzt| ≤
1
t
|s|1−s Szs ds
t
β
|s|1−t Szs ds
≤
1
β
|s|1−s
cs ds
1
β
|s|1−s
cs ds2M.
2.15
Theorem 2.2. The problem 1.1-1.2 has at least a solution for β ∈ β,0 and then β∗ < −18733/105 −0.18733.
Proof. We first prove that the functionzobtained inLemma 2.1is a solution of2.4forβ ∈ β,0.Clearly, we have only to proveSzt ztfort∈β,1, that is,zt≥ctfort∈β,1. First of all, we prove that there existst∈β,1such thatzt> ct. In fact, ifzt≤ct fort∈β,1, then bySzt cβ1−t
cβ
1−β≥zβ
1
β
s1−s Szs ds
1
β
s1−s cs ds
1 2cβ
1−β2. 2.16
This implies thatc2
β≥1β/2≥1−1/5/22/5, which contradictscβ≤
√ 3/12. From the relations
zt −
t
β s
Szsds, z
t − t
Szt, 2.17
we know that z is convex and increasing onβ,0 and concave on0,1. Moreover, since z1 0, there existst∈0,1such thatzt max{zt:t∈β,1}.
Fort ∈ t,1, we haveBzt ≥ Bzt −zt 0. Then, from2.8we deduce that Azt≤zt≤Sztfort∈t,1and hence
Azt−Azt≤t1−t fort∈t,1. 2.18
Integrating the last inequality fortto 1 and usingAz1 0, we know that
Azt2
2 ≤
1
t
s1−sds≤
1
0
s1−sds 1 6.
2.19
And thenzt Azt ≤√3/3.This, together withct≤ cβ ≤
√
3/12 fort∈ 0,1, implies thatSzt≤√3/3 fort∈0,1. Hence
1
0
s1−s Szs ds≥
1
0
s1−s √
3/3 ds √
3
6 . 2.20
Noticing thatSzt≥ctandt1−t<0 fort∈β,0, we obtain
0
β
s1−s Szs ds≥
0
β
s1−s cs ds−
β2
2cβ
. 2.21
Then
zβ
1
β
s1−s Szs ds
0
β
s1−s Szs ds
1
0
s1−s Szs ds≥
√ 3 6 −
β2
By direct computation, we have √3/6−β2/2c
β cβ1−βand thenzβ ≥ cβ. Since z
is convex and increasing onβ,0and concave on0,1withz1 0, we immediately get zt≥ctfort∈β,1. HenceSzzandzis a positive solution of2.4.
Since any positive solution of2.1-2.2is a solution of1.1-1.2 14and2.1-2.2 is equivalent to2.4, hence1.1-1.2has at least a solution forβ∈β, 0and we obtain the desired resultβ∗≤β < −18733/105−0.18733.
Acknowledgments
The author would like to thank very much Professors C. K. Zhong and W. T. Li in Lanzhou University, China, for their guidance and the referees for their valuable comments and suggestions. This research is supported in part by the Training Fund of Sichuan Provincial Academic and Technology Leaders.
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