Skip to main content

On the shear stress function and the critical value of the Blasius problem

Abstract

The Blasius problem has been used to describe the steady two-dimensional flow of a slightly viscous incompressible fluid past a flat plate moving at a constant speed β; and it is well known that there exists the critical value β <0 such that it has at least one solution for each β β and has no positive solution for β< β . The known numerical result shows β 0.3541. In this paper, by the study of the integral equation equivalent to the Blasius problem, we obtain the relation between the velocity function f and the shear stress functions f , upper and lower bounds of f and a new lower bound of β . In particular, 27 4 /9 f 3 /3, β >0.45. Regarding β , previous results presented a lower bound −0.5 and an upper bound −0.18733.

1 Introduction

The Blasius problem [1] arising in the boundary layer problems in fluid mechanics

f (η)+f(η) f (η)=0on [0,)
(1.1)

subject to the boundary conditions

f(0)=0, f (0)=βand f ()=1,
(1.2)

has been used to describe the steady two-dimensional flow of a slightly viscous incompressible fluid past a flat plate. It also arises in the study of the mixed convection in porous media [2], where η is the similarity boundary layer ordinate, f(η) is the similarity stream function, f (η) and f (η) are the velocity and the shear stress functions, respectively. The case of β<0 corresponds to a flat plate moving at a steady speed opposite to that of a uniform mainstream [3].

Regarding the analytic study of the Blasius problem (1.1)-(1.2), Weyl [4] proved that (1.1)-(1.2) has one and only one solution for β=0; Coppel [5] studied the case of β>0; the cases of 0<β<1 [6] and β>1 [7] were also investigated, respectively. Also, see [8]. In 1986, Hussaini and Lakin [9] indicated that there exists a critical value β <0 such that (1.1)-(1.2) has at least a solution for β β and no solution for β< β . A lower bound was presented with β 1/2=0.5 and numerical results showed β 0.3541 [9]. In 2008, Brighi, Fruchard and Sari [10] summarized historical study on the Blasius problem and analyzed the case β<0 in detail, in which the shape and the number of solutions were determined. In 2010, Yang [11] obtained an upper bound β <18733/ 10 6 =0.18733. The Blasius problem is a special case of the Falkner-Skan equation, for β=0, we may refer to [1215] for some recent results on the Falkner-Skan equation.

An open question is: What exactly is β ? And what properties does the shear stress function f (η) have? To our knowledge, there is little study on it. By the study of the integral equation that is equivalent to the Blasius problem, in this paper, we present the relation between f and f , upper and lower bounds of f and new lower bounds of β . In particular, 27 4 /9 f 3 /3, β >0.45.

2 Upper and lower bounds of f

Noticing the basic fact in [10] that if f is a solution of (1.1)-(1.2), then f >0 for η[0,), we use the so-called Crocco transformation [9, 10], which consists of choosing t= f as an independent variable and expressing z= f as a function of t, to change (1.1)-(1.2) to the Crocco equation [10]

d 2 z d t 2 = t z ,βt<1
(2.1)

with the boundary conditions

z (β)=0,z(1)=0.
(2.2)

Integrating (2.1) from β to t, we have by (2.2)

z (t)= β t s z ( s ) dson [β,1).
(2.3)

Integrating (2.3) from t to 1, we obtain the following integral equation that is equivalent to (1.1)-(1.2) [10, 11]:

z(t)= t 1 s ( 1 s ) z ( s ) ds+(1t) β t s z ( s ) ds:=Az(t)+(1t)Bz(t),
(2.4)

where z(t)C[β,1] and z(t)>0 for t[β,1).

Since β 1 2 , our work is restricted to the case of 1 2 β<0 and begins with the following lemma.

Theorem 2.1 Let z be a solution of (2.4), then

  1. (i)

    z h ( β ) ;

  2. (ii)

    β 3 3 z z(0) 3 h ( β ) + β 3 3 h ( β ) , where z=max{z(t):t[β,1]}, h(β)= 1 3 (13 β 2 2 β 3 ).

Proof Let t ˜ [β,1] such that z=z( t ˜ )=max{z(t):t[β,1]}. By (2.3), we know z (t)>0 for t(β,0], and then z(t) is strictly increasing (β,0]. This, together with z( t ˜ )>0, implies 0< t ˜ <1. From (2.1), we know z (t)z(t)=t. Integrating this equality from β to t ˜ , we have

β t ˜ z (s)z(s)ds= β t ˜ sds= t ˜ 2 β 2 2 .

Noticing that z ( t ˜ )= z (β)=0, we obtain

β t ˜ z (s)z(s)ds= z (s)z(s) | β t ˜ β t ˜ z 2 (s)ds= β t ˜ z 2 (s)ds0.

Consequently, |β| t ˜ .

  1. (i)

    From z ( t ˜ )= β t ˜ s z ( s ) ds=Bz( t ˜ ), we know Bz(t)Bz( t ˜ )=0 for t[ t ˜ ,1). This implies Az(t)z(t), t[ t ˜ ,1).

By A z(t)= t ( 1 t ) z ( t ) , we have

Az(t) ( A z ( t ) ) z(t) ( A z ( t ) ) t(1t)for t[ t ˜ ,1).
(2.5)

Noticing that z( t ˜ )=Az( t ˜ ) and z(1)=0, integrating (2.5) from t ˜ to 1, we have by |β| t ˜

z 2 = [ A z ( t ˜ ) ] 2 2 t ˜ 1 s ( 1 s ) d s 2 | β | 1 s ( 1 s ) d s = 1 3 ( 1 3 β 2 2 β 3 ) = h ( β ) .
  1. (ii)

    Integrating (2.3) from β to 0, we have

    z(0)= β 0 Bz(t)dt+z(β) β 0 β t s z dsdt+z(β) β 3 3 z ,

which implies that the left inequality of (ii) holds.

Noticing that |β| t ˜ and utilizing (2.1) and (2.2), we know

z( t ˜ )z(0)= 0 t ˜ t t ˜ s z ( s ) dsdt 0 | β | t | β | s z dsdt= β 3 3 z .

And then z(0)z+ β 3 3 z h ( β ) + β 3 3 h ( β ) = 3 h ( β ) + β 3 3 h ( β ) . Hence, (ii) holds. □

Let

l(t)= { 1 6 h ( β ) ( t + t 2 2 β 3 ) ( 1 t ) , t [ 0 , 1 ] , 1 6 h ( β ) ( 3 β 2 t t 3 2 β 3 ) , t [ β , 0 )

and

u(t)= { 3 h ( β ) ( 2 ln 2 1 + t + ( 1 t ) ln 1 + t 1 t ) , t [ 0 , 1 ] , 6 h ( β ) + 2 β 3 + 3 β 2 t t 3 3 h ( β ) , t [ β , 0 ) .

Utilizing Theorem 2.1, we can obtain upper and lower bounds of z as follows.

Theorem 2.2 Let z be a solution of (2.4), then l(t)z(t)u(t) for t[β,1].

Proof For t[β,0), we have by (2.3) and Theorem 2.1(i)

z (t)=Bz(t) β t s h ( β ) ds= β 2 t 2 2 h ( β ) on [β,0).
(2.6)

From this, we have

z(t) β t β 2 s 2 2 h ( β ) ds+z(β) 1 6 h ( β ) ( 3 β 2 t t 3 2 β 3 ) .

Hence, z(t)l(t) for t[β,0).

For t[0,1], let ε>0, we define a function h ε (t) as follows:

h ε (t)= 1 z + ε ( β 3 3 ( 1 t ) + 0 1 G ( t , s ) s d s ) for t[0,1],

where G(t,s) is the Green function for w (t)=0 with boundary conditions w(0)=0=w(1) defined by

G(t,s)= { t ( 1 s ) , 0 t s 1 , s ( 1 t ) , 0 s t 1 .
(2.7)

Next, we prove z(t) h ε (t) for t[0,1].

In fact, if there exists t 0 [0,1] such that z( t 0 )< h ε ( t 0 ), since h ε (1)=0=z(1), h ε (0)= β 3 3 ( z + ε ) < β 3 3 z z(0) by Theorem 2.1(ii), then t 0 (0,1) and there exists an interval (a,b)(0,1) such that

t 0 (a,b),z(t)< h ε (t)for t(a,b),z(a)= h ε (a),z(b)= h ε (b).

Let φ(t)= h ε (t)z(t), then φ(a)=0=φ(b) and φ( t 0 )>0. Let ξ(a,b) such that φ(ξ)=max{φ(t):t[a,b]}, then φ (ξ)0. On the other hand, we know easily that

φ (t)= h ε (t) z (t)= t z + ε + t z ( t ) >0for t(a,b).

Then φ (ξ)>0, a contradiction.

Taking ε0, we have

z(t) 1 z ( β 3 3 ( 1 t ) + 0 1 G ( t , s ) s d s ) for t[0,1].

Since 0 1 G(t,s)sds= ( t + t 2 ) ( 1 t ) 6 , we have

z(t) 1 6 z ( t + t 2 2 β 3 ) (1t).
(2.8)

Theorem 2.1(i) leads to

z(t) 1 6 h ( β ) ( t + t 2 2 β 3 ) (1t)=l(t)for t[0,1].

Finally, we prove z(t)u(t). For t[β,0), integrating (2.6) from t to 0, we obtain z(0)z(t) t 3 3 β 2 t 6 h ( β ) . Then by Theorem 2.1(ii),

z(t)z(0) t 3 3 β 2 t 6 h ( β ) 3 h ( β ) + β 3 3 h ( β ) t 3 3 β 2 t 6 h ( β ) =u(t).

For t[0,1), since z(t)l(t) 1 6 h ( β ) (t+ t 2 )(1t):= l 0 (t), we have

z ( t ) t 1 s ( 1 s ) z ( s ) d s + ( 1 t ) 0 t s z ( s ) d s t 1 s ( 1 s ) l 0 ( s ) d s + ( 1 t ) 0 t s l 0 ( s ) d s = 6 h ( β ) ( t 1 1 1 + s d s + ( 1 t ) 0 t 1 1 s 2 d s ) = 3 h ( β ) ( 2 ln 2 1 + t + ( 1 t ) ln 1 + t 1 t ) = u ( t ) .

 □

Combining Theorems 2.1 and 2.2, we obtain

Corollary 2.1 Let z be a solution of (2.4), then h ( β ) 4 l ( 3 + 6 β 3 3 ) z h ( β ) . In particular, 27 4 9 z 3 3 .

Proof Let g β (t)=(t+ t 2 2 β 3 )(1t), by (2.8), we have z(t) 1 6 z g β (t) on [0,1]. From g β ( t ˆ )=0, we obtain t ˆ = 3 + 6 β 3 3 , and then max{ g β (t):t[0,1]}= g β ( 3 + 6 β 3 3 ). Hence, z 1 6 z g β ( 3 + 6 β 3 3 ). This, together with g β ( 3 + 6 β 3 3 )=6 h ( β ) l( 3 + 6 β 3 3 ), implies z h ( β ) 4 l ( 3 + 6 β 3 3 ) . The right hand is from Theorem 2.1(i).

Since h(σ)h(0)= 1 3 for σ0, hence z h ( 0 )= 3 3 .

Since g β (t)t(1+t)(1t):= g 0 (t) for t[0,1], by g 0 ( t ˆ )=0, we have t ˆ = 3 3 . Hence, max{ g 0 (t):t[0,1]}= g 0 ( t ˆ )= 2 3 9 . By (2.8), we obtain

zz(t) g 0 ( t ) 6 z on [0,1].

From this, we have z g 0 ( t ˆ ) 6 z = 3 27 z , i.e., z 27 4 9 . □

Based on Theorem 2.2, Corollary 2.1, f (η)=t and f (η)=z(t), we obtain the relation between the velocity function f and the shear stress functions f , upper and lower bounds of f =sup{ f (η):η[0,}.

Theorem 2.3 Let f be a solution of (1.1)-(1.2), then

  1. (i)

    l( f ) f u( f ) for η[0,);

  2. (ii)

    h ( β ) 4 l ( 3 + 6 β 3 3 ) f h ( β ) . Specially, 27 4 9 f 3 3 .

Remark 2.1 There exists very little study on the upper and lower bounds of f (η); Theorem 2.3 fills this gap. Other studies can be found in [9, 10].

3 New lower bound of β

To obtain a better lower bound of β , we first prove

Theorem 3.1 Let z be a solution of (2.4), then

β 1 (1s) z 2 (s)ds 1 1440 h ( β ) ( 60 β 6 192 β 5 + 7 ) .

Proof Firstly, we prove

| z ( t ) | | t 2 t ˜ 2 | 2 h ( β ) on [0,1].
(3.1)

Integrating (2.1) from t to t ˜ , we obtain by z ( t ˜ )=0

z (t)= t t ˜ s z ( s ) ds t t ˜ s h ( β ) ds= t ˜ 2 t 2 2 h ( β ) on [0, t ˜ ).

Integrating (2.1) from t ˜ to t, we have by z ( t ˜ )=0

z (t)= t ˜ t s z ( s ) ds t ˜ t s h ( β ) ds= t ˜ 2 t 2 2 h ( β ) 0on [ t ˜ ,1].

Hence, (3.1) holds.

From (2.6) and (3.1), we know

β 1 ( 1 s ) z 2 ( s ) d s = β 0 ( 1 s ) z 2 ( s ) d s + 0 1 ( 1 s ) z 2 ( s ) d s 1 4 h ( β ) [ β 0 ( 1 s ) ( β 2 s 2 ) 2 d s + 0 1 ( 1 s ) ( s 2 t ˜ 2 ) 2 d s ] .

By

β 0 ( 1 s ) ( β 2 s 2 ) 2 d s = 1 30 ( 5 β 6 16 β 5 ) , 0 1 ( 1 s ) ( s 2 t ˜ 2 ) 2 d s = 1 30 ( 1 5 t ˜ 2 + 15 t ˜ 4 )

and

15 t ˜ 2 +15 t ˜ 4 =15 ( t ˜ 2 1 6 ) 2 + 7 12 7 12 ,

we obtain

β 1 ( 1 s ) z 2 ( s ) d s 1 120 h ( β ) [ ( 5 β 6 16 β 5 ) + 7 12 ] = 1 1440 h ( β ) ( 60 β 6 192 β 5 + 7 ) .

 □

Let

H(β)=73480 β 2 +720 β 4 +192 β 5 380 β 6 ,β [ 1 2 , 0 ] .

Since

H ( β ) = 960 β + 2880 β 3 + 960 β 4 2280 β 5 > 960 β ( 1 3 β 2 ) > 0 on  [ 1 2 , 0 ) , H ( 9 20 ) = 4396387 3200000 < 0 , H ( 11 25 ) = 55395529 48828125 > 0 ,

then there exists a unique β ˜ ( 9 20 , 11 25 )=(0.45,0.44) such that H( β ˜ )=0, H(β)>0 for β( β ˜ ,0) and H(β)<0 for β( 1 2 , β ˜ ).

Theorem 3.2 If β β ˜ , the Blasius problem (1.1)-(1.2) has no solution and then β >0.45.

Proof The proof is by contradiction. If for some β β ˜ , (1.1)-(1.2) has a solution f and then (2.1) has a solution z. Rewrite (2.1) as follows:

( z ( t ) z ( t ) ) +t= z 2 (t).

Integrating this equality from β to t and noticing that z (β)=0, we obtain

z(t) z (t)+ β t sds= β t z 2 (s)ds.

Integrating the last equality from β to 1 and using z(1)=0, we have

1 2 z 2 (β)+ β 1 β t sdsdt= β 1 β t z 2 (s)dsdt= β 1 (1s) z 2 (s)ds.

This, together with z(β)>0 and Theorem 3.1, implies

1 3 β 2 + 2 β 3 6 = β 1 β t sdsdt> 1 1440 h ( β ) ( 60 β 6 192 β 5 + 7 ) .

Since

1 3 β 2 + 2 β 3 6 1 1440 h ( β ) ( 60 β 6 192 β 5 + 7 ) = H ( β ) 1440 h ( β ) ,

then H(β)>0, a contradiction. Hence, β > β ˜ >0.45. □

Remark 3.1 Theorem 3.2 improves the lower bound of β from −0.5 in [9] to −0.45.

References

  1. Blasius H: Grenzschichten in Flüssigkeiten mit kleiner Reibung. Z. Angew. Math. Phys. 1908, 56: 1–37.

    MATH  Google Scholar 

  2. Aly EH, Elliott L, Ingham DB: Mixed convection boundary-layer flow over vertical surface embedded in a porous medium. Eur. J. Mech. B, Fluids 2003, 22(6):529–543. 10.1016/S0997-7546(03)00059-1

    MathSciNet  Article  MATH  Google Scholar 

  3. Weidman PD: New solutions for laminar boundary layers with cross flow. Z. Angew. Math. Phys. 1997, 48(2):341–356. 10.1007/s000330050035

    MathSciNet  Article  MATH  Google Scholar 

  4. Weyl H: On the differential equations of the simplest boundary layer problem. Ann. Math. 1942, 43: 381–407. 10.2307/1968875

    MathSciNet  Article  MATH  Google Scholar 

  5. Coppel WA: On a differential equation of boundary layer theory. Philos. Trans. R. Soc. Lond. 1960, 253: 101–136. 10.1098/rsta.1960.0019

    MathSciNet  Article  MATH  Google Scholar 

  6. Hartman P: Ordinary Differential Equations. Wiley, New York; 1964.

    MATH  Google Scholar 

  7. Belhachmi Z, Brighi B, Taous K: On the concave solutions of the Blasius equations. Acta Math. Univ. Comen. 2000, 69(2):199–212.

    MathSciNet  MATH  Google Scholar 

  8. Oleinik OA, Samokhin VN: Mathematical Models in Boundary Layer Theory. Chapman and Hall/CRC, Boca Raton; 1999.

    MATH  Google Scholar 

  9. Hussaini MY, Lakin WD: Existence and nonuniqueness of similarity solutions of boundary-layer problem. Q. J. Mech. Appl. Math. 1986, 39(1):15–24. 10.1093/qjmam/39.1.15

    MathSciNet  Article  MATH  Google Scholar 

  10. Brighi B, Fruchard A, Sari T: On the Blasius problem. Adv. Differ. Equ. 2008, 13(5–6):509–600.

    MathSciNet  MATH  Google Scholar 

  11. Yang GC:An upper bound on the critical value β involved in the Blasius problem. J. Inequal. Appl. 2010., 2010: Article ID 960365

    Google Scholar 

  12. Yang GC, Lan KQ: The velocity and shear stress functions of the Falkner-Skan equation arising in boundary layer theory. J. Math. Anal. Appl. 2007, 328(2):1297–1308. 10.1016/j.jmaa.2006.06.042

    MathSciNet  Article  MATH  Google Scholar 

  13. Lan KQ, Yang GC: Positive solutions of the Falkner-Skan equation arising in the boundary layer theory. Can. Math. Bull. 2008, 51(3):386–398. 10.4153/CMB-2008-039-7

    MathSciNet  Article  MATH  Google Scholar 

  14. Yang GC: Existence of solutions of laminar boundary layer equations with decelerating external flows. Nonlinear Anal. TMA 2010, 72(3–4):2063–2075. 10.1016/j.na.2009.10.006

    Article  MathSciNet  MATH  Google Scholar 

  15. Yang GC, Lan KQ: Nonexistence of the reversed flow solutions of the Falkner-Skan equations. Nonlinear Anal. TMA 2011, 74(16):5327–5339. 10.1016/j.na.2011.05.017

    MathSciNet  Article  MATH  Google Scholar 

Download references

Acknowledgements

The authors wish to thank the anonymous referees for their valuable comments. This research was supported by the National Natural Science Foundation of China (Grant No. 11171046) and Scientific Research Foundation of the Education Department of Sichuan Province, China.

Author information

Affiliations

Authors

Corresponding author

Correspondence to GC Yang.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors read and approved the final manuscript.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and Permissions

About this article

Cite this article

Yang, G., Xu, Y. & Dang, L. On the shear stress function and the critical value of the Blasius problem. J Inequal Appl 2012, 208 (2012). https://doi.org/10.1186/1029-242X-2012-208

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1029-242X-2012-208

Keywords

  • Blasius problem
  • shear stress function
  • critical value
  • upper and lower bounds
  • Crocco equation