The Study of Micro-Fluid Boundary Layer Theory

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1 Long Li Huasui Zan Te Study of Micro-Fluid Boundary Layer Teory LONG LI Jimei University Scools of Sciences Xiamen 3602 Fujian Province P.R.CHINA HUASHUI ZHAN(corresponding autor) Jimei University Scools of Sciences Xiamen 3602 Fujian Province P.R.CHINA Abstract: Similar to te study of P randtl system by te well-known Oleinik linear metod te paper gets existence uniqueness of te solution for te following initial boundary problem in D = {(t x y) 0 < t < T 0 < x < X 0 < y < } u t + uu x + vu y = U t + UU x + (ν(y)u y ) y u x + v y = 0 u(0 x y) = u 0 (x y) u(t 0 y) = 0 u(t x 0) = 0 v(t x 0) = v 0 (t x) lim u(t x y) = U(t x) y were T is sufficient small and ν(y) is a bounded function. Key Words: Micro-fluid boundary layer Uniqueness Existence Classical solution Introduction As well known P randtl proposed te conception of te boundary layer in 904 []. From ten on te interest for te teory of boundary layer as been steadily growing due to te matematical questions it pose and its important practical applications. According to P randtl boundary layer teory te flow about a solid body can be divided into two regions: a very tin layer in te neigborood of te body (te boundary layer) were viscous friction plays an essential part and te remaining region outside tis layer were friction may be neglected (te outer flow). Tus for fluids wose viscosity is small its influence is perceptible only in a very tin region adjacent to te walls of a body in te flow; te said region according to P randtl is called te boundary layer. Tis penomenon is explained by te fact tat te fluid sticks to te surface of a solid body and tis adesion inibits te motion of a tin layer of fluid adjacent to te surface. In tis tin region te velocity of te flow past a body at rest undergoes a sarp increase: from zero at te surface to te values of te velocity in te outer flow were te fluid may be regarded as frictionless. P randtl derived te system of equations for te first approximation of te flow velocity in te boundary layer. Tis system served as a basis for te development of te boundary layer teory wic as now become one of te fundamental parts of fluid dynamics. Assume tat te motion of a fluid occupying a two-dimensional region is caracterized by te velocity vector V = (u v) were u v are te projections of V onto te coordinate axes x y respectively te P randtl system for a non-stationary boundary layer arising in an axially symmetric incompressible flow past a solid body as te form u t + uu x + vu y = U t + UU x + νu yy (ru) x + (rv) y = 0 u(0 x y) = u 0 (x y) u(t 0 y) = 0 u(t x 0) = 0 v(t x 0) = v 0 (t x) lim y u(t x y) = U(t x) () in a domain D = {(t x y) 0 < t < T 0 < x < X 0 < y < } were ν=const> 0 is te coefficient of kinematic viscosity; U(t x) is called te velocity at te outer edge of te boundary layer U(t 0) = 0 U(t x) > 0 for x > 0; r(x) is te distance from tat point to te axis of a rotating body r(0) = 0 r(x) > 0 for x > 0. In recent decades many scolars ave been carrying out researc in tis field acievements are abundant in literature on teoretical numerical experimental aspects of te teory [2 3]. However some discrepancies were founded between teoretical and experimental results for several important practical problems. P randtl boundary teory does not consider ISSN: Issue 2 Volume 8 December 2009

2 Long Li Huasui Zan bot te influence of wall s properties on te caracteristic of te boundary layer and te interaction of te actual solid wall wit te flow of water. In te absence of cemical reactions and cemical absorption te polar interaction plays te most important role. Under different kinds of solid surface te absorption of water molecule s beavior are different from one to anoter. Tis fact also makes tat te corresponding boundary layer s caracteristics are different from one to anoter. Wen te interfacial adsorption is relatively stronger at tis time experiments sow tat: te boundary layer system embodies te micro-fluid caracteristic. To be specific we sould modify te P randtl system as [4] u t + uu x + vu y = U t + UU x + (ν(y)u y ) y u x + v y = 0 u(0 x y) = u 0 (x y) u(t 0 y) = 0 u(t x 0) = 0 v(t x 0) = v 0 (t x) lim u(t x y) = U(t x) y (2) were (t x y) D and ν(y) > 0. In tis paper we get te following result. Teorem Assume tat U x U t /U U v 0 are bounded functions aving bounded derivatives wit respect to t x in D = {(t x y) 0 < t < T 0 < x < X 0 < y < } were T is sufficient small; ν(y) > 0 is a bounded function of wic te first te second and te tird order derivatives are bounded in D; lim u 0(x y) = U(0 x) u 0 (x 0) = 0; u 0 /U y u 0y /U are continuous in D; u 0y > 0 for y 0 x > 0 K (U(0 x) u 0 (x y)) u 0y (x y) K 2 (U(0 x) u 0 (x y)) wit positive constants K and K 2. Assume also tat tere exist bounded derivatives u 0y u 0yy u 0yyy u 0x u 0xy and te ratios u 0yy u 0y u 0yyyu 0y u 2 0yy u 2 0y are bounded on te rectangle {0 x X 0 y < }. Moreover u 0 v 0 satisfy te following compatibility condition v 0 (0 x)u 0y (x 0) = U t (0 x) + U(0 x)u x (0 x) + (ν(0)u 0y ) y (x 0) (3) and let u 0yx u 0x u 0yy u 0y u 0 u 0yy u 2 0y + U x Uu 0y K 5 (U u 0 (x y)). Ten problem (2) in D as a unique solution (u v) wit te following properties: u/u u y /U are continuous and bounded in D; u y /U > 0 for y 0; lim u y/u = 0 u(t x 0) = 0; v is continuous in y y and bounded for bounded y; te weak derivatives u t u x v y u yt u yx u yy u yyy are bounded measurable functions in D; te first equation in (2) old almost everywere in D; te functions u t u x v y u yy are continuous wit respect to y; moreover u yy u yyyu y u 2 yy (4) u y u 2 y are bounded and te following inequality old: C (U(t x) u(t x y)) u y (t x y) C 2 (U(t x) u(t x y)) (5) u(t x y) U(t x) exp( C 2 y) exp( C y) uyxuy uxuyy uu yy u u y + U 2 y x Uu y C3 (U u) uytuy utuyy uu yy u u y + U 2 y t Uu y C4 (U u). (6) Altoug our metod is similar to Oleinik s [5] owever ν is related to y wic makes te corresponding transform and te calculation become more complicated. Moreover if we permit ν(y) or its derivatives to be a unbounded functions ten Oleinik s linear metod will be invalid in tese circumstances to get te same results seem very difficult. By te way tere are some important progresses to te existence of te global solutions of () in recent recent years one can refer to [7-9] et.al. 2 Some Important Lemmas Consider te following initial boundary problem were u t + uu x + vu y = U t + UU x + (ν(y)u y ) y u x + v y = 0 (t x y) D = {(t x y) : 0 < t < T 0 < x < X 0 < y < } and wit te following initial boundary conditions u(0 x y) = u 0 (x y) u(t 0 y) = 0 u(t x 0) = 0 v(t x 0) = v 0 (t x) (8) lim u(t x y) = U(t x) y (7) ISSN: Issue 2 Volume 8 December 2009

3 Long Li Huasui Zan Definition 2 A solution of problem (7)-(8) is a pair of functions u(t x y) v(t x y) wit te following properties: u(t x y) is continuous and bounded in D; v(t x y) is continuous wit respect to y in D and bounded for bounded y; te weak derivatives u t u x u y u yy v y are bounded measurable functions; equation (7) olds for u v in D and conditions of (8) are satisfied. If we introduce te Crocco transformation τ = t ξ = x = u(t x y) U(t x) we obtain te following equation for w(τ ξ ) = u y (t x y)/u(t x): νw 2 w w τ Uw ξ + Aw + Bw +2ν w 2 w + ν w 3 = 0 (τ ξ ) Ω { w τ=0 = u 0y U = w 0(ξ ) w = = 0 (νww + ν w 2 v 0 w + C) =0 = 0 (0) (9) in te domain Ω = {(τ ξ ) 0 < τ < T 0 < ξ < X 0 < < } were A = ( 2 )U ξ + ( )U τ /U B = U ξ U τ /U C = U ξ + U τ /U. Solutions of problem (9)-(0) are understood in te weak sense. Definition 3 A solution of problem (9) is a function w(τ ξ ) wit te following properties: w(τ ξ ) is continuous in Ω; te weak derivatives w τ w ξ w are bounded functions in Ω w is continuous wit respect to at = 0 and its weak derivative w is suc tat ww is bounded in Ω; equation (9) old almost everywere in Ω for w and conditions of (0) are satisfied. For any function f(τ ξ ) we use te following notation: f () = f( m k) = const > 0. Instead of equation (9) let us consider te following system of ordinary differential equations: ν(w m k + ) 2 w U w w w w m k + A w +B w + 2ν (w m k ) 2 w +ν (w m k ) 2 w = 0 () w 0k = w 0 (k ) w() = 0 (2) [νw m k w + ν (w m k ) 2 v 0 w m k + C ] =0 = 0 (3) were m = 2 [T/]; k = 0 2 [X/]. We take w0 w 0 (ξ ) if w 0 as bounded derivatives w 0ξ w 0 and w 0. If w 0 (ξ ) is not so smoot we take for w0 a certain smoot function (to be constructed below) wic uniformly converges to w 0 in te domain {0 < ξ < X 0 < < } as 0. In wat follows K i M i C i stand for positive constants independent of. Lemma 4 Assume tat A B C v 0 are bounded functions in Ω. Let w0 be continuous in [0 ] and suc tat K ( ) w0 K 2( ) ν(y) is bounded function aving bounded first and second order derivatives in Ω. Ten problem (2)-(3) for ordinary differential equations admits a unique solution for m T 0 and small enoug were T 0 > 0 is a constant wic depends on te data of problem (7). Te solution w of problem (2)-(3) satisfies te following estimate: V (m ) w () V (m ) (4) were V and V are continuous functions in Ω positive for < and suc tat V K 3 ( ) and V K 4 ( ) in a neigborood of =. Proof: Te existence of tis solution follows from its uniqueness wic in its turn can be establised on te basis of te maximum principle and te fact tat tis problem can be reduced wit te elp of te Green function to a F redolm integral equation of te second kind. Indeed let Q be te difference of two solutions w of problem (2)-(3). Ten Q can attain neiter a positive maximum nor a negative minimum at = 0 since oterwise Q (0) 0 (see [6]) wereas te boundary condition [ νw m k w +ν (w m k ) 2 v 0 w m k +C ] =0 = 0 implies tat Q (0) = 0. We also ave Q () = 0 and at te interior points of [0 ] Q can neiter attain a positive maximum nor a negative minimum. Consequently under our assumptions problem (2)-(3) can not ave more tan one solution. Terefore we sall ave a fortiori establis te solvability ISSN: Issue 2 Volume 8 December 2009

4 Long Li Huasui Zan of problem (2)-(3) for m and k suc tat te solutions w m k of problem (2)-(3) admit te following a priori estimate: w m k () V ((m ) ) (5) In order to prove te priori estimate (4) for τ = m it suffices to sow tat tere exist functions V and V wit te properties in Lemma 4 and suc tat ordinary differential equations: L m (V ) ν(w m k + ) 2 V V V m k U V V +A V +ν (w m k ) 2 V 0 + B V + 2ν (w m k ) 2 V λ m (V ) = ν(0)w m k (0)V (0) (6) +ν(0) (w m k (0)) 2 v 0 w m k (0) + C > 0 (7) under te assumption tat L m (V ) 0 λ m (V ) < 0 k = 0 2 [X/] (8) V ((m ) ) w m k () V ((m ) ). (9) Ten te inequalities (4) can be proved by induction wit respect to m. Indeed consider te function q = V (m ) w were w is te solution of problem (2)-(3). We ave L m (q) 0 (20) λ m (V ) λ m (w) = ν(0)w m k (0)q (0) > 0. (2) Moreover by assumption we ave q m k 0 for m m and q () = 0. Let us sow tat q 0. To tis end we introduce new functions by q = e αm S α = const. > 0 Ten L m (q) e αm[ ν(w m k + ) 2 S U S S + A S +2ν (w m k ) 2 S +ν (w m k ) 2 S + B S ( e α )S e α S S m k 0 ] (22) λ m (q) = e αm ν(0)w m k (0)S (0) > 0. (23) It follows tat S (0) > 0 ten S cannot take its maximum positive value at = 0 moreover S () = 0. If S attains te maximum positive value at an interior point of te interval 0 ten at tis point we must ave S 0 (24) S S m k U S S 0 (25) 0 (26) S = 0 (27) [ B + ν (w m k ) 2 ] ( e α ) S (28) < 0 provided tat te constant α is large enoug and is sufficiently small suc tat e α > /2. However tese relations are incompatible wit (22). Terefore Tat is to say q = e αm S 0 (29) V (m ) w. (30) In a similar way we can sow tat (8) and (9) imply te inequality w V (m ). Now let us sow tat tere is a positive T 0 suc tat for m T 0 tere exist functions V and V satisfying te inequalities (6) (7) and (8) under te condition (9). Set V = µκ(α )κ ()e α 2τ were κ is a smoot function suc tat e s for 0 s ; κ(s) = κ 3 for s 3/2; for s 3/2. κ is a smoot function too suc tat for s /4; κ (s) = /2 κ for /4 < s < /2; s for s /2. Te constant µ > 0 is cosen suc tat V (0 ξ ) w0 (ξ ); te positive α α 2 will be specified later. Let us verify te inequalities (6) and (7) for V assuming tat (9) olds. Provided tat α > 0 is ISSN: Issue 2 Volume 8 December 2009

5 Long Li Huasui Zan large enoug and m is suc tat e α 2m > /2 We ave λ m (V ) = ν(0)w m k (0)µα e α 2m +ν (0)(w m k [ (0)) 2 v 0 w m k (0) + C µe α 2(m ) µ(ν(0)α e α 2m +ν (0)e α 2(m ) ) v 0 ] + C > 0 L m (V ) µe α 2m [ ν(w m k + ) 2 (κκ ) (3) +A (κκ ) + B κκ + 2ν (w ] m k ) 2 (κκ ) +ν (w m k ) 2 κκ + κκ α 2 e α 2 (32) were 0 < <. It may be assumed tat κκ = for < σ and small enoug σ > 0. For suc we ave L m (V ) 0 if α 2 is sufficiently large. For < σ we ave V > ρ = const. > 0 and terefore L m (V ) > 0 for large enoug α 2 provided tat e α2m > /2. Set V = Mκ ()κ 2 (β )e β 2m κ 2 is a smoot function suc tat 4 e s for 0 s ; κ 2 (s) = κ 2 3 for s 2; for s 2. Te constant M > 0 is cosen suc tat V (0 ξ ) w0 (ξ ); te positive β β 2 will be specified sortly. In a similar way we can prove tat L m (V ) 0 λ m (V ) < 0 provided tat β > 0 is large enoug and m is suc tat e β2m > /2. Tus if m T 0 and T 0 is suc tat e β 2T 0 2 and e α 2T 0 /2 ten problem (2)-(3) wit small enoug admits a unique solution wic satisfies te inequality (4). In wat follows we takes as w0 (ξ ) te function w 0 (ξ ) if w 0 is bounded in Ω; oterwise we let w0 (ξ ) be a function coinciding wit w 0 for /2 equal to w 0 (ξ ) w 0 (ξ ) for /2 + < and defined on te interval /2 /2 + in suc a way tat for /4 3/4 it as uniformly (in ) bounded derivatives wic are known to be bounded for w 0. Lemma 5 Assume tat te conditions of Lemma 4 are fulfilled and functions A B C v 0 w 0 ave bounded first order derivatives w 0ξ K 5 ( ) w 0 (ξ ) = 0 w 0 w 0 is bounded in Ω ν(y) as bounded tird order derivative and te following compatibility condition is satisfied: ν(0)w 0 w 0 + ν(0) w 2 0 v 0 w 0 + C = 0 (33) Ten and w w w ( + )w w w m k are bounded in Ω for m T T 0 and 0 uniformly wit respect to. Te positive constants T and 0 are determined by te data of problem (7)-(8). Proof: Let us introduce a new unknown function W = w e α in problem (2)-(3) were α is a positive constant wic does not depend on and will be cosen later. We ave and ν(w m k + ) 2 W U W W W W m k + à W + B W + 2ν (w m k ) 2 W +ν (w m k ) 2 W = 0 (34) ν(0)w m k (0)W (0) αν(0)w m k (0)W (0) +ν (0)(W m k (0)) 2 v 0 W m k (0) + C = 0 (35) were à = A 2αν(w m k + ) 2 B = B + α 2 ν(w m k + ) 2 αa 2αν (w m k ) 2. We introduce te functions ρ = W W m k Te function Φ () defined by for k m ; γ = W W. Φ () = (W ) 2 + (ρ ) 2 +(γ ) 2 + K 6 + (36) Φ () = (W ) 2 + (ρ ) 2 + K 6 + (37) for k = 0 m. Te constant K 6 > 0 will be cosen below. Consider Φ (0) for k m. We ave Φ (0) = 2W +2ρ (0)ρ (0)W (0) (0) + 2γ (0)γ (0) + K 6. (38) ISSN: Issue 2 Volume 8 December 2009

6 Long Li Huasui Zan From te boundary condition (36) and te estimates (4) it follows tat W (0) K 7 (39) From (35) and te estimates for w m k establised in Lemma 2. we find tat W (0) = R ρ (0) + R 2 γ (0) + R 3 (40) for 0 m T 0 and R 3 (0) = ρ ρ R (0) = R 2 (0) = ν(w m k + ) 2 U ν(w m k + ) 2 0 [à ν(w m k + ) 2 W + B W +2ν (w m k ) 2 W + ν (w m k ) 2 W ] were R i are bounded uniformly in. Using te boundary condition (36) we find (0) γ (0) for m ; in order to calculate (0) we ave to utilize te compatibility condition (2.) wic olds for w 0 and terefore w 0 = W k 0 e α = w 0k for = 0 since w 0 w 0 for /2 by construction. We ave ρ (0) = αρ (0) + v 0 v m k 0 ν(0) C + ρ m k (0) ν(0)w m k (0)W m 2k (0) ν(0)ρm k (0) ν(0) for m > and ρ k C C m k ν(0)w m 2k (0) (4) (0) = αρ k (0) + vk 0 v 0k 0 Ck C 0k ν(0) ν(0)w 0k (0) Similarly for k γ + (0) = αγ (0) + v 0 v 0 ν(0) C γ m k (0) ν(0)w m k (0)W m k (0) ν(0)γm k (0) ν(0) C C ν(0)w m k (0) (42) Substituting te expressions found for w (0) ρ (0) γ (0) into te rigt-and side of (39) we obtain te following relation for k Φ (0) = K 6 + 2α[ρ (0)] 2 + 2α[γ (0)] 2 +R 4 ρ (0)ρ m k (0) + R 5 ρ (0) +R 6 γ (0)γ m k (0) + R 7 γ (0) + R 8 (43) were R 4 (0) = 2 C ν W m k W m 2k νw m k W m 2k R 5 (0) = 2 v 0 v m k 0 ν 2 C C m k νw m 2k + 2W R R 6 (0) = 2 C ν W m k W m k νw m k W m k R 7 (0) = 2 v 0 v m k 0 2 C C m k ν νw m k R 8 (0) = 2W R 3 R i are bounded uniformly in wile R k 4 = 0. Tese relations imply tat for m > k we ave Φ (0) K 6 + αφ (0) K 8 Φ (0) K 9 Φ m k (0) K 0 (44) were te constant K 0 = K 0 (α). Let us coose α suc tat α/2 > K 8 α/4 > K 9 K 6 > K 0. Ten Φ (0) α 2 Φ (0) α 4 Φm k (0) (45) for m = 2 3 ; k = 2 [X/] m T 0. Note tat γ m k (0) wit m = is uniformly bounded in and since R k 4 = 0 it follows from (44) tat Φ k (0) α 2 Φk (0) (46) if K 6 is sufficiently large. Likewise te inequalities (45) and (46) can be proved for k = 0 m. Let us define te functions Φ 0k. For tis purpose we introduce te functions W k by W 0k W k = ν(w 0k + ) 2 W 0k U 0k W 0k W 0k + Ã0k W 0k + B 0k W 0k +2ν (w 0k ) 2 W 0k + ν (w 0k ) 2 W 0k (47) were à 0k = A 0k 2αν(W 0k e α + ) 2 B 0k = B 0k + α 2 ν(w 0k e α + ) 2 αa 0k 2αν (W 0k e α ) 2. Ten we can define te function Φ 0k for k and k = 0 respectively by (37) and (38). It is easy to ISSN: Issue 2 Volume 8 December 2009

7 Long Li Huasui Zan see tat Φ 0k is bounded uniformly in. Indeed te functions W 0k = (w 0k e α ) = (w 0 (k )e α ) γ 0k = e α w0k w 0k = e α w 0 (k ) w 0 ((k ) ) are bounded uniformly in since te first derivatives of w0 are uniformly bounded in. Te function w 0 w 0 is bounded in Ω and K ( ) w 0 K 2 ( ) Terefore ( +)w0 (ξ ) wic for > /2+ coincides wit te function ( + )w 0 (ξ ) is also uniformly bounded in. It follows tat (w 0k + )W 0k K ( + ) (w 0k e α ) K 2 + K 3 ( + ) w 0k 0 = K 2 + K 3 ( + ) w0 K 4. Consequently te ratio (W 0k W k )/ defined by (48) is uniformly bounded in and Φ 0k K 5. (48) Now we will deduce te equation for Φ () on te interval 0 <. To tis end we differentiate equation (35) in and multiply te result by 2W ; ten we subtract from equation (35) for W equation (35) for W m k and multiply te result by 2ρ /; from equation (35) for W we subtract (35) for W and multiply te result by 2r /. Taking te sum of te tree equations just obtained we get te equation for Φ () m = 2 [T/]; k = 0 2 [X/]. We find te equations for Φ () wit k = 0 m by taking only te first and te second of tese equations. In order to derive te equation for Φ () wit m = we utilize te relation (48) wic determines te values of W k. For m k te equation for Φ () as te form ν(w m k + ) 2 Φ U Φ Φ Φ Φ m k + Ã Φ +2 B Φ + 2ν (w m k ) 2 Φ +2ν (w m k ) 2 Φ + N N 2 = 0 (49) were N 2 is a sum of non-negative terms: 2 = 2νa m k (W ) + 2νa m k (ρ +2νa m k (γ ) 2 U + (γ ) 2 N + (ρ γ m k ρ ρ m k ) 2 + (ρ ) 2 + U ) 2 + U (γ ) 2 + (ρ (γ γ ) 2 ) 2 (50) and N is a linear function wose coefficients are uniformly bounded in and can be expressed in terms of te following quantities: W W ν a m k W W νa m k W W γ W m k ρ ρ m k W γ γ m k (W W γ W W m k ) 2 W m k W m k ρ (W ) 2 ρ γ m k ρ ρ m k γ γ γ ρ ν W m k ρ [(w m k ) 2 (w m 2k ) 2 ] W γ (a m k a m k ) ν w m k w m k W W ν W γ [(w m k ) 2 (w m k ) 2 ] W m k ρ (a m k a m 2k ) γ γ m k ν (w m k ) 2 (W ) 2 ν w m k w m k (W ) 2 ν W m k ρ [(w m k ) 2 (w m 2k ) 2 ] ν W γ [(w m k ) 2 (w m k ) 2 ] ν (w m k ) 2 W W were a = (w + ) 2. Using te inequality 2ab βa 2 + b2 (β > 0) (5) β to estimate te terms tat make up N we obtain from (50) ν(w m k + ) 2 Φ U Φ Φ +2ν (w m k ) 2 Φ + C Φ 0 Φ Φ m k + Ã Φ + 2 B Φ + + 2ν (w m k ) 2 Φ (52) ISSN: Issue 2 Volume 8 December 2009

8 Long Li Huasui Zan were C depend on W m k ρ m k γ m k γ (w m k + w m 2k + 2)W m k (w m k + w m k + 2)W ν (w m k ) 2 ν w m k w m k ν W (w m k + w m k ) ν (w m k ) 2 W W ν W (w m k + w m k ) ν W m k (w m k + w m 2k ) ν W m k (w m k + w m 2k ) ν w m k w m k W C It is easy to see tat for k = te coefficient does not depend on γ since U m0 = 0. Te inequality (53) for Φ m0 is obtained in exactly te same manner as for k. Obviously in tis case te coefficient C depends only on W m k ρ m k γ m k γ (w m k + w m 2k + 2)W m k (w m k +w m k +2)W ν (w m k ) 2 ν w m k w m k ν (w m k ) 2 W W ν W m k (w m k + w m 2k ) ν W m k (w m k + w m 2k ) ν w m k w m k W Now consider te functions Y () = (ρ ) 2 + (γ ) 2 + f() (53) wen k m > 0 Y () = (ρ ) 2 + f() (54) wen k = 0 m > 0. Were f() = κ(β)κ 2 () and te functions κ κ are tose constructed in te proof of Lemma 4; β is a positive constant. Just as we ave proved inequalities (45) and (46) we find tat Y (0) α 2 Y (0) α 4 Y m k (0) (55) for m > k Y (0) α 2 Y (0) (56) for m = k provided tat α β are sufficiently large. We clearly ave Y () = 0. Consider te functions Y 0k. Obviously W 0k W 0k K 6 ( + ) since w 0ξ K 5 ( ) by te assumption of Lemma 4 and te functions W 0k = w 0 (k )eα w 0ξ are uniformly bounded in. It follows from (48) tat W 0k W k K 7 ( + ) since by virtue of (4) and te properties of w0 = w 0k = W 0k e α we ave ν(w 0k + ) 2 W 0k () K 8 ( + ) (57) Ã0k W 0k () K 9 ( + ) (58) B 0k W 0k () K 20 ( + ) (59) ν (w 0k ) 2 W 0k () K 2 ( + ) (60) ν (w 0k ) 2 W 0k () K 22 ( + ) (6) Consequently Y 0k () K 23 ( + ) 2. (62) Let us write out te equation for Y. For m k tis equation as te form ν(w m k + ) 2 Y U Y Y Y Y m k + à Y +2 B Y + 2ν (w m k ) 2 Y +2ν (w m k ) 2 Y ν(w m k + ) 2 f à f 2 B f 2ν (w m k ) 2 f 2ν (w m k ) 2 f + N 3 N 4 = 0 (63) were N 4 is a sum of non-negative terms namely N 4 = 2νa m k (W ) 2 U + ρ m k γ m k + (ρ (γ ) + (ρ ) 2 + 2νa m k (γ ) 2 ρ ) 2 + U (γ γ ) 2 (64) and N 3 is a linear functions wose coefficients are uniformly bounded in and can be expressed in terms ISSN: Issue 2 Volume 8 December 2009

9 Long Li Huasui Zan of te following quantities: ρ γ m k W m k ρ ρ m k ρ W m k ( ) ρ w m k γ γ m k W ρ ρ m k ν γ [(w m k ) 2 (w m k ) 2 ]W ν γ [(w m k ) 2 (w m k ) 2 ]W γ γ m k γ γ ρ w m k ρ ρ m k (w m k + w m 2k + 2)W m k ν ρ [(w m k ) 2 (w m 2k ) 2 ]W m k ν ρ [(w m k ) 2 (w m 2k ) 2 ]W m k γ γ m k (w m k + w m k + 2)W γ W ( ). (65) Using (52) to estimate te terms tat make up N 3 we obtain te following inequality ν(w m k + ) 2 Y U Y Y Y Y m k + Ã Y +2 B Y + 2ν (w m k ) 2 Y +2ν (w m k ) 2 Y + Q Y + Q 2 + Q 3 0 (66) were Q 0 and depends on W m k (w m k + w m 2k + 2)W m k ν (w m k + w m 2k )W m k (w m k + w m k + 2)W W ν (w m k + w m 2k )W m k ν (w m k + w m k )W ν (w m k + w m k )W (67) and Q 2 is a linear combination of te following functions: (γ m k ) 2 (γ ) 2 (ρ m k ) 2 (w m k ) 2 (w ) 2 [(W m k ( )] 2 [(W ( )] 2. (68) We also ave Q 3 K 24 ( + ) 2 ν(w m k + ) 2 f +Ã f + 2 B f +2ν (w m k ) 2 f + 2ν (w m k ) 2 f. (69) It is easy to see tat for k = te function Q 2 does not depend on (ρ m k ) 2. For k = 0 an inequality of te form (67) olds for Y but in tis case Q depends only on W m k (w m k + w m 2k + 2)W m k ν (w m k + w m 2k )W m k ν (w m k + w m 2k )W m k and Q 2 is a linear combination of (ρ m k ) 2 [W m k ( )] 2 (w ) 2. Let us sow by induction tat Φ M Y M 2 ( + ) (70) for m T and some T T 0 te constants T M i being independent of. To sow tis assume tat for m < m and m = m k < k te inequalities (7) old wit constants M M 2 specified below. Let us sow tat if m T te same inequalities are valid for m = m k = k. Note tat under te induction assumptions we can claim tat for m < m or m = m k < k te following inequalities old: (w m k + w m 2k + 2)W m k K 25 ( + ) W m k K 26 ν(w m 2k + ) W m k = K 26 ρ m k + U m k γ m k B m k W m k 2ν (w m 2k ) 2 W m k ν (w m 2k ) 2 W m k (w m 2k + ) K 27 Ãm k W m k In exactly te same manner we get tat (w m k + w m k + 2)W K 28 (7) were te constants K 26 K 27 depend on M and M 2. Terefore if te inequalities (7) old for m < m and m = m k < k ten it can be seen tat in (53) and (67) we ave by C K 29 (M M 2 ) Q K 30 (M M 2 ) Q 2 K 3 ( + ) 2. (72) Let us pass to new functions in (2.25) and (2.32) Φ = Φ e γm (73) Y = Ỹ e γm (74) Te constant γ(m M 2 ) will be cosen later. m m and m = m we ave ν(w m k + ) 2 Φ e γ Φ Φm k +Ã Φ U Φ Φ 2 Φ [ + 2ν (w m k ) + 2ν (w m k ) B + C γe 0 For γ ] Φ (75) ISSN: Issue 2 Volume 8 December 2009

10 Long Li Huasui Zan for 0 < < and also ν(w m k + ) 2Ỹ U Ỹ Ỹ + e γ Ỹ Ỹ m k ÃỸ [ +2ν (w m k ) 2Ỹ + +2 B + 2ν (w m k ) 2 Q γe γ ]Ỹ + +K 32 (M M 2 )( + ) 2 0. (76) m = m k k were te constant K 36 depends neiter on M nor on M 2. If Φ () attains its maximum value at m = 0 we ave already sown tat Φ () max Φ 0k max Φ 0k K 5. It follows tat Φ () max{k 5 K 36 } (83) Coose M > max{2k 5 2K 36 }. Ten Φ () M 2 (84) Let us coose γ(m M 2 ) suc tat for small enoug te following inequalities are valid: were 2ν (w m k ) C γe γ < 0 B 2ν (w m k ) B + γe γ < K 33 (M M 2 ) Q K 33 = 2 K 32 M 2 + K 34 K 34 ( + ) 2 2ν(w m k + ) 2 2Ã ( + ). (77) (78) (79) Consider te point at wic Φ () for 0 m < m or m = m k k attains its largest value. In view of (76) and (78) tis point cannot belong to te interval 0 < < for m. Moreover if Φ () attains its maximum at = 0 m we sould ave Φ () 0 wereas relations (45) and (46) imply tat and terefore 0 Φ (0) α Φ 2 (0) α 4 e γ Φ m k (0) α Φ 2 (0) α Φ 4 m k (0) Φ (0) 2 Φ m k (0) (80) wic is impossible. For = we ave Φ = Φ e γm = e γm[ ] (W ) 2 + K 6 + (8) Estimates (4) for w imply tat W () K 35. (82) Terefore if Φ () attains its maximum value at = we ave Φ () K 36 for m < m and If ten Φ () M 2 eγm (85) e γt 2 m T (86) Φ () M. (87) for m < m and m = m k k as required. Now consider te functions X = Ỹ M 2 2 ( + )2 (88) it follows from (77) and (78) tat ν(w m k + ) 2 X U X X e γ X X m k + Ã [ X +2ν (w m k ) 2 X + 2ν (w m k ) Q γe γ ] X K [ 32 (M M 2 )( + ) 2 M 2 2ν(w m k + ) 2 2Ã ( + )] ( 2 + 2ν (w m k ) B + γe γ ) ( + ) 2] K 32 (M M 2 ] )( + ) [K 2 34 K 33 ( + ) 2 M 2 2 Q B 0 (89) if m < m or m = m k k. Let us sow tat X () 0 for suc m and k. If X () takes positive values ten tere is a point at wic for m < m or m = m k k te function X () attains its largest positive value. Tis point cannot belong to te interval 0 < < for m because of (90). For m = 0 if M 2 /2 > K 23 taking into account of te estimate (90) for Y 0k we find tat [ X () K 23 M 2 2 ] ( + ) 2 0 Since Ỹ () = 0 we ave X () 0. For = 0 m te function X cannot attain its largest ISSN: Issue 2 Volume 8 December 2009

11 Long Li Huasui Zan positive value since inequalities (56) and (57) sow tat X (0) = Ỹ (0) + M 2 ( + ) α 2 Ỹ (0) α 4 Ỹ m k (0) + M 2 ( + ) α 2 X (0) α 4 Xm k (0) + α 8 M 2( + ) 2 + M 2 ( + ) α 2 X (0) α 4 Xm k (0). So X 0 and Ỹ 2 M 2( + ) 2 (90) wen m < m and m = m k k. Terefore if M 2 /2 > K 23 e γt < 2 m T we ave Y () = Ỹ e γm M 2 ( + ) 2. (9) It follows from (7) (74) (75) and (35) tat ( + ) W are uniformly bounded in. Lemma 6 Under te assumptions of Lemmas 4 and 5 problem (9) in Ω wit T = T admits a unique solution w wit te following properties: w is continuous in Ω; C ( ) w C 2 ( ) (92) w as bounded weak derivatives w w ξ w τ ; w ξ C 3 ( ) w τ C 4 ( ) (93) te derivative w is continuous in < ; conditions 0f (8) old for w; te weak derivative w exists and ww is bounded in Ω; equation (9) olds almost everywere in te same domain. Proof: First let us prove te uniqueness of te solution. Assume te contrary namely tat w and w 2 are two solutions of problem (9) wit te properties specified in Lemma 6. Ten almost everywere in Ω te function z = w w 2 satisfies te following equation and te boundary conditions: νw 2z z τ Uz ξ + (A + 2ν w 2)z +[B + 3ν w w 2 + (νw 2 + 2ν w 2 )(w + w 2 )]z +ν z 3 = 0 z τ=0 = 0 z = = 0 [νw z + νw 2 z + ν (w + w 2 )z v 0 z] =0 = 0 (94) Set z = e ατ β z were α β = const. > 0. Ten νw 2 z z τ U z ξ + E z + F z +ν (w w 2 ) 2 z (95) = 0 z τ=0 = 0 z = = 0 (96) [νw z + νw 2 z + ν (w + w 2 ) z (97) βνw z v 0 z] =0 = 0 were E = A + 2ν w 2 2βνw2 F = B + 3ν w w 2 + (νw 2 + 2ν w 2 )(w + w 2 ) +νβ 2 w 2 α Aβ 2ν βw 2 Te constant β is cosen suitable large suc tat [βνw + v 0 νw 2 ν (w + w 2 )] =0 >. (98) Let us multiply equation (96) by z and integrate te result over Ω. Integrating by parts in some of te terms we find tat Ω E z zddξdτ + Ω F z2 ddξdτ + Ω 2 U ξ z 2 ddξdτ + Ω ν (w w 2 ) 2 z 2 ddξdτ 2 Ω νw w z zddξdτ Ω ν w 2 z zddξdτ Ω νw2 z2 ddξdτ 2 z 2 ddξ τ=t U z 2 ddτ νw 2 z zdξdτ = 0 2 ξ=x =0 (99) Using te boundary condition (98) at = 0 we can write te last integral in (00) as w z 2[ βνw + v 0 νw 2 ν (w + w 2 ) =0 ] dξdτ. (00) By our coice of β tis integral is non-negative. Let us estimate te integral over Ω containing z z. Taking into account E < / in Ω we get Ω (E 2νw w ν w 2) z zddξdτ Ω νw2 z2 ddξdτ (0) +C 5 Ω z2 ddξdτ were C 5 = const. Terefore it follows from (00) tat [ Ω F + 2 U ξ + ν (w w 2 ) 2 +C 5 ] z 2 (02) ddξdτ 0 Because C ( ) w i C 2 ( ) w i w i C 6 w i C 7 (i = 2) and assumptions we can coose α β to sure te inequality B + 3ν w w 2 + (νw 2 + 2ν w 2 )(w + w 2 ) +β 2 νw U ξ + (w w 2 ) 2 + +C 5 α Aβ 2βν w 2. (03) ISSN: Issue 2 Volume 8 December 2009

12 Long Li Huasui Zan Ten it follows from (03) tat z 2 ddξdτ 0 (04) Ω Terefore z 0 in Ω and w = w 2 in Ω as required. Now we will prove te existence of te solution of (9)-(0). Te solutions w of problem (2)-(3) sould be linearly extended to te domain Ω. First wen (k ) < ξ k k = 2 k(); k() = [X/] let w m (ξ ) = w m ((k )λ + k( λ) ) = ( λ)w () + λw () Secondly wen (m ) < τ < m m = 2 m(); m() = [T/] let w (τ ξ ) = w ((m )σ + m( σ) ξ ) = ( σ)wk m (ξ ) + σwk m (ξ ) According to Lemma 4 Lemma 5 te functions w (τ ξ) obtained in tis manner form tis family satisfy te Lipscitz condition wit respect to ξ τ and ave uniformly (in ) bounded first derivative in in Ω. By te Arzelà Teorem tere is a sequence i 0 suc tat w i uniformly converge to some w( ξ τ). It follows from Lemma 4 Lemma 5 tat w( ξ τ) as bounded weak derivatives w w ξ w τ and its weak derivative w is suc tat ( )w is bounded since te weak limit of a bounded sequence is bounded by te same constant. Consequently w is continuous in <. Te sequence w i may be assumed suc tat te derivatives w w ξ w τ w in te domain Ω coincide wit weak limits in L 2 (Ω) of te respective functions Denoting w w(τ + i ξ ) w(τ ξ ) i w(τ ξ + i ) w(τ ξ ) i w i w i By te first term of () ν(w m k w U +2ν (w m k +ν (w m k = w (τ ξ ) = w(m k ) + ) 2 w w ) 2 w w ) 2 w = 0 w m k + A w + B w (05) Now suppose tat ϕ(τ ξ ) be a smoot functionwic support set is compact in Ω. Let ϕ () = ϕ(m k ) Let us multiply wit ϕ at te two sides of (06) integrating te resulting equation in from 0 to and taking te sum over k m from to k() m() respectively we obtain +A w w 0 ϕ[ ν(w m k + ) 2 w w m k + B w w U + 2ν (w m k w ) 2 w +ν (w m k ) 2 w ] d = 0 (06) Denote te function f(τ ξ ) on Ω as f(τ ξ ) = f(m k ) wen (k ) < ξ k (m ) < τ < m. and denote ( w ) m = w ) k w m k ( w = w w Ten we can rewrite (07) as [ Ω ν( w m k ( ) w m ϕ + ) 2 w ϕ ) Ū( w k ϕ + Ā w ϕ + B w ϕ ] +2 ν ( w m k ) 2 w ϕ + ν ( w m k ) 2 w ϕ ddξdτ = 0 (07) Because w w w w + w w M + w w wen 0 w uniformly convergent to w i.e. w = w. Just likely ϕ = ϕ Ā ϕ = Aϕ B ϕ = Bϕ Ū ϕ = Uϕ ν ϕ = ν ϕ ν ϕ = ν ϕ ν( w m k + ) 2 = νw 2 At te same time on account of tat ( w ) m w τ ( w ) k w ξ w w in L 2 (Ω). ISSN: Issue 2 Volume 8 December 2009

13 Long Li Huasui Zan So if let 0 in (08) ten Ω (νw2 w w τ Uw ξ + Aw + Bw +2ν w 2 w + ν w 3 )ϕddξdτ = 0. (08) By te arbitrary of ϕ we get ours result. Acknowledgements: Te researc was supported by te NSF of Fujian Province of Cina (No: 2009J0009) and te NSF of Jimei University of Cina. References: [] Prandtl L. Über Flüssigkeisbewegungen bei ser kleine Reibung In: Ver. Int. Mat. Kongr Heidelberg 904. [2] Anderson I.and Toften H. Numerical solutions of te laminar boundary layer equations for powerlaw fluids Non-NewtonFluid Mec pp [3] Sclicing H. and Gersten K. and Krause E. Boundary Layer Teory Springer [4] Suming Wen Micro-Fluid Boundary Layer Teory and Its Applications Metallurgical Industry Press [5] Oleinik O.A and Samkin V.N. Matematical Models in Boundary Layer Teorem Capman and Hall/CRC999. [6] Gilbarg.D and Trudinger.N.S Elliptic Partial Differential Equations of Second Order Springer-Verlag 977. [7] Zang J.W. and Zao J.N. On te Global Existence and Uniqueness of Solutions to Nonstationary Boundary Layer System Science in Cina Ser. A pp [8] E.W.N. Blow up of solutions of te unsteadly Pradtl s equations Comm. pure Appl. Mat. 997 pp [9] Xin Z. and Zang L. On te global existence of solutions to te Pradtl s system Adv. in Mat pp ISSN: Issue 2 Volume 8 December 2009