三维轴对称驻点流系统的一些研究
2011-06-29张明捷党龙飞徐艳芝
张明捷, 党龙飞, 徐艳芝
(成都信息工程学院数学学院,四川成都610225)
1 Introduction
In this paper,we are concerned with the following third order differential equations arising in the boundary layer theory
with boundary conditions
Which has been used to describe the 3D axisymmetric inviscid stagnation flow[1,2].A solution of(1)~(3)is called a similarity solution.The two-dimensional case,λ=g=0,was solved by Hiemenz[3].The axisymmetric stagnation flow towards a plate,λ=1 and g=f,was solved by Homann[4].Howarth[5]studied the case 0<λ<1 which can be applied to the stagnation region of an ellipsoid.Davey[6]investigated the stagnation region near a saddle point(-1<λ<0).For λ≤-1 the vorticity generated is not confined in the boundary layer and the existence of solutions cannot be shown[2].Up to now,there is a little study on the case of λ≤-1.
Utilizing the integral methods[7,8],Du Hauang and Zhang[9]presented a system of two integral equations on the case of λ<0 and obtained some properties and a non-existence result of(1)~ (3)on the case of λ<-4.
In this paper,we shall study(1)~ (3)for the case of λ∈ R and λ≠0,we shall obtain further results on(1)~(3)and a new non-existence result on the case of λ≤-1.
2 Positive solutions of a system of two integral equations
In[9],Du,Hauang,Zhang presented a system of two integral equations
Where G0,1(t,s)denotes the Green function for u″(t)=0 with u(0)=0 and u(1)=0 defined by
They obtained the following results:
(i)For λ<0(1)~(3)has a solution in∑if and only if(4)~(5)has a positive solution
(ii)For λ<-4(1)~(3)has no solution in ∑.
In this paper,let
and
Theorem1 If(λ,z,w)∈(-∞,+∞)×Q is a solution of(4)~(5),then
(iii)w(0)=0,w(1)=1 and w(t)≥-1 for t∈[0,1]if λ<0 and w(t)≤1 for t∈[0,1]if λ>0.
(i)Firstly,we prove z(t)=0.
If z(1)≠0,then z(t)≠0,t∈[0,1].
By the continuity of z(t),then m=min{z(t),t∈[0,1]}>0.From this and λ>0,we obtain by(4)
This implies
a contradiction.Hence,(i)holds.
Since z(t)>0 for t∈(0,1),we have
Integrating this inequality from 0 to ξ,we have
From this we obtain
Letting ξ→1-in the last inequality,we have,which contradicts∞.Thus,(ii)holds.
(iii)Letting t=0 and t=1 in(5),we have w(0)=0 and w(1)=1.
If λ>0 and there exist t0∈[0,1]such that w(t0)>1.Since w(0)=0,and w(1)=1,there must be exist t*∈(0,1)such that w(t*)=max{w(t):t∈[0,1]}and w(t*)>1.
By(5),we have
Reamrk1 Since we do not assume λ<0 and Q ⊆Q*,hence Theorem 1 in[9]is improved.
Utilizing the solutions of(z,w)∈Q*,we may construct the solutions of(1)~(3)and then the use of(4)~(5)is expanded.
Theorem2 If(z,w)∈Q*is a solution of(4)~(5),then(1)~(3)has a solution(f,g).
Proof.Let(z,w)∈Q*be a solution of(4)~(5).By Theorem 1(ii),we have
Let
Then η(t)is strictly increasing on[0,1)and
Let t=h(η)be the inverse function to η=η(t),we define the function
Then
and
From(6),we have
Differentiating(7)with respect to η,we have
Then f″(η)>0 for 0≤η<+∞.
Differentiating(8)with respect to η,we have
Differentiating(4)with respect to t,we have
By setting s=f′(α)and utilizing t=f′(η)and(8),we have
By(8)~(11),we have
Combing(10)and(11),we obtain
This completes the proof.
3 A non-existence result on(1)~(3)
Let
Lemma1 Let λ<0.If(f,g)∈Γis a solution of(1)~(3),then g″(∞)=0.
Proof.Since g′(+∞)=1,we have
Since(1)~(3),we have g‴(0)=-λ,and λ<0.then g‴(0)>0 and g″(η)>0.we have ∃η*,such that g″(η)is increasing on[0,η*].(12)implies that there exists η0∈[η*,+∞)such that g″(η0)<g″(η*).Then we prove that g″(η)is decreasing on(η0,+∞).If there must exist η1,η2∈[ η0,+∞)with η1<η2such that g″(η1)<g″(η2).So let ˜η∈[ η*,η2]such that g″(˜η)=min{g″(η):η∈[ η*,η2]}.This implies g‴(˜η)=0 and g(4)(˜η)≥0.
Differentiating(2)with η,we have
then
By(f,g)∈Γand λ<0,we have g(4)(˜η)≤0,a contradiction.Hence g″(η)is decreasing on[ η0,+∞)and thenexists.By(12),we obtain g″(∞)=0.
Theorem3 If λ≤-1,then(1)~ (3)has no solution in Γ.
Proof.The proof is by a contradiction.If(1)~ (3)has a solution(f,g)in Γ,then g″>0.
Let η:=η(t)=(g′)-1(t)for t∈[0,1)be the inverse function to t=g′(η):[0,∞)→[0,1).It follows that g′is strictly increasing with[0,+∞)and η(t)=(g′)-1(t):[0,1)→[0,∞)with(g′)-1(0)=0
Let x(t)=g″(η)>0 for t∈[0,1),by Lemma 1,x(1)=.This implies that x(t)>0 for t∈[0,1)and x is continuous on[0,1).By Lemma 1,we see that x is continuous from the left at 1.Hence,we have x(t)∈C[0,1]and x(1)=0.
Using the Chain Rule to x(t)=g″(η),we obtainand by the Inverse Function Theorem,we have
This,together with g′(η)=t,implies
Integrating the last equality from 0 to t implies
Let y(t)=f′(η)for t∈[0,1),y(1)=1,then y(t)∈ C[0,1)and 0≤y(t)≤1.Notice thatwe have
Substituting g,g′,g″,g‴and f into(2)implies
Integrating(13)from t to 1,we have
By x(1)=0,then
we have
Reamrk2 Theorem 12 improves non-existence result of(1)~ (3)in[6]from λ<-1 to λ≤-1.
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