A Singular Nonlinear Differential Equation Arising in the Homann Flow

JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS ARTICLE NO.

212, 443]451 Ž1997.

AY975516

A Singular Nonlinear Differential Equation Arising in the Homann Flow Jun Yong Shin Department of Applied Mathematics, Pukyong National Uni¨ ersity, Pusan, 608-737, Korea Submitted by Harlan W. Stech Received May 22, 1995

The existence and uniqueness of a positive solution of the singular nonlinear boundary value problem y 2 y0 y 12 Ž 1 y x 2 . y9 s 0, y9 Ž 0 . s y 21

0 - x - 1, y Ž 1 . s 0,

and

which arises in the Homann flow are studied.

Q 1997 Academic Press

1. INTRODUCTION In the axisymmetric stagnation flow, or Homann flow w10, 11x, the Navier]Stokes equations can be reduced to a single ordinary differential equation for the function f Žh . of the form f - q ff 0 q

1 2

Ž 1 y f 92 . s 0,

Ž 1.1.

with boundary conditions at h s 0,

f s f9 s 0 f9 ª 1

at h ª `.

When there is no third term on the left side of Ž1.1., the boundary value problem f - q ff 0 s 0,

Ž 1.2.

443 0022-247Xr97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.

444

JUN YONG SHIN

with boundary conditions f s f9 s 0 f9 ª 1

at h s 0, at h ª `,

arises in the boundary layer flow and is well known as the classical Blasius system. Letting y s f 0 Žh . and x s f 9Žh ., Ž1.2. can be transformed into the singular nonlinear boundary value problem yy0 q x s 0, 0 - x - 1, y9 Ž 0 . s 0 and y Ž 1 . s 0,

Ž 1.3.

which has been studied by many authors. For a detailed analysis of Ž1.3. or a general form of Ž1.3. one is referred to w1, 2, 5, 8, 9x. However, by letting y s f 0 Žh . and x s f 9Žh ., Ž1.1. can be written as the singular nonlinear boundary value problem y 2 y0 y 12 Ž 1 y x 2 . y9 s 0, y9 Ž 0 . s y

1 2

and

0 - x - 1, y Ž 1 . s 0,

Ž 1.4.

which is very different from Ž1.2. in form because the derivative of y is given. It is not known yet whether Ž1.4. is discussed or not. Problem Ž1.4. has significance in studying a singular nonlinear boundary value problem in the following senses: Ži. Problem Ž1.4. can be used as a model example to study a singular nonlinear boundary value problem; Žii. Problems Ž1.3. and Ž1.4. can be used together as model examples to study a singular nonlinear boundary value problem. The object of this paper is to establish the existence and uniqueness of a positive solution of Ž1.4.. To show the existence of a positive solution of Ž1.4., we will use the constructive method such that the method of upper and lower solutions can be applied. This constructive method can be well adopted to show the existence of a positive solution of singular nonlinear boundary value problems arising in several fields Žsee w4, 5, 7x for examples.. We call a function a g C 2 w0, 1x a positive lower solution of Ž1.4. if

a)0

on Ž 0, 1 . ,

a 2a 0 y 12 Ž 1 y x 2 . a 9 G 0 a 9 Ž 0. G y , 1 2

and

on Ž 0, 1 . ,

a Ž 1 . F 0.

445

SINGULAR NONLINEAR EQUATIONS

By reversing the above inequality signs, a positive upper solution of Ž1.4. is defined. Also similar definitions hold for positive lower and positive upper solutions of a perturbation of Ž1.4. which will be given in Section 2. We call a function y g C w0, 1x l C 2 w0, 1. a positive solution of Ž1.4. if y)0

on Ž 0, 1 . ,

y y0 y Ž 1 y x . y9 s 0 1 2

2

y9 Ž 0 . s y 12 ,

2

and

on Ž 0, 1 . , y Ž 1 . s 0.

2. EXISTENCE OF A UNIQUE POSITIVE SOLUTION For each positive number m, we consider the nonlinear boundary value problem y 2 y0 y 12 Ž 1 y x 2 . y9 s 0, y9 Ž 0 . s y 12

and

0 - x - 1, 1 y Ž 1. s , m

Ž 2.1. m

which may be viewed as a perturbation of Ž1.4.. To prove the existence of a positive solution of Ž1.4., we establish the existence of a positive solution of Ž2.1. m . LEMMA 2.1. Ž2.1. m .

y l m s Ž1r2.Ž1 y x . q 1rm is a positi¨ e lower solution of

Proof. It is clear that y l m Ž x . ) 0 on Ž 0, 1 . ,

yXl m Ž 0 . G y 12 ,

yl m Ž 1. F

1 m

,

and y l2m yYl m y 12 Ž 1 y x 2 . yXl m G 0,

0 - x - 1.

Thus y l m is a positive lower solution of Ž2.1. m . LEMMA 2.2. yu m s 2 1 y x q 1rm is a positi¨ e upper solution of Ž2.1. m , for each m G 1.

'

446

JUN YONG SHIN

Proof. It is clear that yuX m Ž 0 . s y

yu m Ž x . ) 0 on Ž 0, 1 . ,

1

'1 q 1rm

Fy

1 2

yu m Ž 1 . G

,

1 m

,

and 1

yu2m yuY m y

2 1

sy

2

y s

1 2

ž

1 2

Ž 1 y x 2 . yuX m

ž

1yxq

m

y3 r2

?4? 1yxq

/

ž

Ž 1 y x . Ž y1. 1 y x q

ž

2

1yxq

F 0,

1

1 m

1 m

1 m

/

y1 r2

/

y1 r2

/

Ž y3 y x 2 .

0 - x - 1.

Thus yu m is a positive upper solution of Ž2.1. m . The following result is clear from an application of Schauder’s Fixed Point Theorem w6x. LEMMA 2.3. For any m G 1, there exists a positi¨ e solution ym g C 2 w0, 1x of the problem Ž2.1. m such that y l m Ž x . F ym Ž x . F yu m Ž x .

on w 0, 1 x ,

where y l m and yu m are gi¨ en in Lemma 2.1 and Lemma 2.2. LEMMA 2.4.

If ym is a positi¨ e solution of Ž2.1. m for m G 1, then X ym Ž x. - 0

on Ž 0, 1 x .

Proof. It is clear that Y X X ym2 ym ym s 12 Ž 1 y x 2 . Ž ym . G0 2

X 2 X X . is increasing. Since ym Ž0. - 0, ym Ž x . is decreasing and hence and so Ž ym X ym Ž x. - 0

which completes the proof.

on Ž 0, 1 x ,

447

SINGULAR NONLINEAR EQUATIONS

If y 1 and y 2 are two positi¨ e solutions of Ž2.1. m for m G 1,

LEMMA 2.5. then y 1 ' y 2 .

Proof. Let y 1 and y 2 be positive solutions of Ž2.1. m . Suppose that there exists an h g Ž0, 1x such that y 1Žh . / y 2 Žh .. If y 1Ž0. s y 2 Ž0., then by the uniqueness theorem of the initial value problem, y 1 ' y 2 , which is a contradiction. Therefore, without loss of generality, we may assume that y 1Ž0. - y 2 Ž0.. Since y 1Ž1. s y 2 Ž1., by the continuity of y 1 y y 2 , there exists a j g Ž0, 1x such that y1Ž j . s y 2 Ž j .

y 1 Ž x . - y 2 Ž x . on 0, j . .

and

So the function w Ž x . s y 2 Ž x . y y 1Ž x . has a maximum at x s 0 or at the interior of w0, j x. If the function w takes it maximum at x s 0, then we obtain

w 0 Ž 0 . ) 0, which implies that w 9Ž x . ) w 9Ž0. s 0 near x s 0 and so w Ž x . ) w Ž0. near x s 0. This is a contradiction. If the function w takes it maximum at the interior of w0, j x, then there exists a j 1 g Ž0, j . such that w 9Ž j 1 . s 0 and w 0 Ž j 1 . F 0. But since y 1Ž j 1 . - y 2 Ž j 1 . and yX1Ž j 1 . - 0, we obtain 1

1

1 y j 12 . yX1 Ž j 1 . Ž 2

w0 Ž j1. s

y 22

Ž j1.

y

1 y 12

Ž j1.

) 0,

which leads to a contradiction. Thus we complete the proof. Note. If ym is a positive solution of Ž2.1. m for each m G 1, then from Lemma 2.3 and Lemma 2.5 we obtain ymŽ x . G 12 Ž1 y x . on w0, 1x. LEMMA 2.6. obtain

If ym is a positi¨ e solution of Ž2.1. m for m G 1, then we

X ym Gy

1 2

exp

ž

x

H0

2Ž 1 q s . 1ys

ds

/

on w0, 1.. Proof. Since ym is a positive solution of Ž2.1. m , by Lemma 2.3, Lemma 2.5, and Lemma 2.6, we obtain 0s

Y ym

y

1 y x2 1 2

yX ym2 m

F

Y ym

y

1 y x2 1 2

y l2m

X ym .

448

JUN YONG SHIN

Hence we have

ž

1 y s2

x

0 F exp y

H0

1

2

Ž yl m Ž s . .

X ds ym 9,

/

2

which implies X ym Gy

Gy

1 2 1 2

x

exp

ž

exp

žH

H0 x

1 y s2

1

2

Ž yl m Ž s . .

2Ž 1 q s . 1ys

0

ds

/

2

ds

/

on 0, 1 . ,

because y l mŽ s . s Ž1r2.Ž1 y s . q 1rm G Ž1r2.Ž1 y s . on w0, 1.. This completes the proof. LEMMA 2.7. If m1 ) m 2 G 1 and ym 1 and ym 2 are positi¨ e solutions of Ž2.1. m and Ž2.1. m , respecti¨ ely, then we ha¨ e 1 2 0 - ym 1Ž x . - ym 2Ž x . on 0, 1 .

X ym Ž x . - ymX 2Ž x . - 0 on Ž 0, 1 x . 1

and

Proof. This is clear from the fact that ym 2 is a positive upper solution of Ž2.1. m 1 and so on w 0, 1 x .

ym 1Ž x . F ym 2Ž x .

If ym 1Ž0. s ym 2Ž0., then by the uniqueness theorem of the initial value problem, we obtain ym 1Ž x . ' ym 2Ž x . , which is a contradiction. So we may assume that 0 - ym 1Ž0. - ym 2Ž0.. Then we have Y ym Ž 0 . y ymY 1Ž 0 . s y 2

1 4

1

Ž ym Ž 0 . .

2

y

2

1

Ž ym Ž 0 . .

2

) 0,

1

which implies that X ym Ž x . y ymX 1Ž x . ) ymX 2Ž 0 . y ymX 1Ž 0 . s 0, 2

ym 2Ž x . y ym 1Ž x . ) ym 2Ž 0 . y ym 1Ž 0 . ) 0,

for x near to 0.

If there exists a j g Ž0, 1x such that X ym Ž j . y ymX 1Ž j . s 0 2

and

X ym Ž x . y ymX 1Ž x . ) 0, 0 - x - j , 2

449

SINGULAR NONLINEAR EQUATIONS

then we obtain Y ym Ž j . y ymY 1Ž j . s 2

Ž1 y j 2 .

X ym Žj. 1

2

1

Ž ym Ž j . .

2

y

2

1

) 0,

2

Ž ym Ž j . . 1

X Ž j . - 0 and ym Ž j . y ym Ž j . ) 0. Hence we have since ym 1 2 1 X ym Ž x . y ymX 1Ž x . - ymX 2Ž j . y ymX 1Ž j . s 0, 2

0-x-j,

which is a contradiction. This completes the proof. THEOREM 2.8 ŽExistence .. If ym is the positi¨ e solution of Ž2.1. m for each m s 1, 2, 3, . . . , then the sequence  ym 4 con¨ erges to a positi¨ e solution y of Ž1.4.. Proof. To prove this theorem, we prove the following steps: Step 1. Step 2. Step 3.

ym ª y as m ª `. y g C w0, 1x l C 2 w0, 1.. y is a positive solution of Ž1.4..

Our first step is to show that ym ª y as m ª `. From Lemma 2.3 and Lemma 2.7, we know that the sequence  ym 4 is monotone decreasing in m and bounded below by 12 Ž1 y x .. Therefore, ym ª y as m ª `

y Ž x . G 21 Ž 1 y x . on w 0, 1 x .

and

X 4 is Also from Lemma 2.6 and Lemma 2.7, we know that the sequence  ym

monotone decreasing in m and bounded below by yŽ1r2.wexpŽ

x

H0 Ž2Ž1 q

s .rŽ1 y s .. ds .x on w0, 1.. Therefore X ym ª y9

as m ª `

and y9 Ž x . G y 12 exp

Y ym

ž

x

H0

1 y s2

1

2

Ž yl m Ž x . .

2

ds

/

on 0, 1 . .

Our second step is to show that y g C w0, 1x l C 2 w0, 1.. If we integrate from 0 to x, then we have X ym Ž x. s y

1 2

q

x

H0

X 1 Ž 1 y j 2 . ym Žj.

2

Ž ym Ž j . .

2

dj .

Ž 2.2.

450

JUN YONG SHIN

If we integrate both sides of Ž2.2. from 0 to x, then we obtain ym Ž x . y ym Ž 0 . sy

sy

x 2 x 2

q

x

s

H0 H0

Ž ym Ž j . .

2

2

X 1 Ž 1 y j 2 . ym Žj.

x

qx

X 1 Ž 1 y j 2 . ym Žj.

H0

Ž ym Ž j . .

2

2

d j ds

dj y

x

H0

X j Ž 1 y j 2 . ym Žj.

2

Ž ym Ž j . .

dj .

2

Ž 2.3. If we let m ª ` in both sides of Ž2.2. and Ž2.3., then by Lebesgue’s Dominated Convergence Theorem, we obtain y9 Ž x . s y

1 2

q

x

H0

1 Ž 1 y j 2 . y9 Ž j .

Ž yŽ j . .

2

2

dj

and y Ž x . y y Ž 0. sy

x 2

qx

H0

x

1 Ž 1 y j 2 . y9 Ž j . 2

Ž yŽ j . .

2

dj y

x

H0

j Ž 1 y j 2 . y9 Ž j . 2

Ž yŽ j . .

2

dj ,

Ž 2.4. which implies y g C 2 w0, 1.. Since y Ž x . converges to 0 as x approaches 1, y is continuous at 1, which implies y g C w0, 1x l C 2 w0, 1.. Finally, we will show that y is a positive solution of Ž1.4.. It is clear that y9Ž0. s y 12 and y Ž1. s 0. If we take the second derivative of both sides of Ž2.4., then we get y0 Ž x . s

1 Ž 1 y x 2 . y9 Ž x . 2

Ž yŽ x. .

2

which implies that y is a positive solution of Ž1.4.. THEOREM 2.9 ŽUniqueness.. of Ž1.4.. Then y 1 ' y 2 .

Assume that y 1 and y 2 are posit¨ e solutions

Proof. The proof of this theorem is similar to that of Lemma 2.5.

ACKNOWLEDGMENTS The author thanks the referees for their careful reading of the manuscript and useful comments.

SINGULAR NONLINEAR EQUATIONS

451

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