Asymptotically singular boundary value problems

MATHEMATICAL COMPUTER MODELLING PERGAMON

Mathematical

and Computer

Modelling 32 (2000)

541-548 www.elsevier.nl/locate/mcm

Asymptotically Singular Boundary Value Problems W. KELLEY University of Oklahoma Norman, OK 73019, U.S.A. Abstract-we study two point boundary value problems for second-order differential equations that have no solution if a boundary condtion is imposed at z = 0, but that do have a solution of boundary layer type if the boundary condition is given at z = e, for small, positive values of s. The study is motivated by various models for fluid flow past a small object. @ 2000 Elsevier Science Ltd. All rights reserved. Keywords-Boundary solutions.

layer, Boundary

value problem, Low Reynolds number, Upper and lower

1. INTRODUCTION The study of differential ‘equations with a small parameter

multiplying

the highest derivative,

commonly called singular perturbation problems, has produced a substantial body of information on the occurrence and approximation of solutions with boundary layer, shock layer, spike layer, and other nonuniform behavior. There is another class of problems that has been less studied which also involves solutions that exhibit boundary layers. These problems are motivated by the study of flow past a small (or thin) object, and therefore, incorporate a small parameter in the description of the boundary condition. The monograph by Il’in [1] treats linear elliptic models of this type. Lagerstrom [2] formulated a nonlinear model for flow at low Reynolds number that has received much attention. In this paper, we investigate the question of what boundary value problems for second-order ordinary differential equations with left boundary condition given at a small, positive number c have solutions with a boundary layer at the left boundary. The nonuniformity of convergence of the solutions as e --+ 0 is caused in such problems by a singularity in the equation at 0, hence, the name “asymptotically singular”. Singular boundary value problems have been widely studied in recent years (see, for example, [3,4]), sense complimentary to those since the nonexistence of solutions to the singular We distinguish two types of problems. model and is characterized

but the problems considered in this paper are ina real appearance of a boundary layer is associated with the problem. The first, discussed in Section 2, includes the Lagerstrom

by a singularity in the first derivative term in the differential equation.

In Section 3, we analyze a second type, where the singularity occurs in a term involving the unknown solution. For both types, it is possible to have solutions that are bounded or unbounded in the boundary layer as e + 0. A number of examples are given to illustrate the theory. 0895-7177/00/$ - see front matter PII: SO8957177(00)00151-5

@ 2000 Elsevier Science Ltd.

All rights reserved.

Typeset

by k&W

W. KELLEY

542

2. QUASILINEAR The original Lagerstrom

EQUATIONS

model for flow past an object at low Reynolds number is Y” + kY) + cyy’ + b (y’)2 = 0, Y(E) = 0,

0)

Y(W) = 1,

(2)

where k 2 1, c > 0, b 2 0. There have been a variety of formal and rigorous studies of (l),(2) (see [5-g]). In particular, Rosenblath and Shephard [7] showed that 1_

J,” sekews ds $p”s-kc-s ds

(3)

is an asymptotic approximation of the solution of (l),(2), for small E > 0, if c = 1 and b = 0. The graph of the approximation reveals a boundary layer profile for small E > 0. We are interested in finding a class of boundary value problems with similar behavior. Our approach will be to define comparison functions suggested by the expression in equation (3) so that existence of such solutions can be established by the method of upper and lower solutions. Let us first consider problems given on a finite interval Y” + f(x:, Y>Y’+ Sk, Y) + 0, Y) (Y’j2 = 0, y(1) = B. Y(C) = A, THEOREM 1. Assume

(8

the functions

(4) (5)

the following:

f, g,

and h are continuous,

and h,

fv,gv,and h, are bounded for 0 < x 5 1

and y between A and B; (ii) there is a solution z(x) of (4) such that z(1) = B, lim,,oz(e) bounded on (0, 11;

# A, and z and z’ are

(iii) for 0 < x L 1, f(x, Z(X)) 2 k/x, k 2 1; (iv) forO0,~dforO
of z(x), fv(x, y)z’ + gv(x, y) + h,(x, y)(.z’)2 5 0.

Then for small E > 0, (4),(5) has a solution y(x,~) so that y(x,e)

+ z(x) for 0 < x 5 1 as

E ---)0. PROOF. We consider only the case Z(C) > A for small e since the other case is similar. Then z itself is an upper solution for (4),(5) on [E,11, so to obtain the desired solution, we need only produce a lower solution CX(Z,C)< z(z), e I x I 1. Define

s,’S-&(s+ l)O ds a(x, 6) = z(x) - (Z(E) - A) s,’ s_&(s + 11~ds = ‘(x) - 4(x9 ‘)’ where a > 0 is to be chosen below. Note that CY(C,E)= A, a(l,e) = B, and CY(X,E)+ Z(X) for each 0 < x < 1 as c -+ 0. Also, c$(x,E) has the following properties: +(x,E) > 0, c#‘(z,E) < 0, and

for e I x 5 1. To show that a satisfies the requisite differential inequality, we compute a” + f (x, (~)a’ + g(x, a) + h(x, a)@~‘)~ =.Z ” - 4” + [f(x, z> - fy(x, *Ml (2’ - 4’) + g(x, 2) - gv(x, *M + [h(x:, 4 - &(x, *>$I (~‘1~ + h(x, 4 = -4”

- f (xv 44’

[(#I2 - 2z’d’]

- fp(x, *)4 (z’ - 4’) - gv(x, *MJ- h,(x, *I4 (z’)~ - 2h(x, a)&‘,

(6)

Boundary Value Problems

where we used the facts that

equation

(6) is greater

-&o/

than

z solves (4) and h(z,cr)

543

> 0, and * is between

t and cr. Since

or equal to

- f&r> *)dJ (r’ - 9’) - s&,

*)d - h,(G

*)d(z’)2

- 2/&(X,a)&?.

(7)

Choose M > sup {lfY(z,y)I and a = 4M.

Then

Iz(e) - Al + 2 Ih(z, y)l Iz’(z)l : 0 < x I 1, y between

(7) is greater

than

or equal to

-M@ - 9 First,

consider

the interval

A and B)

[r,t,+g,+h, (q2].

c 5 5 2 ,/?. Since (x + I)V

4’ = -

zk s,’ s-~(s

+ l)a ds

and l a-k@ + l)Ods

xk

+ 0

J2 as 2 -+ 0, (7) is greater

than

4

or equal to

M(x + l)Q xk s,’ s-~(s

for small

-

-

+ 1)“ds

&’

gy

1

2 0

11, (t’)2

6, if e 5 x 5 &.

Next, suppose

that

fi

5 x I 1. For small 6, Hypothesis

(iv) implies

that

f&T *)z’ + &Ax, *) + h&3 *) (z’j25 0 on this interval,

We conclude that theorem follows.

so (7) is greater

than or equal to

a is a lower solution

for small

positive

values of 6, and the conclusion

of the

For the sake of simplicity, we have not attempted to state the most general possible hypotheses in Theorem 1. The advantage of having an explicit comparison function is that we can attempt to apply the procedure to problems that do not satisfy one or more hypotheses. In the following elementary example, our approach succeeds, even though Hypothesis (iv) is violated. EXAMPLE

1. Consider

Bessel’s equation

of order 0

y” + Y(f) =

0,

;yr+y=

0,

y(1)

= B # 0.

W. KELLEY

544

Here, we have f(z) = l/z, g(y) = y. We cannot apply Theorem 1 directly since gy = 1. Define z(z) = BJe(z)/Jc(l). If B > 0, then z(z) > 0 for 0 5 z I 1. As in the proof of Theorem 1, set

s,’s-l(s + 1)" ds s,’s-l(s + 1)a ds

a(x, e) = Z(X) - Z(E)

= z- +

From (G),(7),

We need only find a value of a that makes the expression in brackets positive. Choose a = 1 -(x !‘i)m

- ’ = ~(1 _ ,‘_ 1n.r) - ’ ’ O’

for 0 < x < 1. Then z, a, is an upper, lower solution pair on (e, 11, so there is a solution of the problem with a boundary layer at 2 = E (which is easily calculated in terms of Bessel functions). It is interesting to observe that the ample 1 was possible because Bessel’s solutions that are unbounded at zero. problem is unbounded in the boundary

traditional type of boundary layer that we found in Exequation has solutions that are bounded at 0, as well as In the next example, the solution of the boundary value layer as E + 0.

EXAMPLE 2. y” +

iyl- ;y2 = 2,

Y(C)= 0,

y(1) = 1.

Note that z(z) = l/z is a solution of the differential equation, and we cannot apply Theorem because I is unbounded. Nevertheless, we define as before a=---

1

1 s,’ s-~(s -t- 1)” ds E --I

2

c&+(1

+s)ads

2

4

Then we compute as in the proof of Theorem 1 4

a” + ;a’ -

a $+ 3>;E7im’+;($4). x3 -

We can choose a = 0, since we then have

for E 5 z 5 1, and cy is a lower solution. It follows that the boundary value problem has a solution for all 0 < E < 1, so that 1 2 1 - x3 1 ---2

x3 1_63

IY(X~+s;1

for e 5 2 5 1. Finally, observe that Theorem 1 applies directly to the Lagerstrom model on a finite interval y” + iy’

+ cyy’ + b (Y’)~ = 0,

y(e) = A,

y(l) = B,

by choosing Z(X) = B, if B 2 0, B # A, c 1 0, and b 2 0. The next theorem generalizes the Lagerstrom model on the semi-infinite interval.

1

545

Boundary Value Problems THEOREM 2. Assume the following:

(i) the functions f, g, and h are continuous, class C1 in y for x > 0, y between A and B, and far has bounded first derivative with respect to y as x -+ 0; (ii) far(x, y) has at most polynomial growth in x for y near B when x + co; (iii) g(x, B) 5 0, B # A; (iv) f(x, B) 2 k/x + Z/xp f or some k 2 1, 1 > 0, 0 5 p < 1, and all x > 0; (v) h(x, y) 2 0, gV(x, y) 5 0, for x > 0 and y near B. For small c > 0, (4) has a solution y(x, E) so that y(e, E) = A, y(z, e) + B as x -

oo and as E 4 0

for each x > 0. PROOF. We consider only the case B > A.

As in the proof of Theorem

1, we define a lower

solution, in this case, J,”

cr(x,~)=B-(B-A)

s-k(s

+

l)ae+P)-‘d-‘)

&

Jp” s-k(s + l)ae-l(l-p)-‘.+~‘)

ds

E B - 4(x, E).

Since p > 0, the improper integrals converge, and we have 4 > 0, r$’ < 0, cr(0, e) = A, a x --+ 00 and as E + 0 for x > 0. Also, 4 satisfies

A computation like that in the proof of Theorem is at least as big as

B as

1 yields that Q” + f(x, a)a’ t g(x,a) +

h(x,(r)(cr’)2

-&4’ +f&Y

(8)

*M# - s/(x, *M.

Choose A
M>sup{I&(x,y)1(B-A):O
As in the proof of Theorem 1, (8) is nonnegative for E < x 5 & if a 2 4M. Next, consider the values 4 2 x 5 1. Since gY 5 0, we have that (8) is greater than or equal to

#+

(-STf&, -I-

*M

2 0 >

provided only that a 2 2M. Finally, let x 2 1. Choose

iv = sup {I&(x, y)(B - A)(x + l)a-’ lm(,

+ l)as-ke-‘(l-p)-‘s”-P’

dsl : z 2 1, A 5 y I B}

.

Then (8) is at least as big as

1 - se=+ +

iv l)nS-ke-l(l-p)-ls(‘-p)

ds

1

20

for sufficiently small values of 6 Consequently, if a > 4M and e is small, & is a lower solution, B is an upper solution, so the boundary value problem has a solution with the stated properties.

W. KELLEX

546

3. SEMILINEAR

EQUATIONS

In this section, we consider a family of semilinear equations with singularities of a different type. They will have solutions with boundary layer behavior near z = 0 for much the same reason as the equations in Section 2: the nonexistence of a solution to the singular problem at x = 0. Here is an elementary example. EXAMPLE 3. Y” --$y+$=o, Y(E) = 0,

Y(1) = 1.

The constant function z(x) = 1 satisfies both the differential equation and the right boundary condition. The exact solution of the problem is

E)=

Y(X,

&

(x2- x-l) + 1

and Km,,0 y(z, E) = Z(Z), if 0 < x 5 1, so the solution does have a boundary layer near z = 0. More generally, we will study Y” + 9(x, Y) =

Y(E)

= A,

(9)

0,

y(1) = B.

(10)

THEOREM 3. Assume the following: (i) equation (9) has a solution e with z(1) = B, iim,,o z(x) # A, and z bounded on (O,l]; (ii) the function g(s, Y) is continuous, C1 with respect to y, and gv(x, y) 5 --ICxVa, k > 0, a~2,for(z,y)~D~((z,y):O<~~d,YbetweenAandB}U{(z,Y):d~z~l, Jy - z(z)1 5 d}, where d is some positive number. Then for sufficiently small values of E, (9),(10), h as a solution y(z, e) so that y(z, E) + z(z) as e+OforeachO A for E near 0. Define a(x, e) = z(x,~).-- (z(E,E)- A)

,-A .I.”Q”2(8) deE 2(x,

E) -

c#(x, e) 1

The form of a here was suggested by where q(x) = k/ xa and X > 0 is to be determined. We have (Y(E,E) = A, ar(1, E) L B, so if cy satisfies the the Liouville-Green approximation. requisite differential inequality, then Q, z, are a lower, upper solution pair, and we have a solution of (9),(10). Now A)

(z(e)

_

(Z(E)

_ A)

$(x,e) =

( ~)~‘4e-(2X~/(a-2))(c(2-a)‘2-r(l-a)‘2),

if a >

( ;)(2hfi-1)‘2 ,

ifa=2.

Then 4(x, E) --f 0 as e ---)0 for x > 0 if a > 2, or if a = 2 and 2x4 layer behavior. For some * between z(x) and ~(2, E), we have

2,

> 1, so that we have boundary

cr” + g(x, f-X)= z” - #J” + 9(x, z) + 9&l *>(-4)

= (z(e) - A)

( g)1’4e-X.t

q”2 [ $

+ (1 -

X2) q - g]

.

Boundary Value Problems

547

To obtain the desired differential inequality, we need the expression in brackets to be nonnegative for e 5 x 5 1. With q = k/xa, the required condition is a

x2> k > 0 + (1 - ,&.a --

(

>

l-;

G

for E _< x < 1. First, assume a = 2. Recall that we need X > (4k)(-1/2).

Choose A = dw.

(11) Then (11)

is satisfied 2 1 -4x2 0 2

+ (I - (I+

(4Vi))

k = 0

X2

The next case is a(a - 4) < 16k, a > 2. Since there is a X so that 16 (1 -X2)

lc > a~~-~(a - 4),

c < IC< 1, (11) also holds in this case. Finally, suppose that a(a - 4) 2 16k. We need to modify the lower solution: define -y = a - (fJ- C(E), where C(e) > 0 is to be determined. Then

G

y” + g(x, 7) = 4

(1 - ;> + (l -x;2)

[

Ic + -$C(t). I

(12)

Since a > 2, the expression in brackets in equation (12) will be positive if 0 < X < 1 and E is small for E < x < &

For fi

I x 5 c, 4 is exponentially small, and (12) is positive provided

that 1 > C(E) > +(z,E) as E + 0. The result of Theorem 3 is also true if we assume g is of class C2, gy(x, z(x)) < -k/xa, a > 2, and gyy(x, y) is bounded below in D.

k > 0,

EXAMPLE 4.

y” -

$ (y - Py3)

Y(C) =

A

~(1)

= 0,

= B,

where 1 and b are positive constants. We have three constant candidates for the function z in Theorem 3: z = 0, &b-l. Now in the notation of Theorem 3,

9,=-&(1_3b2 2 Y)i-$, if k 5 l(1 - 3b2y2). We need k > 0, so we must have

for y between A and B. Consequently, the only choice of z that will satisfy the hypotheses of Theorem 3 is z = 0. If E is sufficiently small and if IAl < l/(&b), B = 0, the boundary value problem has a solution Y(Z,E) such that y(x,c) -,Oase-+OforeachO
548

W. KELLEY

EXAMPLE 5.

k/!,

y”_by2=

~(4 = A,

Y(l) = 1,

where b > 0. Here, z(x) = 1/x2 satisfies the differential equation, but we cannot apply Theorem 3 since z is unbounded. Define, as in the proof of Theorem 3, cX(x,t) = $-($A)

(;)(2*fi-1)‘2~z-#,

where E is small enough that l/e2 > A. In order that 4(x, E) + 0 as e + 0, we need X > 5/(2&%). Now (y” - ba2 - y

= -4” =

+

$ ,$ -

b@

-$ ; +(1 -X2)

2b - br$2 I

if f + (1 - X2) 2b 2 b (1 - Ae2) ,

(13)

since b(1 - Ac2) 2 x2bq5, if 0 < x 5 1. We have equality in (13) by choosing

if b > 6 and IZis small. Thus, the boundary value problem has a solution 3/(x, e) -+ 1/x2 as 6 -+ 0, x > 0, if e is small and b > 6. No such solution can exist if b L 6, because solutions in this case must be concave up. Our last theorem is a version of Theorem 3 for the semi-infinite interval. THEOREM 4. Assume that g(z, B) = 0 for zc 3 0, B # A, g(x, y) is continuous, C1 in y, and g,(x, y) 5 -kxda for (x, y) E {(x,y) : 0 < x 5 d, y between A and B) U {(x, y) : d 5 x < 1, fj_/- B/ 5 d), d > 0. If e > 0 is small, then (9) has a solution y(x,~), so that y(0,~) = A,

~(x,~)+Basx+ooandas~+Oforeachx>O. PROOF. If B > A, then we define o(x, E) = B - I?(x)r$(x, E) - C(E), where 4 and C are as in the proof of Theorem 3 and I’ is a C2 cutoff function having the value 1 for c I x 5 b and the value 0 for b 5 x, for some positive b. An argument like that of Theorem 3 (applied to each of the intervals [E,b], [b, 2b], and [b, 00 )) s h ows that for an appropriate choice of b and for E small, a is a lower solution, and the theorem follows.

REFERENCES A.M. Ii’in, Matching of Asmptotic Expansions of Soiutions of Boundary Value Problems, AMS‘Translations of Mathematical Monographs, Providence, RI, (1992). 2. P.A. Lagerstrom, Matched Asymptotic Expansions, Springer-Verlag, New York, (1988). 3. D. O’Regan, Theory of Singular Boundary Value Problems, World Scientiilc, Singapore, (1994). 4. D. O’Regan, Existence Theory for Nonlinear Ordinary Differential Equations, Kluwer, Dordrecht, (1997). 5. W.B. Bush, On the Lagerstrom mathematical model for viscous flow at low Reynolds number, SIAM J. Appl. 1.

Math. 20, 279-287, (1971).

6. G.C. Bsaio, Singular perturbations for a no&near differential equation with a smah parameter, SIAM .I. Math. Anal. 4, 283-301, (1973).

7. S. Rosenblat and 3. Shepherd, On the asymptotic solution of the Lagerstrom model equation, SIAM J. Appl. Math. 29, 110-120, (1975).

8. A.D. MscGillivray, On a model equation of Lagerstrom, SIAM J. Appl. Math. 34, 804-812, (1978). 9. J. Kevorkian and J.D. Cole, Multiple Scnle and Singular Perturbation Methods, Springer-Verlag, New York, (1996).