Perturbation problems with quadratic dependence on the first derivative

Nonlinear Analysis 51 (2002) 469–486www.elsevier.com/locate/na

Perturbation problems with quadratic dependenceon the %rst derivative

Walter KelleyDepartment of Mathematics, University of Oklahoma, Norman, OK 73019, USA

Received 26 March 2001; accepted 13 June 2001

Dedicated to the memory of Fred Howes

Keywords: Singular perturbation; Boundary value problem; Asymptotic approximation; Quad-ratic nonlinearity

1. Introduction

In this paper, we study the problem of computing approximations of solutions ofsingularly perturbed two-point boundary value problems that feature the square of the%rst derivative. One such equation is

�x′′ = a(t; x)x′2 + b(t; x)x′ + c(t; x);

where � is a small positive constant. Despite the frequent occurrence of such problemsin applications, only a few studies of them have appeared. Chang and Howes, whoprovided an existence theory for some of these problems in [1], stated on p. 90 that“surprisingly little work has been done on the Dirichlet problem” for this type of dif-ferential equation. Perhaps a partial explanation of the lack of an approximation theoryis that the traditional method for verifying approximations, that is, linearization of thenonlinear problem about the approximation followed by a %xed point argument, doesnot work well for the quadratic case. (See for example, [5], where the quadratic termhad to be assumed small in order for the contraction mapping argument to succeed.)Early work on quadratic problems was done by Vasil’eva and her students (see

[3,7,12,13]). One of the features of these problems that distinguishes them from other

E-mail address: [email protected] (W. Kelley).

0362-546X/02/$ - see front matter c© 2002 Elsevier Science Ltd. All rights reserved.PII: S0362 -546X(01)00840 -9

470 W. Kelley / Nonlinear Analysis 51 (2002) 469–486

problems exhibiting boundary layers is that the derivative of the solution in the bound-ary layer can be as much as exponentially large (see also [14]).The approximation theory presented in this paper is divided into three cases. The %rst

case, developed in Section 2, assumes that the quadratic term is a full participant inthe diMerential equation. Here, two diMerent approximations are computed and veri%ed.The simpler one yields an O(� log �) approximation, while a re%nement of the boundarylayer term leads to an O(�) approximation. In the second case, the quadratic problemis considered as a perturbation of a quasilinear problem. It is shown that an O(�)approximation is valid under assumptions similar to those employed for the quasilinearproblem in studies by Coddington and Levinson [2], Van Harten [11], Howes andO’Malley [8], and Smith [10]. Finally, the third case treats the quadratic problem as aperturbation of the semilinear problem, and under conditions similar to those introducedby Fife [6] for the semilinear problem, we are able to verify a formal approximation.In order to keep the discussion to a reasonable length, we have considered only

%rst order approximations, although higher order approximations can be computed andveri%ed using the same methods. Throughout the paper, the method of super andsub-solutions is used to verify the approximations, giving a uni%ed approach to theanalysis of quadratic, quasilinear, and semilinear problems. For each case considered,we illustrate the theory with an application.

2. The full quadratic problem

First, we consider the case where the small parameter (�¿ 0) multiplies only thesecond derivative term:

�x′′ = a(t; x)x′2 + b(t; x)x′ + c(t; x); (1)

x(0)=A; x(1)=B: (2)

Let w(t)∈C2[0; 1] solve the reduced problem:

a(t; w)w′2 + b(t; w)w′ + c(t; w)= 0; (3)

w(1)=B: (4)

We expect that the dominant terms in the boundary layer at t=0 will be �x′′ anda(t; x)x′2, so the signs of these terms must match. For example, if w(0)¡A, then weshould have x′′ ¿ 0, so we need a(t; x)¿ 0. Consequently, our %rst hypothesis is

(A− w(0))a(t; x)¿ |A− w(0)|�¿ 0 (5)

for some constant �, where (t; x) belongs to a domain containing the reduced solutionand the boundary layer region. In the same domain, we impose a stability condition:

2a(t; x)w′(t) + b(t; x)6− k ¡ 0 (6)

for some constant k ¿ 0.

W. Kelley / Nonlinear Analysis 51 (2002) 469–486 471

We will construct a candidate for a %rst approximation of the solution of (1) and(2) by adding a boundary layer correction to the reduced solution. De%ne �(t; �) to bethe solution of

��′′ = a(0; w(0) + �)�′2

�(0)=A− w(0); �(1)= 0: (7)

Now � is given in implicit form by

t=

∫ A−w(0)� e−1=�

∫ �0 a(0;A−w(0)+s) ds d�∫ A−w(0)

0 e−1=�∫ �0 a(0;A−w(0)+s) ds d�

(8)

for 06 t6 1. Eq. (8) can also be written in the equivalent form:

log(1− t)=−∫ A−w(0)

�

ds∫ s0 e1=�

∫ s� a(0;A−w(0)+s) ds d�

(9)

for 06 t ¡ 1.

Theorem 1. Assume that a; b and c are C1 functions and that hypotheses (5) and(6) are satis1ed. For � su3ciently small; the boundary value problem (1) and (2) hasa solution x(t; �) with

x(t; �)− (w(t) + �(t; �))=O(� log �)

as � → 0; uniformly on [0; 1].

Proof. We will consider only the case that A¿w(0) and a(t; x)¿ 0, since the othercase is similar. The method of proof will be the construction of sub- and super-solutions.First, we de%ne a trial super-solution of the form:

�(t; x; �)=w(t) + �(t; �) + D�f(t; �);

where D¿ 0; f¿ 0; f′ ¡ 0 are to be determined and � is a boundary layer functionde%ned by

��′′ = [a(0; w(0) + �)− d�]�′2;

�(0)=A− w(0); �(1)= 0 (10)

and d is a positive constant to be chosen. Now � has the representation (9) with areplaced by a − d�, so for each %xed t, �(t; �)¿�(t; �) and, consequently, �(t; �)¿w(t) + �(t; �) for 06 t6 1.

Next, we need to show that � satis%es the requisite inequality for a super-solution:

��′′ − a(t; �)�′2 − b(t; �)�′ − c(t; �)

= ��′′ − a(0; w(0) + �)�′2 + O(t)�′2 + DfO(�)�′2 − 2aw′�′ − 2aD�f′�′ − b�′

+ �w′′ + [a(t; w)− a(t; w + �+ D�f)]w′2 − a(t; w)w′2

− 2aw′f′D�− b(t; w)w′ + [b(t; w)− b(t; w + �+ D�f)]w′ − c(t; w)


+[c(t; w)− c(t; w + �+ D�f)] + �2Df′′ − aD2�2f′2 − bD�f′

=− d��′2 + O(t)�′2 + DfO(�)�′2 − [2aw′ + b]�′ − 2aD�f′�′ + �2Df′′

− a2w′2D�f − [2aw′ + b]D�f′ − b2w′D�f − c2D�f + O(�) + O(�); (11)

where we have used Eqs. (3) and (10) and the mean value theorem. Note: whereverthe arguments of a and b are not speci%ed, they are to be evaluated at the original point(t; �), and a2; b2, and c2 are the derivatives of the respective functions with respect totheir second arguments evaluated at the intermediate values given by the mean valuetheorem.Now choose M to be an upper bound for |a2w′2 + b2w′ + c2| in the domain where

(5) and (6) are satis%ed, and de%ne

f(t; �)= e−�(1−t) − 12

to be a solution of

�f′′ + kf′ +Mf=− M2; (12)

where �=−M=k + O(�). Then (6) and Eq. (12) imply that (11) is less than

−d��′2 + DfO(�)�′2 + O(t)�′2 + k�′ + O(�)− MD�2

+ O(�): (13)

Now note that

|t�′(t)| =∫ A−w(0)

�e(1=�)

∫ �� a−d� d�

6∫ A−w(0)

�e(d�−�)(�−�)=� d�=O(�):

So the negative term k�′ in (13) dominates the term O(t)�′2 for small �.Now

|�′(t)| = e(1=�)∫ �0 a−d�

∫ A−w(0)

0e−(1=�)

∫ �0 a−d� d�

¿ e�(�−d�)=�C�

for some C ¿ 0. Also, for any constant N ¿ 0, there is a constant E¿ 0 so that

�e�(�−d�)=� ¿N�

for � suNciently small and �¿E�, so the expression k�′ −MD�=2 dominates the termO(�) for large enough D, and we %nally have from (11) and (13) that

��′′ − a(t; �)�′2 − b(t; �)�′ − c(t; �)6 0

for suNciently large d and small �, 06 t6 1:Our sub-solution will have the form

�(t; �)=w(t) + �(t; �)− D�f;


where � is the solution of

��′′ = [a(0; w(0) + �) + d�]�′2 − l�′; (14)

�(0)=A− w(0); �(∞)= 0 (15)

and l and D are positive constants to be determined below. The solution of (14) and(15) is

t=�l

∫ A−w(0)

�

(∫ s

0e(1=�)

∫ s� a+d� d�

)−1

ds (16)

=�llog

[∫ A−w(0)0 e−(1=�)

∫ �0 a+d� d�∫ �

0 e−(1=�)∫ �0 a+d� d�

]: (17)

Now we check the diMerential inequality for a sub-solution:

��′′ − a(t; �)�′2 − b(t; �)�′ − c(t; �)

= ��′′ − a(0; w(0) + �)�′2 + O(t)�′2 + DO(�)f�′2 − 2aw′�′ + 2aD�f′�′

− b�′ + �w′′ − a(t; w)w′2 + [a(t; w)− a(t; w + �− D�f)]w′2 + 2aw′f′D�

− b(t; w)w′ + [b(t; w)− b(t; w + �− D�f)]w′ − c(t; w)

+ [c(t; w)− c(t; w + �− D�f)]− �2Df′′ + bD�f′ + O(�2)

=d��′2 − l�′ + O(t)�′2 + DfO(�)�′2 − [2aw′ + b]�′ + 2aD�f′�′ − �2Df′′

+[a2w′2 + b2w′ + c2]D�f + [2aw′ + b]D�f′ + O(�) + O(�): (18)

Here, we have used the mean value theorem and Eqs. (3) and (14). Next, choose Dand f as in the construction of the super-solution, choose l so that

l+ 2aw′ + b¿ q¿ 0

for some constant q, and choose d so that the term d� dominates the other O(�) termsin Eq. (18). Then (18) yields:

��′′ − a(t; �)�′2 − b(t; �)�′ − c(t; �)

¿− q�′ + O(t)�′2 + O(�): (19)

Now from Eq. (16),

|�′| = l�

∫ �

0e(1=�)

∫ �� a+d� d�

¿l�

∫ �

0e(�+d�)(�−�)=� d�

=l

�+ d�[e�(�+d�)=� − 1]

¿l��


for small values of �. It follows that the positive term −q�′ in (19) dominates theO(�) term.We still need to compare −q�′ to O(t)�′2 in (19). Using (16) and (17), and the

fact that log x6 x − 1 for x¿ 1, we have

|t�′| =∫ �

0e(1=�)

∫ �� a+d� d� log

∫ A−w(0)0 e−(1=�)

∫ �0 a+d� d�∫ �

0 e−(1=�)∫ �0 a+d� d�

6∫ �

0e(1=�)

∫ �� a+d� d�

[∫ A−w(0)0 e−(1=�)

∫ �0 a+d� d�∫ �

0 e−(1=�)∫ �0 a+d� d�

− 1

]

=∫ A−w(0)

�e−(1=�)

∫ �� a+d� d�=O(�):

Consequently, −q�′ also dominates O(t)�′2 for small �, so (18) is positive on [0; 1] if� is small enough.Next, we claim that

w(t) + �(t; �)¿ �(t; �) (20)

for 06 t6 1: Since

− �llog(1− t)6 t

for 06 t6 1− (�=l), a comparison of (9) and (16) yields �(t; �)6�(t; �) for 06 t61−(�=l). For 1−(�=l)6 t6 1, both � and � are O(�), so (20) is true if D is suNcientlylarge.We have shown that �; � is a sub, super-solution pair for (1) and (2) with

�(t; �)6w(t) + �(t; �)6 �(t; �)

for 06 t6 1 and � small. Consequently, there is a solution x(t; �) so that

x(t; �)− (w(t) + �(t; �))6 �(t; �)− �(t; �)

for 06 t6 1. To complete the proof, we need to show

�(t; �)− �(t; �)=O(� log �); (21)

uniformly for 06 t6 1, as � → 0.First, we claim that

�(t; �)− �(t; �)=O(�) (22)

as � → 0, uniformly for 06 t6 1. Using (8), we get

t=

∫ A−w(0)� e−(1=�)

∫ �0 a d�∫ A−w(0)

0 e−(1=�)∫ �0 a d�

=

∫ A−w(0)� e−(1=�)

∫ �0 a−d� d�∫ A−w(0)

0 e−(1=�)∫ �0 a−d� d�

;


from which it follows that

16

∫ A−w(0)� e−(1=�)

∫ �0 a−d� d�∫ A−w(0)

� e−(1=�)∫ �0 a d�

6 ed(A−w(0))

∫ A−w(0)� e−(1=�)

∫ �0 a d�∫ A−w(0)

� e−(1=�)∫ �0 a d�

:

By Laplace’s method (see [9, Chapter 3, Theorem 7:1]) and the preceding inequality,

ed(w(0)−A)6

∫ A−w(0)� e−(1=�)

∫ �0 a d�∫ A−w(0)

� e−(1=�)∫ �0 a d�

∼ a(0; A− w(0) + �)a(0; A− w(0) + �)

e−(1=�)∫ �0 a

e−(1=�)∫ �0 a

as � → 0, so e−(1=�)∫ �� a is bounded away from zero for all small values of �, and we

can conclude that (22) is valid.Finally, we want to show that

�(t; �)− �(t; �)=O(� log �) (23)

as � → 0, uniformly for 06 t6 1. As in the preceding argument, a change in d resultsin an O(�) change in �. Thus it suNces to take d=0 and set (see Eqs. (8) and (17))

t= h(�; �)=− �llog(1− h(�; �));

where

h(�; �) ≡∫ A−w(0)� e−(1=�)

∫ �0 a d�∫ A−w(0)

0 e−(1=�)∫ �0 a d�

:

By the mean value theorem,

− �llog(1− h(�; �))=

�h(�; �)l(1− c)

;

where 0¡c(�; �)¡h(�; �)¡ 1. By Laplace’s method,

�l(1− c)

=h(�; �)h(�; �)

∼ a(0; A− w(0) + �)e−(1=�)∫ �0 a

a(0; A− w(0) + �)e−(1=�)∫ �0 a

as � → 0, so1�e−(1=�)

∫ �� a

is bounded away from 0 for all small values of �, from which (23) follows immediately.From (22) and (23) we obtain (21), and the proof is complete.

By considering the special case a=1; b= − 1, and c=0 for the boundary valueproblem (1) and (2), one can show that the estimate

x(t; �)− (w(t) + �(t; �))=O(� log �)

is the best possible. However, the proof of Theorem 1 suggests that we can obtain anO(�) approximation by letting the x′ term participate in the boundary layer approxima-tion. De%ne (t; �) to be the solution of

� ′′ = a(0; w(0) + ) ′2 − C ′;

(0)=A− w(0); (∞)= 0;


where

C ≡ −2a(0; w(0))w′(0)− b(0; w(0)):

Then has the representation given by Eqs. (16) and (17) with d=0 and l replacedby C.

Theorem 2. With the same hypotheses as Theorem 1; for all su3ciently small valuesof �; (1); and (2) has a solution x(t; �) so that

x(t; �)− (w(t) + (t; �))=O(�)


Proof. Let $¿ 0, and choose K and L so that

L− $¿− (2aw′ + b)¿K + $¿ 0;

where a and b are evaluated over a domain containing the reduced solution and theboundary layer region. The sub-solution is de%ned as in the proof of Theorem 1.

�(t; �)=w(t) + �(t; �)− D�f;

where � satis%es (14) and (15) with l replaced by L. The upper solution has the form:

�(t; �)=w(t) + '(t; �) + D�f;

where ' satis%es:

�'′′ = [a(0; w(0) + ')− d�]'′2 − K'′;

'(0)=A− w(0); '(∞)= 0:

The veri%cation that � and � satisfy the requisite inequalities is similar to that givenfor the lower solution in the proof of Theorem 1, and we can also show that

�(t; �)− �(t; �)=O(�)

using the method of Theorem 1. The proof is complete.

Example 1. Let y(�) be the distance above the ground (measured in feet) of a fallingobject at time � seconds. At time �=0 the object is observed to be a distance Habove the ground, and at time �=T the object strikes the ground. If we take intoaccount the variable force of gravity and the variable force due to air resistance, thenunder appropriate weather conditions with H6 36; 000; y(�) satis%es the boundaryvalue problem

md2yd�2

=− mgR2

(y + R)2+ k(1− 6:9× 10−6y)4:3

(dyd�

)2y(0)=H; y(T )= 0;

where m is the mass of the object, g is the acceleration of gravity at sea level, R is theradius of the earth, and k is a coeNcient of air resistance that depends on the physical


characteristics of the object. (This information can be found on the NASA webpage.)With the rescaling y=Hx; �=Tt, we obtain the equivalent dimensionless problem:

�x′′ =− Q2

(1 + Hx=R)2+ (1− Ax)4:3x′2;

x(0)= 1; x(1)= 0;

where �=m=Hk; Q2 =mgT 2=H 2k, and A=6:9 × 10−6H . With the restriction H636; 000, we have xH=R¡ 0:0018, so we will for simplicity neglect this term in thepreceding diMerential equation.Then the reduced solution is

w(t)= [− 1(1 + 3:15AQ(t − 1))1=3:15]A−1:

In order to have a boundary layer of the type discussed in this section, we needw(0)¡ 1, or√

mgK

TH

=Q¡1− (1− A)3:15

3:15A:

This restriction speci%es how small T must be in order that the object have a velocitygreater than the terminal velocity at height H .From Theorem 1, we get an O(� log �) approximation w(t)+�(t; �) with the boundary

layer correction � given implicity by

t∫ 1−w(0)

0F(0; �) d0=

∫ 1−w(0)

�F(0; �) d0;

where

F(0; �) ≡ e((1−A(1−w(0)+0))5:3)=5:3A�:

(These integrals can be evaluated in terms of incomplete gamma functions.)From Theorem 2, we get an O(�) approximation w(t) + (t; �) with boundary layer

correction given by

e−Ct=�∫ 1−w(0)

0F(0; �) d0=

∫

0F(0; �) d0;

where C =2(1 − Aw(0))2:15Q. In fact, numerical experiments, for example with H =30; 000 feet, T =101 seconds, k =0:005, and m=4:97, so that �=0:0331; A=0:207,and Q=0:6, reveal that w(t) + (t; �) is a much more accurate approximation of thesolution than w(t) + �(t; �) in the boundary layer.

3. The perturbed quasilinear problem

Now we analyze the problem in which the quadratic term, but not the term that islinear in the %rst derivative, is multiplied by the small parameter:

�(x′′ − a(t; x)x′2)= b(t; x)x′ + c(t; x); (24)

x(0)=A; x(1)=B: (25)


We impose conditions on the coeNcient functions to ensure the existence of a solutionwith boundary layer at the left endpoint. Suppose w∈C2[0; 1] solves

b(t; w)w′ + c(t; w)= 0; (26)

w(1)=B (27)

and (t; �) solves:

�( ′′ − a(0; w(0) + ) ′2)= b(0; w(0) + ) ′ (28)

(0)=A− w(0); (∞)= 0: (29)

The solution of (28) and (29) is

t�=∫ A−w(0)

ds

− ∫ s0 b(0; w(0) + �)e−

∫ �s a(0;w(0)+u) du d�

: (30)

Here, will be a well-de%ned, monotone function of boundary-layer type if we assume:∫ s

0b(0; w(0) + �)e−

∫ �0 a(0;w(0)+u) du d�¡ 0 (31)

for s between 0 and A − w(0). Also, we need a stability assumption for the reducedsolution:

b(t; w(t))6− k ¡ 0 (06 t6 1); (32)

where k is a positive constant. Under conditions (31) and (32), there are positiveconstants C1; C2, and C3 such that

C1e−kt=�6 | (t; �)|6C2e−kt=� (33)

for t¿C3�, and similar estimates hold for � ′ and �2 ′′ with diMerent constants.

Theorem 3. Assume a; b; and c are C1 in a domain containing the reduced solutionand the boundary layer region and that (31) and (32) are satis1ed. For all su3cientlysmall values of �; (24) and (25) has a solution x(t; �) so that

x(t; �)− (w(t) + (t; �))=O(�)


Proof. We will consider only the case A¿w(0). The super-solution will have theform

�(t; �)=w(t) + '(t; �) + �2(t; �) + �f(t; �);

where 2 and f are positive functions to be de%ned below and ' satis%es:

�('′′ − a(0; w(0) + ')'′2)= (b(0; w(0) + ') + d�)'′

'(0)=A− w(0); '(∞)= 0 (34)

for some d¿ 0 to be determined. Note that for small values of d; ' has the sameproperties as those of discussed above. We also have '¿ 0 and '′ ¡ 0.


Let 0= t=�, and let a dot indicate diMerentiation with respect to 0. Then '̇ satis%es:

d2

d02 '̇− (2a'̇+ b+ d�)dd0

'̇− (a2'̇2 + b2'̇)'̇=0; (35)

where all the coeNcients are evaluated at (0; w(0)+'). De%ne 2(0; �) to be the solutionof

R2 − (2a'̇+ b+ d�)2̇ − (a2'̇2 + b2'̇)2=p0'̇;

2(0)= 0; 2(∞)= 0; (36)

where p is a positive constant to be determined. Eq. (35) shows that '̇ is a solutionof the homogeneous part of (36), so by variation of parameters,

2(0)= '̇∫ 0

0('̇(s))−2e

∫ s0 2a'̇+b+d�

∫ s

∞p�'̇2(�)e−

∫ �0 2a'̇+b+d� d� ds

for 0¿ 0.The estimates on '̇ corresponding to (33) imply that 2 is bounded. Moreover,

2̇=− R'∫ 0

0

q(s)

'̇2(s)

∫ ∞

s

p�'̇2(�)q(�)

d� ds+q(0)−'̇(0)

∫ ∞

0

p�'̇2(�)q(�)

d�; (37)

where q(s)= e∫ s0 2a'̇+b+d�. Using (33), we can show that the second term in (37) is less

than (−'̇)(C + D0) for some positive constants C and D and that the absolute valueof the %rst term in (37) is less than | R'|(E + F0 + G02), for some positive constantsE; F and G. It follows that 2̇(0)¡ 0 for 0 suNciently large since R'(0)¿ 0 for 0suNciently large. Also,

|'̇|�|�2̇| (38)

when �02 is small.Now we are ready to check the diMerential inequality for �:

��′′ − �a(t; �)�′2 − b(t; �)�′ − c(t�)

= �'′′ − �a(0; w(0) + ')'′2 − b(0; w(0) + ')'′ + O(�t)'′2

− �2a2'′2(2 + f)− �aw′2 − �3a2′2 − 2�aw′('′ + �2′)

− 2�2a'′(2′ + f′)− 2�2a2(2 + f)w′'′ − 2�3a2(2 + f)'′2′ + O(t)'′

− b(t; w(t))w′ + O(1)'− �b2′ − b(t; w(t))�f′ + O(�)f′'

− �b2(2 + f)(w′ + '′)− b2�2(2 + f)2′ − c(t; w)− �c2(2 + f)

+ �2(2′′ + f′′) + �w′′ + O(�2)

=d�'′ + d�22′ + pt'′ + O(�t)'′2 − �2a2'′2f

− �aw′2 − �3a2′2 − 2a�w′('′ + �2′)− 2�2a'′f′ − 2�2a2(2 + f)w′'′

− 2�3a2(2 + f)'′2′ + O(t)'′ + O(1)'− �b(t; w)f′ + O(�)2′'


− �b2(2 + f)w′ − �b2f'′ − �2b2(2 + f)2′ − �c2(2 + f)

+ �2f′ + �w′′ + O(�2); (39)

where we have used Eqs. (26), (34), and (36), the unspeci%ed arguments of a and bare (t; �), and the unspeci%ed arguments of a2; b2, and c2 are at various intermediatevalues given by the mean value theorem.Now choose p large enough that the negative term pt'′ in the last equation domi-

nates the terms O(�t)'′2 and O(t)'′. Choose

M¿ |b2w′ + c2 + �b22′|for 06 t6 1, and de%ne

f=H (e−�(1−t) − 12 )¿ 0;

where �=−M=k + O(�), to be the solution of

�f′′ + kf′ +Mf=−MH=2;

where H is chosen so thatMH2

¿ |w′′ − c22 − �b222′ − b22w′ − 2�aw′2′ − �2a2′2 − aw′2|for 06 t6 1.

Using the properties of f in the estimation of (39), we have

��′′ − �a(t; �)�′2 − b(t; �)�′ − c(t; �)

6d�'′ + d�22′ − �2a2'′2f − 2�aw′'′ − 2�2a'′f′ − 2�2a2(2 + f)w′'′

− 2�3a2(2 + f)'′2′ + O(1)'+ O(�)'f′ − �b2'′f

¡ 0

for d suNciently large and � suNciently small since by (38) '′(t)��2′(t) when t�√�,

and 2′(t)¡ 0 when t��, and since d�|'′|¿ ' for 06 t6 1, if d is large enough.In a similar way, we can construct a sub-solution of the form

�(t; �)=w(t) + �(t; �)− �2 − �f;

where � is the solution of

�(�′′ − a(0; w(0) + �)�′2)= b(0; w(0) + �)�′ − d��′;

�(0)=A− w(0); �(∞)= 0

and the de%nitions of 2 and f and the veri%cation that � is a sub-solution are similarto those given in the discussion of the super-solution.It remains to show that

�(t; �)¿w(t; �) + (t; �)¿ �(t; �); (40)

�(t; �)− �(t; �)=O(�) (41)


as � → 0; 06 t6 1. First, write (30) in the form

t�=∫ A−w(0)

(t;�)

(∫ s

0−(b(0; w(0) + �) + �)e−

∫ �s a d�

)−1

ds;

diMerentiate with respect to �, and solve for the partial of with respect to �:

@ @�

=

(∫

0−(b+ �)e−

∫ � a d�

)∫ A−w(0)

∫ s0 e

− ∫ �s a d�

(∫ s0 (b+ �)e−

∫ �s a d�)2

ds:

Now (40) follows from the observation that @ =@�¿ 0 for small �. Since∫ s

0e−

∫ �s a d�¿D1s

(∫ s

0(b+ �)e−

∫ �s a d�

)26D2s2

for some positive constants D1; D2, we have

@ @�

=O( log )=O(1);

from (33) and (41) follows immediately. The proof is complete.

Example 2. We consider a very basic predator–prey problem with u(t) denoting thepopulation of prey at time t and v(t) the population of predators. The equations gov-erning the interaction of the two species (see [4]) are:

dud�

= au(1− u=K)− buv;

dvd�

=− cv+ duv:

Let’s assume we observe the prey population at time �=0 to be u(0)=C and at�= a to be u(a)=D. In order to obtain a perturbation problem, we assume that thebirth–death rate a of the prey species is small relative to the corresponding parameterc of the predator species. Then the rescaling u=Kx, v=(a=b)y, �= t=b, yields thedimensionless system

x′ = x(1− x)− xy;

�y′ =− y + $xy;

where �= a=c and $=Kd=c. If we now eliminate y from this system, we obtain theboundary value problem

�(x′′ − x′2

x

)=(−1 + $x − �x)x′ + x − x2 − $x2 + $x3;

x(0)=A; x(1)=B;


where A=C=K and B=D=K . We can apply Theorem 3 to this problem provided thatthe stability conditions (31) and (32) are satis%ed. In particular, (32) holds if thereduced solution w(t) satis%es

$w(t)− 16− k ¡ 0

for some positive constant k. The reduced solution is easily computed to be:

w(t)=B

(1− B)e1−t + B:

For stability, we require B¡ 1=$ in case B¡ 1 and B¡e=($+ e − 1) in case B¿ 1.Let’s assume A¿w(0)=B=(1−Be+B). Then the boundary layer stability condition

(31) is∫ s

0($(w(0) + �)− 1)e−

∫ �0 (w(0)+u)−1 du d�¡ 0

for 0¡s6A− w(0). A short calculation simpli%es this condition to

A− w(0) +1$ln

w(0)A

¡ 0:

Under these conditions, Theorem 3 yields a solution x(t; �) of the boundary valueproblem such that

x(t; �)− (w(t) + (t; �))=O(�)

as � → 0, where the boundary correction is computed from the expression

t�=∫ A−w(0)

ds(w(0) + s)(ln((w(0) + s)=w(0))− $s)

:

4. The perturbed semilinear problem

Next, we consider for small values of �¿ 0:

�(x′′ − a(t; x)x′2 − b(t; x)x′)= c(t; x); (42)

x(0)=A; x(1)=B: (43)

Let w(t)∈C2[0; 1] solve c(t; w(t))= 0 for 06 t6 1. The boundary layer correction (t; �) at t=0 is chosen to satisfy:

�( ′′ − a(0; w(0) + ) ′2)= c(0; w(0) + )

(0)=A− w(0); (∞)= 0:

Similarly, the correction �(t; �) at t=1 satis%es

�(�′′ − a(1; w(1) + �)�′2)= c(1; w(1) + �);

�(1)=B− w(1); �(−∞)= 0:


Then and � can be computed in implicit form. For example, is given by√2�t=∫ A−w(0)

ds√∫ s0 c(0; w(0) + �)e−2

∫ �s a(0;w(0)+u) du d�

; (44)

where we need the boundary layer stability assumption:∫ s

0c(0; w(0) + �)e−2

∫ �0 a(0;w(0)+u) du d�¿ 0 (45)

for 0¡s6A−w(0) or A−w(0)6 s¡ 0. The correction � has a similar representationin which the following assumption is needed:∫ s

0c(1; w(1) + �)e−2

∫ �0 a(1;w(1)+u) du d�¿ 0 (46)

for 0¡s6B−w(1) or B−w(1)6 s¡ 0. Finally, we assume that the reduced solutionis stable:

@c@x

(t; w(t))¿ 0 (47)

for 06 t6 1.

Theorem 4. If hypotheses (45)–(47) are satis1ed; then for small values of �¿ 0; theboundary value problem (42) and (43); has a solution x(t; �) so that

x(t; �)− (w(t) + (t; �) + �(t; �))=O(√�)

uniformly for 06 x6 1 as � → 0.

Proof. We consider the case A−w(0)¿ 0; B−w(0)¿ 0. Our super-solution will havethe form:

�(t; �)=w(t) + �(t; �) + �(t; �) +√�f:

For brevity, we will omit the right boundary layer function � in our calculations sinceit can be treated in a way similar to the left boundary layer function �.Now � will be the solution of

�(�′′ − a(0; w(0) + �)�′2)= c(0; w(0) + �)− d√��;

�(0)=A− w(0); �(∞)= 0;

where d is a positive constant to be chosen later.In order to check the diMerential inequality for �, we will need the following form

of Taylor’s formula:

c(t; w(t) + �+√�f)− c(0; w(0) + �)

= c(t; w(t) + �) + c2(t; w(t) + �)√�f + c22(t; ∗) �f

2

2− c(0; w(0) + �)

= c(t; w(t) + �)− c(0; w(0) + �) + c2(t; w)√�f


+c22(t; ∗∗)√�f�+ c22(t; ∗) �f

2

2

=O(t)�+ O(1)√�f�+ O(1)�f2 + c2(t; w)

√�f; (48)

where ∗ is between w(t)+� and w(t)+�+√�f, and ∗∗ is between w(t) and w(t)+�.

Using (48), we have

�(�′′ − a(t; �)�′2 − b(t; �)�′)− c(t; �)

= �(�′′ − a(0; w(0) + �)�′2)− c(0; w(0) + �) + O(�t)�′2

− �3=2a2f�′2 − 2�aw′�′ − �b(w′ + �′ +√�f′) + O(t)�

+O(1)√�f�+ O(1)�f2 − c2(t; w)

√�f + �3=2f′′ + O(�): (49)

Fife [6, Lemma 2:1] showed that there exist positive constants C1; C2; $; l, so that

C1e(−l−$)t=√

�6 �; |�′|√�6C2e(−l+$)t=√

�

for 06 t6 1 and small values of �. It follows that O(t)�=O(√�), O(�t)�′2 =O(�),

and O(�)�′ =O(√�) uniformly for 06 t6 0, as � → 0.

Note that

�′(t; �)=−√

2�

√∫ �

0(c(0; w(0) + �)− d

√��)e2

∫ �� a(0;w(0)+u) du d�;

so ∫ �

0(c(0; w(0) + �)− d

√��)e2

∫ �� a(0;w(0)+u) du d�=

�2�′2:

Also, for large enough K ¿ 0,

K�¿∫ �

0(c(0; w(0) + �)− d

√��)e2

∫ �� a(0;w(0)+u) du d�

for 06 �6A− w(0), so

K�¿�2�′2

for 06 t6 1:From the preceding estimates and Eq. (49), we have

�(�′′ − a(t; �)�′2 − b(t; �)�′)− c(t; �)

=− d√��− �3=2a2f�′2 + O(1)

√�f�− c2(t; w)

√�f + O(

√�)

¡ 0

for large enough positive constants f and d and small �, since the negative term−d

√�� dominates both −�3=2a2f�′2 and O(1)

√�f�.

The sub-solution �(t; �) is of the form

�(t; �)=w(t) + S�(t; �) + S�(t; �)−√�f;


where, for example, S� satis%es:

�( S�′′ − a(0; w(0) + S�) S�

′2)= c(0; w(0) + S�) + d

√� S�;

S�(0)=A− w(0); S�(∞)= 0

for some d¿ 0. The veri%cation that � satis%es the needed inequality is similar to thatgiven previously for �.From the implicit representation for , we have

S�(t; �)6 (t; �)6 �(t; �)

for 06 t6 1, so

�(t; �)6w(t) + (t; �) + �(t; �)6 �(t; �)

for 06 t6 1.Starting with the representation of � corresponding to (44) and diMerentiating with

respect to d, we have

@�@d

=

√∫ �

0(c − d

√��)e2

∫ �� a d�

√�2

∫ A−w(0)

�

∫ s0 �e2

∫ s� a d�

(∫ s0 (c − d

√�)e2

∫ s� a)3=2

ds

=O(√�);

uniformly for 06 t6 1 as � → 0. Since d ranges over a bounded set as the boundarylayer function ranges from � to S�, we conclude that

�(t; �)− �(t; �)=O(√�)

as � → 0, and the proof is complete.

Example 3. Consider the Weekman–Goring model (see [1,15]) of a chemical reactionon a Uat catalytic surface accompanied by a change in volume. Let x denote thedimensionless concentration of gas, assume that �, the reciprocal of the square of theThiele modulus, is small, and take the order of the reaction to be one. Then

�(x′′ − �

1 + �xx′2)= x(1 + �x):

Here, � is the volume change modulus and the independent variable t is the dimen-sionless distance from the plane of symmetry to the edge of the plate at t=1. Theboundary conditions are:

x′(0)= 0; x(1)= 1:

We cannot apply Theorem 4 immediately since the boundary condition at t=0 is ofNeumann type. For this type of boundary condition, we need an super-solution � with�′(0)6 0; �(1)¿ 1 and a sub-solution � with �′(0)¿ 0; �(1)6 1. The sub-solutionthat we de%ned in the proof of Theorem 4 already has these properties if we omit thefunction S�. For the super-solution, we de%ne

�(t; �)= �(t; �) +√�f(t);


where � satis%es

�(�′′ − �

1 + ��′2)= �(1 + ��) + d

√��

�(−∞)= 0; �(1)= 1

for some d¿ 0. Choose

f(t)= (2− t)C;

so that√�f′(t)=−C

√�¡ 0. Then � satis%es the boundary inequalities. The ver%cation

that � is a super-solution for a suitably chosen C now proceeds as in the proof ofTheorem 4. We can conclude that the Weekman–Gorring model has a solution x(t; �)for small �¿ 0 so that

x(t; �)− (t; �)=O(√�);

uniformly for 06 t6 1 as � → 0. Finally, is given implicitly by√2�(1− t)= �

∫ 1

ds

(1 + �s)√

�s− ln(�s+ 1):

References

[1] K.W. Chang, F.A. Howes, Nonlinear Singular Perturbation Phenomena: Theory and Application,Springer, New York, 1984.

[2] E.A. Coddington, N. Levinson, A boundary value problem for a non-linear diMerential equation with asmall parameter, Proc. Amer. Math. Soc. 3 (1952) 73.

[3] F.W. Dorr, S.V. Parter, L.F. Shampine, Applications of the maximum principle to singular perturbationproblems, SIAM Rev. 15 (1973) 43.

[4] L. Edelstein-keshet, Mathematical Models in Biology, McGraw-Hill, New York, 1988.[5] A. ErdWelyi, Approximate solutions of a nonlinear boundary value problem, Arch. Rational Mech. Anal.

29 (1968) 1.[6] P.C. Fife, Semilinear elliptic boundary value problems with small parameters, Arch. Rational Mech.

Anal. 52 (1973) 205.[7] F.A. Howes, Boundary-interior layer interactions in nonlinear singular perturbation theory, Mem. Amer.

Math. Soc. 203 (1978) 1–108.[8] F.A. Howes, R.E. O’Malley Jr., Singular perturbations of second order semilinear systems, Lecture

Notes in Mathematics, Vol. 827, Springer, Berlin, 1980, 131.[9] F.W.J. Olver, Asymptotics and Special Functions, Academic Press, New York, 1974.[10] D.R. Smith, Singular Perturbation Theory, Cambridge University Press, London, 1985.[11] A. Van Harten, Nonlinear singular perturbation problems: proofs of correctness of a formal

approximation based on a contraction principle in a Banach space, J. Math. Anal. Appl. 65 (1978)126.

[12] A.B. Vasil’Eva, The development of the theory of ordinary diMerential equations with a small parametermultiplying the highest derivative during the period 1966–1976, Russian Math. Surveys 31 (1976) 109.

[13] A.B., Vasil’Eva, Asymptotic behavior of solutions of the problem for ordinary nonlinear diMerentialequations with a small parameter multiplying the highest derivative, Russian Math. Surveys 18 (1963) 13.

[14] M.I. Vishik, L.A. Liusternik, Inital jump for nonlinear diMerential equations containing a small parameter,Soviet Math. Dokl. 1 (1960) 749.

[15] V.W. Weekman, R.L. Gorring, InUuence of volume change on gas-phase reactions in porous catalysts,J. Catalysis 4 (1965) 260.

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