Nonlinear boundary value problems and competition in the chemostat

Nonlinear

Analysis,

Theory,

Methods

& Applications,

Vol. 22, No. II, pp. 1329-1344, 1994 Copyright 0 1994 Elsevier Science Ltd Printed in Great Britain. All rights reserved

Pergamon

0362-546X/94$7.001 .@I

NONLINEAR BOUNDARY VALUE PROBLEMS COMPETITION IN THE CHEMOSTAT J. V. BAxLEYt$ and H. B. TDepartment

THOMPSON~

Science, Wake Forest University, Winston-Salem, NC 27109,U.S.A.; and The University of Queensland, Queensland 4072, Australia

of Mathematics and Computer SDepartment of Mathematics, (Received

AND

9 November

1992; received for publication

4 June 1993)

Key words ond phrases: Positive solutions, shooting, Sturm comparison, eigenvalue, self-adjoint, implicit function theorem, Green’s functions.

bifurcation

from

a simple

1. INTRODUCTION THIS PAPER

is concerned with the existence of positive solutions to boundary value problems for n dimensional systems of second order ordinary differential equations of the form y; + mifi(x~Yl,YZ~

*--9YnlYi =

O9

i = 1,2, . . ..n

(1)

on a finite interval which we take to be 0 I x I 1, with boundary conditions of the form ajYi(O) - U~Yuf(O) = 0, &y,(l) + biyyf(l) = 0,

(2) i= 1,2 ,...,

n,

(3)

+~f>O,bi+bl>O,andai+bi>O,fori= 1,2 ,..., n.Thus,we allow Dirichlet boundary conditions, but not Neumann conditions. The immediate motivation for this study was the paper of So and Waltman [l], which is concerned with a special case of our problem and arose in their investigation of the possibility of long term coexistence of competing organisms in a chemostat, under the assumption that the nutrient and organisms diffuse slowly through the vessel instead of being “well mixed”. The mathematical model consists of a system of three partial differential equations (with boundary and initial conditions) involving the two organisms and the nutrient. So and Waltman show that the study of steady-state solutions can be reduced to the following problem involving nonlinear second order ordinary differential equations Yf +

$ Supported

by a Raybould

m,g,(w(x) - Yl - Y,lY, = 0,

(4)

r; + m,g,(w(x) - Yl - Yz)Yz= 0,

(5)

Y{(O)= Y;(O) = 0,

(6)

Y;(l) + YY,(l) = 0,

(7)

Y;(l) + YYAl) = 0,

(8)

Fellowship

while visiting

the University 1329

of Queensland.

J. V. BAXLEY and H. B. THOMPSON

1330

where g,(S) =

--Eci+s’

fori=

1,2,

with each ci > 0 and where w(x) = (1 + y)/y - x. Since yr and y2 represent the amount of each organism present in the chemostat and w(x) - y, - y, represents the amount of nutrient present, one hopes for solutions yr(x), y2(x) satisfying y,(x) > 0, y2(x) > 0, w(x) - y,(x) - yz(x) > 0 on 0 I x 5 1. So and Waltman then use some basic results of Crandall and Rabinowitz [2, 31 (see also [4, p. 1721) to establish the existence of an unbounded set of parameter values (ml, m,) for which such positive solutions exist. The reader should consult [l] for a more detailed treatment. Our results extend the theorems of [l] by increasing the number of equations (and, hence, the number of organisms) from 2 to n, by allowing more general nonlinearities in the component equations and by allowing more general boundary conditions. We use a combination of Sturm type comparisons and shooting techniques to prepare for application of the following result of Crandall and Rabinowitz [5, lemma 2.11; it applies to operators with null spaces of dimension greater than one and is a generalization of the theorem used in [l], concerning “bifurcation from a simple eigenvalue”. THEOREM 1. Let A, B, , B, be Banach spaces. Consider a mapping F: A x B, + B2), where F, F,, FY, and Fxy are all continuous and F&y)_= Ly f B
(1, z) + Lz + B&y,) is one-to-one and onto. Then the zeros of F in a neighborhood of (A,, 0) in A x Sply,) x Z consist exactly of the trivial solutions (A, 0) and of a Cl-curve s + (&s), s(y, + I,@))), where (II_,J,u): (-6,6) + A x Z and (n(O), v/(O)) = (A,, 0). The proof of theorem 1 depends on the implicit function theorem in Banach spaces and follows the lines of the theorem of bifurcation from a simple eigenvalue. In [5], F is assumed to be C*; an examination of the proof reveals that one only needs the smoothness assumptions of theorem 1. In Section 2 we prove existence of positive solutions of (l)-(3), under very general conditions, by bifurcating from the trivial solution yi = 0, and mi = pi, the smallest eigenvalue of yl + (2)and(3),fori= obtain positive

~ifi(X,

1,2 ,..., n. In Section 3, with addit.ional hypotheses on the functions fi, we solutions of (l)-(3) by starting with any m, > ,u, and a positive solution ii, of Y," + m,f,(x,

and bifurcating

from the solution

1,2 ,...,

0, .... O,y,l~, = 0

(0, 0, . . . , 0, i&J and m, = pi, now the smallest

yl + (2)and(3),fori= problem (4)-(9).

0, . . .) O)yi = 0,

njfi(X,

n - 1. Finally,

eigenvalue

of

0, . . .) 0, ~,)yi = 0, in Section

4, we apply our results to the chemostat

Nonlinear 2. A GENERAL

boundary

1331

value problems

EXISTENCE

THEOREM

We consider the system of second order nonlinear ordinary differential equations (1) with boundary conditions (2), (3). We write Y = (Yi , y,, . . . , y,) and f = (fi , f2, . . . , f,). We assume that f E C’(U, R”) where Uis an open set in [0, 11 x R” containing [0, 11 x (0). We also assume that forOrx5

h(x, 0) > 0

1.

(10)

Replacing U by a smaller set if necessary, we may (and shall) assume that f has been extended to all of [0, l] x R” as a bounded C’ function. Let S denote the set of all points (ml, m2, . . ., m,) E R: for which the boundary value problem (l)-(3) has a solution u = (ui , u2, . . . , u,) with each Ui(X) > 0 on (0, I), and with each Ui positive at any endpoint of (0, 1) where the boundary condition is not a Dirichlet condition. THEOREM 2. Suppose that f E C’([O, l] x R”, R”) is bounded and satisfies (10). Then S # 0. Proof. We use Green’s functions as in [l] to express our equations as equivalent integral equations. Thus, let Gi(x, r) be Green’s functions for the operator -Y” with the ith set of boundary conditions in (2), (3). Then the system (l)-(3) becomes 1 YiW

-

mi

Gi (XT r)fi (~9Y(r))_Yi(5) dt = 0,

i = 1,2, . . . , n.

s0 We seek to apply theorem 1. Let

B, = B, = fi L2(0, 1). i=l

For ,r, = (A, , A,, . . . , L,)ER”andyEB,,let 1 Fit29

Y) = YiCx) - Ai

s0

Gi@* ~).h(~*Y(~)lYi(r) dt,

i=1,2

3 ***,

n

and let F(A,Y) = (F,(J,Y),F,@,Y),

. . ..F.(J,Y)).

Then we easily see that F, Fy , F,, Fxy all belong to C(R” x B, , B,) and F(A, 0) = 0. Let Ai be the positive self-adjoint differential operator in the weighted Hilbert space L2((0, 1); wi) with weight function Wi(X) = 5(x, 0) defined by AiY = -(l/Wi)Y” and the ith set of boundary conditions in (2), (3). It is well known that the spectrum of Ai is discrete (in fact Ai has compact inverse), that the smallest eigenvalue iui is positive and simple, and that the corresponding eigenfunction vi can be chosen positive (with unit norm) on (0, 1). Because our hypotheses imply that solutions to initial value problems are unique, it is also true that Uiis positive at any endpoint where the boundary condition is not a Dirichlet condition. At an endpoint where the condition is a Dirichlet condition, it must be true that ul is nonzero. Since Ai is self-adjoint the range R(Ai - pil) is the orthogonal complement of the null space N(Ai - nil) and so has codimension one. Note that V9 +

~ifi(X,

O)Vi = 0,

ai Vi(O) - al V;(O) = 0, bivi(l)

+ bfuf(l)

= 0,

1332

J. V. BAXLEYand H. B. THOMPSON

and so

s 1

ui(x) - Pi

Now expand

F(il, y) about

Gi(X, T)~(T,

(p, 0) using Taylor’s

F&Y)

O)V~(T) dt = 0.

0

formula

= F,(Lc, 0)Y + F,,(K

to get

O)(A - &Y) + JG,Y),

where p&O) = 0, p,,(,~, 0) = 0, I?~~(P, 0) = 0. Put F,,(fl, O)(A - p, y). Then L is a bounded linear operator and bilinear, mapping R” x B, -+ B,. A straightforward

Ly = F,(p, 0)y and B(L - p,y) = mapping B, + B, and B is bounded calculation produces

1

CLY)i = Yi - Pi

Gi(x, ~lfi(~, OlYi(r) dr*

s0

for

Ly=O if only if Yi E N(Ai - clil) Thus, and ZJ= (ur, ?Jz, . . . . v,) E N(L). We also easily calculate

i = 1,2, . . . . n.

Hence,

1 (WA

-

mu,Y))i

=

-(Ai

-

PiI

Gi@-,

TIJ;:(s, OlYi(~)

dr-

s0 Now put Z = ~

R(Ai - Liz)

i=l

and consider

the mapping

T: R” x Z --* B, defined

by

T(i, z) = Lz + B(;i, v). We shall see that T is one-to-one and onto, and then theorem 1 implies that there exists a neighborhood of (p, 0) in R” x Sp(uJ x Z for which all solutions of F(A, y) = 0 lie either on (,I, 0) or on a Cl-curve (n(s), S(U + w(s))) where (,I, I&: (-6,6)

+ R” x Z

and (A(O), y/(O)) = (& 0). Let U, = u + I,@) so that for SE (0, a), F@(s), su,) = 0. For s sufficiently close to 0, it is not immediately clear, even in the case of nonDirichlet boundary conditions, that u,,i = Ui + v/i(S) is positive on (0, 1) because we only know that wi(S) is small in L’(0, 1). However, since F@(s), su,) = 0, we have 1

us,i(x) = ni(s)

s0

fori=

Gi 6, r)us, i (TX (~9Sus(TIs))d7,

Recalling that eachfi is bounded, it follows from the Schwarz inequality formly bounded as s -+ 0 and satisfies the differential equation Ui,i + tIi(S)fi(X,

SU,(X))U,,i

=

0.

1,2 ,...,

n.

that each u,,i is uni-

Nonlinear

boundary

1333

value problems

Thus, U~,iis also uniformly bounded as s + 0. A boundary condition that is not a Dirichlet condition implies that (Ui,i) is bounded at that endpoint. If both boundary conditions are Dirichlet conditions, then there must be interior points in (0, 1) at which z& vanishes. In either case, integration gives a uniform bound on (U:,i) and Ascoli’s theorem implies that both U,,i, u;,~ converge uniformly to Ui and V: as s -+ 0, since vi(S) converges to zero in L.*(O,1) as s -+ 0. Therefore, for s sufficiently small and positive, SU,,i is positive on (0, l), and also positive at any nonDirichlet endpoint. Note that for s > 0 and sufficiently small, (x, U,(X)) belongs to the original neighborhood U and, thus, U, is a solution to the original problem (l)-(3) with the unextended function f. It remains to verify that T is one-to-one and onto. Note that T(j,z) = 0 if and only if ZiCx)

-

Pi

s 1

1

G~(x, T)fi(t, O)Z~(T)dr = 1;

0

0

Gi 6, r)X (r, O)ui(r) dr

for i = 1,2, . . . . n, which is equivalent to (A, z) satisfying the ith set of boundary conditions in (2), (3) and the differential equation Z/ + /Yif;:(X9 0)zi = -lih(X,

O)Vi 7

or, equivalently (A; - ~iZ)Zi = ~jVi. Since the left-hand side is orthogonal to ui with respect to the weight function wi, it follows that li = 0,

zi = 0.

Thus, the kernel of T is trivial and T is one-to-one. To show that T is onto, let Ui E L*(O, 1) for each i. Then T(l,z) = u is equivalent to 1 zi(x)

- Pi

Gi(X, ~)J;(T, O)zi(r)

do - ~ Ui = Ui,

s0

I

for each i. We decompose U; = iii + (YiUiinto orthogonal components in R(Ai - PiZ) and N(Ai - iuiZ)_ Choosing ii = -pia; to cancel the terms involving Vi, our equation is then equivalent to the differential equation (Zi - ~i)n +

~i~(X,

0)zj = 0

which, after setting yj = zi - i&, can be rewritten as

Since iii E Z?(Ai - p;Z), there exists a solution Ji of this equation. We may certainly choose pi E R(A_i - piZ) and then choose ii = ji + iii. With these choices for i = 1,2, . . . , n, we have found (A, 3 so that T(l,2)= u and T is onto, concluding the proof. 1. Suppose that the hypotheses of theorem 2 are satisfied. The set S contains points (m,, m2, . . . . m,) with each mi arbitrarily close to pi, the smallest eigenvalue of the self-adjoint differential operator Ai in the weighted Hilbert space L*((O, 1); Wi), defined by Aiy = -y”/w, and the ith set of boundary conditions in (2), (3) with wi(x) = fi(X, 0).

COROLLARY

1334

J. V. BAXLEYand H. B. THOMPSON

Remark. t=(t,,&

Note that u in the above proof could be replaced by U, = (t, ,..., t,JER:, ~ti = 1, thus, producing a solution

u1 ,

t2 u2, . . . , t, v,) with

(A, z) = (A(% t), z(s, t)) representing bifurcation along different directions. This solution is a smooth function of the parameters, except that if for some i, ti + 0, the interval (-6, a), which now depends on t, may shrink to zero, since the norm of T-‘, where T is the operator in the proof, becomes unbounded. We conclude this section with a relationship which might mi, if (ml, m2, . . . . m,) E S. First we need the following comparison theorem. mj

LEMMA 3.

Suppose

that

Yi ,

y2 are positive

exist between any of the pairs simple variation on the Sturm

on (x, , x2) and satisfy

Y: + a(x)Y, 2 0,

Y; + b(XIY* 5 0,

for xi I x 5 x2, where a(x) 5 b(x) on (xi, x2) and equality x2 - xi.

holds on a set of measure

less than

(i) Then Y1(X2)Y~(X2)- Y;(x2)Y2(x2) < Y,(X,)YXX,) - Yl(X,)Y,(X,). (ii) If, moreover, a,Y,(x,)

- 4YXxJ

= 0,

a,Y2(x,)

- 4Yl(x,)

= 0,

&Y&2)

+ 4Yl(X2)

= 0,

where a,, ai, bl , 6; , a2, ai are nonnegative, a, + a; > 0,

b1 + b’, > 0,

a2 + ai > 0,

a2a; - ala; 5 0,

then b2Y2(x2) + b;Y;(x2)

< 0

for any pair b2, b; 1 0 such that b2 + bi > 0 and b2 6; - b, 6; I 0. (iii) If, moreover, a2 + b2 > 0, then there exists a point x0 E (x, ,x2) at which b2Y2(G Proof.

Combining

the two differential

+ &Y&G inequalities

x2

and integrating,

we obtain

x2 (YIYi - Y2Y;) k

s x1

= 0.

+

t&4

- &)lY,Y2

do 5 0.

$ XI

Our hypotheses imply that the second integral is positive, and (i) follows after integrating by side of the inequality in (i) is parts. From a2a; - alal I 0, we conclude that the right-hand nonpositive. Thus, Y*(x2)Y;(x2) - Y;(x2)Y2@2) < 0.

Nonlinear boundary value problems

Since blyI(x2)

+ b;y;(x,)

= 0, the desired inequality

in (ii) follows.

&Y,(X,) - &G(Xl) and a2 + & > 0 imply that &y,(x) + &y;(x) intermediate value theorem gives (iii).

1335

Now

= 0

> 0 in some

interval

(xi, x1 + 6) and

the

THEOREM 4. Let (ml, m2, . . . . m,) E S and u = (ui, u2, . . . . u,) be a solution of (l)-(3) with each Ui > 0 on (0, 1). If ajai - aia,! I 0 and bj6: - bib; I 0 for some distinct pair i, j, then there exists x0 E [0, l] for which

Proof. If we assume the contrary, we have mifi(X, u(x)) < mjfj(x, lemma 3 (ii) implies that bj ~~(1) + bJu,!(l) < 0 which is impossible boundary condition at x = 1.

u(x)) on [0, 11. Then since Uj satisfies the

COROLLARY 2. If in theorem 4, the boundary conditions in (2), (3) are identical pair i, j, then there exists a point x0 E [0, l] for which mifi

Proof. Applying

(X0

Y &%)>

=

mjfjbo

for some distinct

9 4X,)).

4 twice, we get xi, x2 in [0, l] such that

theorem

mifi(Xl9

mi.6

and we apply the intermediate

NXJ)

(X23 MX2))

value theorem

1

mj_&

T U(XJ),

5

mjfi(X2,

W2))

to finish the proof.

3. OTHER BIFURCATION

BRANCHES

From theorem 2, we know that S # 0. We would like to prove, as in [l], that S is unbounded. We do this below in theorem 6 but with additional hypotheses, which are satisfied by the chemostat problem (4)-(g). In the proof of theorem 2, we found positive solutions of (l)-(3) by bifurcating from the trivial solution. In the proof of theorem 6, we will bifurcate from a nontrivial solution in much the same way as in [l]. In the following, we use ej to denote the jth standard basis vector in R”: every entry in ej is 0 except the jth, which is 1.

LEMMA 5. In addition

to the hypotheses

of theorem

2, suppose that there exists j (1 I j 5 n) for

which lim sup[max(fj(x, Y-m

Yej): 0 5 X I

111 5: 0.

1336

J. V. BAXLEY and H. B. THOMPSON

Let I, be the smallest eigenvalue of the problem consisting set of boundary conditions in (2), (3). Then: (i) for m > I,, there exists a positive solution ii of

of y” + A&(x, 0)y = 0 and the jth

y” + mfi(x, yej)y = 0 satisfying thejth set of boundary (ii) if, moreover, D,,j.fj(~, yjej) of (11) is unique and the problem

conditions in (2), (3); 0 for 0 5 x s 1 and all yj E R, then the positive consisting of

<

y” + m[&(x, i;ej) + and thejth

set of boundary

(11)

conditions

tiD,,fj(X,

ikj)]Y

=

in (2), (3) has no nontrivial

0 solution.

Proof. We prove (i) by shooting. Replacing x by 1 - x, if necessary, Qj > 0. If uJ! > 0, we let y, be the solution of the initial value problem y” + Y(O) =

solution

we may assume

that

mfj(x,Yejlr = 0, y'(0)

s,

=

5. J

If a; = 0, we let ys be the solution

of the initial

value problem

YN + mfi(X,YejlY = 0, Y(O)= 0,

y’(0) = s.

In both cases, we consider only positive values of the shooting parameter s. Sincefj is bounded, it is well known that solutions to initial value problems continue across the entire interval [0, 11. We first show that for s sufficiently large, in both cases, bjy,(l) + 6jyL(l) > 0. If aJ! > 0, choose k E (0, 7r/2) so that k tan k < aj/aj ; if uJ! = 0, choose k E (0, 7112) so that k tan k < 1. Then, by hypothesis, we may choose y0 > 0 SO that mjj(x, yej) < k2 for y > yO. If a; > 0, let s > y0 and let x1 = 0; then Ujy,(Xi) - a~y~(x,) = 0, Y,(x,) > Y,, &(x1) > 0. If aJ! = 0, we proceed as follows. Since fj is continuous, there exists a constant K for which fj(x, yej) < K for (x, y) E [0, l] x [0, yo]. Choose s L 2y, + mKy, and let x1 5 sup(x E [0, 1121:~; 1 y,,y, Integrating gives y:(x) > s - mKy, > 2y, for 2y,x for 0 5 x 5 xi. Thus, xi < l/2, y,(x,) With x1 and s thus chosen, we claim that y; smallest value in (xi , l] for which y; (x2) = 0

5 y, on [0,x]].

0 5 x I x1 and integrating again gives y,(x) > = yO, and Ay,(x,) - yi(x,) = 0, where A > 1. > 0 on [x1, 11. Otherwise, we choose x, as the and apply lemma 3, using

y” + k’y = 0 as a comparison equation and letting u be the solution satisfying ajU(Xi) - u,!u’(x,) = 0 (or Au(x,) - u’(x,) = 0). Lemma 3 (iii) implies that u’(x) = 0 at some point in (xi, x2). Since U(X) is essentially a translation of cos(kx), then necessarily Uj cos(kx) + aJ!ksin(kx) = 0 (or A cos(kx) + k sin(kx) = 0) at some point in (x2 - x,, 0) c (-l,O). But then ktan(-kx) = aj/aJ! (or k tan(-kx) = A) has a solution in (-1 , 0), which is impossible by our choice of k. Thus, yi > 0 on [x1, l] and bjy,(l) + bjy:(l) > 0.

1337

Nonlinear boundary value problems

Since y, and yj depend continuously on s, we seek s > 0 for which bjy,( 1) + bjyi (1) I 0. Let d = infls > O:y,(x) > 0 on [0, 11). There are two cases to consider. If s^> 0, then y;(x) > 0 on [O, 1) and y,-(l) = 0. Thus, y)(l) < 0 and bjyi(l) + b,!yi(l) I 0. If s^= 0, then continuous dependence implies that small, WZfj(X,v,(X)ej) > Ajfj(X, 0) fj(x9Ys(x)ej) + fjCXv O) as s + 0 and, thus, for s sufficiently for 0 I x I 1. For such s, it follows from lemma 3 (ii) that bjy,(l) + bjyl(1) < 0, completing the proof of (i). For the uniqueness assertion of (ii), suppose that (11) has two solutions u1 and u2. Assume that U, > u2 on some maximal subinterval (xi, x2) of (0, 1). Let a(x) = mf(x, u,(x)ej) and b(x) = m&(x, +(x)ej). Since ui(x) > u2(x) on (x1, x2), the sign hypothesis on Dyjfj implies that a(x) < b(x) on (xi, x2). Now either x1 = 0 or ui(xi) = uz(xl) and u;(x,) 2 u;(x,). In either case,

Thus, lemma 3 (i) implies

Now either ui(xJ = u&) and ui(xz) 2 ui(x,), contradicting the previous x, = 1, in which case the boundary condition gives a contradiction. Passing to the second assertion of (ii), suppose u is nontrivial and satisfies _Y”+

WZ[fj(X, i(X)ej) + ii(X)DyjA(X, li(X)ej)]_Y =

inequality,

or

0

and the jth set of boundary conditions in (2), (3). We may suppose that bjU(X) + b,!u'(X)> 0 on some subinterval (0, x,) of (0, 1) and choose x1 so that this subinterval is as large as possible; it is easily seen that u is positive on (0, xi). Since D satisfies ii” +

mfj(X, fi(X)C?j)ri = 0,

with ti > 0, the sign hypothesis on Dyjfj and lemma 3 (ii) and the concavity of ii imply that ~j ti( 1) + &G’(l) < 0, a contradiction. It is clear that we can allow dependence on y’ in lemma 5 where the limit superior I 0 holds uniformly in y’. Less stringent variants of this dependence are possible. 6. Suppose that all the hypotheses of parts (i) and (ii) of lemma 5 are satisfied. Let ~j be the function whose existence is guaranteed by lemma 5 (i) and suppose that

THEOREM

fi(X, 1;,(X)ei)> 0,

for each i # j

(12)

and x E [0, 11. Then the set S contains points (ml, m2, . , . , m,) with mj arbitrarily large and, thus, S is unbounded. Proof. We assume without loss of generality that j = n. Throughout the proof, we write *e-9 Y,-I), A = @I,&, *e-9 A,_,). Let p,, be the smallest eigenvalue of

Y = (Yl,Y,,

Y,” + Afn(X, O)Y, = 0,

1338

J. V. BAXLEY and H. B. THOMPSON

and the nth set of boundary conditions in (2), (3). Choose and fix an arbitrary m, > p,, . We shall show that S contains a point whose nth coordinate is m,. Then by lemma 5 (i), the problem consisting of Y,”+ m,f, (x,0, YJY, = 0 and the nth set of boundary i= 1,2 ,..., n - l,let

conditions

in (2), (3) has a positive

solution

ii,(x).

For

1 F,tA,Y9

YA

=

YiCx) - li

Gi(x, r)Yi(r)fi(tv

s0

Y(t), Yn(r)) dT

and 1 F,@,Y,YJ

= YnW

-

mi

.r0

G,, (xv T)Y, (Qfn (5, Y(T), in

dr.

Also let F(kY,Y,)

= (F,(A,Y,Y,), ..*9F,(kY,Y,)).

Then F(I,, 0, 2,) = 0 for all A E:R”-‘. proof of theorem 2, we let

We shall apply theorem 1 to this problem. As in the

B, =

B2 = fi L2(0, I), i=l

andalso,fori= 1,2 ,..., n - 1, let Ai be the positive self-adjoint differential weighted Hilbert space L2((0, 1); Wi) with the weight function

operator in the

defined by AiY = -(l/Wi)Y” and the ith set of boundary conditions in (2), (3). Note that by hypothesis, WI(X) > 0 for each i = 1,2, . . . , n - 1. Let pi be the smallest eigenvalue of Ai ; then pi is positive and simple and we choose the corresponding eigenfunction ui with unit norm and positive on (0, 1). Now expand &I, y, y,) about (p, 0, i;,) using Taylor’s formula to get F(kY,Y,)

= &+J(PuI 0, fi,)(Y,Y, - fi,) + Fx&&)(P9 0, &I)(~ - P,Y,Y, - &I) + P@,Y,Y,),

where P@, 0, ii,) = 0, I+,~,#,

0,&J = 0, P~,~~,_,&, 0, ii,) = 0. Put Uk

kz) = qy,&JP* 0, kJ(k

48)

and B(A, z>z,) = fi,(y,y_J~, 0, kJ(k

z, z,).

Then L is a bounded linear operator mapping B, -+ B2 and B is bounded and bilinear, mapping R”-’ x B, * B2. A straightforward computation gives, for i = 1, 2, . . . , n - 1, 1 (Uh9

hd)i

=

hi -

Pi

s0

Gi (X9 rlfi (~vov fin (T))hi(r) dz

Nonlinear

boundary

value problems

1339

and

1 -

n-1 Gn (~3

r)fin

(t)

C i=l

s 0

Dyifn

(~3

0,

iin (r))hi

(T)

dr.

Thus, L(h, h,) = 0 if hi = vi for i = 1,2, . . . , n - 1 and

n-1 =

-m,

iin (X)

C i=l

Dy,fn

(X3 0,

fin (x))ui(x)*

This last equation has a unique solution u,(x) by lemma 5 (ii) and the Fredholm Thus, u = (v,, u2, . . . . v,) E N(L). We also compute for i = 1,2, . . ., n - 1

alternative.

1 (B(Jy

Z9 Z,))i

=

-ii

Gi(X,

t)fi(z,

0,

cn(r)ki(r)

dr

50 and (B& z, z,)),

= 0.

Now put n-1

Z = n R(Ai - I

X

L*(O,1)

i=l

and consider

T:R"-' x Z + B2 defined by

2-k z) We show that T is one-to-one 2, to give a Cl-curve

= Lz + B(I, u).

and onto and then theorem (n(s), s(u + w(G), 4

1 applies,

as in the proof of theorem

+ s(u, + w,(s)))

with (A, v/, I//,): (-&a)

+ R"-' x Z

and Q(O), w(O), v/n(O)) = (A

0, 0)

for which F(A(s), s(u + w(s)), 2, + s(u, + w,(s)))

= 0

for -6 < s < 6. For 0 < s < 6, then (A(s), m,) E S, and A(s) is close to p. We proceed that T is one-to-one. In order for (I, z) E N(T), it is necessary that Z/ +

/lifi(X,

0,

ii&i

=

-;iifi(X,

0,fi,)Ui(X),

to show

1340

J. V. BAXLEY and H. B. THOMPSON

for i = 1,2, . . . . n - 1. But since ui is orthogonal (with respect to the appropriate weight function) to R(Ai - pil), this last equation forces fti = 0, Zi = 0, for i = 1,2, . . . , TV- 1. Since it is then also necessary that z,” + 111,[f, (x, 0, &I) + &JJJ,(x, lemma 5 (ii) forces z, = 0, and T is one-to-one. of theorem 2, and uses lemma 5.

0,

&Jlz, = 0,

The proof that T is onto follows as in the proof

The hypothesis in theorem 6 that (12) holds seems somewhat awkward. So we describe several situations in which it is satisfied. (i) The hypothesis (12) is trivially satisfied if fi(x, _vej) > 0 for all y > 0 and i # j. (ii) If there exists y, such that 6(x, yej) < 0 for Y > ye then ~j(X) I y. for 0 I x 5 1. Otherwise, ~j has a maximum at some x0 with ii, > _Y,. Even if x0 is an endpoint of [0, 11, because of the boundary conditions, fi,!(x,) = 0. But then z$‘(x,) = -mjfi(Xo, fij(xo)ej)tij > 0, contradicting the maximum at x0. Thus, (12) is satisfied if&(x, yej) > 0 for 0 I y I y, and all if j. (iii) More generally, suppose there exists a line I(x) = a + bx for which fj(x, yej) < 0 for y > I(x) and for which UjI(0) - a,!/‘(O) 1 0, bj/(l) Then ~j(X) I I(x) for 0 I some point x0 E [0, l] and

x 5

+ &1’(l)

2 0.

y(x)

Cj(X)

1. Otherwise

=

-

l(x)

has a positive

maximum

at

aj_Y(O) - Q,!y’(O) I 0, bjY(1) + bj_Y’(l) I 0. Therefore,

_Y’(x,) = 0 even if x0 is an endpoint Y”(Xo)

=

ZIy(Xo) =

of [0, 11. But

-mjfi(XO,

ti,.(X&j)ij

>

0,

a contradiction. Hence, (12) is satisfied if fi(x, yej) > 0 for 0 I y I I(x). (iv) In addition to part (iii) above, suppose it is true, as in the chemostat problem (4)-(9) with I(x) = w(x), that fj(x, t(x)ej) = 0 for 0 5 x I 1; then Cj(X) < f(x) for 0 I x I 1. We already = Z’(x,) and since know that Cj(x) 5 I(X). If ij,(Xo) = /(x0) at some x0, then necessarily ii,! r”(x) + nzjfj(x, I(x)ej),(x) = 0, uniqueness gives Gj(X) = 1(x), an impossibility since I(x) cannot satisfy both boundary conditions because aj + bj > 0. (v) As in the remark after corollary 1, we may replace (for n > 2 and i = 1,2, . . . , n - 1) ui by ti ui, where the ti are positive and sum to 1, to vary the direction of bifurcation. We conclude this section with some observations about higher eigenvalues. The eigenvalue problem consisting of y” + 1h(x, and the jth

set of boundary

conditions

0)y = 0

in (2), (3) has a discrete

unbounded

set

a.1 < I, < +** < I, < *aof eigenvalues

with corresponding

eigenfunctions

4k. Lemma

5 (i) can be extended

as follows.

1341

Nonlinear boundary value problems LEMMA 7. With the hypotheses

of theorem

2, suppose

m > A, and the boundary

value problem

Y” + mfj(x, yej)y = 0

(13)

and thejth set of boundary conditions in (2), (3) has a solution fik with k (0 I k < p - 1) zeros in (0, 1). Then this boundary value problem has solutions fii, i = k + 1, . . . , p - 1 with exactly i zeros in (0, 1).

Proof. We show that the existence of iik implies the existence of z&+i and the existence of the remaining iii follows by induction. It is well known that &, has exactly p - 1 zeros in the interval (0, 1). There is nothing to prove if p = 1, so we may assume that p 1 2 and we may let xi < x, < ... < xP_i be the zeros of $P in (0, 1). As in the proof of lemma 5 (i), we may assume that aj > 0 and define Ys in the same way for s > 0. Using continuous dependence, we see that there exists 6 > 0 SO that mfj(x,Y,eJ > ppu,fj(X, 0) for s E (0,6]. For s E (0,6], an application of lemma 3 (iii) shows that Ys has at least one zero in the interval (0, xi) and the Sturm comparison theorem guarantees that Ys has at least one zero in each interval (Xi 9Xi+,), i = 1,2, . . . . xP_,. Thus, for s E (0,6], Y, has at least p - 1 > k zeros in the open interval (0, xp- 1). Define t as the infimum of all those s for which Y, has exactly k zeros in (0, 1). Clearly t > 6, yt has exactly k zeros in (0, l), and Y,(l) = 0. Now define r as the infimum of all those s E (6, t) for which Y, has exactly k + 1 zeros in (0, 1) for all q E (s, t). Using continuous dependence, we see that Y, has k + 1 zeros in (0, 1) and either (a) Y,(l) = 0 or (b) r = 6. The idea here is simply to reduce s below t to the point r at which either Yr has k + 2 zeros in (0, l] or r = 6, whichever occurs first. We shall complete the proof by showing that Y, satisfies the boundary condition bjY,(l) + 6jYj(l) = 0 for some s E [r, t), thus producing the desired solution iik+, . Consider case (a). If the desired boundary condition at x = I is a Dirichlet condition, then Y, is the desired solution. Otherwise, examining the graphs of Y, as s decreases from t to r, it is clear that the sign of Y:(l) is opposite that of Y;(l). Since the sign of bjY,(l) + !$Yb(l) is the same as the sign of Yj(1) for s near t and for s near r, then because of continuous dependence, Ys satisfies the desired boundary condition at x = 1 for some s E (t, t). In case (b), note that k + 1 = p - 1, Y: + b(x)y, = 0 and 4; + a(x)&, = 0 on [xP_i , 11, where b(x) = m&(x, Y,(x)eJ and a(x) = Apfj(X, 0); since these are linear equations, by negating if necessary, we may assume that both yI and &, are positive on (xP_, , 1). Applying lemma 3 (i) on the interval (xP_, , l), we see that

Because

r$P satisfies

4~(l)YXl)

- $Jl)Y,(l)

the boundary

condition bjY,(X,-1)

< -$&I)Yr(x,-l) at x = 1, we conclude

< 0. that

+ b,!YXX,-1) < 0.

But a consideration of the behavior of the graphs of ys as s decreases from t to r shows that the reverse inequality is true for s near t. Again, continuous dependence gives us an s between r and t for which ys is the desired solution iik+i. As an immediate

consequence

Olilp-

set of boundary 1.

5 and 7 we have the following

corollary.

of lemma 5, for m > ,$, there exist solutions of problem (13) conditions in (2), (3) with exactly i zeros in (0, 1) for each i with

COROLLARY 3. With the hypotheses

and the jth

of lemmas

1342

J. V. BAXLEY and H. B. THOMPSON

There is literature giving variants of corollary of Hempel [6]. Hempel considers the problem (@)Y’)’

3. Corollary

3 is closely related

to theorem

3

+ C(xlY - b(x, Y) = 0,

(14)

Y(0) = 0,

(15)

Y(1) = 0,

(16)

using the calculus of variations and genus to obtain his results. tinuously differentiable, c(x) 2 0 and continuous, and (@)Y’)’

He assumes

a(x) > 0 and con-

+ PC(xlr = 0,

(17)

Y(0) = 0,

(18)

Y(1) = 0,

(19)

has distinct eigenvalues 0 < p, < ,+ < ... tending to infinity. Replacing our y” by (a(x)y’)’ and assumingf(x, 0) = c(x) where c is as above instead off(x, 0) > 0, lemmas 5 and 7 and corollary 3 follow. Settingf(x,y)y = c(x)y - b(x,y) his results follow as a special case of ours. Corollary 3 could have been derived from the work of Rabinowitz and Crandall (see, for example, [2, 5, 71 and their bibliographies). They use Rabinowitz’s global bifurcation theorem, an extension of the Krasnoselskii bifurcation theorem, to establish a range of related results. If E = C’[O, l] and & = R x E, it is easy to see from the work of Crandall and Rabinowitz that the set (Ak, yk) of solutions of (11) and the jth set of boundary conditions from (2), (3), with ~~(0) > 0 (y;(O) > 0 if yk(0) = 0), having k - 1 zeros in (0, 1) is unbounded in E. Moreover, from the proof of lemma 5 we see that such solutions have lykl bounded and, hence, bounded in E and that Ak is bounded away from 0, giving the result. 4. APPLICATION

TO THE

CHEMOSTAT

We now return to the problem of steady-state solutions in the chemostat. We consider problem with n species, so that the two dimensional system (4)-(9) is replaced by yf’ + migi

_Yk Yi =

the

09

(20)

0,

(21)

>

Y:(o) =

Y;(l)

+ YYi(l) = Ov

(22)

gi is strictly increasing, bounded, for i = 1,2, . . . . n. We assume only that each function belongs to C’(R), and satisfies g,(O) = 0. If the gi are given explicitly as in (9), before applying our results, we would need to redefine g,(S) for S < 0 so that gi is a bounded C’ function; one such extension is given by gi(S)

= tan-’

2s - + 1

( > ci

- :,

for S < 0.

1343

Nonlinear boundary value problems

that W(X) = (1 + y)/y - x and for i = 1,2, . . ., n, the functions gi are increasing, belong to C’(R), and satisfy gi(O) = 0. Then the set S of n-tuples (m,, m2, . . . . m,) for which the system (20)-(22) has a positive solution u = (u, , u2, . . . , u,) on [0, l] is unbounded. Moreover, any such solution u satisfies

THEOREM 8. Suppose

bounded,

S(x) = w(x) -

i

u,(x)

> 0,

05x5

1,

(23)

k=l

and for each distinct

pair i, j, there exists x0 E [0, l] for which Sj (K%))

mi

mj=gi(s(x,))* In particular, if the functions gi all have the form (9), then each of the fractions the interval between 1 and Ci/cj .

mi/mj

lie in

Proof. We shall verify (23) and the other assertions of the theorem follow immediately from theorem 6 and corollary 2 to theorem 4. If (23) is not true, then S(x) has a nonpositive minimum at some x,, E [0, 11. Since the boundary conditions (22) at x = 1 imply that S’(l) > 0, then the minimum cannot occur at x = 1. Thus, using (21) if necessary, S/(x,) = 0. Note that S satisfies the differential equation S” = Cz= I mkgk(S) for x E [0, 11. If S(x,,) = 0, then uniqueness implies S = 0 on [0, 11, which is impossible since S’(1) > 0. If S(x,) < 0 then the differential equation gives S”(xJ < 0, which is impossible at a minimum. The final result of this last theorem agrees with the numerical observation of So and Waltman [l]. In the case of two species, their computations indicated that the region in the (ml, m2) plane for which coexistence could occur is very “thin”. For c, = 1.5, c, = 2.0, m, = 1.32475, their computations indicated that coexistence would occur only for 1.6689 < m2 < 1.6691. Our bounds give m, < m2 < 4m,/3, or 1.32475 < m2 < 1.76633. Notice that our bounds are very crude, being based first on using the positivity of a function throughout an interval to conclude positivity of the integral, and second on replacing a function by its global maximum and minimum values. However, our result shows in general that the region of coexistence lies very close to the m, = m2 line and becomes “thinner” as c, and cq come closer together. We make one further comment about the chemostat system (20)-(22), in which the gi are given explicitly by (9). Suppose for some distinct pair i, j, it happens that Ci = Cj. Then gi = gj also. In this case the last conclusion of theorem 8 states that mi = mj is necessary for coexistence. Intuitively, there is really only one species that has been subdivided into two groupings i and j. In the two-species case, it is clear that they should be able to coexist in this case. We examine this situation more closely. To illustrate we assume that c,_i = c, in (14) so that m,_, = m,. Lety = (yi, . . . . y,) be a positive solution of (14)-( 16). As the first eigenvalue is simple and our solutions are positive we know that yn = cy,_, for some positive constant c. of the system Thus, z = (z, , . . . , z,_,) = (yi , . . . , Y~_~, (1 + c)y,_,) is a positive solution (14)-(16) for n replaced by n - 1. Conversely, let 0 < t < 1 and z = (zi , . . . , z,_J be a positive solution of the system (20)-(22) for n replaced by n - 1. Then yt = (Zl

9

.

.

.

,

tzZn-1, (1 - tk-l)

1344

J. V. BAXLEY and H. B. THOMPSON

is a positive solution of the system (20)-(22) and m,_ 1 = m, are independent of t. The solutions we have obtained by bifurcation are locally unique so these will be the solutions we would obtain by applying our results directly to the system (20)-(22). More generally, consider the system (l)-(3) with f,- I(x,y) = f,, (x,y) and the same boundary conditions for y,_ 1 and y, . By corollary 2 of theorem 4, m,_, = m, and, by simplicity of the first eigenvalue, we see that we can recover all our positive solutions by considering the system of the first n - 1 equations with fi (x,Y) replaced by fi (x,Y 1, . . . , t~~_~, (1 - t)y,_,). It is also clear from previous work on the chemostat that one species will out compete the other if mi f mj but ci = Cj. The impossibility of coexistence in this last case is also clear from lemma 3; the equation with the larger mi will oscillate faster than the other and, thus, there cannot be two positive solutions of both boundary value problems. REFERENCES 1. So J. & WALTMANP., A nonlinear boundary value problem arising from competition in the chemostat, Appl. maUr. Camp. 32, 169-183 (1989). 2. CIUNDALLM. G. & RABINOWITZP. H., Nonlinear Sturm-Liouville eigenvalue problems and topological degree, J. math. Mech. 19, 1083-1102 (1970). 3. CRANDALLM. G. & RABINOWITZP. H., Bifurcation from simple eigenvalues, J. funct. Analysis 8, 321-340 (1971). 4. SMOLLERJ., Shock Waves and Reuction Diffusion Equations. Springer, New York (1983). 5. CRANDALLM. G. & RAFIINOWITZP. H., Mathematical theory of bifurcation, in Bifurcation Phenomena in Mathematical Physics and Related Topics (Edited by C. BARDOSand D. BESSIS),pp. 3-46. Reidel, Dordrecht (1980). 6. HEMPELJ. A., Multiple solutions for a class of nonlinear boundary value problems, Indiana Math. J. 20, 983-996

(1971). 7. RABINOWITZP. H., Some aspects of nonlinear eigenvalue problems, Rocky Mount. J. Math. 3, 161-202 (1973).