A semilinear elliptic boundary value problem at resonance where the nonlinearity may grow linearly

Nonlinear Anolysrs, Theory. Printed in Great Britain.

Methods

& Applrcomns,

Vol. 16, No.

12, pp.

1159-1168,

1991. 0

0362-546X191 $3.00+ .OO 1991 Pergamon Press plc

A SEMILINEAR ELLIPTIC BOUNDARY VALUE PROBLEM AT RESONANCE WHERE THE NONLINEARITY MAY GROW LINEARLY ADOLFO RUMBOS Department of Mathematics, University of Utah, Salt Lake City, UT 84112, U.S.A. (Received 19 January 1990; received in revised form

1 May 1990; received for publication 6 November

Key words and phrases: Resonance, Landesman-Lazer

condition, Leray-Schauder

1990)

Degree.

1. INTRODUCTION

IN THIS paper

we study the weak solvability

of the semilinear

$24(x) - cYU(X)- g(x, U(X)) = h(x)

boundary

value problem

XEQ (1)

ulan

= 0,

where a is an open and bounded subset of RN, N 2 1, h E L’(sZ), d: is a second order uniformly strongly elliptic differential operator in divergence form

where aij are bounded Caratheodory function, is measurable on a for We shall assume that of the linear boundary

measurable functions on Q with aij = aji; g: Q x R + R is a i.e. the map r -+ g(x, r) is continuous for a.e. x E Q while x -, g(x, 0 all r E R. g(x, <) has at most linear growth in [, and that CYE R is an eigenvalue value problem Gcu(x) = (YU(X)

x E sz, (2)

ulan = 0. We shall not assume that the ratio g(x, O/T is bounded away from zero for large values of ]
+ b(x)

for [ E R and a.e. x E SJ

0 E [0, l), and b E L’(Q) with b 2 0 a.e. in [2] and [3], Ahmad has considered the case in which g(x, {) = g(r) may r, and cr = pr is the first eigenvalue of the linear problem (2). In [3] for shows that if

1160

A. RUMBOS

for some r > 0 and 0 < y < ,Q - ,u, , where ,u2 is the second eigenvalue of the linear problem (2), and if the following Landesman-Lazer condition is satisfied

~-~~~s(x)dr
= limfnfg(0, - m

g-cc = l;mstPg(r), and f3is an eigenfunction corresponding to Q!= pul in (2) with 8 > 0 in Q and au/an < 0 on K? (assuming aQ is smooth enough), where n is the outward unit normal to XJ, then problem (1) has a solution. In both [2] and [3], the fact that the first or principal eigenvalue of the linear part possesses an eigenfunction which is positive almost everywhere on the domain a is crucial for the arguments presented there. By combining one of Ahmad’s ideas in [2] and [3] with Hetzer’s technique in [l], the author has found that problem (1) has a weak solution for CYbeing any eigenvalue of the linear part, regardless of its multiplicity and the structure of its eigenspace. This is true provided that g satisfies a more general Landesman-Lazer type condition similar to that found in [1] and [4], and that g is not allowed to cross the eigenvalue; that is, the ratio g(r)/< is only allowed to approach the eigenvalues of the linear problem from one side only for large ItI. This is known as a one-sided resonance problem. In fact, let lpj]TZ 1 denote the set of eigenvalues of problem (2) and suppose that there is some constant y > 0 such that Y < 3 min,+,IlPj - alI. Let W denote the eigenspace corresponding to cxand V denote the orthogonal complement to W in L2(C2). Suppose that g satisfies the growth condition o _ g(x, 0 -=-52y r

for all l
for some r > 0, and that the following Landesman-Lazer

type condition holds

lim sup h(x)w,(x) dx > 0 g(x, Pn w,(x) + Pn ~,(X))~,(X) dx + n+m U R in 1 for all w, E W with IIw,II = 1 for all n, u,, E I/ with IIv,Ij -+ 0 as n + 00, and pn > 0 with pn + 00 as n + 00. Then problem (1) has a weak solution. This is the content of theorem 1 in Section 3. In Section 2 we shall present some facts from the linear theory that will be needed for the arguments of Section 3. 2. PRELIMINARIES

Let n be a bounded and open subset of RN, n 2 1; and let d: be a second order, symmetric, strongly uniformly elliptic differential operator in divergence form; that is,

A semilinear

elliptic boundary

1161

value problem

where (l;,j E L”(Q), Uij = aji, for all i, j E (1, . . . . NJ; and Ci,jaij(X)rirj 2 clLIZ for all X E Cl, RN, and some constant c > 0 (here, ]
r E

fl For every f E L2(S2), there exists a function solution to the linear problem

w

&,

dx =

h4 dx.

.ia

Tf E HA(C2) such that u = Tf is the unique

=Wx)

=

f(x)

4an

=

0,

weak

XEQ

where T: L2(sZ) + Hi is a bounded, symmetric, and positive linear operator, see for example [4-61. By composing with the compact embedding Hi(Q) q L2(sZ), we can think of T as a compact, symmetric, positive linear operator on L2(sZ). This yields the existence of a discrete set of eigenvalues ( pj}JYE1, such that . ..pj

o<&I~21

5

5

pj+l

. . .

and pn + co as n -+ co for problem (2). If cy is an eigenvalue of problem (2) and y > 0 is such that

then CY+ y is not an eigenvalue

of problem

&U(X) - au(x)

(2). Therefore

=

problem

XEC2

- vu(x) = h(x) ulan

where h E L2(sZ), which is equivalent

the linear

0,

to the problem U=(CX + y)Tu + Th

has a unique solution u = Ah, where A = (1 - (CX+ y)T)-‘T P(Q) to L2(Q). One can estimate the operator norm of A by 1 IIAII 5

is a compact

1

dist(cu + Y, (~jujlj”=1)= minja

linear operator

1

+ 7 - pj] = v’

Denote the eigenspace corresponding to Q!by W. For any w E W we have that Aw hence [IAl] = l/y. Let I/ denote the orthogonal complement to W in L2(sZ). For every w E W and u show that (Au, w) = 0, where (* , *) d enotes the L’(Q) inner product. Hence V under A. Let Al v denote the operator A when restricted to V, then one can show /]A 1~115

rninPjinial+

from

y - ,u~] < t.

= -(l/y)w, E V one can is invariant that

1162

A. RUMBOS 3. THE

SEMILINEAR

PROBLEM

In what follows we shall assume that g: Q x R + R is a Caratheodory r + g(x, r) is continuous for a.e. x E Q, while x + g(x, 0 is measurable that it satisfies a growth condition of the form Ig(x, 01 5 alrl

function, i.e. the map on n for all < E R, and

for r E R and a.e. x + Q

+ b(x)

(4)

for some constant a 1 0, and some function b E L2(sZ) with b L 0 a.e. For functions of this type one can show that the substitution operator u - N(U), given by N(u)(x) = g(x, u(x)) for x E Q, is a continuous map from L2(sZ) to L2(sZ) which maps bounded sets to bounded sets. THEOREM

1. Let 6: be a second order, symmetric, uniformly strongly elliptic differential operator as described in Section 2. Let (Y > 0 be an eigenvalue of the linear boundary value problem (2), and denote its eigenspace by W. Let V denote the orthogonal complement to W in L2(sZ). Let y > 0 be such that Y < tmi+

-

J

41,

where (~j]j”_l is the set of eigenvalues of the linear problem (2). Let h be in L2(sZ), and g: Q x R + R be a Caratheodory function satisfying a growth condition given by (4), and such that o _ g(x, 0 <-522y r for some r > 0. Suppose

for all 151 L r, and for a.e. x E 52,

the following

Landesman-Lazer

lim sup g(x, Pn w,(x) + Pn ~,(X))W,(4 “-CC is D

type condition dx +

for all sequences (w,): c W with I]w,II = 1 for all n, (u,): the boundary value problem (1) has a weak solution. Proof.

Consider

as in [2] and [3] the family

iD

(5) holds

h(x)w,(x)dx 1

> 0

c V with IIu~I] + 0 as n

(6) co. Then

of problems

GU - a!U - yu - t(N(u)

- yu) = h

in Q (770

ulan = 0, for t E [0, 11, where N(u)(x) = g(x, u(x)) for x E &2. The weak solvability of these problems is equivalent to the solvability equations u - (cu + y)Tu = tT(N(u) - yu) + Th inL2(SZ)for05t5

of the operator

l,or u = tA(N(u)

- vu) + Ah

in L2(Q) for 0 I t 5 1, where A = (I - (CY+ y)T)-‘T is the compact previous section. Observe, as in [l], that if u E L2(a) is a solution of u = t(N(A u) - yAu) + h for t E [0, 11, then Au is a solution

of (8, t).

(87 t> operator

described

in the

(930

1163

A semilinear elliptic boundary value problem

Let H: [0, l] x L2(sZ) + L*(Q) be given by H(t, 2.4)= 24 - t(N(Au)

- yAz.4) - h.

Then H is a homotopy of compact perturbations of the identity. The result will then follow from the homotopy invariance of the Leray-Schauder degree, if we can show the existence of an a priori bound for solutions of in L*(Q) H(t, u) = 0 independently of t. Suppose, by way of contradiction, that there exists a sequence (u,) of solutions of H(t, u) = 0 in L’(Q), with a corresponding sequence (t,) in [0, 11, such that 11~~11 -+ 00 as n + 00. Let P be the orthogonal projection of L2(sZ) onto W, and write u, = Pu, + (I - P)u, = unO+ unl, where uno E W and u,r E V for all n E N. We have the following two cases: Case 1: there exists a 6 > 0 and a subsequence (nj) of (n) for which I]u,,,]l I Sllz~,~, II for allj. Case 2: I]u,J #O for all n, and

II~nIlI , () Il~noll

asn

+ 00.

Let q be a continuous real valued function on R such that 0 5 q 5 1, q = 1 on [-r, r], and q(r) = 0 for I<] 2 2r. Put g, = qg and g, = (1 - q)g. Then g = g, + g,, where g, and g, are Caratheodory functions on fi x R. Writing N,(u)(x)

= gr(x, u(x))

for a.e. x E Sz

N,(u)(x)

=

g2k

for a.e. x E fi

= N,(u)

+

u(x))

and u E L2(sZ), we see that Mu) Since

for all 24E L*(Q).

N2(u)

klcc 0 5 w(ml + I

for all l E R, and a.e. x E Sz

for some a L 0, and b E L2(sZ) with b 1 0 a.e., Igr(x, 0

5 k + b(x)

for some constant

k 2 0. Hence,

and some constant Using condition

IINr(u)ll 5 K K 2 0. (5), we have that o _ g2k <-52y

t-1

we see that

for all c E R, and a.e. x E Sz

since 0 is bounded, for all u E L*(Q)

for all l
r Defining

g2(x, <)I< to be 0 when r = 0 for all x E Q we may assume o _ g&G 0 <-52y r

that

for all r E R, and a.e. x E Sz.

A. RUMBOS

1164

This may be written

in the form _y I &(X 0 p-y
for all [ E R, and a.e. x E Sz

or I”‘“, ‘-y

i:

/

5y

for all r E R, and a.e. x E C2.

Hence I&(X> 5) - rrl 5 rlrl which implies

that II&(4

In case 1, by replacing we obtain

- Y4 5

for all r E R, and a.e. x E Q

Yll4

for all 2.4E L’(Q).

u by u,,, and t by tn, in (9, t), and taking

where a = l/y = IIAII. H ere we have used condition The last inequality shows that

(10)

the inner product

(10).

lim inf ! IIAUnJli > 1 n-m a llUqll . Let b = IIA I yII where A I v denotes Now, since both

the operator

W and V are invariant

A when restricted

subspaces

(11) to V.

of A, we may write

ll~unjl12 = Il~~njol12 + l14j,l12. Thus

IIAunjl12 5 ~*ll~njclll* + ~2114j1112.

Hence

llAUn;llI ~~II~,iol12~ll~,i~l12~ + @*/a*))“* (dl~njol1241wz,,l12) + 1Y2 . a Ilwz,ll 1

Since for allj, we may assume,

passing

to a subsequence

if necessary,

II%Iioll~ ‘5 II~n,III

asj

with u,,.,

that + co

A semilinear

where r E [0,6].

elliptic boundary

1165

value problem

Hence lim inf 1 ll~&J j-a

0

I (r2 + (b2/a2)Y2 (t2 + 1)“2

IIu,;II I

*

Therefore

1 IIAUn,ll <

inf

lim

1

j-m

a IlU,,ll

since a > b by (3). This contradicts (11). t ,, < 1, then there exists a number 8, In case 2, we first show that t, -+ 1. If lim inf,,, 0 I 8 < 1, and a subsequence (t,) with t,,, 5 0 for all j. Replacing u by u,,, and t by tnj in (9, t), and taking the inner product with u,,,, we obtain, using condition (lo), II~njl12= 5

LjW2w4tj)

-

w4tj)9

%J)

+

t,w1w,),

hi)

+

v,

%/I

ell~n,ll” + W + Il~ll)ll~~,ll.

This shows that

IIU“J.I1I

K1-e.+ l”l t, = 1.

But this contradicts the assumption that IIu,J + co. Thus lim,,, Now, taking the inner product with Us,,, we obtain that 0 = 11%112- ~,(W(A&J

- YA4A

= ll~nol12- t,WWkA

kJ

= II%#

+ A%J,

- mW4lo

+ Yt,W,,

+ AhI

Thus,

dividing

by l]u,J

- (h %lo)

+ 4 Yb%

3&IO) - (A, 40)

%o)

9 %lcJ - @, ~,o)

= II~nol12- +-;40 = (1 - cl)ll~,0112 - t,

%)

+ Au,,), N -;40 ((

ho)

- fn@nO,40) - (k u,o)

+ A&z,

)

,%I0 - (k &o). )

and putting

we have that 0 = (1 - tn)ll&lOII + t,NPnW,

+ PnhAWn)

where w, E W with 1)w,,II = 1 for all n, u,, E V with pn -+ 00 as n + co. We then have that for all n, t,w(PnWn

+ Pn%),

w,)

+ @, WA

I]u,II +O

+ (h,

w,)

as n -+ to, and p,, > 0 with

5 0.

Hence lim SUP(UP,W, n-m

which contradicts

condition

+ P~uJ,

(6). This completes

w,)

+ V,

the proof.

w,N

n

5 0.

1166

A. RUMBOS

Remarks. (1) This result represents an improvement over the results of [2] and [3] by not requiring the simplicity assumption on the eigenvalue CY,as well as the assumption that it has an eigenfunction which is strictly of one sign on Q. This is obtained, however, at the expense of placing more restrictions on the nonlinearity g. Condition (5) allows the ratio g(x, o/r to interact with the eigenvalue (Yfrom the right side only, i.e. g(x, <)/r is not allowed to “cross” the eigenvalue, but rather approach it from one side only. Problems of this type are known in the literature as one-sided resonance problems. (2) If condition (5) is replaced by the condition -2y

g(x, 0 < 0 I r -

for all I<] 2 r and for a.e. x E Q

for some r > 0, we obtain a similar existence result provided and limsup is replaced by liminf.

that in (6) the inequality

is reversed

(3) Example 1. The folloGng result also follows from a more general result found in [3]. Let Q be an open and bounded subset of RN with smooth boundary XJ, and d: be a uniformly elliptic symmetric differential operator as described in Section 2. Let (Y = ,ui be the first eigenvalue of the linear problem (2). It can be shown, as in [3], that pi is a simple eigenvalue and that it has an eigenfunction 13with B(x) > 0 for almost every x E Sz. Let h E L*(Q) and let g: R -+ R be a continuous function satisfying o&L2y

for all Ill 2 r, for some r > 0, r

where

p2 being the second

eigenvalue

of the linear

problem

(2). Suppose

(12)

g-~~~B(x)~<~~-h(x)e(x)dx
and g,,

are as before.

~W

Then the boundary

- Pl4e

- g(uW) =

value problem

w

XEL-2

4an = 0, has a weak solution.

This result will follow

PROPOSITION 2. If (12) holds,

from theorem

1 if we can prove the following.

then (6) also holds.

Proof. Without loss of generality we may assume that ]lf?]l = 1. Since ,~i is simple, then W = span(e). Suppose (12) holds. If (6) does not hold, then there are sequences w, E W, u,~~,andp,>O,suchthatIIw,II = lforalln,u,~OinL*(~)asn~oo,p,~ooasn~co, and an n, E N such that if n 2 n, then

D

g(AIw,

+ PnhAWfl +

hw, I 0. n

A semilinear

Write C

*k

=

elliptic boundary

1167

value problem

w, = c,8 for all n. Then Ic,j = 1 for all n. If lnkl is a subsequence 1 for all k, then for nk I n,

I Hence,

by Fatou’s

n

g(AZ, 0 + PQ %,)~ +

n

he I 0.

lemma .I n

-h(x)Qx)

dx 2 g+m

which contradicts (12). On the other hand, if (n,] is a subsequence

for nk 2 n,,

i

of (n) such that

in

e(x) dx,

of (n) such that cnr = -1 for all k, then

or

!

g(-h,(e

3

he 2 0,

- u,,))e +

n

R

for +Zn,. Passing to a subsequence if necessary, we may assume that v,, -+ 0 a.e. k + 03. Thus 0 - u+ > 0 a.e. for sufficiently large k. Hence, by Fatou’s lemma again, g--i’n which contradicts

e(x)dx

as

2 ie -h(x)&x)dx, n

(12). This proves the assertion.

(4) Example 2. We now consider the case in which CYis a simple eigenvalue, and every eigenfunction corresponding to CYchanges sign. This is the case, for example, of higher eigenvalues of a two-point boundary value problem of the Sturm-Liouville type. Suppose that g(x, 0) = g(r) for all x E fi and < E R, and that o&Q~2y r

for all J 0,

where y > 0 is such that

Suppose,

in addition,

that g is bounded

below on R and that

g, = lip+Ef then (6) holds and therefore, by solution. To see why (6) holds consider IIw,(I = 1 for all n, IIu,II + 0, and so that )(t9(( = 1. Then w, = c, B Q2- = lx E Sz I e(x) < 0). If (n,]~,,

ia

g(Pnrwnk

+ PnkQJW”k

theorem

g(t)

= 00,

1, the boundary

value

problem

(1) has a weak

sequences (w,)f c W, (v,)yc V, and (p,): c R+, with p,, + COas n + co. Let W = span(e) where 0 is normalized where (c,I = 1 for all n. Let Q2+ = lx E Q I e(x) > 0) and is a subsequence of (n)~ such that c,,~ = 1 for all k, then =

.i fit

g(p,, 8 + pnku,,)e +


dp,,

f3 + pnku,,)e.

A. RUMBOS

1168

Since v, + 0 in L2(sZ), we may assume, passing to another subsequence, that vnk + 0 a.e. in Q. Thus, for large k, since g is bounded below, and g(c) I 0 for r < 0, we have that

1 n-

g(p,,Q + Pn,,vn*)e 2 -m

for some positive constant m. Therefore

.i a

g(P?zkwrzk+ PnkVnk)Wnk 2

g(p,,(e .i Cl+

+ v,,))e - m

for large k. Hence, by Fatou’s lemma and the assumption li$mirf g(r) = 00, we get that lim inf k-tm

s fl

g(P,, W?zk+ Pnpnkwnk

= O”

which implies (6). If, on the other hand, {n,);, 1 is a subsequence of In): with c,, = - 1 for all k, then

1

dPn,

wn,+ Pq vnk)wnk =-

l-l

.iR+

g(-Pn,(e -

v,,w+

i a-

g(p,,(lel + v,,))lel~

which implies again that lim inf k+-

.! D

‘!dP,,

WrQ + PQ %JWnk

= * 9

and so (6) follows. A similar argument shows that if g is bounded above and g-,

= lit; su_pg(<) = -00,

then problem (1) has a weak solution. Acknowledgements-This research is part of the author’s work towards the completion of a Ph.D. degree under the supervision of Professor E. Landesman, to whom the author wishes to express his most sincere gratitude for being a constant source of encouragement and ideas throughout the author’s graduate studies at the University of California, Santa Cruz. The author would also like to thank Professor Shair Ahmad, who refereed this paper, for helpful comments and suggestions, and for suggesting the example presented in remark 4 of Section 3.

REFERENCES 1. HETZER G., On semilinear operator equations at resonance, Houston J. Math. 6, 277-285 (1980). 2. AHMAD S., A resonance problem in which the nonlinearity may grow linearly, Proc. Am. math. Sot. 92, 381-384 (1984). 3. AHMAD S., Nonselfadjoint resonance problems with unbounded perturbations, Nonlinear Analysis 10,147-156 (1986). 4. DE FIGUEIREDO D. G., The Dirichlet problem for nonlinear elliptic equations: a Hilbert space approach, in Partial Differential Equations and Related Topics, Lecture Notes in Mathematics, No. 4, Springer, New York (1977). 5. LANDESMANE. M. & LAZER A. C., Nonlinear perturbations of a linear elliptic boundary value problem at resonance, J. Math. Mech. 19, 609-623 (1970). 6. LANDESMANE. M. & LAZER A. C., Linear eigenvalues and a nonlinear boundary value problem, Pacif. J. Math. 33,

311-328 (1970).