A unified monotone iterative technique for semilinear elliptic boundary value problems

Nonlinear Analysis 51 (2002) 567 – 586

www.elsevier.com/locate/na

A unied monotone iterative technique for semilinear elliptic boundary value problems S. K'oksal ∗ , V. Lakshmikantham Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, FL 32901, USA Received 20 January 2001; received in revised form 22 February 2001; accepted 15 May 2001

Abstract A unied approach to monotone iterative techniques is developed for elliptic boundary value problems when the nonlinear term involved admits a splitting of a di2erence of two monotone functions. The results obtained, using both the classical and variational methods, include several known results as well as some interesting new ones. c 2002 Elsevier Science Ltd. All rights reserved.
Keywords: Semilinear elliptic problems; Unied monotone iterative technique

1. Introduction It is well known [1–5] that the method of lower and upper solutions, coupled with the monotone iterative technique provides an e2ective and 9exible mechanism that o2ers theoretical as well as constructive existence results for nonlinear problems in a closed set which is generated by the lower and upper solutions. The lower and upper solutions serve as bounds for solutions which are improved by a monotone iterative process. The ideas imbedded in this technique have proved to be of immense value and have played an important role in unifying a variety of nonlinear problems [3]. Let
⊂ Rn be a bounded domain with the boundary @
. We consider the following semilinear elliptic boundary value problem (BVP) in nondivergence form n n

Lu = − ai; j (x)uxi xj + bi (x)uxi + c(x)u i; j

i=1

= F(x; u) in
; Bu =  on @
; ∗ Corresponding author. E-mail address: skoksal@t.edu (S. K'oksal).

c 2002 Elsevier Science Ltd. All rights reserved. 0362-546X/02/$ - see front matter  PII: S 0 3 6 2 - 5 4 6 X ( 0 1 ) 0 0 8 0 9 - 4

(1.1)

568

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

I R]; c(x) ¿ 0 in
;  ∈ C 1;  [
; I R]; F ∈ C  where we assume that ai; j ; bi ; c ∈ C  [
; [
I × R; R] and the ellipticity condition n
ai; j (x)i j ¿ ||2 i; j=1

holds in
with  ¿ 0. Moreover, we let p; q ∈ C 1;  [@
; R+ ] with p(x) ¿ 0;  be the unit outer normal on @
, and Bu = p(x)u + q(x) @u @ ;

I R]: u ∈ C 1 [
;

Assume that @
∈ C 2;  . A well known result [2,3,5] in monotone iterative technique is the following theorem relative to the BVP (1.1). Theorem 1.1. Assume that: I R] with 0 (x) 6 0 (x) in
and satisfy (A1 ) 0 ; 0 ∈ C 2 [
; L0 6 F(x; 0 ) in
;

B0 6  on @
;

L0 ¿ F(x; 0 ) in
;

B0 ¿  on @
;

(1.2)

(A2 ) F ∈ C  [
I × R; R] and for some M ¿ 0; F(x; u) + Mu is nondecreasing in u for I x ∈
. I R] such that Then there exist monotone sequences {n (x)}; {n (x)} ∈ C 2;  [
; 2 I n (x) → (x); n (x) → r(x) in C [
; R] and (; r) are the minimal and maximal solutions of (1:1); respectively. If F(x; u) satises a one-sided Lipschitz condition, namely F(x; u1 ) − F(x; u2 ) ¿ − M (u1 ; u2 );

(1.3)

where u1 ¿ u2 ;

M ¿ 0;

x∈


then F(x; u) + Mx is nondecreasing in u for x ∈
. The functions 0 ; 0 verifying (1.2) are known as lower and upper solutions of (1.1). The special case when F(x; u) is nondecreasing in u is covered in Theorem 1.1 when M = 0. However, the other case, when F(x; u) is nonincreasing in u is not included in Theorem 1.1 and is of special interest. Under somewhat special conditions, one can prove that when F(x; u) is nonincreasing in u, a single iteration procedure yields an alternative sequence which forms two monotone sequences bounding solutions of (1.1). The iteration scheme in the present case is simply either Ln+1 = F(x; n ) in
;

Bn+1 =  on @


(1.4)

Ln+1 = F(x; n ) in
;

Bn+1 =  on @
;

(1.5)

or n = 0; 1; 2; : : : . In this case, the following results are valid.

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

569

Theorem 1.2. Suppose that F(x; u) is nonincreasing in u for x ∈
and F ∈ C  [
I × R; R]. Then: (i) the iterates {n (x)} satisfy the relation 0 6 2 6 4 6 · · · 6 2n 6 2n+1 6 · · · 6 3 6 1 on
I I Moreover; the alternating sequences {2n (x)}; provided 0 6 2 on
. I R] to (; r); respectively; {2n+1 (x)} converge in C 2 [
; and (ii) the iterates {n (x)} verify 1 6 3 6 · · · 6 2n+1 6 2n 6 · · · 6 2 6 0 on
I I Furthermore; the alternating sequences {2n+1 (x)}; if we suppose that 260 on
. I R] to (r ∗ ; ∗ ); respectively. {2n (x)} converge in C 2 [
; Note that we did not assume condition (A1 ) of Theorem 1.1. In fact, one can show that lower and upper solutions exist satisfying (A1 ) in this case. A natural question that arises is whether it is possible to obtain the monotone sequences {n (x)}; {n (x)}, when F(x; u) is nonincreasing in u without the additional assumptions on the iterates, I The answer is positive if we assume the existence namely 0 6 2 and 2 6 0 on
. of coupled lower and upper solutions. In fact, one can prove the following result in this direction. Theorem 1.3. Assume that: I R] with 0 (x) 6 0 (x) in
and (A1 ) 0 ; 0 ∈ C 2 [
; L0 6 F(x; 0 ) in
;

B0 6  on @
;

L0 ¿ F(x; 0 ) in
;

B0 ¿  on @
;

(A2 ) F ∈ C [
I × R; R] and F(x; u) is nonincreasing in u for x ∈
. Then the conclusion of Theorem 1:1 is valid. 

In this paper, we shall investigate the situation when F(x; u) admits a splitting of a di2erence of two monotone functions or equivalently, F(x; u) = f(x; u) + g(x; u) where f(x; u) is nondecreasing and g(x; u) is nonincreasing in u for x ∈
. As we shall see, this simple setting unies and covers several known results as well as provides some interesting new results. We shall rst discuss the BVP (1.1) in this new setting using the classical method and then investigate the BVP in the divergence form via the variational approach. 2. The classical method In this section, we shall discuss in a unied setting, the monotone iterative technique by the classical method for the BVP Lu = f(x; u) + g(x; u) in
;

570

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

Bu =  on @
;

(2.1)

where f; g ∈ C  [
I × R; R]. I R] are said to be Denition 2.1. Relative to BVP (2.1), the functions ;  ∈ C 2 [
; (a) natural lower and upper solutions if L 6 f(x; ) + g(x; ) in
;

B 6  on @
;

L ¿ f(x; ) + g(x; ) in
;

B ¿  on @
;

(2.2)

(b) coupled lower and upper solutions of type I if L 6 f(x; ) + g(x; ) in
;

B 6  on @
;

L ¿ f(x; ) + g(x; ) in
;

B ¿  on @
;

(2.3)

(c) coupled lower and upper solutions of type II if L 6 f(x; ) + g(x; ) in
;

B 6  on @
;

L ¿ f(x; ) + g(x; ) in
;

B ¿  on @
;

(2.4)

(d) coupled lower and upper solutions of type III if L 6 f(x; ) + g(x; ) in
;

B 6  on @
;

L ¿ f(x; ) + g(x; ) in
;

B ¿  on @
:

(2.5)

Whenever  6  in
, we note that the lower and upper solutions dened in (2.2) and (2.5) also verify (2.4) and hence it is enough to consider cases (2.3) and (2.4), which is precisely what we plan to do. We need the following known results [3] to proceed further. I R] satisfying Lemma 2.1. For any p ∈ C 2 [
; Lp 6 0 in
;

Bp 6 0 on @
;

I we have p 6 0 in
. Lemma 2.2. Consider the linear BVP Lu = h in
;

Bu =  on @
;

I R]. Then (2:6) has a unique solution u ∈ C where h ∈ C [
; 

(2.6) 2; 

I R]. [
;

We are now in a position to prove the rst main result. Theorem 2.1. Assume that I R] are the coupled lower and upper solutions of type I with (A1 ) 0 ; 0 ∈ C 2 [
; 0 (x) 6 0 (x) in
; (A2 ) f; g ∈ C  [
I × R; R]; f(x; u) is nondecreasing in u and g(x; u) is nondecreasing in u for x ∈
.

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

571

I R] such that Then there exist monotone sequences {n (x)}; {n (x)} ∈ C 2;  [
; I R] and (; r) are the coupled minimal and n (x) → (x); n (x) → r(x) in C 2 [
; maximal solutions of type I of (2:1); respectively; that is; (; r) satisfy L = f(x; ) + g(x; r) in
;

B =  on @
;

Lr = f(x; r) + g(x; ) in
;

Br =  on @
:

Proof. Consider the following linear BVPs for each n = 1; 2; : : : ; Ln+1 = f(x; n ) + g(x; n ) in
;

Bn+1 =  on @


(2.7)

Ln+1 = f(x; n ) + g(x; n ) in
;

Bn+1 =  on @
:

(2.8)

and In order to conclude the existence of the unique solutions of the BVPs (2.7) and (2.8) I R] with 0 6 " 6  6 0 ; for each n ¿ 1, we need to show that for any ";  ∈ C 2 [
; I R] and h2 (x) ∈ C  [
; I R], where h1 (x) ∈ C  [
; h1 (x) = f(x; ") + g(x; ) and h2 (x) = f(x; ) + g(x; "): I R], then ";  ∈ W 2;  [
; I R] in view of the boundedness We note that if ";  ∈ C 2 [
; 2;  I of
and @
∈ C [
; R]. The imbedding Theorem A:3:5 of [2] then shows that I R]. Consequently, we have ";  ∈ C 1;  [
; |f(x; "(x)) − f(y; "(y))| 6 K1 [ x − y  + |"(x) − "(y)| ] 6 K1 [ x − y  + |"|C 1 [
;I R] x − y  ] 6 L1 x − y  ;

where L1 = K1 [1 + |"|C 1 [
;I R] ]:

Similarly, we get |g(x; (x)) − g(y; (y))| 6 L2 x − y  ;

where L2 = K2 [1 + ||C 1 [
;I R] ]:

As a result, we nd that, because of the denition of h1 (x), |h1 (x) − h1 (y)| = |f(x; "(x)) + g(x; (x)) − f(y; "(y)) − g(y; (y))| 6 f(x; "(x)) − f(y; "(y))| + |g(x; (x)) − g(y; (y))| 6 C x − y  ;

where C = L1 + L2 :

I R]. In a very similar way, we can show that h2 (x) ∈ C  [
; I R]. Hence, h1 (x) ∈ C  [
; 2;  I Consequently, for each n ¿ 1, there exist unique solutions n ; n ∈ C ]
; R] of the BVPs (2.7) and (2.8) by Lemma 2.2 provided that 0 6 n 6 n 6 0 . Therefore, our aim now is to show that 0 6 1 6 2 6 · · · 6 n 6 n 6 · · · 6 2 6 1 6 0

in
:

(2.9)

572

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

I For this, let p = 0 − 1 so that Bp 6 0 on @
We rst claim that 1 ¿ 0 in
. and Lp = L0 − L1 6 f(x; 0 ) + g(x; 0 ) − f(x; 0 ) − g(x; 0 ) = 0

in
:

I Similarly, we I which implies 0 6 1 in
. Hence by Lemma 2.1, p(x) 6 0 in
, I I can show that 1 ¡ 0 in
. We next prove that 1 6 1 in
. Consider p = 1 − 1 so that Bp = 0 on @
and Lp = f(x; 0 ) + g(x; 0 ) − f(x; 0 ) − g(x; 0 ) 6 0

in


using the monotone nature of f; g. Thus we get by Lemma 2.1, p 6 0 in
I which I As a result, it follows that yields 1 6 1 , in
. I 0 6 1 6 1 6 0 in
: I Then we show that Assume that for some k ¿ 1; k−1 6 k 6 k 6 k−1 in
. I k 6 k+1 6 k+1 6 k in
. To do this, let p = k+1 − k so that Bp = 0 on @
and because of monotone character of f; g we get Lp = f(x; k ) + g(x; k ) − f(x; k−1 ) − g(x; k−1 ) 6 0

in
:

I Similarly, we can show that k+1 6 k in Lemma 2.1 then implies k 6 k+1 in
. I I
. Now to prove k+1 6 k+1 in
, consider  = k+1 − k+1 and note that Bp = 0 on @
. Moreover Lp = f(x; k ) + g(x; k ) − f(x; k ) − g(x; k ) 6 0 in
using the assumption and monotone nature of f; g. Thus we have by Lemma 2.1, k+1 6 k in
and as a result, it follows that k 6 k+1 6 k+1 6 k

for any k ¿ 1:

Hence by induction, we see that (2.9) is valid for all n = 1; 2; : : : . I R] for n = 1; 2; : : :. Since C 2;  [
; I R] ⊂ W 2; q [
; I R] for We recall that n ; n ∈ C 2;  [
; q ¿ 1, by Theorem A:3:3 [3], we have

n W 2; q [
;I R] 6 C[ hn Lq [
;I R] +  W 2; q [
;I R] ];

(2.10)

where hn (x) = f(x; n−1 ) + g(x; n−1 ). I R]. Since The continuity of hn implies that {hn (x)} is uniformly bounded in C[
; q I q I I C[
; R] is dense in L [
; R], {hn (x)} is also uniformly bounded in L [
; R]. This I R]. For together with (2.10) shows that {n (x)} is uniformly bounded in W 2; q [
; 2; q I q = 2=(1 − ), n ∈ W [
; R] and hence by imbedding Theorem A:3:5 in [3]

n C 1;  [
;I R] 6 C n W 2; q [
;I R]

for n = 1; 2; : : :

for some positive constant C independent of the elements of W 2; q . Thus {n (x)} I R]. This implies that {hn (x)} is uniformly bounded is uniformly bounded in C 1;  [
;

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

573

I R]. Consequently, by Schauder’s estimate given in A.3.5 of [3], we nd that in C  [
;

n C 2;  [
;I R] 6 C[ hn C  [
;I R] +  C 1;  [
;I R] ] for all n; I R]. As a result, we have which implies the uniform boundedness of {n (x)} in C 2;  [
; 2 I {n (x)} is relatively compact in C [
; R] that yields the existence of a subsequence I R]. Let ∗ ∈ C 2 [
; I R] be the limit of {nk (x)}. By the {nk } which converges in C 2 [
; I But the convermonotone nature of n (x); {n (x)} converges pointwise to (x) in
. 2 I gence of {nk (x)} in C [
; R] implies pointwise convergence and thus ∗ (x) = (x) in I This shows that the entire sequence {n (x)} converges in C 2 [
; I R] to (x), that is,
. 2 I I limn→∞ n (x) = (x) in C [
; R] and 0 6  6 0 in
. I R] and 0 6  6 r 6 0 Similar arguments show that limn→∞ n (x) = r(x) in C 2 [
; I in
. Thus the limits lim Ln = L;

n→∞

lim Bn = B;

n→∞

lim Ln = Lr;

n→∞

lim Bn = Br;

n→∞

lim {f(x; n ) + g(x; n )} = f(x; ) + g(x; r)

n→∞

and lim {f(x; n ) + g(x; n )} = f(x; r) + g(x; )

n→∞

I We then see immediately that  and r are the solutions of the exist uniformly in
. semilinear BVPs L = f(x; ) + g(x; r) in
;

B =  on @
;

Lr = f(x; r) + g(x; ) in
;

Br =  on @
:

Finally, we claim that  and r are the coupled minimal and maximal solutions of (2.1), that is, if u is any solution of (2.1) such that 0 (x) 6 u(x) 6 0 (x), then I 0 (x) 6 (x) 6 u(x) 6 r(x) 6 0 (x) in
: I Setting p = n+1 − u; Bp = 0 on @
, we Suppose that for some n; n 6 u 6 n in
. I Similarly, u 6 n+1 in
. I Hence obtain p 6 0 by Lemma 2.1, implying u ¿ n+1 in
. I I n+1 6 u 6 n+1 in
. By induction n 6 u 6 n , in
for all n ¿ 1. Now, taking the I This completes the proof. limit as n → ∞, we get  6 u 6 r in
. Corollary 2.1. In addition to the assumptions of Theorem 2:1; if f and g satisfy for u1 ¿ u2 f(x; u1 ) − f(x; u2 ) 6 N1 (u1 − u2 ) and g(x; u1 ) − g(x; u2 ) ¿ − N2 (u1 − u2 ); where N1 ¿ 0; N2 ¿ 0; then  = u = r is the unique solution of (2:1) provided c(x) − (N1 + N2 ) ¿ 0 in
.

574

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

Proof. Since we have  6 r, it is enough to show that r 6 . Setting p = r − , we have Bp = 0 on @
, and Lp = f(x; r) + g(x; ) − f(x; ) − g(x; r) 6 N1 (r − ) + N2 (r − )

in
:

Then Lp − (N1 + N2 )p 6 0 in
; which implies, by Lemma 2.1 p 6 0 since c(x) − (N1 + N2 ) ¿ 0. This proves that p = r − u is the unique solution of (2.1). Several remarks are now in order. Remark 2.1. (i) In Theorem 2.1, suppose that g(x; u) ≡ 0. Then Theorem 1.1 results with M = 0. (ii) If f(x; u) ≡ 0 in Theorem 2.1, then we get Theorem 1.3. (iii) If g(x; u) ≡ 0 and f(x; u) is not nondecreasing in u but f(x; u) + Mu, M ¿ 0 is nondecreasing in u, then one can consider the BVP ˜ u) in
; ˜ = Lu + Mu = f(x; u) + Mu = f(x; Lu Bu =  on @
:

(2.11)

˜ u) is nondecreasIt is clear that 0 , 0 are the lower and upper solutions of (2.11), f(x; ing in u and c(x) + M ¿ 0 in
. As a result, Theorem 2.1 yields the same conclusion as in Theorem 1.1 for the BVP (1.1). (iv) If f(x; u) ≡ 0 and g(x; u) is not nonincreasing in u but g(x; ˜ u) = g(x; u) − Nu; N ¿ 0 is nonincreasing in u, we consider the BVP ˜ u) + g(x; Lu = f(x; ˜ u) in
;

Bu =  on @
;

(2.12)

˜ u) = Nu and assume that the coupled lower and upper solutions of type I where f(x; of (2.12) exist. Then Theorem 2.1 implies the same conclusions as Theorem 1.1 for the BVP (2.12). In addition, if c(x) − (N2 + N ) ¿ 0 in
, where N2 is as in Corollary 2.1, then the BVP (2.12) has a unique solution. (v) If g(x; u) is nonincreasing in u and f(x; u) is not nondecreasing in u but ˜ u) = f(x; u) + Mu; M ¿ 0 is nondecreasing in u, then we consider the BVP f(x; ˜ u) + g(x; u) in
; ˜ = Lu + Mu = f(x; Lu

Bu =  on @
:

(2.13)

It is easy to see that 0 ; 0 are coupled lower and upper solutions of type I of (2.13). Hence Theorem 2.1 applied to (2.13) yields the same assertion of Theorem 2.1 for the BVP Lu = f(x; u) + g(x; u) in
;

Bu =  on @
;

where f(x; u) is not nondecreasing in u. (vi) If in BVP (2.1), f(x; u) is nondecreasing in u and g(x; u) is not nonincreasing in u but g(x; ˜ u) = g(x; u) − Nu; N ¿ 0 is nonincreasing in u, then consider

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

the BVP ˜ = Lu − Nu = f(x; u) + g(x; Lu ˜ u) in
;

575

Bu =  on @
;

(2.14)

and assume that the coupled lower and upper solutions of type I of (2.14) exist. Then, the assertion of Theorem 2.1 is guaranteed for the BVP (2.14). In addition, if the uniqueness condition, c(x) − (N1 + N2 + N ) ¿ 0 in
, where N1 and N2 are as in Corollary 2.1, is satised, then Theorem 2.1 applied to the BVP (2.14) implies the same conclusions for the BVP Lu = f(x; u) + g(x; u) in
, Bu =  on @
, where g(x; u) is not nonincreasing in u. (vii) If both functions f(x; u); g(x; u) do not satisfy the required monotone character in Theorem 2.1 but f(x; u) + Mu; g(x; u) − Nu; M; N ¿ 0 are nondecreasing and nonincreasing in u respectively, we consider the BVP ˜ u) + g(x; ˜ = Lu + Mu − Nu = f(x; Lu ˜ u) in
; Bu =  on @
; (2.15) ˜ u) = f(x; u)+(M )u and g(x; where f(x; ˜ u) = g(x; u)−(N )u and assume that the coupled lower and upper solutions of type I of (2.15) exist. Clearly, the conditions of Theorem ˜ g˜ and 0 ; 0 . Hence Theorem 2.1 shows that the conclusion of 2.1 are fullled by f; Theorem 2.1 is valid for the BVP (2.15). In addition, if c(x) − (N1 + N2 + M + N ) ¿ 0 in
where N1 and N2 are as in Corollary 2.1, then Theorem 2.1 shows that the conclusion of Theorem 2.1 is valid for the BVP Lu = f(x; u) + g(x; u) in
; Bu =  on @
, where f(x; u) and g(x; u) are not monotone functions. Let us next consider utilizing the coupled lower and upper solutions of type II. We prove the following result. Theorem 2.2. Assume that (A2 ) of Theorem 2:1 holds. Then for any solution u(x) of I we have the iterates {n (x)}; {n (x)} satisfying (2:1) with 0 6 u 6 0 in
; I (2.16) 0 6 2 6 · · · 6 2n 6 u 6 2n+1 6 · · · 6 3 6 1 ; in
; 1 6 3 6 · · · 6 2n+1 6 u 6 2n 6 · · · 6 2 6 0 ;

in
I

(2.17)

I where the iteration schemes are given by provided 0 6 2 and 2 6 0 in
; Ln+1 = f(x; n ) + g(x; n ) in
;

Bn+1 =  on @
;

Ln+1 = f(x; n ) + g(x; n ) in
;

Bn+1 =  on @
:

Moreover; the monotone sequences {2n }; {2n+1 }; {2n }; {2n+1 } ∈ C I R]; respectively; and they verify verge to ; r; ∗ ; r ∗ in C 2 [
; Lr = f(x; ∗ ) + g(x; ) in
; ∗

L = f(x; r ) + g(x; r) in
; L∗ = f(x; r) + g(x; r ∗ ) in
;

B∗ =  on @
:

I Also;  6 u 6 r; r 6 u 6  in
.

I R] con[
;

B =  on @
; Br ∗ =  on @
;

∗

(2.19) 2; 

Br =  on @
;

Lr ∗ = f(x; ) + g(x; ∗ ) in
; ∗

(2.18)

(2.20)

576

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

Proof. We shall rst show that the coupled lower and upper solutions of type II of I R] for the BVP (2.1) exist. Since f(x; 0) + g(x; 0) ∈ C  [
; Lz = f(x; 0) + g(x; 0) in
;

Bz =  on @
:

Dene 0 = z − R0 ; 0 = z + R0 by choosing R0 ¿ 0 suOciently large so that we have I Then using the monotone character of f(x; u) and g(x; u), 0 (x) 6 0 6 0 (x) in
. we get L0 = Lz − c(x)R0 = f(x; 0) + g(x; 0) − c(x)R0 6 f(x; 0 ) + g(x; 0 ) in
; B0 = Bz − c(x)R0 6  on @
: Similarly, we obtain L0 ¿ f(x; 0 ) + g(x; 0 ) in
, B0 ¿  on @
. Thus, 0 ; 0 are the coupled lower and upper solutions of type II for the BVP (2.1). Following the arguments used in Theorem 2.1, we see that the linear BVPs (2.18), I R] for each n = 1; 2; : : : . Our aim is (2.19) have unique solutions n+1 ; n+1 ∈ C 2;  [
; therefore to prove relations (2.16) and (2.17). Let u be any solution of (2.1) such that I We shall show that 0 6 u 6 0 in
. I 0 6 2 6 u 6 3 6 1 in
;

I 1 6 3 6 u 6 2 6 0 in
:

(2.21)

Setting p = u − 1 , we get Bp = 0 on @
and Lp = f(x; u) + g(x; u) − f(x; 0 ) − g(x; 0 ) 6 0 in
using the monotone nature of f; g and the fact 0 6 u 6 0 in
. This implies by I Similarly, we can show that Lemma 2.1, p 6 0 in
I and this implies u 6 1 in
. I Next let p = 2 − u so that Bp = 0 on @
and Lp = f(x; 1 ) + g(x; 1 ) − u 6 1 in
. I A similar f(x; u)−g(x; u) 6 0 in
. Hence by Lemma 2.1, it follows that 2 6 u on
. I Considering p = 3 − 1 , we see that Bp = 0 on @
and, argument yields u 6 2 on
. as before Lp = f(x; 2 ) + g(x; 2 ) − f(x; 0 ) − g(x; 0 ) 6 0 in
; I In the same way, we get 1 6 3 in
. I which implies by Lemma 2.1, 3 6 1 on
. I proving (2.21). Also, using similar reasoning, we obtain u 6 3 ; 3 6 u on
, Now assuming for some k ¿ 2, the inequalities 2n−4 6 2n−2 6 u 6 2n−1 6 2n−3 ; 2n−3 6 2n−1 6 u 6 2n−2 6 2n−4 in
I hold, it can be shown, by employing similar arguments, that 2n−2 6 2n 6 u 6 2n+1 6 2n−1 ;

I 2n−1 6 2n+1 6 u 6 2n 6 2n−2 in
:

Thus by induction, (2.16) and (2.17) are valid for all n = 0; 1; 2; : : : . Since n ; n ∈ C 2;  [
; R] for all n, employing a similar reasoning as in Theorem 2.1, we conclude that the limits lim 2n = ;

n→∞

lim 2n+1 = r;

n→∞

lim 2n+1 = r ∗

n→∞

and

lim 2n = ∗

n→∞

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

577

I R]. Thus the limits exist in C 2 [
; lim L2n = L;

n→∞

lim B2n = B;

n→∞

lim L2n+1 = Lr;

n→∞

lim B2n+1 = Br;

n→∞

lim L2n+1 = Lr ∗ ;

n→∞

lim L2n = ∗

n→∞

lim B2n+1 = Br ∗ ;

n→∞

lim B2n = B∗ ;

n→∞

lim [f(x; 2n−1 ) + g(x; 2n−1 )] = f(x; r ∗ ) + g(x; r);

n→∞

lim [f(x; 2n ) + g(x; 2n )] = f(x; ∗ ) + g(x; );

n→∞

lim [f(x; 2n ) + g(x; 2n )] = f(x; ) + g(x; ∗ );

n→∞

lim [f(x; 2n−1 ) + G(x; 2n−1 )] − f(x; r) + g(x; r ∗ ):

n→∞

We then nd that ; r; ∗ ; r ∗ satisfy L = f(x; r ∗ ) + g(x; r) in
;

B =  on @
;

∗

Lr = f(x;  ) + g(x; ) in
; ∗

Br =  on @
;

∗

L = f(x; r) + g(x; r ) in
;

B∗ =  on @
;

Lr ∗ = f(x; ) + g(x; ∗ ) in
;

Br ∗ =  on @


and from (2.16) and (2.17), it follows that I  6 u 6 r and r ∗ 6 u 6 ∗ in
: The proof is therefore complete. Corollary 2.2. Under the assumption of Theorem 2:2; if f(x; u); g(x; u) satisfy the relations f(x; u1 ) − f(x; u2 ) 6 N1 (u1 − u2 ); g(x; u1 ) − g(x; u2 ) ¿ − N2 (u1 − u2 );

u1 ¿ u2 ; N1 ¿ 0; x ∈
; u1 ¿ u2 ; N2 ¿ 0; x ∈


and c(x) − (N1 + N2 ) ¿ 0; then u =  = r = ∗ = r ∗ is the unique solution of (2:1). I It then follows Proof. Let v1 = r −, v2 = ∗ −r ∗ so that we have v1 ¿ 0, v2 ¿ 0 on
. that Bv1 = 0, Bv2 = 0 on @
and Lv1 6 N1 (∗ − r ∗ ) + N2 (r − ); Lv2 6 N1 (r − ) + N2 (∗ − r ∗ ) and therefore L(v1 + v2 ) 6 (N1 + N2 )(v1 + v2 ) in
, B(v1 + v2 ) = 0 on @
. Lemma I which implies  = r, ∗ = r ∗ in
. I We claim that 2.1 now yields v1 + v2 6 0 in
, ∗ ∗ I If not, assuming contrary and proceeding as above, we are lead  = r , r =  in
. to a contradiction. Hence u =  = ∗ = r = r ∗ is the unique solution of (2.1) and the proof is complete. Theorem 2:2 contains several special cases which we give below.

578

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

Remarks 2.2. (i) In Theorem 2:2, suppose that f(x; u) ≡ 0. Then we get Theorem 1.2. (ii) If g(x; u) ≡ 0 in Theorem 2:2, then we obtain a new result dual to Theorem 1.1 with M = 0. ˜ u) = f(x; u) + Mu, (iii) If g(x; u) ≡ 0 and f(x; u) is not nondecreasing in u but f(x; M ¿ 0, is nondecreasing in u, then we consider the BVP ˜ u) in
; Bu =  on @
: ˜ = Lu + Mu = f(x; Lu (2.22) Clearly there exist 0 ; 0 as in Theorem 2:2, such that ˜ 0 ) in
; B0 6  on @
; ˜ 0 6 f(x; L ˜ 0 ) in
; B0 ¿  on @
˜ 0 ¿ f(x; L I Hence Theorem 2:2 yields a new result dual to Theorem 1.1. with 0 6 0 6 0 on
. (iv) If f(x; u) ≡ 0 and g(x; u) is not nonincreasing in u but g(x; ˜ u) = g(x; u) − Nu, N ¿ 0 is nonincreasing in u and c(x) − N ¿ 0, then we consider the BVP ˜ = Lu − Nu = g(x; Lu ˜ u) in
;

Bu =  on @
:

(2.23)

Evidently, 0 ; 0 exist verifying ˜ 0 6 g(x; L ˜ 0 ) in
;

B0 6  on @
;

˜ 0 ¿ g(x; L ˜ 0 ) in
;

B0 ¿  on @
:

Hence Theorem 2:2 provides a new result which is an extension of Theorem 1.2. (v) If g(x; u) is nonincreasing in u and f(x; u) is not nondecreasing in u but ˜ u) = f(x; u) + Mu; M ¿ 0 is nondecreasing in u for x ∈
, then we consider the f(x; BVP ˜ u) + g(x; u) in
; Bu =  on @
: ˜ = Lu + Mu = f(x; Lu (2.24) It is easy to see that the coupled lower and upper solutions 0 ; 0 of type II exist I Theorem 2:2 applied to BVP (2.24) gives for the BVP (2.24) such that 0 6 0 on
. the same conclusion of Theorem 2:2 for the BVP Lu = f(x; u) + g(x; u) in
;

Bu =  on @
;

where f(x; u) is not nondecreasing in u, when uniqueness assumptions of Corollary 2.2 are satised. (vi) If, in BVP (2.1), f(x; u) is nondecreasing and g(x; u) is not nonincreasing in u but g(x; ˜ u) = g(x; u) − Nu; N ¿ 0 is nonincreasing in u for x ∈
and c(x) − N ¿ 0, then consider the BVP ˜ = Lu − Nu = f(x; u) + g(x; Lu ˜ u) in
;

Bu =  on @
:

(2.25)

As before, 0 ; 0 exist as coupled lower and upper solutions of type II for the BVP (2.25) and therefore Theorem 2:2 yields the same assertion of Theorem 2:2 for the BVP Lu = f(x; u) + g(x; u) in
;

Bu =  on @


when g(x; u) is not nondecreasing in u, provided uniqueness conditions hold as in Corollary 2.2.

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

579

˜ u) = f(x; u) + (vii) If, in Theorem 2:2, both f(x; u); g(x; u) are not monotone but f(x; Mu; M ¿ 0 is nondecreasing, g(x; ˜ u) = g(x; u) − Nu; N ¿ 0 is nonincreasing in u for x ∈
and c(x) + (M − N ) ¿ 0, consider the BVP ˜ u) + g(x; ˜ = Lu + (M − N )u = f(x; Lu ˜ u) in
;

Bu =  on @
:

(2.26)

As before, 0 ; 0 exist as coupled lower and upper solutions of type II for the BVP (2.26) and consequently, Theorem 2:2 is applicable to (2.26) to provide the same assertion as in Theorem 2:2 for the BVP Lu = f(x; u) + g(x; u) in
;

Bu =  on @


when f; g are not monotone, provided the uniqueness conditions of Corollary 2.2 are satised. 3. The variational method In this section, we shall consider results parallel to Theorems 2.1 and 2:2 relative to the BVP Lu = f(x; u) + g(x; u) in
; u = 0 on @
; in the sense of trace;

(3.1)

where L now denotes the second-order partial di2erential operator in the divergence form n
(ai; j (x)uxi )xj + c(x)u Lu = − i; j

and
is an open, bounded subset of Rn . n We assume that aij ; c ∈ L∞ (
); i; j = 1; 2; : : : ; n; aij = aji , i; j=1 aij (x)i j ¿ ||2 for x ∈
, a.e.  ∈ Rn with  ¿ 0 and c(x) ¿ 0. We shall always mean that the boundary condition is in the sense of trace and hence we shall not repeat it to avoid monotony. Also, f :
I ×R → R and g :
I ×R → R, f(x; u) and g(x; u) are Caratheordory functions, that is, f(·; u) and g(·; u) are measurable for all u ∈ R, and f(x; ·) and g(x; ·) are continuous a.e. x ∈
. The bilinear form B[ ; ] associated with the operator L is   
n B[u; v] =  aij (x)uxi vxj + c(x)uv dx (3.2)


i; j=1

for u; v ∈ H01 (
). Denition 3.1. The function u ∈ H01 (
) is said to be a weak solution of (3.1) if f(x; u) ∈ L1 (
), g(x; u) ∈ L1 (
), f(x; u)u ∈ L1 (
), g(x; u)u ∈ L1 (
) and B[u; v] = (f + g; v) for all

v ∈ H01 (
)

where ( ; ) denotes inner product in L2 (
).

We need the following known results for our discussion.

(3.3)

580

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

Lemma 3.1. For p ∈ H 1 (
) satisfying   
n  aij (x)pxi vxj + c(x)pv dx 6 0


i; j=1

for each v ∈ H01 (
); v ¿ 0 a.e. in
, and p 6 0 on @
; we have p(x) 6 0 a.e. in
provided c(x) ¿ 0. Lemma 3.2. Consider the linear BVP Lu = h(x)

in
;

u = 0 on @
:

(3.4)

There exists a unique solution u ∈ H01 (
) for the linear BVP (3:4) provided 0 ¡ c∗ 6 c(x) a.e. in
and h ∈ L2 (
). In view of the discussion in Section 2, with respect to the coupled lower and upper solutions of various types and the conclusion, it is suOcient to dene, in the present setup, only the following. Denition 3.2. Relative to the BVP (3.1), the functions 0 ; 0 ∈ H 1 (
) are said to be (i) weakly coupled lower and upper solutions of type I if B[0 ; v] 6 (f(x; 0 ) + g(x; 0 ); v); B[0 ; v] ¿ (f(x; 0 ) + g(x; 0 ); v); for each v ∈ H01 (
); v ¿ 0 a.e. in
; (ii) weakly coupled lower and upper solutions of type II if B[0 ; v] 6 (f(x; 0 ) + g(x; 0 ); v); B[0 ; v] ¿ (f(x; 0 ) + g(x; 0 ); v); for each v ∈ H01 (
); v ¿ 0, a.e. in
. We are now in a position to prove the rst result parallel to Theorem 2.1. Theorem 3.1. Assume that: (A1 ) 0 ; 0 ∈ H 1 (
) are the weak-coupled lower and upper solutions of type I with 0 (x) 6 0 (x) a.e. in
; (A2 ) f; g :
I × R → R are Caratheodory functions such that f(x; u) is nondecreasing in u; g(x; u) is nonincreasing in u for x ∈
; a.e.; (A3 ) c(x) ¿ N ¿ 0 a.e. in
; and for any "; . ∈ H 1 (
) with 0 6 "; . 6 0 ; the function h(x) = f(x; ") + g(x; .) ∈ L2 (
).

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

581

Then there exist monotone sequences {n (x)}; {n (x)} ∈ H01 (
) such that n → ; n → r weakly in H01 (
) as n → ∞ and (; r) are weak coupled minimal and maximal solutions of (3:1); respectively; that is L = f(x; ) + g(x; r) in
;

 = 0 on @
;

Lr = f(x; r) + g(x; ) in
;

r = 0 on @
:

Proof. Consider the linear BVPs Ln+1 = f(x; n ) + g(x; n ) in
;

n+1 = 0 on @


(3.5)

Ln+1 = f(x; n ) + g(x; n ) in
;

n+1 = 0 on @
:

(3.6)

The variational forms associated with (3.5) and (3.6) are  B[n+1 ; v] = [f(x; n ) + g(x; n )]v d x; 
[f(x; n ) + g(x; n )]v d x B[n+1 ; v] =

(3.7)

and

(3.8)




for all v ∈ H01 (
); v ¿ 0 a.e. in
. We shall show that the weak solutions n ; n of (3.5) and (3.6) are uniquely dened and satisfy 0 6 1 6 2 6 · · · 6 n 6 n 6 · · · 6 2 6 1 6 0

a:e: in
:

(3.9)

For each n ¿ 1; by assumption (A3 ); h1 (x) ≡ f(x; n )+g(x; n ) ∈ L2 (
) and h2 (x) ≡ f(x; n ) + g(x; n ) ∈ L2 (
) provided 0 6 n 6 n 6 0 a.e. in
. Therefore, Lemma 3.2 implies that BVPs (3.5) and (3.6) have unique solutions n and n in view of c(x) ¿ N ¿ 0 for each n ¿ 1. To prove (3.9), we rst claim that 1 ¿ 0 a.e. in
. Now, let p = 0 − 1 so that p 6 0 on @
and for v ∈ H01 (
); v ¿ 0 a.e. in
;    n
 B[p; v] = aij (x)pxi vxj + c(x)pv dx


i; j=11

 6




 [f(x; 0 ) + g(x; 0 )]v d x −




[f(x; 0 ) + g(x; 0 )]v d x = 0:

Hence by Lemma 3.1, p(x) 6 0; that is, 0 6 1 a.e. in
. Similarly, we can show that 1 6 0 a.e. in
. Assume that, for some xed n ¿ 1; n 6 n+1 and n ¿ n+1 a.e. in
. Now consider p = n+1 − n+2 . Note that p = 0 on @
; and using the monotone nature of f; g; we get  B[p; v] = [f(x; n ) − f(x; n+1 ) + g(x; n ) − g(x; n+1 )]v d x 6 0:


This implies by Lemma 3.1 that n+1 6 n+2 a.e. in
.

582

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

Also, by setting p = n+2 − n+1 ; one can easily show in the same way, that n+2 6 n+1 a.e. in
. Hence, using the induction argument, we have n ¿ n+1 and n ¿ n+1 a.e. in
for all n ¿ 1. We now show that 1 6 1 a.e. in
. Consider p = 1 − 1 and note that p = 0 on @
; and  B[p; v] = [f(x; 0 ) + g(x; 0 ) − f(x; 0 ) − g(x; 0 )]v d x 6 0


because of the fact f; g are monotone by assumption. Hence Lemma 3.1 yields 1 6 1 a.e. in
. Employing similar arguments, it is easy to show that if we assume for some xed n ¿ 1; we have n 6 n a.e. in
; then it follows that n+1 6 n+1 a.e. in
. Consequently, using the induction argument, it is clear that (3.9) holds for all n ¿ 1. By the monotone character of {n }; {n } there exist pointwise limits lim n (x) = (x) a:e: in


n→∞

and

lim n (x) = r(x) a:e: in
:

n→∞

Moreover, since 0 6 n 6 n 6 0 a.e. in
; it follows by Lebesgue’s dominated convergence theorem that n →  and n → r

in L2 (
):

For each n ¿ 1; we note that n satises for each v ∈ H01 (
); v ¿ 0; a.e. in
    n
 aij (x)(n )xi vxj + c(x)n v d x = hn−1 (x)v d x;





i; j=1

where hn−1 (x) = f(x; n−1 ) + g(x; n−1 ). We now use the ellipticity condition and the fact that c(x) ¿ N ¿ 0 with v = n to get   2 2 [|(n )x | + N |n | ] d x 6 hn−1 (x)v d x:





Since the integrand on the right-hand side belongs to L2 (
); we obtain the estimate sup n H01 (
) ¡ ∞: n

Hence there exists a subsequence {nk } which converges weakly to (x) in H01 (
). A similar argument implies that supn n H01 (
) ¡ ∞. Hence there exist subsequences {nk }; {nk } which converge weakly in H01 (
) to (; r) ∈ H01 (
), respectively. Since n and n satisfy (3.7) and (3.8), for a xed v ∈ H01 (
); v ¿ 0 a.e. in
by taking the limit as n → ∞; we see that  B[; v] = [f(x; ) + g(x; r)]v d x;


and

 B[r; v] =




[f(x; r) + g(x; )]v d x:

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

583

Finally, we claim that  and r are the weak coupled minimal and maximal solutions of (3.1) that is, if u is any weak solution of (3.1) such that 0 (x) 6 u(x) 6 0 (x) a.e. in
; then 0 (x) 6 (x) 6 u(x) 6 r(x) 6 0

a:e: in
:

(3.10)

Suppose that for some xed n ¿ 1; n 6 u 6 n a.e. in
. Setting p = n+1 − u; we have p = 0 on @
and employing the monotone character of f and g; it follows that  B[p; v] = [f(x; n ) + g(x; n ) − f(x; u) − g(x; u)]v d x 6 0;


which implies, by Lemma 3.1, n+1 6 u a.e. in
. In a similar way, we obtain u 6 n+1 a.e. in
so that n+1 6 u 6 n+1 a.e. in
. By induction n 6 u 6 n a.e. in
for all n ¿ 1. Now taking the limit as n → ∞; we get (3.10), completing the proof. To show the uniqueness, we have the following corollary. Corollary 3.1. Assume; in addition to the conditions of Theorem 3:1; that f and g satisfy f(x; u1 ) − f(x; u2 ) 6 N1 (u1 − u2 ) and g(x; u1 ) − g(x; u2 ) ¿ − N2 (u1 − u2 ); where u1 ¿ u2 ; N1 ¿ 0; N2 ¿ 0 and c(x) − (N1 + N2 ) ¿ 0 a.e. in
. Then  = u = r; is the unique weak solution of (3:1). Proof. Since we have  6 r; it is enough to show that r 6 . Setting p = r − ; we have p = 0 on @
and using the assumptions on f; g; it follows that  B[p; v] = [f(x; r) + g(x; ) − f(x; ) − g(x; r)]v d x


 6




[N1 (r − ) + N2 (r − )]v d x = ((N1 + N2 )p; v):

This implies by Lemma 3.1, p 6 0 a.e. in
. Hence u =  = r is the unique weak solution of (3.1). Now consider the linear iteration scheme Ln+1 = f(x; n ) + g(x; n ) in
;

n+1 = 0 on @
;

(3.11)

Ln+1 = f(x; n ) + g(x; n ) in
;

n+1 = 0 on @


(3.12)

and the associated variational forms  B[n+1 ; v] = [f(x; n ) + g(x; n )]v d x; 
[f(x; n ) + g(x; n )]v d x B[n+1 ; v] =


(3.13) (3.14)

584

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

for all v ∈ H01 (
); v ¿ 0 a.e. in
. Then, for the weak solutions of (3.1) we have the following theorem which is analogous to Theorem 2:2. Theorem 3.2. Assume that the conditions (A2 ) and (A3 ) of Theorem 3:1 hold. Then for any weak solution u(x) of (3:1) such that 0 (x) 6 u(x) 6 0 (x) a.e. in
; we have the iterates {n (x)}; {n (x)} satisfying 0 6 2 6 · · · 6 2n 6 u 6 2n+1 6 · · · 6 3 6 1 ;

a:e: in
;

(3.15)

1 6 3 6 · · · 6 2n+1 6 u 6 2n 6 · · · 6 2 6 0 ;

a:e: in


(3.16)

provided 0 6 2 and 2 6 0 a.e. in
; where the iterative schemes are given by (3:11) and (3:12); respectively. Moreover; the monotone sequences {2n }; {2n+1 }; {2n }; {2n+1 } ∈ H01 (
) converge weakly in H01 (
) to ; r; ∗ ; r ∗ ; respectively; and they satisfy the relations Lr = f(x; ∗ ) + g(x; ) in
;

r = 0 on @
;

L = f(x; r ∗ ) + g(x; r) in
;

 = 0 on @
;

Lr ∗ = f(x; ) + g(x; ∗ ) in
;

r ∗ = 0 on @
;

L∗ = f(x; r) + g(x; r ∗ ) in
;

∗ = 0 on @
:

Also  6 u 6 r and r ∗ 6 u 6 ∗ ; a.e. in
. Proof. Using an argument similar to that employed in Theorem 2:2 with suitable modications, it is easy to show that there exist 0 ; 0 ∈ H01 (
) with 0 6 0 a.e. in
satisfying the weak coupled lower and upper solutions of (3.1) of type II. Also, as in Theorem 3.1, using condition (A3 ); we have the existence of unique weak solutions n (x); n (x) of (3.11), (3.12) respectively. We shall therefore proceed to establish inequalities (3:15) and (3:16). In view of the assumptions 0 6 2 and 2 6 0 a.e. in
; we shall rst show that 0 6 2 6 u 6 3 6 1

and

1 6 3 6 u 6 2 6 0

a:e: in
:

(3.17)

Setting p = u − 1 ; we get because of the monotone character of f and g;  B[p; v] = [f(x; u) + g(x; u) − f(x; 0 ) − g(x; 0 )]v d x 6 0


v ∈ H01 (
);

v ¿ 0 a.e. in
. Since p = 0 on @
; this implies, by Lemma 3.1 for each that u 6 1 a.e. in
. Hence we have 0 6 u 6 1 a.e. in
. Similarly, one can obtain 1 6 u 6 0 a.e. in
. Now we set p = 3 − 1 and get, by using a similar reasoning  [f(x; 2 ) + g(x; 2 ) − f(x; 0 ) − g(x; 0 )]v d x 6 0 B[p; v] =


and p = 0 on @
so that by Lemma 3.1, it follows that 3 6 1 a.e. in
. Also setting p = 2 − u and p = u − 2 yields 2 ¿ u and u ¿ 2 a.e. in
. In addition, if p = u − 3 and p = 3 − u; we obtain 3 ¿ u and 3 6 u a.e. in
.

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

585

Thus, the inequalities in (3.17) hold a.e. in
. Now, for some xed n ¿ 1; assuming 2n−4 6 2n−2 6 u 6 2n−1 6 2n−3 and 2n−3 6 2n−1 6 u 6 2n−2 6 2n−4 ;

a:e:

hold a.e. in
; if we let p be 2n−2 − 2n ; 2n − u; 2n−1 − 2n+1 ; 2n+1 − u; u − 2n+1 ; 2n+1 − 2n−1 ; u − 2n ; and 2n − 2n−2 ; and use the monotone nature of f and g; and Lemma 3.1, respectively, we obtain 2n−2 6 2n 6 u 6 2n+1 6 2n−1 and 2n−1 6 2n+1 6 u 6 2n 6 2n−2

a:e: in
:

Hence, by induction, we obtain inequalities (3:15) and (3:16) a.e. in
. We note that 2n ; 2n+1 ∈ H01 (
) and 2n ; 2n+1 ∈ H01 (
); and hence arguing as in the proof of Theorem 3.1 with appropriate modications, we see that weakly in H01 (
);

2n → 

weakly in H01 (
);

2n+1 → r

weakly in H01 (
);

2n+1 → r ∗ 2n → ∗

weakly in H01 (
):

Now passing the limit in the variational forms (3.13) and (3.14), we see that (; r) and (r ∗ ; ∗ ) satisfy  B[; v] = [f(x; r ∗ ) + g(x; r)]v d x;


 B[r; v] = B[∗ ; v] = B[r ∗ ; v] =




[f(x; ∗ ) + g(x; )]v d x;











[f(x; r) + g(x; r ∗ )]v d x; [f(x; ) + g(x; ∗ )]v d x

for v ∈ H01 (
); v ¿ 0 a.e. in
. Also, from (3:15) and (3:16), we get  6 u 6 r and r ∗ 6 u 6 ∗ a.e. in
. This completes the proof. Corollary 3.2. In addition to Theorem 3:2 assume that f and g are as in Corollary 3:1. Then; u =  = r = ∗ = r ∗ a.e. in
is the unique solution of (3:1).

586

S. K)oksal, V. Lakshmikantham / Nonlinear Analysis 51 (2002) 567 – 586

Proof. Let p1 = r −  and p2 = ∗ − r ∗ . By Theorem 3.1, p1 ¿ 0 and p2 ¿ 0 a.e. in
. Then, utilizing the conditions and the monotone character of f; g; it follows that  B[p1 ; v] 6 [N1 (∗ − r ∗ ) + N2 (r − )]v d x;


 B[p2 ; v] 6




[N1 (r − ) + N2 (∗ − r ∗ )]v d x

and

 B[p1 + p2 ; v] 6




[(N1 + N2 )(p1 + p2 )]v d x

for v ∈ H01 (
); v ¿ 0 a.e. in
. Consequently, we have    n
 aij (x)(p1 + p2 )xi vxj + (c(x) − (N1 + N2 )) (p1 + p2 )v dx 6 0


i; j=1

which yields, by Lemma 3.1, (p1 + p2 ) 6 0 a.e. in
. This implies p1 6 0 and p2 6 0 a.e. in
; that is,  = r; ∗ = r ∗ a.e. in
. This, in turn, implies u =  = r = ∗ = r ∗ a.e. in
is the unique solution of (3.1). This completes the proof. Remark 3.1. We can make remarks similar to Remarks 2.1 and 2.2 relative to Theorems 3.1 and 3.2. We are not repeating them in order to avoid monotony. References [1] P. Drabek, J. Hernandez, Existence and uniqueness of positive solutions for some quasilinear elliptic problems, Nonlinear Anal. Ser. A, to appear. [2] H.B. Keller, Elliptic BVPs suggested by nonlinear di2usion processes, Arch. Rat. Mech. Anal. 35 (1969) 363–385. [3] G.S. Ladde, V. Lakshmikantham, A.S. Vatsala, Monotone Iterative Techniques for Nonlinear Di2erential Equations, Pitman Publishing, London, 1985. [4] V. Lakshmikantham, A.S. Vatsala, Generalized quasilinearization and semilinear elliptic boundary value problems, JMAA 249 (2000) 199–220. [5] D.H. Sattinger, Monotone methods in nonlinear elliptic and parabolic BVPs, Indiana University Math. J. 31 (11) (1972) 979–1000.