Multiple positive solutions for a semipositone fourth-order boundary value problem
Ruyun Ma*
(Received July 16, 2002) (Revised January 7, 2003)
Abstract. We consider the nonlinear fourth order boundary value problem uð4ÞðxÞ ¼lfðx;uðxÞ;u0ðxÞÞ
uð0Þ ¼u0ð0Þ ¼u00ð1Þ ¼u000ð1Þ ¼0
where f :½0;1 ½0;yÞ ½0;yÞ ! ðy;yÞ is continuous with fðx;u;pÞbM for some positive constantM. We show the existence and multiplicity of positive solutions by using a fixed point theorem in cones.
1. Introduction
The deformations of an elastic beam are described by a fourth-order two- point boundary value problem [6]. The boundary conditions are given accord- ing to the controls at the ends of the beam. For example, the nonlinear fourth order problem
uð4ÞðxÞ ¼lfðx;uðxÞ;u0ðxÞÞ
uð0Þ ¼u0ð0Þ ¼u00ð1Þ ¼u000ð1Þ ¼0 ð1:1Þ describes the deformations of an elastic beam whose one end fixed and the other end free.
The existence of solutions of (1.1) has been studied by Gupta [6]. But to the best of our knowledge, there are no any results concerning the existence of positive solutions of (1.1). In this paper, we will study the existence and multiplicity of positive solutions of (1.1).
2000 Mathematics Subject Classification. 34B18
Key words and phrases. Fourth order boundary value problem, positive solution, cone, fixed point.
* Supported by the NSFC (No. 10271095), GG-110-10736-1003, NWNU-KJCXGC-212 and the Foundation of Major Project in Science and Technology of Chinese Education Ministry.
We will make the following assumptions:
(A1) f :½0;1 ½0;yÞ ½0;yÞ ! ðy;yÞis continuous and there exists M>0 such that
fðx;u;pÞbM; for ðx;u;pÞA½0;1 ½0;yÞ ½0;yÞ; ð1:2Þ (A2) There exists a subinterval ½a;bHð0;1Þ with a<b such that
p!limy
fðx;u;pÞ
p ¼y ð1:3Þ
holds uniformly for ðx;uÞA½a;b ½0;yÞ;
(A3)
fðx;u;0Þ>0; ðx;uÞA½0;1 ½0;yÞ: ð1:4Þ Remark 1. It is easy to see that (A3) implies that there exist two constants a;bAð0;yÞ such that
fðx;u;pÞbb; ðx;u;pÞA½0;1 ½0;a ½0;a:
Very recently, Anuradha, Hai and Shivaji [1] studied the existence of positive solutions for second order boundary value problem
ðpðtÞu0ðtÞÞ0þlfðt;uðtÞÞ ¼0; r<t<R
auðrÞ bpðrÞu0ðrÞ ¼0; cuðRÞ þdpðRÞu0ðRÞ ¼0 ð1:5Þ under some superlinear semipositone conditions when l>0 is small enough.
Motivated by their work, we study the existence and multiplicity of positive solutions for fourth order problems (1.1). The main results of this paper are the following
Theorem 1. Assume (A1) and (A2) hold. Then the problem (1.1) has at least one positive solution if l>0 is small enough.
Theorem 2. Assume (A1), (A2) and (A3) hold. Then the problem (1.1) has at least two positive solutions if l>0 is small enough.
The proofs of above theorems are based upon the following Guo- Krasnoselskii fixed point theorem
Theorem 3. [5] Let E be a Banach space, and let KHE be a cone.
Assume W1;W2 are open and bounded subsets of E with 0AW1, W1HW2, and let
A:KVðW2nW1Þ !K be a completely continuous operator such that
( i ) kAukakuk, uAKVqW1, and kAukbkuk, uAKVqW2; or (ii) kAukbkuk, uAKVqW1, and kAukakuk, uAKVqW2.
Then A has a fixed point in KVðW2nW1Þ.
For the results concerning the existence and multiplicity of positive solu- tions of fourth-order ordinary di¤erential equations with other di¤erent con- ditions and nonnegative nonlinearities, one may refer, with further references therein, to Del Pino and Mana´sevich [2], Dunninger [3], Graef and Yong [4], Ma and Wang [7] and Zhang and Kong [10].
2. The preliminary lemmas
To prove Theorem 1 and Theorem 2, we need several preliminary results.
Lemma 1. For yAC½0;1, the problem
uð4ÞðxÞ ¼ yðxÞ; xAð0;1Þ
uð0Þ ¼u0ð0Þ ¼u00ð1Þ ¼u000ð1Þ ¼0 ð2:1Þ is equivalent to the integral equation
uðxÞ ¼ ðx
0
ðt 0
ð1 r
ð1 s
yðtÞdt
ds
dr
dt: ð2:2Þ
Moreover, if yb0 on ½0;1, then ( i ) uðxÞb0, xA½0;1;
( ii ) u0ðxÞb0, xA½0;1;
(iii) u00ðxÞb0, xA½0;1;
(iv) u000ðxÞa0, xA½0;1.
Proof. It is easy to check that (2.1) is equivalent to (2.2).
If yb0 on ½0;1, then (2.2) implies ub0 on ½0;1. Moreover u0ðxÞ ¼
ðx 0
ð1 r
ð1 s
yðtÞdt
ds
drb0; xA½0;1
u00ðxÞ ¼ ð1
x
ð1 s
yðtÞdt
dsb0; xA½0;1 and
u000ðxÞ ¼ ð1
x
yðtÞdta0; xA½0;1:
In the following, we will use the Banach space C½0;1 and its sup norm k k0.
Lemma 2. If yAC½0;1 and yb0, then the unique solution u of the problem (2.1) satisfies
u0ðxÞbku0k0qðxÞ;
where
qðxÞ:¼x; xA½0;1: ð2:3Þ
Proof. By Lemma 1, we know that u000ðxÞa0 on ½0;1. So, the graph of u0 is concave down. This together with the facts that u0ð0Þ ¼ ðu0Þ0ð1Þ ¼0 and u00b0 imply
u0ðxÞbku0k0qðxÞ:
Lemma 3. The boundary value problem uð4ÞðxÞ ¼1; xAð0;1Þ
uð0Þ ¼u0ð0Þ ¼u00ð1Þ ¼u000ð1Þ ¼0 ð2:4Þ has a solution
wðxÞ ¼x2 4 x3
6 þx4
24: ð2:5Þ
Moreover,
wðxÞa1
8qðxÞ; xA½0;1 ð2:6Þ
and
w0ðxÞa1
2qðxÞ; xA½0;1: ð2:7Þ
Proof. (2.5) is an immediate consequence of Lemma 1. Since the graph ofw is concave upward andkwk0¼1
8, we know that (2.6) holds. By (2.5), we have that
w0ðxÞ ¼x 2x2
2 þx3
6 ð2:8Þ
w00ðxÞ ¼1
2xþx2
2 ð2:9Þ
and
w000ðxÞ ¼ 1þxa0; xA½0;1: ð2:10Þ
(2.10) implies that the graph of w0 is concave downward. Therefore w0ðxÞa w00ð0ÞqðxÞ ¼1
2qðxÞ for xA½0;1.
3. Proof of the theorems Set
C01½0;1 ¼ fujuAC1½0;1;uð0Þ ¼u0ð0Þ ¼0g:
We furnish the set C01½0;1 with the norm
kuk ¼supfju0ðxÞj:xA½0;1g ¼ ku0k0; by which C01½0;1 is a Banach space.
Proof of Theorem 1. Let
z¼lMw ð3:1Þ
where w is defined by (2.5). Then (1.1) has a positive solution u if uþz:¼uu~ is a solution of
uð4Þ¼lgðx;uz;u0z0Þ; xAð0;1Þ
uð0Þ ¼u0ð0Þ ¼u00ð1Þ ¼u000ð1Þ ¼0 ð3:2Þ and uu~0ðxÞ>z0ðxÞ for xAð0;1Þ, where g:½0;1 RR! ½0;yÞ is defined by
gðx;u;pÞ ¼
fðx;u;pÞ þM; ðx;u;pÞA½0;1 ½0;yÞ ½0;yÞ fðx;u;0Þ þM; ðx;u;pÞA½0;1 ½0;yÞ ðy;0Þ fðx;0;pÞ þM; ðx;u;pÞA½0;1 ðy;0Þ ð0;yÞ fðx;0;0Þ þM; ðx;u;pÞA½0;1 ðy;0Þ ðy;0Þ 8>
>>
<
>>
>:
ð3:3Þ
Let us denote
K¼ fujuAC01½0;1;uðxÞb0 on ½0;1;
u0ðxÞb0 on ½0;1;u0ðxÞbkukqðxÞg; ð3:4Þ where qðxÞ is defined by (2.3). It is obvious that K is a cone in C01½0;1.
For vAK, denote by Av the unique solution of uð4Þ¼lgðx;vz;v0z0Þ
uð0Þ ¼u0ð0Þ ¼u00ð1Þ ¼u000ð1Þ ¼0; ð3:5Þ then
AvðxÞ ¼l ðx
0
ðt 0
ð1 r
ð1 s
gðt;vz;v0z0Þdt
ds
dr
dt: ð3:6Þ By Lemma 2, we know that AðKÞHK.
Let
lAð0;LÞ ð3:7Þ
be fixed, where
L¼min 12 M1
; 1 M
ð3:8Þ M1¼maxfgðx;u;pÞ j0axa1;0aua2;0apa2g: ð3:9Þ ChooseW1¼ fuAC01½0;1 j kuk<2g. Then foruAKVqW1, we have from (3.8), (3.9), (3.3) and the facts zb0 and z0b0 that
ðAuÞ0ðxÞ ¼l ðx
0
ð1 r
ð1 s
gðt;uz;u0z0Þdt
ds
dr
alM1 ð1
0
ð1 r
ð1 s
1dt
ds
dr
¼lM11 6
<2: ð3:10Þ
Therefore,
kAukakuk; uAKVqW1: We note that
aasabmin qðsÞ ¼a: ð3:11Þ
Choose a real number N>0, such that Nla1
2 1
2ðbaÞ 1
2ðb2a2Þ þ1
6ðb3a3Þ
b1 ð3:12Þ
(We note that 1
2ðbaÞ 1
2ðb2a2Þ þ1
6ðb3a3Þ is always positive for all 0<a<b<1). Choose R>2, such that hb1
2Ra implies gðx;u;hÞ
h bN; for ðx;uÞA½a;b ðy;yÞ; ð3:13Þ
and
1lM 2R b1
2: ð3:14Þ
Let
W2¼ fuAC01½0;1 j kuk<Rg: ð3:15Þ Then for uAKVqW2, we have that
z0ðsÞ ¼lMw0ðsÞalM1
2qðsÞa1
2lMu0ðsÞ kuk ¼lM
2Ru0ðsÞ: ð3:16Þ Thus
u0ðsÞ z0ðsÞb 1lM 2R
u0ðsÞ: ð3:17Þ
Combining (3.17) with (3.14) and (3.11), we conclude that u0ðsÞ z0ðsÞb1
2u0ðsÞb1
2kukqðsÞb1
2Ra; sA½a;b: ð3:18Þ This together with (3.13) implies
gðx;uz;u0z0ÞbNðu0z0ÞbNRa
2 ; sA½a;b: ð3:19Þ Thus from (3.12) we get
ðAuÞ0ð1Þ ¼l ð1
0
ð1 r
ð1 s
gðt;uz;u0z0Þdt
ds
dr
blNRa 2
ðb a
ð1 r
ð1 s
1dt
ds
dr
blNRa 2
1
2ðbaÞ 1
2ðb2a2Þ þ1
6ðb3a3Þ
bR¼ kuk ð3:20Þ
foruAKVqW2. Therefore, it follows from the first part of Theorem 3 thatA has a fixed point uu~ in KVðW2nW1Þ such that
2ak~uukaR: ð3:21Þ
Moreover, by combining (3.21) with (3.7) and (3.8) and using Lemma 2 and Lemma 3, we know that
u~
u0ðxÞbkuukqðxÞ~ b2qðxÞ>2lMqðxÞb2lMw0ðxÞ12z0ðxÞ; xAð0;1Þ: ð3:22Þ So
u0ðxÞ ¼~uu0ðxÞ z0ðxÞb1
2uu~0ðxÞ; xAð0;1Þ ð3:23Þ and moreover, since
uðxÞ ¼uuðxÞ ~ zðxÞ
¼ ðx
0
ð~uu0ðsÞ z0ðsÞÞds
b1 2
ðx 0
u~ u0ðsÞds
b ðx
0
z0ðsÞds>0; xAð0;1Þ ð3:24Þ we get a positive solution uðxÞ ¼uuðxÞ ~ zðxÞ of (1.1).
Proof of Theorem 2. From (3.23), we have that (1.1) has a positive solution u1 satisfying
ku1kb1
2kuuk~ b1: ð3:25Þ
To find the second positive solution of (1.1), we set
fðx;u;pÞ ¼
fðx;u;pÞ; forðx;u;pÞA½0;1 ½0;a ½0;a fðx;a;pÞ; forðx;u;pÞA½0;1 ða;yÞ ½0;a fðx;u;aÞ; forðx;u;pÞA½0;1 ½0;a ða;yÞ fðx;a;aÞ; forðx;u;pÞA½0;1 ða;yÞ ða;yÞ: 8>
>>
<
>>
>:
ð3:26Þ
Then fðx;u;pÞbb for ðx;u;pÞA½0;1 ½0;a ½0;a, where a;b are given in Remark 1.
Now, we consider the auxiliary equation
uð4Þ¼lfðx;u;u0Þ; xAð0;1Þ
uð0Þ ¼u0ð0Þ ¼u00ð1Þ ¼u000ð1Þ ¼0: ð3:27Þ It is easy to check that (3.27) is equivalent to the fixed point problem
u¼Fu ð3:28Þ
where
FuðxÞ:¼l ðx
0
ðt 0
ð1 r
ð1 s
fðt;uðtÞ;u0ðtÞÞdt
ds
dr
dt: ð3:29Þ It is easy to check that F :K!K is completely continuous and FðKÞHK.
Set
H ¼minf0:9;ag; ð3:30Þ
and
L1¼min 6H M2;L
ð3:31Þ and fix
lAð0;L1Þ; ð3:32Þ
where
M2¼maxffðx;u;pÞ j0axa1;0auaH;0apaHg: ð3:33Þ Choose W3¼ fuAC01½0;1 j kuk<Hg. Then for uAKVqW3, we have that
ðFuÞ0ðxÞ ¼l ðx
0
ð1 r
ð1 s
fðt;u;u0Þdt
ds
dr alM2
ð1 0
ð1 r
ð1 s
1dt
ds
dr
alM2
1 6
aH: ð3:34Þ
Therefore
kFukakuk; uAKVqW3: ð3:35Þ From (A3) and Remark 1, we know that
p!0limþ
fðx;u;pÞ
p ¼ þy ð3:36Þ
uniformly for ðx;uÞA½0;1 ½0;a. This means that there exists a constant r0 ðr0<HÞ, such that
fðx;u;pÞbhp; for ðx;u;pÞA½0;1 ½0;r0 ½0;r0 where
1
8hlb1: ð3:37Þ
Then for uAK and kuk ¼r0, we have from (3.11) and (3.37) that ðFuÞ0ð1Þ ¼l
ð1 0
ð1 r
ð1 s
fðt;u;u0Þdt
ds
dr
bl ð1
0
ð1 r
ð1 s
hu0ðtÞdt
ds
dr
blh ð1
0
ð1 r
ð1 s
qðtÞdt
ds
drkuk
blh ð1
0
r2
2 qðrÞdrkuk
¼lh1 8kuk
bkuk: ð3:38Þ
Thus, we may let W4¼ fuAC01½0;1 j kuk<r0g so that
kFukbkuk; uAKVqW4: ð3:39Þ By the second part of Theorem 3, it follows that (3.27) has a positive solution u2 satisfying
r0aku2kaH: ð3:40Þ
Combining this with (3.26) and (3.30), we find thatu2 is also a solution of (1.1).
From (3.30), (3.25), (3.31) and (3.40), we know that (1.1) has two distinct positive solutions u1 and u2 for lAð0;L1Þ.
Acknowledgment
The author is very grateful to the referee for his or her valuable suggestions.
References
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Ruyun Ma
Department of Mathematics Northwest Normal University
Lanzhou 730070 Gansu, P. R. China e-mail: [email protected]
