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Applied Mathematics Letters 17 (2004) 329-336
ELSEVIER
Applied Mathematics Letters www.elsevier.com/locate/aml
Multiple Positive Solutions for Higher-Order Boundary Value Problems with Sign Changing Nonlinearities YANPING Guo College of Sciences, Hebei University of Science and Technology Shijiazhuang, 050018, Hebei, P.R. China and College of Mathematics and Information Science, Hebei Normal University Shijiazhuang, 050016, Hebei, P.R. China
guoyanping65©sohu,c o m
YUJUN ZHU College of Mathematics and Information Science, Hebei Normal University Shijiazhuang, 050016, Hebei, P.R. China
JIQING QIu College of Sciences, Hebei University of Science and Technology Shijiazhuang, 050018, Hebei, P.R. China
(Received April 2002; accepted February 2003) Abstract--For
the 2nth-order boundary value problem
O
O
where f is allowed to cha~ge sign, ~(v) --- IvIP-2v, p > 1, growth conditions are imposed on f which yield the existence of at least two symmetric positive solutions by using the fixed-point theorem in double cones. @ 2004 Elsevier Ltd. All rights reserved.
Keywords--Fixed-point problem.
theorem, Positive symmetric solution, Higher-order boundary value
1. I N T R O D U C T I O N In this paper, we are concerned with the existence of multiple positive solutions for the 2nth-order boundary value problem with p-Laplacian operator The project is supported by the Natural Science Foundation of China (10371030), the Natural Science Foundation of Hebei Province (603384), and the Scientific Research Foundation of the Education Department of Hebei Province (2003216). 0893-9659/04/$ - see front matter @ 2004 Elsevier Ltd. All rights reserved. doi: 10.1016/S0893-9659(04)00015-1
Typeset by ~4A/~S-TEX
Y. Guo et al.
330
((~ (y(2n--1)))'(t) -~ f (t,y(t),y H (t),...,y (2(n-1)) (t)),
0 <~t < 1, (1)
c~iy(2i) (0) - / 3 i y (2i+1) (0) = aiy (2i) (1) +/Siy (2~+1) (1) = 0,
0 < i < n - 1,
where our nonlinear t e r m f is allowed to change sign, and ~(v) = ]v]P-2v, p > 1, c~,/~ > O, 0 < i < n - 1. We will impose growth conditions on f which ensure the existence of at least two positive solutions for (1). There is much current attention focused on positive solutions of b o u n d a r y value problems for ordinary differential equations, as well as for finite-difference equations; see [1-10], to name a few. In [11], Davis, f l o e and Henderson imposed conditions on f which yield at least three symmetric positive solutions to the 2rnth-order Lidstone boundary value problem y(2~) = f ( y ( t ) , y " ( t ) , . . . , y (2('~-t)) ( t ) ) ,
t • [0,1], (2)
y(~) (0) = y(~) (1) = o,
0
where ( - 1 ) m f : R "~ ~ [0, oc) is continuous, using the Leggett-Williams fixed-point theorem [12] and the s y m m e t r y of the associated Green,s function and then Davis, Henderson and Wong [13] established similar results by applying the five functionals fixed-point theorem [14]. Note that [11] and [13] are the only two works in which f depends on higher-order derivatives of y. This paper is the only work which allow f to change sign as far as we know. The paper is divided into three sections. In Section 2, we prove the fixed-p0int theorem in double cones. In Section 3, we impose growth conditions on f which allow us to apply the theorem in obtaining two symmetric positive solutions of (1).
2. T H E F I X E D - P O I N T
THEOREM
IN D O U B L E
CONES
For a cone K in a Banaeh space X with norm N' II and a constant r > 0, let K r = {x • K : I[xN < r}, OK~ = {x E K : IIx[[ = r}. Suppose a : K --~ R + is a continuously increasing functional, i.e., a is continuous and c~(Ax) < c~(x) for A • (0, 1). Let K (b) = {x • K : ~ (~) < b},
O K (b) = {x • K : ~ (~) = b}
and K~(b) = {x • K : a < [[x[I, a ( x ) < b}. The origin in X is denoted by e. THEOREM 1. Let X be a r e d Banach space with norm [[. [[ and K, K ' C X t w o solid cones with K' C K. Suppose T : K -+ K and T' : K' --+ K'
are
two completely continuous operators and
a : K ' --+ R + a continuously increasing functional satisfying a ( x ) < ]]x[I _< M a ( x ) for a11 x in K ' , where M ~ 1 is a constant. If there are constants b > a > 0 such that
IlTxll < a, for x • 0t(~; (c2) I]T'~:II < a, for x • a K i and c4T'x) > b for x • OK'(b); (C3) T x = T ' x , for x • K ' ( b ) n { ~ : T ' ~ = ~}, (c1)
then T has in K at least two fixed points Yl and Y2 such that
0 _< llyll] < a < lly=PI,
~ (y2) < b.
(3)
PROOF. Let the symbol deg K denote the degree on cone K [15]. Then condition (C1) implies deg K { I - T, K a , 0} = 1, and therefore, T has in Ka a fixed point Yl with the norm 0 _< []Yl[] < a. Similarly, the first part of Condition (C~) implies
degK, {I - T', K', 0} = 1.
Multiple Positive Solutions
331
We now prove that deg K, { I - T', K t ( b ) , 0} = O. The relation s ( x ) < Ilxll < M s ( x ) and the second part of Condition (C2) imply infxeaK;(b) IIT'xH > b > 0. Let T : / ~ ' ( b ) -* K be an extension of T'[aK,(b) : OK'(b) ~ K . Dugundji's extension theorem [16] ensures that T is completely continuous and T ( / ( ' ( b ) ) C ~ T ' ( O K ' ( b ) ) . Therefore, inf Tx >b>0. x~R'(b) Take a homotopy H ( x , A) = x - )~Tx. We claim H(x,A)#0,
for a l l x e O K ' ( b ) ,
),_>1.
If it is not in the case, then there exist )~0 >_ 1 and xo E OKt(b) such t h a t x0 - ),0Tx0 = 0, i.e., xo = )~oTxo.
And it follows that b = s (xo)
~
S
(~o~o)
= s (~oT'~o)_> s (T'~o)> b,
a contradiction. Hence,
When A > M , for all x c / ~ t ( b ) , one has IIxIl <_ M b ,
ATx
= A Tx
>_ )~b > M b ,
and therefore, x - ATx = 0 has no solution in/~'(b), which implies degK,{f -- AT, K ' ( b ) , 0} = 0 for A _> 1. It follows that deg K, {I T', K ' (b), 0} = 0, -
and then degK,: {I -- T', K i (b), O} = deg K, {I - T', K ' (b), 0} - deg K, {I - T', K~, 0} = - 1 . So T ' has in K'~(b) a fixed point Y2: IlY2[[ > a, s ( y 2 ) < b. Condition (C3) implies Y2
="
Try2 = Ty2 C K ~ C K .
Theorem 1 is now proved. 3. T W O
SYMMETRIC
POSITIVE
SOLUTIONS
In this section, we will impose growth conditions on f which allow us to apply Theorem 1 to obtain two symmetric positive solutions of (1). Let g~(t, s) (0 < i < n - 1), be the Green's function for y" = 0,
t e [0, 1],
(4)
s i y (0) - ~ y ' (0) = s G (1) + ~iy' (1) = O, g~ (t, 8)
__1 J" (s~ +/3i - s~t)(/3i + s i s ) ,
(si + 9i
s~s) (9~ + s~t),
0 < s < t < 1,
0 < t < s < 1,
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332
where pi = ai(ai + 2/3i). Let Gl(t, s) = 9~-2(t, s), then for 2 _< j < n - 1 we recursively define
aj (t, ~)
:
/01
g~-j-x (t, ~) aj_~ (~, ~) d~.
As a result, Gj(t, s) (1 _< j _< n - 1) is the Green's function for the b o u n d a r y value problem y(2j) (t) = 0,
0 < t < 1,
o~n_j+i_lY (2i) (0) -- ~n_j+i_lY (2i+1) (0) : ~n_j+i_lY (2i) (1) + ~n_j+i_lY (2i+1) (1) -~ 0, 0
1
Ajv(s) : ]o Vj (s,-r) v ('1") dT. By the construction of Aj, we see that, for each 1 ___j <_ n - 1,
(Ajv) (2j) (t) = v (t) ,
O
a ~ - j + i - 1 (Air) (2i) (0) - ~n-j+i-1 (Ajv) (2i+i) (0) = O,
O
a,~-y+{-1 (Ajv) (2~) (1) + fl~-j+i-1 (Ajv) (2i+1) (1) = O,
0
Therefore, (1) has a solution if and only if the b o u n d a r y value problem (~(vi))'(t) : f(t,A~_lv(t),A~_2v(t),...,Alv(t),v(t)),
0 < t < 1,
O~n_lV (0) -- Zn_l v' (0) = O~n_lV (1) ~- ~n_l v' (1) = 0
(5)
has a solution. If y is a solution of (1), then v = y(2(,-1)) is a solution of (5). Conversely, if v is a solution of (5), then y : An_iv is a solution of (1). In addition if (--1)n--Iv(t) _> 0 ( ~ 0) on [0, 1], then y : A~_lv is a positive solution of (1). For our main result, we will need to make use of various properties of gi(t, s), namely, for 1
l~ = °
f~-~ [g~ (t, s)[ ds : a
1 ,s ) [ ds = ~~i + -~, 1
(1 - 25)2a~ (Z~ + 5~)
Let X = C[0,1], K : {x E X : ( - 1 ) ~ - l x ( t ) _> 0, x(t) = x(1 - t), t E [0,1]}, K ' : {x • X : ( - 1 ) " - I x ( t ) _> 0, x(t) = x(1 - t ) , ( - - 1 ) n - l x are c onc a ve on [0, 1]}. Obviously, K , K ' C X are two cones with K ' c K. For x • K and 6 • (0, 1/2), we define
a(x)=
min
[x(t)l,
_ f~ ®-1 (¢:/2 s (~, A~_lx (~), An_2x
+ 1 (~), .,Aix(~-),X(T)) d~-) ds , O<_t<_-~,
(Tx) (t) =
.., Alx (~-) ,x (T)) d~-) Otn--1
1 <_t < 1 , .,Alx(~-),x(~')) d'r) ds ) + , -~ _
Multiple Positive Solutions
333
where (B) + = ( - 1 ) ~-~ m a x ( ( - 1 ) ~ - ~ B , 0}.
d'r)
• ~n--1~--1 (fl/2 f(7, An_lX(~-),An_lx(~-),...,Alx(7),x(~-)) O~n-- I
-- fo ~)--1 (f:/2 f (7. An_lX(7. ) ,An_2X(7- ) ,... ,AlX(q- ) ,x(f)) dT) ds,
O
1 -
(Ax)(t)=
__~n--i C~--1 (fll2 f (T, An_lX(q_),An_2X(.F),...,AlX(T),X(T)
)
dT)
O~n-- 1
- ft ~ ~ - 1
/~ f (% A~_~x (~), A~_2x ( T ) , . . . , Amx (~-) , x (~)) d~- ds,
For x E X, define 0 : X ~ K by (Ox)(t) x C K ~, let
/
~n--I
= (--1) n-1
1 ~
m a x { ( - 1 ) ~ - l x ( t ) , 0 } , then T = 0 o A. For
¢_1 s+(.,
, 1
O~n--$ 1
where f + (t, u ~ _ l , . . . , u0) = ( - 1 ) ~ m a x ( ( - 1 ) " f ( t , u ~ - l , We will suppose the following conditions are satisfied
1,
., u0), 0}.
(H1) f : [0, 1] x R ~ --+ R is continuous and for each ( u , - 1 , . . . , uo) C R ~, f ( t , U n - l , . . . , u0) is symmetric about t = 1/2; (H2) ( - 1 ) ~ f ( t , 0 , 0 , . . . , 0 ) > 0 ( ~ 0) for t • [0,11. LEMMA 1. Suppose A : K ~ X is completely continuous, then OoA : K --+ K is also a completely continuous operator. PROOF. The complete continuity of A implies that A is continuous and relatively compact. Therefore, given arbitrary D C K, bounded, and Vs > 0, there are y~, i -- 1 , . . . , m such that A D C U i ~ l B ( y i , s ) , where B(y~,s) = {x • X : I I x - y i U < s}. Then, V~ • ( O o A ) ( D ) , there is y • A D such t h a t ~(t) = ( - 1 ) ~ - l m a x { ( - 1 ) ~ - l y ( t ) , 0 } . We choose one Yi • {Yl,-.. ,Y,~} such t h a t max0<, 0, there is 5 > 0 such that HAy - Axl[ < s, if HY- x]] < and then
11(0 o A ) y - (0 o A)xll = m a x II (fiY) (t) - (fix) (t)l I 0
< m a x II( A y ) ( t ) - (Ax)(t)il --
O
= l l A y - A x H < s, when IlY - xll < 6, which means that 0 o A is continuous on K. Therefore, 0 o A is completely continuous• (H1) implies that A and T t are well defined. Based on the continuity of f~ it is easy to see that A : K -~ X and T / : K t ~ K / are completely continuous. So T : K --+ / ( is completely continuous by using L e m m a 1. For x e K t, we have Ix(t)l >_ 61Ix Hfor t e [6, 1 - 6 ] by the concavity of ( - 1 ) n x Therefore, _< B]xll < THEOREM 2. Suppose (HI) and (H2) hold. In addition there exist nonnegative numbers a, b, and d such that d < (an--1/2~n--1 + 1)d < a < 6b < b and f ( t , u,~-l, u ~ - 2 , . . . , uz, uo) satisfies the foiiowing growth conditions:
Y. Cuo et al.
334
(H3) (--1)nf(t, Un--l,. U0) _> 0 for (t, lUn-ll,.. i/J+1 L b] [d, b]; i=2 n--z j X • . ,
. ,
I~01) E
r l-[j+l [0,1] X zxj=n_lLiii=21n_id, ]--[1
(H4) (--1)nf(t, un_i,... ,Uo) < 2~(20Ln-1/(O~n_1 Jr- 2~n--1)a) for (t, lUn_il,..., I~ol) ~ [0,1] x 1 rZ j=n1 [0,1-I/+1 i=2 Ln-ia] x [0, a]; (Hh) ( - 1 ) n f ( t , l t n - 1 , . . . , tz0) > (25/(1 - 2 5 ) ) ~ ( ( a , - i h / ( f l ~ - i + an-l(~))b) for (t, I ~ - ~ 1 , ' " , [uo[) e [0, 1] x I-Ij=n--1 1 [I-TJ+I LI 1i=2 rn --i~b, I~ j+l i=2 L~_tb] x [hb, b].
Then the boundary value problem (1) has at least two positive solutions x 1 and x2 such that
0 < X~2(n-l)) < a < x(22(n-l)) ,
mill
t615,1--5]
X~2(n-1)) (t) < 5b.
PROOF. First, we show that T has a fixed point Yl E K with 0 < IlY~[I - a. In fact, Vx E OKa, we h a v e l ] x l l = a . For l _ < j _ < n - 1 , j+l
j+i
IIAjxH <- H L~_~ IlxH = H n~_~a.
i=2
i=2
From (H4), we obtain
[]Tx]] = 0
,-'8
O
0 (//
f (7, An_Ix @), An-2X ( 7 ) , . . . , A l x (T) ,x (w)) dz
)
f ('r, A n - i x ('r), An-2X (9-),..., A l x (~-) ,x (~-)) dT ds
O~n_ 1
f (T, A n - i x (T), An-2x @ ) , . . . , A l x (~-), x (T)) d~-
;o (r
,/8
)))
f (% A n - i x (-c), An-2x (~-),..., A i x ('c) ,x (7)) dT ds
,0
(21___ / 20~n-1 "~) ~ (2 / 20~n-1 kl) < Zn--l(]}--loLn_l "2~ ~kOln--1-]-2]~n--laJ "~ C£5-I "2~ ~kO~n--l"[-2Zn--la/ = (~n-1 + \OLn--1
1)
2 a = a. OLn--1 + ~n--1
The existence of yl is proved by using (Ci) of Theorem 1. Obviously, Yi is a solution of (5) if and only if Yi is a fixed point of A. Next, we need to prove that Yi is a solution of (5). Suppose the contrary, i.e., there is to E (0, 1) such that yi(to) (Ayl)(t0). It must be (-1)~-i(Ayi)(to) < 0 = yi(to). Let ( a , ~ ) be the interval that to E (a,~), and (-1)~-i(Ayi)(t) < 0 for Vt E (a,Z). Obviously, [a, fl] ¢ [0, 1] by (H2). Suppose without loss of generality that fl < 1. Then yi(t) - 0 for t E [a, fl] and (-1)n-l(Ayi)(t) < 0 for t E (a,~), (Ayl)(~) = 0. This and (H2) implies (-1)n-i(~((Ayi)'))'(t)= ( - 1 ) ~ - i f ( t , 0, 0 . . . , 0) <_ 0 for t E [a, ~], so a = 0 and ( - 1 ) n - i ( A y l ) ' ( 0 ) > 0. On the other hand, c~n-i(Ayi)(O)-~n-i(Ayi)'(O) = 0 implies (Ayl)'(0) _< 0, a contradiction. We now show that (C2) of Theorem 1 is satisfied. For x E OK'~, i.e., Ilxll = a. For 1 < j _< n - l , ]IAjxl I < rlJ+l Ln-ia. --
1 1 i = 2
Multiple Positive Solutions
~-i
IIT'xll = O
335
f+ (7, An_ix (7), An_2x (7) ... Alx (7.) ,x (7)) dr
O~n_ 1
~
'
f+ (r, A,~-lx (r), An-2X ( 7 ) , . . . , Alx (7) ,x (r)) d7 < O~n_ l#n-lff2'l(ffO1/2(-1)nf+(%An-lx(T)'An-2x(r)'''''Alx(r)'x(r))dT
1(/?
(-1)n f+ @,An_lx@),An_2x(7),...,Alx(r),x(7)) dr
~_ (~--1
)
)
a))=a
< Whereas for 6b _< [x(t)l _< b,
x e OK'(6b),
i.e.,
a(x) = 6b.
For 6 < t < 1 - 6, 1 _< j _< n - 1, we have
j+l
]Ajx (t)[
n,~-ib,
< II i=2
f l - ~ tGj(t,s)l ds >_J+~ IAjx(t)[ = fL 1 Gj(t,s)x(s) ds >_Sbj~ dO
i=2
We may use condition (Hs) to obtain
a(T'x)=
~
An_iX(T , An-2X(~-),... , Alx(r),x(T))d7)
fot~ -1 f l / 2 f + (T, An-lX(7),An_2x(7) . ,Alx(r) x(7)) d7) ds JS
-
~n-i~-1 [112( - 1 ) n f + (T, An-ix (7), A~_2x ( 7 ) , . . . ,
O~n--1
"~-
dO
(--1) n f + (% An-ix (T),
I1}--1
Alx (7) ,x @)) dr
An-2X ( T ) , . . . , AlX (r) ,x (7)) dr
) ds
J8
> ~n-~e-~ OLn-- i
+
-
~5-1
\an-1
,]6
J5
(-1Ff+(r, An-lx(7) An_2x(7),... A~x('~),x(7))d7 '
'
(-1)nf+(7, A~-lx(7),An-2X(7),...,Alx(7),x(r)) d7 ds \Sa,~_i
+/3n_l b
=
5b.
Finally, we show that (Ca) of Theorem 1 is also satisfied. Let x E K~(6b) n {u : T'u = u}, then ( - 1 ) n - I x ( I / 2 ) = []xll > a > ( a n - 1 / 2 f n - 1 ÷ 1)d. Next, we will prove (--1)n--Ix(0) _> d. Suppose this is not true, then there exists to e (0, 1/2) such that ( - 1 ) n - l x ' ( t 0 ) > (an-1//~n-1)d. It follows from the concavity of ( - 1 ) n - i x that ( - 1 ) n-1 x' (0) > (--1) ~-1 x' (to) > O~n--I /hn-I d "
Therefore,
O=(-1)n-l(o~n_lX(O)-fln_lX'(O))
~
-- W n - - i
c~n-ld = O, ~n--I
Y. Guo et al.
336
a contradiction. Thus, d < Ix(t)] < b for t E [0, 1]. For 1 < j < n .
.
.
.
1, 0 < t < 1, ~ 1j i+=l2l n _ ~ d < J-
l j - I - 1 L n - i b . C o n d i t i o n (Ha) ensures t h a t f ( t , A n - I X ( t ) , A n - 2 x ( t ) , . I A j x ( t ) l __ < y11i=2 , .A l X.( t ) , . x ( t ) ) f+(t, An-lx(t),An-2X(t),... , A l x ( t ) , x ( t ) ) , t h e n T x = T ' x for x E K~(Sb) A { u : T ' u = u}. A n
a p p l i c a t i o n of T h e o r e m 1 yields t h e result. M o r e o v e r , xg(t) = A n - ~ y ~ ( t ) = fo1 a n - l ( t , s)y (s) ds,
i = 1,2.
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