Large Solutions To Non-Monotone Quasilinear Elliptic Systems
Qin Li
y, Zuodong Yang
zReceived 5 December 2011
Abstract
We present the existence of entire large positive radial solutions for the non- monotonic system
4pu=m(jxj)g(v) onRN; 4qv=n(jxj)f(u) onRN;
whereN 3andp q 2. The functionsf andgsatisfy a Keller-Osserman type condition while nonnegative functionsmandnare required to satisfy some decay conditions. Furthermore,mandnare such thatminfm; ngdoes not have compact support.
1 Introduction
In this paper, we consider the quasilinear elliptic system 4pu=m(jxj)g(v) forx2RN;
4qv=n(jxj)f(u) forx2RN; (1) forN 3,p q 2. We are concerned with existence of entire large solutions of (1), i.e. solutions such that u(jxj) ! 1and v(jxj) ! 1as jxj ! 1. Such problems arise in the theory of quasi-regular and quasi-conformal mappings as well as in the study of non-Newtonian ‡uids. In the latter case, the pair (p; q) is a characteristic of the medium. Media with (p; q) > (2;2) are called dilatant ‡uids and those with (p; q)<(2;2)are called pseudo-plastics. If (p; q) = (2;2), they are Newtonian ‡uids.
Whenp=q= 2, the system
4u=m(jxj)g(v) forx2RN;
4v=n(jxj)f(u) forx2RN (2) have received much attention recently. We list here, for example, [3, 4, 8, 11–13, 15, 16] and the references therein.
Mathematics Sub ject Classi…cations: 35J65, 35J50.
yInstitute of Mathematics, School of Mathematical Sciences, Nanjing Normal University, Jiangsu Nanjing 210023, China
za. Institute of Mathematics, School of Mathematical Sciences, Nanjing Normal University, Jiangsu Nanjing 210023, China; b. School of Teacher Education, Nanjing Normal University, Jiangsu Nanjing 210097, China
1
Whenp=q= 2,m(jxj)g(v) =p(jxj)v ,n(jxj)f(u) =q(jxj)u , system (2) becomes 4u=p(jxj)v forx2RN;
4v=q(jxj)u forx2RN; (3) for which existence results for entire positive solutions can be found in [17]. Lair and Wood established that all positive entire radial solutions of (3) are explosive provided
that Z 1
0
tp(t)dt=1and Z 1
0
tq(t)dt=1. If, on the other hand,
Z 1
0
tp(t)dt <1and Z 1
0
tq(t)dt <1;
then all positive entire radial solutions of (3) are bounded. In addition, [17] shows the set of central values, f(u(0); v(0))g, such that the system has an entire solution is a closed, bounded, and convex subset of R+ R+.
Jesse D. Peterson and Aihua W. Wood [3], Cirstea and Radulescu [16] extended the above results to a large class of systems (2).
Carcia-Melian and Rossi [18] studied the existence and non-existence results for boundary blow-up solutions, which can be obtained to the elliptic system
4u=um1vn1 forx2 2RN;
4v=um2vn2 forx2 2RN: (4) Wu and Yang [19] extended the above results to a class of systems
( div (jrujp 2ru) =um1vn1 forx2 ; div (jrvjq 2rv) =um2vn2 forx2 :
In [20], author discussed the existence of positive entire solutions of the quasilinear elliptic system
div(jrujp 2ru) =a(r)v forx2RN;
div(jrvjq 2rv) =b(r)u forx2RN; (5) whereaandbsatisfya(r) Kr1 andb(r) Kr2 forr r0>0, andK1,K2are positive constants and
>q 1(q ) +pand >
p 1(p ) +q,
then (5) has in…nitely many spherically symmetric positive entire solutions.
In [6, 21], Yang showed the existence of entire explosive radial solutions for quasi- linear elliptic system (1), wheref andg are positive and non-decreasing functions on (0;1)satisfying the Keller-Osserman condition. Other results can be seen in [5, 7, 10, 14].
A key feature common to all results known to us is that the functions f and g are required to be non-decreasing. This condition is necessary to construct monotonic
sequences of functions converging to solutions of (1). The combination of the absence of a meaningful maximum principle for systems and a lack of monotonicity of the functions has been the major hurdle in understanding the solution structures for this type of system. This is re‡ected by the vacuum between results for systems and those for the corresponding single equations. Motivated by the results of the above cited papers, we further study the existence of large solutions for problem (1), the results of the semilinear problem are extended to the quasilinear ones. We can …nd the related results for p=q= 2 in [3]. This paper marks our …rst step toward …lling this gap. In particular, we prove similar results forf andg being non-monotonic, rather "banded"
between monotonic functions.
Given system (1), we de…ne
G(t) := min inf
t s(f(s));inf
t s(g(s)) ; satisfying the following properties:
(G1) G(0) = 0,
(G2) G(s)>0fors >0,
(G3) the Keller-Osserman condition Z 1
1
Z s 0
G(t)dt
1 p
ds <1: Let
(jxj) = minfm(jxj); n(jxj)gand (jxj) = maxfm(jxj); n(jxj)g:
Notice (jxj) m(jxj) (jxj) and (jxj) n(jxj) (jxj). It is also clear that and both satisfy
Z 1
0
t1 N Z t
0
sN 1 (s)ds
1 p 1
dt <1and Z 1
0
t1 N Z t
0
sN 1 (s)ds
1 q 1
dt <1 whenmand nsatisfy
Z 1
0
t1 N Z t
0
sN 1m(s)ds
1 p 1
dt <1and Z 1
0
t1 N Z t
0
sN 1n(s)ds
1 q 1
dt <1: Given system (1), we de…ne the following function
H(t) = max max
0 s t(f(s)); max
0 s t(g(s)) :
We can see that H and Gare nonnegative, non-decreasing, and continuous functions satisfyingG(0) =H(0) = 0 andH(t)>0 whent >0. Furthermore,
G(t) f(t) H(t)andG(t) g(t) H(t):
Notice the functionH necessarily satis…es the Keller-Osserman condition whileGsat- is…es this requirement by hypothesis.
Inspired by [2, 3, 6, 24], we establish the following main theorems.
THEOREM 1. Supposem; n2C(RN)are nonnegative functions such that Z 1
0
t1 N Z t
0
sN 1m(s)ds
1 p 1
dt <1, Z 1
0
t1 N Z t
0
sN 1n(s)ds
1 q 1
dt <1; (6) and minfm; ng does not have compact support. Suppose f, g 2 C[0;1) are locally Lipschitz continuous on (0;1) such that G(t) satis…es (G1), (G2), and the Keller- Osserman condition (G3). Then there are in…nitely many entire positive solutions of system (1).
REMARK 1. IfN 3 andN > p, then condition (6) can be replaced by 0<
Z 1
1
rp11m(r)p11dr <1if1< p 2; (A) 0<
Z 1
1
r(pp2)N+11 m(r)dr <1ifp 2; (B)
0<
Z 1
1
rq11n(r)q11dr <1if1< q 2; (C) 0<
Z 1
1
r(q q2)N+11 n(r)dr <1ifq 2: (D)
We let
J(r) = Z r
0
t1 N Z t
0
sN 1 (s)ds
1 p 1
dt:
In fact, if1< p 2, then J(r) C1+
Z r 1
t1p N1 Z t
0
sN 1 (s)ds
1=(p 1)
dt:
Using the assumptionN 3in the computation of the …rst integral above and Jensen’s inequality to estimate the last one, we obtain
J(r) C2+C3
Z r 1
t3pN1p Z t
1
sNp 11 (s)p11dsdt:
Computing the above integral, we obtain J(r) C2+C4
Z r 1
tp11 (t)p11dt:
Applying (A) in the integral above, we infer that H1 = limr!1J(r) <1. On the other hand, if p 2, set
H(t) = Z t
0
sN 1 (s)ds
and note that either,H(t) 1fort >0orH(t0) = 1for somet0>0. In the …rst case, Hp11 1, and hence,
J(r) = Z r
0
t1pN1H(t)p11dt C5+ Z r
1
t1pN1dt
so thatJ(r)has a …nite limit becausep < N. In the second case,H(s)p11 H(s)for s s0 and hence,
J(r) C6+ Z r
1
t1p N1 Z t
0
sN 1 (s)dsdt:
Estimating and integrating by parts, we obtain J(r) C6+ p 1
N p
Z 1 0
tN 1 (t)dt +p 1
N p
Z r 1
t(pp2)N+11 (t)dt rpp N1 Z r
0
tN 1 (t)dt C7+C8
Z r 1
t(pp2)N+11 (t)dt:
By (B), we obtain thatH1= limr!1J(r)<1.
Using the notationR+ [0;1)and de…ning the set of central values S=f(a; b)2R+ R+:u(0) =a; v(0) =bg
where (u; v)is an entire solution of (1), we have the following theorem.
THEOREM 2. Given the hypotheses of Theorem 1, the setS is a closed bounded subset ofR+ R+.
Finally, de…ning the edge set
E=f(a; b)2@S:a >0; b >0g; we have an existence theorem for entire large solutions of (1).
THEOREM 3. Given the hypotheses of Theorem 1, any entire positive radial solution of system (1) with central value (u(0); v(0))2E is large.
LEMMA 1. Under the hypotheses of Theorem 1, for any central values (c; d) 2 R+ R+ where c > 0, d > 0; system (1) has a solution in a …nite ball of radius centered at 0 (denoted byB(0; )) RN:
PROOF. Note that radial solutions of (1) are solutions of the system ( (ju0p 2u0)0+Nr1ju0p 2u0 =m(r)g(v);
(jv0q 2v0)0+Nr1jv0q 2v0=n(r)f(u): (7)
Thus, solutions of (1) are simply …xed points of the operatorT :C[0;1) C[0;1) ! C[0;1) C[0;1)given by
T(u; v) = (bu;bv) = (
c+Rr
0(t1 NRt
0sN 1m(s)g(v)ds)p11dt for0 r < , d+Rr
0(t1 NRt
0sN 1n(s)f(u)ds)q11dt for0 r < : To prove such a …xed point exists, we will apply a version of Schauder’s Fixed Point Theorem. We consider the Banach spaceC[0; ] C[0; ];where >0 with norm
k(u; v)k= maxfkuk;kvkg: De…ne the subset
X =f(u; v)2C[0; ] C[0; ] :k(u; v) (c; d)k1 minfc; dgg:
ClearlyX is closed, bounded, and convex inC[0; ] C[0; ]. Further,T is a compact operator (refer to [1]). If we can show T maps X into X, then Schauder’s Fixed Point Theorem will guarantee we have a solution. Let (u; v) 2 X be arbitrary and k(u; v)k1 Q. We have the estimate
Z r 0
t1 N Z t
0
sN 1m(s)g(v(s))ds
1 p 1
dt Z r
0
t1 N Z t
0
sN 1 (s)H(v(s))ds
1 p 1
dt Hp11(v(r))
Z r 0
t1 N Z t
0
sN 1 (s)ds
1 p 1
dt Hp11(Q)
Z
0
1 p 1(s)ds:
Similarly, we may show Z r
0
t1 N Z t
0
sN 1n(s)f(u(s))ds
1 q 1
dt Hq11(Q) Z
0
1 q 1(s)ds:
Then choose >0 small enough so that max Hp11(Q)
Z
0
1
p 1(s)ds; Hq11(Q) Z
0
1
q 1(s)ds <minfc; dg: Doing so, we then haveT(u(r); v(r)) = (bu(r);bv(r));where
c bu(r) =c+ Z r
0
t1 N Z t
0
sN 1m(s)g(v(s))ds
1 p 1
dt c+ minfc; dg and similarly, d v(r)b d+ minfc; dg for all 0 r . Thus (bu(r);bv(r)) 2 X. Therefore a …xed point of T, a solution to (1) in the ballB(0; ), exists.
From [9, 22, 23], we give the following lemma
LEMMA 2. (Weak comparison principle) Let be a bounded domain inRN(N 2) with smooth boundary @ and : (0;1)!(0;1) is continuous and nondecreasing.
Letu1; u22W1;p( ) satisfy Z
jru1jp 2ru1r dx+ Z
(u1) dx Z
jru2jp 2ru2r dx+ Z
(u2) dx;
for all non-negative 2W01;p( )satisfyu1 u2 on@ . Thenu1 u2 on :
2 Proof of Theorem 1
LetT :C[0;1) C[0;1) !C[0;1) C[0;1)be the same as in Lemma 1. We have shown thatT has a …xed point inC([0; ]). Next, we prove thatT has a …xed point in C([0;1)). De…ne
(uk; vk) =T(uk; vk) = 8>
<
>: c+Rr
0
h t1 NRt
0sN 1m(s)g(vk)dsip11 dt;
d+Rr 0
h t1 NRt
0sN 1n(s)f(uk)dsiq11 dt:
Just as in the proof of Lemma 1, we examine (uk, vk) over [0;1] [0;1] and then we have fukg and fvkg are each uniformly bounded on [0;1]. The nonnegative se- quences of derivativesfuk0gandfvk0gare also bounded on[0;1]implying the sequence has a uniformly convergent subsequence. Call this subsequence f(wk1; zk1)g, and let (w1k; z1k) !(cu1;vb1)as k ! 1, uniformly on[0;1] [0;1]. Notice that (cu1;vb1)is a solution of (1) in B(0;1)with central value(a0; b0)2@S. Likewise, the subsequences fw1kg andfz1kg are each uniformly bounded and equicontinuous on[0;2], so there ex- ists a subsequencef(w2k; zk2)g off(w1k; z1k)gsuch that(w2k; z2k) !(cu2;vb2)ask ! 1, uniformly on [0;2] [0;2]. Sincef(wk2; zk2)g f(wk1; zk1)g, we see(uc2;vb2) = (uc1;vb1)on [0;1] [0;1]. Continuing, we obtain a sequence f(cuk;vbk)g, each a solution of (1) in B(0; k)with central value(a0; b0)2@S, and each solution satis…es
(cuk;vbk) = (cu1;vb1) forr2[0;1];
(cuk;vbk) = (cu2;vb2) forr2[0;2];
...
(cuk;vbk) = (u[k 1;v[k 1) forr2[0; k 1]:
Thus,(uck;vbk)converges to(u; v)where
(u(r); v(r)) = (uck;vbk) if0 r k:
This convergence is uniform on any bounded set. Thus(u; v)2C[0;1) C[0;1)and satis…es(T u; T v) = (u; v). The proof of Theorem 1 is completed.
3 Proof of Theorem 2
To prove that S is bounded, we …rst assume to the contrary that S is not bounded.
Therefore, since[0; a] [0; b] Swhenever(a; b)2S, we must have either[0;1) f0g S or f0g [0;1) S. Lethbe a positive radial solution of
4ph= (r)G(h2) 0 r <1;
limr !1 h(r) =1: (8)
(see [2] for the proof of existence). Let(u; v)be any solution which exists by hypotheses, to the system
8>
<
>:
u(r) =a+Rr 0
h t1 NRt
0sN 1m(s)g(v(s))ds ip11
dt forr >0;
v(r) =b+Rr 0
h t1 NRt
0sN 1n(s)f(u(s))ds iq11
dt forr >0
(9)
with a > h(0)and b= 0. Without loss of generality, we assume that a 1. We now show thath u+v for all r 0, which, if proven, will contradict with the fact that u+v exists for all r 0. Clearly, h(0) < a u(0) +v(0). Thus, there exists > 0 such thath(r)< u(r) +v(r)for allr2[0; ). Let
R0= supf >0jh(r)< u(r) +v(r)forr2[0; )g:
We claim that R0 = 1. Indeed, suppose that R0 < 1. Since h(r) < u(r) +v(r) in [0; R0),
g(v(r)) G(v(r)) G(u(r) +v(r)
2 )ifv(r) u(r);
f(u(r)) G(u(r)) G(u(r) +v(r)
2 )ifu(r) v(r);
and elementary estimates, we observe that h(0) +
Z R0
0
t1 N Z t
0
sN 1 (s)G(h(s) 2 )ds
1 p 1
dt
< h(0) + Z R0
0
t1 N Z t
0
sN 1 (s)G(u(s) +v(s)
2 )ds
1 p 1
dt
< h(0) + Z R0
0
t1 N Z t
0
sN 1 (s)(g(v) +f(u))ds
1 p 1
dt
< h(0) + Z R0
0
t1 N Z t
0
sN 1( (s)g(v) + (s)f(u))
1 p 1
dt
< h(0) + Z R0
0
t1 N Z t
0
sN 1(m(s)g(v) +n(s)f(u))
1 p 1
dt
< h(0) + Z R0
0
t1 N Z t
0
sN 1m(s)g(v)
1 p 1
dt
+ Z R0
0
t1 N Z t
0
sN 1n(s)f(u)
1 q 1
dt
< u(0) +v(0) + Z R0
0
t1 N Z t
0
sN 1m(s)g(v)
1 p 1
dt +
Z R0
0
t1 N Z t
0
sN 1n(s)f(u)
1 q 1
dt
< u(R0) +v(R0):
Thus, since h(R0) < u(R0) +v(R0), there exists >0 such that h(r) < u(r) +v(r) in [0; R0+ ). This contradicts with the fact that R0 is a superemum. Thus, R0= 1, establishing the boundness of the setS.
To show S is closed, we will show the set contains its boundary. Let h1(r) and h2(r)be positive solutions of
4ph1= (r)G(h1) for0 r < R;
limr !R h1(r) =1; (10)
and
4qh2= (r)G(h2) for0 r < R;
limr !R h2(r) =1: (11)
where Ris an arbitrary positive real number.
It is easy to show that by Lemma 2,u h1 andv h2. Thus, u+v h1+h2 in [0; R]. Leth1+h2= . Thenu+v in [0; R]. Now, let(a0; b0)2@S and consider a positive, increasing sequence fRkg such thatRk ! 1 ask ! 1, and (jxj)>0 for jxj = Rk. We can …nd a sequence, since is radial and does not have compact support, and then we have B((a0; b0),1=Rk)\S6=?. For eachk 1, we denote the arbitrary point (ak0; bk0) 2 B((a0; b0);1=Rk)\S. Notice that (ak0; bk0) ! (a0; b0) as k ! 1. We de…ne the sequence
(uk; vk) = 8>
<
>:
ak0+Rr 0
h t1 NRt
0sN 1m(s)g(vk(s))ds ip11
dt ifr 0;
bk0+Rr 0
h t1 NRt
0sN 1n(s)f(uk(s))dsiq11
dt ifr 0;
where each (uk; vk) is an entire solution of (1). These solutions exist since each (ak0; bk0) 2 S and each uk+vk . Examining (uk; vk) over [0;1] [0;1], we have uk +vk (1) < 1 for 0 r 1. Thus, fukg and fvkg are each uniformly bounded on [0;1]. The nonnegative sequences of derivatives fu0kg and fv0kg are also bounded on[0;1]implying the sequence has a uniformly convergent subsequence. Call this subsequence f(wk1; zk1)g, and let (w1k; zk1) ! (cu1;vb1) as k ! 1, uniformly on [0;1] [0;1]. Notice that (uc1;vb1) is a solution to (1) in B(0;1) with central value (a0; b0) 2 @S. Likewise, the subsequences fw1kg and fzk1g are each uniformly bounded and equicontinuous on [0;2], so there exists a subsequence f(wk2; zk2)g of f(wk1; zk1)g such that (w2k; zk2) ! (uc2;vb2) uniformly on [0;2] [0;2] as k ! 1. Since f(w2k; zk2)g f(wk1; zk1)g, we see(cu2;vb2) = (cu1;vb1) on[0;1] [0;1]. Continuing,
we obtain a sequence f(cuk;vbk)g, each a solution to (1) in B(0; k) with central value (a0; b0)2@S, and each solution satis…es
(cuk;vbk) = (cu1;vb1) forr2[0;1];
(cuk;vbk) = (cu2;vb2) forr2[0;2];
...
(cuk;vbk) = (u[k 1;v[k 1) forr2[0; k 1]:
Thus,(uck;vbk)converges to(u; v)where
(u(r); v(r)) = (cuk;vbk) for0 r k:
This convergence is uniform on bounded sets, and thus(u; v)is an entire solution with central value(a0; b0)2@S. We consider that @S S, implyingS is closed.
4 Proof of Theorem 3
Letun andvn be de…ned as positive solutions of 8>
<
>:
un=u(0) + 1n+Rr 0
h t1 NRt
0sN 1m(s)g(vn(s))dsip11 dt;
vn =v(0) +n1 +Rr 0
h t1 NRt
0sN 1n(s)f(un(s))dsiq11 dt
(12)
where (u(0); v(0))2E. We denote thatu0n(r) 0 andv0n(r) 0. Also, since(u(0) +
1
n; v(0) +n1)*S, for eachn= 1;2;3; :::, there existsRn<1such that lim
r !Rn
un(r) =1, lim
r !Rn
vn(r) =1, andR1 R2 R3 : From (12) and the fact that vn0(r) 0, we get
vn(r) v(0) + 1
n+Hq11(un(r)) Z 1
0
t1 N Z t
0
sN 1n(s)ds
1 q 1
dt C1un(r) +C2Hq11(un(r));
where valueC1 is any upper bound on v(0)+n1
u(0)+1n and C2=
Z 1
0
t1 N Z t
0
sN 1n(s)ds
1 q 1
dt <1:
Next, we de…neh(t) =H(C1t+C2Hq11(t)). It is an easy matter to show thath(0) = 0, h(s)>0fors >0, andhsatis…es the Keller-Osserman condition
Z 1
1
Z s 0
h(t)dt
1 p
ds <1:
De…ne
F(s) = Z 1
s
dt hp11; which is well de…ned for s >0:Notice that
F0(s) = 1
hp11 <0 andF00(s) = h0(s)
(p 1)hpp1 >0:
We now have
div(jrunjp 2run) = m(r)g(vn(r)) m(r)H(vn(r)) m(r)H
h
C1un(r) +C2Hq11(un(r)) i
= m(r)h(un(r)):
Then, we may calculate
div(jrF(un)jp 2rF(un))
= jF0(un)jp 1div(jrunjp 2run) + (p 1)jrunjpjF0(un)jp 2F00(un) jF0(un)jp 1div(jrunjp 2run) jF0(un)jp 1m(r)h(un)
= 1
hp11(un)
!p 1
m(r)h(un) = m(r):
Hence,
div(jrF(un)jp 2rF(un)) m(r):
Rewriting the Laplacian in radial form and multiplying byrN 1, we obtain (rN 1jF0(un)jp 2F0(un))0N 1m(r):
Integrating over [0; r];where0< r < Rn; we get d
drF(un) r1 N Z r
0
sN 1m(s)ds
1 p 1
:
Next, we integrate over[r; Rn]. Note thatF(un(r)) !0asr !Rn. This integration yields,
F(un(r))
Z Rn
r
t1 N Z r
0
sN 1m(s)ds
1 p 1
dt:
That is,
F(un(r))
Z Rn
r
t1 N Z r
0
sN 1m(s)ds
1 p 1
dt:
SinceF0(s)<0fors >0, we have un(r) F 1
Z Rn
r
t1 N Z r
0
sN 1m(s)ds
1 p 1
dt
! :
Now, we letn ! 1, so Rn !R, andun !u, we have F 1
Z R r
t1 N Z r
0
sN 1m(s)ds
1 p 1
dt
!
u(r):
Finally, let r !R, we have
rlim!RF 1 Z R
r
t1 N Z r
0
sN 1m(s)ds
1 p 1
dt
!
=1 lim
r !Ru(r):
However, u(jxj) and v(jxj) have a central value (u(0); v(0)) 2 E and entire. Thus R=1, and our proof is complete.
References
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