Applied Mathematics E-Notes, 15(2015), 1-13 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Large Solutions To Non-Monotone Quasilinear
Elliptic Systems
Qin Liy, Zuodong Yangz
Received 5 December 2011
Abstract
We present the existence of entire large positive radial solutions for the nonmonotonic system
4p u = m(jxj)g(v) on RN ;
4q v = n(jxj)f (u) on RN ;
where N
3 and p
q
2. The functions f and g satisfy a Keller-Osserman
type condition while nonnegative functions m and n are required to satisfy some
decay conditions. Furthermore, m and n are such that minfm; ng does not have
compact support.
1
Introduction
In this paper, we consider the quasilinear elliptic system
4p u = m(jxj)g(v) for x 2 RN ;
4q v = n(jxj)f (u) for x 2 RN ;
(1)
for N 3, p q 2. We are concerned with existence of entire large solutions of (1),
i.e. solutions such that u(jxj) ! 1 and v(jxj) ! 1 as 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.
When p = q = 2, the system
4u = m(jxj)g(v) for x 2 RN ;
4v = n(jxj)f (u) for x 2 RN
(2)
have received much attention recently. We list here, for example, [3, 4, 8, 11–13, 15,
16] and the references therein.
Mathematics Subject Classi…cations: 35J65, 35J50.
of Mathematics, School of Mathematical Sciences, Nanjing Normal University, Jiangsu
Nanjing 210023, China
z a. 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
y Institute
1
2
Large Solutions to Elliptic Systems
When p = q = 2, m(jxj)g(v) = p(jxj)v , n(jxj)f (u) = q(jxj)u , system (2) becomes
for x 2 RN ;
for x 2 RN ;
4u = p(jxj)v
4v = q(jxj)u
(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
Z 1
tp(t)dt = 1 and
0
tq(t)dt = 1.
0
If, on the other hand,
Z
0
1
tp(t)dt < 1 and
Z
0
1
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 = um1 v n1
4v = um2 v n2
for x 2
for x 2
2 RN ;
2 RN :
(4)
Wu and Yang [19] extended the above results to a class of systems
(
div (jrujp 2 ru) = um1 v n1 for x 2 ;
div (jrvjq
2
rv) = um2 v n2
for x 2 :
In [20], author discussed the existence of positive entire solutions of the quasilinear
elliptic system
div(jrujp 2 ru) = a(r)v
for x 2 RN ;
(5)
q 2
div(jrvj rv) = b(r)u
for x 2 RN ;
1
2
where a and b satisfy a(r) K
and b(r) K
r for r
r
constants and
>
(q
) + p and >
q 1
p
r0 > 0, and K1 , K2 are positive
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 quasilinear elliptic system (1), where f and g 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
Q. Li and Z. D. Yang
3
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 for f and g being non-monotonic, rather "banded"
between monotonic functions.
Given system (1), we de…ne
G(t) := min inf (f (s)); inf (g(s)) ;
t s
t s
satisfying the following properties:
(G1) G(0) = 0,
(G2) G(s) > 0 for s > 0,
(G3) the Keller-Osserman condition
Z
1
1
Z
1
p
s
ds < 1:
G(t)dt
0
Let
(jxj) = minfm(jxj); n(jxj)g and (jxj) = maxfm(jxj); n(jxj)g:
Notice (jxj)
m(jxj)
and both satisfy
Z
1
t
1 N
0
Z
(jxj) and
(jxj)
dt < 1 and
Z
dt < 1 and
Z
1
t
p
N
s
1
1
(s)ds
0
n(jxj)
1
(jxj). It is also clear that
t
1 N
0
Z
1
t
q
N
s
1
1
dt < 1
(s)ds
0
when m and n satisfy
Z
0
1
t
1 N
Z
1
t
p
N
s
1
1
m(s)ds
0
1
t
1 N
0
Z
1
t
q
N 1
s
0
n(s)ds
1
dt < 1:
Given system (1), we de…ne the following function
H(t) = max
max (f (s)); max (g(s)) :
0 s t
0 s t
We can see that H and G are nonnegative, non-decreasing, and continuous functions
satisfying G(0) = H(0) = 0 and H(t) > 0 when t > 0. Furthermore,
G(t)
f (t)
H(t) and G(t)
g(t)
H(t):
4
Large Solutions to Elliptic Systems
Notice the function H necessarily satis…es the Keller-Osserman condition while G satis…es this requirement by hypothesis.
Inspired by [2, 3, 6, 24], we establish the following main theorems.
Z
THEOREM 1. Suppose m; n 2 C(RN ) are nonnegative functions such that
1
0
t
1 N
Z
Z
1
t
p
N
s
1
1
m(s)ds
dt < 1,
0
1
1 N
t
Z
1
t
q
N
s
1
n(s)ds
0
0
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 KellerOsserman condition (G3). Then there are in…nitely many entire positive solutions of
system (1).
REMARK 1. If N
3 and N > p, then condition (6) can be replaced by
Z 1
1
1
r p 1 m(r) p 1 dr < 1 if 1 < p 2;
0<
(A)
1
0<
0<
Z
Z
1
1
1
(p
r
rq
1
1
2)N +1
p 1
n(r) q
m(r)dr < 1 if p
1
dr < 1 if 1 < q
1
1
0<
Z
1
r
(q
2)N +1
q 1
J(r) =
Z
r
t1
N
0
In fact, if 1 < p
2, then
J(r)
C1 +
Z
(B)
2;
n(r)dr < 1 if q
1
We let
2;
(C)
2:
(D)
1
t
p
sN
1
1
dt:
(s)ds
0
Z
r
1
tp
Z
N
1
1
1=(p 1)
t
sN
1
dt:
(s)ds
0
Using the assumption N 3 in the computation of the …rst integral above and Jensen’s
inequality to estimate the last one, we obtain
Z r
Z t
3 N p
N 1
1
p
1
J(r) C2 + C3
t
s p 1 (s) p 1 dsdt:
1
1
Computing the above integral, we obtain
J(r)
C2 + C4
Z
r
tp
1
1
(t) p
1
1
dt:
1
Applying (A) in the integral above, we infer that H1 = limr!1 J(r) < 1. On the
other hand, if p 2, set
Z t
H(t) =
sN 1 (s)ds
0
Q. Li and Z. D. Yang
5
and note that either, H(t)
1
1, and hence,
Hp 1
1 for t > 0 or H(t0 ) = 1 for some t0 > 0. In the …rst case,
J(r) =
Z
r
t
1 N
p 1
H(t)
1
p
1
dt
C5 +
0
Z
r
1
N
1
tp
dt
1
so that J(r) has a …nite limit because p < N . In the second case, H(s) p
s s0 and hence,
Z t
Z r
1 N
sN 1 (s)dsdt:
tp 1
J(r) C6 +
1
1
H(s) for
0
1
Estimating and integrating by parts, we obtain
Z
p 1 1 N 1
J(r)
C6 +
t
(t)dt
N p 0
Z r
(p 2)N +1
p 1
t p 1
+
(t)dt
N p 1
Z r
(p 2)N +1
(t)dt:
C7 + C8
t p 1
r
p N
p 1
Z
r
tN
1
(t)dt
0
1
By (B), we obtain that H1 = limr!1 J(r) < 1.
Using the notation R+
[0; 1) and de…ning the set of central values
S = f(a; b) 2 R+
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 set S is a closed bounded
subset of R+ 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)) 2 E 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 by B(0; )) RN :
PROOF. Note that radial solutions of (1) are solutions of the system
(
(ju0p 2 u0 )0 + N r 1 ju0p 2 u0 = m(r)g(v);
(jv 0q
2 0 0
v) +
N
1
r
jv 0q
2 0
v = n(r)f (u):
(7)
6
Large Solutions to Elliptic Systems
Thus, solutions of (1) are simply …xed points of the operator T : C[0; 1)
C[0; 1) C[0; 1) given by
(
Rr
Rt
1
c + 0 (t1 N 0 sN 1 m(s)g(v)ds) p 1 dt for 0
Rr 1 N Rt N 1
T (u; v) = (b
u; vb) =
1
d + 0 (t
s
n(s)f (u)ds) q 1 dt for 0
0
C[0; 1) !
r< ,
r< :
To prove such a …xed point exists, we will apply a version of Schauder’s Fixed Point
Theorem. We consider the Banach space C[0; ] C[0; ]; where > 0 with norm
k(u; v)k = maxfkuk ; kvkg:
De…ne the subset
X = f(u; v) 2 C[0; ]
C[0; ] : k(u; v)
(c; d)k1
minfc; dgg:
Clearly X is closed, bounded, and convex in C[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
Z
r
t
1 N
0
Z
p
N
1
N
1
s
1
m(s)g(v(s))ds
dt
0
Z
r
t
1 N
0
1
t
p
s
1
dt
(s)H(v(s))ds
0
1
Z
(v(r))
Z
1
p
1
(Q)
H
H
1
t
p
1
r
t
1 N
0
Z
1
t
p
N
s
1
1
(s)ds
dt
0
1
p
1
(s)ds:
0
Similarly, we may show
Z
r
t
0
1 N
Z
1
t
q
N
1
s
n(s)f (u(s))ds
1
dt
Hq
1
1
(Q)
Z
1
q
1
(s)ds:
0
0
Then choose
> 0 small enough so that
Z
Z
1
1
1
p
1
p
1
q
1
max H
(Q)
(s)ds; H
(Q)
0
1
q
1
Doing so, we then have T (u(r); v(r)) = (b
u(r); vb(r)); where
c
u
b(r) = c +
Z
0
(s)ds
< minfc; dg:
0
r
1 N
t
Z
1
t
p
N
s
1
m(s)g(v(s))ds
1
dt
c + minfc; dg
0
and similarly, d
vb(r)
d + minfc; dg for all 0
r
. Thus (b
u(r); vb(r)) 2 X.
Therefore a …xed point of T , a solution to (1) in the ball B(0; ), exists.
Q. Li and Z. D. Yang
7
From [9, 22, 23], we give the following lemma
LEMMA 2. (Weak comparison principle) Let be a bounded domain in RN (N 2)
with smooth boundary @ and : (0; 1) ! (0; 1) is continuous and nondecreasing.
Let u1 ; u2 2 W 1;p ( ) satisfy
Z
Z
Z
Z
p 2
p 2
jru1 j
ru1 r dx +
(u1 ) dx
jru2 j
ru2 r dx +
(u2 ) dx;
for all non-negative
2
2 W01;p ( ) satisfy u1
u2 on @ . Then u1
u2 on
:
Proof of Theorem 1
Let T : C[0; 1) C[0; 1) ! C[0; 1) C[0; 1) be the same as in Lemma 1. We have
shown that T has a …xed point in C([0; ]). Next, we prove that T has a …xed point in
C([0; 1)). De…ne
8
i 1
R h
R
>
< c + r t1 N t sN 1 m(s)g(vk )ds p 1 dt;
0
0
(uk ; vk ) = T (uk ; vk ) =
i 1
R
R h
>
: d + r t1 N t sN 1 n(s)f (uk )ds q 1 dt:
0
0
Just as in the proof of Lemma 1, we examine (uk , vk ) over [0; 1] [0; 1] and then
we have fuk g and fvk g are each uniformly bounded on [0; 1]. The nonnegative sequences of derivatives fuk 0 g and fvk 0 g are also bounded on [0; 1] implying the sequence
has a uniformly convergent subsequence. Call this subsequence f(wk1 ; zk1 )g, and let
(wk1 ; zk1 ) ! (c
u1 ; vb1 ) as k ! 1, uniformly on [0; 1] [0; 1]. Notice that (c
u1 ; vb1 ) is a
solution of (1) in B(0; 1) with central value (a0 ; b0 ) 2 @S. Likewise, the subsequences
fwk1 g and fzk1 g 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 (wk2 ; zk2 ) ! (c
u2 ; vb2 ) as k ! 1,
uniformly on [0; 2] [0; 2]. Since f(wk2 ; zk2 )g f(wk1 ; zk1 )g, we see (c
u2 ; vb2 ) = (c
u1 ; vb1 ) on
[0; 1] [0; 1]. Continuing, we obtain a sequence f(c
uk ; vbk )g, each a solution of (1) in
B(0; k) with central value (a0 ; b0 ) 2 @S, and each solution satis…es
(c
uk ; vbk ) = (c
u1 ; vb1 )
(c
uk ; vbk ) = (c
u2 ; vb2 )
..
.
(c
uk ; vbk ) = ([
uk 1 ; v[
k 1)
Thus, (c
uk ; vbk ) converges to (u; v) where
(u(r); v(r)) = (c
uk ; vbk )
for r 2 [0; 1];
for r 2 [0; 2];
for r 2 [0; k
if 0
r
1]:
k:
This convergence is uniform on any bounded set. Thus (u; v) 2 C[0; 1)
satis…es (T u; T v) = (u; v). The proof of Theorem 1 is completed.
C[0; 1) and
8
3
Large Solutions to Elliptic Systems
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] S whenever (a; b) 2 S, we must have either [0; 1) f0g
S or f0g [0; 1) S. Let h be a positive radial solution of
4p h = (r)G( h2 )
limr !1 h(r) = 1:
0
r < 1;
(8)
(see [2] for the proof of existence). Let (u; v) be any solution which exists by hypotheses,
to the system
8
i 1
R h
R
>
< u(r) = a + r t1 N t sN 1 m(s)g(v(s))ds p 1 dt for r > 0;
0
0
(9)
i 1
R h
R
>
: v(r) = b + r t1 N t sN 1 n(s)f (u(s))ds q 1 dt for r > 0
0
0
with a > h(0) and b = 0. Without loss of generality, we assume that a 1. We now
show that h 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 that h(r) < u(r) + v(r) for all r 2 [0; ). Let
R0 = supf > 0jh(r) < u(r) + v(r) for r 2 [0; )g:
We claim that R0 = 1. Indeed, suppose that R0 < 1. Since h(r) < u(r) + v(r) in
[0; R0 ),
u(r) + v(r)
g(v(r)) G(v(r)) G(
) if v(r) u(r);
2
u(r) + v(r)
f (u(r)) G(u(r)) G(
) if u(r) v(r);
2
and elementary estimates, we observe that
h(0) +
Z
R0
t
1 N
0
<
h(0) +
Z
<
h(0) +
R0
t
1 N
<
h(0) +
R0
t
1 N
<
h(0) +
R0
t
1 N
<
h(0) +
0
h(s)
)ds
(s)G(
2
N
1
u(s) + v(s)
)ds
(s)G(
2
N
1
N
1
N
1
N
1
s
dt
1
t
s
p
1
dt
1
t
p
s
1
(s)(g(v) + f (u))ds
dt
Z
1
t
p
s
1
( (s)g(v) + (s)f (u))
dt
0
R0
t
1 N
0
Z
Z
1
1
0
0
Z
Z
p
N
0
0
Z
1
t
0
0
Z
Z
Z
1
t
p
s
(m(s)g(v) + n(s)f (u))
0
R0
t
1 N
Z
0
1
t
p
s
m(s)g(v)
1
dt
1
dt
Q. Li and Z. D. Yang
+
Z
9
R0
t
1 N
0
<
Z
1
t
q
N
1
s
Z
R0
1 N
t
0
+
0
<
dt
0
u(0) + v(0) +
Z
1
n(s)f (u)
R0
t
1 N
Z
Z
1
t
p
N
s
1
m(s)g(v)
1
dt
0
t
1
q
N
s
1
n(s)f (u)
1
dt
0
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 set S.
To show S is closed, we will show the set contains its boundary. Let h1 (r) and
h2 (r) be positive solutions of
4p h1 = (r)G(h1 ) for 0
limr !R h1 (r) = 1;
r < R;
4q h2 = (r)G(h2 ) for 0
limr !R h2 (r) = 1:
r < R;
(10)
and
(11)
where R is an arbitrary positive real number.
It is easy to show that by Lemma 2, u h1 and v h2 . Thus, u + v h1 + h2 in
[0; R]. Let h1 + h2 = . Then u + v
in [0; R]. Now, let (a0 ; b0 ) 2 @S and consider
a positive, increasing sequence fRk g such that Rk ! 1 as k ! 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 ) \ S 6= ?. For each k 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
8
i p1 1
R
R h
>
< ak0 + 0r t1 N 0t sN 1 m(s)g(vk (s))ds
dt if r 0;
(uk ; vk ) =
1
h
i
>
: bk + R r t1 N R t sN 1 n(s)f (u (s))ds q 1 dt if r 0;
k
0
0
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, fuk g and fvk g are each uniformly
bounded on [0; 1]. The nonnegative sequences of derivatives fu0k g and fvk0 g are also
bounded on [0; 1] implying the sequence has a uniformly convergent subsequence. Call
this subsequence f(wk1 ; zk1 )g, and let (wk1 ; zk1 ) ! (c
u1 ; vb1 ) as k ! 1, uniformly
on [0; 1] [0; 1]. Notice that (c
u1 ; vb1 ) is a solution to (1) in B(0; 1) with central
value (a0 ; b0 ) 2 @S. Likewise, the subsequences fwk1 g and fzk1 g 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 (wk2 ; zk2 ) ! (c
u2 ; vb2 ) uniformly on [0; 2] [0; 2] as k ! 1.
Since f(wk2 ; zk2 )g f(wk1 ; zk1 )g, we see (c
u2 ; vb2 ) = (c
u1 ; vb1 ) on [0; 1] [0; 1]. Continuing,
10
Large Solutions to Elliptic Systems
we obtain a sequence f(c
uk ; vbk )g, each a solution to (1) in B(0; k) with central value
(a0 ; b0 ) 2 @S, and each solution satis…es
(c
uk ; vbk ) = (c
u1 ; vb1 )
for r 2 [0; 1];
(c
uk ; vbk ) = (c
u2 ; vb2 )
for r 2 [0; 2];
..
.
(c
uk ; vbk ) = ([
uk 1 ; v[
k 1)
for r 2 [0; k
Thus, (c
uk ; vbk ) converges to (u; v) where
(u(r); v(r)) = (c
uk ; vbk )
for 0
r
1]:
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, implying S is closed.
4
Proof of Theorem 3
Let un and vn be de…ned as positive solutions of
8
i 1
R h
R
>
< un = u(0) + 1 + r t1 N t sN 1 m(s)g(vn (s))ds p 1 dt;
n
0
0
iq11
Rrh 1 N Rt N 1
>
1
:
dt
vn = v(0) + n + 0 t
s
n(s)f
(u
(s))ds
n
0
(12)
where (u(0); v(0)) 2 E. We denote that u0n (r) 0 and vn0 (r) 0. Also, since (u(0) +
1
1
n ; v(0) + n ) * S, for each n = 1; 2; 3; :::, there exists Rn < 1 such that
lim un (r) = 1,
lim vn (r) = 1, and R1
r !Rn
From (12) and the fact that vn0 (r)
vn (r)
C1 un (r) + C2 H q
1
:
1
t
1 N
0
Z
1
t
1 N
0
Z
1
t
q
N
s
1
n(s)ds
1
dt
0
(un (r));
1
where value C1 is any upper bound on
C2 =
R3
0, we get
1
1
v(0) + + H q 1 (un (r))
n
Z
R2
r !Rn
Z
1
v(0)+ n
1
u(0)+ n
and
1
t
q
N
s
1
1
n(s)ds
0
1
dt < 1:
Next, we de…ne h(t) = H(C1 t+C2 H q 1 (t)). It is an easy matter to show that h(0) = 0,
h(s) > 0 for s > 0, and h satis…es the Keller-Osserman condition
Z
1
1
Z
0
1
p
s
h(t)dt
ds < 1:
Q. Li and Z. D. Yang
11
De…ne
F (s) =
Z
1
dt
hp
s
;
1
1
which is well de…ned for s > 0: Notice that
1
F 0 (s) =
hp
h0 (s)
< 0 and F 00 (s) =
1
(p
1
1)h p
> 0:
p
1
We now have
div(jrun jp
2
run )
=
=
m(r)g(vn (r)) m(r)H(vn (r))
i
h
1
m(r)H C1 un (r) + C2 H q 1 (un (r))
m(r)h(un (r)):
Then, we may calculate
div(jrF (un )jp
=
0
p 1
jF (un )j
jF 0 (un )jp
1
1
=
h
1
p
1
(un )
2
rF (un ))
div(jrun jp
div(jrun jp
!p 1
2
2
1)jrun jp jF 0 (un )jp
run ) + (p
jF 0 (un )jp
run )
m(r)h(un ) =
1
2
F 00 (un )
m(r)h(un )
m(r):
Hence,
div(jrF (un )jp
2
rF (un ))
m(r):
Rewriting the Laplacian in radial form and multiplying by rN
(rN
1
jF 0 (un )jp
2
F 0 (un ))0N
1
1
, we obtain
m(r):
Integrating over [0; r]; where 0 < r < Rn ; we get
d
F (un )
dr
r
1 N
Z
1
r
N
s
1
p
1
m(s)ds
:
0
Next, we integrate over [r; Rn ]. Note that F (un (r)) ! 0 as r ! Rn . This integration
yields,
1
Z r
Z Rn
p 1
1 N
N 1
F (un (r))
t
s
m(s)ds
dt:
r
That is,
Z
F (un (r))
0
Rn
t
1 N
r
un (r)
F
Z
r
1
r
p
N
1
N
1
s
1
m(s)ds
dt:
0
Since F 0 (s) < 0 for s > 0, we have
1
Z
Rn
t
1 N
Z
0
1
r
s
p
m(s)ds
1
!
dt :
12
Large Solutions to Elliptic Systems
Now, we let n ! 1, so Rn ! R, and un ! u, we have
!
1
Z
Z
R
F
1
r
t1
N
sN
r
lim F
r !R
Z
r
p
1
dt
m(s)ds
u(r):
0
Finally, let r ! R, we have
1
1
R
t
1 N
Z
0
1
r
N
s
1
p
m(s)ds
1
!
dt
=1
lim u(r):
r !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.
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