Existence and Uniqueness of Fast Decay Entire Solutions of Quasilinear Elliptic Equations

(4.12)

or

Now we distinguish two cases. Case (i). There is a subsequence of [! 2k&1 ] such that (4.11) holds. For simplicity, denote the subsequence again by [! 2k&1 ]. By (2.2) and the fact that u 1 =u 2 at t=! 2k&1 we obtain m n (n&m) u$1 |u$1 | m&2 u 1 ! n&1 2k&1 +(m&1) |u$1 | ! 2k&1 m n >(n&m) u$2 |u$2 | m&2 u 2 ! n&1 2k&1 +(m&1) |u$2 | ! 2k&1 ,

UNIQUENESS OF ENTIRE SOLUTIONS

171

leading to (n&m) u 1(u$1 |u$1 | m&2 &u$2 |u$2 | m&2 )+(m&1)(|u$1 | m & |u$2 | m ) ! 2k&1 >0. Observe that (4.8) 1 implies u$1 |u$1 | m&2 &u$2 |u$2 | m&2 >0

at

t=! 2k&1 ;

thus (n&m) u 1 +(m&1) ! 2k&1 }

|u$2 | m & |u$1 | m >0. u$2 |u$2 | m&2 &u$1 |u$1 | m&2

Writing h k = |u$2(! 2k&1 )||u$1(! 2k&1 )| >1, we then obtain (n&m) u 1 +(m&1) ! 2k&1 u$1 }

hm k &1 >0. h k m&1 &1

Observe that m h k m &1 > , m&1 hk &1 m&1 since the function h(x)=(x m &1)(x m&1 &1), m>1, is strictly increasing for x>1 and h(x) a m(m&1) as x a 1; we finally get (n&m) u 1 +m! 2k&1 u$1 >0, or equivalently, u 1 +m! (n&1)(m&1) u$1 >0; (n&m) ! (n&m)(m&1) 2k&1 2k&1 yielding lim sup ((n&m) v 1 +mw 1 )0. tÄ

However, applying (4.2)(4.3) directly, we have lim ((n&m) v 1 +mw 1 )=(n&m) c 1 &

tÄ

a contradiction.

m(n&m) n&m } c 1 =& } c 1 <0, m&1 m&1

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MOXUN TANG

Case (ii). There is a subsequence of [! 2k ] such that (4.12) holds. For simplicity of notation, denote the subsequence again by [! 2k ]. By interchanging u 1 and u 2 , and replacing ! 2k&1 with ! 2k , we can then use exactly the same argument as in Case 1 to obtain (n&m) u 2 +m! 2k u$2 >0, which again yields a contradiction. Now Step 2 is complete. Step 3. u 1 and u 2 intersect more than once in (0, ). If this is not true, then by Step 1 u 1 and u 2 intersect exactly once in (0, ), that is, r(u)=t 1(u)&t 2(u) has exactly one zero, say, u=u I , in (0, : 1 ). Clearly r>0

r(u I )=0,

on (0, u I ),

and

r<0 on (u I , : 1 ).

(4.13)

Applying the maximum principle of Peletier and Serrin and noticing that r$(u)<0 for u slightly smaller than : 1 , one sees that either r$<0

on

(0, : 1 ),

(4.14)

or there exists some u c # (0, u I ) such that r$>0

r$(u c )=0,

on (0, u c ),

and

r$<0 on (u c , : 1 ).

(4.15)

Next we show that both (4.14) and (4.15) are impossible. Suppose for contradiction that (4.14) holds. Then from Lemma 3.2 follows S(u)>S(')

for 0
S(u)
for '
Now using (3.3) and (3.8) leads to 9(') =

|

'

n8(u)[Q 1(u) Q 2(')&Q 2(u) Q 1(')] du :1

|

'

n8(u)[S(u)&S(')] Q 2(u) Q 2(') du0,

(4.17)

:1

since for '
|

'

n8(u)[S(u)&S(')] Q 2(u) Q 2(') du<0, 0

since for 0S(') by (4.16 1 ). Thus we get a desired contradiction.

UNIQUENESS OF ENTIRE SOLUTIONS

173

Now suppose for contradiction that (4.15) holds. Then r(u c )>0 since u c # (0, u I ) on which r>0 by (4.13 1 ); consequently from Lemma 3.3 follows 9(u c )<0.

(4.18)

Also by (4.15) and Lemma 3.2 there holds S(u)
0
for

(4.19)

Now if u c >', then using exactly the same argument as for (4.17) we obtain 9(u c )0, contradicting (4.18). While if u c ', then by (3.9) 9(u c )=

|

uc

n8(u)[S(u)&S(u c )] Q 2(u) Q 2(u c ) du>0,

0

since for 0
r$(u)<0

and

for

u c
(4.20)

Applying Lemma 3.2 we then have S(u)
for

u c
Also from (H2) follows 8(u)0 for u c '. Now using (3.8) we find 9(u c )

|

uc

n8(u)[S(u)&S(u c )] Q 2(u) Q 2(u c ) du0;

:1

together with (4.20) and Lemma 3.3, yielding 9(u c )>0 and r(u c )<0. But then using (3.2) we get r"(u c )>0, contradicting (4.20). Thus r(u) has no critical numbers in (', : 1 ). Now all possible critical numbers of r are in (0, '), and we let u c be the smallest one; thus either r$(u c )=0

and

r$(u)<0

for

0
(4.21)

174

MOXUN TANG

or r$(u c )=0

and

r$(u)>0

for

0
(4.21$)

If (4.21) holds, then by Lemma 3.2 S(u)>S(u c )

for

0
Combining this with (3.9) and noting that 8(u)0 on (0, u c ) by (H2), we obtain 9(u c )=

|

uc

n8(u)[S(u)&S(u c )] Q 2(u) Q 2(u c ) du0;

0

thus 9(u c )<0 and r(u c )>0 by Lemma 3.3. But using (3.2) again we get r"(u c )<0, contradicting (4.21). Thus (4.21) is impossible. By a similar argument (4.21$) can be ruled out and so r has no critical points at all in (0, : 1 ), leading to a contradiction. Thus Step 4 is established; and so Theorem 4.1 is proved. K

5. EXISTENCE OF CROSSING SOLUTIONS In this section, we shall follow the approach of [GST] to prove Theorem 5.1. Let (H1) and (H3) hold. Then problem (1.2) has at least one crossing solution with 00, which makes their argument much more delicate. The following lemma is similar to Lemma 3.1 of [GST]; the statement here is simpler since F(u)>0 for 00,

Let (H1) hold. If u is an entire solution of (1.2)(1.3), then

t<

Cu , (F(u)) 1m

C=(n&1)

\

m m&1

+

(m&1)m

.

(5.1)

Proof. Let u be an entire solution of (1.2)(1.3), and E(t) be its energy defined in (2.3); clearly lim t Ä  E(t)=0. For any fixed T>0, let U=u(T )

175

UNIQUENESS OF ENTIRE SOLUTIONS

and M=sup [T, ) |u$(t)| = |u$(T 1 )|, where T 1 # [T, ); therefore formula (2.3) at t=T together with (2.4) integrated on (T, ) gives F(U )


|

T

(n&1)

M

m&1

T



|

|u$(t)| m dt t |u$(t)| dt=(n&1) M m&1

T

U . T

(5.2)

Similarly, (2.3) at t=T 1 together with (2.4) integrated on (T 1 , ) yields m&1 m M
|



T1

|u$(t)| m U dt(n&1) M m&1 , t T

and so M<

m(n&1) U ; m&1 T

together with (5.2) this leads to F(U )<(n&1)

U m(n&1) U T m&1 T

\

Now (5.1) readily follows.

+

m&1

=(n&1) m

\

m m&1

m&1

U T

m

+ \ +.

K

Remark. By (5.1), it is evident that if (1.2) has an entire solution, then necessarily lim inf F(u)u m =0. ua0

Proof of Theorem 5.1. If #<, then u(t, :) Ä # as : A # uniformly on every bounded subset of [0, ) (see Section 2.2 of [FLS] for a proof ). Now, if :<# and sufficiently close to #, then the unique value t satisfying u(t, :)=#2 can be sufficiently large; while the right hand side of the inequality (5.1) with u=#2 is a finite constant independent of :. Thus (5.1) is violated and u must be a crossing solution. Now we assume that #= and (H2$) holds, and suppose for contradiction that u(t, :) is an entire solution for any :>0. Choose :>+*, and define t * by u(t * )=*:. Writing Eq. (1.2) in the form (t n&1u$ |u$| m&2 )$=&t n&1f (u),

176

MOXUN TANG

and integrating this identity over [0, t] leads to, for any t # (0, t * ), u$(t) &

\

f (* 2 :) n

+

1(m&1)

t 1(m&1),

where f (* 2 :)=max u # [*:, :] f (u), ** 2 1. Integrating this over [0, t * ] now yields t * c *

\

: m&1 f (* 2 :)

+

1m

,

_

c * =n 1m (1&*)

where

m m&1

&

(m&1)m

.

(5.3)

Define t + by u(t + )=+. Then t + >t * for *:>+, and by (5.1) t +
C+ . (F(+)) 1m

(5.4)

Define 8 + =min 0&. By (H2$) we have 8(* 1 :) 8(u(t)) 0 & |8 + |

for 0
{

for

(5.5)

t>t + .

Now applying the identities (2.2) and (2.5) together with (5.3) and (5.5) yields, for t>t + , t nE(t)>P(t)=

\|

t* 0

+

t+

| |

t

+

t*

t+

+ n{

n&1

8(u({)) d{

8(* 1 :) t n* & |8 + | t n 8(* 1 :) c n*(: m&1f (* 2 :)) nm & |8 + | t n. Thus for any t +
\

n

+ 8(* :)(: 1

m&1

 f (* 2 :)) nm & |8 + | &F(+)

by (5.4). Applying condition (1.9) we can certainly make the right hand side larger than + m if : is sufficiently large, resulting |u$(t)| >+ for t +
6. PROOF OF THEOREMS 1 AND 2 Proof of Theorem 1. The first part of Theorem 1 follows directly from Lemma 2.4 and Theorem 4.1.

177

UNIQUENESS OF ENTIRE SOLUTIONS

Now assume that (H1), (H2), and (H2$) hold, and define :* to be the supremum of : such that (0, :)/I +. Then by Lemma 2.4 :*>'; by Theorem 5.1 :*<#; and finally from Lemma 2.4 it follows that u(t, :*) must be a fast decay solution. Finally, if :*<:<#, then u must be a crossing solution by Lemma 2.4 and Theorem 4.1 again. Thus the second part of Theorem 1 follows. K Proof of Theorem 2. Let f (u)=u p &u q, and _
u 2p } g(z), nmf 2(u)

_

g(z)=q*z 2 & p*+q*+

nm( p&q) 2 z+ p*, ( p+1)(q+1)

&

where z=u q& p, p*=n&m&

nm >0 p+1

q*=n&m&

and

nm >0. q+1

Since g(0)= p*>0,

g(1)=&

nm( p&q) 2 <0, ( p+1)(q+1)

there exists a number z ' # (0, 1) such that g(z)>0

in (0, z ' ),

and

g(z)<0 in (z ' , 1).

p) , and so the second part of It follows that (H2) is fulfilled with '=z 1(q& ' Theorem 1 (or the first part of Theorem 2) is valid. To obtain a better lower bound for :*, we compute

8(u)=

1 (q*u q+1 & p*u p+1 ). nm

Thus 8(u)<0 for 0'* in view of the last paragraph of the proof of Lemma 2.4. Now, any function u with :*
178

MOXUN TANG

solution when u(0)=:*, we find that lim sup : a :* b(:)=lim sup : A 1 b(:) =. Let R*=inf :*<:<1 b(:). Then 0
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[E]

[ET1] [ET2] [ET3] [ET4] [FLS] [GST]

[GMS]

[GNN]

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