Variational Settings

In this section, we give the variational setting for Equation (E ) following [25], and we establish the compactness conditions. Let

X = u2 D^1;2 R^N j Z

R^N

V⁺u²dx <1

be equipped with the inner product and norm hu; vi =

Z

R^N

rurv + V⁺uvdx; kuk = hu; ui¹⁼²: For > 0; we also need the following inner product and norm

hu; vi = Z

R^Nrurv + V⁺uvdx; kuk = hu; ui¹⁼²:

It is clear that kuk kuk for 1:Furthermore, by the Hardy-Sobolev inequality and 0 < V ¹

and since V⁺ b then ^V_b⁺ 1, so

We use the variational methods to …nd positive solutions of equation (E ) : Associated

Because the energy functional J ; is not bounded below on X; it is useful to consider the functional on the Nehari manifold

M _; =n

u2 Xn f0g j D

J⁰_; (u) ; uE

= 0o : and de…ne the notation of ; as following

; = inf

Note that M ; contains every non-zero solution of equation (E ) :

Next, we de…ne the Palais–Smale (simply by (PS)) sequences, (PS)–values, and (PS)–

conditions in X for J ; as follows.

De…nition 2.22 (i) For 2 R; a sequence fuⁿg is a (PS) –sequence in X for J ^; if J ; (un) = + o(1) and J⁰_; (un) = o(1) strongly in X ¹ as n ! 1:

(ii) 2 R is a (PS)–value in X for J ^; if there exists a (PS) –sequence in X for J ; : (iii) J _; satis…es the (PS) –condition in X if every (PS) –sequence in X for J ; con-tains a convergent subsequence.

We have the following results.

Lemma 2.23 The energy functional J ; is coercive and bounded below on M ; : Fur-thermore, if u0 is a solution of Equation (E ) ; then

Proof. (i)If u 2 M ^; ; then

Thus, J ; is coercive and bounded below on M ; : This completes the proof.

Remark 2.1 Proof that (21) (22) :

Proof. We let constants A; B and x = kuk ;where

A = (p 2)(N 2)² 4

(N 2)² and B = (p q) V⁺< b

q(2 p)

2 p S ^qkf⁺kL^q ;

then (21) become _2p^Ax² _pq^Bx^q and (22) become _2p¹ _pq¹ A^{2 qq}B^22q. That is we just proof that

A This completes the proof.

The Nehari manifold M ; is closely linked to the behavior of the function of the form h_u : t! J ^; (tu) for t > 0: Such maps are known as …bering maps and were introduced by Drábek and Pohozaev in [24]. They are also discussed in Brown and Zhang [20] and Brown and Wu [18, 19]. If u 2 X; we have points of hu correspond to points on the Nehari manifold.

In particular, h⁰_u(1) = 0 if and only if u 2 M ^; : Thus, it is natural to split M ;

into three parts corresponding to local minima, local maxima and points of in‡ection.

Accordingly, we de…ne

M⁺_; = fu 2 M ^; j h⁰⁰u(1) > 0g ; M⁰_; = fu 2 M ^; j h⁰⁰u(1) = 0g ; M _; = fu 2 M ^; j h⁰⁰u(1) < 0g : We now derive some basic properties of M⁺_; ; M⁰_; and M _; :

Lemma 2.24 Suppose that u0 is a local minimizer for J ; on M ; and that u0 2 M= ⁰; : Then, J⁰_; (u₀) = 0 in X ¹:

Proof. If u0 is a local minimizer of J ; on M ; then u0 is a solution of the optimization problem

minimize J ; subject to =D

J⁰_; (u) ; uE

= 0 Hence, by the Lagrange multipliers,there exists 2 R such that

J⁰_; (u₀) = ⁰ (u₀) in X ¹

This implies D

J⁰_; (u₀) ; u₀E

= h ⁰ (u₀) ; u₀i since u0 2 M ^; ; that is ku⁰k² R

R^NV u²₀dx R

R^Nfju⁰j^qdx R

R^N gju⁰j^pdx = 0:

Hence,we have h ⁰ (u₀) ; u₀i

= 2 ku⁰k² Z

R^N

V u²₀dx q Z

R^N

fju⁰j^qdx p Z

R^N

gju⁰j^pdx

= 2 Z

R^N

fju⁰j^qdx + Z

R^N

gju⁰j^pdx q Z

R^N

fju⁰j^qdx p Z

R^N

gju⁰j^pdx

= (2 q) Z

R^N

fju⁰j^qdx + (2 p) Z

R^N

gju⁰j^pdx6= 0:

Since h ⁰ (u₀) ; u₀i 6= 0; thus = 0:i.e.J⁰_; (u₀) = 0: This completes the proof.

Note: The proof is essentially the same as that in Brown and Zhang [20, Theorem 2.3] (or see Binding et al. [15]).

For each u 2 M ^; ;we have

Lemma 2.25 Suppose that the functions f; g and V satisfy the conditions (F 1) ; (G1) and (V 1) (V 4) : Then, for each ₀; we have M⁰_; =;:

Proof. Suppose the contrary. Then, there exists 0 such that M⁰_; 6= ;: Then, for u2 M⁰; (i.e.h⁰⁰_u(1) = 0);by (23) we have

Similarly, using (24) and the Hölder and Sobolev inequalities, we have (N 2)² 4

(N 2)² kuk²

Z

R^Njruj²+ V _; u²dx

= p q

2 q Z

R^N

gjuj^pdx p q

2 q g⁺

1

Z

R^N juj^pdx (p q)kg⁺k₁jfV⁺< bgj^{22 p}

S^p(2 q) kuk^p; which implies

kuk

"

S^p(2 q) (N 2)² 4

(p q) (N 2)²kg⁺k₁jfV⁺ < bgj^{22 p}

#1=(p 2)

: (27)

Hence, combine (26) and (27) implies that

V⁺ < b ¹

p

2 S (N 2)² 4 ¹⁼²

(p q)¹⁼²(N 2)

"

2 q kg⁺k₁

2 q p 2

kf⁺kL^q

p 2#1=2(p q)

which is a contradiction with (V 4). This completes the proof.

Lemma 2.26 (i) For any u2 M⁺; [ M⁰; ; we have R

R^Nfjuj^qdx > 0:

(ii) For any u2 M ; ; we have R

R^Ngjuj^pdx > 0:

Proof. (i) For any u 2 M⁺; [ M⁰; ;since M⁰_; = ; implies that u 2 M⁺; that is h⁰⁰_u(1) > 0:

By (23) ; we have

h⁰⁰_u(1) = (2 p) Z

R^Njruj²+ V _; u²dx (q p) Z

R^N

fjuj^qdx > 0

=) (2 p) Z

R^N

jruj²+ V _; u²dx > (q p) Z

R^N

fjuj^qdx then by (18) ; implies that

Z

R^N

fjuj^qdx > p 2 p q

Z

R^Njruj²+ V _; u²dx

p 2

p q

(N 2)² 4 (N 2)²

! kuk²

Hence, R then by (18) ; implies that

Z

To obtain a better understanding of the Nehari manifold and …bering maps, we consider the function mu : R⁺! R de…ned by

implies that

and clearly muis strictly increasing on (0; tmax; (u))and strictly decreasing on (tmax; (u) ;1) with lim_t!1m_u(t) = 1. Moreover, if the functions f; g and V satisfy the conditions (F 1) ; (G1) and (V 1) (V 3) ; and 0;then

Thus, we have the following lemma.

Lemma 2.27 Suppose that the functions f; g and V satisfy the conditions (F 1) ; (G1) and (V 1) (V 4) : Then, for each ₀ and u 2 Xn f0g withR

(ii) If R

R^N fjuj^qdx > 0; then there are unique

0 < t⁺ = t⁺(u) < t_max; (u) < t = t (u)

such that t⁺u 2 M⁺; , t u 2 M ; , hu is decreasing on (0; t⁺), increasing on (t⁺; t ) and decreasing on (t ; 1). Moreover,

J _; t⁺u = inf

0 t tmax; (u)

J _; (tu) ; J _; t u = sup

t t⁺

J _; (tu) : (30)

Proof. Fix u 2 Xn f0g with R

R^Ngjuj^pdx > 0:

(i) SupposeR

R^N fjuj^qdx 0:

Then, mu(t) =R

R^Nfjuj^qdx has a unique solution t > tmax; (u).

such that

m_u(t ) = Z

R^N

fjuj^qdx() t u 2 M ^; () h⁰u t = 0

=) t ^{q 1}m⁰_u t = h⁰⁰_u t < 0

and by h⁰⁰_{t u}(1) = (t )²h⁰⁰_u(t ) < 0: then we know that t u 2 M ; and h⁰_u(t ) = 0.

Hence,hu has a unique critical point at t = t and h⁰⁰_u(t ) < 0. Thus (29) holds.

(ii) SupposeR

R^N fjuj^qdx > 0.

Because mu(t_max; (u)) > R

R^Nfjuj^qdx, the equation mu(t) = R

R^Nfjuj^qdx has ex-actly two solutions t⁺ < t_max; (u) < t

CaseI:

For t⁺2 (0; t^max; (u)) and mu(t) is strictly increasing on (0; tmax; (u)),that is m_u(t⁺) =

Z

R^N

fjuj^qdx() t⁺u2 M ^; () h⁰u t⁺ = 0

=) t^{+ q 1}m⁰_u t⁺ = h⁰⁰_u t⁺ > 0

and by h⁰⁰_t+u(1) = (t⁺)²h⁰⁰(t⁺) > 0: then we know that t⁺u2 M⁺; : CaseII:

For t 2 (t^max; (u) ;1) and m^u(t)is strictly decreasing on (tmax; (u) ;1),that is m_u(t ) =

Z

R^N

fjuj^qdx() t u 2 M ^; () h⁰u t = 0

=) t ^{q 1}m⁰_u t = h⁰⁰_u t < 0

thus, t u 2 M ; and by h⁰⁰_{t u}(1) = (t )²h⁰⁰_u(t ) < 0: then we know that t u 2 M ; : Hence, there has critical points at t = t⁺and t = t with h⁰⁰_u(t⁺) > 0and h⁰⁰_u(t ) < 0:

Thus, hu is decreasing on (0; t⁺) ; increasing on (t⁺; t ) and decreasing on (t⁺;1) : Therefore, (30) must hold. This completes the proof.

We remark that it follows from Lemma 2.25 that M _; = M⁺_; [ M ;

for all > ₀. Furthermore, by Lemma 2.27, it follows that M⁺_; and M _; are non-empty and, by Lemma 2.23, we may de…ne

+

Moreover, we have the following result.

Theorem 2.28 Suppose that the functions f; g and V satisfy the conditions (F 1) ; (G1) and (V 1) (V 4) : Then, for each ₀; there exists bC₀ > 0 such that ⁺_; < 0 < bC₀ <

then by (24) u 2 M ; (i.e. h⁰⁰_u(1) < 0) ;we have

and by(18) and (33) and the Sobolev inequality, (2 q) (N 2)² 4 Then we substitute (34) into (32) ; we have

J _; (u)

Thus, using the condition (V 4) and if 0;then

; > bC₀ for some bC₀ > 0: This completes the proof.

Lemma 2.29 For each u 2 M ^; , there exists > 0 and a di¤erentiable function : B (0; ") Xn f0g ! R such that (0) = 1; the function (v) (u v)2 M ^; and

h ⁰(0) ; vi = 2 R

R^Nrurv + V ^; uvdx qR

R^Nfjuj^{q 2}uvdx pR

R^Ngjuj^{p 2}uvdx (2 q) R

R^Njruj²+ V _; u²dx + (q p)R

R^Ngjuj^pdx ; (35) for all v 2 Xn f0g :

Proof. For u 2 M ^; ;de…ned a function F : R Xn f0g ! R by F_u( ; w)

= J⁰_; ( (u w)) ; (u w)

= ²

Z

R^Njr (u w)j²+ V _; (u w)²dx ^q Z

R^N

fju wj^qdx ^p Z

R^N

gju wj^pdx

Then Fu(1; 0) = J⁰_; (u) ; u = 0 and d

d F_u(1; 0) = 2 Z

R^N

jruj² + V _; u²dx q Z

R^N

fjuj^qdx p Z

R^N

gjuj^pdx

= (2 q) Z

R^N

jruj² + V _; u²dx + (q p) Z

R^N

gjuj^pdx6= 0 and using L⁰Hôpital’s rule

limt!0

R

R^Nfju (w + tv)j^qdx R

R^Nfju wj^qdx t

= lim

t!0q Z

R^N

fju (w + tv)j^{q 2}[u (w + tv)] ( v) dx

= q Z

R^N

fju wj^{q 2}(u w) vdx; (36)

Similarly,

limt!0

R

R^Ngju (w + tv)j^pdx R

R^Ngju wj^pdx t

= p Z

R^N

gju wj^{p 2}(u w) vdx; (37)

then using Gâteaux derivative, (36) and (37)

According to the implicit function theorem , there exists > 0 and a di¤erentiable function : B (0; ") Xn f0g ! R such that (0) = 1; which is equivalent to

J⁰_; ( (v) (u v)) ; (v) (u v) = 0 for all v 2 B (0; ") ; that is, (v) (u v)2 M ^; :

Lemma 2.30 For each u 2 M ; ;there exist > 0 and a di¤erential function : B (0; ") Xn f0g ! R such that (0) = 1; the function (v) (u v)2 M ; and

Proof. Similar to the argument in Lemma 2.29, there exist > 0 and a di¤erential function : B (0; ") Xn f0g ! R such that (0) = 1; the function (v) (u v)2 M _; for all v 2 Xn f0g. Since u 2 M ; means that h⁰⁰_u(1) < 0;that is

h⁰⁰_u(1) =kuk² Z

R^N

V u²dx (q 1) Z

R^N

fjuj^qdx (p 1) Z

R^N

gjuj^pdx < 0:

Thus, by the continuity of the function , then we have h⁰⁰ _{(v)(u v)}(1) = (v) (u v) ²

Z

R^N

V (v) (u v) ²dx (q 1)

Z

R^N

f (v) (u v) ^qdx (p 1) Z

R^N

g (v) (u v) ^pdx

< 0

if su¢ ciently small,this implies that (v) (u v)2 M ; : Proposition 2.31 For 0; then

(i) there exists a minizing sequence fuⁿg M ; such that J _; (u_n) = ⁺_; + o (1) ; J⁰_; (u_n) = o (1) in X ¹;

(ii) there exists a minizing sequence fuⁿg M _; such that J _; (u_n) = _; + o (1) ; J⁰_; (u_n) = o (1) in X ¹:

Proof. (i) By Lemma 2.23 and Eklend variational principle.There exists a minizing sequence fuⁿg M _; such that

J _; (u_n) < ⁺_; + 1

n (39)

and

J _; (u_n) < J _; (w) + 1

n kw u_nk for each w 2 M ^; (40) By taking n large, from Theorem 2.28 , we have

J _; (u_n) = p 2 2p

Z

R^N

jruⁿj²+ V _; u²_ndx p q pq

Z

R^N

fjuⁿj^qdx

< ⁺_; + 1 n < ^;

2 < 0: (41)

This implies Consequently, un6= 0 and by(42) we have

kuk > pq

and putting together (41) (42) and (18), we obtain

p 2 Now we will show that

J⁰_; (u_n)

and by mean value theorem , we have D

then (45) become to D

J⁰_; (u_n) ; w E

+ ( _n(w ) 1)D

J⁰_; (u_n) ; (u_n w )E 1

n u_n + o u_n (46)

From _n(w ) (u_n w )2 M ^; we have

J⁰_; ( _n(w ) (u_n w )) ; _n(w ) (u_n w ) = 0

implies that D

J⁰_; ; (u_n w )E

= 0 substitute into (46) ; it follows that

J⁰_; (u_n) ; u

kuk + ( _n(w ) 1)D

J⁰_; (u_n) J⁰_; ; (u_n w )E 1

n u_n + o u_n :

Thus,

J⁰_; (u_n) ; u kuk

un

n + o un

+( _n(w ) 1) D

J⁰_; (u_n) J⁰_; ; (u_n w )E

; (47) since

un = k n(w ) (un w ) unk

= k n(w ) (u_n) _n(w ) w u_nk

= kuⁿ[ _n(w ) 1] _n(w ) w k j n(w ) 1j kuⁿk + j n(w )j ; and

lim!0

j n(w ) 1j = lim

!0

j n(w ) _n(0)j lim

!0

j n(0 + ) _n(0)j = ⁰_n(0) :

If we let ! 0 in (47) for a …xed n, then by (44) we can …nd a constant C > 0; which is independent of ; such that

J⁰_; (u_n) ; u kuk

C

n 1 + ⁰_n(0) :

We are done once we show that ⁰_n(0) is uniformly bounded in n. By(35) ; (44) and Hölder inequality, we have

h ⁰(0) ; vi bkvk

(2 q) R

R^Njruj²+ V _; u²dx + (q p)R

R^Ngjuj^pdx for some b > 0:

We only need to show that (2 q) for some c > 0 and n large.We argue by way of contradiction.Assume that there exists a subsequence fuⁿg such that

(2 q) Combing (49) with (43) ;we can …nd a suitable constant d > 0 such that

Z

Let I ; : M _; ! R and K (p; q) = ^{2 q}_{p q}

However,by (50)and (52). We get

I _; (u_n) = K (p; q)

J⁰_; (u_n) ; u kuk

C

n 1 + ⁰_n(0) :

We have that it contradicts to (53). Consequently,this show that fuⁿg is a (PS) ⁺ -sequence for J ; .

(ii) Similarly, we can use Lemma 2.30 to prove (ii). We will omit deltailed proof here.

3 Proof of Theorem 1.1

Theorem 3.1 Suppose that 2 < p < 2 and the functions f; g and V satisfy the con-ditions (F 1) ; (G1) and (V 1) (V 5) : Then for each ₀; the functional J ; has a minimizer u⁺_; 2 M⁺; M _; and it satis…es

(i) J _; u⁺ = ⁺ = inf_{u2M ;} J _; (u) < 0;

(ii) u⁺ is a positive solution of equation (E ; ) :

Proof. By Theorem 2.28 and the Ekeland variational principle [26] (or see Wu [40, Proposition 1]), there exists fuⁿg M⁺_; such that it is a (PS) ⁺–sequence for J ; . Moreover, Lemma 2.23, fuⁿg is bounded in X : Therefore, there exist a subsequence fuⁿg and u⁺ in X such that

u_n * u⁺ weakly in X ;

u_n ! u⁺ strongly in L^r_loc R^N ; for 2 r < 2 :

Moreover, J⁰_; u⁺ = 0: First, we claim that u⁺ 6 0: Suppose the contrary, then by (18) ; the condition (F 1) ; the Egorov theorem and the Hölder inequality, we have

Z

R^N

fjuⁿj^qdx! 0 as n ! 1;

which implies that Z

R^Njruⁿj²+ V _; u²_ndx = Z

R^N

gjuⁿj^pdx + o (1) and

J _; (u_n) = 1 2

Z

R^N

jruⁿj²+ V _; u²_ndx 1 q

Z

R^N

fjuⁿj^qdx 1 p

Z

R^N

gjuⁿj^pdx (p 2) (N 2)² 4

2p (N 2)² kuⁿk² + o (1) ;

this contradicts lim_n!1J _; (u_n) = _; < 0: Thus, by Lemma 2.26 (i) ;R

R^N f u^{+ q} > 0:

In particular, u⁺ is a nontrivial solution of Equation (E ) : Now we prove that un! u⁺ strongly in X : Suppose the contrary, then by (18) ;

Z

Proposition 3.2 Suppose that 2 < p < 2 and the functions f; g and V satisfy the conditions (F 1) ; (G1) and (V 1) (V 2) : Then for each D > 0 there exists 0 = (D) > 0 such that J ; satis…es the (PS) –condition in X for all < D and > ₀:

Proof. Let fuⁿg be a (P S) -sequence with < D: By Lemma 2.23, there exists C such that kuⁿk C : Thus, there exist a subsequence fuⁿg and u⁰ in X such that

u_n * u₀ weakly in X ;

u_n ! u⁰ strongly in L^r_loc R^N for 2 r < 2 :

Moreover, J ; (u₀) = 0: Thus, by the condition (F 1), Hölder inequality and Egorov’s theorem, then

Hence, we have Z

R^N Then, by the Hölder and Sobolev inequalities, we have

Z

and by (54) ; we let hn= fjuⁿ u₀j^q (a) hn ! 0;

(b) let _n:= 2^{q 1}f (ju⁰j^q juⁿj^q)! := 2^pfju⁰j^q; we have _n! ; (c) hn n :

Since (a) to (c) and by Generalized Lebesgue Dominated Convergence Theorem,we have Z

jhⁿ 0j ! 0 as n ! 1;

that implies Z

R^N

fjvⁿj^qdx! 0:

By the conditions (F 1) ; (G1) and Brezis-Lieb Lemma, we have

n!1lim Z

juⁿj^p+ Z

juⁿ u₀j^p = Z

ju⁰j^p; implies

J _; (u₀) = lim

n!1[J _; (u_n) J _; (v_n)] : Hence,

J _; (v_n) = J _; (u_n) J _; (u₀) + o (1) and J ; (v_n) = o (1) : Consequently, by this together with (54) and Lemma 2.23, we obtain

1 2

1 p

Z

R^N

gjvⁿj^p = 1 2

1 p

Z

R^N

gjvⁿj^pdx 1 q

1 2

Z

R^N

fjvⁿj^qdx + o (1)

= J _; (v_n) 1 2

D

J⁰_; (v_n) ; v_nE

+ o (1)

= J _; (u₀) + o (1) D + K₀+ o (1) ; where

K₀ = 2 q 2pq

(N 2)²

(N 2)² 4 (p 2)

!q=(2 q)0

@(p q)jfV⁺ < bgj

q(^{2 p})

2 p kf⁺kL^q

S^q

1 A

2=(2 q)

and so Z

R^N

gjvⁿj^p 2p

p 2(D + K₀) + o (1) : (56)

Because J⁰_; (v_n) ; v_n = o (1) ; it follows from (18) ; (55) and (56) that

Note that if the functions f; g and V satisfy the conditions as in Theorem 1.1, then by Lemma 2.27, we may choose 2 C0¹( _g) ;which the function

this implies D0 for 0: Moreover, we have the following result.

Theorem 3.3 Suppose that the functions f; g and V satisfy the conditions as in Theo-rem 1.1. Then for each > ₀; the functional J ; has a minimizer u _; in M _; and it satis…es

(i) J _; u _; = _; ;

(ii) u _; is a positive solution of equation (E ; ) :

Proof. By the Ekeland variational principle [26] (or see Wu [40, Proposition 1]), there exists fuⁿg M _; such that it is a (PS) –sequence for J ; . Then, by Proposition 3.2, there exists a subsequence fuⁿg and u 2 M ; is a non-zero solution of equation (E ) ; such that un ! u strongly in X and J ^; u = : Because J ; u = J _; u and u 2 M ; ; by Lemma 2.24 we may assume that u is a positive solution of equation (E ) :

We can now complete the proof of Theorem 1.1: by Theorems 2.28, 3.1 and 3.3, Equation (E ) has two positive solutions, u⁺_; ; u _; ; such that u⁺_; 2 M⁺; and u _; 2 M ; with

J _; u⁺_; = ⁺_; < 0 < bC₀ < J _; u _; = _; : This completes the proof of Theorem 1.1.

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在文檔中 奇異橢圓方程正解的多解性 (頁 26-55)