G(u k )/ρk = u k

˜ ’2 F(x, u k )d x/ρk ’ 0.

2 2 2

D

Since

|F(x, u k )|/ρk ¤ C(|u k (x)|2 + W3 (x)/ρk ),

2 2

W3 ∈ L1 ( ),

˜

(8.44)

by (8.30), and the right-hand side of (8.44) converges to C|u(x)|2 in L 1 ( ) and

˜

2F(x, u k (x))/ρk ’ ±+ (x)(u + )2 + ±’ (x)(u ’ )2

˜ ˜

2

(8.45) a.e.,

we see that the convergence in (8.45) is not only pointwise a.e., but also in L 1 ( ).

Since u k D = 1, (8.44) implies

˜

{±+ (u + )2 + ±’ (u ’ )2 }d x = 1.

˜ ˜

(8.46)

Also,

(G (u k ), v)/ρk = 2(u k , v) D ’ 2( f (u k ), v)/ρk ’ 0

˜

for each v ∈ D. This implies

(u, v) D = (±+ u + ’ ±’ u ’ , v),

˜˜ ˜ ˜ v ∈ D.

Consequently, u is a solution of (8.34). By hypothesis, u ≡ 0. But this contradicts

˜ ˜

(8.46). Hence, ρk ¤ C. The theorem now follows from Theorem 3.4.1 of [122].

8.4 Unbounded domains

‚ Rn to be unbounded. Assume hypothesis (A), and

Now we allow the domain

assume that

H (x, t) = 2F(x, t) ’ t f (x, t) ≥ ’W3 (x) ∈ L 1 ( ), x∈ , t ∈ R,

(8.47)

72 8. Semilinear Problems

and

H (x, t) ’ ∞ a.e. as |t| ’ ∞.

(8.48)

We have

Theorem 8.7. Assume that the spectrum of A consists of isolated eigenvalues of ¬nite

multiplicity

0 < »0 < »1 < · · · < »k < · · · ,

(8.49)

and let be a nonnegative integer. Take N to be the subspace of D spanned by the

eigenspaces of A corresponding to the eigenvalues »0 , »1 , . . . , » . We take M = N ⊥ ©

D. Assume that there are functions W1 , W2 ∈ L 1 ( ) and numbers a1 , a2 such that

± < a1 ¤ a2 and

(8.50) a1 (t ’ )2 + γ (a1 )(t + )2 ’ W1 (x) ¤ 2F(x, t)

¤ a2 (t ’ )2 + (a2 )(t + )2 + W2 (x), x∈ , t ∈ R,

and that (8.47) and (8.48) hold. Then (8.13) has at least one solution.

Proof. First, we note that

sup G ¤ B1 , inf G ≥ ’B2 , Bj = W j (x)d x.

(8.51)

M

N

To see this, note that by (8.28) we have

¤ a1 v ’ + γ (a1 ) v + 2 ,

v v ∈ N.

2 2

(8.52) D

By (8.29) we have

a2 w ’ (a2 ) w+

+ ¤w D, w ∈ M.

2 2 2

(8.53)

Hence,

G(v) ¤ B1 , v ∈ N,

and

G(w) ≥ ’B2 , w ∈ M,

by (8.50). By Theorem 3.19, we conclude that for any sequence Rk ’ ∞, there is a

sequence {u k } ‚ D such that

B1 + B2

G(u k ) ’ c, ’B2 ¤ c ¤ B1 , (Rk + u k D) G (u k ) ¤ .

(8.54)

ln(4/3)

In particular, we have

’2 F(x, u k ) d x ’ c

2

(8.55) uk D

8.4. Unbounded domains 73

and

’ ( f (·, x k ), u k ) ¤ K .

2

(8.56) uk D

Consequently,

H (x, u k ) d x ¤ K .

(8.57)

If ρk = u k D ’ ∞, let u k = u k /ρk . Then u k D = 1. Consequently, there is a

˜ ˜

renamed subsequence such that u k ’ u weakly in D, strongly in L 2 ( ), and a.e. in

˜ ˜

. In view of (8.55), we have

1’2 F(x, u k )/ρk d x ’ 0.

2

(8.58)

But by (8.50), we have

2F(x, u k )/ρk ¤ a2 (u ’ )2 + (a2 )(u + )2 + W2 (x)/ρk .

˜k ˜k

2 2

(8.59)

In the limit this implies

1 ¤ a2 u ’ (a2 ) u + 2 .

˜ + ˜

2