|H (x, st)| ¤ (|V1 (x)s|q1 + W1 (x))r1 (t), s ∈ R, t ≥ 1.

(8.70)

(d) There is a ψ ∈ such that

ψ(t) ¤ tr0 (t), tψ(t) ¤ r1 (t), t ≥ 1,

(8.71)

and

ri (t) ’ ri as t ’ ∞, i = 0, 1,

(8.72)

with

r0 < ∞

(8.73)

and

r1 ¤ ∞.

(8.74)

II. There are positive constants ±, β such that

2F(x, t) ¤ »0 (1 ’ ±)t 2 , t 2 < β.

(8.75)

III. (a) If r0 = 0, it is assumed that there is a measurable function γ (x, a) on —R

such that

f (x, st)/tr0 (t) ’ γ (x, a) a.e. as s ’ a, t ’ ∞.

(8.76)

(b) If r0 = 0 and r1 = 0, it is assumed that there is a measurable function (x, a)

on — R such that

H (x, st)/r1(t) ’ (x, a) a.e. as s ’ a, t ’ ∞.

(8.77)

If r1 = ∞, then (8.77) is required to hold only for a = 0.

IV. If r0 = 0 and r1 = 0, then we assume that there does not exist a function

u ∈ D\{0} and b > 0 such that

1

(Au, u) = 1,

(8.78) a(u) :=

2

’1

r0 Au = γ (x, u) a.e.,

(8.79)

(x, u) d x = ’b/r1 .

(8.80)

V. If r0 = 0 and r1 = ∞, we assume that all solutions u ≡ 0 of (8.79) are nonzero

a.e. We have the following.

8.5. Further applications 77

Theorem 8.14. Under hypotheses I to V, there is a solution of

Au = f (x, u)

(8.81)

in D. If there is a u 0 ∈ D\{0} such that

G(u 0 ) ¤ 0,

(8.82)

where

G(u) = a(u) ’

(8.83) F(x, u) d x,

then (8.81) has a nonzero solution.

Proof. Under hypothesis I(b), it is easily checked that G(u) given by (8.83) is continu-

ously Fr´ chet differentiable with

e

(G (u), v) = 2a(u, v) ’ f (x, u)vd x, u, v ∈ D.

(8.84)

Also by hypothesis I(b),

2 2 2

+ V0 u + Wu ¤ Ca(u), u ∈ D.

(8.85) Vu q q q

Thus, by (8.75),

1

G(u) ≥ ±a(u) + (1 ’ ±)a(u) ’ »0 (1 ’ ±) u 2

(8.86)

2

|V0 u|(|V u|q’1 + β (1’q)/2|W u|q’1 ) d x

’C

u 2 >β

≥ a(u)(±C1 a(u)(q/2)’1).

Hence, there are positive constants ρ, δ such that

a(u) = δ 2 .

G(u) ≥ ρ,

(8.87)

We consider two cases.

Case 1. G(u) ≥ 0 for all u ∈ D. In this case 0 is a minimum point and we must have

G (0) = 0. By (8.84), we see that 0 is a solution of (8.81).

Case 2. There is a u 0 ∈ D\{0} such that (8.82) holds. In this case the hypotheses of

Theorem 2.18 are satis¬ed. In particular, there is a sequence {u k } ‚ D such that

G(u k ) ’ b

(8.88)

and

G (u k )/ψ(rk ) ’ 0,

(8.89)

78 8. Semilinear Problems

where b is given by (8.80), tk = a(u k ), and ψ(t) ∈

2 is the function satisfying (8.71).

We shall show that this sequence satis¬es

a(u k ) ¤ C.

(8.90)

If so, we can ¬nd a renamed subsequence that converges weakly in D to a function u

and such that V0 u k ’ V0 u in L q ( ) and a.e. in while V u k ’ V u in L q’1 (K ) for

each compact subset K of . By (8.84) and (8.89),

2a(u k , v) ’ ( f (x, u k ), v) ’ 0, v ∈ D.

(8.91)

∞

If v ∈ C0 ( ), then, by hypothesis I(b),

| f (x, u k )v| ¤ C(|V (x)u k | + W (x)q’1 )|V0 v|,

and the right-hand side converges in L 1 ( ) to

C(|V u|q’1 + W q’1 )|V0 u|.

Thus,

∞

( f (x, u k ), v) ’ ( f (x, u), v), v ∈ C0 ( ).

(8.92)

Since u k ’ u weakly in D, we have

∞

2a(u, v) = ( f (·, u), v), v ∈ C0 ( ),

(8.93)

which shows that u is a solution of (8.81).

I claim that u = 0. To see this, note that b > 0 by (8.87). Moreover, by (8.70),

|H (x, u k )| ¤ C(|V1 u k |q1 + W1 ),