C(|V1 u k |q1 + W1 ).

Since

H (x, u k ) ’ H (x, u) a.e.,

we have

H (x, u k ) d x ’ H (x, u) d x.

But for any u ∈ D,

1

(G (u), u) = G(u) + H (x, u) d x.

(8.94)

2

Hence, (8.88) and (8.89) imply that

H (x, u) d x = ’b = 0.

Thus, u = 0 since H (x, 0) ≡ 0.

8.5. Further applications 79

It therefore remains only to prove (8.90). Assume that tk ’ ∞, and let u k = u k /tk .

˜

Then

a(u k ) = 1

˜

(8.95)

and there is a renamed subsequence {u k } 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 (8.69),

| f (x, u k )|/tk r0 (tk ) ¤ (|V u k |q’1 + W q’1 )|V0 u k |.

˜ ˜

Thus,

q’1 q’1

f (x, u k ) d x /tk r0 (tk ) ¤ ( V u k

˜ +W q) ˜ q.

(8.96) V0 u k

q

Moreover, by (8.71) and (8.89),

2a(u k )/r0 (tk ) ’ ( f (x, u k )/tk r0 (tk ), u k ) ’ 0.

˜ ˜

(8.97)

This shows that u(x) ≡ 0. Otherwise, we would have V0 u k ’ 0 in L q ( ) and con-

˜ ˜

sequently the left-hand side of (8.96), which is equal to the second term of (8.97), will

converge to 0. But this would mean that a(u k ) ’ 0, contradicting (8.95).

˜

If r0 = 0, we obtain another contradiction, for the second term in (8.97) is bounded

by (8.96) while the ¬rst becomes in¬nite by (8.95) if tk ’ ∞. Hence, r0 = 0 implies

(8.90), and the proof is complete in this case.

It remains to consider the case when r0 = 0. By (8.69),

| f (x, u k )v|/tk r0 (tk ) ¤ (|V u k |q’1 + W q’1 )|V0 v|.

˜

∞

When v ∈ C0 ( ), the right-hand side converges in L 1 ( ) to

(|V u|q’1 + W q’1 )|V0 v|

˜

by hypothesis II(b). By (8.76), the left-hand side converges a.e. to γ (x, u). Thus,

˜

f (x, u k )v(x)d x/tk r0 (tk ) ’ γ (x, u)v(x) d x

˜

(8.98)

∞

for each v ∈ C0 ( ). By (8.71) and (8.89),

2a(u k /t0 (tk ), v) ’ ( f (x, u k )/tk r0 (tk ), v) ’ 0

˜

(8.99)

for each v ∈ D. Thus, by (8.72) and (8.98),

∞

2a(u/r0 , v) = (γ (x, u), v),

˜ v ∈ C0 ( ).

˜

This shows that u is a solution of (8.79). By (8.70),

|H (x, u k )|/r1 (tk ) ¤ |V1 u k |q1 + W1 ,

˜

(8.100)

80 8. Semilinear Problems

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

|V1 u|q1 + W1

˜

by hypothesis II(c). If r1 = 0, this produces a contradiction, for by (8.88), (8.89), and

(8.94),

H (x, u k ) d x/r1(tk )

(8.101)

converges to ’b/r1. Since b = 0, this means that expression (8.101) becomes in¬nite.

Thus, the proof is complete for the case when r1 = 0.

It therefore remains to consider the case when r0 = 0, r1 = 0. If r1 = ∞, the

integrand of (8.101) converges a.e. to (x, u) by hypothesis III(b). If r1 = ∞, this

˜

convergence takes place only for those points x where u(x) = 0. But by hypothesis V,

˜

u(x) = 0 a.e. since it is a solution of (8.79). Hence, the convergence is a.e. in all cases,

˜

and expression (8.101) converges to

(x, u) d x.

˜

(8.102)

Since it also converges to ’b/r1 by (8.88), (8.89) and (8.94), we see that u is a solution

˜

of (8.80) as well as (8.79). As before, (8.97) holds by (8.71) and (8.89). In view of

(8.76), this implies that

2/r0 = (γ (x, u), u).

˜˜

˜

If we combine this with (8.79), we see that u satis¬es (8.78) as well as (8.79) and

(8.80). A contradiction is now provided by hypothesis IV. Hence, (8.90) holds, and the

proof is complete.

8.6 Special cases

In this section we present some consequences of Theorem 8.14.

Theorem 8.15. Assume that

|H (x, t)| ¤ W (x) ∈ L 1 ( ),

(8.103)

H (x, t) ’ H± (x) as t ’ ±∞ a.e.,

(8.104)

2t ’2 F(x, t) ’ b± (x) as t ’ ±∞ a.e.,

(8.105)

| f (x, t)| ¤ W1 (x) ∈ L 1 ( ), |t| ¤ 1,

(8.106)