Au = f (x, u), u ∈ D(A).

(10.39)

Proof. Let

a(u, v) = (Au, v), a(u) = (Au, u), u, v ∈ D,

(10.40)

and

G(u) = a(u) ’ 2 u ∈ D,

(10.41) F(x, u) d x,

10.4. Some applications 115

where

t

F(x, t) = f (x, s)ds.

(10.42)

0

From (10.33) it is readily veri¬ed that G is continuously differentiable on D and

(G (u), v)/2 = a(u, v) ’ ( f (u), v), u, v ∈ D,

(10.43)

where we write f (u) in place of f (x, u) (cf., e.g., [112]). We note that G (u) satis¬es

(10.5) on E = D : for, if u k ’ u weakly in D, then a(u k , v) ’ a(u, v) for each

v ∈ D, and there is a renamed subsequence such that V u k ’ V u in L 2 ( ). Since

V (x) > 0 in , there is a renamed subsequence such that u k ’ u a.e. in . Thus,

f (x, u k )v ’ f (x, u)v

(10.44) a.e.

Since

| f (x, u k )v| ¤ |V u k V v| + W |V v|

and the right-hand side converges to |V uV v|+W |V v| in L 1 ( ), we see that ( f (u k ), v)

’ ( f (u), v). Thus, G (u) satis¬es (10.5).

I claim that

G(v) ’ ’∞ as v ’ ∞, v ∈ N,

(10.45) D

G(w) ’ ∞ as w ’ ∞, w ∈ M,

(10.46) D

where u D = |A|1/2u . To prove (10.45), let {v k } be a sequence such that

ρk = v k D ’ ∞, and let v k = v k /ρk . Then v k D = 1, and there is a renamed

˜ ˜

subsequence that converges weakly in D and a.e. in to a function v ∈ D (again using

˜

the properties of V ). I claim that

F(x, v k )d x/ρk ’ ±(v)

˜

2

(10.47) 2

and

G(v k )/ρk ’ ’1 ’ ±(v) = ( v

˜ ˜ ’ 1) ’ ( v

˜ + ±(v)) ¤ 0.

˜

2 2 2

(10.48) D D

To see this, note that

2F(x, v k ) 2F(x, v k )

v k ’’ ±+ (v + )2 + ±’ (v ’ )2 a.e.

= ˜2 ˜ ˜

ρk vk

2 2

Moreover, by (10.33),

|F(x, v k )|/ρk ¤ C(|V v k |2 + W |V v k |/ρk ) ’’ C|V v|2

˜ ˜ ˜

2

in L 1 ( ). This implies (10.47) and (10.48). Now, the only way the right-hand side of

(10.48) can vanish is if

v

˜ =1

(10.49) D

116 10. Weak Linking

and

(Av, v) = ±(v).

˜˜ ˜

(10.50)

Let

(v) = (Av, v) ’ ±(v).

Then

(v) ¤ 0, v ∈ N.

If (v) = 0, then v is a maximum point for on N. This means that

˜ ˜ (v) = 0, i.e.,

˜

that

(Av, v) ’ ±(v, v) = 0, v ∈ N.

˜ ˜

Thus, v is a solution of (10.38). However, this contradicts (10.49). Hence, the right-

˜

hand side of (10.48) must be negative. This proves (10.45). The proof of (10.46) is

similar. Note that (10.46) implies

b0 = inf G > ’∞,

M

for if G(wk ) ’ b0 , {wk } ‚ M, then wk D ¤ C by (10.46). But then (10.33) implies

that G is bounded on bounded sets in D. From (10.45) we see that there is an R such

that (10.1) holds with A = N © ‚ B R , B = M, Q = N © B R , and F = P, the

orthogonal projection of D onto N, p = 0. If S is a ¬nite-dimensional subspace of

D such that P S = {0}, let S0 = P S. Clearly, (10.6) is satis¬ed. We can now apply

Theorem 10.2 to conclude that there is a sequence {u k } ‚ D such that

G(u k ) ’ c, b0 ¤ c ¤ a1 , G (u k ) ’ 0.

(10.51)

Assume ¬rst that ρk = u k D ’ ∞, and write u k = u k /ρk . Then u k D = 1.

˜ ˜

Consequently, there is a renamed subsequence such that u k ’ u weakly in D and a.e.

˜ ˜

in (using the properties of V ). Since

| f (x, u k )v|/ρk ¤ |V u k V v| + W |V v|/ρk

˜

and the right-hand side converges in L 1 ( ), we see by (10.34) that

( f (u k ), v)/ρk ’ ±(u, v).

˜

(10.52)

Thus,

(G (u k ), v)/2ρk ’ a(u, v) ’ ±(u, v),

˜ ˜ v ∈ D.

˜

Since this limit vanishes by (10.51), we see that u is a solution of (10.38). Hence,

u ≡ 0 by hypothesis. Let u k = v k + wk , where v k ∈ N, wk ∈ M. Arguments similar

˜ ˜ ˜ ˜ ˜ ˜

to that given in the proof of (10.47) show that