j

β 2 ’ „k ’ (μ ’ 1)/4 + O(β ’2 )

= 2

j j

where „k = 2kπ R/T . (We may assume k ≥ 0.) Now

1 1

β j ’ „k = π j+ ν ’ ’ ak/4b = π[(4 j + n ’ 3)b ’ ak]/4b.

(12.22)

2 4

Since the expression in the brackets is an integer, we see that either β j = „k or

|β j ’ „k | ≥ π/4b.

(12.23)

Thus,

lim » j k = ’(μ ’ 1)/4R 2 = »0

(12.24)

j,|k|’∞

β j =„k

144 12. Rotationally Invariant Solutions

and

lim |» j k | = ∞.

(12.25)

j,|k|’∞

β j =„k

If n ’ 3 is not a multiple of (4, a), then

β j ’ „k = π((4 j + n ’ 3) ’ ak/b)/4

(12.26)

can never vanish. To see this, note that if (b, k) = b, then ak/b is not an integer.

Hence, β j = „k . If b = (b, k), then

(n ’ 3) = ak ’ 4 j ∀ j, k = k/b.

(12.27)

Thus, in this case we always have β j = „k and |» j k | ’ ∞ as j, k ’ ∞. On the other

hand, if n ≡ 3(mod(4, a)), then there are an in¬nite number of positive integers j, k

such that

n ’ 3 = ak ’ 4 j.

(12.28)

Hence, the point »0 is a limit point of eigenvalues. Consequently, it is in σe (L). This

completes the proof.

12.3 The nonlinear case

We now turn to the problem of solving (12.1)“(12.3). If one is searching for rotationally

invariant solutions, the problem reduces to

Lu = f (t, r, u), u ∈ D(L)

(12.29)

where L is the self-adjoint extension of the operator L 0 given in Theorem 12.2. Under

the hypotheses of that theorem, the spectrum of L is discrete. We assume that

f (t, r, s) = μs + p(t, r, s),

(12.30)

where μ is a point in the resolvent set of L and p(t, r, s) is a Carath´ odory function on

e

— R such that

| p(t, r, s)| ¤ C(|s|θ + 1), s ∈ R,

(12.31)

for some number θ < 1. We have

Theorem 12.3. Let f (t, r, s) satisfy (12.30) and (12.31), and assume the hypotheses

of Theorem 12.2. If

n ≡ 3 (mod(4, a)),

(12.32)

make no further assumptions. If

n≡3 (mod(4, a))

(12.33)

and »0 < μ, assume that p(t, r, s) is nondecreasing in s. If (12.33) holds and

μ < »0 , assume that p(t, r, s) is nonincreasing in s. Then (12.29) has at least one

weak solution.

12.3. The nonlinear case 145

Proof. Since μ is in the resolvent set of L, there is a δ > 0 such that

|» j k ’ μ| ≥ δ ∀ j, k,

(12.34)

where the » j k are given by (12.19). Each u ∈ L 2 ( , ρ) can be expanded in the form

u= ± j k ψ j k (t, r ),

(12.35)

where the ψ j k are given by (12.18). Let N0 be the subspace of those u ∈ L 2 ( , ρ) for

which ± j k = 0 if β j = „k (cf. the proof of Theorem 12.2). For u ∈ N0 ,

u= ± j k ψ j k (t, r ),

(12.36)

[0]

where summation is taken over those j, k for which β j = „k . Let E be the subspace of

L 2 ( , ρ) consisting of those u for which

= |» j k ’ μ||± j k |2

2

(12.37) u E

is ¬nite. With this norm, E becomes a separable Hilbert space. Note that E ‚

D(|L|1/2 ) and the embedding of E N0 into L 2 ( , ρ) is compact [we use (12.25)

for this purpose]. Let

G(u) = ([L ’ μ]u, u) ’ 2 u ∈ E,

(12.38) P(t, r, u)ρdtdr,

where

s

P(t, r, s) = p(t, r, σ )dσ,

(12.39)

0

and then scalar product is that of L 2 ( , ρ). One checks readily that G is a

C 1 -functional on E with

(G (u), v)/2 = ([L ’ μ]u, v) ’ ( p(u), v), u, v ∈ E,

(12.40)

where we write p(u) in place of p(t, r, u). This shows that u is a weak solution of

(12.29) iff G (u) = 0. Let N be the subspace of E spanned by the ψ j k corresponding

to those » j k < μ and let M denote the subspace of E spanned by the rest. Thus,

M = N ⊥ in E. Assume ¬rst that N © N0 = {0}. Then

= (μ ’ » j k )|± j k |2 , u ∈ N.

2

(12.41) u E

Thus,

2

= ’v ’2 P(t, r, v)ρdtdr

G(v) E

¤ ’v +C (|v|1+θ + |v|)ρdtdr

2

E

¤ ’v +C ( v + v ) ’ ’∞, v ’ ∞, v ∈ N.

2 1+θ

E

E

146 12. Rotationally Invariant Solutions

If w ∈ M,

G(w) ≥ δ w ’ C( w + w ) ≥ ’K , w ∈ M.

2 1+θ

(12.42)