Schechter Minimax Systems and Critical Point Theory (Springer, 2009)

Assume this for the moment. Then there is an R > 0 sufficiently large that (3.4) holds with A = N −1 ∩ ∂ BR and B = S,1. Moreover, for this choice of A and B, A links B [hm] by Lemma 11.10. We can now apply Theorem 3.4 to conclude that there is a

Let ± = {x : ±u˜(x) > 0}, 0 = {x : u˜ (x) = 0}. Then uk (x) → ±∞ for x ±. By (11.9) and (11.13),

−c = lim H (x, uk) dx

≤H+(x) dx

+

+H−(x) dx

−

+W0(x) dx

0

< − B1,

contradicting (11.66). Thus, (11.70) does not hold. Once we know that the sequence {uk } is bounded in D, we can use standard techniques [122] to obtain a solution of

(11.75) G(u) = c, −B0 ≤ c ≤ B1, G (u) = 0,

which is a solution of (11.2).

It thus remains to prove (11.64). Let {vk } N −1 be a sequence such that ρk =vk D → ∞. Let v˜k = vk /ρk . Then v˜k D = 1 and there is a renamed subsequence

→ v˜ 2D − a v˜− 2 − b v˜+ 2

≤(λ −1 − a) v˜− 2 + (λ −1 − b) v˜+ 2.

Since both a and b are greater than λ −1, this will always be positive since v˜ ≡ 0. Thus, (11.64) holds, and the proof of Theorem 11.1 is complete.

where A = M ∩ ∂ BR and B = S,2. This is not quite (3.4). To correct the situation, we let G1 = −G. Then (11.82) becomes

We know from Lemma 11.11 that A links B [hm]. Consequently, we may apply Theorem 3.4 to conclude that there is a sequence {uk } D such that (11.65) holds with G replaced by G1. Now c satisfies the estimates

By (11.14), (11.56), and (11.80), we see that c satisfies (11.66) as well. Thus, for the given sequence, (11.65), (11.67)–(11.69) hold with c replaced by −c, while c satisfies (11.66). If we assume that (11.70) holds, we can reason as in the proof of Theorem 11.1 that there is a renamed subsequence of {u˜k } converging in D to a function u˜ D and a.e. in . Moreover, the function u˜ satisfies (11.72) and (11.73). Also, (11.74) holds with c replaced by −c. But then (11.15) and (11.16) imply

(11.86)

This contradiction shows that the ρk are bounded, and we can now employ standard techniques to obtain a solution of

This produces the desired result.

It therefore remains only to prove (11.81). Let {wk } M be a sequence such that

which is positive since both a and b are less than λ +1. This completes the proof of the theorem.

11.4 Notes and remarks

The presentation given here is from [124]. Further work was done in [102].

Chapter 12

Rotationally Invariant Solutions

12.1 Introduction

In this chapter (and the next) we study periodic solutions of the Dirichlet problem for the semilinear wave equation. In this chapter we study radially symmetric solutions for the problem

Our basic assumption is that the ratio R/ T is rational. Thus, we can write

(12.5) 8R/ T = a/b,

where a, b are relatively prime positive integers. We show that

(12.6) n ≡ 3 (mod(4, a))

implies that the linear problem corresponding to (12.1)–(12.3) has no essential spectrum. If

then the essential spectrum of the linear operator consists of precisely one point:

M. Schechter, Minimax Systems and Critical Point Theory, DOI 10.1007/978-0-8176-4902-9_12, © Birkhäuser Boston, a part of Springer Science+Business Media, LLC 2009

This shows that the spectrum has at most one limit point. We can then consider the nonlinear case

for some number θ < 1. Our main theorem is

Theorem 12.1. If (12.6) holds, then (12.1)–(12.3) has a weak rationally invariant solution. If (12.7) holds and λ0 < μ, assume, in addition, that p(t, r, s) is nondecreasing in s. If μ < λ0, assume that p(t, r, s) is nonincreasing in s. Then (12.1)–(12.3) has a weak rotationally invariant solution.

Our proof of this theorem will make use of Theorem 10.2. For the definition of essential spectrum, cf., e.g., [126] and [106].

12.2 The spectrum of the linear operator

In proving Theorem 12.1, we shall need to calculate the spectrum of the linear operatorapplied to periodic rotationally symmetric functions. Specifically, we shall need

Theorem 12.2. Let L0 be the operator

where = [0, T ] × [0, R]. Then L0 is symmetric on L2( , ρ), where ρ = r n−1. Assume that 8R/ T = a/b, where a, b are relatively prime integers [i.e., (a, b) = 1]. Then L0 has a self-adjoint extension L having no essential spectrum other than the point λ0 = −(n − 3)(n − 1)/4R2. If n ≡ 3 (mod(4, a)), then L has no essential spectrum. If n ≡ 3 (mod(4, a)), then the essential spectrum of L is precisely the point λ0.

Proof. Let ν = (n − 2)/2 and let γ be a positive root of Jν (x) = 0, where Jν is the Bessel function of the first kind. Set

It is easily checked that the functions ψ j k , when normalized, form a complete orthonormal sequence in L2( , ρ). We shall show that the corresponding eigenvalues (12.19) are not dense in R. It will then follow that L0 has a self-adjoint extension L with spectrum equal to the closure of the set {λ j k } (cf., e.g., [126]). Now

=β2j − τk2 − (μ − 1)/4 + O(β−j 2)

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

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

| |→∞

β j =τk

where the λ j k are given by (12.19). Each u L2( , ρ) can be expanded in the form

(12.35) u = α j k ψ j k (t, r ),

where the ψ j k are given by (12.18). Let N0 be the subspace of those u L2( , ρ) for which α j k = 0 if β j = τk (cf. the proof of Theorem 12.2). For u N0,

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

≤− v 2E + C (|v|1+θ + |v|)ρdtdr

≤ − v 2E + C ( v 1+θ + v ) → −∞, v E → ∞, v N.

Thus, G (un ) → G (u) weakly in E. Hence, all of the hypotheses of Theorem 10.2 are satisfied, and we can conclude that there is a sequence {uk } satisfying (10.3). Write uk = vk + wk + yk , where vk N, wk M N0, yk N0. Then

If N0 = {0}, then it follows from (12.45) and (12.46) that uk E is bounded and consequently there is a renamed subsequence that converges weakly in E and strongly in L2( , ρ) to a function u. Thus, G (uk ) → G (u) weakly. But G (uk ) → 0. Consequently G (u) = 0 and the proof for this case is complete. If N0 = {0}, we note that

(12.47) yk 2E ≤ G (uk ) yk ) E /2 + C yk ||( uk θ + 1)

as well. Again, this together with (12.45) and (12.46) implies that uk ||E is bounded and has a renamed subsequence that converges weakly in E and such that uk = vk +wk converges strongly in L2( , ρ). Now