Existence and Multiplicity of Periodic Solutions for Dirichlet–Neumann Boundary Value Problem of a Variable Coefficient Wave Equation

We assume that

where

The function y∈L2⁢(Ω) is called a weak solution of (2.1) if it satisfies

The function y∈L2⁢(Ω) is called a weak solution to problem (2.3) if it satisfies

Lemma 2.3 ([12])

Let λ02<λ12<⋯ and z0,z1,… denote the eigenvalues and real orthonormal eigenfunctions of (2.6), respectively. Then

where ρ1=2π⁢∫0πηu⁢(x)⁢𝑑x<34.

The operator A is reversible, its reverse A-1 is compact on L2⁢(Ω), and

Proof.

By the definition of A, it is easy to see that A⁢y=f if and only if

where ym⁢n and fm⁢n are the Fourier coefficients of y and f in L2⁢(Ω), respectively, i.e.,

with ψm and φn given by (2.4) and (2.5).

By (2.4) and Lemma 2.3, it is obvious that λn2-μm2≠0 for any m∈ℤ and n∈ℕ0. Furthermore, we can also obtain

This implies that the null space N⁢(A) of A is trivial which, in combination with Parseval’s formula

shows that A is reversible with

In what follows, we shall prove that A-1 is compact on L2⁢(Ω). Define the finite dimensional operator AN-1 by

where N∈ℕ, and fm⁢n denotes the Fourier coefficient of f∈L2⁢(Ω). We only need to prove

Note that

so it is sufficient to prove

Since

the convergence of

implies that

is convergent. Therefore, we have (2.7), which shows that A-1 is compact on L2⁢(Ω). ∎

Assume that u satisfies Hypothesis 1.1 and that the function g is bounded. Then, for any given f∈L2⁢(Ω), problem (1.1)–(1.3), with a1,a2,b1,b2 satisfying (1.4), has at least one solution y∈L2⁢(Ω).

Proof.

By the definition of operator A, we know that y∈L2⁢(Ω) is the solution of problem (1.1)–(1.3) if and only if it satisfies the operator equation

For α∈[0,1], define Tα:L2⁢(Ω)→L2⁢(Ω) by

It is obvious that the zeros of T1 correspond to the solutions of (3.1), which are also the solutions of problem (1.1)–(1.3).

Since T0=I (where I denotes the identity map), we have deg⁡(T0,BR⁢(0),0)=1 for any R>0, where BR⁢(0)={y∈L2⁢(Ω):∥y∥<R}. Furthermore, we can prove that there exists R0>0 (independent of α) such that 0∉Tα⁢(∂⁡BR⁢(0)) for all α∈[0,1] and R>R0. In fact, if y∈L2⁢(Ω) is the solution of Tα⁢y=0 for some α∈[0,1], then the boundedness of g shows that

for some R0>0 (independent of α) large enough.

Thus, by the homotopy invariance property of the Leray–Schauder degree, we have

Therefore, problem (1.1)–(1.3), with a1,a2,b1,b2 satisfying (1.4), has at least one solution in BR⁢(0). ∎

The function g is Lipschitz continuous with Lipschitz constant L>0, i.e.,

In addition, g satisfies g⁢(0)=0 and is Gâteaux differentiable at 0 with Gâteaux derivative d⁢g⁢(0,y)=γ⁢y for some γ>0.

Assume that g satisfies Hypothesis 3.2 and that K⊂L2⁢(Ω) is a compact set. Then there exists a function h:(0,∞)→[0,∞), satisfying h⁢(s)→0 as s→0, such that

Proof.

Define the function h:(0,∞)→[0,∞) by

Then (3.2) is automatically satisfied. Thus, we only need to prove that h⁢(s)→0 as s→0. To this end, we shall prove ∥1s⁢g⁢(s⁢ϕ)-γ⁢ϕ∥→0 uniformly on K as s→0.

Denote

Then, for each ϕ∈K, gs⁢(ϕ)→0 as s→0. In fact, it is obvious that gs⁢(0)=0. In addition, for each ϕ∈K with ϕ≠0, by Hypothesis 3.2, we know that g satisfies g⁢(0)=0 and is Gâteaux differentiable at 0 with Gâteaux derivative d⁢g⁢(0,y)=γ⁢y, thus it is easy to show that gs⁢(ϕ)→0 as s→0.

Next, we shall prove that the family {gs} is equicontinuous on K. For any ε>0, if ϕ1,ϕ2∈K satisfy ∥ϕ1-ϕ2∥<εL+γ, then the Lipschitz continuity of g yields

Thus, {gs} is equicontinuous and converges pointwise on K. Then the Arzelà–Ascoli theorem shows that gs⁢(ϕ)→0 uniformly on K as s→0. ∎

The operator AH:H→H is reversible, its reverse AH-1 is compact and ∥AH-1∥=1d. Moreover, AH⁢y=f if and only if

where ym⁢n and fm⁢n are the Fourier coefficients of y and f in H, respectively, i.e.,

with ϑm=cos⁡(μm⁢t)=cos⁡(2⁢m⁢t) and φn given by (2.5).

Proof.

The conclusions are easily drawn from the properties of operator A. ∎

Assume 74-ρ<γ<154-ρ1. Then we have

where BrH⁢(0)={y∈H:∥y∥<r}.

Proof.

By the properties of A (see the proof of Proposition 2.4), it is easy to show that Aℋ-1 can be approximated in operator norm by the finite dimensional operator AℋN-1:ℋN→ℋN given by

where ℋN=span⁡{ϑm⁢φn:m≤N,n≤N} and f=∑m≤N,n≤Nfm⁢n⁢ϑm⁢φn.

By the definition of the Leray–Schauder degree, it is known that

for N∈ℕ large enough. Note that I+γ⁢AℋN-1 can be identified with an (N+1)2×(N+1)2 diagonal matrix whose entries are 1+γλn2-μm2. Thus, we have

A simple analysis shows that if 74-ρ<γ<154-ρ1, then the only negative value of 1+γλn2-μm2 occurs when n=0 and m=1, which is simple because of our restriction to the subspace ℋ. Thus, we get

Assume that u satisfies Hypothesis 1.1 and that g satisfies Hypothesis 3.2 with 74-ρ<γ<154-ρ1. If g is bounded and f∈H satisfies ∥f∥<ε0 for some ε0>0 small enough, then problem (1.1)–(1.3), with a1,a2,b1,b2 satisfying (1.4), has at least two solutions in H.

Proof.

Since the operator Aℋ:ℋ→ℋ denotes the restriction of A to the subspace ℋ, it is obvious that y∈ℋ is the solution of problem (1.1)–(1.3) if and only if it satisfies the operator equation

For β∈[0,1], define Tβℋ:ℋ→ℋ by

It is obvious that zeros of T1ℋ correspond to the solutions of (3.3), which are also the solutions of problem (1.1)–(1.3).

In what follows, we shall prove that there exists sufficiently small r>0 such that Tβℋ⁢y≠0 for all β∈[0,1] and y∈∂⁡Brℋ⁢(0)={y∈ℋ:∥y∥=r}. On the contrary, we assume Tβ0ℋ⁢y0=0, i.e.,

for some β0∈[0,1] and y0∈ℋ with ∥y0∥=r.

Since 74-ρ<γ<154-ρ1, it is easy to verify that -γ is not an eigenvalue of Aℋ which, in combination with the compactness of Aℋ-1, shows that

Set y¯0=y0r, then y0=r⁢y¯0 and ∥y¯0∥=1. Thus, by (3.5), it is easy to see that

On the other hand, (3.4) implies that

Thus, we have

which is equivalent to

This shows that

which is a compact set in ℋ⊂L2⁢(Ω) by Lemma 3.4. Therefore, by Lemma 3.3, there exists a function h:(0,∞)→[0,∞), satisfying h⁢(r)→0 as r→0, such that

Hence, for sufficient small r>0 and ε0>0, we have

which, in view of (3.4), contradicts (3.6).

Thus, by the homotopy invariance property of the Leray–Schauder degree and Lemma 3.5, there exists sufficiently small r>0 such that

Therefore, problem (1.1)–(1.3), with a1,a2,b1,b2 satisfying (1.4), has at least one solution in Brℋ⁢(0).

Furthermore, similar to the proof of Theorem 3.1, we can also get