c Allerton Press, Inc., 2007. ISSN 1066-369X, Russian Mathematics (Iz. VUZ), 2007, Vol. 51, No. 2, pp. 44–52.  c I. A. Rudakov, 2007, published in Izvestiya Vysshykh Uchebnykh Zavedenii. Matematika, 2007, No. 2, pp. 46–55. Original Russian Text 

Periodic Solutions of a Nonlinear Wave Equation with Neumann and Dirichlet Boundary Conditions I. A. Rudakov Bryansk State University, ul. Bezhitskaya 14, Bryansk, 241036 Russia E-mail: [email protected] Received April 22, 2005

DOI: 10.3103/S1066369X07020065

1. INTRODUCTION Consider the problem utt − uxx + g(u) = f (x, t), 0 < x < π, t ∈ R, u(x, t + 2π) = u(x, t), 0 < x < π, t ∈ R.

(1) (2)

We will study the following two types of boundary conditions: u(0, t) = u (π, t) = 0, t ∈ R;

(3)



u (0, t) = u(π, t) = 0, t ∈ R.

(4)

Here f (x, t) is a given function, 2π-periodic with respect to time. Problem (1), (2) with zero Dirichlet boundary conditions has been studied thoroughly in [1]–[6]. In [7], problem (1), (2) with homogeneous boundary conditions of 3rd kind was studied. In [8]–[10], the existence of generalized periodic solutions of a wave equation with nonconstant with respect to x coefficients has been proved. The main result of this paper is the theorem on existence of a classical solution of problems (1)–(3) and (1), (2), (4) in the case when the summand g(u) satisfies the nonresonance condition. 2. THE LINEAR PART OF THE EQUATION Denote Ω = [0, π] × [0, 2π]. In the study of boundary conditions (4), we will denote Ω1 = [−π, π] × [0, 2π]. In the study of boundary conditions  (3), we will denote Ω1 =[0, 2π] × [0, 2π]. For functions f, g ∈ L2 (Ω), we use the notation (f, g) = f (x, t)g(x, t)dx dt, f 2 = f 2 (x, t)dx dt. We will seek the Ω

Ω

solutions of problems (1)–(3) and (1), (2), (4) in the form of a Fourier series. To construct orthonormal in L2 (Ω) systems of eigenfunctions of the linear parts of these problems, we consider the following Sturm– Liouville problems: ϕ + λ2 ϕ = 0, ϕ(0) = ϕ (π) = 0;

(5)

ϕ + λ2 ϕ = 0, ϕ (0) = ϕ(π) = 0.

(6)

Problems (5), (6) have the same eigenvalues λ2n = (n + 12 )2 . The eigenfunctions of problems (5), (6) are of the form Xn = sin(n + 12 )x and Xn = cos(n + 12 )x respectively. Here n ∈ Z+ ≡ N ∪ {0}. Denote √ √   ϕnm = π2 Xn cos mt, ψnm = π2 Xn sin mt, Λ = {ϕnm , ψnm | n ∈ Z+ , m ∈ N} ∪ π1 Xn | n ∈ Z+ . The system Λ is a complete orthonormal in L2 (Ω) system of eigenfunctions of the d’Alembert operator ∂tt − ∂xx with boundary conditions (3) or (4). 44

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45

  N  M   We define the operator A0 : L2 (Ω) → L2 (Ω) for which D(A0 ) = u = (anm ϕnm + bnm ψnm )  n=0 m=0  anm , bnm ∈ R; N, M ∈ N and A0 u = utt − uxx ∀u ∈ D(A0 ). Denote by A the operator in L2 (Ω) which is the closure with respect to the graph of the operator A0 . The system Λ is the system of eigenfunctions of the operators A and A0 with eigenvalues μnm = (n + 12 )2 − m2 . The function ∞  ∞  ∞ ∞   (anm ϕnm + bnm ψnm ) belongs to D(A) if and only if μ2nm (a2nm + b2nm ) < ∞. In u= n=1 m=0

addition, Au =

∞  ∞ 

n=1 m=0

μnm (anm ϕnm + bnm ψnm ). It is easily seen that the kernel N (A) = {0}.

n=1 m=0

The following properties of the operator A are proved in the standard manner: 1) A is a self-adjoint ∞  ∞  1 is operator, 2) A−1 : L2 (Ω) → C(Ω) is a completely continuous operator, 3) the series μ2 n=0 m=0

nm

convergent. Let T = 2π. Denote by C ∞ , C ∞ the spaces of infinitely differentiable real functions on Ω and Ω1 respectively, T -periodic with respect to t. Let Hk , H k denote the Sobolev spaces which are, respectively, k k   Dσ ϕ2 , ϕ2H = Dσ ϕ2L2 (Ω1 ) , the closures of C ∞ , C ∞ with respect to the norms ϕ2k = |σ|=0

k

|σ|=0

∂ σ1 +σ2 ∂xσ1 ∂tσ2 ,

= σ1 , σ2 , k ∈ N ∪ {0}. We will denote H0 = L2 (Ω), H0 = σ = (σ1 , σ2 ), |σ| = σ1 + σ2 , L2 (Ω1 ). Let C k , C k be the spaces of T -periodic with respect to t and k times continuously differentiable k  sup |Dσ ϕ|. functions on Ω and Ω1 with the standard norm ϕk = Dσ

|σ|=0 (x,t)∈Ω(Ω1 )

Studying the operator A with boundary conditions (4), for a function f ∈ L2 (Ω), we will denote by f ∈ L2 (Ω1 ) an even with respect to x function such that f (x, t) = f (−x, t) = f (x, t) ∀(x, t) ∈ Ω. Studying the operator A with boundary conditions (3), for a function f ∈ L2 (Ω), we will denote by f (x, t) a function from L2 (Ω1 ) such that f (x, t) = f (2π − x, t) = f (x, t) ∀(x, t) ∈ Ω. For the operator A, the following analogue of theorem 4 of [11] holds: Theorem 10. For any k ∈ N ∪ {0} and any function f ∈ Hk such that f ∈ Hk , the inclusions ∈ Hk+1 ∩ C(Ω), A−1 f ∈ Hk+1 ∩ C(Ω1 ) hold, and there exists a constant Ck such that

A−1 f

A−1 f k+1 ≤ Ck f k ,

A−1 f Hk+1 ≤ Ck f k

∀f ∈ Hk .

(7)

Proof. The first case. Let A be the operator corresponding to problem (1), (2), (4). Consider the auxiliary linear problem (8) ωtt − ωxx = h(x, t), −π < x < π, t ∈ R; ω(−π, t) = ω(π, t) = 0, t ∈ R; (9) ω(x, t + 2π) = ω(x, t), t ∈ R. (10) Here h is a given 2π-periodic with respect to t function from L2 (Ω1 ), where Ω1 = [−π, π] × [0, 2π]. A generalized solution of problem (8)–(10) can be found with the use of Fourier series with respect to the trigonometric system 



1 1 x cos mt, cos k + x sin mt . Λ1 = sin nx cos mt, sin nx sin mt, cos k + 2 2 n∈N, k,m∈N∪{0} Let H be the set of even functions from L2 (Ω1 ). Then any function h ∈ H has an expansion in a Fourier series in terms of the cosines of the system Λ1 : 

+∞ +∞

1 x(hkm cos mt + hkm sin mt), cos k + h= 2 m=0 k=0

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where hkm , hkm are the Fourier coefficients. Denote by S : H → H the operator which determines a generalized solution of problem (8)–(10) for even h: 

+∞ +∞

1 1 x(hkm cos mt + hkm sin mt). Sh =    cos k + 1 2 2 2 − m k + k=0 m=0 2 One can easily see that A−1 f = Sf ∀f ∈ Hk . One can easily verify that Theorem 4 for the segment [0, π] of the paper [11] holds for problem (8)–(10). From this theorem, we obtain Sf Hk+1 ≤ Ck f Hk , where Ck does not depend on f ∈ Hk , which implies (7). The second case. Let A be the operator corresponding to problem (1)–(3). Consider the auxiliary problem ωtt − ωxx = h(x, t), 0 < x < 2π, t ∈ R; ω(0, t) = ω(2π, t) = 0, t ∈ R; ω(x, t + 2π) = ω(x, t), 0 < x < 2π, t ∈ R.

(11) (12) (13)

Here h ∈ L2 (Ω1 ) (Ω1 = [0, 2π] × [0, 2π]). Denote by H the set of functions from L2 (Ω1 ) such that h(x, t) = h(2π − x, t) ∀(x, t) ∈ Ω1 . Denote by S : H → H the operator which determines a generalized solution of problem (11)–(13) for functions h ∈ H: 

+∞ +∞

1 1 x(hkm cos mt + hkm sin mt), sin k + Sh =    1 2 2 2 k+ 2 −m k=0 m=0 where hkm , hkm are the coefficients of the Fourier series of h ∈ H with respect to the system 



1 1 x cos mt, sin k + x sin mt . sin k + 2 2 k,m∈N∪{0} One can easily see that A−1 f = Sf ∀f ∈ Hk . Then, as in the first case, Theorem 4 of the paper [11] implies (7). Corollary 1. Let the conditions of Theorem 1 be fulfilled, k ≥ 2, f ∈ Hk , f ∈ Hk . Then A−1 f ∈ −1 1 ), and, for for the function u = A f , conditions (3) or (4) are fulfilled in the classical sense.

C 1 (Ω

Corollary 1 follows from the Sobolev embedding theorem and from the fact that the function u is even with respect to x relative to 0 or π. 3. OPERATOR EQUATION IN A HILBERT SPACE Let H be a real Hilbert space with the scalar product ( , ) and the norm  · . For any subset M ⊂ H, by M and L(M ) we will denote, respectively, the closure of M with respect to the norm of H and the set of finite linear combinations of elements of M . Property I. Let A : H → H be a linear self-adjoint operator such that D(A) is everywhere dense in H. Let there exist a complete orthonormal in H system Λ = {e1 , e2 , . . . , en , . . . } of eigenvectors of the operator A, and let {λn } be a sequence of eigenvalues such that Aen = λn en , n ∈ N. We will say that an operator A satisfying all these conditions possesses property I. Let N (A) and R(A) denote, respectively, the kernel and the image of an operator A. Property II. Let Λ = Λ1 ∪ Λ2 ∪ Λ3 , where the subsets Λ1 , Λ2 , Λ3 are pairwise disjoint, the subsets Λ2 , Λ3 are infinite, and 1) L(Λ1 ) = N (A), 2) there exist positive constants a and b such that, for any e ∈ Λ2 , the corresponding eigenvalue λ belongs to [a, b],

1 < ∞. (14) 3) λ2i ei ∈Λ3

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In the above formula, the summation is taken over all the eigenvalues of the operator A from Λ3 . Note that if condition (14) holds for an operator A, then all eigenvectors of A belonging to Λ3 are of finite multiplicity. We will say that property III holds for a self-adjoint operator A if A satisfies property I, Λ = Λ1 ∪ Λ3 , Λ1 ∩ Λ3 = ∅, Λ3 is an infinite set, and conditions 1), 3) from property II hold. Denote N1 = N (A), N2 = L(Λ2 ), N3 = L(Λ3 ). The sets N1 , N2 , N3 are pairwise disjoint orthogonal closed subspaces of H. Let P1 , P2 , and P3 be orthogonal projectors onto N1 , N2 , and N3 respectively. For any element u ∈ H, we will denote ui = Pi u, i ∈ {1, 2, 3}. Therefore, u = u1 + u2 + u3 , ui ∈ Hi , i ∈ {1, 2, 3}. Note that if property II or property III holds, then the inverse operator A−1 : N3 → N3 is completely continuous. Denote σ(A) = {λn | n ∈ N}. In all applications, the spectrum σ(A) is bounded neither from above nor from below. Lemma 9. For any number λ ∈ σ(A), the operator (A − λ)−1 : N3 → N3 is completely continuous.  1 Proof. It suffices to prove the convergence of the series I = (λi −λ)2 . From (14) it follows that the ei ∈Λ3

set {λi | ei ∈ Λ3 , |λi | < 2|λ|} is finite since the eigenvalues λi ∈ Λ3 are of finite multiplicity. Therefore, there exists a constant C1 such that

2 −1

1

1

λ ≤ C1 < ∞.  1 − I= 2 λi λi λ2i ei ∈Λ3

ei ∈Λ3

As a consequence of Lemma 1, we obtain the following Corollary 2. If property III holds for an operator A and dim N (A) < ∞, then, for any λ ∈ / σ(A), the operator (A − λ)−1 : H → H is completely continuous. Let B : H → H be a (nonlinear) operator for which there exist constants C, γ ∈ (0, +∞) and λ ∈ R such that 1 (15) (B(u) − λu, u) ≥ B(u) − λu2 − C ∀u ∈ H. γ Consider in H the equation Au − B(u) = f.

(16)

Theorem 11. If property III holds for an operator A : H → H, dim N (A) < ∞, and B : H → H is a demicontinuous operator (i. e., B is continuous as an operator from H with the strong topology to H with the weak topology) for which relation (15) holds, where λ ∈ (λ, λ),

γ ∈ (0, λ − λ),

λ ∈ σ(A),

(λ, λ) ∩ σ(A) = ∅,

(17)

then, for any f ∈ H, equation (16) has a solution in H, i. e., R(A − B) = H. Proof. Let us rewrite equation (16) in the following form u = T (u),

(18)

λ)−1 (B(u)

− λu + f ). From the assumptions of the theorem and from Corollary 2 where T (u) = (A − it follows that the operator T : H → H is completely continuous. We will prove the existence of a fixed point of the operator T using the Leray–Schauder principle. For this, we consider the equation u = μT (u) with the parameter μ ∈ (0, 1]. We transform this equation to the form 1 B1 (u) = − (−A + λ)(u − μω), μ RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 51 No. 2 2007

(19)

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RUDAKOV

where ω = (A − λ)−1 f, B1 (u) = B(u) − λu. One can easily see that η = λ − λ is the greatest negative eigenvalue of the operator −A + λ. Let us take the scalar product in H of each of the sides of (19) and u − μω and then make use of the standard [1] inequality with the greatest negative eigenvalue η: 1 (B1 (u), u − μω) = − ((−A + λ)(u − μω), u − μω) μ μ 1 (−A + λ)(u − μω)2 = B1 (u)2 . ≤ μ(λ − λ) λ−λ From this inequality, using (15), we derive the inequality

 1 1 − B1 (u)2 − B1 (u) ω − C ≤ 0. γ λ − λ) Then, from (17) and (19), we obtain the estimates B1 (u) ≤ C2 , (−A + λ)(u − μω) ≤ C2 , where C2 does not depend on μ. From the latter inequality and (17), we derive the estimate u ≤ C3 , where C3 does not depend on μ. The Leray–Schauder principle implies the existence of a solution of equation (18). Theorem 12. If conditions I, III hold for an operator A : H → H, dim N (A) = ∞, and B : H → H is a demicontinuous monotone operator for which relation (15) holds, where λ ∈ (λ, λ),

γ ∈ (0, λ − λ),

λ ≥ 0,

−λ, −λ ∈ σ(A),

(−λ, −λ) ∩ σ(A) = ∅,

(20)

then, for any f ∈ H, the equation Au + B(u) = f

(21)

has a solution in H. Note that, in [1], Theorems 2, 3 are proved for the case when λ (and, for Theorem 3, respectively, −λ) is the greatest negative eigenvalue of A. Theorems 2, 3 are generalizations of the results obtained in [1] to the case of arbitrary neighbor eigenvalues λ and λ (or −λ, −λ in Theorem 3) of the operator A. A result equivalent to Theorem 3 was obtained in [5]. Proof. Consider first, as in [1], the auxiliary equation εu1ε + Auε + B(uε ) = f.

(22)

Here ε > 0, u1ε = P1 uε . Projecting equation (22) to N1 and N3 and omitting, for the sake of simplicity of notation, the index ε, we obtain the two equations: εu1 + B1 (u1 + u3 ) = f1 , Au3 + B3 (u1 + u3 ) = f3 ,

(23) (24)

where B1 (u) = P1 B(u), B3 (u) = P3 B(u), u = u1 + u3 , f1 + f3 = f . For a fixed u3 ∈ N3 , we denote by Su3 : N1 → N1 the operator acting as follows: Su3 (v) = εv + B1 (v + u3 ) ∀v ∈ N1 . From the assumptions of the theorem it follows that Su3 is a strongly monotone demicontinuous operator. Consequently, for each fixed u3 ∈ N3 , equation (23) has a unique solution u1 = u1 (u3 ). Let us prove that the operator u1 : N3 → N1 (u3 → u1 (u3 )) is continuous in H. We take a sequence {u3n } ⊆ N3 such that u3n → u3 in H. Show that u1 (u3n ) −→ u1 (u3 ) in H. Denote u1n = u1 (u3n ), u1 = u1 (u3 ). We have εu1n + B1 (u1n + u3n ) = f1 , (25) εu1 + B1 (u1 + u3 ) = f1 . Subtracting the second equality from the first one, taking the scalar product in H of the difference and u1n − u1 , and making use of the fact that B is a monotone operator, we obtain RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 51 No. 2 2007

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0 = εu1n − u1 2 + (B(u1n + u3n ) − B(u1 + u3 ), u1n + u3n − u1 − u3 ) − (B(u1n + u3n ) − B(u1 + u3 ), u3n − u3 ) ≥ εu1n − u1 2 − (B(u1n + u3n ) + B(u1 + u3 ))u3n − u3 . (26) In what follows we will denote constants as follows: C1 , C2 , C3 , . . . From (15), one can easily derive the estimate B(u) ≤ (λ + γ)u + C1 ∀u ∈ H.

(27)

The above estimate and (26) imply that the sequence {u1n } is bounded. Then from (26) it follows that u1n −→ u1 . Let us substitute u1 = u1 (u3 ) into (24): Au3 + B3 (u1 (u3 ) + u3 ) = f3 . We rewrite this equation in the form (28)

u3 = T (u3 ), λ)−1 (−B3 (u1 (u3 ) +

u3 ) + λu3 + f3 ). From the assumptions of the theorem, where T (u3 ) = (A + Lemma 1, and the fact that u1 (u3 ) is continuous it follows that the operator T (u3 ) is continuous. Show that the operator u1 (u3 ) is bounded. Let u3  ≤ M . Taking the scalar product in H of each of the sides of (25) and u1 and making use of the monotonicity of B and (27), we obtain εu1 2 = (f1 , u1 ) − (B(u1 + u3 ), u1 ) = (f1 , u1 ) − (B(u1 + u3 ), u1 + u3 ) + (B(u1 + u3 ), u3 ) ≤ f1  u1  + M ((λ + γ)(u1  + M ) + C1 ).  Therefore, {u1 (u3 )  u3  ≤ M } is a bounded set. Consequently, the operator T : N3 → N3 is completely continuous. The existence of a solution of equation (28) will be proved with the use of the Leray–Schauder principle. For this, we consider the equation (29)

u3 = μT (u3 )

with parameter μ ∈ (0, 1]. To verify the Leray–Schauder condition, we rewrite equation (29) in the form 1 (A + λ)(u3 − μω) + B3 (u) − λu3 = 0. μ

(30)

Here u = u1 (u3 ) + u3 , ω = (A + λ)−1 f3 . One can easily see that λ − λ is the greatest negative eigenvalue of the operator A + λ. Denoting B(u) = B(u) − λu, B1 (u) = P1 B(u) and taking the scalar product in H of (30) and (u3 − μω), we obtain 1 μ (A + λ)(u3 − μω)2 = (B(u), u) − B3 (u)2 μ(λ − λ) λ−λ 

1 1 B(u)2 − (B1 (u), u1 ) − ω B(u) − C. − − μ(B(u), ω) − (B1 (u), u1 ) ≥ γ λ−λ

0 ≥ (B(u), u3 − μω) −

Taking the scalar product in H of each of the sides of (23) and u1 , we obtain the expression (B1 (u), u1 ) = −(λ + ε)u1 2 + (f1 , u1 ). Substituting this expression into the latter inequality, we obtain 

1 1 B(u)2 + (λ + ε)u1 2 − ω B(u) − f1  u1  − C ≤ 0. − γ λ−λ Then condition (20) implies the estimates B(u) ≤ C2 ,

u1  ≤ C2 .

(31)

From (30) and (31), we obtain (A + λ)u3  ≤ C2 + f3 ,

u3  ≤ C3 .

(32)

The constants C2 , C3 do not depend on μ and ε. Let us pass to the limit in equality (22) for ε → 0 using the method from [1]. By virtue of (31) and (32), there exists a sequence εn → 0 such that uεn → u weakly in H, Auεn → η weakly in H. From (14) it follows that the operator A−1 : N3 → N3 is completely RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 51 No. 2 2007

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continuous. Therefore, u3εn → A−1 η, A−1 η = u3 , η = Au3 = Au and (Auεn , uεn ) = (Auεn , u3εn ) → (Au, u3 ) = (Au, u). The monotonicity of B implies the inequality (B(uεn ) − B(ξ), uεn − ξ) ≥ 0 ∀ξ ∈ H. From this inequality and (22) it follows that (f − εn u1εn − Auεn − B(ξ), uεn − ξ) ≥ 0 ∀ξ ∈ H. Passing to the limit for n → ∞, we obtain (f − Au − B(ξ), u − ξ) ≥ 0 ∀ξ ∈ H. Substituting ξ = u + τ ψ, where τ > 0, ψ ∈ H, into this inequality, cancelling out τ , and tending τ to zero, we obtain (f − Au − B(u), ψ) ≤ 0 ∀ψ ∈ H, which implies (21). 4. NONLINEAR WAVE EQUATION

 Denote σ(A) = {μnm | n, m ∈ Z+ }. One can easily see that σ(A) = l + 1/4 | l ∈ Z}. We will assume that the function g(u) satisfies the condition g(u) g(u) ≤ lim ≤ β, where α ≤ β. u→∞ u u u→∞

α ≤ lim

(33)

Definition 3. A generalized solution of problem (1)–(3) or (1), (2), (4) is a function u ∈ H0 such that   u(ϕtt − ϕxx )dx dt = (g(u) + f (x, t))ϕ dx dt ∀ϕ ∈ D(A0 ). (34) Ω

Ω

Theorem 13. Let the function g(u) be continuous on R and satisfy condition (33), where [α, β] ∩ σ(A) = ∅.

(35)

Then, for any f ∈ H0 , each of problems (1)–(3) and (1), (2), (4) has a generalized solution u ∈ H1 . If in addition g ∈ C k (R) and f ∈ Hk , where k ∈ N, then u ∈ Hk+1 . If f ∈ H3 , g ∈ C 3 (R), then the generalized solution is a classical solution. Proof is based on Theorem 1. Denote H = H0 , B(u) = g(u) ∀u ∈ H. From (33) it follows that B : H → H. A function u ∈ H is a generalized solution of problems (1)–(3) and (1), (2), (4) if and only if u satisfies (16). Let us check that the conditions of Theorem 2 are fulfilled. From the definition and properties 1), 3) of the operator A it follows that property III holds. From (33) and (35) it follows that there exist positive constants C1 , C2 , α1 , β1 , λ, γ such that [α, β] ⊆ (α1 , β1 ), [α1 , β1 ] ∩ σ(A) = ∅, λ ∈ (α1 , β1 ), γ ∈ (0, β1 − λ)

(36)

and g(u) = λu + p(u), where p(u)u ≥ −C1 , |p(u)| ≤ γ|u| + C2 ∀u ∈ R. Consequently,   p(u)u dx dt = |p(u)u + C1 |dx dt − 2π 2 C1 (B(u) − λu, u) = Ω Ω  1 1 1 ≥ p(u)2 − |p(u)|dx dt − 4π 2 C1 ≥ p(u)2 − C3 p(u) − 4π 2 C1 γ γ Ω γ 

1 1 − ε p(u)2 − 4π 2 C1 − C4 . ≥ γ ε Here C3 , C4 , ε ∈ (0, +∞). Then, for a sufficiently small ε, (33) and (36) imply (15) and (17). The conditions of Theorem 2 are fulfilled. From this theorem it follows that equation (16) has a solution. Let f ∈ Hk . From (16), we obtain u = A−1 (g(u) + f ). Then, from Theorem 1, we obtain u ∈ Hk+1 . The last statement of the theorem follows from the Sobolev embedding theorem.  Consider the question of uniqueness of solutions. Proposition. In addition to the assumptions of Theorem 4, let α(u − v)2 ≤ (g(u) − g(v))(u − v) ≤ β(u − v)2 ∀u, v ∈ R.

(37)

Then each of problems (1)–(3) and (1), (2), (4) has a unique generalized solution. RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 51 No. 2 2007

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Proof. Denote h(u) = g(u) − αu. From (37) it follows that 0 ≤ (p(u) − p(v))(u − v) ≤ (β − α)(u − v)2 ∀u, v ∈ R. Let u1 and u2 be solutions of equation (16). Then p(u1 ) + (α − A)u1 = −f and p(u2 ) + (α − A)v2 = −f . Subtracting the second equality from the first one and multiplying the result in L2 (Ω) by u1 − u2 , we obtain 0 = (p(u1 ) − p(u2 ), u1 − u2 ) + ((α − A)(u1 − u2 ), u1 − u2 ) 1 1 p(u1 ) − p(u2 )2 − (α − A)(u1 − u2 )2 ≥ β−α λ−α 

1 1 − (α − A)(u1 − u2 )2 . = β−α λ−α Here λ ∈ σ(A), λ > α and (α, λ) ∩ σ(A) = ∅. Consequently, u1 = u2 . Remark 3. Condition (37) means that, for u = 0 and for sufficiently small δ, the graph of the function y = g(u) lies between the lines y = (λ + δ)u and y = (λ − δ)u, where λ, λ are neighbor eigenvalues of A. In just the same way as in [6], one can show that if this condition is not fulfilled, then problems (1)–(3) and (1), (2), (4) may have two solutions. Remark 4. Conditions (33) and (35) mean that, for large values of |u|, the graph of the function y = g(u) does not meet the lines y = λu, λ ∈ σ(A). If this is not fulfilled, problems (1)–(3) and (1), (2), (4) may have no solutions. Example. Consider the equation 3 5 (38) utt − uxx = u + sin u + 3 sin x cos t. 4 2 Problems (38), (2), (3) and (38), (2), (4) have no solutions. To prove this fact, it suffices to take the scalar product in L2 (Ω) of each of the sides of equation (38) and sin 32 x cos t. In this example, we deal with the resonance case since the graph of the function g(u) = 54 u + sin u meets the line y = μ11 u infinitely many times. 5. THE CASE OF AN ARBITRARY PERIOD Consider the problem on periodic solutions of the nonlinear wave equation with boundary conditions (3) or (4) and an arbitrary period T comparable with the length of the string, b (39) T = 2π , a where a and b are mutually prime natural numbers. Let us write the periodicity condition in the form u(x, t + T ) = u(x, t) = 0,

0 < x < π, t ∈ R.

(40)

In this case, the eigenfunctions of the d’Alembert operator are of the form 2 am 2 am Xn cos t, ψnm = √ Xn sin t. b b πT πT The operators A0 and A are defined in just the same way as in Section 2. The eigenvalues of A0 and A  2 2  can be represented in the form μnm = n + 12 − ab m . One can easily show that, for an odd number b, the kernel N (A) = {0}. If b is an even number, then dim N (A) = ∞. In fact, in this case, μnm = 0 if and b(2l+1) , where l ∈ Z+ . only if n = la + a−1 2 ,m= 2 Denote Ω = [0, π] × [0, T ], Ω1 = [−π, π] × [0, T ] for conditions (4) or Ω1 = [0, 2π] × [0, T ] for conditions (3). Let C ∞ , C ∞ , Hk , Hk be the function spaces introduced in Section 2 which correspond to T = 2π ab . For an odd b, the statement of Theorem 1 remains valid. ϕnm = √

Definition 4. A function u ∈ H0 is called a generalized solution of problem (1), (40), (3) ((1), (40), (4)) if it satisfies relation (34). RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 51 No. 2 2007

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Let σ(A) = {μnm | n, m ∈ Z+ }. Theorem 14. Let b be an odd number, relation (39) hold, and let g(u) be a continuous function on R satisfying (33), where [α, β] ∩ σ(A) = ∅. Then, for any function f ∈ H0 , problem (1), (40), (3) ((1), (40), (4)) has a generalized solution u ∈ H1 . If in addition g ∈ C k (R) and f ∈ Hk , where k ∈ N, then u ∈ Hk+1 . If f ∈ H3 and g ∈ C 3 (R), then the generalized solution is a classical solution. The proof of Theorem 5 follows, in fact, verbatim the proof of Theorem 4. Let b be an even number. For convenience, we write equation (1) in the form utt − uxx + g(u) = f (x, t),

0 < x < π, t ∈ R.

(41)

Definition 5. A function u ∈ H0 is called a generalized solution of problem (41), (40), (3) or (41), (40), (4) if    u(ϕtt − ϕxx )dx dt + g(u)ϕ dx dt = f (x, t)ϕ dx dt ∀ϕ ∈ D(A0 ). Ω

Ω

Ω

Theorem 15. Let b be an even number, relation (39) hold, and let g(u) be a continuous function which is nondecreasing on R and satisfies (33), where α, β ∈ (0, +∞), [−β, −α] ∩ σ(A) = ∅.

(42)

Then, for any function f ∈ H0 , problem (41), (40), (3) ((41), (40), (4)) has a generalized solution u ∈ H0 . Proof. Denote H = L2 (Ω), B(u) = g(u) ∀u ∈ H. A function u ∈ H is a generalized solution of problem (41), (40), (3) or (41), (40), (4) if and only if relation (21) holds. The proof of the fact that the 1 is convergent is a matter of direct verification. This implies that conditions I, III for the series μ2 μkl =0

kl

operator A are fulfilled. From (42), in just the same way as in the proof of Theorem 4, we derive (15) and (20). The conditions of Theorem 3 are fulfilled, which implies the existence of a solution of equation (21).

REFERENCES 1. H. Brezis and L. Nirenberg, “Characterizations of the Ranges of Some Nonlinear Operators and Applications to Boundary Value Problems,” Ann. Scuola Norm. Sup. Pisa 5 (2), 225–325 (1978). 2. H. Brezis and L. Nirenberg, “Forced Vibration for a Nonlinear Wave Equations,” Comm. Pure Appl. Math. 31 (1), 1–30 (1978). 3. P. Rabinowitz, “Free Vibration for a Semilinear Wave Equation,” Comm. Pure Appl. Math. 31 (1), 31–68 (1980). 4. P. I. Plotnikov, “Existence of a Countable Set of Periodic Solutions of the Problem on Forced Vibrations for a Weakly Nonlinear Wave Equation,” Matem. sborn. 136 (4), 546–560 (1988). 5. I. A. Rudakov, “Nonlinear Vibrations of a String,” Vestnik Mosk. Univ., Ser. 1. Matem. Mekhan. No. 2, 9–13 (1984). 6. I. A. Rudakov, “A Problem on Free Periodic Vibrations of a String with Nonmonotonic Nonlinearity,” Usp. Mat. Nauk 40 (1), 215–216 (1985). 7. I. A. Rudakov, “A Time-Periodic Solution of the Equation of Forced Vibrations of a String with Homogeneous Boundary Conditions,” Differents. Uravneniya 39 (11), 1556–1561 (2003). 8. V. Barby and N. H. Pavel, “Periodic Solutions to Nonlinear One-Dimensional Wave Equation with xdependent Coefficients,” Trans. Amer. Math. Soc. 349 (5), 2035–2048 (1997). 9. I. A. Rudakov, “A Time-Periodic Solution of a Nonlinear Wave Equation with Nonconstant Coefficients,” Fundament. i Prikl. Matem. 8 (3), 877–886 (2002). 10. I. A. Rudakov, “Periodic Solutions of a Nonlinear Wave Equation with Nonconstant Coefficients,” Matem. Zametki 76 (3), 427–438 (2004). 11. P. Rabinowitz, “Periodic Solutions of Nonlinear Hyperbolic Partial Differential Equations,” Comm. Pure Appl. Math. 20, 145–205 (1967).

RUSSIAN MATHEMATICS (IZ. VUZ) Vol. 51 No. 2 2007