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Research | Open | Published:

Bounded solutions for the nonlinear wave equation

Abstract

We investigate the number of periodic weak solutions for the wave equation with nonlinearity decaying at the origin. We get a theorem which shows the existence of a bounded weak solution for this problem. We obtain this result by approaching the variational method and applying the critical point theory for the indefinite functional induced from the invariant subspaces and the invariant functional.

MSC:35L05, 35L70.

1 Introduction and statement of the main result

Let g(x,t,u) be a C 1 function from [0,π]×R×R to R and T-periodic in t. In this paper we are concerned with the number of weak periodic solutions of the following wave equation with boundary and periodic conditions:

u t t u x x = g ( x , t , u ) , u ( 0 , t ) = u ( π , t ) = 0 , u ( x , t + T ) = u ( x , t ) , u t ( x , 0 ) = u t ( x , T ) x [ 0 , π ] ,
(1.1)

where T is a rational multiple of π. We assume that g satisfies the following conditions:

  1. (g1)

    g C 1 ([0,π]×R×R,R) is T-periodic in t,

  2. (g2)

    g(x,t,0)=0, g(x,t,ξ)=o(|ξ|) uniformly with respect to x[0,π] and tR,

  3. (g3)

    there exists C>0 such that |g(x,t,ξ)|<C x[0,π], tR, ξR.

Nonlinear problem of this type has been considered by many authors (cf. [15]).

The purpose of this paper is to show the existence of T-periodic weak solutions of problem (1.1).

Our main result is as follows.

Theorem 1.1 Assume that g satisfies (g1)-(g3). Then problem (1.1) has at least one bounded solution.

For the proof of our main result, we approach the variational method and apply the critical point theory induced from the invariant subspaces and the invariant functional. The outline of the proof of Theorem 1.1 is as follows. In Section 2, we introduce two Banach spaces H and E of functions satisfying some symmetry properties, stable by A (Au= u t t u x x ), g such that the intersection of H with the kernel of A is reduced to 0. The search of a solution of problem (1.1) in the space H reduces the problem to a situation where A 1 is a compact operator. In Section 3, we introduce a functional I defined on E whose critical points and weak solutions of (1.1) possess one-to-one correspondence. We prove that I C 1 (E,R) and satisfies the Palais-Smale condition. By a critical point theorem for indefinite functionals (cf. [6]), we prove that there exists at least one solution of (1.1) which is bounded, and we prove Theorem 1.1.

2 Invariant Banach space

Let Ω=(0,π)×(0,T); T is a rational multiple of π

T= 2 π b a ,

where a and b are coprime integers. Let A be the operator defined by

D ( A ) = { u C 2 ( Ω ¯ ) | u ( 0 , t ) = u ( π , t ) = 0 t [ 0 , T ] , D ( A ) = { u C 2 ( Ω ¯ ) | u ( x , 0 ) = u ( x , T ) x [ 0 , π ] , D ( A ) = { u C 2 ( Ω ¯ ) | u t ( x , 0 ) = u t ( x , T ) x [ 0 , π ] } , A u = u t t u x x .

Let A be the adjoint of A in L 2 (Ω). We investigate the weak solutions of

Au=g(x,t,u).

We note that the eigenvalues of A are λ j k = j 2 ( 2 π k T ) 2 , j=1,2, and k=0,1,2, , and the corresponding eigenfunctions are

sinjxsin 2 π k t T andsinjxcos 2 π k t T .

We also note that the set of functions sinjxsin 2 π k t T , sinjxcos 2 π k t T is an orthogonal base for L 2 (Ω). Let u be a function of L 2 (Ω). Then there exists one and only one function of L 2 ([0,π]×R) which is T-periodic in t and equals u on Ω. We shall denote this function by u. Let us denote an element u, in L 2 (Ω), by

u= j > 0 , k u j k sinjxexpik a b t

with

u j , k = u ¯ j , k .

We assume that b is even and a is odd. Let H be the closed subspace of L 2 (Ω) defined by

H= { u L 2 ( Ω ) | u ( x , t ) = u ( x , t + T 2 )  a.e.  x ( 0 , π ) , t R } .

Then H is invariant under shift: Let hH and τ be a real number. If v(x,t)=u(x,t+τ), then vH. H is invariant under g: Let uH such that g(u) L 2 (Ω). Then g(u)H. Let u ˜ (x,t)=u(x,t+ T 2 ). Then

u= j > 0 , k u j k ( 1 ) k sinjxexpik a b t.

Therefore

uH u j , k =0for any even k.
(2.1)

Let A 1 be a linear operator of H defined by

D ( A 1 ) = D ( A ) H , A 1 u = A u .

Then A 1 is self-adjoint in H and HN(A)={0}, where N(A) is the kernel of A. In fact, let uHN(A). Then

u = u j , k sin j x exp i k a b t , j 2 k 2 a 2 b 2 0 u j , k = 0 .

Let j and k be such that

j 2 k 2 a 2 b 2 =0.

Since b is even and a is odd, k is even. Using (2.1), we have u j , k =0, and therefore HN(A)={0}. The eigenvalues of A 1 are j 2 ( 2 π k T ) 2 , where j is odd and k is even and HN(A)={0}. Given uH, we write

u= j > 0 u j , k sinjxexpi 2 π k t T ,j is odd,k is even

with

u j , k = u ¯ j , k .

Let

E = { u H | j , k | j 2 a 2 k 2 b 2 | | u j , k | 2 < + } , ( u , v ) = j , k | j 2 a 2 k 2 b 2 | u j , k v ¯ j , k for  u , v E ,

where (,) is a scalar product on E. With this scalar product, E is a Hilbert space with a norm

u= ( u , u ) 1 2 ,uE.

Let

u r = ( Ω | u | r ) 1 r ,r1.

By the classical theorem of Riesz (cf. [[7], p.525]), we have

u r ( π T 2 ) 1 r ( j , k | u j , k | r ) 1 r ,r2, 1 r + 1 r =1.

Since for every ϵ>o

j odd , k even 1 | j 2 a 2 k 2 b 2 | 1 + ϵ <,

it follows that for every r[2,+), there is c r R such that

u r c r u.
(2.2)

Let

E + = { u | u E , u j , k = 0  if  j 2 a 2 k 2 b 2 < 0 } , E = { u | u E , u j , k = 0  if  j 2 a 2 k 2 b 2 > 0 } .

Then E= E + E . Let P + be the orthogonal projection from E onto E + and P be the orthogonal projection from E onto E . We can write u= P + u+ P u for uE.

3 Critical point theorem and the proof of Theorem 1.1

Now, we are going to seek a function u in E such that

A 1 u=g(x,t,u).
(3.1)

We consider the associated functional of (3.1),

I ( u ) = 1 2 Ω [ | u t | 2 + | u x | 2 ] d x d t Ω G ( x , t , u ) d x d t = 1 2 ( P + u 2 P u 2 ) Ω G ( x , t , u ) d x d t ,
(3.2)

where

G(x,t,u)= 0 u g(x,t,s)ds.

By (g1), I is well defined.

We recall the critical point theorem for the indefinite functional (cf. [6]).

Let

B r = { u H | u r } , B r = { u H | u = r } .

Theorem 3.1 (Critical point theorem for the indefinite functional)

Let E be a real Hilbert space with E= E 1 E 2 and E 2 = E 1 . Suppose that I C 1 (E,R) satisfies (PS), and that

  1. (I1)

    I(u)= 1 2 (Lu,u)+bu, where Lu= L 1 P 1 u+ L 2 P 2 u and L i : E i E i is bounded and self-adjoint, i=1,2,

  2. (I2)

    b is compact, and

  3. (I3)

    there exist a subspace E ˜ E and sets SE, Q E ˜ and constants α>ω such that

    1. (i)

      S E 1 and I | S α,

    2. (ii)

      Q is bounded and I | Q ω,

    3. (iii)

      S and ∂Q link.

Then I possesses a critical value cα.

The eigenvalues of A 1 are λ j k = j 2 ( 2 π k T ) 2 , where j is odd and k is even, j=1,2, and k=0,1,2, . Thus A 1 1 is a compact operator.

It is convenient for the following to rearrange the eigenvalues λ j k , where j is odd and k is even, by increasing magnitude: from now on we denote by ( η i ) i 1 = λ j k <0 the sequence of negative eigenvalues, by ( η i + ) i 1 = λ j k >0 the sequence of positive ones, so that

η i η 2 η 1 < η 1 + η 2 + η i + .
(3.3)

We note that each eigenvalue has a finite multiplicity and that η i , η i + + as i.

We shall show that the functional I satisfies the geometrical assumptions of the critical point theorem for the indefinite functional.

By the following lemma, I(u) C 1 (E,R) and the solutions of (3.1) coincide with the nonzero critical points of I(u).

Lemma 3.1 Assume that g satisfies (g1)-(g3). Then I(u) is continuous and Fréchet differentiable in E with the Fréchet derivative

I ( u ) h = Ω [ u t h t + u x h x g ( x , t , u ) h ] d x d t = ( P + u , P + h ) ( P u , P h ) Ω g ( x , t , u ) h d x d t

for all hE. Moreover, if we set

F(u)= Ω G(x,t,u)dxdt,

then F (u) is continuous with respect to weak convergence, F (u) is compact, and

F (u)h= Ω g(x,t,u)hdxdtfor allhE.

This implies that I C 1 (E,R) and F(u) is weakly continuous.

For the proof of Lemma 3.1, refer to [6].

Lemma 3.2 Assume that g satisfies (g1)-(g3). The problem

u t t u x x in E , u ( 0 , t ) = u ( π , t ) = 0 , u ( x , t + T ) = u ( x , t ) , u t ( x , 0 ) = u t ( x , T ) x [ 0 , π ]
(3.4)

has only a trivial solution.

Proof Let Au= u t t u x x . Since HN(A)={0}, EN(A)={0}. Thus (3.4) has only a trivial solution. □

Lemma 3.3 Assume that g satisfies (g1)-(g3). Then I(u) satisfies the Palais-Smale condition: If for a sequence ( u m ), I( u m ) is bounded from above and I ( u m )0 as m, then ( u m ) has a convergent subsequence.

Proof Let ( u m ) be a sequence with I( u m )M and I ( u m )0 as m. It suffices to show that ( u m ) is bounded. By contradiction, we suppose that u m as m. Let w m = u m u m . Then w m =1 and w m converges weakly to an element, say w. Then we have

M u m 2 I ( u m ) u m 2 = 1 2 Ω [ | ( w m ) t | 2 + | ( w m ) x x | 2 ] Ω G ( x , t , u m ) u m 2 .
(3.5)

Since G(x,t,u) is bounded, letting m∞ in (3.5), we have

0lim P + w m lim P w m = Ω [ | w t | 2 + | w x x | 2 ] .
(3.6)

On the other hand, we have

0 I ( u m ) u m 2 = Ω [ | ( w m ) t | 2 + | ( w m ) x x | 2 ] Ω g ( x , t , u m ) u m 2 .
(3.7)

Letting m in (3.7), we have

0= lim m P + w m 2 lim m P w m 2 = Ω [ | w t | 2 + | w x x | 2 ] = P + w 2 P w 2 .
(3.8)

By (3.6) and (3.8), lim m P + w m 2 = P + w 2 and lim m P w m 2 = P w 2 . Thus lim m w m =w, so w m converges strongly to w and by (3.8), w is a weak solution of the problem

w t t w x x =0in E.

By Lemma 3.2, w=0, which is absurd for the fact that w=1. Thus ( u m ) is bounded. □

Let

Q=( B ¯ R E ){re|e B 1 E + 0<r<R}.

Lemma 3.4 Assume that g satisfies the conditions (g1)-(g3). There exist a subspace E ˜ E and sets B ρ E, Q E ˜ and constants ρ>0 and α>0 such that

  1. (i)

    B ρ E + and I | B ρ α,

  2. (ii)

    there are e B 1 E + and R>ρ such that if Q=( B ¯ R E ){re|0<r<R}, then I | Q 0,

  3. (iii)

    there exists u 0 EQ such that u 0 >R and I( u 0 )0,

  4. (iv)

    B ρ and ∂Q link.

Proof (i) Let u E + . Since G(x,t,u) is bounded, there exists a constant C>0 such that

I ( u ) = 1 2 P + u 2 1 2 P u 2 Ω G ( x , t , u ) 1 2 P + u 2 C

for C>0. Then there exist a constant ρ>0 and a constant α>0 such that if u B ρ E + , then I(u)α.

(ii) Let us choose an element e B 1 E + . Let u( B ¯ r E ){re|0<r}. Then u=v+w, v B r E , w=re. We note that

if v B r E , then  Ω [ | v t | 2 + | v x | 2 ] dxdt η i v L 2 ( Ω ) 2 <0.

Thus we have

I ( u ) = 1 2 r 2 1 2 v 2 Ω G ( x , t , v + r e ) 1 2 r 2 + 1 2 η 1 v L 2 ( Ω ) 2 + C 1

for C 1 >0. Then there exists R>ρ>0 such that if uQ=( B ¯ R E ){re|0<r<R}, then I(u) | Q 0.

(iii) Let us choose u 0 EQ. By (ii), I( u 0 )0.

(iv) S and Q link at the set {e}. □

Proof of Theorem 1.1 By Lemma 3.1, I(u) is continuous and Fréchet differentiable in E. By Lemma 3.3, the functional I satisfies the (PS) condition. We note that I(0)=0. By Lemma 3.4, there are constants ρ>0, α>0 and a bounded neighborhood B ρ of 0 such that I | B ρ E + α, and there are e B 1 E + and R>ρ such that if Q=( B ¯ R E ){re|0<r<R}. Let us set

Γ= { γ C ( Q ¯ , H ) | γ = id  on  Q } .

By Theorem 3.1, I possesses a critical value cα. Moreover, c can be characterized as

c= inf γ Γ max u Q I ( γ ( u ) ) .

Thus we prove that I has at least one nontrivial critical point, we denote by u ˜ a critical point of I such that I( u ˜ )=c. We claim that c is bounded. In fact, we have

c max 0 τ 1 I(τ u 0 ),

and by (g3),

I ( τ u 0 ) = τ 2 ( 1 2 P + u 0 2 1 2 P u 0 2 ) Ω G ( x , t , τ u 0 ) d x d t τ 2 u 0 2 Ω G ( x , t , τ u 0 ) d x d t C 1 τ 2 u 0 2 + C 1

for some constant C 1 >0. Since 0τ1, c is bounded:

c< C ˜ .
(3.9)

We claim that u ˜ is bounded. In fact, by contradiction, u ˜ t t + u ˜ x x =g(x,t, u ˜ ) and, for any K>0, max Ω | u ˜ (x,t)|>K imply that

c=I( u ˜ )= 1 2 ( P + u ˜ 2 P u ˜ 2 ) Ω G(x,t, u ˜ )dxdt

is not bounded, which is absurd to the fact that c=I( u ˜ ) is bounded. Thus u ˜ is bounded, so (1.1) has at least one bounded weak solution, and hence we prove Theorem 1.1. □

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Acknowledgements

This work (Choi) was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (KRF-2013010343).

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Correspondence to Q-Heung Choi.

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Keywords

  • wave equation
  • critical point theory
  • invariant function
  • invariant subspace
  • ( P . S . ) c condition
  • eigenvalue problem