Electronic Journal of Qualitative Theory of Differential Equations 2012, No. 7, 1-13; http://www.math.u-szeged.hu/ejqtde/

Electronic Journal of Qualitative Theory of Differential Equations 2012, No. 7, 1-13; http://www.math.u-szeged.hu/ejqtde/

Observation problems posed for the Klein-Gordon equation

ANDRÁS SZIJÁRTÓ 1

, JENŽHEGED–S

(SZTE,Bolyai Institute,Szeged, Hungary)

szijartomath.u-szeged.hu, hegedusjmath.u-szeged.hu

Abstrat.Transversal vibrations

u = u(x, t)

^{ofa stringof length}

l

^{withxed ends}

are onsidered, where

u

^{is governed by the} Klein-Gordonequation

u tt (x, t) = a ² u xx (x, t) + cu(x, t), (x, t) ∈ [0, l] × R , a > 0, c < 0.

Suient onditions are obtained that guarantee the solvability of eah of four

observation problems with given state funtions

f, g

^{at two distint time instants}

−∞ < t 1 < t 2 < ∞

^{. The essential onditions are the following: smoothness of}

f, g

as elements of a orresponding subspae

D ^s+i (0, l)

^{(introdued in [2℄) of a Sobolev}

spae

H ^s+i (0, l)

^,where

i = 1, 2

^{dependingonthe typeofthe} observation problem,and the representability of

t 2 − t 1

^{as a rational multiple of}

2l

a

^{. The} reonstrution of the unknown initialdata

(u(x, 0), u t (x, 0))

^{astheelementsof}

D ^s+1 (0, l) × D ^s (0, l)

^{are given}

by means of the method of Fourier expansions.

2010 AMS Subjet Classiations: Primary 35Q93, 81Q05;Seondary 35L05,

35R30,42A20

Key words: Observation problems, Klein-Gordonequation, generalizedsolutions,

method of Fourierexpansions.

1. BACKGROUND AND KNOWN RESULTS

In ontrol theory - whih is losely related to the subjet of this paper - numer-

ous monographies and artiles dealt with the aessability of a nal state (position

and speed) of osillations (in partiular string osillations) in the time interval

0 ≤ t ≤ T < ∞

^{; see for example, [1℄ - [10℄. Although, only the short ommunia-}

tion [11℄ dealt with observability of the string osillations on the interval

0 ≤ x ≤ l

^,

and ittreated just the ase whenthe observation instants

t 1

^and

t 2

^{are small, namely}

0 ≤ t ₁ ≤ t ₂ ≤ 2l

a

^{, where}

a

^{is the speed of the wave} propagation. Furthermore, it is assumed in [11℄ that the initial data are known on some subinterval

[h ₁ , h ₂ ] ⊂ [0, l]

^.

We reonstruate the initial data in eah of the four observation problems related to

the Klein-Gordon equation for arbitrarylarge

t 1

^and

t 2

^{. Our} preassumptions are only that

(t 2 − t 1 ) a

2l

^{isrationaland thegiven statefuntions aresmooth enough. Theases}

f, g ∈ D ^s

^{with arbitrary}

s ∈ R

are also admitted.

1

Correspondingauthor

Let

Ω = {(x, t) : 0 < x < l, t ∈ R }

^{. Consider the problem(atrst inthe lassial}

sense)ofthe vibrating

[0, l]

^{stringwith xedends whenthereisanelasti} withdrawing fore proportional to the transversal deetion

u(x, t)

^{of the point}

x

^{of the string at}

theinstantdenoted by

t

^{.Thisphenomenon isdesribedbythe} Klein-Gordonequation asfollows:

(1) u tt (x, t) = a ² u xx (x, t) + cu(x, t), (x, t) ∈ Ω, a, c ∈ R , 0 < a, 0 > c,

with the initialonditions

(2) u(x, 0) = ϕ(x), u t (x, 0) = ψ(x), 0 ≤ x ≤ l,

and the homogeneousboundaryonditions of the rst kind

(3) u(0, t) = 0, u(l, t) = 0, t ∈ R .

We reall, that the funtion

u

^{is said to be a lassial solution of this problem, if}

u ∈ C ² (Ω)

^{and onditions}

(1) − (3)

^{are satised.}

It is wellknown that if

(4) ϕ ∈ C ² [0, l], ψ ∈ C ¹ [0, l]

^and

ϕ(0) = ϕ(l) = ϕ ^′′ (0) = ϕ ^′′ (l) = ψ(0) = ψ(l) = 0,

then the Fourier method gives the lassial solution

u

^{of the problem}

(1) − (3)

^posed

forthe Klein-Gordonequation, whih isof the following form:

(5) u(x, t) =

∞

X

n=1

[α n cos (tω n ) + β n sin (tω n )] sin( nπ

l x), (x, t) ∈ Ω,

where

(6) ω n =

r ( nπ

l a) ² − c, n ∈ N ,

(7) ϕ(x) = u(x, 0) =

∞

X

n=1

α n sin( nπ

l x) ⇒ α n = 2 l

Z l

0

ϕ(x) sin( nπ

l x)dx, n ∈ N ,

(8) ψ(x) = u t (x, 0) =

∞

X

n=1

ω n β n sin( nπ

l x) ⇒ β n = 1 ω n

2 l

Z l

0

ψ(x) sin( nπ

l x)dx, n ∈ N .

The uniqueness of the solution isa onsequene of the lawof onservation of energy.

Tohaveawiderlass offuntionsfor

ϕ, ψ

^and

f, g

^{,weshallonsider ertaingener-}

alizedsolutionsoftheproblem

(1)−(3)

^{.Namely,byusingthe}suggestionsofthereferee,

we introdue the spaes

D ^s (0, l)

^,

s ∈ R

mentioned in the abstrat (see [2℄). Given an arbitrary real number

s

^{, on the linear span}

D

^{of the funtions}

sin ^nπ _l x

^,

n = 1, 2, ...,

onsider the followingEulidean norm:

∞

X

n=1

c n sin( nπ l x)

s

:=

∞

X

n=1

n ^2s |c n | ²

! ¹ ₂ .

Completing

D

^{withrespettothisnorm,weobtainaHilbertspae}

D ^s

^{.Oneanreadily}

verify that for

s ≥ 0

^,

D ^s

^{is a losedsubspae of the Sobolevspae}

H ^s (0, l)

^{, namely}

D ^s = {u ∈ H ^s (0, l) : u ⁽²ⁱ⁾ (0) = u ⁽²ⁱ⁾ (l) = 0, i = 0, 1, ..., [(s − 1)/2]}.

If we identify

D ⁰ = L ² (0, l)

^{with its dual, then}

D ^−s

^{is the dual spae of}

D ^s

^{. Some}

of the results of [2℄ (see Setion 1.1-1.3) and [10℄ say that for arbitrary

s ∈ R

with

(ϕ, ψ) ∈ D ^s+1 × D ^s

^the generalized mixed problem

(1) − (3)

^{has a unique solution}

u

satisfying

u ∈ C( R , D ^s+1 ) ∩ C ¹ ( R , D ^s ) ∩ C ² ( R , D ^s−1 )

givenbytheFourierseries

(5)

^withoeients

α n , β n

^denedby

(7)

^and

(8)

^.Hereand

belowallFourierexpansions for

ϕ, ψ, f

^,

g

^and

u

^{are understood inthe spaes}

D ^s (0, l)

^.

2. NEW RESULTS

Denition 1. The observation problem posed for the Klein-Gordon equation is the

following. The initial funtions

ϕ

^,

ψ

^{are unknown, but suh funtions}

f(x)

^and

g(x)

are given for whih one of the followingfour onditions holds:

(9) u(x, t 1 ) = f(x), u(x, t 2 ) = g(x), 0 ≤ x ≤ l;

(10) u t (x, t 1 ) = f (x), u(x, t 2 ) = g(x), 0 ≤ x ≤ l;

(11) u(x, t 1 ) = f (x), u t (x, t 2 ) = g(x), 0 ≤ x ≤ l;

(12) u t (x, t ₁ ) = f(x), u t (x, t ₂ ) = g(x), 0 ≤ x ≤ l.

Here

u

^{is the solution of the} generalized problem

(1) − (3)

^{, and the given funtions}

f, g

^{are said to be the partial state of the string at distint time instants}

t 1

^and

t 2

^,

−∞ < t 1 < t 2 < ∞

^{. Now the problem is to nd the initialfuntions}

ϕ

^,

ψ

^{interms of}

f(x)

^,

g (x)

^.

(13) f ∈ D ^s+2 , g ∈ D ^s+2 , where s ∈ R ,

(14) t ₂ − t ₁ = p

q 2l

a ,

where

p, q

^{are positive integers and they are relative primes. Inaddition, suppose that}

(15) sin

(t 2 − t 1 ) r

( nπ

l a) ² − c

6= 0, ∀n ∈ N .

Then the observation problem

(1) − (3)

^{under ondition}

(9)

^{has a unique solution for}

(ϕ, ψ) ∈ D ^s+1 × D ^s

^{. They are} represented by their Fourier expansions in the proof below.

Theorem 2. Suppose that

(16) f ∈ D ^s+1 , g ∈ D ^s+2 , where s ∈ R ,

ondition

(14)

^{holds and}

(17) cos

(t 2 − t 1 ) r

( nπ

l a) ² − c

6= 0, ∀n ∈ N .

Thenthe observation problem

(1) − (3)

^{underondition}

(10)

^{has a unique solution for}

(ϕ, ψ) ∈ D ^s+1 × D ^s

^{. They are} represented by their Fourier expansions in the proof below.

Theorem 3. Suppose that

(18) f ∈ D ^s+2 , g ∈ D ^s+1 , where s ∈ R ,

and onditions

(14)

^and

(17)

^{hold. Thenthe} observation problem

(1) − (3)

^{under on-}

dition

(11)

^{has a uniquesolution for}

(ϕ, ψ) ∈ D ^s+1 × D ^s

^{. They are} represented bytheir Fourier expansionsin the proof below.

Theorem 4. Suppose that

(19) f ∈ D ^s+1 , g ∈ D ^s+1 , where s ∈ R ,

and onditions

(14)

^and

(15)

^{hold. Thenthe} observation problem

(1) − (3)

^{under on-}

dition

(12)

^{has a uniquesolution for}

(ϕ, ψ) ∈ D ^s+1 × D ^s

^{. They are} represented bytheir Fourier expansionsin the proof below.

Lemma 1. If ondition

(14)

^{holds, then there exist}

N ∈ N

and

m ∈ R

suh that

1

|sin(ω n (t ₂ − t ₁ ))| < n

m , ∀n > N.

Proof. First,we deal with the denominator of the left-hand side of the inequality

(20) sin(ω n (t 2 − t 1 )) = sin

(t 2 − t 1 ) nπ

l a + (t 2 − t 1 ) h

ω n − nπ l a i

=

= sin

(t 2 − t 1 ) nπ

l a + (t 2 − t 1 ) ω _n ² − ( ^nπ _l a) ² ω n + ^nπ _l a

= sin

(t 2 − t 1 ) nπ

l a + (t 2 − t 1 ) −c ω n + ^nπ _l a

.

It follows from the ondition

(14)

^that

(t 2 − t 1 ) nπ

l a = p q 2nπ,

and that ittakeson atmost

q

^{dierent values}

(

^mod

2π)

^as

n

^{varies. Let}

z n := (t 2 − t 1 ) nπ

l a

^and

d 1 := min

n, sin z n 6=0 {|sin (z n )|}.

Dueto the absolutevalue bars, there isa real number

d 2

^{suh that}

sin(d 2 ) = d 1 , 0 < d 2 ≤ π 2 .

It iseasy tosee, that

(t ₂ − t ₁ ) −c

ω n + ^nπ _l a = O( 1

n )

^as

n → ∞.

Therefore, there existonstants

N ∈ N

,

m ∈ R ⁺

suh that

(21) πm

2n <

(t 2 − t 1 ) −c ω n + ^nπ _l a

< d 2

2

^and

m

n < sin d 2

2

, ∀n > N.

So, if

sin

(t 2 − t 1 ) nπ l a

6= 0

^{, then}

sin

(t ₂ − t ₁ ) nπ

l a + (t ₂ − t ₁ ) −c ω n + ^nπ _l a

>

sin

d ₂ − d ₂ 2

= sin d ₂

2

> m n ,

whenever

n > N

^{,by virtue of}

(21)

^.

On the other hand, if

sin

(t 2 − t 1 ) nπ l a

= 0

^{, then}

sin

(t 2 − t 1 ) nπ

l a + (t 2 − t 1 ) −c ω n + ^nπ _l a

=

sin

(t 2 − t 1 ) −c ω n + ^nπ _l a

>

> 2 π

(t 2 − t 1 ) −c ω n + ^nπ _l a

> m

n , ∀n > N,

due to

(21)

^{and the inequality}

(22) |sin t| > 2

π |t| ,

^if

0 < |t| < π 2 .

Combining the two ases just above, weget that

sin

(t 2 − t 1 ) nπ

l a + (t 2 − t 1 ) −c ω n + ^nπ _l a

> m

n , ∀n > N.

Lemma 2. If ondition

(14)

^{holds, then there exist}

N ∈ N

and

m ∈ R

suh that

1

|cos(ω n (t 2 − t 1 ))| < n

m , ∀n > N.

Proof. Similarlyto

(20)

^{inthe proof of Lemma 1,now we obtainthat}

cos(ω n (t 2 − t 1 )) = cos

(t 2 − t 1 ) nπ

l a + (t 2 − t 1 ) −c ω n + ^nπ _l a

.

Let

z n := (t 2 − t 1 ) nπ

l a

^and

d 1 := min

n, cos z n 6=0 {|cos z n |}.

Dueto the absolutevalue bars, there isa real number

d 2

^{suh that}

cos(d 2 ) = d 1 , 0 ≤ d 2 < π 2 .

Similarly to

(21)

^{in the proof of Lemma 1, there exist onstants}

N ∈ N

and

m ∈ R ⁺

suh that

(23) πm 2n <

(t 2 − t 1 ) −c ω n + ^nπ _l a

<

π 2 − d 2

2

^and

m n < cos

π 2 + d 2

2

, ∀n > N.

In this manner, if

cos

(t 2 − t 1 ) nπ l a

6= 0

^{, we obtainagain that}

cos

(t 2 − t 1 ) nπ

l a + (t 2 − t 1 ) −c ω n + ^nπ _l a

>

cos

d 2 +

π 2 − d 2

2

= cos π

2 + d 2

2

> m n ,

whenever

n > N

^{,by virtue of}

(23)

^.

On the other hand, inthe ase when

cos

(t 2 − t 1 ) nπ l a

= 0

^{, we get}

cos

(t 2 − t 1 ) nπ

l a + (t 2 − t 1 ) −c ω n + ^nπ _l a

=

sin

t 2 − t 1 ) −c ω n + ^nπ _l a

>

> 2 π

(t 2 − t 1 ) −c ω n + ^nπ _l a

> m

n , ∀n > N,

due to

(22)

^and

(23)

^.

cos

(t ₂ − t ₁ ) nπ

l a + (t ₂ − t ₁ ) −c ω n + ^nπ _l a

> m

n , ∀n > N.

4.PROOFSOF THE THEOREMS

1 − 4

Proof of Theorem1. Sineanyofthesolutions

u

^ofproblem

(1)−(3)

^hasrepresentation

(5)

^{with some oeients}

α n , β n ; n ∈ N

, the observation problem an be redued to the problemof the appropriatehoies of

α n

^and

β n

^suhthat

(9)

^{is satised. Forthis}

reason,we substitute

t 1

^and

t 2

^into

(5)

^{, and use the two onditionsin}

(9)

^{. Asa result,}

weget the followingneessary onditions for

α n , β n

^:

(24) f(x) = u(x, t 1 ) =

∞

X

n=1

[α n cos(ω n t 1 ) + β n sin(ω n t 1 )] sin( nπ

l x), x ∈ [0, l],

(25) g (x) = u(x, t 2 ) =

∞

X

n=1

[α n cos(ω n t 2 ) + β n sin(ω n t 2 )] sin( nπ

l x), x ∈ [0, l],

where

ω n

^{isdened in}

(6)

^.

The assumption

(13)

^{guarantees that the oeients of the sine Fourier expansions of}

the funtions

f(x), g(x)

^are unambiguously determined and omparing these Fourier series with

(24)

^and

(25)

^{, for}

α n , β n

^{we get the followingonditions:}

(26)

α n cos(ω n t 1 ) + β n sin(ω n t 1 ) = 2 l

Z l

0

f(x) sin( nπ

l x)dx, n ∈ N , α n cos(ω n t 2 ) + β n sin(ω n t 2 ) = 2

l Z l

0

g(x) sin( nπ

l x)dx, n ∈ N .

The linearsystem

(26)

^{an be uniquely solved forthe unknown oeients}

α n

^and

β n

due toassumption

(15)

^:

(27)

α n = sin(ω n t 2 ) ² _l R l

0 f (x) sin( ^nπ _l x)dx − sin(ω n t 1 ) ² _l R l

0 g(x) sin( ^nπ _l x)dx

sin(ω n (t 2 − t 1 )) ,

β n = − cos(ω n t 2 ) ² _l R l

0 f (x) sin( ^nπ _l x)dx + cos(ω n t 1 ) ² _l R l

0 g(x) sin( ^nπ _l x)dx

sin(ω n (t 2 − t 1 )) .

So the unknown initial funtions

ϕ

^and

ψ

^{are uniquely determined and found in the}

form of

(7)

^and

(8)

^{. It remains to show that}

ϕ, ψ

^{are from the lasses}

D ^s+1 , D ^s

^,

respetively, i.e. toshow that the followinginequality holds:

(28) max{kϕk ² _s+1 , kψk ² _s } < ∞.

D n := 2 l

Z l

0

f (x) sin( nπ l x)dx, E n := 2

l Z l

0

g(x) sin( nπ l x)dx.

Sine

(f, g) ∈ D ^s+2 × D ^s+2

^{, we havethe following} inequality:

(29)

∞

X

n=1

n ^2s+4 max{|D n | ² , |E n | ² } < ∞.

Byusing Lemma 1,for every

n > N

^{we get}

|α n | =

sin(ω n t 2 )D n − sin(ω n t 1 )E n

sin(ω n (t 2 − t 1 ))

<

n m D n

+

n m E n

,

|β n | =

− cos(ω n t 2 )D n + cos(ω n t 1 )E n

sin(ω n (t 2 − t 1 ))

<

n m D n

+

n m E n

,

whih meansthat

(30) max{|α n |, |β n |} < c ₁ n max{|D n |, |E n |} n ∈ N ,

with asuitable onstant

c 1

^.

Let

M ≥ 1

^{be aonstant suh that}

ω n < Mn, ∀n ∈ N

. Combining

(29)

^,

(30)

^{and the}

denition of the norm

k.k s

^{we get the desired inequality}

(28)

^:

max{kϕk ² _s+1 , kψk ² _s } = max{

∞

X

n=1

n ^2s+2 |α n | ² ,

∞

X

n=1

n ^2s |ω n β n | ² } ≤

≤

∞

X

n=1

M ² n ^2s+2 max{|α n | ² , |β n | ² } < c ² ₁ M ²

∞

X

n=1

n ^2s+4 max{|D n | ² , |E n | ² } < ∞.

Remark 1. In the lassial ase when the given state funtions are ontinuously dif-

ferentiable, aording to Theorem 1, the initialfuntions are also ontinuously dier-

entiable.More preisely, if

u(x, t 1 ) = f (x) ∈ C ⁴ [0, l], u(x, t 2 ) = g(x) ∈ C ⁴ [0, l], f, g| _0,l = f ^′′ , g ^′′ | _0,l = 0,

then

f, g ∈ D ⁴

^{and the} observation problemhas a unique lassialsolution

u(x, 0) = ϕ(x) ∈ D ³ ⊂ C ² , u t (x, 0) = ψ(x) ∈ D ² ⊂ C ¹ .

Remark2. Takingintoaount

(20)

^,ondition

(15)

^{anbewrittenintothefollowing}

form:

(31) sin ((t 2 − t 1 )ω n ) = sin (t 2 − t 1 ) nπ

l a + (t 2 − t 1 ) −c

p ( ^nπ _l a) ² − c + ^nπ _l a

! 6= 0

forall

n ∈ N

.AnalysingtheproofofLemma1,itiseasytoseethattheaboveondition isertainly satisedfor all

n

^{large enough, say}

n > N

^.

Ifwe want toget aneasily veriableondition insteadof

(15)

^{, whih is not neessary}

then

(32) (t 2 − t 1 ) −c

p ( ^π _l a) ² − c + ^π _l a < π q

is suh a suient ondition. We justify this laim as follows. The rst term in the

argument of the sine funtion in

(31)

^{is either}

0 (

^mod

2π)

^{, or its distane is at least}

π

q

^{fromitszeroes, and the seond term inthe argument of the sine funtionin}

(31)

^is

positiveandmonotone dereasingfuntionof

n

^{.So, ifweassumethat theseond term}

isalready smaller than

π

q

^for

n = 1

^{, whihis atually the ase in}

(32)

^{,then ondition}

(31)

^{issatised for eah}

n ≥ 1

^.

Nevertheless, wean see fromthis simplerondition

(32)

^{,thatifthe parameters}

|c|

and

a

^inequation

(1)

^{aresuhthateither}

c

^issmallor

a

^{isgreatenough, thenondition}

(31)

^{is always satised. Similar} observations an be made in the following Theorems

2 − 4

^.

Proof of Theorem 2. In an analogous way asin the proof of Theorem 1, nowwestart

with the following equalities:

f (x) = u t (x, t ₁ ) =

∞

X

n=1

[−α n ω n sin(ω n t ₁ ) + β n ω n cos(ω n t ₁ )] sin( nπ

l x), x ∈ [0, l],

g(x) = u(x, t 2 ) =

∞

X

n=1

[α n cos(ω n t 2 ) + β n sin(ω n t 2 )] sin( nπ

l x), x ∈ [0, l].

Hene we get the followingneessary onditions for the oeients

α n

^,

β n

^:

−α n ω n sin(ω n t 1 ) + β n ω n cos(ω n t 1 ) = 2 l

Z l

0

f (x) sin( nπ

l x)dx, n ∈ N , α n cos(ω n t 2 ) + β n sin(ω n t 2 ) = 2

l Z l

0

g(x) sin( nπ

l x)dx, n ∈ N .

Thelinear equations justreeived an beuniquely solved for the unknown oeients

α n

^and

β n

^{,due to assumption}

(17)

^:

α n = − sin(ω n t ₂ ) ² _l R l

0 f (x) sin( ^nπ _l x)dx + cos(ω n t ₁ )ω n 2 l

R l

0 g(x) sin( ^nπ _l x)dx

ω n cos(ω n (t 2 − t 1 )) ,

β n = cos(ω n t 2 ) ² _l R l

0 f (x) sin( ^nπ _l x)dx + sin(ω n t 1 )ω n 2 l

R l

0 g(x) sin( ^nπ _l x)dx ω n cos(ω n (t ₂ − t ₁ )) .

So the unknown initial funtions

ϕ

^and

ψ

^{are uniquely determined and found in the}

form of

(7)

^and

(8)

^{. It remains to show that}

ϕ, ψ

^{are from the lasses}

D ^s+1 , D ^s

^,

respetively. To this eet, it isenough toshow that

(28)

^holds.

Again, let

D n := 2 l

Z l

0

f (x) sin( nπ l x)dx, E n := 2

l Z l

0

g(x) sin( nπ l x)dx.

Sine

(f, g) ∈ D ^s+1 × D ^s+2

^{, we havethat the inequality}

(29 ^′ )

^holds:

(29 ^′ )

∞

X

n=1

n ^2s+4 max{| 1

n D n | ² , |E n | ² } < ∞.

Byusing Lemma 2,for every

n > N

^{we have}

|α n | =

− sin(ω n t 2 )D n + cos(ω n t 1 )ω n E n

ω n cos(ω n (t 2 − t 1 ))

<

1 ω n

n m D n

+

n m E n

,

|β n | =

cos(ω n t 2 )D n + sin(ω n t 1 )ω n E n

ω n cos(ω n (t 2 − t 1 ))

<

1 ω n

n m D n

+

n m E n

,

whih meansthat

(30 ^′ ) max{|α n |, |β n |} < c 2 n max{| 1

n D n |, |E n |} n ∈ N ,

with asuitable onstant

c 2

^.

Combining

(29 ^′ )

^,

(30 ^′ )

^{andthedenition ofthenorm}

k.k s

^{wegetthedesiredinequality}

(28)

^:

max{kϕk ² _s+1 , kψk ² _s } = max{

∞

X

n=1

n ^2s+2 |α n | ² ,

∞

X

n=1

n ^2s |ω n β n | ² } ≤

≤

∞

X

n=1

M ² n ^2s+2 max{|α n | ² , |β n | ² } < c ² ₂ M ²

∞

X

n=1

n ^2s+4 max{| 1

n D n | ² , |E n | ² } < ∞.

Proof of Theorem3. Thisproofgoesalongthesamelines asthatofTheorem2,exept

that here wehave tointerhange the roles of the oeients

α n

^and

β n

^.

f (x) = u t (x, t 1 ) =

∞

X

n=1

[−α n ω n sin(ω n t 1 ) + β n ω n cos(ω n t 1 )] sin( nπ

l x), x ∈ [0, l], g(x) = u t (x, t ₂ ) =

∞

X

n=1

[−α n ω n sin(ω n t ₂ ) + β n ω n cos(ω n t ₂ )] sin( nπ

l x), x ∈ [0, l],

whene the neessary onditions for the oeients

α n

^,

β n

^{are the following:}

−α n ω n sin(ω n t 1 ) + β n ω n cos(ω n t 1 ) = 2 l

Z l

0

f (x) sin( nπ

l x)dx, n ∈ N ,

−α n ω n sin(ω n t 2 ) + β n ω n cos(ω n t 2 ) = 2 l

Z l

0

g(x) sin( nπ

l x)dx, n ∈ N .

Thelinear equations justreeived an beuniquely solved for the unknown oeients

α n

^and

β n

^{,due to assumption}

(15)

^:

α n = cos(ω n t 2 ) ² _l R l

0 f(x) sin( ^nπ _l x)dx − cos(ω n t 1 ) ² _l R l

0 g(x) sin( ^nπ _l x)dx ω n sin(ω n (t ₂ − t ₁ )) , β n = sin(ω n t 2 ) ² _l R l

0 f(x) sin( ^nπ _l x)dx − sin(ω n t 1 ) ² _l R l

0 g(x) sin( ^nπ _l x)dx ω n sin(ω n (t 2 − t 1 )) .

So the unknown initial funtions

ϕ

^and

ψ

^{are uniquely determined and found in the}

form of

(7)

^and

(8)

^{. It remains to show that}

ϕ, ψ

^{are from the lasses}

D ^s+1 , D ^s

^,

respetively. To this eet, it isenough toshow that

(28)

^holds.

Again, let

D n := 2 l

Z l

0

f (x) sin( nπ l x)dx, E n := 2

l Z l

0

g(x) sin( nπ l x)dx.

Sine

(f, g) ∈ D ^s+1 × D ^s+1

^{, we havethat the inequality}

(29 ^′′ )

^holds:

(29 ^′′ )

∞

X

n=1

n ^2s+2 max{|D n | ² , |E n | ² } < ∞.

Byusing Lemma 1,for every

n > N

^{we get}

|α n | =

cos(ω n t 2 )D n − cos(ω n t 1 )E n

ω n sin(ω n (t 2 − t 1 ))

<

1 ω n

n m D n

+

1 ω n

n m E n

,

|β n | =

sin(ω n t 2 )D n − sin(ω n t 1 )E n

ω n sin(ω n (t 2 − t 1 ))

<

1 ω n

n m D n

+

1 ω n

n m E n

,

whih meansthat

(30 ^′′ ) max{|α n |, |β n |} < c 4 max{|D n |, |E n |} n ∈ N ,

with asuitable onstant

c 4

^.

Combining

(29 ^′′ )

^,

(30 ^′′ )

^{andthedenitionofthenorm}

k.k s

^{wegetthedesiredinequality}

(28)

^:

max{kϕk ² _s+1 , kψk ² _s } = max{

∞

X

n=1

n ^2s+2 |α n | ² ,

∞

X

n=1

n ^2s |ω n β n | ² } ≤

≤

∞

X

n=1

M ² n ^2s+2 max{|α n | ² , |β n | ² } < c ² ₄ M ²

∞

X

n=1

n ^2s+2 max{|D n | ² , |E n | ² } < ∞.

ACKNOWLEDGEMENTS

We would like to sinerely thank Professor Feren Móriz for his enouragement

andonstanthelponapreviousdraft.Wealsothanktheanonymousreferee forareful

reading of the previous draft and his/her suggestions whih resulted in the present

version of this paper.

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(Reeived Marh30, 2011)