Selçuk J. Appl. Math. Selçuk Journal of Vol. 10. No. 1. pp. 33-44, 2009 Applied Mathematics

The Justification of The Fourier Method For The Equations of Vi-bration With Impulsive Conditions

¸

Serife Faydaoglu

Faculty of Engineering Dokuz Eylul University, 35160, Buca, Izmir, Turkey e-mail: serife.faydaoglu@ deu.edu.tr

Received: March 5, 2008

Summary. This study examines an eigenvalue problem which arises when the Fourier method is applied to a vibration equation. Obtained results are used for constructing a solution of the vibration equation.

Key words: Impulsive boundary value problem, Fourier method, eigenvalue, eigenvector.

2000 Mathematical Subject Classification: 35R12. 1. Introduction

An equation for vibration in a solid 0 ≤ ≤ , composed of a layer 0 ≤
of material in contact with a layer ≤ of another material is given by
(see [2], [8], [9])
(1) ()
2_{( )}
2 =
2_{( )}
2 − ()( ) ( ∈ [0 ) ∪ ( ] 0)

We assume that () and () are real-valued and piecewise continuous functions on [0 ) ∪ ( ].

The function () has the form () =

½

2_{1} _{0 ≤ }
2_{2} _{ ≤ }
where _{1}and _{2} are positive constants.

(2) _{( − 0 ) = ( + 0 ) }( − 0 ) = ( + 0 )

and at the end faces = 0 and = we take the conditions

(3) (0 ) = 0 ( ) = 0

The initial conditions are given by

(4) ( 0) = () ( 0) = 1() ∈ [0 ) ∪ ( ]

The Fourier method is classical technique that is eﬀective in solving several types of partial diﬀerential equations. In many problems, the Fourier method is more convenient and more useful. We need separated solutions that satisfy certain additional conditions, which are suggested by the physics of the problem. They may be in the form of boundary conditions or conditions of boundedness. The conditions (2) represent an impulse phenomenon at = (See [1], [4], [5], [9]). The end conditions in (3) are homogeneous. Unfortunately, the conditions in (4) are non-homogeneous; we must deal with two non-homogeneous boundary conditions. The method of separation of variables (Fourier method) involves superposition of solutions satisfying the homogeneous conditions to obtain a solution that also satisfies a single non-homogeneous boundary condition. Let us describe, formally, a scheme of solution by the Fourier method. We seek a nontrivial solution of equation (1) which has the form

(5) ( ) = (√ + √)() _{ ∈ [0 ) ∪ ( ]}

where is a complex constant, and () is a function not depending on which may depend on and which we desire to be nontrivial. Substituting (5) into (1)-(3), we obtain

(6) _{−}00+ () = () _{ ∈ [0 ) ∪ ( ]}

(7) _{( − 0) = ( + 0) 0( − 0) = 0( + 0)}

Therefore, the function (5) is a nontrivial solution of problem (1)-(3) if and only if is an eigenvalue and () is a corresponding eigenfunction of problem (6)-(8) (see [4]). Where and are positive real numbers.

Let us signify all the eigenvalues of problem (6)-(8) by 1 2 , and the

cor-responding eigenfunctions by 1() 2() . Then by the linearity of problem

(1)-(3) the function (9) ( ) = ∞ X =1 ( p + p )() ∈ [0 ) ∪ ( ]

will also be a solution of problem (1)-(3), where 1 2 and 1 2 are

ar-bitrary constants (not depending on and ). Now we are required to try to choose the constants so that (9) will satisfy the initial condition (4) as

well: (10) ∞ X =1 () = (); ∞ X =1 p () = 1()

Let () be a function given in the interval ∈ [0 )∪( ] and let it be required to expand this in the form

(11)

∞

X

=1

() = ()

Thus, the problem of possibility to expand a given function () in eigenfunc-tions 1() 2() arises. If the solution of equation (6) is denoted by ( )

satisfying the initial conditions

(0 ) = 0 0(0 ) = 1

then the eigenvalues 1 2 can be calculated as the roots of equation ( ) =

0, and ( ) will be an eigenfunction of (6)-(8) corresponding to the

eigen-value . For the eigenfunctions ( ) the ‘orthogonality’ relation

Z 0 ()( )( ) + Z ()( )( ) = 0 ( 6= )

= Z 0 ()2( ) + Z ()2( )

Then the functions

() =

( )

√_{}

( = 1 2 ) form an ‘orthonormal system’ of eigenfunctions:

Z
0
()()() +
Z
()()() =
½
1 =
0 _{ 6= }
The coeﬃcients in (11) are defined by

(12) = Z 0 ()()() + Z ()()()

We have on replacing () by () and 1() in (11) and (12):

(13)
=
R
0
() ()() +
R
() ()()
= √1_{}_{}
(
R
0
()1()() +
R
()1()()
)

The justification of Fourier’s method involves considerable diﬃculties in view of the impulse conditions at x=a. We shall justify Fourier’s method here from the point of view generalized solutions of the vibration equation.

2. Uniformly Convergent Expansions And Justification of The Fourier Method

Let and () where = 1 2 be the eigenvalues and orthonormal

eigen-functions of the problem. Here, we introduce the formal diﬀerential operator and consider the eigenvale problem

= −00+ () = () _{ ∈ [0 ) ∪ ( ] }

(0) = () = 0

By Theorem 6.4 (This Theorem was obtained earlier in [5].) for function () the uniformly convergent expansion formula

(14) () =

∞

X

=1

() ∈ [0 ) ∪ ( ]

holds, where the coeﬃcients ( = 1 2 ) are found by the formula

(15) = Z 0 ()()() + Z ()()()

Consider the formal solution of (1)-(3)

(16) ( ) = ∞ X =1 (cos p + sin √ )() ∈ [0 ) ∪ ( ]

We can prove that the series (16) can be diﬀerentiated term by term twice with respect to and twice with respect to , and the function ( ) defined by (16) satisfies equation (1) and conditions (2), (3), (4).

Let us set = 2. It follows from Theorem 5.6 and (5.15) (see [5], [6]) that, as

→ ∞
=
∙
1 + (1
)
¸
(17) =
2_{}2
2
∙
1 + (1
)
¸

Therefore, we obtain from Theorem 5.3 (see [5]),
( ) =
_{1}
_{1}
+ (
1
2) ∈ [0 )
( ) =
2_{1}_{2}[
1()
−
2()
] + (
1
2) ∈ ( ]

Hence for (see [5]) = Z 0 ()2( ) + Z ()2( )

We find the asymptotic formula =

1

2+ (

1

3) as → ∞

where 1is a positive constant. Consequently, for

() =
( )
√_{}
we obtain
(18) _{|}()| ≤ ∈ [0 ) ∪ ( ]

where is a positive constant not depending on and . Further, from (13) we have

=_{}1_{}
R
0
() [()] +_{}_{}
R
() [()]
= 1
32_{}
(
R
0
1() [()] +
R
1() [()]
)

Hence, integrating by parts and using the boundary and impulse conditions for (), 1() and (), we find the formula

= _{}1_{}{
R
0
[ ()] () +
R
[ ()] ()}
= 1
32_{} {
R
0
[1()] () +
R
[1()] ()}

We get from the last formula, by virtue of (17) and (18),

(19) _{|}| ≤

1

2 || ≤

2

3

where 1 and 2 are positive constants not depending on .

(20) ¯¯_{¯cos}p
¯
¯
¯ ≤ 1 ¯¯¯sinp
¯
¯
¯ ≤ 1
for all and .

It follows from (18), (19), and (20) that series (16) is uniformly convergent for ∈ [0 ] and ∈ [0 ) ∪ ( ] Its sum is therefore a continuous function for these values of arguments.

Consequently, we have ( 0) = lim→+0( ) = ∞ P =1 () = () ∈ [0 ) ∪ ( ] ( 0) = lim→+0( ) = ∞ P =1 √ () = 1() ∈ [0 ) ∪ ( ]

by (14). This proves that the initial condition (4) is fulfilled. The boundary conditions (3) and the first of impulse conditions (2) are fulfilled by virtue of the fact that all the functions () satisfy these conditions. It remains to verify

the second condition of (2) and equation (1) for any . Each term of series (16) satisfies the second condition of (2) and equation (1). Therefore we only have to show that series (16) can be diﬀerentiated term by term twice with respect to and twice with respect to i.e. We split (16) into two series and first consider

(21) ∞ X =1 cos p ()

We only need to show that the series

∞ X =1 p sin p () ∞ X =1 cos p () ∞ X =1 cos p 0() (22) ∞ X =1 cos p 00()

In view of the fact that 1 for all suﬃciently large k, we can claim that

√

for all suﬃciently large k.

If, with certain conditions imposed on () and () we can prove that the series

(23)

∞

X

=1

|()|

is uniformly convergent for ∈ [0 ) ∪ ( ] we can prove all our previous assertions regarding the term by term diﬀerentiation of series (16) (see [5], [10]). This is obvious for series (21) itself, since → ∞, and for the series obtained

by diﬀerentiation with respect to t it follows from the fact that√ for

all suﬃciently large k.

Hence, taking into account (18) and the first of (19) we find that the first and second series of (22) are uniformly convergent for ∈ [0 ) ∪ ( ] We investigate the next series using equation

() = −00() + ()() = ()() ∈ [0 ) ∪ ( ]

Hence

(24) _{}00() = ()() − ()() ∈ [0 ) ∪ ( ]

and therefore by (18),

(25) _{|}_{}00_{()| ≤ |()| }3+ || 4

for all and ∈ [0 ) ∪ ( ].
Since 0_{}(0) = √1_{}
we have from (24),
(26) 0_{}() = _{√}1
+
Z
0
[()() − ()()] ∈ [0 )
(27) _{}0() = _{}0( + 0) +
Z
[()() − ()()] ∈ ( ]

From (26) it follows that

(28) _{|}_{}0_{()| ≤ }5+ || 6 ∈ [0 )

and from (27) we find using 0

( + 0) = 10( − 0) that

(29) _{|}_{}0_{()| ≤ }7+ || 8 x ∈ ( ]

Now from (25), (28), (29) and the first of (19) it follows that the third and fourth series of (22) are uniformly convergent for ∈ [0 ) ∪ ( ]

We now consider the second part of the series of (16)

∞ X =1 sin p () ∈ [0 ) ∪ (]

where the are given by the second of equation (13). As shown above, it is

suﬃcient to prove the uniform convergence of the series

∞

X

=1

|()| ∈ [0 ) ∪ (]

i.e. of the series

∞
X
=1
p
|0()| ∈ [0 ) ∪ (]
where
0_{}=
Z
0
()1()() +
Z
()1()()

This proves that the function ( ) defined by (16) has the relevant partial derivatives and satisfies the second condition of (2) and equation (1) for any . Thus, we have established the existence of solution ( ) of problem (1)-(4) in the form (16). It is also proved that this solution is continuous for any and ∈ [0 ) ∪ ( ] and that given any the solution has a continuous derivatives and and derivatives and such that is continuous and is

integrable on [0 ) ∪ ( ] We shall now prove the uniqueness of this solution under the assumption () ≥ 0

We need to prove that the solution 0( ) of (1)-(3) with the above mentioned

properties, satisfying the homogeneous initial condition

(30) 0( 0) = 0

0( 0)

= 0 ∈ [0 ) ∪ ( ] is identically zero.

We introduce the function (see [10]):

(31) ( ) =

Z

0

0( ) ∈ [0 ) ∪ ( ]

It has continuous derivatives , , , , for the values of the variables

indicated. We write equation (1) for 0( ) and integrate with respect to

over the interval from = 0 to = . Taking the homogeneous boundary conditions and (31) into account, we obtain

()

2_{( )}

2 =

2_{( )}

2 − ()( ) ( ∈ [0 ) ∪ ( ])

we replace by in this equations, multiply both sides by ( ) and integrate with respect to from = 0 to = . Taking the homogeneous boundary conditions and (31) into account we obtain

1
2()
2
( ) =
Z
0
( )
2_{( )}
2 −
1
2()
2_{( )}

We integrate both sides with respect to over [0 ) and change the order of integration in the iterated integral:

1
2
Z
0
()_{}2( ) =
Z
0
{
Z
0
( )
2_{( )}
2 } −
1
2
Z
0
()2( )

We integrate by parts in the inner integral and observe that the term outside the integral vanishes the homogeneous conditions and (31):

1
2
R
0
()2
( ) =
R
0
( + 0 )( + 0 )
−
R
0 {
R
0
( )()_{} } −1_{2}
R
0
()2( )

Again changing the order of integration, performing the integration with respect to and observing that ( 0) = 0, we get:

(32) 1 2 R 0 ()2 ( ) = R 0 ( + 0 )( + 0 ) −12 R 0 2( ) −12 R 0 ()2( ) Similarly we get (33) 1 2 R ()2( ) = − R 0 ( + 0 )( + 0 ) −12 R 2( ) −12 R ()2( )

Now we multiply (33) by and add the result to (32). We obtain

(34) 1 2{ R 0 ()2 ( ) + R ()2 ( )} = −12{ R 0 2 ( ) + R 2 ( )} −12{ R 0 ()2( )+ R ()2( )}

Whence it follows that
1
2{
Z
0
()2_{}( ) +
Z
()2_{}_{( )} ≤ 0}

so that ( ) = 0 for ∈ [0 ) ∪ ( ] and −∞ ∞. Taking (31) into

account, we find that 0( ) = 0, which is the desired result.

3. Conclusions

Impulsive diﬀerential equations have become more important in recent years in some mathematical models of real world phenomena, especially in the biological or medical domain (see [1], [3], [7], [11]). This paper is an extension to impulsive diﬀerential equations of the results [5].

The Fourier method was applied to solve a boundary value problem with impulse [5]. An expansion formula in the eigenfunctions was established. Here, obtained

results were used for constructing a solution of the vibration equation for two-layered composite. Note that, such problems were examined in the case of absence of the impulse in literature (see [10]). In this paper, the problem in the case of the impulse phenomenon is investigated.

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