vibrating string with variable density

vibrating string with variable density

The unidimensional wave’s problem may be stated as

∂

2

⁡

u

∂

⁡

t

2

-

c

2

⁢

(

x

)

⁢

∂

2

⁡

u

∂

⁡

x

2

=

f

⁢

(

x

,

t

)

,

x

∈

(

0

,

l

)

,

t

>

0

,

with

initial conditions

{

u

⁢

(

x

,

0

)

=

f

⁢

(

x

)

,

∂

⁡

u

∂

⁡

t

⁢

(

x

,

0

)

=

g

⁢

(

x

)

,

and

boundary conditions

{

u

⁢

(

0

,

t

)

=

0

,

u

⁢

(

l

,

t

)

=

0

,

which may be specialized to a string’s motion if we physically interpret

c

2

⁢

(

x

)

=

T

0

/

ρ

⁢

(

x

)

as the ratio between the string’s initial tension and its linear

density

. We will discuss the free string’s vibrations (i.e.

f

⁢

(

x

,

t

)

≡

0

) given by the string’s problem

∂

2

⁡

u

∂

⁡

t

2

-

(

1

+

x

)

2

⁢

∂

2

⁡

u

∂

⁡

x

2

=

0

,

x

∈

(

0

,

1

)

,

t

>

0

,

(1)

initial conditions

{

u

⁢

(

x

,

0

)

=

f

⁢

(

x

)

,

initial string’s form

,

∂

⁡

u

∂

⁡

t

⁢

(

x

,

0

)

=

0

,

string starts from the rest

,

and boundary conditions

{

u

⁢

(

0

,

t

)

=

0

,

string’s left end fixed

,

u

⁢

(

1

,

t

)

=

0

,

string’s right end fixed

.

Without loss of generality, we assume

unitary

the natural undeformed string’s length. The solution of this problem approaches to a string’s motion whose linear density is proportional to

(

1

+

x

)

-

2

. The

method of

separation of variables

(i.e.

u

⁢

(

x

,

t

)

=

X

⁢

(

x

)

⁢

T

⁢

(

t

)

) gives the equations

X

′′

+

λ

(

1

+

x

)

2

⁢

X

=

0

,

(2)

with boundary conditions

X

⁢

(

0

)

=

X

⁢

(

1

)

=

0

, and

T

′′

+

λ

⁢

T

=

0

,

(3)

with initial conditions

T

⁢

(

0

)

=

1

,

T

′

⁢

(

0

)

=

0

. In these equations,

λ

is a constant parameter.

In (2), we are dealing with a Sturm-Liouville problem. To find out the

eigenvalues

, one searches the solution of (2) on the form

X

⁢

(

x

)

=

(

1

+

x

)

a

, as we realize that (2) outcomes the associated

characteristic equation

a

(

a

-

1

)

+

λ

=

0

.

That is

,

a

=

1

2

(

1

±

1

-

4

⁢

λ

)

.

In order to satisfying

X

⁢

(

0

)

=

0

, let us choose

X

⁢

(

x

)

=

(

1

+

x

)

1

2

⁢

(

1

+

1

-

4

⁢

λ

)

-

(

1

+

x

)

1

2

⁢

(

1

-

1

-

4

⁢

λ

)

.

Thus, the boundary condition

X

⁢

(

1

)

=

0

becomes

2

1

2

⁢

(

1

+

1

-

4

⁢

λ

)

-

2

1

2

⁢

(

1

-

1

-

4

⁢

λ

)

=

0

,

or

2

1

-

4

⁢

λ

=

1

.

We next study all the possible cases for the eigenvalue

λ

in the last above equation.

1.

λ

<

1

/

4

. Then

1

-

4

⁢

λ

is real, and the equation does not have solution.

2.

λ

=

1

/

4

. Then the pair of solutions, above indicated, will not be independent. Indeed the functions

(

1

+

x

)

1

/

2

and

(

1

+

x

)

1

/

2

⁢

log

⁡

(

1

+

x

)

are

linearly independent

solutions of (2), in

(

0

,

1

)

. Nevertheless, although the last one satisfies the boundary condition at

x

=

0

, it does not vanish at

x

=

1

. Hence,

λ

=

1

/

4

is not an eigenvalue.

3.

λ

>

1

/

4

. Then

1

-

4

⁢

λ

is

imaginary

. We may even get two solutions by setting (

i

=

-

1

)

Z

⁢

(

x

)

=

X

1

⁢

(

x

)

+

i

⁢

X

2

⁢

(

x

)

=

(

1

+

x

)

1

2

⁢

(

1

+

i

⁢

4

⁢

λ

-

1

)

=

(

1

+

x

)

1

2

⁢

e

1

2

⁢

i

⁢

4

⁢

λ

-

1

⁢

log

⁡

(

1

+

x

)

=

(

1

+

x

)

1

2

⁢

{

cos

⁡

(

λ

-

1

4

⁢

log

⁡

(

1

+

x

)

)

+

i

⁢

sin

⁡

(

λ

-

1

4

⁢

log

⁡

(

1

+

x

)

)

}

,

being the real and

imaginary parts

of

Z

⁢

(

x

)

two linearly independent solutions. For satisfying

X

⁢

(

0

)

=

0

one sets

X

⁢

(

x

)

=

(

1

+

x

)

1

2

⁢

sin

⁡

(

λ

-

1

4

⁢

log

⁡

(

1

+

x

)

)

,

then the boundary condition

X

⁢

(

1

)

=

0

gives

2

1

2

⁢

sin

⁡

(

λ

-

1

4

⁢

log

⁡

2

)

=

0

.

Therefore,

λ

-

1

/

4

⁢

log

⁡

2

must be an

integral

multiple

of

π

, i.e.

λ

-

1

/

4

⁢

log

⁡

2

=

n

⁢

π

, or

λ

↦

λ

n

=

n

2

⁢

π

2

log

2

⁡

2

+

1

4

,

n

∈

ℤ

+

.

To these eigenvalues correspond the

eigenfunctions

X

n

⁢

(

x

)

=

(

1

+

x

)

1

2

⁢

sin

⁡

(

n

⁢

π

⁢

log

⁡

(

1

+

x

)

log

⁡

2

)

,

and these form a

complete system

, whenever we impose to

f

⁢

(

x

)

certain conditions that we shall see later. Moreover,

f

⁢

(

x

)

may be

expanded

in

Fourier series

f

⁢

(

x

)

∼

∑

n

=

1

∞

c

n

⁢

X

n

⁢

(

x

)

,

where

c

n

=

∫

0

1

f

⁢

(

x

)

⁢

(

1

+

x

)

-

3

2

⁢

sin

⁡

(

n

⁢

π

⁢

log

⁡

(

1

+

x

)

log

⁡

2

)

⁢

𝑑

x

∫

0

1

(

1

+

x

)

-

1

⁢

sin

2

⁡

(

n

⁢

π

⁢

log

⁡

(

1

+

x

)

log

⁡

2

)

⁢

𝑑

x

=

2

log

⁡

2

⁢

∫

0

1

f

⁢

(

x

)

⁢

(

1

+

x

)

-

3

2

⁢

sin

⁡

(

n

⁢

π

⁢

log

⁡

(

1

+

x

)

log

⁡

2

)

⁢

𝑑

x

.

The completeness above mentioned and the Fourier series

converges absolutely

and uniformly to

f

⁢

(

x

)

in

(

0

,

1

)

, only if

f

⁢

(

x

)

∈

𝒫

⁢

𝒞

1

⁢

(

0

,

1

)

,

f

⁢

(

0

)

=

f

⁢

(

1

)

=

0

and

∫

0

1

f

′

(

x

)

2

d

x

is

finite

.

1

1

A result due to Green-Parseval-Schwarz (GPS) and Bessel’s

inequality

.

On the other hand, for satisfying (3) and its initial conditions, we choose the eigenfunction

T

⁢

(

t

)

↦

T

n

⁢

(

t

)

=

cos

⁡

λ

n

⁢

t

.

Thus, a solution of (1) is given by

u

n

⁢

(

x

,

t

)

=

X

n

⁢

(

x

)

⁢

T

n

⁢

(

t

)

=

(

1

+

x

)

1

2

⁢

sin

⁡

(

n

⁢

π

⁢

log

⁡

(

1

+

x

)

log

⁡

2

)

⁢

cos

⁡

λ

n

⁢

t

,

and the

general solution

of (1) may be determined as a linear (

infinite

)

combination

of these eigenfunctions, that is

u

⁢

(

x

,

t

)

∼

∑

n

=

1

∞

c

n

⁢

u

n

⁢

(

x

,

t

)

=

∑

n

=

1

∞

c

n

⁢

X

n

⁢

(

x

)

⁢

T

n

⁢

(

t

)

.

So that, the string’s problem (1) has the formal solution

u

⁢

(

x

,

t

)

=

∑

n

=

1

∞

c

n

⁢

(

1

+

x

)

1

2

⁢

sin

⁡

(

n

⁢

π

⁢

log

⁡

(

1

+

x

)

log

⁡

2

)

⁢

cos

⁡

(

n

2

⁢

π

2

log

2

⁡

2

+

1

4

⁢

t

)

.

(4)

This series converges uniformly, and hence satisfies the initial and boundary conditions, as the series for

f

⁢

(

x

)

converges uniformly. However, in order to assure

continuous

derivatives

and the

partial differential equation

(1) to be satisfied, we need suppose that the series for

f

′′

⁢

(

x

)

converges uniformly, i.e. we must suppose that

f

⁢

(

x

)

to be

regular

2

2

i.e.

f

⁢

(

x

)

∈

𝒞

2

⁢

(

0

,

1

)

.

, that

f

⁢

(

0

)

=

f

⁢

(

1

)

=

f

′′

⁢

(

0

)

=

f

′′

⁢

(

1

)

=

0

, and that

∫

0

1

f

′′′

(

x

)

2

d

x

to be finite.

3

3

By GPS, again

.

Title

vibrating string with variable density

Canonical name

VibratingStringWithVariableDensity

Date of creation

2013-03-22 17:26:42

Last modified on

2013-03-22 17:26:42

Owner

perucho (2192)

Last modified by

perucho (2192)

Numerical id

9

Author

perucho (2192)

Entry type

Example

Classification

msc 35L05

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