solving the wave equation due to D. Bernoulli

solving the wave equation due to D. Bernoulli

A string has been strained between the points

(

0

,

0

)

and

(

p

,

0

)

of the

x

-axis.  The vibration of the string in the

x

⁢

y

-plane is determined by the one-dimensional

wave equation

∂

2

⁡

u

∂

⁡

t

2

=

c

2

⋅

∂

2

⁡

u

∂

⁡

x

2

(1)

satisfied by the ordinates

u

⁢

(

x

,

t

)

of the points of the string with the abscissa

x

on the time

t

(

≧

0

)

.
The

boundary conditions

are thus

u

⁢

(

0

,

t

)

=

u

⁢

(

p

,

t

)

=

0

.

We suppose also the

initial conditions

u

⁢

(

x

,

0

)

=

f

⁢

(

x

)

,

u

t

′

⁢

(

x

,

0

)

=

g

⁢

(

x

)

which give the initial position of the string and the initial velocity of the points of the string.

For trying to separate the variables, set

u

⁢

(

x

,

t

)

:=

X

⁢

(

x

)

⁢

T

⁢

(

t

)

.

The boundary conditions are then

X

⁢

(

0

)

=

X

⁢

(

p

)

=

0

,  and the

partial differential equation

(1) may be written

c

2

⋅

X

′′

X

=

T

′′

T

.

(2)

This is not possible unless both sides are equal to a same constant

-

k

2

where

k

is positive; we soon justify why the constant must be negative.  Thus (2) splits into two ordinary linear differential equations of second order:

X

′′

=

-

(

k

c

)

2

⁢

X

,

T

′′

=

-

k

2

⁢

T

(3)

The solutions of these are, as is well known,

{

X

=

C

1

⁢

cos

⁡

k

⁢

x

c

+

C

2

⁢

sin

⁡

k

⁢

x

c

T

=

D

1

⁢

cos

⁡

k

⁢

t

+

D

2

⁢

sin

⁡

k

⁢

t

(4)

with integration constants

C

i

and

D

i

.

But if we had set both sides of (2) equal to

+

k

2

, we had got the solution

T

=

D

1

⁢

e

k

⁢

t

+

D

2

⁢

e

-

k

⁢

t

which can not present a vibration.  Equally impossible would be that

k

=

0

.

Now the boundary condition for

X

⁢

(

0

)

shows in (4) that

C

1

=

0

,  and the one for

X

⁢

(

p

)

that

C

2

⁢

sin

⁡

k

⁢

p

c

=

0

.

If one had

C

2

=

0

,  then

X

⁢

(

x

)

were identically 0 which is naturally impossible.  So we must have

sin

⁡

k

⁢

p

c

=

0

,

which implies

k

⁢

p

c

=

n

π

(

n

∈

ℤ

+

)

.

This means that the only suitable values of

k

satisfying the equations (3), the so-called eigenvalues, are

k

=

n

⁢

π

⁢

c

p

(

n

=

1

,

2

,

3

,

…

)

.

So we have infinitely many solutions of (1), the

eigenfunctions

u

=

X

⁢

T

=

C

2

⁢

sin

⁡

n

⁢

π

p

⁢

x

⁢

[

D

1

⁢

cos

⁡

n

⁢

π

⁢

c

p

⁢

t

+

D

2

⁢

sin

⁡

n

⁢

π

⁢

c

p

⁢

t

]

or

u

=

[

A

n

⁢

cos

⁡

n

⁢

π

⁢

c

p

⁢

t

+

B

n

⁢

sin

⁡

n

⁢

π

⁢

c

p

⁢

t

]

⁢

sin

⁡

n

⁢

π

p

⁢

x

(

n

=

1

,

2

,

3

,

…

)

where

A

n

’s and

B

n

’s are for the time being arbitrary constants.  Each of these functions satisfy the boundary conditions.  Because of the linearity of (1), also their sum series

u

⁢

(

x

,

t

)

:=

∑

n

=

1

∞

(

A

n

⁢

cos

⁡

n

⁢

π

⁢

c

p

⁢

t

+

B

n

⁢

sin

⁡

n

⁢

π

⁢

c

p

⁢

t

)

⁢

sin

⁡

n

⁢

π

p

⁢

x

(5)

is a solution of (1), provided it converges.  It fulfils the boundary conditions, too.  In order to also the initial conditions would be fulfilled, one must have

∑

n

=

1

∞

A

n

⁢

sin

⁡

n

⁢

π

p

⁢

x

=

f

⁢

(

x

)

,

∑

n

=

1

∞

B

n

⁢

n

⁢

π

⁢

c

p

⁢

sin

⁡

n

⁢

π

p

⁢

x

=

g

⁢

(

x

)

on the

interval

[

0

,

p

]

.  But the left sides of these equations are the

Fourier sine series

of the functions

f

and

g

, and therefore we obtain the expressions for the

coefficients

:

A

n

=

2

p

⁢

∫

0

p

f

⁢

(

x

)

⁢

sin

⁡

n

⁢

π

⁢

x

p

⁢

d

⁢

x

,

B

n

=

2

n

⁢

π

⁢

c

⁢

∫

0

p

g

⁢

(

x

)

⁢

sin

⁡

n

⁢

π

⁢

x

p

⁢

d

⁢

x

.

References

1

K. Väisälä:

Matematiikka IV

.  Hand-out Nr. 141. Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1967).

Title

solving the wave equation due to D. Bernoulli

Canonical name

SolvingTheWaveEquationDueToDBernoulli

Date of creation

2013-03-22 16:31:41

Last modified on

2013-03-22 16:31:41

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Example

Classification

msc 35L05

Synonym

vibrating string

Related topic

ExampleOfSolvingTheHeatEquation

Related topic

EigenvalueProblem

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