separation of variables method of separation of variables is one of the most widely used techniques...

Separation of VariablesMethod of separation of variables is one of the most widely used techniques to solve PDE. It is based on the assumption that the solution of the equation is separable, that is, the final solution can be represented as a product of several functions, each of which is only dependent upon a single independent variable.

Ex. String displacement function u(x,t)= X(x)T(t), is a product of two functions X(x) & T(t), where X(x) is a function of only x, not t. On the other hand, T(t) is a function of t, not x.

By substituting the new product solution form into the original PDE one can obtain a set of ordinary differential equations (hopefully), each of which involves only one independent variable.

Example: String Vibrations

2 22 2

2 2

The string vibration problem can be modeled by the one-dimensional

wave equation: , .

The variables T and represent tension and linear density of the

string, respectively. The sol

u u Tc c

t x

ution, u(x,t), is the deflection of the string.

Since the string is fixed at both ends, the boundary conditions are:

u(x=0,t)=u(x=L,t)=0 for all t.

To solve the wave equation, method of the separation of

variables

is used by assuming the solution can be written in this form:

, ( ) ( )

where X is a function of x only, and T is a function of t only.

u x t X x T t

2

2

Substitute the solution into the wave equation:

, where ( ) corresponds to differentiation

with respect to either x or t , .

, where must be a constant since the l

XT c X T

dX dTX T

dx dtT X

k kc T X

2

eft-hand

side of the equality is a function of t only and the other side

of the equality is a function of x only.

Therefore, we obtain two ordinary differ

If k i

ential equations:

0, a 0

s

nd T kc T X kX , the general solution of the second

equation is

a positive

: ( )

numberk x k xX x Ae Be

The boundary conditions: (0, ) (0) ( ) 0 and

( , ) ( ) ( ) 0mean that (0) ( ) 0 for all t.

k must

( ) , (0)

be

0

( ) 0

0, and X(x)=0 is a trivial solution.

Therefo re,

k x k x

kL kL

u t X T t

u L t X L T t X X L

X x Ae Be X A B

X L Ae Be

A B

2 2

.

Assuming - 0

( ) cos( ) sin( )

. . ' (0) 0

( ) sin( ) 0 ,

( ) sin sin , set 1 (can I do that

a negative number

?)n n n

k p X p X

X x A px B px

B C s X A

nX L B pL pL n p

Ln x n x

X x B BL L

2

22 2

2

are eigenfunctions of the equation 0,

and , 1,2,.. are eigenvalues of the equation.

The other equation: 0

Define = 0, The general solution is

n

n

n n

n

X X p X

np n

L

ncT c p T T T

L

ncT T

L

T

*

*

( ) cos( ) sin( ).

The total solution is ( , ) ( ) ( )

( , ) cos( ) sin( ) sin , 1,2,...

n n n n

n n n n n

t B t B t

u x t X x T t

n xu x t B t B t n

L

*

0

*0 0 0 0

At any time t and space x, the general solution is of the form

( , ) cos( ) sin( ) sin , 1,2,...

At a given time t=t

( , ) cos( ) sin( ) sin ( )sin

n n n n n

n n n n n n

n xu x t B t B t n

L

n x n xu x t B t B t t

L L

1 0 1 0

2 0 2 0

For example, n=1: ( , ) ( )sin

22 : ( , ) ( )sin ,...

xu x t t

L

xn u x t t

L

n=1

x=0 x=L

n=2 n=3

Possible spatial modes of the vibrating string (ch. 11.3)

01

Recall the Fourier analysis: any function f(x) can be represented

by a set of trignometric functions

( ) cos sin

In this case: all cosine terms vanish because it is an o

n nn

n x n xf x a a b

L L

0 0

0 01 1

dd function

( , ) ( )sin

and ( ) ( , ) ( )sin .

As we have learned earlier, the function can be decomposed into

its Fourier components. Similarly, we can say that

n

n nn n

n xu x t t

L

n xf x u x t t

L

the vibrational

motion of a string at a given instant is the combination of many

harmonic motions in x. These harminc components are characterized

nby their corresponding eigenvalues , 1,2,..

Ln

This superposition is possible since the governing equation

(the wave equation) is a linear and homogeneous equation and

the total solution is the sum of all possible solutions

(eigenfunctions as corr

0

* 00

espond to their respective eigenvalues.)

Wave motion at a given position :

( , ) cos( ) sin( ) sin , 1,2,...

It is clear that the string is going through a sinusoidal motion i

n n n n n

x x

n xu x x t B t B t n

L

n

*

1

n

time with characteristic circular frequencies of , 1,2..

These are eigenvalues of the vibrating string in time. is the

corresponding eigenfunctions.

( , ) ( , ) cos( ) sin(

n

n n n n nn

n cn

Lu

u x t u x t B t B

1

) sinn

n xt

L

n

Normal Mode decomposition:

When the string is under a specific excitation that corresponds to one of

its fundamental frequency , one can expect that the spatial

distribution of the string at each

n c

L

th

2

*2 0 0

instant can be represented by the

harmonic component given in the previous slide. This particular motion

is called the n normal mode of the string.

2 cExample: n=2 =

L

( , ) cos( ) sin(n n n nu x t t B t B

0

2) sin . The spatial

distribution of this mode is shown in the following figure:

xt

L

x

2sin 0,at

2

called node where the string

does not move

x Lx

L

node

String vibration setup

Loudspeaker with adjustablefrequency control

Hanging massto adjust tension

Fundamental mode

Second normal modeWith one node

Third normal modewith two nodes

Node points: not moving