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Chapter 3   Series Solutions of Differential Equations

3.1   Introduction

These are functions defined by series such as
finite series:
infinite series:
We could consider the functions (eg. sinx , cosx , ex , ...) as being defined as the solutions to particular differential equations.

To demonstrate this we will seek series solutions to the linear second order different equation
d2y
dx2
+y=0
we already know thought that the solutions are sinx and cosx however this helps us to understand this technique of using power series to obtain our solutions.

We assume that the solution(s) are given by
y=
¥
å
n=0
anxn
how are we to find the values of an ?

y=a0+a1x+a2x2+a3x3+a4x4+...+anxn
we need to differentate this twice
Þ
d2y
dx2
=0+0+(2.1.a2)+(3.2.a3x)+(4.3.a4x2)+...+((n+2)(n+1)an+2xn
we require
d2y
dx2
+y=
¥
å
n=0
[ (a+2)(n+1)an+2+an ] xn=0

For this to hold for all x the coefficient of xn must be zero for all n .
even n :
odd n :
this can be reduced to
even n :
odd n :
you will notice that the odd values are not related to the even values, we can already see two solutions emerging.

The general pattern for the even values of n is
a2n=
(-1)n
2n!
a0
The general pattern for the odd values of n is
a(2n+1)=
(-1)n
(2n+1)!
a1
Note that the general solutions contain no relation between a0 and a1 .

y=
¥
å
n=0
anxn
Þ y=a0 é
ê
ê
ë		 1-
x2
2!
+
x4
4!
-
x6
6!
+... ù
ú
ú
û		 +a1 é
ê
ê
ë		 x-
x3
3!
+
x5
5!
-
x7
7!
+... ù
ú
ú
û

From the knowledge of our Maclarine Expansion's we deduce
y=a0cosx+a1sinx
where a0 and a1 are arbitrary constants.

Note that we will see later that we will have to consider anxn where n is not a positive integer.

3.2   Some of The Differential Equations Of Mathematical Physics - 2 Dimesions

Examples of these types of equations are
Helmhotz's Equation:
Ñ2u+l u=0
Poussin's Equation:
Ñ2u=f(x,y,z)
Schrödinger's Equation (time independent):
Ñ2+a (E-V(x,y,z)u)=0
Schrödinger's Equation (time dependent):
u/ t=-2/2m2u/ x2+V(x)u
Telegraph Equation:
2u/ x2-2/ t2- u/ t=0
Fokker Planck Equation:
u/ t=a2/ x2+b (x,y)/ x

3.2.1   The Wave Equation

Summary

Ñ2u=
1
c2
2u
t2
recalling that
Ñ2=
2
x2
+
2
y2
+
2
z2
normally we would have to consider in real life applications that u=u(x,y,z,t) however we will really only consider two variable equations, for example
2u
x2
=
1
c2
2u
t2

Deriving - Wave on a String

Lets consider a wave on a string in one dimension

Figure 3.1 - Wave on a String
The transverse force is Tsinq which we can approximate to Tq for small q . We also have tanq= u/ x~q , again for small q .
F (force)~ T
u
x

Figure 3.2 - Considering a Small Section of the String
F(x+d x)=F(x)+
dF
dx
.d x+...
F(x+d x)=F(x)+T
x
æ
ç
ç
è
u
x
ö
÷
÷
ø		 d x+...
F(x+d x)~ F(x)+T
2u
x2
d x
net force=T
2u
x2
d x
the mass of an element of length d x is rd x where r is the linear density of the string.
mass× acceleration=T
2u
x2
d x
rd x×
2u
t2
=T
2u
x2
d x
Þ
2u
x2
=
r
T
2u
t2

An Extra Note on Waves on a String (stretched)


Figure 3.3 - The Tension ( T ) in a String

Figure 3.4 - The Components of Tension in the String
This is true for when qP and qQ are small and also when qP¹qQ

Figure 3.5 - Some on the Tension Components Cancel
this is because TcosqP~ TcosqQ~ T

Figure 3.6 - When the String is Cut
When the string is cut we get an acceleration occuring and so
r dx=mass
2u
t2
=acceleration
Þ T
2u
x2
d x

Deriving - Maxwell's Equation

ÑnE=-
B
t
,  ÑD=r
ÑnH=
D
t
+J,  Ñ.B=0
D=ere0E,  B=µrµ0H
in an non-conductor r=0 and J=0
Ñn(ÑnE)=Ñ(Ñ.E)-Ñ2E-Ñ2E
as Ñ(Ñ.E)=0 we get
-Ñ n æ
ç
ç
è
B
t
ö
÷
÷
ø		 =-µrµ0 æ
ç
ç
è		 Ñn
H
t
ö
÷
÷
ø		 =-µrµ0
(ÑnH)
t
=-µrµ0ere0
2E
t2
=-Ñ2E
Þ Ñ2E=µrµ0ere0
2E
t2
from here we can determine that:
µrµ0ere0=
1
v2

3.2.2   (Heat) Diffusion Equation

Summary

Ñ2u=
1
D
u
t
®
2
x2
=
1
D
u
t

Deriving


Figure 3.7 - Heat Dispersion
The heat contained in the volume d v is
d h=T(r,t)r(r)c(r)d v
where r(r) is the density and c(r) is the heat capacity. Therefore the total heat contained is
H= ó
õó
õó
õ
 


V
T(r,t)r(r)c(r)dv
H
t
=- ó
õó
õ
 


S
q.ds =- ó
õó
õó
õ
 


V
dwqdv
however dwqdv=Ñ.q and also that
H
t
=
t
é
ê
ê
ê
ê
ë		 ó
õó
õó
õ
 


V
T(r,t)r(r)c(r)dv ù
ú
ú
ú
ú
û		 = ó
õó
õó
õ
 


V
r(r)c(r)
T(r,t)
t
dv
ie.
ó
õó
õó
õ
 


V
r(r)c(r)
t
[T(r ,t)]dv=- ó
õó
õó
õ
 


V
Ñ .q dv
or
ó
õó
õó
õ
 


V
æ
ç
ç
è		 r (r)c(r)
T(r,t)
t
+Ñ.q ö
÷
÷
ø		 dv=0
this applies for V and all subvolumes.

We now assume that
q-KÑT
where the minus sign is as heat transfer is from hot to cold and the K is the conductivity of the volume.

so for any subvolume
r (r)c(r)
T
t
+Ñ.(-KÑT)
r (r)c(r)
T
-
KÑ2T=0
Þ Ñ2T=
r c
K
T
t

3.2.3   Laplace's Equation

Summary

Ñ2u=0®
2u
x2
+
2u
y2
=0

Deriving

Thus
Ñ.Ñf =0
Ñ2f =0
2f
x2
+
2f
y2
+
2f
z2
=0
and in a two dimensional system
2f
x2
+
2f
y2
=0


Figure 3.8 - We Need An Initial Condition (for example U(X,0) )

An Example of the boundary conditions we can use, in this case for Laplaces Equation
2u
x2
+
2u
y2
=0

Figure 3.9 - Boundary Conditions for Laplaces Equation
or we could

Figure 3.10 - A Different Boundary Condition
where the normal boundary condition is u(x,y)/ n

3.2.4   A Note About Boundary Conditions and Initial Conditions

The wave and heat equations depend upon time so we expect to have to specify some initial conditions. For the Wave, Heat and Laplaces equations, all of which depend upon space coordinates and so we have to specify conditions of spatial boundary.

Example - A Wave on a String


Figure 3.11 - A Wave on a String
Usually from here we would specify the shape of the sting at t=0
u(x,0)=f(x)  (0£ x£ L)
also we could specify the strings velocity at t=0
u
t
(x,0)=g(x)  (0£ x£ L)
note that if g(x)=0 then the string is stationary at t=0 .

These are conditions of general boundary conditions. For spatial conditions we have
fixed ends:
these are homogenous boundary conditions
sliding fixture:
an alternative type of boundary condition can be the sliding fixture

Figure 3.12 - The Sliding Fixture Boundary Conditions
so the boundary condition becomes
u(0,t)
x
=0
For all these specifications we would need four boundary conditions as we have a pair of second order differential equations. What we choose as the specification is up to us and makes no difference.

Example - The Heat Equation


Figure 3.13 - A Thin Insulated Rod
u(0,t)=U0;  u(L,t)=UL

Other examples of the heat equation and possible boundary conditions are:
Particle/Dye Diffusion Uses Same Heat Equation:

Figure 3.14 - Particle/Dye Diffusion
u(0,t)=0;  u(L,t)=0
these are known as absorbing boundary conditions
Thin Insulated Rod With Insulators at Either End:

Figure 3.15 - Thin Insulated Rod With Insulators at Either End
u
x
=0  at  x=0, x=L
u(0,t)
x
=0;  
u(L,t)
x
=0
there is no heat transfer across the insulator boundaries, there are called reflective boundaries
Partial Boundaries (aka Partial Reflective):

Figure 3.16 - Thin Insulated Rod With Partial Boundaries
a u(L,t)+b
u(L,t)
x
=0  at  x=L

3.3   Method of Seperation of Variables

This can be applied to linear homogenous equations. The technique is quite general, we follow two steps
  1. seperate the variables, this produces a differential equation of some order that needs solving.
  2. apply the boundary conditions

3.3.1   An Example

2u
x2
=
1
c2
2u
t2
where u=u(x,t) . We assume that we can obtain a solutions of the form
u(x,t)=X(x)T(t)

we now subsitute into the partial differential equation
u
x
=
X(x)
x
.T(t)=
dX
dx
.T(t)=X'T
u
t
=X(x)
T
t
=X(x).
dT
dt
=XT'
2u
t2
=X
d2T
dt2
=XT''
so
X''T=
1
c2
XT''

we now divide by u(x,y)=X(x)T(t)
X''T
XT
=
1
c2
XT''
XT
Þ
X''(x)
X(x)
=
1
c2
T''(t)
T(t)
we now have seperated the indepedent variables x and t . So this equation must hold for all values of x and t .

Suppose we choose a particular value of x then the left hand side takes on a value and the right hand side has to equal this value for all t . So for example the left hand side is equal to the right hand side which equals a constant.

More formally if we look at
d
dt
æ
ç
ç
è
X''(x)
X(x)
ö
÷
÷
ø		 =0=
1
c2
d
dt
æ
ç
ç
è
T''(t)
T(t)
ö
÷
÷
ø
and this is true of the vice versa

now we let the left hand side to equal the right hand side to both equal a seperation constant. We will see later that we need to distinguish between the constant being zero, positive or negative. So we choose the constant to be either 0 , -k2 or +k2 .

We now put the seperation constant to equal:
zero:
X''
X
=0=
1
c2
T''
T
Þ X''=
d2X
dx2
=0,  T''=
d2T
dt2
=0
dX
dx
=A,  
dT
dt
=C
Þ X=Ax+B,  T=Ct+D
where A , B , C and D are constants.
u(x,t)=(Ax+B)(Ct+D)
-k2 :
X''
X
=-k2
X''+k2X=0
X=Acos(kx)+Bsin(kx)
this solution looks like a solution to simple harmonic motion so also
1
c2
T''
T
=-k2
T''+k2c2T=0
T=Ccos(kct)+Bsin(kct)
+k2 :
X''
X
=+k2
X''-k2X=0
we now try X=emx
(m2-k2)emx=0
m2=k2
Þ m=± k
X=Aekx+Be-kx
this solution does not look like a wave like solution. so also
1
c2
T''
T
=+k2
T''-k2c2T=0
T=Aekct+Be-kct
however this can be re-written as
coshx=
ex+e-x
2
,  sinhx=
ex-e-x
2
ex=coshx+sinhx,  e-x=coshx-sinhx
by using these we can obtain
X=A'cosh(kx)+B'sinh(kx)

Figure 3.17 - A Plot of X=A'cosh(kx)+B'sinh(kx)
Now we have the following solutions
type 1 ( k=0 ):
u(x,t)=(Ax+B)(Ct+D)
type 2 ( -k2 ):
u(x,t)=(Acos(kx)+Bsin(kx))(Ccos(kct)+Dsin(kct))
type 3 ( +k2 ):
u(x,t)=(Acosh(kx)+Bsinh(kx))(Ccosh(kct)+Dsinh(kct))
Note that to proceed any further we need to apply boundary (or initial) conditions. Also a boundary condition is said to be homogenous if all terms contain u or u/ x , etc. eg. u=0 is a homogenous boundary condition. u/ x=0 is also homogenous. u(x,0)=f(x) is not homogenous, this is known as inhomogenous.

Note we apply the homogenous boundary conditions first.

Figure 3.18 - For a Stretched String
now we try the type one solution
u(x,t)=(Ax+B)(Ct+D)
where (Ax+B)=X(x) so
X(0)=0=A0+B=B Þ B=0
X(L)=0=AL Þ A=0
this is the trivial solution u(x,t)=0 . This solution type is usually trivial however we will see in a later example that it is not always trivial.

now we try the type two solution
u(x,t)=(Acos(kx)+Bsin(kx))(Ccos(kct)+Dsin(kct))
where (Acos(kx)+Bsin(kx))=X(x) so
X(0)=0=A+B0=A Þ A=0
X(L)=Bsin(kL)=0
from the X(L) answer we have two choices
  1. we could set B=0 and so we are back to the trivial solution from the type one solution
  2. make sin(kL)=0 which provides several repeating solutions
kL=np  (n=integer)
k=
np
L
this provides us with a discrete set of values for k , these are called eigenvalues. Thus
X(x)=Bnsin
æ
ç
ç
è
np x
L
ö
÷
÷
ø
the sin component is called an eigenfunction which is associated with the eigenvalue's k=np/L .
u(x,t)=Bnsin
æ
ç
ç
è
np x
L
ö
÷
÷
ø
æ
ç
ç
è		 Cncos
æ
ç
ç
è
np ct
L
ö
÷
÷
ø
+Dnsin
æ
ç
ç
è
np ct
L
ö
÷
÷
ø
ö
÷
÷
ø
Note that we would normally write (eg. for n=1 )
cos
æ
ç
ç
è
p ct
L
ö
÷
÷
ø
=cos
æ
ç
ç
è		 2p
ct
2L
ö
÷
÷
ø
=cos(2pn t)
where n is the frequency:
n=
c
2L
=
c
l
  (where l=2L)

now we try the type three solution
X(x)=u(x,t)=Acosh(kx)+Bsinh(kx)
we get
X(0)=0=1A+B0=A Þ A=0
X(L)=0=Bsinh(kL)
B=0 Þ trivial solution

If we quickly summarise what we have done, for the solutions of
2u
x2
=
1
c2
2u
t2
which satisies the homogenous boundary conditions (which are fixed ends of a string)
u(0,t)=0,  u(L,t)=0
all these combine to produce
u(x,t)=Bnsin
æ
ç
ç
è
np x
L
ö
÷
÷
ø
æ
ç
ç
è		 Cncos
æ
ç
ç
è
np ct
L
ö
÷
÷
ø
+Dnsin
æ
ç
ç
è
np ct
L
ö
÷
÷
ø
ö
÷
÷
ø

we now apply the inital conditions to obtain values for Bn , Cn and Dn .

Figure 3.19 - The Initial Conditions of Our Fixed Ended String
also another initial condition that tells use the string starts off stationary
u(x,0)
t
=0
therefore
u
t
=Bnsin
æ
ç
ç
è
np x
L
ö
÷
÷
ø
é
ê
ê
ë		 Cn æ
ç
ç
è
np
L
ö
÷
÷
ø		 æ
ç
ç
è		 -sin
æ
ç
ç
è
np ct
L
ö
÷
÷
ø
ö
÷
÷
ø		 +Dn æ
ç
ç
è
np c
L
ö
÷
÷
ø		 æ
ç
ç
è		 cos
æ
ç
ç
è
np ct
L
ö
÷
÷
ø
ö
÷
÷
ø		 ù
ú
ú
û		 =0  (at t=0)
0=Bnsin
æ
ç
ç
è
np x
L
ö
÷
÷
ø
é
ê
ê
ë		 Dn æ
ç
ç
è
np c
L
ö
÷
÷
ø		 ù
ú
ú
û		 Dn=0
Hence ( BnCn® Bn )
u(x,t)=Bnsin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
Cncos
æ
ç
ç
è
np ct
L
ö
÷
÷
ø
Þ u(x,t)=Bnsin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
cos
æ
ç
ç
è
np ct
L
ö
÷
÷
ø
how do we find a solution which matches the shape of the string at t=0 when we release it? Suppose we arranged so that at t=0 the string was
u(x,0)=3sin
æ
ç
ç
è
4p x
L
ö
÷
÷
ø
  (some how?)
we now have the solution that takes the form
u(x,t)=Bnsin
æ
ç
ç
è
np x
L
ö
÷
÷
ø
cos
æ
ç
ç
è
np ct
L
ö
÷
÷
ø
By inspection Bn=3 and n=4
u(x,t)=3sin
æ
ç
ç
è
4p x
L
ö
÷
÷
ø
cos
æ
ç
ç
è
4p ct
L
ö
÷
÷
ø
which is valid 0<x<L and t>0

Figure 3.20 - Plots of Solutions in Regional Space
However in practice we are more likely to

Figure 3.21 - Some Possible Practical Situations
note the solution obtained (at t=0 ) is of the form Bnsin(np x/L)

Figure 3-22 - Plausible Solutions
We note that we can can use any linear combination of the form
B1sin
æ
ç
ç
è
p x
L
ö
÷
÷
ø
cos
æ
ç
ç
è
p ct
L
ö
÷
÷
ø
+B2sin
æ
ç
ç
è
2p x
L
ö
÷
÷
ø
cos
æ
ç
ç
è
2p ct
L
ö
÷
÷
ø
+B3sin
æ
ç
ç
è
3p x
L
ö
÷
÷
ø
cos
æ
ç
ç
è
3p ct
L
ö
÷
÷
ø
+...
will satisfy the original differential equation and will also satisify the homeogenous boundary conditions that u=0 at x=0 and x=L . We also note that at t=0
u(x,0)=B1sin
æ
ç
ç
è
p x
L
ö
÷
÷
ø
+B2sin
æ
ç
ç
è
2p x
L
ö
÷
÷
ø
+B3sin
æ
ç
ç
è
3p x
L
ö
÷
÷
ø
+...
Notice how this looks similar to a Fourier Series.

3.4   Reminder about Fourier Series

For a periodic function f(x)=f(x+nx0) where the period is x0 . Any periodic function can be represented by an infinte series of the form
f(x)=
a0
2
+
¥
å
n=1
é
ê
ê
ë		 ancos
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
+bnsin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
ù
ú
ú
û
To obtain the constants an , a0 and bn we do the following
an :
multiply by cos(2p px/x0) and integrate in the range -x0/2 to x0/2

Figure 3.23 - Calculating the Fourier Constants
Noting these results
ó
õ
x0
2


-
x0
2
cos
æ
ç
ç
è
2p px
x0
ö
÷
÷
ø
cos
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
dx=0  (p¹ n)
ó
õ
x0
2


-
x0
2
cos
æ
ç
ç
è
2p px
x0
ö
÷
÷
ø
cos
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
dx=
x0
2
  (p=n)
also that
ó
õ
x0
2


-
x0
2
cos
æ
ç
ç
è
2p px
x0
ö
÷
÷
ø
sin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
dx=0
ó
õ
x0
2


-
x0
2
cos
æ
ç
ç
è
2p px
x0
ö
÷
÷
ø
dx=0
so by doing all the integrations
ó
õ
x0
2


-
x0
2
f(x)cos
æ
ç
ç
è
2p px
x0
ö
÷
÷
ø
=ap
x0
2
so by changing p to n we can say
an=
2
x0
ó
õ
x0
2


-
x0
2
f(x)cos
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
a0 :
from our work calculating an we obtain
a0=
2
x0
ó
õ
x0
2


-
x0
2
f(x)dx
bn :
multiply by sin(2p px/x0) and integrate in the range -x0/2 to x0/2
ó
õ
x0
2


-
x0
2
sin
æ
ç
ç
è
2p px
x0
ö
÷
÷
ø
sin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
dx=0  (p¹ n)
ó
õ
x0
2


-
x0
2
sin
æ
ç
ç
è
2p px
x0
ö
÷
÷
ø
sin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
dx=
x0
2
  (p=n)
also that
ó
õ
x0
2


-
x0
2
sin
æ
ç
ç
è
2p px
x0
ö
÷
÷
ø
dx=0
and we obtain
bn=
2
x0
ó
õ
x0
2


-
x0
2
f(x)sin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
The validities of this technique are
  1. periodic function f(x) is At a discontinuity the fourier series tends to
    f-(x)+f+(x)
    2

    Figure 3.27 - Discontinuity in Function
    At a discontinuity in the gradient

    Figure 3.28 - Discontinuity in Gradient
    At a finite number of discontinuities

    Figure 3.29 - Finite Number of Discontinies
If f(x) is not period but we choose a range (say) -x0/2<x<x0/2 we can construct a Fourier Series the standard way which is valid only for this range.

Figure 3.30 - Part of the Function f(x) as a Fourier Series
The Fourier Series over the complete range of x -¥<x<¥ , represents the functions f(x) in the range -x0/2<x<x0/2 periodically extended.

Special Results are obtained if f(x) is
even:
f(-x)=f(x)

Figure 3.31 - An Even Function
consider
bn=
2
x0
ó
õ
x0
2


-
x0
2
f(x)sin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
dx=0
this is because the function is even so there are no contributing parts of sin
Þ feven(x)even=
a0
2
+
¥
å
n=1
ancos
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
odd:
f(-x)=-f(x)

Figure 3.32 - An Odd Function
consider
an=
2
x0
ó
õ
x0
2


-
x0
2
f(x)cos
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
dx=0
also note that a0=0 too.
Þ fodd(x)=
¥
å
n=1
bnsin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø

Figure 3.33 - Back to Our String Problem
u(x,0)=
2hx
L
  0<x<
L
2
u(x,0)=
2h
L
(L-x)  
L
2
<x<L
this is inhomogeneous

We are now going to use
u(x,t)=Bnsin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
cos
æ
ç
ç
è
p nct
L
ö
÷
÷
ø
so the Fourier Series must only conatin sin terms

Figure 3.34 - The String as a Fourier Series
f(x)=
a0
2
+
¥
å
n=1
é
ê
ê
ë		 ancos
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
+bnsin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
ù
ú
ú
û
we have constructed and odd function from -L<x<L , x0=2L and a0=0 , an=0 .
bn=
2
x0
ó
õ
x0
2


-
x0
2
fodd(x)sin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
dx=2 é
ê
ê
ê
ê
ê
ê
ê
ë
2
x0
ó
õ
x0
2


0
fodd(x)sin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
dx ù
ú
ú
ú
ú
ú
ú
ú
û
the factor of two is there because there are two halves

For this case
bn=
2
x0
ó
õ
x0
2


-
x0
2
f(x)sin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
dx=
2
2L
ó
õ
L


-L
foddf(x)sin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
dx
bn=2 é
ê
ê
ê
ê
ê
ê
ê
ë
1
L
ó
õ
L
2


0
æ
ç
ç
è
2hx
L
ö
÷
÷
ø		 sin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
dx+
1
L
ó
õ
L


L
2
æ
ç
ç
è
2h
L
(L-x) ö
÷
÷
ø		 sin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
dx ù
ú
ú
ú
ú
ú
ú
ú
û
bn= é
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ë		 é
ê
ê
ê
ê
ê
ë
2hx
L
-cos
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
æ
ç
ç
è
p n
L
ö
÷
÷
ø
ù
ú
ú
ú
ú
ú
û
L
2






0
- ó
õ
L
2


0
2h
L
-cos
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
æ
ç
ç
è
p n
L
ö
÷
÷
ø
dx- ó
õ
L


L
2
2h
L
-cos
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
æ
ç
ç
è
p n
L
ö
÷
÷
ø
dx ù
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
û
bn=
2
L
é
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ê
ë
-2h
L
L
2
cos
æ
ç
ç
è
p n
L
L
2
ö
÷
÷
ø
æ
ç
ç
è
p n
L
ö
÷
÷
ø
-0+ é
ê
ê
ê
ê
ê
ê
ê
ê
ë
2h
L
sin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
æ
ç
ç
è
p n
L
ö
÷
÷
ø
2



 
ù
ú
ú
ú
ú
ú
ú
ú
ú
û
L
2









0
+0--
2h
L
L
2
cos
æ
ç
ç
è
p n
L
L
2
ö
÷
÷
ø
æ
ç
ç
è
p n
L
ö
÷
÷
ø
- é
ê
ê
ê
ê
ê
ê
ê
ê
ë
2h
L
sin
æ
ç
ç
è
p nhx
L
ö
÷
÷
ø
æ
ç
ç
è
p n
L
ö
÷
÷
ø
2



 
ù
ú
ú
ú
ú
ú
ú
ú
ú
û
L









L
2
ù
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
ú
û
bn=
2
L
é
ê
ê
ë
2h
L
.2sin
æ
ç
ç
è
np
2
ö
÷
÷
ø
.
L2
n2p2
ù
ú
ú
û
bn=
8h
p2
sin
æ
ç
ç
è
np
2
ö
÷
÷
ø
n2
n sin(np/2)
1 1
2 0
3 -1
4 0
bn=2 é
ê
ê
ê
ê
ê
ê
ê
ë
1
L
ó
õ
L
2


0
æ
ç
ç
è
2hx
L
ö
÷
÷
ø
 



L
sin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
dx+
1
L
ó
õ
L


L
2
æ
ç
ç
è
2h
L
(L-x) ö
÷
÷
ø		 sin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
dx ù
ú
ú
ú
ú
ú
ú
ú
û
Bn=bn
u(x,t)=
8h
p2
¥
å
n=1
1
n2
sin
æ
ç
ç
è
np
2
ö
÷
÷
ø
sin
æ
ç
ç
è
np x
L
ö
÷
÷
ø
cos
æ
ç
ç
è
np ct
L
ö
÷
÷
ø
this is only valid for 0<x<L and for t³ 0

3.5   Wave on a String With Fixed Ends


Figure 3.35 - Wave on a String
u(x,t)=
8h
p2
¥
å
n=1
1
n2
sin
æ
ç
ç
è
np
L
ö
÷
÷
ø
sin
æ
ç
ç
è
p ct
L
ö
÷
÷
ø
cos
æ
ç
ç
è
p nct
L
ö
÷
÷
ø
Þ u(x,t)=
8h
p2
é
ê
ê
ë
1
12
× 1×sin
æ
ç
ç
è
p x
L
ö
÷
÷
ø
cos
æ
ç
ç
è
p ct
L
ö
÷
÷
ø
+0-
1
32
sin
æ
ç
ç
è
3p x
L
ö
÷
÷
ø
cos
æ
ç
ç
è
3p ct
L
ö
÷
÷
ø
+0+
1
52
sin
æ
ç
ç
è
5p x
L
ö
÷
÷
ø
cos
æ
ç
ç
è
5p ct
L
ö
÷
÷
ø
... ù
ú
ú
û
n ct=0 ct=1/2 ct=L
1
3
5
and at the inbetween intervals
ct=L/4 ct=3L/4
We note that in practice the loses are higher for higher order modes (frequencies) therefore this leaves only the fundemental or n=1 mode.

Figure 3.36 - The Fundemental Mode Remains
so why do we get

Figure 3.37 - The Strings Center is Flattened
we recall that
sinA+sinB=2sin
æ
ç
ç
è
A+B
2
ö
÷
÷
ø
cos
æ
ç
ç
è
A+B
2
ö
÷
÷
ø
Þ sinA+sinB=2sinCcosD
where C+D=A and C-D=B
sinCcosD=
1
2
[ sin(C+D)+sin(C-D) ]
Þ sin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
cos
æ
ç
ç
è
p nct
L
ö
÷
÷
ø
=
1
2
é
ê
ê
ë		 sin
æ
ç
ç
è
p n
L
(x+ct) ö
÷
÷
ø
+sin
æ
ç
ç
è
p n
L
(x-ct) ö
÷
÷
ø
ù
ú
ú
û
u(x,t)=
¥
å
n=1
Bnsin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
cos
æ
ç
ç
è
p nct
L
ö
÷
÷
ø
Þ u(x,t)=
1
2
¥
å
n=1
Bnsin
æ
ç
ç
è
p n
L
(x+ct) ö
÷
÷
ø
+
1
2
¥
å
n=1
Bnsin
æ
ç
ç
è
p n
L
(x-ct) ö
÷
÷
ø
Note that the Fourier Series
f(x)=
¥
å
n=1
Bnsin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø

Figure 3.38 - Conversion to a Fourier Series

Figure 3.39 - Summation of Waves

3.6   d'Alemerts Solution to the Wave Equation

D. Alemberts (1717-1783) solution to the wave equation was (which we will be solving)
2u
x2
=
1
c2
2
t2
this is subject to u(x,0)=f(x) which describes the shape and the following equation is the transverse velocity
u
t
|t=0=g(x)
so the general solution we obtain is
u(x,t)=F(x+ct)+G(x-ct)
where F and G are arbitrary functions
u(x,0)=F(x)+G(x)=f(x)
u
t
=cF'(x+ct)-cG'(x-ct)
we now put in that t=0
F'(x)-G'(x)=
1
c
g(x)
and now integrate with respect to x (the contents of the bracket)
ó
õ
x


a
F'(s)ds- ó
õ
x


a
G'(s)ds=
1
c
ó
õ
x


a
g(s)ds
F(x)-F(a)-[G(x)-G(a)]=
1
c
ó
õ
x


a
g(s)ds
Thus we have
F(x)+F(x)=f(x)
F(x)-G(x)=
1
c
ó
õ
x


a
g(s)ds+[F(a)-G(a)]
F(x)=
1
2
f(x)+
1
2c
ó
õ
x


a
g(s)ds+
1
2
[F(a)-G(a)]
G(x)=
1
2
f(x)-
1
2c
ó
õ
x


a
g(s)ds+
1
2
[F(a)-G(a)]
we find that
u(x,t)=F(x+ct)+G(x-ct)
and the general solution is
u(x,t)=
1
2
[f(x+ct)+G(x-ct)]+
1
2
ó
õ
x+ct


x-ct
g(s)ds
where the first term describes the shape and the second describes the velocity

we found that to satisfy the follow we require that f(x) is odd with a period of 2L
u(0,t)=0,  u(L,t)=0
so the general solution gives the vibration of afixed string we took g(s)=0

3.7   Vibration of a Rectangular Drum

2u
x2
+
2u
y2
=
1
c2
2u
t2

Figure 3.40 - A Rectangular Drum
u(x,y,t)=V(x,y)T(t)
u
x
=
V
x
× T,  
2u
x2
=
2V
x2
× T
likewise
2u
y2
=
2V
y2
× T
and
u
t
=VT'
2u
t2
=VT''
we now substitute
æ
ç
ç
è
2V
x2
+
2V
y2
ö
÷
÷
ø		 T=
1
c2
VT''
we divide by u=VT
1
V
æ
ç
ç
è
2V
x2
+
2V
y2
ö
÷
÷
ø		 =
1
c2
×
T''
T
=seperation constant
From experience we need the seperation constant to be -k2 as we require a harmonic time variation for a solution
1
c2
×
T''
T
=-k2
T''+k2c2T=0
T(t)=Ecos(kct)+Fsin(kct)
and for the left hand side
1
V
æ
ç
ç
è
2V
x2
+
2V
y2
ö
÷
÷
ø		 =-k2
2V
x2
+
2V
y2
+k2V=0
this is known as the Helmholtz Equation.

We now assume that V(x,y)=X(x)Y(y) so
X''Y+XY''+k2XY=0
divide by V=XY
X''
X
+
Y''
Y
+k2=0
the variables are seperable so
d
dx
æ
ç
ç
è
X''
X
ö
÷
÷
ø		 +
d
dx
æ
ç
ç
è
Y''
Y
ö
÷
÷
ø		 +
d
dx
(k2)=0
sp we can see that X''/X=constant and Y''/Y=constant so we choose the constants
-kx2-ky2+k2=0
Þ k2=kx2+ky2
X(x)=Acos(kxx)+Bsin(kxx)
Y(y)=Ccos(kyy)+Dsin(kyy)
T(t)=Ecos(kct)+Fsin(kct)
now we apply our homogenous boundary conditions

Figure 3.41 - The Boundary Condition of the Rectangular Drum
U=X(x)Y(y)T(t)
for when
x=0Þ u=0 so X(0)=0
x=aÞ u=0 so X(a)=0
therefore
X(0)=A+0Þ A=0
X(a)=0+Bsin(kxa)Þ A=0
so B=0 which is the trivial solution or the more interesting solution which is
kxa=mpÞ kx=
mp
a
  (m=integer)
likewise for the boundary condition Y(0)=0 and Y(b)=0 so
C=0Þ ky=
np
b
  (n=integer)
U(x,y,t)=Bmsin
æ
ç
ç
è
p mx
a
ö
÷
÷
ø
Dnsin
æ
ç
ç
è
p ny
b
ö
÷
÷
ø
× é
ê
ê
ë		 Ecos
æ
ç
ç
è		 p
m2
a2
+
n2
b2
ct ö
÷
÷
ø
+Fsin
æ
ç
ç
è		 p
m2
a2
+
n2
b2
ct ö
÷
÷
ø
ù
ú
ú
û
If at t=0 we obtain u/ t=0 which implies F=0 so now we have
u(x,y,t)=Bm,nsin
æ
ç
ç
è
p mx
a
ö
÷
÷
ø
sin
æ
ç
ç
è
p ny
b
ö
÷
÷
ø
×cos
æ
ç
ç
è		 p
m2
a2
+
n2
b2
ct ö
÷
÷
ø
where Bm,n=BmDnE . We can now say that T(t) has the frequency of oscillation ( cos(w t) )
2pn =p
m2
a2
+
n2
b2
c
Þ n =
c
2
m2
a2
+
n2
b2

Figure 3.42 - The m=1 and n=1 Solution

Figure 3.43 - The m=1 and n=2 Solution

Figure 3.44 - The m=1 and n=3 Solution

Figure 3.45 - The m=2 and n=1 Solution

Figure 3.46 - The m=3 and n=1 Solution

Figure 3.47 - The m=2 and n=3 Solution
These are normal modes of oscillation. The eigenfunctions from which we can construct a solution to satisfy the inhomogenous boundary conditions at t=0 . This involves so
f(x,y)=
¥
å
n=1
¥
å
m=1
Bm,nsin
æ
ç
ç
è
p mx
a
ö
÷
÷
ø
sin
æ
ç
ç
è
p ny
b
ö
÷
÷
ø
To obtain the Fourier Series conatining only sin terms we need to extend f(x,y) as an odd function into the ranges -a<x<0 and -b<y<0 . So we put
Cm(y)=
¥
å
n=1
Bm,nsin
æ
ç
ç
è
p ny
b
ö
÷
÷
ø
thus we obtain
f(x,y)=
¥
å
m=1
Cm(y)sin
æ
ç
ç
è
p mx
a
ö
÷
÷
ø
Þ Cm(y)=2 é
ê
ê
ê
ê
ë
2
2a
ó
õ
a


0
f(x,y)sin
æ
ç
ç
è
p mx
a
ö
÷
÷
ø
dx ù
ú
ú
ú
ú
û
The factor of 2 is used as the range is half and so needs doubling. Then
Bm,n=2 é
ê
ê
ê
ê
ë
2
2b
ó
õ
b


0
Cm(y)sin
æ
ç
ç
è
p ny
b
ö
÷
÷
ø
dy ù
ú
ú
ú
ú
û
where 2/2b=y0 and so when we combine these we get
Bm,n=
4
ab
ó
õ
b


0
ó
õ
a


0
f(x,y)sin
æ
ç
ç
è
p mx
a
ö
÷
÷
ø
sin
æ
ç
ç
è
p ny
b
ö
÷
÷
ø
dxdy

3.8   The Heat Equation With Inhomogenous Boundary Conditions


Figure 3.48 - A Rod With a Temperature Distribution Across It
2u
x2
=
1
D
u
t
if u(0)=0 and u(L)=0 then the seperable solutions are
u(x,t)=Bnsin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
e
-n2p2Dt
 

we now construct our solution
u(x,t)=
¥
å
n=1
Bnsin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
e
-
n2p2Dt
L2
 
where
Bn=2 é
ê
ê
ê
ê
ë
2
2L
ó
õ
L


0
f(x)sin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
dx ù
ú
ú
ú
ú
û

Figure 3.49 - A Different Set of Boundary Conditions
the inhomogenous boundary conditions are
u(0,t)=U0¹ 0
u(L,t)=UL¹ 0
u(x,t)=v(x)+w(x,t)
where v is the steady state function whilst w is the time varying one.
2u
x2
=
1
D
u
t
2v
x2
+
2w
x2
=0
ie.
2v
x2
=0
2w
x2
=
1
D
w
t
v(x)=Gx+H
the steady state part of the solution is
v(0)=U0=G.0+H  (H=U0)
v(L)=UL=G.L+U0
G=
UL-U0
L
v(x)= æ
ç
ç
è
UL-U0
L
ö
÷
÷
ø		 x+U0

Figure 3.50 - Plot Of The Steady State
we have
w(x,t)=u(x,t)-v(x)
we know that
u(0,t)=U0
u(L,t)=UL
w(0,t)=U0-U0=0
w(L,t)=UL-UL=0
w(x,t) satisifies the differential equation with the homogeneous boundary conditions
w(0,t)=0
w(L,t)=0
but u(x,0)=f(x) so
w(x,0)=u(x,0)-v(x)
Þ w(x,0)=f(x)- æ
ç
ç
è
UL-U0
L
ö
÷
÷
ø		 x-U0
w(x,t)=
¥
å
n=1
Bnsin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
e
-
p2n2Dt
L2
 
where
Bn=2 é
ê
ê
ê
ê
ë
2
2L
ó
õ
L


0
æ
ç
ç
è		 f(x)- æ
ç
ç
è
UL-U0
L
ö
÷
÷
ø		 x-U0 ö
÷
÷
ø		 sin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
dx ù
ú
ú
ú
ú
û

3.9   Summary of Seperation of Variables

solving for example
2u
x2
=
1
D
u
t
  1. u(x,t)=X(x)T(t)
    Þ X''T=
    1
    D
    		XT'
  2. divide by U=XT (seperates the variables)
    Þ
    X''
    X
    		=
    1
    D
    T'
    T
  3. choose the seperation constant ( 0 , ± k2 )
    Þ
    X''
    X
    		=0, ± k2
    Þ
    T'
    T
    		=0, ± k2
  4. look at the physics to help choose your seperation constant (we choose 0 and -k2 )
    -k2® e
    -k2Dt
     
    Þ X=Ax+B
    Þ X=Acos(kx)+Bsin(kx)
  5. apply the homogenous boundary conditions. eg.
    X(0)=0
    X(L)=0
    Þ A=0  so  k=
    p n
    L
    where n is an integer and k is a discrete set of eigenvalues with spacing p/L

    therefore out solution is (the bit after Bn is the eigenfunction)
    u(x,t)=Bnsin
    æ
    ç
    ç
    è
    p nx
    L
    		ö
    ÷
    ÷
    ø
    		e
    -
    p2n2Dt
    L2
     
  6. we construct a linear combination of these eigenfunctions to satisify the inhomogenous boundary conditions. eg. u(x,0)=f(x)
  7. use a half range Fourier Series with f(x) extended as an odd function in the range -L<x<0

    Figure 3.51 - Half Range Fourier Series
    the period of the Fourier Series is 2L and the series represents f(x) extended as an odd function periodically extended for all x
The proceedure works because the function
sin
æ
ç
ç
è
p nx
L
ö
÷
÷
ø
  (n=1®¥ )
are orthogonal over the range -L<x<L

3.10   Orthogonal Functions

A set of functions Un(x) are said to be orthogonal over a range a<x<b if
ó
õ
b


a
Um(x)Un(x)r(x)dx=0  m¹ n (or m=n¹ 0)
where r(x) is the weight function

Note that for sin(2p nx/x0) , etc we have r(x)=2 .

We can now represent f(x) in the range a£ x£ b as a sum of the orhogonal function Un
f(x)=
¥
å
n=0
anUn(x)
for example the Fourier Series
Un®sin
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
, cos
æ
ç
ç
è
2p nx
x0
ö
÷
÷
ø
, 1
This is an example, there are many others such as

3.11   Partial Differential Equations (Other Than The Cartesian Coordinate System)

Ñ2u=
2u
x2
+
2u
y2
+
2u
z2
the above is of the cartesian coordinate system, instead of this we will be looking at

3.11.1   Plane Polar Coordinates

2u
x2
+
2u
y2
® ?
(x,y)® (r,q )
0£ r<¥
-p£q£p

Figure 3.52 - Plane Polar Coordinates
Þ x=rcosq,  y=rsinq
r2=x2+y2,  tanq=
y
x
so we can obtain
u
x
=
u
r
r
x
+
u
q
q
x
u
y
=
u
r
r
y
+
u
q
q
y
so from r2=x2+y2
2r
r
x
=2x
Þ
r
x
=
x
r
=cosq
and also for
2r
r
y
=2y
Þ
r
y
=sinq
and from tanq=y/x
sec2q
q
x
=
-y
x2
q
x
=
-cos2q
x2
y=
-cos2q× rsinq
r2cos2q
Þ
q
x
=
-sinq
r
and the same is true for
sec2q
q
y
=
1
x
q
y
=
cos2q
rcosq
Þ
q
y
=
cosq
r
now substituting into the original equation
u
x
=cosq
u
r
-
sinq
r
u
q
we make this an operator
Þ
x
=cosq
r
-
sinq
r
q
and its a similar process for y
u
y
=sinq
u
r
+
cosq
r
u
q
Þ
y
=sinq
r
+
cosq
r
q
we now need
2u
x2
= æ
ç
ç
è		 cosq
r
-
sinq
r
q
ö
÷
÷
ø		 æ
ç
ç
è		 cosq
u
r
-
sinq
r
u
q
ö
÷
÷
ø
2u
x2
=cosq é
ê
ê
ë		 cosq
2u
r2
-
sinq
r
2u
rq
-
cosq
r
u
q
-
sinq
r
2u
q r
ù
ú
ú
û
Þ
2u
x2
=cos2q
2u
r2
-
2sinqcosq
r
2u
q r
+
2sinqcosq
r2
u
q
+
sin2q
r
u
r
+
sin2q
r2
2u
q2
also you can show that
2u
y2
= æ
ç
ç
è		 sinq
r
+
cosq
r
q
ö
÷
÷
ø		 æ
ç
ç
è		 sinq
u
r
+
cosq
r
u
q
ö
÷
÷
ø
Þ
2u
y2
=sin2q
2u
r2
+
2cosqsinq
r
2u
q r
-
2cosqsinq
r2
u
q
+
cos2q
r
u
r
+
cos2q
r2
2u
q2
so back to our first equation
2u
x2
+
2u
y2
=
1
r
u
r
+
1
r2
2u
q2

3.11.2   Cylindrical Polar Coordinates


Figure 3.53 - Cylindrical Polar Coordinates
Ñ2u=
2u
r2
+
1
r
u
r
+
1
r2
2u
q2
+
2u
z2

3.11.3   Spherical Polar Coordinates


Figure 3.54 - Spherical Polar Coordinates
x,y,z® r,q ,f
z=rcosq
r =rsinq
x=rcosq
y=rsinq
from these results we can obtain
2u
z2
+
2u
y2
=
2u
r2
+
1
r
u
r
+
1
r2
2u
q2
and by replacing x with z and y with r we obtain 2u/ z2+2u/r2 . Now by using our result from plane polar coordinates
y
=sinq
r
+
cosq
r
q
and through similar means like before we obtain
u
r
=sinq
u
r
+
cosq
r
u
q
we also have
x=rcosf,  y=rsinf
and by using 2u/ x2+2u/ y2 we get
2u
x2
+
2u
y2
=
2u
r2
+
1
r
u
r
+
1
r2
2u
q2
if we substitute for u/r
2u
x2
+
2u
y2
=
2u
r2
+
1
rsinq
æ
ç
ç
è		 sinq
u
r
+
cosq
r
u
q
ö
÷
÷
ø		 +
1
r2sin2q
2u
f2
now if we add this to 2u/ z2+2u/f2
Ñ2u=
2u
x2
+
2u
y2
+
2u
z2
Þ Ñ2=
2u
r2
+
2
r
2u
q2
+
cotq
r2
u
q
+
1
r2sin2q
2u
f2
also we can get as an alternative version of Ñ2u
Ñ2u=
1
r2
r
æ
ç
ç
è		 r2
u
r
ö
÷
÷
ø		 +
1
r2sin2q
q
æ
ç
ç
è		 sinq
u
q
ö
÷
÷
ø		 +
1
r2sin2q
2u
f2

3.12   Laplace's Equation in Cyclindrical Polar Coordinates

Ñ2u=0
where u=u(r,f ,z)
Ñ2u=
2u
r2
+
1
r
u
r
+
1
r2
2u
f2
+
2u
z2
=0
We now assume that we can find a solution of the form
u(r,f ,z)=R(r)F (f )Z(z)
R''F Z+
R'
R
F Z+
R
r
F ''Z+RF Z''=0
by dividing by U=RF Z
R''
R
+
R'
rR
+
1
r2
F ''
F
+
Z''
Z
=0
we consider the case of no z variation and we multiply by r2 and so seperate the variables
r2
R''
R
+r
R'
R
+
F ''
F
=0  (r+f =0)
we could choose the seperation constant, 0 and ± k2 , so we look at the physics of the problem and decide we want to use -k2 .
F ''
F
=-k2Þ F ''+k2F =0
F =Acos(kf )+Bsin(kf )
from symmetry
u(r,f ,z)=u(r,f +2p ,z)
so we obtain
F (f )=F (f +2p )
it is because of this cyclic symmetry we choose the seperation constant to be -k2
F =Acos[k(f +2p )]+Bsin[k(f +2p )]=Acos(kf )+Bsin(kf )
Þ k=n  (integer)
now for the r terms
r2
R''
R
+r
R'
R
=n2   (+k2)+(-k2)=0
Þ r2R''+rR'-n2R=0
by substituting R=rm
Þ r2m(m-1)rm-2+rmrm-1-n2rm=0
Þ m(m-1)rm+mrm-n2rm=0
Þ (m(m-1)+m-n2)rm=0
(m2-n2)rm=0
m=± n
R(r)=Crn+
D
rn
we discard 1/rn as it tends to infinity as r® 0
u(r,f )=rn(Acos(nf )+Bsin(nf ))

u(r,f )=rn ( Ancos(nf )+Bnsin(nf ) )
but what about k=0
F ''+kF =0®F ''=0
F '=Cf +D
but we require that the solution is symmetric
F (f)=F (f +2p )® C=0
Þ u(r,f )=A+
¥
å
n=1
rn ( Ancos(nf )+Bnsin(nf ) )


Figure 3.55 - A Split Conductor (aka capacitor)

Figure 3.56 - u(r,f )® u(a,f )=f(f )
As f(f ) is odd when we construct the Fourier Series we get A0=0 and An=0 so we obtain
f(f )=
¥
å
n=1
bnsin(nf )  (standard fourier series)
as the period is 2p , f0=2p
bn=
2
f0
ó
õ
f0
2


-
f0
2
f(f )sin
æ
ç
ç
è
2p nf )
f0
ö
÷
÷
ø
df
bn=2 é
ê
ê
ê
ê
ë
2
2p
ó
õ
p


0
Vsin(nf )df ù
ú
ú
ú
ú
û		 =
2V
p
é
ê
ê
ë
-cos(nf )
n
ù
ú
ú
û
p



0
=
2V
p
é
ê
ê
ë
-cos(np )
n
-
-1
n
ù
ú
ú
û
bn=
2V
p
é
ê
ê
ë
1-(-1)n
n
ù
ú
ú
û
Bnan=bn
Þ u(r,f )=
2V
p
¥
å
n=1
1-(-1)n
n
æ
ç
ç
è
r
a
ö
÷
÷
ø
n



 
sin(nf )

3.12.1   Vibration of a Circular Drum

Ñ2u=
1
c2
2u
t2
we choose our function of u to be u=u(r,f ,t) (in plane polar coordinates)
2u
r2
+
1
r
u
r
+
1
r2
2u
f2
=
1
c2
2u
t2
u(r,f ,t)=R(r)F (f)T(t)
R''F T+
R'
r
F T+
RF ''T
r2
=
1
c2
RF T
now by dividing by U=RF T
R''
R
+
R'
rR
+
F ''
r2F
=
1
c2
T''
T
We know we are looking for a oscilating solution and so we use the seperation constant -k2=1/c2T''/T
T''+k2c2T=0
T(t)=Ecos(kct)+Fsin(kct)
we now notice that w =kc if we are looking of a oscilating solution (of which we are). This leaves
R''
R
+
R'
rR
+
F ''
r2F
+k2=0
now by multiplying r2
r2R''
R
+
rR'
R
+
F ''
F
+r2k2=0
we can now easily seperate the variables, so we put
F ''
F
=-k
2
 
f
F ''+k
2
 
f
F =0
F (f )=Ccos
(k
 
f
F )
+Dsin
(k
 
f
F )
As we require that there is symmetry (ie. F (f )=F (f +2p) )
k
 
f
=n  (integer)
So far our solution is
u(r,f ,t)=R(r) ( Ccos(nf )+Dsin(nf ) ) × ( Ecos(kct)+Fsin(kct) )
r2
R''
R
+r
R'
R
-n2+k2r2=0
r2
d2R
dr2
+r
dR
dr
+(k2r2-n2)R=0

--------

Note that if we differentiated, so
r
dR
dr
®
x
k
.
dR
d æ
ç
ç
è
x
k
ö
÷
÷
ø
=
x
k
dR
1
k
dx
® xy'

--------

where k2 is associated with the variation of time and n2 is the ??germination?? This is Bessel's Equation, however we are used to seeing it in terms of x and y . So if we put R=y and rk=x
Þ x2y''+xy'+(x2-n2)y=0
n is the general Bessel Equation and doesn't have to be an integer, however in our case n =n=integer .

This has solutions (we expect two solutions due to the differential equation being second order)
y=AJ
 
n
(x)+BJ
 
-n
(x)
so when n is equal to an integer (ie. n ) then
J-n(x)=(-1)nJn(x)
so our solutions are linearly dependent and so we only can directly obtain one solution. So to get our second solution we use altering equation
y(x)=AJn(x)+BYn(x)
where AJn(x) is the first kind solution whilst BYn(x) is the second kind solution.


Figure 3.57 - Plots of Our Solutions

For a circular drum of radius a
includegraphics[scale=0.5]figures/eps/3-58.eps
Figure 3.58 - A Circular Drum ( R(a)=0 )
we know that R(r)=Jn(kr)
u=Jn(kr)(Ccos(nf )+Dsin(nf ))(Ecos(kct)+Fsin(kct))
we remember also that kc=w . At r=a
Jn(ka)=0

For example
m=1 m=2 m=3 m=4
J0=0  (n=0) 2.404 5.520 8.654 11.792
J1=0  (n=1) 3.832 7.016 10.173 13.323
Let the zeros of Jn(x) be an,m (eg. a0,1=2.404 , etc) where n is the order of the Bessel Function and m is the number to identify which zero is being referenced to.

We require that
ka=an,m
k=
anm
a
®w=
can,m
a
this is because w=kc .

We now consider that n=0 which implies that there is no angular variation. We recall that our Bessel function is doing and and we note that J0(a0,m) are orthogonal.

Figure 3.59 - Recalling Our Bessel Function
so when m=1

Figure 3.60 - The Drum Solution For When n=0 and m=1
when m=2

Figure 3.61 - The Drum Solution For When n=0 and m=2
when m=3

Figure 3.62 - The Drum Solution For When n=0 and m=3
now we look at n=1 so when m=1

Figure 3.63 - The Drum Solution For When n=1 and m=1
and when m=2

Figure 3.64 - The Drum Solution For When n=1 and m=2
In general we need a mixture of these modes to satisify the initial conditions, we use a double series (using k=an,m/a )
u(r,f ,t)=
¥
å
m=1
¥
å
n=0
Jn æ
ç
ç
è		 an,m
r
a
ö
÷
÷
ø		 (Cn,mcos(nf )+Dn,msin(nf ))(En,mcos(kct)+Fn,msin(kct))
we now use u(r,f ,0 and dU/dT|t=0 to find the constants Cn,m , Dn,m , En,m and Fn,m . This requires we know that
ó
õ
a


0
rJn æ
ç
ç
è		 an,m'
r
a
ö
÷
÷
ø		 Jn æ
ç
ç
è		 an,m
r
a
ö
÷
÷
ø		 dr=0  (m¹ m')
ó
õ
a


0
rJn æ
ç
ç
è		 an,m'
r
a
ö
÷
÷
ø		 Jn æ
ç
ç
è		 an,m
r
a
ö
÷
÷
ø		 dr=
a2
2
[ Jn+1(an,m) ]   (m=m')

3.13   Spherical Polor Coordinates

Lapaces Equation Ñ2u=0 and u=u(r,q ,f )
1
r
r
æ
ç
ç
è		 r
u
r
ö
÷
÷
ø		 +
1
r2sin2q
q
æ
ç
ç
è		 sinq
u
q
ö
÷
÷
ø		 +
1
r2sin2q
2u
f2
=0
in the spherical case where there is no angular variation
1
r2
d
dr
æ
ç
ç
è		 r2
du
dr
ö
÷
÷
ø		 =0
We now integrate with repect to r
r2
du
dr
=0
du
dt
=
A
r2
® U(r)=-
A
r
+B
An example of this is electrical/gravitational potential
V=
q
4pe0r

In the general case
u(r,q ,f )=R(r)Y(q ,f )
1
r2
r
æ
ç
ç
è		 r2
R
r
ö
÷
÷
ø		 Y+
R
r2sinq
q
æ
ç
ç
è		 sinq
Y
q
ö
÷
÷
ø		 +
R
r2sin2q
2Y
q2
=0
Now by dividing by u=RY
1
Ysinq
q
æ
ç
ç
è		 sinq
Y
q
ö
÷
÷
ø		 +
1
Ysin2q
2Y
q2
=-
1
R
r
æ
ç
ç
è		 r2
R
R
ö
÷
÷
ø
we now choose our seperation constant to be -l(l+1) (this is not obvious why however it will prove convenient)
Þ
r
æ
ç
ç
è		 r2
R
r
ö
÷
÷
ø		 =l(l+1)R
r2
2R
r2
+2r
R
r
-l(l+1)R=0
we try R=rm where m=l or m=-l(l+1)
R(r)= é
ê
ê
ë		 Arl+
B
rl+1
ù
ú
ú
û
note that you can obtain a second solution if it tends to infinity when r® 0 .

The angular part becomes
1
sinq
q
æ
ç
ç
è		 sinq
Y
q
ö
÷
÷
ø		 +
1
sin2q
2Y
f2
+l(l+1)Y=0
we now put Y(q ,f )=Q (q)F (f ) then substitute and divide by Y to obtain
1
Q
1
sinq
q
æ
ç
ç
è		 sinq
Q
q
ö
÷
÷
ø		 +
1
sin2q
F ''
F
+l(l+1)=0
and multipy by sin2q
sin
q
 
Q
d
dq
æ
ç
ç
è		 sinq
dQ
dq
ö
÷
÷
ø		 +l(l+1)sin2q=-
F ''
F
we put that F ''/F=-m2
Þ F ''+m2F=0
we note that if m=0 then we get an unsuitable solution ( F =CF +D ) so we ignore it. However when m 0 then we obtain
F =Cme
mf
 
+Dme
- mf
 
as we require symmetry ( F (f )=F (f +2p ) ) we discover that m must be an integer.

we now substiture -m2 into out q equation and divide by sin2q
1
sinq
d
dq
æ
ç
ç
è		 sinq
dQ
dq
ö
÷
÷
ø		 + é
ê
ê
ë		 l(l+1)-
m2
sin2q
ù
ú
ú
û		 Q =0
We put x=cosq therefore dx/dq=-sinq
d
dx
=-
1
sinq
d
dq
dQ
dq
=
dQ
dx
dx
dq
=-sinq
dQ
dx
we note that sinq=1-cos2q=1-x2
d
dx
é
ê
ê
ë		 (1-x2)
dQ
dx
ù
ú
ú
û		 + é
ê
ê
ë		 l(l+1)-
m2
1-x2
ù
ú
ú
û		 Q =0

If m=0 then we obtain the Legendre Equation
(1-x2)
d2Q
dx2
-2x
dQ
dx
+l(l+1)Q =0
and when m 0 then we obtain the Associate Legendre Equation
(1-x2)
d2Q
dx2
-2x
dQ
dx
+ é
ê
ê
ë		 l(l+1)-
m2
1-x2
ù
ú
ú
û		 Q =0

3.13.1   Legendre Equation

The solution for an integer l exists in the range -1£ x£ 1 . The solutions are then polynominals ( Pl(x) ), for example with the Associate Legendre Equation
Plm(x)=(1-x2)
m
2
 
dm
dxm
(Pl(x))
we therefore have for the spherical problem
Yl,m(q ,f )=Plm(cosq )e
mf
 
Yl,-m(q ,f )=Pl-m(cosq )e
- mf
 

So our total solution is
u(r,q ,f )= æ
ç
ç
è		 Arl+
B
rl+1
ö
÷
÷
ø		 ( CYl,m(q ,f )+DYl,-m(q ,f ) )
these are orthogonal eigenfunctions.

We now integrate this from x=a to x=b
ó
õ
b


a
d
dx
[p(unum'-umun')]dx+(lm-ln) ó
õ
b


a
r(x)um(x)un(x)dx
The first half of this integral becomes
p[unum'-umun']ab
This relates to the value of this expression at the ends of the range ( x=a and x=b ). We can choose this part of the integral to equal zero.
p(b)[un(b)um'(b)-um(b)un'(b))]-p(a)[un(a)um'(a)-um(a)un'(a))]=0
If this is satisified then
(lm-ln) ó
õ
b


a
r(x)um(x)un(x)dx=0
we now assume that lmln
Þ ó
õ
b


a
r(x)um(x)un(x)dx=0
this shows that the eigenfunctions um and un are orthogonal (note that this only makes them orthogonal over the range a£ x£ b and subject to the weight function r(x) ). This only works when
p(b)[un(b)um'(b)-um(b)un'(b))]-p(a)[un(a)um'(a)-um(a)un'(a))]=0
There are several ways of ensuring this condition
  1. make um(a)=um(b)=0 for all values of m (and also n ), this is known as a periodic boundary condition
  2. make um'(a)=um'(b)=0 again for all values of m (and n )
  3. make um'(a)+a u(a)=0 and um'(b)+b u(b)=0 for all values of m (and n )
Þ un(a)um'(b)-um(a)un'(a)=un(a)(-a un(a))-um(a)(-a un(a)=0
and this process can be repeated for the b extreme.

Subject to these conditions of the value of u and u' at the end of the range a to b , we can represent any function.
f(x)=
¥
å
n=1
anun(x)
The values of an are derived using the orthogonality relationship. To do this we multiply this equation by um(x)r(x) (must include the weight function) and then we integrate across the range a to b .
ó
õ
b


a
f(x)r(x)um(x)dx=
¥
å
n=1
an ó
õ
b


a
um(x)un(x)r(x)dx
we now look for òabum(x)un(x)r(x)dx=0 for when m n so we get
ó
õ
b


a
f(x)r(x)um(x)dx=an ó
õ
b


a
um(x)un(x)r(x)dx
and from here we can obtain
an=
ó
õ
b


a
f(x)un(x)r(x)dx
ó
õ
b


a
un(x)un(x)r(x)dx
We can arrange that that the functions un are normalised by
ó
õ
b


a
um(x)un(x)r(x)dx=0,1  (m n;m=n)
(see the problem sheet to establish that if um is complex then the eignenvalues lm are real ( lm*=lm )).

3.14   Series Solution of Differential Equations

Note that we already have done this for y''+y=0 .

3.14.1   Legendre's Equation

(1-x2)y''-2xy'+l(l+1)y=0
we expect that something special will occur at x=± 1 as the first term will disappear. We firstly assume that
y=a0+a1x+a2x2+a3x3+... =
¥
å
n=0
anxn
Þ y'=0+a1+2a2x+3a3x2+... =
¥
å
n=1
nanxn-1
Þ y''=0+0+2a2+3.2a3x+... =
¥
å
n=2
n(n-1)anxn-2
we now substitute this into the differential equation.
(1-x2)y''-2xy'+l(l+1)y=0
¥
å
n=2
n(n+1)anxn-2-
¥
å
n=2
n(n-1)anxn-2
¥
å
n=1
nanxn+l(l+1)
¥
å
0
anxn=0
for this to hold for different values of x the coefficients in say xp must be zero so at p=0 where we obtain the x0
Þ 2(2-1)a2+0-0+l(l+1)a0=0
and at p=1 where we obtain the x1
Þ 3(3-1)a3+0-2.1a1+l(l+1)a1=0
and when p³ 2
Þ (p+2)(p+1)ap+2-p(p-1)ap-2pap+l(l+1)ap=0
(p+2)(p+1)ap+2=[p(p-1)+2p-l(l+1)]ap
(p+2)(p+1)ap+2=[p2-l2+p-l]ap
(p+2)(p+1)ap+2=-(l-p)(l+p+1)ap
we now have
a2=-
-l(l+1)
2!
a0
also
a3=-
2+l2+l
3!
a1=-
(l-1)(l+2)
3!
a1
and similarly
a4=-
(l-2)(l+3)
4.3
a2
we now substitute for a2
a4=
(l-2)(l)(l+1)(l+3)
4.3.2.1
a0
and also for a5 we obtain
a5=-
(l-3)(l+4)
5.4
a3=-
(l-3)(l-1)(l+2)(l+4)
5.4.3.2.1
a1
so now we can see a pattern develop
y(x)=a0 é
ê
ê
ë		 1-
l(l+1)
2!
x2+
(l-2)(l)(l+1)(l+3)
4!
x4+... ù
ú
ú
û		 +a1 é
ê
ê
ë		 x-
(l-1)(l+2)
3!
x3+
(l-3)(l-1)(l+2)(l+4)
5!
x5+... ù
ú
ú
û
we write this as
y(x)=a0y1(x)+a1y2(x)
These are linearly independent solutions. We now need to investigate the range of values of x for which these two solutions y1(x) and y2(x) converge.

We use the Ratio Test for any series where for the series å up we look at
 
lim
p®¥
½
½
½
½
up+1
up
½
½
½
½
however we need to consider
 
lim
p®¥
½
½
½
½
up+2
up
½
½
½
½
so we look at
 
lim
p®¥
½
½
½
½
ap+2xp+2
apxp
½
½
½
½		 =
 
lim
p®¥
½
½
½
½
(l-p)(l+p+1)
(p+2)(p+1)
x2 ½
½
½
½		 <1  (to converge)
 
lim
p®¥
® 1x2<1
the series solution is only valid for x2<1 (ie. -1<x<1 ). More sophisticated tests show that the series converge fo x± 1 only if l is an integer. This is required in most physical situations (such as quantum mechanics). Then one solution becomes a polynominal (Legendre Polynominal), which is finite when -1£ x£ 1 . The other series diverges, however it does not match a physical situation.

We now set up the polynominal Pl(x) such that Pl(1)=1 so that when
l=0Þ y1=1,  P0(x)=1
l=1Þ y2=x,  P1(x)=x
l=2Þ y1=1-3x,  P2(x)=
1
2
(3x2-1)
l=3Þ y2=x-
5x2
3
,  P3(x)=
1
2
(5x3-3x)
and as a follow on example
P4=
1
8
(35x4-302+3)
these are only valid for -1£ x£ 1 . The properties of this are

3.15   The Gamma (or Factorial) Function

we define the gamma function ( G ) as
G (x)= ó
õ
¥


0
e-ttx-1dt
so when x=x+1 we obtain
G (x+1)= ó
õ
¥


0
e-ttxdt= [ -e-ttx ]
¥
 
0
- ó
õ
¥


0
-e-txtx-1dt=0+xG (x)
so we can say
G (x+1)=xG (x)
also we notice that
G (1)= ó
õ
¥


0
e-tdt= [ -e-t ]
¥
 
0
=--1
Þ G (1)=1
and the same sort of thing applies for x=2,3,4,...
G (2)=G (1+1)=1G (1)=1
G (3)=G (2+1)=2G (2)=2.1
G (4)=G (3+1)=3G (3)=3.2.1
so the general patteren is
G (n+1)=n!
so it follows that
G (0+1)=0!=1

Figure missing.2 - Plot of the Gamma Function

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