THE METHOD OF SEPARATION OF VARIABLES AND THE EIGENVALUE PROBLEM FALL 2011 “ The universe has always been presented by mathematical problems “ The Minh Tran 1... Explicitly show ther
Trang 1THE METHOD OF SEPARATION OF VARIABLES AND THE EIGENVALUE
PROBLEM
FALL 2011
“ The universe has always been presented by mathematical problems “
The Minh Tran
1 Solution :
Solve the eigenvalue problem :
2
dx
We have equation form :
( )
"
2
0
We have auxiliary equations r
Case
x Ae Be
A B Ae Be
A B Ae Be
There are not exist
λ λ
φ
π
λ
−
−
−
<
⇔
( )
2
1 2
Case
x A Bx
λ
φ
=
Trang 2( ) ( )
( )
2
0
B B
B
x A A
π
π φ
⇔
=
=
∀
We have not had eigenvalue
( )
2 2
2 2
2
Case
A B Cos
Cos Because A B
λ
π λ
π λ
>
⇔
2 Explicitly show there are no negative eigenvalues for
2
dx
λφ
Solution :
We have equation form :
Trang 3( )
( )
"
2
0
0
,
We have auxiliary equations r
When
x Ae Be
d
dx
d
dx
Thus there are no negative eigenvalues
λ λ
φ
φ
λ φ
π
−
<
=
3 Solve the heat equation :
2
x L t subject to
t x
Solution :
a u x
L
π
= +
We have infinite series solution
2 2 2
1
kn t L n n
n x
u x t A e
L
π
π
=
∑
So
Trang 4( )
1
0
0
0
0
3
0
3
6 4 cos cos
6 4 cos 6
0
n n L
n
L
L
x
A
Others
π
∞
=
=
∑
∫
∫
∫
b u x
L
π
= −
We have infinite series solution
2 2 2
1
kn t L n n
n x
u x t A e
L
π
π
=
∑
So
Trang 5( )
1
0
0
0
0
8
0
8
0
8
cos cos
cos 0
cos cos
0
n n L
n
L
L L
x
A
x
Others
π
π
∞
=
=
∑
∫
∫
∫
∫
c u x
L
π
= −
We have infinite series solution
2 2 2
1
kn t L n n
n x
L
π
π
=
∑
So
1
0
0
0
4
sin cos
2 sin
n n L
n
L
x A
π
π
∞
=
∑
∫
∫
4 For the following PDEs, what ODEs are implied by the method of
separation of variables ?
Solution :
We can use the method of separation of variables to find the following ODEs
Trang 6( ) ( )
4 4
4
)
( , )
( , )
a
t x
From the equation u x t x G t
where x is only a function of x and G t is a function of t
We will get first partial derivetive of u x t with respect to t and the fourth partial derivative with respect to x
dG t u
x
u d
x
φ φ
φ
=
=
∂
=
∂
∂
=
∂
4
4 4
( )
( ) (1)
G t dx
dG t d
dt dx
φ
φ φ
( )
4
We can separate iable by dividing both sides of by x G t
dG t d
φ φ
φ
=
We can give the both sides equal the same constant
( )
( )
4 4
4
4
dG t d
where is an arbitrary constant
Thus we have ODEs one for G t and one for x
dG t
G dt
d
dx
φ
φ
φ λ
φ
λφ
= −
= −
Trang 7( ) ( )
2
2 2
)
( , )
From the equation u x t x G t
where x is only a function of x and G t is a function of t
We will get ond partial derivetive of u x t with respect to t and the ond partial derivative with respect to x
d G t u
x
φ φ
φ
=
=
∂
=
∂
2
( )
( ) (1)
u d
G t
x dx
d G t d
d t dx
φ
φ φ
∂
=
∂
( ) ( )
2
We can separate iable by dividing both sides of by x G t
c
φ φ
φ
=
We can give the both sides equal the same constant
( )
( )
2
2
2
2
2 2
c where is an arbitrary constant
Thus we have ODEs one for G t and one for x
d G t
G
d t
d
c
dx
φ
φ
φ λ
φ
λφ
= −
= −
Trang 8( ) ( )
)
( , )
( , )
In the method of separation of iables we will find solutions in the product form
u r t r G t
where r is only a function of r and G t is a function of t
We will get the first partial derivetive of u r t with respec
φ
φ
=
( ) ( )
( ) ( ) ( ) ( )
sec
(1)
t to t and the ond partial derivative with respect to r
dG t
u
r
dG t kG t d d
φ
φ
φ φ
∂
=
∂
=
( ) ( )
We can separate iable by dividing both sides of by k r G t
r
kG dt r dr dr
φ φ
φ
We can give the both sides equal the same constant
( )
1 dG t 1 d d
r where is an arbitrary constant
kG dt r dr dr
φ
φ
( )
Thus we have ODEs one for G t and one for r
dG t
kG dt
dr dr
φ λ
φ
λφ
= −
= −