AMS 212A Applied Mathematical Methods I
Lecture 06
Copyright by Hongyun Wang, UCSC
Recap of Lecture 5
Classification of boundary conditions
Dirichlet
Neumann
Mixed
Adjoint operator, self-adjoint operator
Sturm-Liouville problem (in self-adjoint form)
L (u ) = r ( x ) u ,
u ( a ) + u ( a ) = 0
u b + u b = 0
( )
( )
x ( a, b )
where the linear differential operator is
L ( u ) ( p ( x ) u x )x + q ( x ) u
The inner product is
b
u, v u ( x ) v ( x ) r ( x ) dx
a
Results of Sturm-Liouville theory:
u,
1
1
L (v) =
L (u ), v
r ( x)
r ( x)
(that is,
1
L ( i ) is self-adjoint).
r ( x)
All eigenvalues are real.
For each eigenvalue, we can find a real eigenfunction.
Eigenfunctions for different eigenvalues are orthogonal to each other.
Results of Sturm-Liouville theory (continued)
Since eigenvalues and eigenfunctions are both real, we only need to work with real functions.
All eigenvalues are simple.
(i.e., one corresponding eigenfunction for each eigenvalue.)
-1-
AMS 212A Applied Mathematical Methods I
Suppose
u(x) and v(x) are eigenfunctions corresponding to eigenvalue .
Claim:
u(x) and v(x) are proportional to each other.
Proof:
L (u ) = r ( x ) u
L (v) = r ( x ) v
==>
u L ( v ) vL ( u ) = u( r ( x ) v ) v( r ( x ) u ) = 0
On the other hand, using the differential form of L(•), we write
(
) (
u L ( v ) vL ( u ) = u ( pvx )x + qv v ( pu x )x + qu
)
= u( pvx )x v( pu x )x
(Using Lemma 1 of Lecture 5)
= ( p ( uvx vu x ))x
Combining these two results, we have
( p (uv
==>
x
vu x ))x = 0
p ( uvx vu x ) = const
Since both u(x) and v(x) satisfy the boundary conditions, we have
( u v vu )
=0
(Lemma 2 of Lecture 5)
x=b
==>
p ( u v vu ) = 0 ,
==>
u v vu = 0 ,
==>
v
= 0 ,
u
==>
v
= const
u
x [ a, b ]
x [ a, b ]
x [ a, b ]
End of proof
Eigenvalue sequence
Eigenvalues form an unbounded strictly increasing infinite sequence
0 < 1 < 2 < < n < -2-
AMS 212A Applied Mathematical Methods I
lim n = +
n+
We will prove this after we describe the completeness of eigenfunctions.
Completeness of eigenfunctions
The set of eigenfunctions {v0 ( x ) , v1 ( x ) , v2 ( x ) , …, vn ( x ) , …} is a complete basis for all
piecewise continuous functions on [a, b].
That is, any piecewise continuous function ƒ(x) can be expanded as
f ( x ) = bn vn ( x )
n=0
where coefficient bn is given by
b
f , vn
bn =
=
vn , vn
f ( x ) v ( x ) r ( x ) dx
n
a
b
( v ( x )) r ( x ) dx
2
n
a
Below we do 3 things:
prove the eigenvalue sequence,
introduce the Rayleigh quotient, and then
prove the completeness
Proof of eigenvalue sequence
Eigenvalue and eigenfunction u(x) satisfy the Sturm-Liouville equation
( p ( x)u )
x x
x ( a, b )
+ q ( x )u = r ( x )u ,
and boundary conditions
u ( a ) + u ( a ) = 0
u (b ) + u (b ) = 0
The key step of the proof is Prufer substitution:
p ( x ) u x ( x ) = ( x ) cos ( ( x ))
u ( x ) = ( x ) sin ( ( x ))
Before the Prufer substitution, we have
a second order linear ODE for u(x).
After the Prufer substitution, (as we will see) we have
-3-
AMS 212A Applied Mathematical Methods I
a first order non-linear ODE system for (x) and (x).
Let us look at the advantages of Prufer substitution.
Advantage #1:
( x ) 0 , x [ a, b ]
Suppose ( x0 ) = 0 at some point x0 in [a, b].
==>
p ( x0 ) u x ( x 0 ) = 0
u ( x0 ) = 0
p(x0) 0
==>
u x ( x 0 ) = 0
u ( x0 ) = 0
That is, u(x) satisfies the zero initial conditions at x0.
The Sturm-Liouville equation is a second order ODE of u(x)
==>
u ( x) 0
(due to the zero initial conditions at x0)
In the Sturm-Liouville problem, we are only interested in non-trivial solutions.
Advantage #2:
Boundary conditions affect only Boundary condition at x = a:
u ( a ) + u ( a ) = 0
(a)
cos ( ( a )) = 0
p (a)
==>
( a ) sin ( ( a )) + ==>
p ( a ) sin ( ( a )) + cos ( ( a )) = 0
==>
sin ( ( a ) ) = 0
where satisfies
cos ( ) =
p (a)
( p ( a ))
2
+ 2
,
sin ( ) =
( p ( a ))
2
+ 2
The boundary condition becomes
(a) = n ==>
(a) = + n We can shift the whole function (x) to write the boundary condition at x = a as
-4-
AMS 212A Applied Mathematical Methods I
BC at x = a : ( a ) = ,
0 <
After the shifting, we write the boundary condition at x = b as
BC at x = b : ( b ) = μ + n ,
0< μ Note: we use shifting and pick suitable value of n to make 0 < and 0 < μ ,
which is important in the proof of Lemma 1 and proof of eigenvalue sequence.
Advantage #3:
The evolution equation of (x) is independent of (x).
(x) satisfies the first order non-linear ODE
x = ( r ( x ) + q ( x )) sin 2 +
1
cos 2 p( x)
Key steps in derivation: 2 equations for (x) and (x):
Equation 1:
( p ( x ) u ) + ( r ( x ) + q ( x )) u = 0 (the Sturm-Liouville equation)
==> ( cos ) + ( r ( x ) + q ( x )) sin = 0
x x
x
Equation 2:
cos = p ( x ) u x = p ( x ) ( sin )x
…
(See Exercise #6 for derivation).
Lemma 1:
Let (x, ) be the solution of
x = ( r ( x ) + q ( x )) sin 2 +
1
cos 2 p( x)
(a) = , 0 < where p(x) > 0 and r(x) > 0 for x in [a, b] (from the Sturm-Liouville problem).
Then, (x, ) has the 3 properties below. For x > a,
a) (x, ) is a strictly increasing function of .
b)
c)
lim ( x, ) = +
+
lim ( x, ) = 0
(For proof of Lemma 1, see Appendices of Lecture 6 in a separate PDF file).
-5-
AMS 212A Applied Mathematical Methods I
Now we use the results of Lemma 1 to prove the eigenvalue sequence.
lim ( b, ) = 0 < μ
(since 0 < μ )
==>
μ 0 < ( b, ) < μ
for negative and large.
As increases to infinity, (b, ) also increases to infinity.
At some value, 0, we will have (b, 0) = μ.
==>
0 is the smallest value of for which (b, 0) = μ + n is satisfies.
==>
0 is the smallest eigenvalue.
As increases further, at some value, 1, we will have (b, 1) = μ + ;
==>
1 is the smallest eigenvalue after 0.
At some value, 2, we will have (b, 2) = μ + 2;
==>
2 is the smallest eigenvalue after 1.
At some value, 3, we will have (b, 3) = μ + 3;
==>
3 is the smallest eigenvalue after 2.
…
Thus, we have a strictly increasing sequence of eigenvalues
0 < 1 < 2 < < n < Next we prove lim n = + .
n+
Since { n} increases monotonically, we have either lim n = + or lim n = finite .
n+
Suppose lim n = ( c) = finite . We have
n+
( c ) > n
==>
==>
for any n
( )
( b, ( ) ) = + ,
b, ( c) > ( b, n ) = μ + n c
for any n
which contradicts with ( c) = finite .
End of proof of eigenvalue sequence
Rayleigh quotient
Definition:
-6-
n+
AMS 212A Applied Mathematical Methods I
Rayleigh quotient is defined as
R (u ) =
u,
1
L (u )
r ( x)
u, u
b
=
u L ( u ) dx
a
b
u r ( x ) dx
2
a
Rayleigh quotient on eigenfunctions:
Suppose
0 < 1 < 2 < < n < is the eigenvalue sequence and
{v ( x ), v ( x ), v ( x ), …, v ( x ), …}
0
1
2
n
is the sequence of corresponding eigenfunction.
It is straightforward to verify that Rayleigh quotient satisfies
R ( vn ) = n ,
n = 0, 1, 2, …
N
R bn vn =
n=0
N
b
2
vn , vn
n n
n=0
N
b
2
n
vn , vn
n=0
N
0 = min R bn vn {bn } n = 0 (See Exercise #6 for derivation).
This last result motivates the Rayleigh principle below.
Rayleigh principle:
Let
u ( a ) + u ( a ) = 0
2
C BC
= u ( x ) u C 2 and satisfies
u (b ) + u (b ) = 0
2
Consider the minimization of R(u) over C BC
.
Let
u0 = arg min R ( u )
2
uCBC
, u0
Claim 1:
u0 is an eigenfunction corresponding to eigenvalue 0 and
0 =
min
2
uCBC
, u0
R (u )
Proof of Claim 1:
Consider a function of a scalar variable s
-7-
AMS 212A Applied Mathematical Methods I
g ( s ) = R ( u0 + s u )
2
where u C BC
2
By definition of u0, we have g ( 0 ) =0 for any u C BC
The derivatives of numerator and denominator of R(u0 + s u) are
d
u0 + s u, u0 + s u
ds
= 2 u, u0
s=0
1
d
u0 + s u,
L ( u0 + s u )
r ( x)
ds
= 2 u,
s=0
1
L ( u0 )
r ( x)
(See Exercise #6 for derivation).
The derivative of R(u0 + s u) has the expression
2 1
d
R ( u0 + s u )
=
L ( u0 ) + R ( u0 ) u, u0 u,
u0 , u0 r ( x)
ds
s=0
=
2
u0 , u0
u,
1
L ( u0 ) + R ( u0 ) u0
r ( x)
(See Exercise #6 for derivation).
2
The derivative of R(u0 + s u) is zero for any u C BC
.
d
2
R ( u0 + s u )
= 0 for any u C BC
ds
s=0
==>
u,
1
2
L ( u0 ) + R ( u0 ) u0 = 0 for any u C BC
r ( x)
==>
1
L ( u0 ) + R ( u0 ) u0 = 0
r ( x)
==>
L ( u0 ) = R ( u0 ) r ( x ) u0
==>
R(u0) is an eigenvalue and u0 is a corresponding eigenfunction.
Next we show R ( u0 ) = 0 .
By definition of u0, we have
R ( u0 ) = min R ( u )
2
uCBC
==>
R ( u0 ) R ( v0 ) = 0
(v0 is an eigenfunction corresponding to 0).
Since 0 is the smallest eigenvalue and R(u0) is an eigenvalue, we have
-8-
AMS 212A Applied Mathematical Methods I
0 R ( u 0 )
Combining these two results, we conclude
0 = R ( u0 ) = min R ( u )
2
uCBC
End of proof of Claim 1
2
Next we look at the minimum of Rayleigh quotient over subset of C BC
.
Let
WN = span {v0 , …, vN }
{
2
WN = u u C BC
and u WN
}
Consider the minimization of R(u) over WN .
Let
u N +1 = arg min R ( u )
uWN , u0
Claim 2:
uN+1 is an eigenfunction corresponding to eigenvalue N+1 and
N +1 =
min
uWN , u0
R (u )
(For proof of Claim 2, see Appendices of Lecture 6 in a separate PDF file)
Rayleigh principle provides a way of approximating the lowest few eigenvalues, especially the
smallest eigenvalue.
Example:
u = u
u ( 0 ) = 0 , u (1) = 0
This is a Sturm–Liouville problem with
p ( x ) = 1 , r ( x ) = 1 , q ( x ) = 0 , L ( u ) = u We solved it previously. The eigenvalues are
n = ( n + 1) 2 ,
2
n = 0, 1, 2, …
Now we use Rayleigh principle to approximate the smallest eigenvalue.
(Skip the details in Lecture)
We try
w ( x ) = x (1 x )
(it has to satisfy the boundary conditions)
-9-
AMS 212A Applied Mathematical Methods I
The numerator and denominator of Rayleigh quotient are
1
1
1
1
L ( w ) = w ( x ) w ( x ) dx = 2 x (1 x ) dx =
r ( x)
3
0
0
w,
1
w, w =
( w ( x ))
1
2
0
dx = x 2 (1 x ) dx =
2
0
1
L (w)
r ( x)
w,
==>
R (w) =
==>
0 = min R ( u ) R ( w ) = 10
w, w
1
30
1
3 = 10
1
30
=
2
uCBC
The true value is 0 = 2 9.87 .
Proof of the completeness of eigenfunctions
For a piecewise continuous function ƒ(x), we show that
N
lim f bn vn = 0
N n=0
where
f , vn
vn , vn
bn =
b
u, v = u ( x ) v ( x ) r ( x ) dx
a
u =
u, u
The proof consists of results 1-4 below.
Result 1:
N
f
bn vn v j
n=0
for 0 j N
This follows directly from the definition of bn.
Result 2:
N
N
f bn vn = min f cn vn
n=0
{cn }
n=0
- 10 -
AMS 212A Applied Mathematical Methods I
This follows directly from Result 1 above
N
N
N
N
f
b
v
v
,
f
b
v
+
n n n n n n + n vn
n=0
n=0
n=0
n=0
N
N
= f bn vn , f bn vn +
n=0
n=0
==>
N
N
n=0
n=0
N
n vn ,
n=0
N
v
n n
n=0
f cn vn f bn vn
Result 3:
We only need to show that for a piecewise continuous function ƒ(x), we have
N
for any > 0 there exists N and {cn} such that f cn vn < (***)
n=0
Result 4:
2
is dense in the set of piecewise continuous functions, we only need to show
Since C BC
2
that for f ( x ) C BC
, we have
N
for any > 0 there exists N and {cn} such that f cn vn < n=0
For proof of (***), see Appendices of Lecture 6 in a separate PDF file.
- 11 -
(***)