Physics 307
Mathematical Physics
Luis Anchordoqui
Monday, October 24, 16
1
ORDINARY DIFFERENTIAL EQUATIONS III
3.1 Setting the Stage ✔
3.2 Initial Value Problem ✔
Picard’s existence and uniqueness theorem
Systems of first-order linear differential equations
Green matrix as a generalized function
3.3 Boundary Value Problem
Self-adjointness of Sturm-Liouville operator
Green function of Sturm-Liouville operator
Series solutions to homogeneous linear equations
3.4 Fourier Analysis
Fourier series
Fourier transform
Monday, October 24, 16
2
BOUNDARY VALUE PROBLEM
Self-adjointness of Sturm-Liouville operator
So far we have seen differential equations with initial conditions
For a linear equation of second order
we have seen that integration constants are determined
from values of unknown function and its first derivative at
t = t0
We will begin to study problems of second order
in which constants of integration are determined
not by initial conditions ☛
but by boundary conditions
In these problems
range of variation of variable is restricted to a certain range
and integration constants are determined
from values of unknown function
or its derivative in extreme points of interval
Monday, October 24, 16
3
Definition 3.3.1
General form of a second order linear equation reads
d2 u
du
+ A(x)
+ B(x)u = F (x)
dx2
dx
A(x), B(x) and F (x) are continuous
(3.3.167.)
functions
Preceding equation can be rewritten as

d
du
p(x)
+ q(x)u = f (x)
dx
dx
f (x) =
p(x)F (x), q(x) =
Note that
p0 (x) = A(x)p(x)
and
p(x)B(x)
and
(3.3.168.)
p(x) = e
Eq. (3.3.168.) is generally written as
where L =
Monday, October 24, 16

A(x)dx
(3.3.169.)
(pu0 )0 = pu00 + p0 u0 = p[u00 + A(x)u0 ]
so (3.3.168.) reduces to (3.3.167.) multiplied by
R
p(x)
L[u(x)] = f (x)
(3.3.170.)
d
d
p(x)
+ q(x) ☛ is Sturm-Liouville (SL) operator
dx
dx
(3.3.171.)
4
This operator acts on functions u(x)
defined in a given real interval a  x  b
and is only completely defined
after specifying values of unknown function or its first derivative
or linear combinations of them at boundaries
These conditions are known as boundary conditions
(a, b)
and the functions that satisfy them
constitute domain of Sturm-Liouville operator
In this class ☛ we study the case in which
Next class we will study the case
in which
p(x) is
non-zero in [a, b]
p(x)
vanishes at one or both ends of the interval
that leads to study of so-called special functions
Throughout we let
C (n) ([a, b]) denotes
[a, b]
be a bounded interval in
the space of functions
with derivatives of
1
L2 ([a, b])
R
n -th order continuous up to
p(x) 2 C ([a, b])
and
q(x) 2 C 0 ([a, b])
endpoints
is subspace of functions that vanish near endpoints
Monday, October 24, 16
5
Theorem 3.3.1.
8u, v 2 L2 ([a, b])
☛
L
is most general second order real operator
hv, L[u]i = hL[v], ui (3.3.172.)
Z b
product hv, ui =
v ⇤ (x) u(x) dx
which can be formally self-adjoint
hv, ui
denotes usual inner
a
Proof.
Integration by parts leads to
Z b
hv, L[u]i
hL[v], ui =
=
Z
[v(pu0 )0
u(pv 0 )0 ]dx
[(vpu0 )0
(upv 0 )0 ] dx
a
b
a
= p [vu
For
L
(3.3.173.)
0
0 a
uv ]b
to be self-adjoint ☛ we need to impose conditions on
u, v
at the endpoints to make right hand side of (3.3.173.) vanish
Monday, October 24, 16
6
Definition 3.3.2.
A boundary condition B is an expression of the form
Bu = ca u(a) + cb u(b) + da u0 (a) + db u0 (b)
for real constants ☛
(3.3.174.)
c a , c b , da , db
Definition 3.3.3.
Boundary conditions
if
B 1 , B2
are self-adjoint for
L
8u, v 2 C 2 ([a, b]) satisfying B1 u = B2 u = B1 v = B2 v = 0
Z
b
L[u] v dx =
a
vanishing of Bj u and
Z
b
u L[v] dx
a
(3.3.175.)
Bj v
implies right-hand side of (3.3.173.) vanishes
Monday, October 24, 16
7
Definition 3.3.4.
Local boundary conditions are those establishing a relationship
between unknown function and its derivative in each edge
separately
We say that
if
B1
and
B1 u = 0 and B2 u = 0 are local
B2 are independently chosen
or separated b.c
to guarantee that right hand side of (3.3.173.) vanish
➤Dirichlet conditions
➤Neumann conditions
B1 u = u(a) & B2 u = u(b)
0
B1 u = u0 (a) & B2 u = u (b)
0
➤Robin-Cauchy conditions ca u(a) + da u (a) = 0 & cb u(b) + db u
0
(b) = 0
These are separated boundary conditions
B1 is a condition at a and B2 is a condition at b
Any pair of separated conditions is self-adjoint for general
Monday, October 24, 16
L
8
Definition 3.3.5.
Non-local boundary conditions establish a relationship between
value of unknown function and its derivative in one and other edge
Most common examples of non-separated boundary conditions are
➤Periodic conditions
B1 u = u(b)
u(a)
&
B2 u = u0 (b)
➤Anti-Periodic conditions
B1 u = u(b) + u(a)
These are self-adjoint for
if
L
&
u0 (a)
B2 u = u0 (b) + u0 (a)
p(b) = p(a)
Next ☛ we discuss whether homogeneous equation
L[u] = 0
with Dirichlet boundary conditions has non-trivial solutions u(x)
6= 0
Later ☛ we will see that
the non-existence of such solutions (a.k.a. zero modes)
is a necessary and sufficient condition for inhomogeneous equation
to have a unique solution via Green's function
Monday, October 24, 16
9
Example 3.3.1.
Consider case with
p(x) = 1
q(x) = 0
and
General solution of homogeneous
Therefore ☛ if
d2
(3.3.177.)
L=
dx2
equation is u(x) = cx + d
(3.3.178.)
u(a) = u(b) = 0 ) c = d = 0
and homogeneous equation only has trivial solution
Example 3.3.2.
d2
2
For
L=
k
dx2
general solution of homogeneous equation can be written as
u(x) = ceikx + de ikx
(3.3.179.)
or equivalently
0
0
(3.3.180.)
u(x) = c sin[k(x
u(a) = 0 ) d0 = 0
The condition u(b) = 0
a)] + d cos[k(x
a)]
If
leads to
which has a non-trivial solution
Otherwise ☛
Monday, October 24, 16
c0 = 0
c0 sin[k(b
a)] = 0
(c0 6= 0) , k(b
a) = n⇡, n 2 Z
and only solution is
u(x) = 0
10
Definition 3.3.6.
Fix a positive weight function
⇢(x) 2 C 2 ([a, b])
c > 0 for x 2 [a, b]
so that ⇢(x)
and consider the Sturm-Liouville eigenvalue problem
L u = ⇢ u, with B1 u = B2 u = 0
(3.3.181.)
We say that
is an eigenvalue of
L
u 2 C 2 ([a, b]) of (3.3.181.)
and we call u an eigenfunction
if there is a non-zero solution
Lemma 3.3.1.
Let
(L, B1 , B2 , ⇢)
be a self-adjoint Sturm-Liouville system
Eigenfunctions associated to different eigenvalues are orthogonal
in the inner product ☛ hu, vi⇢
(i) eigenfunction
(ii)
h⇢
1
=
Z
b
u⇤ (x) v(x) ⇢(x) dx
(3.3.182.)
a
1
is an eigenvector for operator ⇢ L
i.e. ⇢ 1 Lu
hu, ⇢ 1 Lvi⇢
2
1
is self-adjoint in domain C ([a, b])
u
Lu, vi⇢ =
so that ⇢ L
= u
with respect to inner product
Monday, October 24, 16
h,i
11
Proof.
If L[ui (x)]
=
0 = huj , L[ui ]i⇢
and so
Z
i ui (x) and
L[uj (x)] =
hL[uj ], ui i⇢ = (
j)
i
j uj (x)
Z b
we have
ui (x) uj (x) ⇢(x)dx
a
b
ui (x) uj (x) ⇢(x) dx = 0 if
a
i
6=
j
(3.3.184.)
Theorem 3.3.2.
(L, B1 , B2 , ⇢) there exists:
eigenvalues
1  2  ···  n  ...
(i) For any self-adjoint SL system
a countable set of
with associated eigenfunctions
which satisfy
u1 (x), u2 (x), . . . , un (x) . . .
L[un (x)] = ⇢(x)un (x)
and are orthogonal with respect to inner product (3.3.182.)
For local boundary conditions ☛
i < j if i 6= j
because there cannot be two l.i. solutions of L[u] =
for same
⇢u
satisfying (3.3.181.)
(as the Wronskian would be null)
Monday, October 24, 16
12
(ii) Any function f (x)
2 C 2 ([a, b]) that
satisfies boundary conditions
can be written in terms of eigenfunctions
un (x)
as absolutely uniformly convergent series
f (x) =
1
X
cn un (x)
(3.3.185.)
n=1
Set of all eigenfunctions forms a basis
of vector space of differentiable functions to second order
that satisfy
boundary conditions on interval
In other words ☛ eigenfunctions of
Monday, October 24, 16
L
[a, b]
form a complete set
13
Proof.
Note that space
L2⇢ ([a, b])
consisting of eigenfunctions for Sturm-Liouville problem
is space of measurable
kuk2L2⇢ =
u
Z
on
b
[a, b] such that
|u(x)|2 ⇢(x) dx < 1
(3.3.186.)
a
Since ⇢(x) is bounded from above and below
L2 ([a, b])
product h , i⇢ are different
p
2
L : k ⇢ ukL2 = kukL2⇢
this is the same space of functions as
u!
p
but norm and inner
⇢u
is unitary map of
In particular
L⇢
onto
p
2
1
is an orthonormal basis for L⇢ , { ⇢ uj }j=1
is an orthonormal basis for
Coefficients cn are given by
{uj }1
j=1
cn =
Monday, October 24, 16
Rb
a
⇢(x) un (x) f (x)dx
hun , f i⇢
=
Rb
2 (x)dx
hun , un i⇢
⇢(x)
u
n
a
L2
(3.3.187.)
14
Indeed ☛ if we multiply (3.3.185.) by
Z
b
⇢(x) un (x) f (x) dx =
a
1
X
cm
m=1
= cn
Z
Z
⇢(x)un (x)
and integrate
b
⇢(x)um (x) un (x)
a
b
a
(3.3.188.)
⇢(x) u2n (x)dx
where we have used orthogonality of eigenfunctions
For a given
f
☛ coefficients of expansion are unique
1
X
cn un (x) = 0 ) cn = 08n
If
n=1
Corollary 3.3.2.
Let
⇢ : [a, b] ! R
The space ☛
⇢(x) c > 0
p
2
L⇢ ([a, b]; C) = {f : [a, b] ! C such that ⇢f 2 L2 }
be an arbitrary function such that
equipped with inner product
p
hf, gi⇢ = h ⇢f,
p
⇢giL2 =
Z
b
⇢(x) f (x) g ⇤ (x)dx
a
is a Hilbert space ☛ which we denote by
Monday, October 24, 16
(L2⇢ ([a, b]; C), h, i⇢ )
15
Green function of Sturm-Liouville operator
Definition 3.3.7.
The Green function is defined as as solution of the equation
Lx [G(x, x0 )] = (x
x0 )
(3.3.191.)
which satisfies (in first variable) Robin-Cauchy boundary conditions
dG
ca G(a, x ) + da
|(x=a,x0 ) = 0 ,
dx
dG
0
cb G(b, x ) + db
|(x=b,x0 ) = 0
dx
0
i.e
with
a<x
and
(3.3.192.)
x0 < b
subscript in operator indicates that it acts
on the first variable of the argument of the Green's function
Monday, October 24, 16
16
Theorem 3.3.3.
(i) Solution of (3.3.191.) exists and is unique
if and only if trivial solution
is only solution of
L[u(x)] = 0
u(x) = 08x 2 [a, b]
subject to RC boundary condition
i.e. there are no zero modes
(ii) Solution of inhomogeneous equation with RC boundary conditions
is given by ☛
u(x) =
Proof.
Z
b
G(x, x0 ) f (x0 ) dx0
(3.3.193.)
a
We first show that (ii) holds
if
G(x, x0 )
L[u(x)] =
Z
b
a
exists then (3.3.193.) is a solution of
Lx [G(x, x0 )] f (x0 ) dx 0=
Z
b
(x
(3.3.170.)
x0 ) f (x0 ) dx0 = f (x)
a
Note that L acts on first variable ☛ which is unaffected by integral
Monday, October 24, 16
17
In addition ☛
because
G
u
satisfies boundary condition
satisfies this condition
in first variable which is unaffected by integral
0
ca u(a) + da u (a) =
and
0
cb u(b) + db u (b) =
Z
b
a
Z
b
"
"
dG
ca G(a, x ) + da
dx
0
dG
cb G(b, x ) + db
dx
(x=a,x0 )
0
a
(x=b,x0 )
To show (i) we now construct Green function
Let
u1 (x)
and
u2 (x)
#
#
f (x0 )dx0 = 0
f (x0 )dx0 = 0
be two solutions of homogeneous equation
L[u(x)] = 0
each satisfying one of boundary conditions
ca u1 (a) + da u01 (a) = 0
For
x < x0
we have
and
cb u2 (b) + db u02 (b) = 0
L[G(x, x0 )] = 0
G(x, x0 ) = c1 (x0 )u1 (x)
condition at x = a
with
satisfying boundary
Monday, October 24, 16
(3.3.194.)
18
In an analogous manner ☛ if
Therefore ☛
G(x, x0 ) =
x0 +✏
⇢
x0 ✏
d
[p G0 (x, x0 )]dx +
dx
c1 (x0 )u1 (x)
c2 (x0 )u2 (x)
Z
leads to
0
[p G (x, x
0
x=x0 +✏
)]x=x0 ✏
Due to continuity of
G(x, x0 )
p
+
and
then
G(x, x0 ) = c2 (x0 )u2 (x)
satisfying boundary condition at x
Integration of (3.3.191.) over
Z
x > x0
[x0
x0 +✏
x < x0
x > x0
x0 ✏
Z
q
(3.3.195.)
✏, x0 + ✏]
q(x) G(x, x0 )dx =
x0 +✏
=b
(with
Z
x0 +✏
(x
✏>0
)
x0 ) dx
x0 ✏
q(x) G(x, x0 )dx = 1
(3.3.196.)
x0 ✏
this equation can only be satisfied if:
is continuous
its derivative has a discontinuity of magnitude
Monday, October 24, 16
1/p(x0 )
at
x = x0
19
Indeed
lim
✏!0
because
q
and
G
Z
p(x )
or equivalently
x0 ✏
q(x) G(x, x0 ) dx ! 0
are continuous functions
0
and thus ☛
x0 +✏
dG
dx
(
dG
dx
dG
dx
x0 +✏
dG
dx
x!x0 +
x0 ✏
)
=
x!x0
(3.3.197.)
=1
(3.3.198.)
1
p(x0 )
(3.3.199.)
in summary
p(x)G0 (x, x0 )
must be of the form
⇥(x
x0 ) + (x)
with
so that
Monday, October 24, 16
( pG0 )0
a continuous function at
contains a
(x
x0 )
x = x0
term
20
We impose
G
requirements on (3.3.195.) to obtain
c1 (x0 )u1 (x0 )
c1 (x
0
)u01 (x0 )
c2 (x0 )u2 (x0 ) = 0
c2 (x
0
)u02 (x0 )
=
1
p(x0 )
(3.3.200.)
which determines
c1 (x0 ) =
u2 (x0 )/C
and
c2 (x0 ) =
u1 (x0 )/C
(3.3.201.)
where
C
=
p(x0 )[u1 (x0 )u02 (x0 )
=
[pW (u1 , u2 )]x=x0
with
W (u1 , u2 ) =
and
u1
u01
u2
u02
u2 (x0 )u01 (x0 )]
(3.3.202.)
(3.3.203.)
u0 (x) = du/dx
Monday, October 24, 16
21
The solution exists only if
C 6= 0
that is ☛ only if Wronskian
Note that this is satisfied if
W (u1 , u2 )
is non-zero
u1 (x) and u2 (x)
are two linearly independent solutions of
In such a case ☛ C
[p(u1 u02 u2 u01 )]0
is a constant
= p0 (u1 u02
= u1 (pu02 )0
= q(u1 u2
Therefore ☛ if
G(x, x ) =
C=0
⇢
u2 u001 )
u2 (pu01 )0
(3.3.204.)
u2 u1 ) = 0
C 6= 0 we have
0
If
u2 u01 ) + p(u1 u002
L[u] = 0
u1 (x)u2 (x0 )/C
u1 (x0 )u2 (x)/C
x  x0
x x0
(3.3.205.)
Green function does not exist
u1 and u2 are linearly dependent
i.e. u2 (x) = cu1 (x) ☛ u1 (x) satisfies boundary conditions at
In this case solutions
both ends
This implies that if C = 0 ☛ there is a non-trivial solution u1 6= 0
satisfying L[u1 ] = 0 and RC boundary conditions
Monday, October 24, 16
22
Green function exists
, the
only solution of homogeneous equation
L[u] = 0
that satisfies RC conditions is
u=0
This concludes proof of (i)
Additionally ☛ (3.3.205.) gives explicit expression for Green function
Z b
(3.3.206.)
G(x, x0 ) u(x0 ) dx0
Linear operator G[u(x)] =
a
is then inverse of L operator and it is sometimes denoted also as L 1
Let us note that:
(i) inverse of differential linear operator L is integral linear operator
( G(x, x0 ) is known as kernel of the integral operator)
(ii)
G
depends not only on coefficients
p(x) and q(x) of L
but also on the boundary condition
(iii) symmetry of (3.3.205.) yields
G(x, x0 ) = G(x0 , x)
(3.3.207.)
and it allows to state theorem that follows
Monday, October 24, 16
23
Theorem 3.3.4. [Reciprocity of Green function]
0
a point source in x
0
is identical to response of system in x to a point source in
even if p and q depend on
Response of system in
x to
This is due to self-adjointness of
x
x
L
hLx G(x, x0 ), G(x, x00 )i = hG(x, x0 ), Lx G(x, x00 )i
(3.3.208.)
Using differential equation that satisfies Green function
Z
b
(x
and definition of Dirac distribution we have
Z b
0
00
x0 ) G(x, x00 ) dx =
x ) dx
(3.3.209.)
a
a
G(x0 , x00 ) = G(x00 , x0 )
which leads to
Inverse operator
hv, G[u]i =
G(x, x ) (x
Z
Monday, October 24, 16
b
a
Z
G
b
a
(3.3.210.)
is also self-adjoint
v(x) G(x, x0 ) u(x0 ) dx dx0 = hG[v], ui
(3.3.211.)
24
Note that Green's function (3.3.205.)
is not invariant under space translations (due to boundary conditions)
Translational invariance is broken ☛ even if
G(x, x0 ) 6= G(x
Therefore ☛
p
and
q
are constants
x0 )
Solution (3.3.193.) can be rewritten as
u(x) =
"
1
u2 (x)
C
Z
x
u1 (x0 ) f (x0 ) dx0 + u1 (x)
a
and one can explicitly verify that
Z
b
x
L[u] = f
It is always possible to write solution in form or
up
uh
u2 (x0 ) f (x0 ) dx0
#
u(x) = up (x) + uh (x)
☛ particular solution of inhomogeneous equation
☛ solution of homogeneous equation
(L[up ] = f )
(L[uh ] = 0)
subject to boundary condition
Monday, October 24, 16
25
Example 3.3.3.
Consider case p(x) = 1 and
q(x) = 0
L=
i.e. ☛
Let us set
d
dx2
a = 0, b > 0 and u(0) = u(b) = 0
that u1 (x) = x and u2 (x) = x
b with C = x
It follows
Then we obtain
0
G(x, x ) =
⇢
x(b x0 )/b x  x0
x0 (b x)/b x x0
(3.3.212.)
(x
b) = b
(3.3.213.)
Solution of inhomogeneous equation
2
d u
= f (x) (0  x  b)
2
dx
with u(a) = u(b) = 0 is then
u(x) =
Z
=
Monday, October 24, 16
b
G(x, x0 ) f (x0 ) dx
a
1
b
(3.3.214.)
"Z
x
0
x (b
0
0
0
x) f (x ) dx +
Z
b
x (b
x
x0 ) f (x0 ) dx0
#
26
If f (x)
= x2 we
have
4
3
x
x
b
(3.3.215.)
x3 ) =
+
12
12
x4 /12
consists of particular solution
3
solution of homogeneous equation xb /12
such that u(0) = u(b) = 0
1
u(x) =
x(b3
12
that
and
Example 3.3.4.
Let
d2
2
(3.3.216.)
L=
!
2
dx
with a = 0, b > 0
In this case ☛ for u(a) = u(b) = 0
we have ☛ u1 (x)
= sin(!x) and u2 (x) = sin (!(x
b)) (3.3.217.)
which lead to
C = ! [sin(!x) cos (!(x
Monday, October 24, 16
b))
cos(!x) sin (!(x
b))] = ! sin(!b)
27
Green function exists only if
in such a case we have
1
0
G(x, x ) =
! sin(!b)
⇢
sin(!b) 6= 0 that
is
sin(!x) sin (!(b x0 ))
sin(!x0 ) sin (!(b x))
! 6= n⇡/b
with
x  x0
x x0
n2Z
(3.3.218.)
! ! 0 we recover (3.3.213.)
If ! = ik with k 2 R ☛ Green function exists 8k 6= 0
G(x, x0 ) follows from (3.3.218.) with substitution ! ! k & sin ! sinh
Note that for
Example 3.3.5.
Let us consider again
L=
d
dx2
operator
with boundary condition
u0 (a) = u0 (b) = 0
Green function does not exist because
u1 (x) = c1
and u2 (x) = c2
yielding
This is due to constant solution
non-zero solution of
Monday, October 24, 16
(3.3.219.)
C=0
u(x) = c 6= 0
L[u] = 0
and satisfies
u0 (a) = u0 (b) = 0
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In this case ☛ solution of inhomogeneous problem
if it exists
is not unique
For a given a solution one can always add up an arbitrary constant
which satisfies homogeneous equation and the boundary condition
Example 3.3.6.
Finally ☛ consider
L=
d2
dx2
!2
operator
with boundary condition
It follows that
u0 (a) = u0 (b) = 0
u1 (x) = cos(!x) and u2 (x) = cos (!(x b)) (3.3.220.)
with C = ! sin(!b)
Green function exists only if sin(!b) 6= 0 that is ! 6= n⇡/b
with n 2 Z
In such a case we have
1
G(x, x ) =
! sin(!b)
0
If
!!0
☛ |G(x, x
On other hand ☛ if
Monday, October 24, 16
⇢
cos(!x) cos (!(b x0 ))
cos(!x0 ) cos (!(b x))
x  x0
(3.3.221.)
x x0
0
)| ! 1
! = ik with k 2 R, G(x, x0 )
exists
8k 6= 0
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Monday, October 24, 16
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