Sturm-Liouville operators
Sturm-Liouville operators have form (given p(x) > 0, q(x))
L=
d d p(x)
+ q(x),
dx
dx
(notation means Lf = (pf 0 )0 + qf )
Sturm-Liouville operators
Sturm-Liouville operators have form (given p(x) > 0, q(x))
L=
d d p(x)
+ q(x),
dx
dx
(notation means Lf = (pf 0 )0 + qf )
For example,
L=
d2
− 4,
dx 2
Lf (x) = f 00 (x) − 4f (x).
Therefore L sin(kx) = (−k 2 − 4) sin(kx)
Sturm-Liouville operators
Sturm-Liouville operators have form (given p(x) > 0, q(x))
L=
d d p(x)
+ q(x),
dx
dx
(notation means Lf = (pf 0 )0 + qf )
For example,
L=
d2
− 4,
dx 2
Lf (x) = f 00 (x) − 4f (x).
Therefore L sin(kx) = (−k 2 − 4) sin(kx)
Typical vector space of functions L acts on: C0∞ [a, b] so that
f (a) = f (b) = 0.
Inner products and self adjointness
One possible inner product for C0∞ [a, b] is
Z b
hf , g i =
f (x)g (x)dx, L2 inner product
a
Inner products and self adjointness
One possible inner product for C0∞ [a, b] is
Z b
hf , g i =
f (x)g (x)dx, L2 inner product
a
Adjoint of Sturm-Liouville operator: compute by moving operator
around using integration by parts. For any two functions f , g in
C0∞ [a, b],
Z b
df d p(x)
g (x) + q(x)f (x)g (x)dx
hLf , g i =
dx
a dx
Z b
df dg
=
−p(x)
+ q(x)f (x)g (x)dx
dx dx
a
Z b
d dg =
p(x)
f (x) + q(x)f (x)g (x)dx
dx
a dx
= hf , Lg i.
Thus S-L operators are self-adjoint on C0∞ [a, b].
More examples of operator adjoints
Example 2: Use inner product on C0∞ [0, R]
Z R
rf (r )g (r )dr ,
hf , g i =
0
More examples of operator adjoints
Example 2: Use inner product on C0∞ [0, R]
Z R
rf (r )g (r )dr ,
hf , g i =
0
−1
r (rf 0 )0
For linear operator Lf =
integration by parts gives
Z R df −1 d
hLf , g i =
r r
r
g (r )dr
dr dr
0
Z R
df dg
dr , (IBP)
=−
r
dr dr
0
Z R dg d
r
f (r ) dr , (IBP again)
=
dr
0 dr
Z R dg −1 d
=
r r
r
f (r ) dr
dr dr
0
= hf , Lg i.
Thus L is self adjoint with respect to the weighted inner product
(but not with respect to the L2 one!)
More examples of operator adjoints
Example 3: L = d/dx acting on C0∞ [a, b].
Z
b
hLf , g i =
a
Therefore L† = −L.
df
g (x) = −
dx
Z
a
b
dg
f (x)dx = hf , −Lg i.
dx
Eigenvalue problems for differential operators
Problem: find eigenfunction v (x) and eigenvalue λ solving
Lv (x) = λv (x),
Eigenvalue problems for differential operators
Problem: find eigenfunction v (x) and eigenvalue λ solving
Lv (x) = λv (x),
Remarks:
Sometimes written Lv (x) + λv (x) = 0
Constant times v (x) is also eigenfunction.
Infinite dimensions implies infinite number of eigenvalues and
eigenfunctions (usually).
If eigenfunctions span required vector space, called complete.
Sturm-Liouville eigenvalue problems
Self-adjoint operators (like S-L) have nice properties:
1 The eigenvalues are real.
2 The eigenfunctions are orthogonal to one another (with
respect to the same inner product used to define the adjoint).
Sturm-Liouville eigenvalue problems
Self-adjoint operators (like S-L) have nice properties:
1 The eigenvalues are real.
2 The eigenfunctions are orthogonal to one another (with
respect to the same inner product used to define the adjoint).
Sturm-Liouville eigenvalue problem: find v (x) ∈ C0∞ [a, b] so that
d dv p(x)
+ q(x)v (x) + λv (x) = 0,
dx
dx
Sturm-Liouville eigenvalue problems
Self-adjoint operators (like S-L) have nice properties:
1 The eigenvalues are real.
2 The eigenfunctions are orthogonal to one another (with
respect to the same inner product used to define the adjoint).
Sturm-Liouville eigenvalue problem: find v (x) ∈ C0∞ [a, b] so that
d dv p(x)
+ q(x)v (x) + λv (x) = 0,
dx
dx
Facts about this problem:
1 The real eigenvalues can be ordered λ1 < λ2 < λ3 . . . so that
there is a smallest (but not largest) eigenvalue.
2 The eigenfunctions vn (x) corresponding to each eigenvalue
λn (x) form a complete set, i.e. any f ∈ C0∞ [a, b], we can
write f as a (infinite) linear combination
f =
∞
X
n=1
cn vn (x).
Example: A Sturm-Liouville eigenvalue problem
Consider operator L = d 2 /dx 2 on the vector space C0∞ [0, π]. The
eigenvalue problem reads
d 2v
+ λv = 0,
dx 2
v (0) = 0,
v (π) = 0,
Example: A Sturm-Liouville eigenvalue problem
Consider operator L = d 2 /dx 2 on the vector space C0∞ [0, π]. The
eigenvalue problem reads
d 2v
+ λv = 0,
dx 2
v (0) = 0,
v (π) = 0,
If λ > 0, ODE has general solution
√
√
v (x) = A cos( λx) + B sin( λx).
Left boundary condition implies A cos(0) + B sin(0) = 0, so that
A = 0.
Example: A Sturm-Liouville eigenvalue problem
Consider operator L = d 2 /dx 2 on the vector space C0∞ [0, π]. The
eigenvalue problem reads
d 2v
+ λv = 0,
dx 2
v (0) = 0,
v (π) = 0,
If λ > 0, ODE has general solution
√
√
v (x) = A cos( λx) + B sin( λx).
Left boundary condition implies A cos(0) + B sin(0) = 0, so that
A = 0.
√
√
Then other b.c. implies B sin(π λ) = 0 so π λ is a multiple of π
or
λn = n2 , n = 1, 2, 3, . . . .
The corresponding eigenfunctions are
√
vn (x) = sin( λx) = sin(nx), n = 1, 2, 3, . . .
Example: A Sturm-Liouville eigenvalue problem, cont.
Case λ = 0 leads to v (x) = Ax + B, which needs A = B = 0 to
satisfy the boundary conditions.
Example: A Sturm-Liouville eigenvalue problem, cont.
Case λ = 0 leads to v (x) = Ax + B, which needs A = B = 0 to
satisfy the boundary conditions.
p
p
Case λ < 0 gives v (x) = A exp( |λ|x) + B exp(− |λ|x).
Boundary
pA + B = 0 and
p conditions imply
A exp( |λ|π) + B exp(− |λ|π) = 0 or
1
1
p
p
exp( |λ|π) exp(− |λ|π)
A
B
=
0
0
.
Example: A Sturm-Liouville eigenvalue problem, cont.
Case λ = 0 leads to v (x) = Ax + B, which needs A = B = 0 to
satisfy the boundary conditions.
p
p
Case λ < 0 gives v (x) = A exp( |λ|x) + B exp(− |λ|x).
Boundary
pA + B = 0 and
p conditions imply
A exp( |λ|π) + B exp(− |λ|π) = 0 or
1
1
p
p
exp( |λ|π) exp(− |λ|π)
A
B
=
0
0
.
p
p
Determinant is exp( |λ|π) − exp(− |λ|π) 6= 0 so that the only
solution is A = B = 0.
Fourier series
Big payoff: completeness of eigenfunctions means that any smooth
function with f (0) = 0 = f (π) can be written
f (x) =
∞
X
n=1
Bn sin(nx),
“Fourier sine series”
Fourier series
Big payoff: completeness of eigenfunctions means that any smooth
function with f (0) = 0 = f (π) can be written
f (x) =
∞
X
Bn sin(nx),
“Fourier sine series”
n=1
Computing coefficients in an orthogonal expansion is just a matter
of taking inner products:
Rπ
f (x) sin(nx)dx
hf (x), sin(nx)i
= 0R π 2
Bn =
.
hsin(nx), sin(nx)i
0 sin (nx)dx
Other Fourier series
Series type
Space of functions
Fourier
f (x) : [−L, L] → R
f (−L) = f (L)
f 0 (−L) = f 0 (L)
Orthogonal expansion
for f (x)
P
A0
nπx
+ ∞
n=1 An cos( L
2P
∞
nπx
n=1 Bn sin( L )
Coefficients
)
+
RL
An = L1 −L
f (x) cos( nπx
)dx
L
RL
f (x) sin( nπx
)dx
Bn = L1 −L
L
Sine
f (x) : [0, L] → R
f (0) = 0 = f (L)
P∞
Bn sin( nπx
)
L
RL
Bn = L2 −L
f (x) sin( nπx
)dx
L
Cosine
f (x) : [0, L] → R
f 0 (0) = 0 = f 0 (L)
A0
2
P∞
An cos( nπx
)
L
RL
An = L2 −L
f (x) cos( nπx
)dx
L
Complex
f (x) : [−L, L] → C
f (−L) = f (L)
f 0 (−L) = f 0 (L)
P∞
)
exp( −inπx
L
1 R L f (x) exp( inπx )dx
cn = 2L
−L
L
n=1
+
n=1
n=−∞ cn
Linear Integral operators
“Hilbert-Schmidt” integral operators
Z
Lu(x) =
k(x, y )u(x)dx.
Ω
obey linearity
Z
L(c1 u1 + c2 u2 ) = c1
Z
k(x, y )u1 (x)dx + c2
Ω
= c1 Lu1 + c2 Lu2 .
k(x, y )u2 (x)dx
Ω
Linear Integral operators
“Hilbert-Schmidt” integral operators
Z
Lu(x) =
k(x, y )u(x)dx.
Ω
obey linearity
Z
L(c1 u1 + c2 u2 ) = c1
Z
k(x, y )u1 (x)dx + c2
Ω
k(x, y )u2 (x)dx
Ω
= c1 Lu1 + c2 Lu2 .
Often inverses of differential operators: Lu = g has solution
u = L−1 g where L−1 is HS type
Linear Integral operators
“Hilbert-Schmidt” integral operators
Z
Lu(x) =
k(x, y )u(x)dx.
Ω
obey linearity
Z
L(c1 u1 + c2 u2 ) = c1
Z
k(x, y )u1 (x)dx + c2
Ω
k(x, y )u2 (x)dx
Ω
= c1 Lu1 + c2 Lu2 .
Often inverses of differential operators: Lu = g has solution
u = L−1 g where L−1 is HS type
Adjoints are similar. Example: using the L2 inner product
Z
Z
hLu, v i = v (y ) k(x, y )u(x) dxdy
ZΩ
ZΩ
= u(x) k(x, y )v (y ) dydx = hu, L† v i
Ω
where
L†
Ω
is an integral operator with kernel k(y , x).