sturm liouville.problem.for.fourier.series.transform

8/2/2019 Sturm Liouville.problem.for.Fourier.series.transform

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Sturm Liouville problem leading to complete Fourier Series

Q: (i) Identify the corresponding Sturm-Liouville operator (SLO) for the Sturm-Liouville

problem (SLP) appearing below. (ii) Solve SLP, identify the dimension of each eigenspace

of SLO, and identify an orthonormal basis for each eigenspace of SLO. (iii) Identify the

corresponding Sturm-Liouville transform (SLT).

ODE X(x) = X(x), x (L, L)BC X(L) = X(L), X(L) = X(L)

Solution: (i) In this case, the Sturm-Liouville operator (SLO) is the linear transformation

T defined by T[X] =

X on a vector space domainT. Here domainT is a suitable vector

space of functions X : (L, L) R such that both X and T[X] satisfy the BC in somesuitable sense. Also domainT is a subspace of the Hilbert space of all square integrable

functions on (L, L) with respect to the inner product

(f, g) LL

f(x)g(x) dx

and corresponding norm

f

L

L

f(x)2 dx

Since SLP is regular, solving SLP consists of finding the eigenvalues and corresponding

eigenspaces E of T. Solving singular Sturm-Liouville problems, calls for finding the gen-

eralized eigenvalues, eigenfunctions, and eigenspaces of the corresponding singular Sturm-

Liouville operator. These generalized eigenvalues and eigenfunctions are suitable limits of

eigenvalues and eigenfunctions of regular Sturm Liouville problems which are formed by

restrictin the singular Sturm Liouville problem to closed proper subintervals of the basic

interval (

L, L).

(ii) Now we solve SLP.

Suppose that < 0, and X is a non trivial solution of ODE and BC. Then there exist

A, B R such thatX(x) = Aex

+ Bex

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Hence

X(L) = AeL + Be(L)

= x(L) = AeL

+ BeL

and

X(L) = AeL(

) + Be(L)() =

= X(L) = AeL(

) + BeL(

)

Hence

A

eL eL

+ B

e(L) eL

= 0

and

A

eL() eL() + B e(L)() eL() = 0Hence, since

= 0,

A

eL eL

+ B

e(L) eL

= 0

and

A

eL eL

+ Be(L)

+ eL

= 0

Thus, since X is not identically zero and A and B are not both zero, we have

eL eL

e(L) eL

eL

eL

e(L) + eL

=

=

eL eL

e(L) eL

1 1

1 1

=

=

eL

eL

e(L)

eL

(2) = 0

Thus either eL = eL

or e(L)

= eL

Thus since for x, y R we have that ex = ey if and only

if x = y either L = L or (L) = L Thus since tacitly assumed that L > 0 either = 0 or = 0 Thus = 0 which is a contradiction. So 0 <

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Next suppose that = 0 and that X is a non trivial solution of ODE and BC. Then

there exist A, B R such that not both A and B are zero and X(x) = A + Bx HenceX(L) = A+B(L) = x(L) = A+BL and X(L) = B = X(L) = B Hence 0A2BL = 0and 0

A + 0

B = 0 Thus B = 0. Thus A

= 0 and X(x)

A.

Thus if = 0, every possible solution of ODE and BC is a constant function on [L, L].Also, every such constant function clearly satisfies the ODE with = 0 and the BC.

If A R then LL A2dx = 2LA2 = 1 if and only if A = 12L . Thus the functionX0(x)

12L is a normalized eigenfunction of SLP for = 0

Thus

E0 = {AX0 : A R}

and we see that dim E0 = 1Finally suppose that > 0, and X is a non trivial solution of ODE and BC. Then there

exist A, B R such that

X(x) = A cos(x

) + B sin(x

)

Hence

X(L) = A cos(L

) + B sin(L

) = x(L) = A cos(L

) + B sin(L

)

and

A cos(L

) B sin(L

) = A cos(L

) + B sin(L

)

and

B sin(L

) = 0

Also

X(x) = A(1) sin(x

)

+ B cos(x

)

and

X(L) = A(1) sin(L

)

+ B cos(L

)

=

= X(L) = A(1) sin(L

)

+ B cos(L

)

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and

A sin(L

) + B cos(L

) = A(1) sin(L

) + B cos(L

)

and

A sin(L) = 0Thus, since X is not identically zero and A and B are not both zero, we have

sin(L

) 0

0 sin(L

)

= sin2(L

) = 0

Thus there exists n Z such that L

= n. Since L > 0 and

> 0, there exists

n {1, 2, 3,...} such that L

= n and = nL

2. Thus

n

L

2: n {1, 2, 3,...}

We conclude that every possible strictly positive eigenvalue of SLP is an element of the set

displayed above.

Now choose n {1, 2, 3,...} and note that cosnx

L

and sin

nxL

are both non-

trivial solutions of ODE for =nL

2. Thus every element of the set displayed above is an

eigenvalue of SLP. Thus the set of all eigenvalues of SLP is

n

L

2: n {0, 1, 2, 3,...}

Now fix n {1, 2,...} and set

En

X : X(x) =n

L

2X(x) for all x (L, L) and

X(

L) = X(L) and X(

L) = X(L)

}So En is the

nL

2eigenspace of SLO. By ode theory we know that dim En 2. From the

above we know that cosnx

L

and sin

nxL

are both elements of En.

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We now show that these functions are independent elements of En. Thus suppose that

A, B R and that A cosnx

L

+ B sin

nxL

= 0 for all x (L, L). Then

A cosnxL

+ B sinnxL

x=0

= A = 0

and

A cos

nxL

+ B sin

nxL

= A(1) sin

nxL

nL

+ B cosnx

L

nL

= 0

for all x (L, L), and

A(1)sinnx

L

nL

+ B cosnx

L

nL

x=0

= Bn

L= 0

Thus A = B = 0. Thus cos

nxL

and sin

nx

L

are independent on (L, L) and hence

independent elements of En.

Thus dim En = 2.

Now we construct an orthonormal basis of En; here orthonormal refers to the inner

product of the Hilbert space which is the source and target of the SLO T. In general we

would have to use the Gram-Schmidt procedure starting with a basis for En. But in our case

the basis at hand is already orthogonal. To see this note that

LL

cosnx

L

sin

nxL

dx =

substitute w = cos

nx

L

and dw = (1)sin

nx

L

n

Ldx

=n

L

1 11

w dw = 0

So we only need to normalize these basis functions. We have

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LL

cos2nx

L

dx = (1/2)

LL

1 + cos

2nx

L

dx =

= (1/2)

x + sin2nx

L2n

L1

L

L=

= (1/2)

L + sin

2nL

L

2n

L

1 (1/2)

L + sin

2n(L)

L

2n

L

1=

= (1/2) (L + 0) (1/2) (L + 0) = L

Also

LL

sin2

nxL

dx = (1/2)

LL

1 cos2nxL

dx =

= (1/2)

x sin

2nx

L

2n

L

1L

L

=

= (1/2)

L sin

2nL

L

2n

L

1 (1/2)

L sin

2n(L)

L

2n

L

1=

= (1/2) (L 0) (1/2) (L 0) = L

This checks since the sum of the two integrals is clearly 2 L

Thus we define

Xn,1(x) 1L

cosnx

L

and Xn,2(x) 1

Lsin

nxL

for all x [L, L], and conclude that (Xn,1, Xn,2) is an orthonormal basis for En. Thiscompletes the solution of (ii).

(iii) The Sturm-Liouville transformation corresponding to T is transformation T whose

domain is the Hilbert space of all square integrable real functions on [L, L] and whose rangeis the set of square summable real functions on the set {0} {(n, 1) : n = 1, 2,...} {(n, 2) :n = 1, 2,...} and which is defined by the rules

T [f](0) LL

f(x)X0(x) dx =

LL

f(x)12L

dx =12L

LL

f(x) dx

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and

T [f](n, 1) LL

f(x)Xn,1(x) dx =

LL

f(x)1L

cosnx

L

dx =

=

1

LLL f(x)cos

nxL

dx

and

T [f](n, 2) LL

f(x)Xn,2(x) dx =

LL

f(x)1L

sinnx

L

dx =

=1L

LL

f(x)sinnx

L

dx

By the Sturm-Liouville expansion theorem we have, in the sense explained earlier,

f(x) = T [f](0)X0(x) +n=0

(T [f](n, 1)Xn,1(x) + T [f](n, 2)Xn,2(x)) =

= T [f](0)12L

+n=0

T [f](n, 1)

1L

cosnx

L

+ T [f](n, 2)

1L

sinnx

L

Thus to use a more familiar notation, if we define

A0

1

2L L

Lf(x) dx

and

An 1L

LL

f(x)cosnx

L

dx

and

Bn 1L

LL

f(x)sinnx

L

dx

we have that

f(x) = A0 12L

+

n=0

An 1

Lcos

nxL

+ Bn 1

Lsin

nxL

By comparison, the theory of the fourier series transform states that if

a0 1L

LL

f(x) dx

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and

an 1L

LL

f(x)cosnx

L

dx

and

bn 1

LLL f(x)sin

nxL

dx

we have that

f(x) =a02

+n=0

an cos

nxL

+ bn sin

nxL

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sturm liouville.problem.for.fourier.series.transform

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