Bernatz R. Fourier Series and Numerical Methods for Partial Differential Equations

Подождите немного. Документ загружается.

72 STURM-LIOUVILLE PROBLEMS

3.3.1 Eigenfunction Orthogonality

Theorem 3.1

IfX

and X

are

arbitrary,

distinct eigenvalues for

the

Sturm-Liouville

problem

(3.6),

and X

(x) and X

(x) are the respective corresponding eigenfunc-

tions, then X

and X

are orthogonal with respect to the weight function p(x) [as

given in the ODE of Problem (3.6)] on the interval a < x < b. That is,

/ p(x)X

(x)X

(x)dx = 0

for m φ η.

Proof:

By definition of eigenvalue-eigenfunction pairs it follows that:

[r(x)X'

(x)]

+ [q(x) + X

p(x)]X

(x) = 0 (3.7)

and

[r(x)X'

(x)Y + \q(x) + X

p(x)]X

(x) = 0 (3.8)

Multiplying Equation (3.7) by X

, Equation (3.8) by

(x),

and subtracting the

results gives

{x)[r{x)X'

{x))'

- X

(x)[r(x)X'

(x)}'+

p(x)X

(x)X

(x) - \

p(x)X

(x)X

(x) = 0 (3.9)

Rearranging Equation (3.9) and expanding the right-hand side results in

(Xm - X„)p(x)X

(x)Xn(x) = X

{x)[r{x)X'

{x)]' - Xn{x)[r{x)X'

(x)]'

= X

(x)[r'(x)X'

(x) + r(x)X';(x)} -

(x)[r'(x)X'

(x) + r(x)X'^(x)}

= \X

(x)X'

(x) - Xn(x)X

(x)V(x) +

(x)X'¿(x) - Xn(x)X

(x)]r(x) (3.10)

Using the fact that

(x)X'

(x) - X

{x)X'

{x)]' = X

(x)X'

(x) + X

(x)Xn(x) -

(x)x'

(x)

Xn(x)x

(x)

= X

(x)X'¿(x) - X

(x)X^(x) (3.11)

Equation (3.10) becomes

(Xm - X

)p(x)X

(x)X

(x) = [(Xm(x)X'

(x) ~ X

(x)X'

(x))r(x)}'

= ~[(X

(x)X'

(x) - X

(x)X'

(x))r(x)}

(3.12)

PROPERTIES 73

Taking the definite integral of both sides in Equation (3.12) gives

(Xm-K)J p(x)X

(x)Xn(x)dx = j ^ [(X

(x)X'

(x)

-X

(x)X

(x))r(x)}dx

r(x)A(x)\

r(b)A{b)-r(a)A{a) (3.13)

where A(x) is defined as

A(x) = X

{x)X'

(x) - X

{x)X'

{x)

(x)

XUx)

(x)

xZ(x)

(3.14)

Note: When the first boundary condition of the Sturm-Liouville problem is consid-

ered, we may write

aiX

(a) + a

(a) = 0

(a)+a

XM = 0

(

)

Because a\ and a

cannot both be zero, the system of equations given in Equations

(3.15) has a nontrivial solution that means the determinant, given by the function

A(x) calculated at x = a, must be zero. That is, Δ(α) = 0. Similarly, the second

boundary condition of the Sturm-Liouville problem and requirement that not both

bi and b

are zero leads us to accept that A(b)

—

0 as well. Consequently,

- A

) / p(x)X

(x)X

(x)dx = r(b)A(b) - r(a)A(a)

= 0 (3.16)

Because the eigenvalues A

and λ

are unique, we must conclude from Equation

(3.16) that the integral inner product with weight function p(x) is zero, so eigen-

functions X

(x) and X

(x) are orthogonal with respect to p(x) on the interval

a < x < b.

3.3.1.1 Periodic Boundary Conditions Suppose the boundary conditions of

Problem (3.6) are replaced with periodic boundary conditions. That is, suppose the

problem to solve is

Í [r(x)X(x)]

+ [q(x) + Xp(x)} X(x) = 0 (a < x < b)

(3 {J)

\ X(a) = X{b) ^'

Under these boundary conditions,

have Δ(b) = Δ(α). If, in addition, r(a) = r(b),

the result in Equation (3.16) would be the same, so X

(x) and X

(x) would be

orthogonal here as well.

This is just one instance in which the eigenfunctions of distinct eigenvalues of a

singular Sturm-Liouville problem are orthogonal.

74 STURM-LIOUVILLE PROBLEMS

3.3.2 Real Eigenvalues

Now we show that eigenvalues of Problem (3.6) are real numbers. First, a formal

statement of the result is given.

Corollary 3.1 If X is an eigenvalue of the Sturm-Liouville problem

(3.6),

then X is a

real number.

Proof Suppose an arbitrary eigenvalue λ of Problem (3.6) is complex. That is,

suppose

= a + iß

where a and ß

are

real numbers and β φ 0, is an eigenvalue and X(x)

—

u(x)+iv(x)

is the corresponding eigenfunction. Note: Both component functions u(x) and v(x)

of X(x) are real-valued, and v(x) is not necessarily nonzero. Next, we establish that

the conjugate pair

λ = a

—

iß and X(x) = u(x)

—

iv(x)

is also an eigenvalue-eigenfunction pair of Problem (3.6). To show this, we begin

with the fact that λ and X(x) solve Problem (3.6) to write

[r(x)X'(x)Y + (q(x) + '\p(x))X(x) = 0 (3.18)

Taking the conjugate of both sides of Equation (3.18) and simplifying using properties

of conjugate algebra we have

=^>

r(x)X'(x)]' + (q(x) + Xp(x))X{x) = 0

r{x)X'(x)Y + (q{x) + Xp(x))X(x) = 0

r(x)X'(x)] +{q{x) + Xp(x))X(x) = 0

r(x)X'(x)} + (q(x) + Xp{x))X(x) = 0

r{x)X

(x)} +(q(x) + \p(x))X(x) = 0

The last line in the sequence of manipulations shows that λ

—

X(x) is a solution to

Problem (3.6).

Now, if

take β φ 0, the conjugate pair λ and λ are distinct. Invoking the result

of Theorem 3.1 we may write

p(x)X(x)X(x)dx = 0 (3.19)

Using the fact that

X(x)X(x) = [u(x) + iv(x)][u(x) - iv(x)} = u

(x) + v

(x) = \X(x)

PROPERTIES

Equation (3.19) implies

p(x)\X(x)\

= 0

(3.20)

Because p{x) is not identically zero and

X(x)

is continuous on a

< x <

Equation

(3.20) implies that

|X(Z)|

EEO

which contradicts the fact that

X(x)

as an

eigenfunction

Problem (3.6),

is a

nontrivial function. Consequently, we must reject the assumption that

is complex.

3.3.3 Eigenfunction Uniqueness

In this section, we address the uniqueness issue of eigenfunctions for a given eigen-

value of Problem (3.6). The proof of the main result here is based on an existence

and uniqueness result

for

second-order, linear, homogeneous ordinary differential

equations. It is stated below, as a lemma, for sake of reference.

Lemma 3.1 Given the

second-order,

linear,

ordinary differential equation

y"(x)

P{x)y'(x)

Q(x)y(x)

= 0 (a < x <

with

prescribed initial conditions

y{xo)

and

y'(xo)

y'o

for xo in a

< x <

b. If functions

P(x)

and Q(x) are continuous on

x <b,

then

there is one and only one function y, continuous together with its derivative y', that

satisfies the ODE and prescribed initial conditions.

The proof

this statement will not be presented here. Instead, the reader

encouraged to see Coddington [12] (Chapter 6), or Utz [30] (Chapter 8). This result

is stated as a theorem below.

Theorem 3.2

IfX

and

are eigenfunctions for a regular Sturm-Liouville problem

(3.6) corresponding to an arbitrary eigenvalue λ, then

Y(x)

= CX{x)

(a<x<b) (3.21)

where

is a nonzero constant.

Proof.

A minor preliminary consideration in the proof is the fact that for

A, B,

(A and

not both zero) and

nonzero,

Y(x) = CX(x) & AX{x) - BY(x) = 0

The right-hand version of this last line will be the primary objective in establishing

the proof to Theorem 3.2.

76 STURM-LIOUVILLE PROBLEMS

Suppose for a regular Sturm-Liouville problem (3.6) the eigenvalue λ has X(x)

and Y(x) as eigenfunctions. Our objective will be to show there exist constants A

and B, such that

AX(x) - BY(x) = 0 (3.22)

In an effort to determine possible values for A and B, we will examine the first of

the boundary conditions in Problem (3.6). Because both X and Y are solutions of

Problem (3.6), it follows that:

αχΧ(α) 4- a

X'{a) = 0

a{Y(a) 4- a

Y\a) = 0

(3.23)

with ai and a

both not zero. For this system to have a nontrivial solution, it must

be that the corresponding system coefficient matrix has zero determinant. That is,

X(a)Y'(a) - X'{a)Y(a) = 0 (3.24)

Comparison of Equations (3.22) and (3.24) suggests letting A = Y'{a) and B =

X'{a).

Define function U as

U(x) =

Y'{a)X{x)

- X'{a)Y{x) (3.25)

Our objective is to show that U(x) = 0.

that

end,

because U is

linear combination

of solutions to the ODE in the Sturm-Liouville problem, U itself

a solution. Now,

U(a)

—

0 by Equation (3.24). Furthermore,

U'(a) =

Y'(a)X'(a)

- X'{a)Y'{a) = 0 (3.26)

The ODE of the Sturm-Liouville problem, together with the initial conditions for U

prescribed above, define the IVP

[r(x)F(x)]' + [q(x) 4- λρ(χ)1 F(x) =0 (a < x < b)

n 9T

F(o) = 0 and F'(a) = 0 ^

,z/;

for which U{x) defined above is a solution. Because the trivial function O(x) Ξ 0

also solves IVP (3.27), it follows from Lemma 3.1 that

U(x) =

Y'(a)X(x)

- X'(a)Y{x) = 0 (3.28)

We have reached our objective provided that not both X'(a) and Y'(a) are

zero.

First,

note that if one of these values is zero than the other must be. Suppose, for instance,

that X'(a) = 0. Combining Equations (3.25) and (3.28) gives

(a)X{x)

= 0. If

Y'(a) φ 0, then X(x) = 0, which violates the hypothesis that X is an eigenfunction

of Problem (3.6). Similar arguments pertain to the case if Y'(a)

—

Now we know that either both X'{a) and Y'(a) are zero, or both are nonzero. If

neither X'(a) nor Y'{a) are zero, the fact that U(x) = 0 and the definition of U(x)

allows one to write

PROPERTIES 77

and our objective is attained.

Suppose, however, that both X'(a) and Y

(a) are zero. In this case, it follows that

neither X(a) and Y (a) are zero. For example, ΊίΧ(ά) were zero, and X'(a)

—

0 as

assumed, then X(x) = 0 by Lemma (3.1) applied to IVP (3.27).

So,

in the event both X'(a) and Y'{a) are zero, because we know neither X(a)

nor Y (a) are zero, define the function W(x) as

W(x) = Y(a)X(x) - X(a)Y(x) (3.29)

Using arguments similar to those used for U(x), it can be shown that W(x) = 0.

Because neither X(a) nor Y (a) is zero, It follows, by the definition of W(x), that

X{a)

A secondary, yet important result of Theorem 3.2 is presented here as a corollary.

Corollary 3.2 Each eigenfunction X of a given Sturm-Liouville problem (3.6) may

be made real-valued by multiplying by an appropriate nonzero constant.

Proof.

Let X be the complex eigenfunction associated with the (real-valued)

eigenvalue λ of the Sturm-Liouville problem (3.6). Write X in its real and imaginary

component parts as

X(x) = U(x) + iV(x)

It can be shown that both real-valued functions U and V are eigenfunctions associated

with the eigenvalue λ. Consequently, by Theorem (3.2) we know there exists a

nonzero constant C, such that

V(x) = CU(x)

Then,

X(x) = U(x) + iCU(x) = (1 4- iC)U(x)

Solving the last equation for U(x) gives

rhc

x{x)

Because U is real-valued, we have transformed the complex eigenfunction X into a

real-valued eigenfunction by multiplication by the constant factor

1 + iC

3.3.4 Non-negative Eigenvalues

The last fact to be established in this section is that, under certain additional cir-

cumstances, the eigenvalues of regular Sturm-Liouville are all non-negative. The

statement of the theorem and its proof follow.

78 STURM-LIOUVILLE PROBLEMS

Theorem 3.3 Suppose λ is an eigenvalue of the regular Sturm-Liouville problem

(3.6).

If the following conditions are satisfied

q(x) < 0 (0 < x < b) and a

< 0, 6

> 0

then λ > 0.

Proof.

For an arbitrary eigenvalue λ of the regular Sturm-Liouville problem we

know, by Theorem 3.2, that there exists a real-valued eigenfunction X that satisfies

the ODE

[r(x)X'(x)]' + [q(x) + Xp(x)]X(x) = 0 (3.30)

Multiplying each term in Equation (3.30) by X and integrating each term from a to

b gives

pb pb pb

I X(x)[r(x)X'(x)}'dx+ q(x)X

(x)dx + \ p(x)X

(x)dx = 0 (3.31)

Ja Ja Ja

A slight rearrangement gives

pb pb pb

X p(x)X

(x)dx = - X(x)[r(x)X'(x)}'dx+ -q(x)X

(x)dx = 0

Ja Ja Ja

(3.32)

Using integration by parts results in

/ X(x)[r{x)X\x)]

dx = r{x)X(x)X'(x)\

- Í r(x)[X'(x)]

Ja Ja

= r(b)X(b)X'(b) - r(a)X(a)X'{a)

r(x)[X'(x)}

dx (3.33)

Using the result of Equation (3.33) in Equation (3.32) gives

X p(x)X

(x)dx = r{a)X{a)X'{a)-r{b)X{b)X'{b)

+ [ r(x)\X'(x)Ydx

■

/ r(x)[X'(x)Y

+ / -q(x)X

(x)dx (3.34)

Because q(x) < 0 on [a,

b],

the second integral on the right-hand side of Equation

(3.34) is non-negative. Regularity of the Strum-Liouville problem means r(x) > 0

on [a,

b].

Consequently, the first integral on the right-hand side of Equation (3.34) is

non-negative as well.

Next, we will consider the term r(a)X(a)X'(a). If either a\ or a

is zero, then

X'(a)

or X(a) is zero to satisfy the first boundary condition of (3.6). Therefore,

EXAMPLES 79

r(a)X(a)X'(a) is zero if at least one of a\ or a^ is zero. On the other hand, if neither

a\ nor

Ü2

is zero, it follows from this same boundary condition in Problem (3.6) that

r(a)X(a)X'(a) =

(«)[«i*(«)]

—α\(ΐ2

The summary conclusion on r(a)X(a)X'(a) is that it is non-negative in all instances.

The arguments of the preceding paragraph may be applied in a similar way to

conclude that r{b)X{b)X

{b) is non-negative in all instances.

Because all terms on the right-hand side of Equation (3.34) are non-negative, it

follows that

λ / p(x)X

(x)dx > 0 (3.35)

The fact that the integral in Equation (3.35) must be positive implies that

λ>0

3.4 EXAMPLES

Solutions to Sturm-Liouville problems resulting from Neumann and periodic bound-

ary conditions are presented in this section. The case for Dirichlet boundary condi-

tions is left as an exercise.

3.4.1 Neumann Boundary Conditions on [0, c]

Consider the following Sturm-Liouville problem

/ X"(x) + XX(x) =0

\ X'(0) = 0*ndX'(c) = 0

36)

which results, for example, in the case of ID heat transfer on the interval

[0,

for a

medium that is insulated (no heat transfer) at the end points of x = 0 and x

—

c. The

initial boundary value problem (3.3) is one example of such an application.

Comparing the Sturm-Liouville problem stated above with the general Sturm-

Liouville problem (3.6), we see that in Equation (3.36) r(x) = 1, q(x) = 0, and

p(x) Ξ 1. In terms of the interval, a = 0 and b — π. For the boundary conditions,

α,ι = bi = 0, and

a<i

— &2 = 1· This is an example of a regular Sturm-Liouville

problem because both parts of the regularity requirements (continuity of p,

<?,

r, and

r', as well as value requirements of p and r) are satisfied by the problem stated

above. As a consequence of regularity, we know for this Sturm-Liouville problem

there are an infinite number of discrete, real, eigenvalues having unique, orthogonal

eigenf unctions. Because the products α\α,2 and &1&2 or both zero, Property 5 of

Sturm-Liouville problems implies all eigenvalues are non-negative. Next, a case

analysis based on values of λ will provide the set of all possible eigenvalues and

eigenfunctions.

80 STURM-LIOUVILLE PROBLEMS

Case

= 0

For λ = 0, we have X"{x) = 0, which means X(x) = A-\-Bx represents the general

solution to the ODE. Applying the boundary conditions gives X'(0) — B · 0 = 0

and X'{c) = B · 1 = 0 => B — 0. These conditions give no restrictions on A,

so Xo(x) = 1 is the simplest eigenfunction corresponding to λ =. Any other eigen-

f unction corresponding to the eigenvalue of 0 is just

constant multiple of Xo (x) = 1.

Case

> 0

If λ > 0, then let

= k

(k > 0) and the ODE becomes

X"(x)

+ k

X(x) = 0,

which, as a second order, linear, homogeneous ODE, has the general solution

X(x) = A cos kx + B sin kx

Now, X'{x) = —kA sin kx + kB cos kx. Invoking the boundary conditions give

X'(0)

= 0=> -kAsink-0 + kBcosk-0 = 0=ïB = 0

and

X\c)

=0=> -kAsmk = 0

For nontrivial solutions (i.e., A φ. 0), it must be that

Tí

sin k = 0 => fc

= —, n = 1,2,3,...

So,

= fc^ = (^

)

, n = 1,2,3,... are eigenvalues for this case with corre-

sponding eigenfunctions X

(x) = cos (^^F) , n = 1,2,3,...

In summary, the eigenvalues for this problem are A

= (^)

, n

—

0,1,2,...

with corresponding eigenfunctions XQ{X) — 1 and X

(x) = cos(

zlÄ

^) , n =

1,2,3,...

3.4.2 Robin and Neumann BCs

The are some heat transfer applications in which energy is allowed to move across the

boundary. The resulting boundary condition involves both the boundary temperature

and its normal derivative (see Section 4.2 for details). The resulting Regular Sturm-

Liouville problem is solved for this case to determine the resulting eigenvalues and

associated eigenfunctions.

The Sturm-Liouville problem under consideration is

X"{x)

+ \X{x) = 0 (3.37)

ftX(0) - X'(0) = 0 and X'(c) = 0 (3.38)

EXAMPLES 81

As with the two previous cases already considered, we determine eigenvalues and

eigenfunctions using case analysis on λ. In the regular Sturm-Liouville problem

presented in Equations (3.38), q(x) — 0, a\a

— — h < 0, and &162 = 0, so that

Property 5 assures us that all eigenvalues for this example are non-negative.

Case

= 0

For λ = 0, we have X"(x)

—

0, which means X(x) = A+Bx represents the general

solution to the ODE. Applying the boundary condition at x = c gives X'{c) = 0 =>

B — 0. Applying the boundary condition at x — 0 gives hX(0)

—

X'(0) = 0 =>

h A

— 0

= 0

A = 0. Consequently, this not a nontrivial solution for the case λ = 0

Case

> 0

If λ > 0, then let λ = k

(k > 0), and the ODE becomes

X"{x)

+ k

X(x) = 0

which, as a second-order, linear, homogeneous ODE, has the general solution

X(x) = A cos kx + B sin kx

Invoking the boundary condition at x

—

0 first gives

hX(0) -X'(0) = 0 =¿> h(Acosk0 + Bsink0) -

(-kAsinkO

+ kB cos

kO)

= 0

=> hA - kB = 0

h_B_

^ k~~Ä

Next, the boundary condition at x = c gives

X'(c)

= 0 => -&.AsinA;c + kB cos kc = 0

=> A:( — A sin kc-\- B cos

fee)

= 0

sin fcc 5

cos kc A

tan kc— —

Combining the results of the boundary conditions gives

tan kc—

—

So the eigenvalues for this case are given by the x locations for which the graphs of

y = tan ex and y — ^ intersect, as shown in Figure 3.1. Therefore, the eigenvalues