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728

Chapter 9 Eigenvalue Problems and Fourier Series

way as initial-value problems except that the value of the function and its derivatives are given at two values of the independent variable instead of one. The general form of a second-order (two-point) boundary-value problem is

d2y

dy

a0 x y

a1 x

dx2

dx

dy

k1 y a k2

a

Α, h1 y b

dx

a2 x

f x,a

dy

b

Β

h2

dx

(9.1)

where k1 , k2 , Α, h1 , h2 , and Β are constants and at least one of k1 , k2 and at least one

of h1 , h2 is not zero.

Note that if Α Β 0, then we say the problem has homogeneous boundary

conditions. We also consider boundary-value problems that include a parameter

in the differential equation. We solve these problems, called eigenvalue problems,

in order to investigate several useful properties associated with their solutions.

EXAMPLE 9.1.1: Solve

y

y

y 0

0, 0 < x < Π

0, y Π

0.

SOLUTION: Because the characteristic equation is k2 1 0 with roots

i, a general solution of y

y 0 is y c1 cos x c2 sin x and it

k1,2

c1 sin x c2 cos x. Applying the boundary conditions,

follows that y

c2 0. Then, y c1 cos x. With this solution, we have

we have y 0

c1 sin Π 0 for any value of c1 . Therefore, there are inﬁnitely

y Π

many solutions, y c1 cos x, of the boundary-value problem, depending

on the choice of c1 . In this case, we are able to use DSolve to solve the

boundary-value problem

In[1823]:= sol

Out[1823]=

DSolve y x

y x

0 ,y x ,x

y Π

y x

C 1 Cos x

0, y 0

0

We conﬁrm that the boundary conditions are satisﬁed for any value of

C[1] by graphing several solutions with Plot in Figure 9-1.

In[1824]:= toplot

grays

Table y x /. sol /. C 1

i, 5 , 5

> i,

Table GrayLevel i , i, 0, 0.5, 0.5/10

In[1825]:= Plot Evaluate toplot , x, 0, Π ,

PlotStyle grays

9.1 Boundary-Value Problems, Eigenvalue Problems, Sturm–Liouville Problems

729

4

2

0.5

1

1.5

2

2.5

3

-2

-4

Figure 9-1 The boundary-value problem has inﬁnitely many solutions

From the result in the example, we notice a difference between initial-value problems

and boundary-value problems: an initial-value problem (that meets the hypotheses of

the Existence and Uniqueness Theorem) has a unique solution while a boundaryvalue problem may have one solution, more than one solution, or no solution.

EXAMPLE 9.1.2: Solve

y

y

0, 0 < x < Π

0, y Π

y 0

1.

SOLUTION: Using the general solution obtained in the previous

c2

0, so

example, we have y

c1 cos x c2 sin x. As before, y 0

c1 sin Π 0 1, the boundary

y c1 cos x. However, because y Π

conditions cannot be satisﬁed with any choice of c1 . Therefore, there is

no solution to the boundary-value problem.

As indicated in the general form of a boundary-value problem, the boundary conditions in these problems can involve the function and its derivative. However,

this modiﬁcation to the problem does not affect the method of solution.

EXAMPLE 9.1.3: Solve

y

y 0

y

0, 0 < x < 1

3y 0

0, y 1

y1

1.

See Chapter 4 and

Theorem 2.

730

Chapter 9 Eigenvalue Problems and Fourier Series

SOLUTION: The characteristic equation is k2

1

0 with roots

1. Hence, a general solution is y

c1 ex c2 e x with derivak1,2

tive y

c1 ex c2 e x . Applying y 0

3y 0

0 yields y 0

3y 0

4c1 2c2 0. Because y 1 y 1

1,

c1 c2 3 c1 c2

y 1

c1 e1

y1

c2 e

1

c1 e1

c2 e

1

2c1 e

1,

1

and c2

2e

unique solution y

1

. Thus, the boundary-value problem has the

e

1 x 1 x 1 x 1

e

e

e x 1 , which we conﬁrm with

2e

2e

e

Mathematica. See Figure 9-2.

so c1

In[1826]:= sol

Out[1826]=

DSolve

y 1

1

y x

2

y x

y 1

1 x

2

y x

0, y 0

1 ,y x ,x

3y 0

0,

2x

In[1827]:= Plot y x /.sol, x, 0, 1 ,

AspectRatio Automatic

9.1.2 Eigenvalue Problems

If a value of the parameter

leads to the trivial solution,

then the value is not

considered an eigenvalue of

the problem.

We now consider eigenvalue problems, boundary-value problems that include

a parameter. Values of the parameter for which the boundary-value problem has

a nontrivial solution are called eigenvalues of the problem. For each eigenvalue,

the nontrivial solution that satisﬁes the problem is called the corresponding eigenfunction.

0.3

0.2

0.1

0.2

0.4

0.6

0.8

1

-0.1

Figure 9-2 The boundary-value problem has a unique solution

9.1 Boundary-Value Problems, Eigenvalue Problems, Sturm–Liouville Problems

EXAMPLE 9.1.4: Solve the eigenvalue problem y

subject to y 0

0 and y p

0.

Λy

0, 0 < x < p,

SOLUTION: Notice that the differential equation in this problem

differs from those solved earlier because it includes the parameter Λ.

However, we solve it in a similar manner by solving the characteristic

equation k2 Λ 0. Of course, the values of k depend on the value of

the parameter Λ. Hence, we consider the following three cases.

0 with

1. (Λ

0) In this case, the characteristic equation is k2

0, which indicates that a general solution is y

roots k1,2

0 yields

c1 x c2 . Application of the boundary condition y 0

y0

c1 0 c2 0, so c2 0. For the second condition, y p

0, so c1

0 and y

0. Because we obtain the trivial

c1 p

solution, Λ 0 is not an eigenvalue.

2. (Λ < 0) To represent Λ as a negative value, we let Λ

Μ2 < 0.

2

2

Μ

0, so k1,2

Μ.

Then, the characteristic equation is k

Therefore, a general solution is, y

c1 eΜx c1 e Μx (or y

c1 cosh Μx c1 sinh Μx). Substitution of the boundary condition

c1 . Because y p

0

y0

0 yields y 0

c1 c2 0, so c2

0, substitution gives us

indicates that y

c1 eΜp c1 e Μp

the equation y p

c1 eΜp c1 e Μp c1 eΜp e Μp . Notice that

Μp

Μp

e

0 only if eΜp e Μp which can only occur if Μ 0

e

02

0 which contraor p

0. If Μ

0, then Λ

Μ2

dicts the assumption that Λ < 0. We also assumed that p > 0, so

c1 eΜp e Μp implies that c1 0, so

eΜp e Μp > 0. Hence, y p

c1 0 as well. Because Λ < 0 leads to the trivial solution

c2

y 0, there are no negative eigenvalues.

3. (Λ > 0) To represent Λ as a positive value, we let Λ

Μ2 >

2

2

Μ

0 with

0. Then, we have the characteristic equation k

Μi. Thus, a general solution is

complex conjugate roots k1,2

c1 cos Μ 0 c2 sin Μ 0 c1 ,

y c1 cos Μx c2 sin Μx. Because y 0

0. Hence,

the boundary condition y 0

0 indicates that c1

0 yields y p

c2 sin Μp, so

y c2 sin Μx. Application of y p

either c2 0 or sin Μp 0. Selecting c2 0 leads to the trivial

solution that we want to avoid, so we determine the values of

Μ that satisfy sin Μp 0. Because sin nΠ 0 for integer values

731

In Example 9.1.5 we solve the

same differential equation but

use the boundary conditions

y 0

0 and y p

0.

732

Chapter 9 Eigenvalue Problems and Fourier Series

of n, sin Μp 0 if Μp nΠ, n 1, 2, . . . Solving for Μ, we have

Μ nΠ/ p, so the eigenvalues are

Λ

Μ2

Λn

nΠ

p

2

,n

1, 2, . . . .

Notice that the subscript n is used to indicate that the parameter

depends on the value of n. (Notice also that we omit n

0,

because the value Μ

0 was considered in Case 1.) For each

eigenvalue, the corresponding eigenfunction is obtained by

substitution into y c2 sin Μx. Because c2 is arbitrary, we choose

nΠ/ p 2 , n 1, 2, . . . has

c2 1. Therefore, the eigenvalue Λn

corresponding eigenfunction

yx

yn x

sin

nΠx

,n

p

1, 2, . . . .

We did not consider negative values of n because sin nΠx/ p

sin nΠx/ p ; the negative sign can be taken into account in the

constant; we do not obtain additional eigenvalues or eigenfunctions by using n

1, 2, . . . .

We will ﬁnd the eigenvalues and eigenfunctions in Example 9.1.4 quite useful in

future sections. The following eigenvalue problem will be useful as well.

In Example 9.1.4 we solve the

same differential equation but

use the boundary conditions

y0

0 and y p

0.

EXAMPLE 9.1.5: Solve y

0.

y p

Λy

0, 0 < x < p, subject to y 0

0 and

SOLUTION: Notice that the only difference between this problem and

that in Example 9.1.4 is in the boundary conditions. Again, the characteristic equation is k2 Λ 0, so we must consider the three cases Λ 0,

Λ < 0, and Λ > 0. Note that a general solution in each case is the same

as that obtained in Example 9.1.4. However, the ﬁnal results may differ

due to the boundary conditions.

c1 . Therefore, y 0

c1 0,

1. (Λ 0) Because y c1 x c2 , y

0

so y c2 . Notice that this constant function satisﬁes y p

for all values of c2 . Hence, if we choose c2 1, then Λ 0 is an

1.

eigenvalue with corresponding eigenfunction y y0 x

c1 ΜeΜx

2. (Λ < 0) If Λ

Μ2 < 0, then y c1 eΜx c2 e Μx and y

c1 Μ c2 Μ

c2 Μe Μx . Applying the ﬁrst condition results in y 0

Μp

Μp

c1 Μe

c1 Μe =0 which is not

0, so c1 c2 . Therefore, y p

9.1 Boundary-Value Problems, Eigenvalue Problems, Sturm–Liouville Problems

possible unless c1

0, because Μ

0 and p

0. Thus, c1

c2 0, so y 0. Because we have the trivial solution, there are

no negative eigenvalues.

c1 cos Μx c2 sin Μx and y

3. (Λ > 0) By letting Λ

Μ2 , y

c2 Μ 0, so c2 0. Conc1 Μ sin Μx c2 Μ cos Μx. Hence, y 0

c1 Μ sin Μp 0 which is satisﬁed if Μp nΠ,

sequently, y p

n 1, 2, . . . Therefore, the eigenvalues are

Λ

Λn

nΠ

p

2

,n

1, 2, . . . .

c1 cos Μx

0 in y

Note that we found c2

corresponding eigenfunctions are

y

if we choose c1

yn

cos

nΠx

,n

p

c2 sin Μx, so the

1, 2, . . .

1.

EXAMPLE 9.1.6: Consider the eigenvalue problem y Λy 0, y 0

0,

2

0. (a) Show that the positive eigenvalues Λ Μ satisfy the

y1 y 1

relationship Μ

tan Μ. (b) Approximate the ﬁrst eight positive eigenvalues. Notice that for larger values of Μ, the eigenvalues are approxi2n 1 Π/ 2 2 , n 1,

mately the vertical asymptotes of y tan Μ, so Λn

2, . . . .

SOLUTION: In order to solve the eigenvalue problem, we consider

the three cases.

0, y 0

0, y 1

y 1

1. (Λ

0) The problem y

solution y 0, so Λ 0 is not an eigenvalue.

In[1828]:= DSolve y x

y 1

y 1

Out[1828]=

y x

0

0, y 0

0,

0 ,y x ,x

2. (Λ < 0) Similarly, y

Μ2 y

0, y 0

0, y 1

y 1

solution y 0, so there are no negative eigenvalues.

In[1829]:= DSolve y x

y 1

y 1

Out[1829]=

y x

0

0 has the

Μˆ2 y x

0, y 0

0 ,y x ,x

0 has

0,

733

734

Chapter 9 Eigenvalue Problems and Fourier Series

20

15

10

5

5

Figure 9-3

y

tan x

10

15

20

The eigenvalues are the x-coordinates of the points of intersection of y

x and

3. (Λ > 0) If Λ Μ2 > 0, we solve y Μ2 y 0, y 0

0, y 1 y 1

0. In this case, the result returned by DSolve is incorrect.

In[1830]:= DSolve y x

y 1

y 1

Out[1830]=

y x

0

Μˆ2 y x

0, y 0

0 ,y x ,x

0,

A general solution of y

Μ2 y

0 is y

A cos Μx B sin Μx. Applying

0

y0

0 indicates that A 0, so y B sin Μx. Applying y 1

y 1

ΜB cos Μx yields B sin Μ ΜB cos Μ

0. Because we want to

where y

avoid requiring that B

0, we note that this condition is satisﬁed if

sin Μ

Μ cos Μ or tan Μ

Μ. To approximate the ﬁrst eight positive

roots of this equation, we graph y

tan x and y x simultaneously

in Figure 9-3. (We only look for positive roots because tan Μ

tan Μ,

meaning that no additional eigenvalues are obtained by considering

negative values of Μ.) The eigenfunctions of this problem are y sin Μx

where Μ satisﬁes tan Μ Μ.

In[1831]:= Plot

Tan x , x , x, 0, 24 ,

PlotRange

0, 24 , PlotStyle

Dashing 0.01

, AspectRatio

GrayLevel 0 ,

1

In Figure 9-3, notice that roots are to the right of the vertical asymptotes of y

tan x which are x

2n 1 Π/ 2, n any integer. We use

9.1 Boundary-Value Problems, Eigenvalue Problems, Sturm–Liouville Problems

FindRoot to obtain approximations to the roots using initial guesses

near the asymptotes. Here, we guess 0.1 unit to the right of 2n 1 Π/ 2

for n 1, 2, . . . , 8.

In[1832]:= kvals

Table FindRoot

Tan x

x,

2n 1 Π

x,

0 . 1 , n , 1, 8

2

x 2.02876 , x 4.91318 , x 7.97867 ,

x 11.0855 , x 14.2074 , x 17.3364 ,

x 20.4692 , x 23.6043

Out[1832]=

Therefore, the ﬁrst eight roots are approximately 2.02876, 4.91318,

7.97867, 11.0855, 14.2074, 17.3364, 20.4692, and 23.6043. As x increases,

the roots move closer to the value of x at the vertical asymptotes of

y

tan x. We can compare the two approximations by ﬁnding a for

the ﬁrst eight vertical asymptotes, x a.

1 Π

, n , 1, 8

2

1.5708, 4.71239, 7.85398, 10.9956, 14.1372,

17.2788, 20.4204, 23.5619

2n

In[1833]:= Table N

Out[1833]=

The ﬁrst eight eigenvalues are approximated by squaring the elements of kvals. We call this list evals.

In[1834]:= evals Table kvals j, 1, 2 ˆ2, j, 1, 8

Out[1834]= 4.11586, 24.1393, 63.6591, 122.889, 201.851,

300.55, 418.987, 557.162

9.1.3 Sturm–Liouville Problems

Because of the importance of eigenvalue problems, we express these problems in

the general form

a2 x y

a1 x y

a0 x

Λ y

(9.2)

0, a < x < b,

where a2 x

0 on a, b and the boundary conditions at the endpoints x

x b can be written as

k1 y a

k2 y a

0

and

h1 y b

h2 y b

0

a and

(9.3)

for the constants k1 , k2 , h1 , and h2 where at least one of h1 , h2 and at least one of k1 ,

k2 is not zero. Equation (9.2) can be rewritten by letting

px

e

a1 x / a2 x dx

,

qx

a0 x

px,

a2 x

and

sx

px

.

a2 x

(9.4)

735

736

Chapter 9 Eigenvalue Problems and Fourier Series

By making this change, equation (9.2) can be rewritten as the equivalent equation

d

dy

px

q x Λs x y 0,

(9.5)

dx

dx

which is called a Sturm–Liouville equation and along with appropriate boundary

conditions is called a Sturm–Liouville problem. This particular form of the equation is known as self-adjoint form, which is of interest because of the relationship

of the function s x and the solutions of the problem.

EXAMPLE 9.1.7: Place the equation x2 y

adjoint form.

SOLUTION: In this case, a2 x

px

e

and s x

d 2 dy

x

dx

dx

a1 x / a2 x dx

px

a2 x

Λy

e

x2

x2

2x/ x2 dx

x2 , a1 x

2 ln x

e

2xy

Λy

0, x > 0, in self-

2x, and a0 x

0. Hence,

a0 x

px

0,

x , qx

a2 x

2

1, so the self-adjoint form of the equation is

0. We see that our result is correct by differentiating.

Solutions of Sturm–Liouville problems have several interesting properties, two of

which are included in the following theorem.

Theorem 33 (Linear Independence and Orthogonality of Eigenfunctions). If ym x

and yn x are eigenfunctions of the regular Sturm–Liouville problem

dy

d

px

q x Λs x y 0

dx

dx

0, h1 y b h2 y b

k1 y a k2 y a

(9.6)

0.

where m n, ym x and yn x are linearly independent and the orthogonality condition

b

s x ym x yn x dx 0 holds.

a

Because we integrate the product of the eigenfunctions with the function s x in

the orthogonality condition, we call s x the weighting function.

EXAMPLE 9.1.8: Consider the eigenvalue problem y

Λy

0, 0 <

x < p, subject to y 0

0 and y p

0 that we solved in Example 9.1.4.

sin Πx/ p and y2

sin 2Πx/ p are

Verify that the eigenfunctions y1

linearly independent. Also, verify the orthogonality condition.

9.2 Fourier Sine Series and Cosine Series

737

SOLUTION: We can verify that y1 sin Πx/ p and y2

linearly independent by computing the Wronskian.

sin 2Πx/ p are

In[1835]:= Clear x, p

caps

ws

Out[1835]=

Sin

Πx

2Πx

, Sin

p

p

Simplify Det

Πx

p

2 Π Sin

caps,

x caps

3

p

We see that the Wronskian is not the zero function by evaluating it for

a particular value of x; we choose x p/ 2.

p

2

In[1836]:= ws/.x

2Π

Out[1836]=

p

Because W y1 , y2 is not zero for all values of x, the two functions are

Λy 0,

linearly independent. In self-adjoint form, the equation is y

p

with s x

1. Hence, the orthogonality condition is 0 ym x yn x dx 0,

m n, which we verify for y1 and y2 .

p

In[1837]:=

Sin

0

Πx

p

Sin

2Πx

p

Of course, these two

properties hold for any

choices of m and n, m n.

x

Out[1837]= 0

9.2 Fourier Sine Series and Cosine Series

9.2.1 Fourier Sine Series

Recall the eigenvalue problem

y

y0

Λy

0

0, y p

0

that was solved in Example

nΠ/ p 2 , n

1, 2, . . . , with

9.1.4. The eigenvalues of this problem are Λ

Λn

sin nΠx/ p , n 1, 2, . . . .

corresponding eigenfunctions Φn x

We will see that for some functions y f x , we can ﬁnd coefﬁcients cn so that

cn sin

f x

n 1

nΠx

.

p

(9.7)

738

Chapter 9 Eigenvalue Problems and Fourier Series

A series of this form is called a Fourier sine series. To make use of these series, we

must determine the coefﬁcients cn . We accomplish this by taking advantage of the

orthogonality properties of eigenfunctions stated in Theorem 33.

Λy

0 is in self-adjoint form, we have

Because the differential equation y

p

that s x

1. Therefore, the orthogonality condition is 0 sin nΠx/ p sin mΠx/ p dx,

m n. In order to use this condition, multiply both sides of f x

n 1 cn sin nΠx/ p

by the eigenfunction sin mΠx/ p and s x

1. Then, integrate the result from x 0 to

x p (because the boundary conditions of the corresponding eigenvalue problem

are given at these two values of x). This yields

p

f x sin

0

p

mΠx

dx

p

cn sin

0

n 1

mΠx

nΠx

sin

dx.

p

p

Assuming that term-by-term integration is allowed on the right-hand side of the

equation, we have

p

f x sin

0

p

mΠx

dx

p

cn sin

0

n 1

mΠx

nΠx

sin

dx.

p

p

Recall that the eigenfunctions Φn x , n

1, 2, . . . are orthogonal, so

sin mΠx/ p dx 0 if m 0. On the other hand if m n,

p

sin

0

p

mΠx

nΠx

sin

dx

p

p

p

sin nΠx/ p

nΠx

dx

p

2nΠx

1 cos

dx

p

2nΠx p p

p

sin

.

2nΠ

p 0 2

sin2

0

1 p

2 0

1

x

2

In[1838]:=

p

0

nΠx 2

x

p

p Sin 2 n Π

4nΠ

Sin

0

p

Out[1838]=

2

p

Therefore, each term in the sum n 1 cn 0 sin nΠx/ p sin mΠx/ p dx equals zero

p

1

except when m n. Hence, 0 f x sin nΠx/ p dx

2 cn p, so the Fourier sine series

coefﬁcients are given by

cn

where we assume that y

2

p

p

f x sin

0

nΠx

dx,

p

f x is integrable on 0, p .

(9.8)

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