Eigenvalues Fourier Series

8/10/2019 Eigenvalues Fourier Series

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Ma 221 Eigenvalues and Fourier Series

Eigenvalue and Eigenfunction ExamplesExample Find the eigenvalues and eigenfunctions for

y′′

− 12 y′

47 y 0 y0 y5 0Solution: The characteristic equation is

r 2 − 12r 47 0

so

r 12 144 − 447

2 6 2 2 −

Thus we have 3 cases to deal with, 2 − 0, 2 − 0, and 2 − 0.

Case I: 2 − 0. Let 2 − 2 where ≠ 0. The the general homogeneous solution is

y x C 1e62 x C 2e6−2 x

The BCs imply

C 1 C 2 0

C 1e625 C 2e6−25 0

, Solution is: C 2 0, C 1 0. Thus y 0 and there are no eigenvalues for this case.

Case II: 2. Then

y x C 1e6 x C 2 xe6 x

The BCs imply

C 1 0

C 25e30 0 C 2 0

Therefore 2 is not an eigenvalue.

Case III: 2 − 0. Let 2 − −2 where ≠ 0. Then r 6 2i. The solution to the DE is

y x C 1e6 x sin2 x C 2e6 x cos2 x

The BCs imply

y0 C 2 0

y5 C 1e30 sin10 0

Thus

10 n, n 1,2, …

or

n10

n 1,2, …

and the eigenvalues are

2 2 2 n22

100 n 1,2, …

1



The eigenfunctions are

yn x Ane6 x sin n5

x

Example

y′′ y 0 y y2 0

Solution: There are 3 cases to consider. 0, 0,, and 0 .

I. 0. Let −2 where ≠ 0. Then the differential equation becomes

y′′ − 2 y 0

and has the general solution

y x c1e x c 2e− x.

Then

y c1e c 2e− 0

y2 c1e2 c 2e−2 0

Thusc2 −c1e2

and the second equation implies

c1 e2 − 1 0

Hence c 1 0 and thus c2 0, so y 0 is the only solution. There are no negative eigenvalues.

II. 0. Then we have y ′′ 0 so

y x c1 x c 2

y c1 c 2 0

y2 2c1 c 2 0Therefore c1 c2 0 and y 0, so 0 is not an eigenvalue.

III. 0. Let 2 The DE becomes

y′′ 2 y 0

so

y x c1 sin x c 2 cos x

The initial conditions yield

y c1 sin c 2 cos 0

y2 c

1 sin2 c

2 cos2

0This system will have a non-trivial solution if and only if

sin cos

sin2 cos2 0

That is if and only if

sincos2 − cossin2 sin − 2 − sin 0

2



Thus we must have

n n 1,2,3, …

or

n n 1,2,3, …

Hence the eigenvalues are 2 n2 n 1,2,3, …

The two equations above for c 1 and c 2 become

c1 sin n c 2 cos n 0

c1 sin2n c 2 cos2n 0

Thus c 2 0 and c 1 is arbitrary. The eigenfunctions are

yn x an sin nx

3



Ma 221 Inner Products

Remark. If u and v are 2 vectors, then u v u v 0

u x1, . . . , xn v y1, . . . , yn As n → u v → xi yi.

Definition. Let f x, g x be two continuous functions on a, b. We define the inner product of f and gin an interval a ≤ x ≤ b, denoted by f , g , by

f , g a

b f xg xdx.

Definition. Two functions f and g are said to be orthogonal on a, b if

f , g 0.

Example. 0

sin x cos xdx sin2 x

2 |0

0. Therefore sin x and cos x are orthogonal on 0,.

Definition. The set of functions f 1, f 2, . . . is called an orthogonal set if

f i, f j 0 i ≠ j.

Example. 1,cos x L

,cos 2 x L

, . . . ,cos n x L

, . . . is an orthogonal set on 0, L, because

0

Lcos i

L x cos

j L

x dx 0 for i ≠ j

Remark. For vectors we have the following: if u u1, . . . , un then the length of

u ‖u ‖ ∑ui2

1

2 u u . Motivated by this we have the following definition.

Definition. Let f x be a continuous function on a, b. Then the norm of f is defined by

‖ f ‖ f , f a

b f 2 xdx .

Example. 0, 1

‖ x2 ‖2

x2, x2 0

1 x4dx x5

5 |0

1 15

‖ x2 ‖ 1

5

.

Remark. Let y x2

‖ x2 ‖ 5 x2 ‖ y‖ 5 ‖ x2 ‖ 1.

Definition. If ‖ f ‖ 1, then f is said to be normalized .

Definition. A set of functions 1,2, . . . is called orthonormal if

(1) the set is orthogonal, and

(2) each has norm 1. Therefore 1,2, . . . is an orthonormal set

4



i, j ij 0 i ≠ j

1 i j

Example sinnx sin x,sin2 x,sin3 x, . . . on 0, is an orthogonal set since

sinmx,sinnx 0

sin mx sin nxdx 1

2

0

cosm − n x − cosm n xdx m ≠ n

12

sinm − n xm − n −

sinm n xm − n

0

12

sinm − nm − n −

sinm nm n 0 m ≠ n

since m and n are integers.

Now

sin nx,sin nx 0

sin2nxdx

12

0

1 − cos 2nxdx

12

x − sin2nx2n

|0

2 .

Therefore

‖sin nx‖ sin nx,sin nx 12

2

this set is not orthonormal. We can make an orthonormal set from these functions by dividing each

element in the original by

2 2

sin nx is orthonormal set n 1,2, ....

Properties of the inner product.

1. f , g g, f since a

b f xg xdx

a

bg x f xdx

2. f g, h f , h g, h since f ghdx fhdx ghdx

3. a. f , f 0 if f 0

b. f , f 0 if f ≠ 0

Remarks. (1) It will be necessary when dealing with partial differential equations to “expand” anarbitrary function f x in terms of an orthogonal set of functions n.

(2) Recall that in 3 space, if u 1 1,0,0, u 2 0,1,0, and u 3 0,0,1 thenv 1,2,3 1u1 2u2 3u3.

Note that

5



u 1, v u 1 v

u 1,1 u 1 2 u 2 3 u 3 u 1,1 u 1 u 1,2 u 2 u 1,3 u 3

1 u 1, u 1 2 u 1, u 2 3 u 1, u 3 1

Also u2

, v 2

and u3

, v 3

.

Suppose we are given a set of orthogonal functions n on 0, L, and we desire to expand a function f x given on 0, L in terms of them. Then we want

f x ∑n1

nn x.

Question. What does k ?

Consider

k , f x k ,∑1

nn

k ,11 22

1 k ,1 k k ,k k 1 k ,k 1

But k , j 0 if j ≠ k since the set k is orthogonal.

k , f x k k ,k k ‖k ‖2

Therefore

k

0

L f xk xdx

‖k ‖2

0

L f xk xdx

0

Lk x

2dxk 1 ,2, . . . ∗

∗ is the formula for the coefficients in the expansion of a function f x in terms of a set of orthogonalfunctions.

6



Ma 221 Fourier Series

Fourier Series Formulas

Suppose we want to write a function f x given on 0, L as a Fourier sine series

f x ∑1

k sin k x L

then from ∗ above

k 2 L

0

L f x sin k x

L dx,

since

0

Lk x

2dx L2

.

Note that the orthogonal functions sin k x L , k 1,2,3, … are the eigenfunctions of the eigenvalue problem

y′′ y 0 y0 y L 0.

These formulas are for the Fourier sine series for f x on 0 x L.

The Fourier sine series converges to function F x where

F x f x 0 x L

− f − x − L x 0F x 2 L F x

since sine is an odd function.

Suppose that the graph of the function f x is given by the figure below.

7



Then the Fourier sine series generates a function F x defined on - x whose graph is given below.

Example Find the Fourier sine series of

f x 1 0 x

2

0 2

x

8



Now

f x ∑ n sin n x L

∑1

n sin nx,

since 2 L 2 L .

The formula above for the coefficients in the Fourier sine series implies

n 2 L

0

L f x sin n x

L dx 2

0

f x sin nxdx

n 2 0

2 1 sin nxdx 2

2

0 sin nxdx − 2

cos nx

n |0

2

− 2n cos n

2 − 1

n

2n n odd

−2n −1

n2

− 1 n even

Therefore

f x ∑1

n sin nx

2 sin x 2

2 sin2 x 1

3 sin3 x 0 sin 4 x 1

5 sin5 x 2

6 sin6 x

Note that our function f x on 0 ≤ x ≤ is extended to the following on − x .

9



Fourier Cosine Series

We write

f x 0 ∑1

n cos n x L

Proceeding as above in our derivation of the constants in the Fourier Sine series, we get for theconstants in the Fourier Cosine series

n 2 L

0

L f xcos n x

L dx n 1,2,3, … 0 1

L

0

L f xdx

Note the book writes

f x~ a 0

2 ∑

1

an cos n x L

and an 2

L

0

L f xcos n x

L dx n − 0,1,2, …

Thus

0 a0

2

10



Again the Fourier series is periodic with period 2 L. However, now f − x f x since cosine is an evenfunction. Here the Fourier Cosine series extends f x which is given on 0, L to a function F x whichis defined on − x as

F x f x 0 x L

f − x − L x 0F x

2 L F x

.

If the graph of f x looked as below

then F x, the even extension of f x, would look like

Example (a) Find the first four nonzero terms of the Fourier cosine series for the function

f x x on 0 x 1

Solution:

f x 0 ∑1

n cos n x L

where

11



0 1 L

0

L f xdx and n 2

L

0

L f xcos n x

L dx n 1,2,3, …

Here L 1 so

f x 0 ∑1

n cos n x

0 11

0

1 xdx 1

2

n 21

0

1 x cos n xdx 2

n22 cos n x n x sin n x|0

1

2n22

cos n − 1 2n22

−1n − 1 n 1,2,3, …

Hence 1 − 42

, 2 0, 3 − 492

, 4 0, 5 − 4252

Therefore

f x 12 − 42 cos x − 492 cos3 x − 4252 cos5 x

Note: The book gives the formulas

f x 0

2 ∑

1

n cos n x L

where

n 2 L

0

L f xcos n x

L dx n 0,1,2,3, …

Using this formula we get

0 2

1

0

1 xdx 1

Therefore, the first term in the book’s formula for the Fourier cosine series is 0

2 1

2 as before.

(b) Sketch the graph of the function represented by the Fourier cosine series in (a) on −3 x 3.

x

12



-3 -2 -1 0 1 2 3

0.5

1.0

x

y

Example (a) Find the Fourier sine series for the function

f x x on 0 x 1

Solution:

f x ∑1

k sin k x L

where

k 2 L

0

L f xsin k x

L dx, k 1,2,3, …

Here L 1 so

f x ∑1

k sink x

where

k 2 0

1 f xsink xdx, k 1,2,3, …

Thus

k 2 0

1 x sink xdx 2 1

k 2 sin k x − k x cos k x

0

1

−

2 1

k cos k 2

k −

1k 1 k 1,2,3, …

Thus

f x 2 ∑

1

−1k 1

k sink x

(b) Sketch the graph of the function represented by the Fourier sine series in 5 (a) on −3 x 3.

Solution:

13



1

-3 -2 -1 1 2 3

-1.0

-0.8

-0.6

-0.4

-0.2

0.2

0.4

0.6

0.8

1.0

x

y

Ma 221 Fourier Cosine Series Example

5. (a) (15 pts.) Find the first five nonzero terms of the Fourier cosine series for the function

f x 2 − x; 0 x 2

Be sure to give the Fourier series with these terms in it.

Solution:

f x b0 ∑n1

bn cos n x

L

b0 1 L

0

L f xdx bn 2

L

0

L f xcos n x

L dx n 1,2, …

Here L 2 so

f x b0 ∑n1

bn cos n x2

b0 12

0

2 f xdx 1

2

0

22 − xdx 1

2 2 x − x2

2 0

2 1

bn 2 L

0

L f xcos n x

L dx 2

2

0

22 − xcos n x

2

dx n 1,2, …

so

bn 2n 2 − x sin n x

2 − 4

n22 cos n x

2 0

2 4n22

−cosn 1 n 1,2, …

Thus

14



b1 42

1 1 82

b2 442

−1 1 0

b3

4

92 1 1 8

92

b4 4162

−1 1 0

b5 4252

1 1 8252

b6 4362

−1 1 0

b7 4492

1 1 8492

and

f x 1 82 cos

x

2 892 cos

3 x

2 8252 cos

5 x

2 8492 cos

7 x

2

(b) (10 pts.) Sketch the graph of the function represented by the Fourier cosine series in 5 (a) on−2 x 6.

2 − x

-2 -1 0 1 2 3 4 5 6

0.5

1.0

1.5

2.0

x

y

15

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Eigenvalues Fourier Series