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, …
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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
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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
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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
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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
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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.
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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.
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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
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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 .
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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
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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
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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
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-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:
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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
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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
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