Loading in 5 sec....

Orthogonal Functions and Fourier SeriesPowerPoint Presentation

Orthogonal Functions and Fourier Series

- 162 Views
- Uploaded on
- Presentation posted in: General

Orthogonal Functions and Fourier Series

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

CHAPTER 12

Orthogonal Functions and Fourier Series

- 12.1 Orthogonal Functions
- 12.2 Fourier Series
- 12.3 Fourier Cosine and Sine Series
- 12.4 Complex Fourier Series
- 12.5 Strum-Liouville Problems
- 12.6 Bessel and Legendre Series

DEFINITION 12.2

Orthogonal Function

Two functions f1 and f2 are said to be orthogonal on an

interval [a, b] if

DEFINITION 12.1

Inner Product of Function

The inner product of two functions f1 and f2on an

interval [a, b] is the number

- The function f1(x) = x2, f2(x) = x3 are orthogonal on the interval [−1, 1] since

DEFINITION 12.3

Inner Product of Function

A set of real-valued functions {0(x), 1(x), 2(x), …}

is said to be orthogonalon an interval [a, b] if

(2)

- The expression (u, u) = ||u||2 is called the square norm. Thus we can define the square norm of a function as(3)If {n(x)} is an orthogonal set on [a, b] with the property that ||n(x)|| = 1 for all n, then it is called an orthonormal set on [a, b].

Show that the set {1, cos x, cos 2x, …} is orthogonal on [−, ].

SolutionLet 0(x) = 1, n(x) = cos nx, we show that

and

Find the norms of each functions in Example 1.

Solution

- Recalling from the vectors in 3-space that(4)we have(5)Thus we can make an analogy between vectors and functions.

- Suppose {n(x)} is an orthogonal set on [a, b]. If f(x) is defined on [a, b], we first write as(6)Then

- Since {n(x)} is an orthogonal set on [a, b], each term on the right-hand side is zero except m = n. In this casewe have

In other words,(7)(8)Then (7) becomes (9)

DEFINITION 12.4

- Under the condition of the above definition, we have (10)(11)

A set of real-valued functions {0(x), 1(x), 2(x), …}

is said to be orthogonal with respect to a weight

function w(x) on [a, b], if

Orthogonal Set/Weight Function

- An orthogonal set is complete if the only continuous function orthogonal to each member of the set is the zero function.

- Trigonometric SeriesWe can show that the set(1)is orthogonal on [−p, p]. Thus a function f defined on [−p, p] can be written as(2)

- Now we calculate the coefficients.(3)Since cos(nx/p) and sin(nx/p) are orthogonal to 1 on this interval, then (3) becomes Thus we have(4)

- In addition, (5)by orthogonality we have

- andThus (5) reduces to and so (6)

- Finally, if we multiply (2) by sin(mx/p) and useand we find that (7)

DEFINITION 12.5

Fourier Series

The Fourier series of a function f defined on the

interval (−p, p) is given by(8)where (9) (10) (11)

Expand (12)in a Fourier series.

SolutionThe graph of f is shown in Fig 12.1 with p = .

←cos n = (-1)n

From (11) we haveTherefore(13)

THEOREM 12.1

Let f and f’ be piecewise continuous on the interval (−p, p); that

is, let f and f’ be continuous except at a finite number of points

in the interval and have only finite discontinuous at these points.

Then the Fourier series of f on the interval converge to f(x) at a

point of continuity. At a point of discontinuity, the Fourier

series converges to the averagewhere f(x+) and f(x－) denote the limit of f at x from the right

and from the left, respectively.

Criterion for Convergence

- Referring to Example 1, function f iscontinuous on (−, ) except at x = 0. Thus the series (13) will converge to at x = 0.

- Fig 12.2 is the periodic extension of the function f in Example 1. Thus the discontinuity at x = 0, 2, 4, …will converge toand at x = , 3, … will converge to

- Sequence of Partial SumsReferring to (13), we write the partial sums asSee Fig 12.3.

- Even and Odd Functions
- even if f(−x) = f(x)
- odd if f(−x) = −f(x)

THEOREM 12.2

(a) The product of two even functions is even.

(b) The product of two odd functions is even.

(c) The product of an even function and an odd function is odd.

(d) The sum (difference) of two even functions is even.

(e) The sum (difference) of two odd functions is odd.

(f) If f is even then

(g) If f is odd then

Properties of Even/Odd Functions

- If f is even on (−p, p) thenSimilarly, if f is odd on (−p, p) then

DEFINITION 12.6

(i) The Fourier series of an even function f on the interval (−p, p) is the cosine series(1)where(2)(3)

Fourier Cosine and Sine Series

DEFINITION 12.6

(ii) The Fourier series of an odd function f on the interval (−p, p) is the sine series(4)where(5)

Fourier Cosine and Sine Series

Expand f(x) = x, −2 < x < 2 in a Fourier series.

SolutionInspection of Fig 12.6, we find it is an odd function on (−2, 2) and p = 2. Thus(6)Fig 12.7 is the periodic extension of the function in Example 1.

- The functionshown in Fig 12.8 is odd on (−, ) with p = .From (5), and so (7)

- Fig 12.9 shows the partial sums of (7). We can see there are pronounced spikes near the discontinuities. This overshooting of SN does not smooth out but remains fairly constant even when N is large. This is so-called Gibbs phenomenon.

- If a function f is defined only on 0 < x < L, we can make arbitrary definition of the function on −L < x < 0.
- If y = f(x) is defined on 0 < x < L,
- reflect the graph about the y-axis onto −L < x < 0; the function is now even. See Fig 12.10.
- reflect the graph through the origin onto −L < x < 0; the function is now odd. See Fig 12.11.
- define f on −L < x < 0 by f(x) = f(x + L). See Fig 12.12.

Expand f(x) = x2, 0 < x < L, (a) in a cosine series, (b) in a sine series (c) in a Fourier series.

SolutionThe graph is shown in Fig 12.13.

(a) Then (8)

(b)Hence (9)

(c) With p = L/2, n/p = 2n/L, we have Therefore(10)The graph of these periodic extension are shown in Fig 12.14.

- Consider the following physical system(11)where (12)is a half-range sine expansion.

- Referring to (11), m = 1/16 slug, k = 4 lb/ft, the force f(t) with 2-period is shown in Fig 12.15. Though f(t) acts on the system for t > 0, we can extend the graph in a 2-periodic manner to the negative t-axis to obtain an odd function. With p = 1, from (5) we haveFrom (11) we have(13)

To find a particular solution xp(t), we substitute (12) into (13). Thus Therefore(14)

- Euler’s formulaeix= cos x + i sin xe-ix= cos x i sin x(1)

- From (1), we have (2)Using (2) to replace cos(nx/p) and sin(nx/p), then(3)

where c0 = a0/2, cn = (an ibn)/2, c-n = (an+ ibn)/2. When the function f is real, cn and c-n are complex conjugates.We have(4)

(5)

(6)

DEFINITION 12.7

The Complex Fourier Series of function f defined on

an interval (p, p)is given by(7)where (8)

Complex Fourier Series

- If f satisfies the hypotheses of Theorem 12.1, a complex Fourier series converges to f(x) at a point of continuity and to the average at a point of discontinuity.

Expand f(x) = e-x, < x <, in a complex Fourier series.

Solutionwith p = , (8) gives

Using Euler’s formulaHence (9)

The complex Fourier series is then (10)The series (10) converges to the 2-periodic extension of f.

- The fundamental period is T = 2p and then p = T/2. The Fourier series becomes (11)where = 2/T is called the fundamental angular frequency.

- If f is periodic and has fundamental period T, the plot of the points (n, |cn|) is called the frequency spectrum of f.

- In Example 1, = 1, so that n takes on the values 0, 1, 2, … Using , we see from (9) thatSee Fig 12.17.

- Find the spectrum of the wave shown in Fig12.18. The wave is the periodic extension of the function f:

SolutionHere T = 1 = 2p so p = ½. Since f is 0 on (½, ¼) and (¼, ½), (8) becomes

It is easy to check that Fig 12.19 shows the frequency spectrum of f.

- Eigenvalue and EigenfunctionsRecall from Example 2, Sec 3.9(1)This equation possesses nontrivial solutions only when took on the values n = n22/L2, n = 1, 2, 3,…called eigenvalues. The corresponding nontrivial solutions y = c2sin(nx/L) or simply y = sin(nx/L) are called the eigenfunctions.

- It is left as an exercise to show the three possible cases: = 0, = 2 < 0, = 2 > 0, ( > 0), that the eigenvalues and eigenfunctions for (2)are respectively n = n2 = n22/L2, n = 0, 1, 2, …and y = c1 cos(nx/L), c1 0.

- Let p, q, r and r be real-valued functions continuous on [a, b], and let r(x) > 0 and p(x) > 0 for every x in the interval. ThenSolve(3)Subject to(4)(5)is said to be a regular Sturm-Liouville problem. The coefficients in (4), (5) are assumed to be real and independent of .

THEOREM12.3

Properties of the Regular

Strum-Liouville Problem

There exist an infinite number of real eigenvalues that can be arranged in increasing order 1 < 2 < 3 < … < n < … such that n → as n → .

For each eigenvalue there is only one eigenfunction (except for nonzero constant multiples).

Eigenfunctions corresponding to differenteigenvalues are linearly independent.

The set of eigenfunctions corresponding to the setof eigenvalues is orthogonal with respect to the weight function p(x) on the interval [a, b].

Proof of (d)Let ymand ynbe eigenfunctions corresponding to eigenvalues m and n. Then(6)(7)From (6)yn (7)ym we have

Integrating the above equation from a to b, then (8)Since all solutions must satisfy the boundary condition (4) and (5), from (4) we have

For this system to be satisfied by A1 and B1 not both zero, the determinant must be zeroSimilarly from (5), we haveThus the right-hand side of (8) is zero.Hence we have the orthogonality relation(9)

Solve(10)

SolutionYou should verify that for = 0 and < 0, (10) possesses only trivial solution. For = 2 > 0, > 0, the general solution is y = c1cos x + c2sin x. Now the condition y(0) = 0 implies c1= 0, thus y = c2sin x. The second condition y(1) + y(1) = 0 implies c2sin + c2 cos. = 0.

Choosing c2 0, we have(11)From Fig 12.20, we see there are infinitely many solution for > 0. It is easy to get the values of > 0. Thus the eigenvalues are n = n2, n = 1, 2, 3, …and the corresponding eigenfunctions are yn = sin nx.

- There are several import conditions of (3)
- r(a) = and a boundary condition of the type given in (5) is specified at x = b;(12)
- r(b) = 0 and a boundary condition of the type given in (4) is specified at x = a.(13)
- r(a) = r(b) = 0 and no boundary condition is specified at either x = a or at x = b;(14)
- r(a) = r(b) and boundary conditions y(a) = y(b), y’(a) = y’(b).(15)

- Equation (3) satisfies (12) and (13) is said to be a singular BVP.Equation (3) satisfies (15) is said to be a periodic BVP.

- By assuming the solutions of (3) are bounded on [a, b], from (8) we have
- If r(a) = 0, then the orthogonality relation (9) holds with no boundary condition at x = a; (16)
- If r(b) = 0 , then the orthogonality relation (9) holds no boundary condition at x = b;(17)
- If r(a) = r(b) = 0, then the orthogonality relation (9) holds with no boundary conditions specified at either x = a or x = b;(18)
- If r(a) = r(b), then the orthogonality relation (9) holds wuth peroidic boundary conditions y(a) = y(b), y’(a) = y’(b).(19)

- In fact (3) is the same as(20)Thus we can write the Legendre’s differential equation as (21)Here we find that the coefficient of y is the derivative of the coefficient of y.

- In addition, if the coefficients are continuous and a(x) 0 for all x in some interval, then any second-order differential equation(22)can be recast into the so-called self-adjoint form (3).
- To understand the above fact, we start from a1(x)y + a0(x)y = 0Let P = a0/a1, = exp( Pdx), = P,theny + Py = 0, y + Py = 0, Thus d(y)/dx = 0.

- Now for (22), let Y = y, and the integrating factor be e [b(x)/a(x)] dx. Then (22) becomes In summary, (22) can become

- In addition, (23) is the same as (3)

- In Sec 5.3, we saw that the general solution of the parametric Bessel differential equationDividing the Bessel equation by x2 and multiplying the resulting equation by integrating factor e [(1/x)] dx = eln x = x, we have

Now r(0) = 0, and of the two solutions Jn(x) and Yn(x) only Jn(x) is bounded at x = 0. From (16), the set {Jn(ix)}, i = 1, 2, 3, …, is orthogonal with respect to the weight function p(x) = x on [0, b]. Thus (24)provided the i and hence the eigenvalues i = i2 are defined by a boundary condition at x = b of the type given by (5):A2Jn(b) + B2Jn(b) = 0(25)

- From (21), we identify q(x) = 0, p(x) = 1 and = n(n + 1). Recall from Sec 5.3 when n = 0, 1, 2, …, Legendre’s DE possesses polynomial solutions Pn(x). We observer that r(−1) = r(1) =0 together with that fact that Pn(x) are the only solutions of (21) that are bounded on [−1, 1], to conclude that the set {Pn(x)}, n = 0, 1, 2, …, is orthogonal w.s.t. the weight function p(x) = 1 on [−1, 1]. Thus

- Fourier-Bessel SeriesWe have shown that {Jn(ix)}, i = 1, 2, 3, …is orthogonal w.s.t. p(x) = x on [0, b] when the i are defined by (1)This orthogonal series expansion or generalized Fourier series of a function f defined on (0, b) in terms of this orthogonal set is(2)where(3)

The square norm of the function Jn(ix) is defined by(4)The series (2) is called a Fourier-Bessel series.

- Recalling from (20) and (21) in Sec 5.3, we have the differential recurrence relations as(5)(6)

- The value of (4) is dependent on i = i2. If y = Jn(x) we haveAfter we multiply by 2xy’, then

- Integrating by parts on [0, b], thenSince y = Jn(x), the lower limit is 0 for n > 0, because Jn(0) = 0. For n = 0, at x = 0. Thus (7)where y = Jn(x).

- Now we consider three cases for the condition (1).
- Case I: If we choose A2 = 1 and B2 = 0, then (1) is(8)There are an infinite number of positive roots xi = ib of (8) (see Fig 5.3), which defines i = xi/b. The eigenvalues are positive and then i = i2 = (xi/b)2. No new eigenvalues result from the negative roots of (8) since Jn(−x) = (−1)nJn(x).

The number 0 is not ab eigenvalue for any n since Jn(0) = 0, n= 1, 2, 3, … and J0(0) = 1. When (6) is written as xJn(x) = nJn(x) – xJn+1(x), it follows from (7) and (8)(9)

- Case II: If we choose A2 = h 0 and B2 = b, then (1) is(10)There are an infinite number of positive roots xi = ib for n = 1, 2, 3, …. As before i = i2 = (xi/b)2. = 0 is not an eigenvalue for n = 1, 2, 3, …. Substituting ibJn(ib) = – hJn(ib) into (7), then(11)

- Case III: If h = 0 and n = 0 in (10), i are defined from the roots of(12)Though (12) is a special case of (10), it is the only situation fro which = 0 is an eigenvalue. For n = 0, the result in (6) implies J0(b) = 0 is equivalent to J1(b) = 0.

Since x1 = 1b = 0 is a root of the last equation and because J0(0) = 1 is nontrivial, we conclude that from 1 = 12 = (x1/b)2 that 1 is an eigenvalue. But we can not use (11) when 1= 0, h = 0, n = 0, and n = 0. However from (4) we have(13)For i> we can use (11) with h = 0 and n = 0:(14)

DEFINITION 12.8

The Fourier-Bessel series of a function f defined on

the interval (0, b) is given by

(i)

(15)

(16)

where the i are defined by Jn(b) = 0.

Fourier-Bessel Series

DEFINITION 12.8

(ii)

(17)

(18)

where the i are defined by hJn(b) + bJ’n(b) = 0.

Fourier-Bessel Series

DEFINITION 12.8

(iii)

(19)

(20)

where the i are defined by J’0(b) = 0.

Fourier-Bessel Series

THEOREM 12.4

If f and f’ are piecewise continuous in the open interval

(0, b), then a Fourier-Bessel expansion of f converges

to f(x) at any point where f is continuous ant to the

converge [f(x+) + f(x-)] / 2 at a point where f is

discontinuous.

Conditions for Convergence

Expand f(x) = x, 0 < x < 3, in a Fourier-Bessel series, using Bessel function of order one satisfying the boundary condition J1(3) = 0.

SolutionWe use (15) where ci is given by (16) with b = 3:

Let t = ix, dx = dt/i, x2 = t2/i2, and use (5) in the form d[t2J2(t)]/dt = t2J1(t):

- If the i in Example 1 are defined by J1(3) + J1(3) = 0, the only thing changing in the expansion is the value of the square norm. Since 3J1(3) + 3J1(3) = 0 matching (10) when h = 3, b = 3and n = 1. Thus (18) and (17) yield in turn

DEFINITION 12.9

The Fourier-Legendre series of a function f defined

on the interval (－1, 1) is given by

(i)

(21)

(22)

where the i are defined by Jn(b) = 0.

Fourier-Legendre Series

THEOREM 12.5

If f and f’ are piecewise continuous in the open interval

(－1, 1), then a Fourier-Legendre series (21) converges

to f(x) at a point of continuity ant to the converge

[f(x+) + f(x-) / 2 at a point of discontinuous.

Conditions for Convergence

Write out the first four nonzero terms in the Fourier-Legendre expansion of

Solution From page 269 and (22):

- If we let x = cos, the x = 1implies = 0, x = −1implies = . Since dx = −sin d, (21) and (22) become, respectively,(23)(24)where f(cos ) has been replaced by F().

Thank You !