2375-5-Orthogonal - Lecture notes 5 PDF

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Professor Alexander Mikhailov...

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University of Leeds

MATH 2375

Linear Differential Equations and Transforms Chapter 5 Orthogonal Functions and Sturm-Liouville Eigenvalue Problems

Contents 1 Introduction

1

2 A Simple BVP with Trigonometric Solutions 2.1 Orthogonality Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

1

3 The Sturm-Liouville Form of an Equation 3.1 Putting an Equation into Sturm-Liouville Form . . . . . . . . . . . . . . . . . 3.2 Sturm-Liouville Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . .

4 5

3

4 Legendre polynomials 6 4.1 The Generating Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 4.2 The Recurrence Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 4.3 The formula for hPn , Pn i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 4.4 Rodrigues’ Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 4.5 Expansions in terms of Legendre Polynomials. . . . . . . . . . . . . . . . . . . 10 5 Hermite polynomials 5.1 The Generating Function . . . . 5.2 The Recurrence Relation . . . . . 5.2.1 The formula for hHn , Hn i 5.3 Rodrigues’ Formula . . . . . . . .

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1

Introduction

2

A Simple BVP with Trigonometric Solutions

12 12 14 14 15

In Section 2.2.2 of the previous Chapter 4 we studied the Boundary Value Problem y′′ + λy = 0,

y(0) = 0, y(1) = 0.

(1)

We have proved the following (Lemma 3): Lemma 1 (Eigenvalues and Eigenfunctions) The BVP (1) has nontrivial solutions only for an infinite sequence of values of λ, labelled λn . For each eigenvalue λn there is a unique (up to overall numerical factor) eigenfunction yn , given by λn = n2 π 2

and

yn = sin(nπx), 1

for n ≥ 1.

(2)

2.1

Orthogonality Relations

Inner Product Space. On this space of piecewise smooth functions on the interval 0 ≤ x ≤ 1 we define the inner product hf, gi =

Z

1

f (x)g(x)dx.

0

We see that if n 6= m Z Z 1 1 1 sin(πnx) sin(πmx) dx = (cos(π(n − m)x) − cos(π(n + m)x)) dx = hym , yn i = 2 0 0 =



1 1 1 sin(π(n − m)x) − sin(π(n + m)x)) = 0 . 2π(n − m) 2π(n + m) 0

Furthermore, we can use the trigonometric identity sin2 A = to show kyn k2 = hyn , yn i =

Z

1

1 (1 − cos 2A) 2

 1 sin(2nπx) 1 1 x − = . sin (nπx) = 2nπ 2 2 0 2

0

We can incorporate both this and the orthogonality in the single formula hym , yn i =

1 δm,n , 2

where δm,n is the Kroneker delta symbol. Remark 2 (Trigonometry versus Function Theory) We have appealed to a knowledge of Trigonometry, which is a bit of a “cheat” when we are trying to develop a theory of differential equations! However, recall that in Section 7 of Chapter 1 we showed how to derive these identities from the fact that the functions sin x and cos x are solutions of the differential equation y′′ + y = 0. The functions yn (x) form a basis in the linear space L, so, given any f ∈ L, there exist an , such that ∞ X an yn . f= n=1

Orthogonality then implies

hf, ym i =

∞ X

n=1

an hyn , ym i =

∞ X am an δm,n = 2 2 n=1

so am = 2 hf, ym i . This is the abstract form of our Fourier coefficient formula (Chapter 4).

2

We now derive orthogonality relations for the eigenfunctions yn (x), using the Lagrange identity, together with (1): d ′′ ′ − ym y′n ) = yn ym − ym yn′′ = (λn − λm )yn ym . (yn ym dx Integrating this over the interval (0, 1) and using the boundary conditions of (1), leads to (λn − λm )

Z

1 0

1 ′ yn ym dx = [yn ym − ym y′n ]0 = 0.

Therefore λn 6= λm

3

⇒

Z

1

yn ym dx = 0, 0

for m 6= n.

The Sturm-Liouville Form of an Equation

In Chapter 2 we encountered several equations of the form A(x)

d2 y dy + B(x) + (C(x) + λD(x))y = 0, dx2 dx

(3)

where A(x), B(x), C (x), D(x) were some functions (usually polynomials) and λ some parameter. Some important examples were 1. Legendre’s equation (1 − x2 )

d2 y dy − 2x + ℓ(ℓ + 1)y = 0, dx2 dx

which was discussed in Example 14 of Chapter 2, with A = (1 − x2 ),

B = −2x, C = 0, D = 1,

λ = ℓ(ℓ + 1).

2. Hermite’s equation

d2 y dy − x + sy = 0, 2 dx dx which appeared on Example Sheet 1, with A = 1, B = −x, C = 0, D = 1, λ = s.

3. Bessel’s equation

x2 y′′ + xy′ + (x2 − ν 2 )y = 0,

which was discussed in Section 5 of Chapter 2, with A = x2 , B = x,

C = x2 , D = −1,

λ = ν 2.

The significance of λ was not always evident, but we did find that: 1. Legendre’s equation has polynomial solutions for ℓ a positive integer. 2. Hermite’s equation has polynomial solutions for s a positive integer. 3

3. Bessel’s equation has solutions which can be written in terms of elementary (trigonometric) functions for ν a “half-integer”. In order to compare two equations of the form (3), we need to transform the equations into some “canonical” (standard) form. In the next section, we show how to transform equation (3) into Sturm-Liouville form:   dy d + qy + λσy = 0. (4) p dx dx This form of the equation can be related to a more general (Sturm-Liouville) eigenvalue problem (see Section 3.2) than the simple one considered in Section 2. We derive general orthogonality relations for the eigenfunctions. The polynomial solutions of Legendre’s and Hermite’s equations will arise as eigenfunctions of the corresponding Sturm-Liouville eigenvalue problems, rather than as an ad hoc observation of the truncation of infinite series for certain special values of a coefficient. Since they generally satisfy orthogonality relations, they are called orthogonal polynomials. We have systematic, algebraic techniques for building these polynomial eigenfunctions.

3.1

Putting an Equation into Sturm-Liouville Form

Any second order linear equation of the form (3), where λ is some parameter, can be put into Sturm-Liouville form (6). Multiplying (3) by an integrating factor R(x), it can be written as     d dy d dy (R(x)A(x)) + R(x)(C(x) + λD(x))y = 0. + R(x)B(x) − R(x)A(x) dx dx dx dx We then choose R(x) in order to eliminate the coefficient of y′ (x): d R(x)B(x) − (R(x)A(x)) = 0 dx

⇒

B A′ R′ = − R A A

⇒

1 R(x) = exp A(x)

We are then left with d dx

  dy + R(x)(C (x) + λD(x))y = 0, R(x)A(x) dx

giving p(x) = R(x)A(x),

q(x) = R(x)C(x),

σ (x) = R(x)D(x),

with R(x) given as above. Example 3 (Legendre’s equation) Legendre’s equation, d2 y dy − 2x + ℓ(ℓ + 1)y = 0, dx2 dx has no need of an integrating factor, since it can already be written as   d 2 dy + λy = 0. (1 − x ) dx dx (1 − x2 )

We therefore have p = (1 − x2 ),

q = 0,

σ = 1, 4

λ = ℓ(ℓ + 1).

Z

 B dx . A

Example 4 (Hermite’s Equation) This is the equation d2 y dy − x + λy = 0, dx2 dx

(5)

which appeared in Example Sheet 1. It was found that, for the eigenvalue λn = n, the series solution truncated to a polynomial. With an appropriate normalisation, these are called Hermite polynomials. In this case we find that R(x) satisfies R′ (x) + xR(x) = 0

⇒

R = e−

x2 2

.

With this choice, (5) takes the form   x2 x2 dy d + λe− 2 y = 0, e− 2 dx dx giving p = e−

3.2

x2 2

,

q = 0,

x2

σ = e− 2 .

Sturm-Liouville Eigenvalue Problems

We now want to consider the eigenvalue problem (boundary value problem (BVP)) associated with the general equation of Sturm-Liouville type (4):   dy d + qy + λσy = 0, a < x < b, (6) p dx dx with σ(x) > 0, for almost all x in the range a ≤ x ≤ b, and λ interpreted as the eigenvalue. Remark 5 We allow the infinite ranges (0, ∞) and (−∞, ∞). We generalise the manipulations of Section 2.1. Let ym and yn be eigenfunctions of (6), with corresponding eigenvalues λm 6= λn for m 6= n. Then       dyn dyn dym dym dyn dym dym dyn d d d p − p − ym p p + p = yn yn p − ym dx dx dx dx dx dx dx dx dx dx dx     dyn dym d d p − ym p = yn dx dx dx dx = yn (−qym − λm σym ) − ym (−qyn − λn σyn ) =

(λn − λm )σym yn .

Integrating over the interval (a, b): b   dyn dym − ym σym yn dx = p yn . (7) dx dx a a   dyn m − y The right hand side vanishes if p yn dy vanishes on the boundary. It happens m dx dx if (λn − λm )

Z

b

5

1. the eigenfunctions satisfy boundary conditions of the form α1 y(a) + α2 y′ (a) = 0,

and β1 y(b) + β2 y′ (b) = 0,

for some choices of αi , βi . Or 2. the function p(x) satisfies p(a) = p(b) = 0. Or 3. A mixture of these conditions (condition of type 1. on one boundary and condition of type 2. on another). We then have λm 6= λn

Z

⇒

b

σ(x)ym yn dx = 0. a

Inner Product Space. Again we interpret this as an inner product on the linear space of functions L which are smooth (at least C 2 ) and satisfy (for positive function σ(x)) Z b σ(x)(f (x))2 dx < ∞. a

σ(x) is called a weight function. On this space we define the inner product Z b σ(x)f (x)g(x)dx. hf, gi = a

Given a specific BVP of the form (6), we construct the eigenfunctions yn (x), eigenvalue λn , satisfying λm 6= λn , for m 6= n. We then have hym , yn i = 0 for m 6= n.

4

Legendre polynomials

We have already seen (see Example 3) that Legendre’s equation has Sturm-Liouville form   d dy + λy = 0, −1 < x < 1, (8) (1 − x2 ) dx dx so, σ = 1 > 0, for all x, and (a, b) = (−1, 1), to avoid the singular points at x = ±1. We then have the inner product Z 1 hf, gi = f (x)g(x)dx. (9) −1

Our task is to build a sequence of polynomial eigenfunctions. We found in Example 14 of Chapter 2, that, for λ = n(n + 1) (n a positive integer), one of the series solutions truncated to a polynomial. These are defined up to an overall numerical factor. We now give a direct construction of these polynomials, without any appeal to mysterious and miraculous truncations of infinite series! 6

4.1

The Generating Function

This is a remarkable device which we have to “pull out of a hat”! The Legendre polynomial Pn is the coefficient of tn in the following expansion: G(x, t) = (1 − 2tx + t2 )−1/2 =

∞ X

Pn (x)tn .

(10)

n=0

The first few terms are easily computed: 1 1 1 G(x, t) = 1 + xt + (3x2 − 1)t2 + (5x3 − 3x)t3 + (35x4 − 30x2 + 3)t4 + · · · , 8 2 2 giving the first few Legendre polynomials as 1 1 P0 = 1, P1 = x, P2 = (3x2 − 1), P3 = (5x3 − 3x), 2 2 1 1 P4 = (35x4 − 30x2 + 3), P5 = (63x5 − 70x3 + 15x), 8 8

(11)

which are plotted in Figure 1. Notice that Pn has exactly n zeros in the interval (−1, 1) and

Figure 1: Plots of Pn for n = 2, . . . , 5. that Pn (1) = 1 (for all n) and Pn (−1) = ±1 (for even and odd n respectively). These last two formulae follow from G(±1, t) = (1 ∓ t)−1

and (1 − t)−1 =

∞ X

n=0

tn , (1 + t)−1 =

∞ X (−1)n tn . n=0

The binomial expansion of G(x, t) generates this series, which is guaranteed to have coefficients which are polynomial in x, but how do we know that they are Legendre polynomials? Theorem 6 The coefficients Pn (x) in (10) are Legendre Polynomials. Proof: Equation (10) gives two different formulae for G(x, t). Using the second (and knowing the formula we wish to confirm), note  ∞  ∞ X d2 d X (1 − x2 ) 2 − 2x Pn (x)tn = ((1 − x2 )Pnxx − 2xPnx)tn , dx dx n=0 n=0 7

and d2 t 2 dt

t

∞ X

Pn (x)t

n=0

n

!

=

∞ X (n + 1)nPn (x)tn . n=0

If we can prove that the sum of these two is zero, ie ∞ X

n=0

((1 − x2 )Pnxx − 2xPnx + (n + 1)nPn )tn = 0

for all t,

then we have that each coefficient must vanish and hence Pn satisfies Legendre’s equation with λ = n(n + 1). To prove this we just use the first formula given for G(x, t) in (10): (1−x2 )Gxx−2xGx = t(t(x2 +3)−2x(t2 +1))G5

and t

d 2 (tG) = −t(t(x2 +3)−2x(t2 +1))G5 , dt2

which confirms the above formula.  We thus have a sequence of polynomial functions Pn (x), which satisfy Legendre’s equation, each with its specific eigenvalue λn = n(n + 1): (1 − x2 )Pnxx − 2xPnx + (n + 1)nPn = 0. Since this is (8) for λ = λn , then formula (7), with inner product (9) is Z 1 1 ′ (λn − λm ) Pm Pn dx = [(1 − x2 )(Pn Pm − Pm Pn′ )] −1 = 0, −1

orthogonality being a result of 1 − x2 = 0 at x = ±1 (note that Pn (±1) 6= 0). In particular, since P0 = 1, we have Z 1 hP0 , Pn i = Pn dx = 0, for n = 6 0. −1

4.2

The Recurrence Relation

For a computer, the expansion of the generating function is possibly the simplest way to compute the Legendre polynomials. However, with “pen and paper” it is rather laborious. In this section, we manipulate the generating function to obtain a simple recurrence relation. Starting again with the generating function (10), we have (x − t) (x − t)G ⇒ (1 − 2tx + t2 )Gt = (x − t)G. = (1 − 2tx + t2 ) (1 − 2tx + t2 )3/2 P∞ Now substitute the series n=0 Pn tn into this equation: Gt =

(1 − 2tx + t2 )(P1 + 2P2 t + · · · + (n + 1)Pn+1 tn + · · · )

= (x − t)(P0 + P1 t + · · · + Pn tn + · · · ),

and equate coefficients of tn , n ≥ 0, giving the recurrence relation P1 = xP0 ,

(n + 1)Pn+1 = (2n + 1)xPn − nPn−1 , 8

n ≥ 1.

(12)

This is a second order recurrence relation for Pn , with initial conditions P0 = 1,

P1 = xP0 = x.

We then obtain P2 =

1 1 (3xP1 − P0 ) = (3x2 − 1), 2 2

P3 =

1 1 (5xP2 − 2P1 ) = (5x3 − 3x), 3 2

so we see that the list starts correctly. This is already much easier that expanding G(x, t).

4.3

The formula for hPn , Pn i

We have

hPm , Pn i = 0, for m 6= n, R1 2 but what about hPn , Pn i = −1 P n dx? For low values of n, we can explicitly calculate this to find  3 1 Z 1 Z 1 Z Z 1 1 1 x 2 2 2 dx = P P02dx = 2, P12dx = (9x4 − 6x2 + 1)dx = . = , 2 5 4 3 3 −1 −1 −1 −1 −1 We can use the recurrence relation to provide an inductive step, which allows us to prove the general formula: 2 hPn , Pn i = , for all n ≥ 0. (13) 2n + 1 Notice that the inner product (9) satisfies Z 1 Z hx f, gi = (x f (x))g (x)dx = −1

1 −1

f (x)(x g(x))dx = hf, x g i .

Using this and the 3−point recurrence relation (12), we have   n−1 2n − 1 Pn−2 xPn−1 − hPn , Pn i = Pn , n n 2n − 1 = hPn , xPn−1 i (since hPn , Pn−2 i = 0) n 2n − 1 = hxPn , Pn−1 i n   n 2n − 1 n + 1 Pn+1 + Pn−1 , Pn−1 = 2n + 1 n 2n + 1 2n − 1 = hPn−1 , Pn−1 i (since hPn+1 , Pn−1 i = 0). 2n + 1 We now use induction. Suppose hPn−1 , Pn−1 i = hPn , Pn i =

2n − 1 hPn−1 , Pn−1 i 2n + 1

2 2(n−1)+1

⇒

=

2 . 2n−1

hPn , Pn i =

Then 2 . 2n + 1

Since this formula holds for n = 0, 1, 2, we have shown that it holds for all n ≥ 0. 9

4.4

Rodrigues’ Formula

The Legendre Polynomials can also be generated by the formula Pn =

1 dn (x2 − 1)n , 2n n! dxn

known as Rodrigues’ formula. This is straightforward to apply when n is small. For P2 we have P2 =

4.5

1 d2 2 1 1 d (4x(x2 − 1)) = (3x2 − 1). (x − 1)2 = 2 8 dx 8 dx2

Expansions in terms of Legendre Polynomials.

A given piecewisesmooth function f (x) can be approximated by the Legendre polynomials fN (x) =

N X an Pn (x).

(14)

n=0

In order to find the coefficients an we use the orthogonality relations hPn , Pm i = We have hf (x), Pm i = giving am =

∞ X

n=0

2 δm,n . 2n + 1

an hPn , Pm i = am hPm , Pm i =

2m + 1 2m + 1 hf, Pm i = 2 2

Z

2am . 2m + 1

1

f (x)Pm (x)dx.

−1

It follows from Theorem 15 Chapter 3, Section 3.3 that the coefficients of the form an =

hf, Pn i . hPn , Pn i

in (14) results in the best in the mean approximation of the function f (x) Example 7 (f (x) = sin 5x) We must calculate an =

2n + 1 2n + 1 hf, Pn i = 2 2

We define fN (x) =

N X

Z

1

Pn (x) sin 5x dx. −1

an Pn (x),

n=0

and plot f3 , f5 , f7 and f11 , together with sin 5x (Figure 2).

10

Figure 2: Approximation to sin 5x using Legendre polynomials Example 8 (piecewise smooth function) Let us consider approximations by the Legendre polynomials a piecewise smooth function  0, −1 ≤ x < 0 f (x) = 1 − x, 0...