Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 7 docx

We would like to write this problem in a form that can be used to obtain qualitative information about the problem.First we will write the operator in self-adjoint form.. Thus the eigenv

for later use.

Z 2π

0

xy dt =

Z 2π 0

=

Z 2π 0

2

+ dyds

2

= 1

In terms of t = 2πs/L this is

 dxdt

2

+ dydt

2

= L2π

2

We integrate this identity on [0, 2π]

L22π =

Z 2π 0

 dxdt

2

+ dydt

2!dt

We assume that the curve is parametrized so that the area is positive (Reversing the orientation changes the sign

of the area as defined above.) The area is

A =

Z 2π 0

xdy

dt dt

=

Z 2π 0

implies that all the coefficients except a0, c0, a1, b1, c1 and d1 are zero The constraint,

sin2(nπx) dx = 1

2.The coefficients in the expansion are

an= 2

Z 1 0

Thus the Fourier sine series is

The Fourier cosine series has the form

12dx = 1,

Z 1 0

cos2(nπx) dx = 1

2.The coefficients in the expansion are

a0 =

Z 1 0

x(1 − x) dx = 1

6,

an = 2

Z 1 0

-1 -0.5 0.5 1

-0.2 -0.1

0.1 0.2

-0.2 -0.1

0.1 0.2

Figure 28.13: The odd and even periodic extension of x(1 − x), 0 ≤ x ≤ 1

The Fourier sine series converges to the odd periodic extension of the function Since this function is C1,continuously differentiable, we know that the Fourier coefficients must decay as 1/n3 The Fourier cosine seriesconverges to the even periodic extension of the function Since this function is only C0, continuous, the Fouriercoefficients must decay as 1/n2 The odd and even periodic extensions are shown in Figure28.13 The sine series

is better because of the faster convergence of the series

2 (a) We substitute x = 0 into the cosine series

(b) We substitute x = 1/2 into the cosine series.

∞

X

n=1

(−1)n(2n − 1)3 = −π

3

32

Chapter 29

Regular Sturm-Liouville Problems

I learned there are troubles

Of more than one kind

Some come from ahead

And some come from behind

But I’ve bought a big bat

I’m all ready, you see

Now my troubles are going

To have troubles with me!

-I Had Trouble in Getting to Solla Sollew

-Theodor S Geisel, (Dr Suess)

Consider the eigenvalue problem on the finite interval [a b],

p2(x)y00+ p1(x)y0+ p0(x)y = µy,

subject to the homogeneous unmixed boundary conditions

α1y(a) + α2y0(a) = 0, β1y(b) + β2y0(b) = 0

Here the coefficient functions pj are real and continuous and p2 > 0 on the interval [a b] (Note that if p2 werenegative we could multiply the equation by (−1) and replace µ by −µ.) The parameters αj and βj are real

We would like to write this problem in a form that can be used to obtain qualitative information about the problem.First we will write the operator in self-adjoint form We divide by p2 since it is non-vanishing

α1y(a) + α2y0(a) = 0, β1y(b) + β2y0(b) = 0

This is known as a Regular Sturm-Liouville problem We will devote much of this chapter to studying the properties ofthis problem We will encounter many results that are analogous to the properties of self-adjoint eigenvalue problems

Example 29.1.1

ddx



ln xdydx

+ λxy = 0, y(1) = y(2) = 0

is not a regular Sturm-Liouville problem since ln x vanishes at x = 1

Result 29.1.1 Any eigenvalue problem of the form

p2y00 + p1y0+ p0y = µy, for a ≤ x ≤ b,

α1y(a) + α2y0(a) = 0, β1y(b) + β2y0(b) = 0, where the pj are real and continuous and p2 > 0 on [a, b], and the αj and βj are real can be written in the form of a regular Sturm-Liouville problem,

(py0)0 + qy + λσy = 0, on a ≤ x ≤ b,

α1y(a) + α2y0(a) = 0, β1y(b) + β2y0(b) = 0.

Self-Adjoint Consider the Regular Sturm-Liouville equation

L[y] ≡ (py0)0 + qy = −λσy

We see that the operator is formally self-adjoint Now we determine if the problem is self-adjoint.

hv|L[u]i − hL[v]|ui = hv|(pu0)0+ qui − h(pv0)0+ qv|ui

= [vpu0]ba− hv0|pu0i + hv|qui − [pv0u]ba+ hpv0|u0i − hqv|ui

= [vpu0]ba− [pv0u]ba

= p(b) v(b)u0(b) − v0(b)u(b)
+ p(a) v(a)u0(a) − v0(a)u(a)

= p(b)

v(b)



−β1

β2

u(b) −



−β1

β2

v(b)u(b)



+ p(a)

v(a)



−α1

α2

u(a) −



−α1

α2

v(a)u(a)



= 0Above we used the fact that the αi and βi are real

Thus L[y] subject to the boundary conditions is self-adjoint

Real Eigenvalues Let λ be an eigenvalue with the eigenfunction φ We start with Green’s formula

hφ|L[φ]i − hL[φ]|φi = 0hφ| − λσφi − h−λσφ|φi = 0

−λhφ|σ|φi + λhφ|σ|φi = 0(λ − λ)hφ|σ|φi = 0Since hφ|σ|φi > 0, λ − λ = 0 Thus the eigenvalues are real

Infinite Number of Eigenvalues There are an infinite of eigenvalues which have no finite cluster point Thisresult is analogous to the result that we derived for self-adjoint eigenvalue problems When we cover the Rayleighquotient, we will find that there is a least eigenvalue Since the eigenvalues are distinct and have no finite cluster point,

λn → ∞ as n → ∞ Thus the eigenvalues form an ordered sequence,

Unique Eigenfunctions Let λ be an eigenvalue Suppose φ and ψ are two independent eigenfunctions sponding to λ

corre-L[φ] + λσφ = 0, L[ψ] + λσψ = 0

We take the difference of ψ times the first equation and φ times the second equation

ψL[φ] − φL[ψ] = 0ψ(pφ0)0− φ(pψ0)0 = 0(p(ψφ0− ψ0φ))0 = 0p(ψφ0− ψ0φ) = const

In order to satisfy the boundary conditions, the constant must be zero

p(ψφ0− ψ0φ) = 0

Since p > 0 the second factor vanishes.

dx

 φψ



= 0φ

ψ = const

φ and ψ are not independent Thus each eigenvalue has a unique, (to within a multiplicative constant), eigenfunction

Real Eigenfunctions If λ is an eigenvalue with eigenfunction φ, then

φ = (const)φ

Since φ and φ only differ by a multiplicative constant, the eigenfunctions can be chosen so that they are real-valuedfunctions

Rayleigh’s Quotient Let λ be an eigenvalue with the eigenfunction φ.

hφ|L[φ]i = hφ| − λσφihφ|(pφ0)0 + qφi = −λhφ|σ|φi

φpφ0b

a− hφ0|p|φ0i + hφ|q|φi = −λhφ|σ|φi

λ = −pφφ0b

a+ hφ0|p|φ0i − hφ|q|φihφ|σ|φi

This is known as Rayleigh’s quotient It is useful for obtaining qualitative information about the eigenvalues

Minimum Property of Rayleigh’s Quotient Note that since p, q, σ and φ are bounded functions, the Rayleighquotient is bounded below Thus there is a least eigenvalue If we restrict u to be a real continuous function thatsatisfies the boundary conditions, then

where λ1 is the least eigenvalue This form allows us to get upper and lower bounds on λ1

To derive this formula, we first write it in terms of the operator L

λ1 = min

u

−hu|L[u]ihu|σ|uiSince u is continuous and satisfies the boundary conditions, we can expand u in a series of the eigenfunctions

−hu|L[u]ihu|σ|ui = −

∞ n=1cnφn L [P∞

m=1cmφm]

∞ n=1cnφn σ P

∞ m=1cmφm

= −

∞ n=1cnφn −P∞

m=1cmλmσφm

∞ n=1cnφn ... class="text_page_counter">Trang 27< /span>

29 .4 Exercises

Exercise 29.1

Find the eigenvalues and eigenfunctions of

y00+...

Find the eigenfunctions for this problem and the equation which the eigenvalues must satisfy

To this, consider the eigenvalues and eigenfunctions for,

y00+ λy... a

 p

 dx 142 8