Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 8 potx

Since this is a Sturm-Liouville problem, there are only real eigenvalues... λx .The solution that satisfies the left boundary condition is y = c sin√ λx.For nontrivial solutions we must

x2α+1y0
0+ λx2α−1y = 0, y(a) = y(b) = 0Now we verify that the Sturm-Liouville properties are satisfied.

• The eigenvalues

λn = α2+

nπln(b/a)

2

, n ∈ Zare real

• There are an infinite number of eigenvalues

λ1 < λ2 < λ3 < · · · ,

α2+

πln(b/a)

2

< α2+

2πln(b/a)

2

< α2+

3πln(b/a)

2

,but there is no greatest eigenvalue, (λn→ ∞ as n → ∞)

• For each eigenvalue, we found one unique, (to within a multiplicative constant), eigenfunction φn We were able

to choose the eigenfunctions to be real-valued The eigenfunction

φn = x−αsin



nπln(x/a)ln(b/a)



has exactly n − 1 zeros in the open interval a < x < b

• The eigenfunctions are orthogonal with respect to the weighting function σ(x) = x2α−1.

Z b a

φn(x)φm(x)σ(x) dx =

Z b a

x−αsin



nπln(x/a)ln(b/a)



x−αsin



mπln(x/a)ln(b/a)



x2α−1dx

=

Z b a

sin



nπln(x/a)ln(b/a)

sin



mπln(x/a)ln(b/a)

 1

xdx

= ln(b/a)π

Z π 0

sin(nx) sin(mx) dx

= ln(b/a)2π

Z π 0

• The eigenvalues can be related to the eigenfunctions with the Rayleigh quotient.

λn =

−pφndφdxn

b

a+Rb a

dx



x2α+1x−α−1ln(b/a)nπ cosnπln(x/a)ln(b/a)− α sinnπln(x/a)ln(b/a)2

dx

Rb a





nπ ln(b/a)





nπ ln(b/a)

2

cos2(x) − 2αln(b/a)nπ cos(x) sin(x) + α2sin2(x)

dx

Rπ

0 sin2(x) dx

= α2+

nπln(b/a)

,where

f (x)xα−1sin



nπln(x/a)ln(b/a)

dx

Solution 29.4

y00− y0+ λy = 0, y(0) = y(1) = 0

The factor that will put this equation in Sturm-Liouville form is

Thus we see that the eigenfunctions will be orthogonal with respect to the weighting function σ = e−x

Substituting y = eαx into the differential equation yields

α2− α + λ = 0

α = 1 ±

√

1 − 4λ2

ex/2cos(pλ − 1/4 x), ex/2sin(pλ − 1/4 x)

The left boundary condition gives us

y = c ex/2sin(pλ − 1/4 x)

The right boundary condition demands that

p

λ − 1/4 = nπ, n = 1, 2, Thus we see that the eigenvalues and eigenfunctions are

f (x) e−x/2sin(nπx) dx

Solution 29.5

Consider the eigenvalue problem

y00+ λy = 0 y(0) = 0 y(1) + y0(1) = 0

Since this is a Sturm-Liouville problem, there are only real eigenvalues By the Rayleigh quotient, the eigenvalues are

λ =

−φdφ dx

1

0

+R1 0



dφ dx

2dx

R1

λx

.The solution that satisfies the left boundary condition is

y = c sin√

λx.For nontrivial solutions we must have

λx In Figure 29.1

we plot the functions x and − tan(x) and draw vertical lines at x = (n − 1/2)π, n ∈ N

From this we see that there are an infinite number of eigenvalues, λ1 < λ2 < λ3 < · · · In the limit as n → ∞,

λn → (n − 1/2)π The limit is approached from above

Now consider the eigenvalue problem

y00+ y = µy y(0) = 0 y(1) + y0(1) = 0

From above we see that the eigenvalues satisfy

p

1 − µ = − tanp1 − µand that there are an infinite number of eigenvalues For large n, µn≈ 1 − (n − 1/2)π The eigenfunctions are

φn = sinp1 − µnx

Figure 29.1: x and − tan(x).

To solve the inhomogeneous problem, we expand the solution and the inhomogeneity in a series of the eigenfunctions

Solution 29.6

Consider the eigenvalue problem

y00+ y = µy y(0) = 0 y(1) + y0(1) = 0

From Exercise 29.5 we see that the eigenvalues satisfy

p

1 − µ = − tan

p

1 − µ

and that there are an infinite number of eigenvalues For large n, µn≈ 1 − (n − 1/2)π The eigenfunctions are

y = cx satisfies the boundary conditions Thus λ = 0 is an eigenvalue.

Now consider negative real λ The general solution is

y = c1cosh√

−λx+ c2sinh√

−λx.The solution that satisfies the left boundary condition is

y = c sinh√

−λx.For nontrivial solutions of the boundary value problem, there must be negative real solutions of

y = c sin√

λx.For nontrivial solutions of the boundary value problem, there must be positive real solutions of

The difficulty with the boundary conditions, y(0) = 0, y0(0) − y(1) = 0 is that the problem is not self-adjoint Wedemonstrate this by showing that the problem does not satisfy Green’s identity Let u and v be two functions thatsatisfy the boundary conditions, but not necessarily the differential equation.

hu, L[v]i − hL[u], vi = hu, v00i − hu00, vi

= [uv0]10− hu0, v0i − hu0, v0i − [u0v]10+ hu0, v0i − hu0, v0i

= u(1)v0(1) − u0(1)v(1)Green’s identity is not satisfied,

hu, L[v]i − hL[u], vi 6= 0;

The problem is not self-adjoint

Solution 29.8

First we write the equation in formally self-adjoint form,

L[y] ≡ (xy0)0 = −λxy, |y(0)| < ∞, y(1) = 0

Let λ be an eigenvalue with corresponding eigenfunction φ We derive the Rayleigh quotient for λ

hφ, L[φ]i = hφ, −λxφi

hφ, (xφ0)0i = −λhφ, xφi[φxφ0]10− hφ0, xφ0i = −λhφ, xφi

We apply the boundary conditions and solve for λ

λ = hφ0, xφ0i

hφ, xφiThe Bessel equation of the first kind and order zero satisfies the problem,

y00+ 1

xy

0

+ y = 0, |y(0)| < ∞, y(r) = 0,

where r is a positive root of J0(x) We make the change of variables ξ = x/r, u(ξ) = y(x) to obtain the problem

6 ≈ 2.4494 (The smallest zero of J0(x) is approximately2.40483.)

at z = l

y1(l) = y2(l), y10(l) = y02(l)

c1l = c2(π − l), c1 = −c2

Since there is only the trivial solution, c1 = c2 = 0, λ = 0 is not an eigenvalue.

Now consider λ 6= 0 For 0 ≤ z ≤ l a set of linearly independent solutions is

ncos(√aλz), sin(√

aλz)o.The solution which satisfies y(0) = 0 is

y1 = c1sin(√

aλz)

For l < z ≤ π a set of linearly independent solutions is

ncos(√bλz), sin(√

bλz)o.The solution which satisfies y(π) = 0 is

√bλ(π − l)), c1

√

aλ cos(

√aλl) = −c2

√

bλ cos(

√bλ(π − l))

We divide the second equation byp(λ) since λ 6= 0 and write this as a linear algebra problem

sin(√aλl) − sin(√bλ(π − l))

λ(l√

a − (π − l)√

b)+ (√

b +√a) sin√

λ(l√

a + (π − l)√

b)= 0

Clearly this equation has an infinite number of solutions for real, positive λ However, it is not clear that this equationdoes not have non-real solutions In order to prove that, we will show that the problem is self-adjoint Before going on

to that we note that the eigenfunctions have the form

φn(z) =

(sin √

aλnz

0 ≤ z ≤ lsin √

bλn(π − z)

l < z ≤ π

Now we prove that the problem is self-adjoint We consider the class of functions which are C2 in (0 π) except

at the interior point x = l where they are C1 and which satisfy the boundary conditions y(0) = y(π) = 0 Note thatthe differential operator is not defined at the point x = l Thus Green’s identity,

hu|q|Lvi = hLu|q|vi

is not well-defined To remedy this we must define a new inner product We choose

hu|vi ≡

Z l 0

uv dx +

Z π l

uv dx

This new inner product does not require differentiability at the point x = l

The problem is self-adjoint if Green’s indentity is satisfied Let u and v be elements of our class of functions In

addition to the boundary conditions, we will use the fact that u and v satisfy y(l−) = y(l+) and y0(l−) = y0(l+).

hv|Lui =

Z l 0

vu00dx +

Z π l

vu00dx

= [vu0]l0−

Z l 0

v0u0dx + [vu0]πl −

Z π l

v0u0dx

= v(l)u0(l) −

Z l 0

v0u0dx − v(l)u0(l) −

Z π l

v0u0dx

= −

Z l 0

v0u0dx −

Z π l

v0u0dx

= − [v0u]l0+

Z l 0

v00u dx − [v0u]πl +

Z π l

v00u dx

= −v0(l)u(l) +

Z l 0

v00u dx + v0(l)u(l) +

Z π l

v00u dx

=

Z l 0

v00u dx +

Z π l

We use integration by parts and utilize the homogeneous boundary conditions.

hv, L[u]i − hL[v], ui = 0,and is thus self-adjoint

Let vk and vm be eigenfunctions corresponding to the distinct eigenvalues λk and λm We start with Green’sidentity

hvk, L[vm]i − hL[vk], vmi = 0

hvk, λmsvmi − hλksvk, vmi = 0

(λm− λk)hvk, svmi = 0

hvk, svmi = 0The eigenfunctions are orthogonal with respect to the weighting function s

3 From part (a) we know that there are only positive eigenvalues The general solution of the differential equationis

φ = c1cos(λ1/4x) + c2cosh(λ1/4x) + c3sin(λ1/4x) + c4sinh(λ1/4x)

Applying the condition φ(0) = 0 we obtain

φ = c1(cos(λ1/4x) − cosh(λ1/4x)) + c2sin(λ1/4x) + c3sinh(λ1/4x)

The condition φ00(0) = 0 reduces this to

Chapter 30

Integrals and Convergence

Never try to teach a pig to sing It wastes your time and annoys the pig

-?

30.1 Uniform Convergence of Integrals

Consider the improper integral

Z ∞ c

f (x, t) dt

The integral is convergent to S(x) if, given any  > 0, there exists T (x, ) such that

Z τ c

f (x, t) dt − S(x)

≤ 2

λ.

We will prove the Riemann-Lebesgue lemma for the case when f (x) has limited total fluctuation on the interval(a, b) We can express f (x) as the difference of two functions

f (x) = ψ+(x) − ψ−(x),where ψ+ and ψ− are positive, increasing, bounded functions

From the mean value theorem for positive, increasing functions, there exists an x0, a ≤ x0 ≤ b, such that

Z b a

ψ+(x) sin(λx) dx

=

≤ |ψ+(b)|2

λ.Similarly,

... variables ξ = x/r, u(ξ) = y(x) to obtain the problem

6 ≈ 2 .44 94 (The smallest zero of J0(x) is approximately2 .40 483 .)

at z = l

y1(l) = y2(l),... class="text_page_counter">Trang 8< /span>

Figure 29.1: x and − tan(x).

To solve the inhomogeneous problem, we expand the solution and the inhomogeneity... c3sin(λ1 /4< /sup>x) + c4< /sub>sinh(λ1 /4< /sup>x)

Applying the condition φ(0) = we obtain

φ = c1(cos(λ1 /4< /sup>x) − cosh(λ1 /4< /sup>x)) + c2sin(λ1 /4< /sup>x)