Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 2 pps

23.3 Irregular Singular PointsIf a point z0 of a differential equation is not ordinary or regular singular, then it is an irregular singular point.. At leastone of the solutions at an ir

Example 23.2.4 Consider a series expansion about the origin of the equation

n + α)(n + α − 1)an+ (n + α − 1)an− (n + α − 1)an−1izn = 0

Equating powers of z to zero,

Thus the first solution is

n

= 2a0

1z

We see that in this case there is no trouble in defining a2(α2) The second solution is

w2 = a0z

∞

X

n=0

znn! =



23.3 Irregular Singular Points

If a point z0 of a differential equation is not ordinary or regular singular, then it is an irregular singular point At leastone of the solutions at an irregular singular point will not be of the Frobenius form We will examine how to obtainseries expansions about an irregular singular point in the chapter on asymptotic expansions

If we want to determine the behavior of a function f (z) at infinity, we can make the transformation ζ = 1/z andexamine the point ζ = 0

Example 23.4.1 Consider the behavior of f (z) = sin z at infinity This is the same as considering the point ζ = 0 ofsin(1/ζ), which has the series expansion

sin 1ζ

Thus we see that the point ζ = 0 is an essential singularity of sin(1/ζ) Hence sin z has an essential singularity at

In order to classify the point at infinity of a differential equation in w(z), we apply the transformation ζ = 1/z,u(ζ) = w(z) We write the derivatives with respect to z in terms of ζ.

z = 1ζ

dz = − 1

ζ2dζd

dz = −ζ

2 ddζ

Thus we see that the differential equation for w(z) has an irregular singular point at infinity.

1 Find two linearly independent solutions in the form of power series about x = 0.

2 Compute the radius of convergence of the series Explain why it is possible to predict the radius of convergencewithout actually deriving the series

3 Show that if α = 2n, with n an integer and n ≥ 0, the series for one of the solutions reduces to an evenpolynomial of degree 2n

4 Show that if α = 2n + 1, with n an integer and n ≥ 0, the series for one of the solutions reduces to an oddpolynomial of degree 2n + 1

5 Show that the first 4 polynomial solutions Pn(x) (known as Legendre polynomials) ordered by their degree andnormalized so that Pn(1) = 1 are

6 Show that the Legendre equation can also be written as

((1 − x2)y0)0 = −α(α + 1)y

Note that two Legendre polynomials Pn(x) and Pm(x) must satisfy this relation for α = n and α = m respectively

By multiplying the first relation by Pm(x) and the second by Pn(x) and integrating by parts show that Legendrepolynomials satisfy the orthogonality relation

0

+1 − z

z2 w = 0,such that the series are real-valued on the positive real axis Do not calculate the coefficients in the expansions

Exercise 23.5

Find the series expansions about z = 0 for

w00+ 54zw

0

+z − 18z2 w = 0

1 Show that x = 0 is a regular singular point Determine the location of any additional singular points and classifythem Include the point at infinity.

2 Compute the indicial equation for the point x = 0

3 By solving an appropriate recursion relation, show that one solution has the form

y1(x) = 1 + ax

b +

(a)2x2(b)22! + · · · +

(a)nxn(b)nn! + · · ·where the notation (a)n is defined by

(a)n= a(a + 1)(a + 2) · · · (a + n − 1), (a)0 = 1

Assume throughout this problem that b 6= n where n is a non-negative integer

4 Show that when a = −m, where m is a non-negative integer, that there are polynomial solutions to this equation.Compute the radius of convergence of the series above when a 6= −m Verify that the result you get is in accordwith the Frobenius theory

5 Show that if b = n + 1 where n = 0, 1, 2, , then the second solution of this equation has logarithmic terms.Indicate the form of the second solution in this case You need not compute any coefficients

Hint, Solution

Equating coefficients gives us a difference equation for an:

(n + 2)(n + 1)an+2− 2nan+ 2λan = 0

an+2 = 2 n − λ

(n + 1)(n + 2)an.

The first two coefficients, a0 and a1 are arbitrary The remaining coefficients are determined by the recurrence relation.

We will find the fundamental set of solutions at x = 0 That is, for the first solution we choose a0 = 1 and a1 = 0; forthe second solution we choose a0 = 0, a1 = 1 The difference equation for y1 is

an+2 = 2 n − λ

(n + 1)(n + 2)an, a0 = 1, a1 = 0,which has the solution

a2n = 2

nQn k=0(2(n − k) − λ)(2n)! , a2n+1 = 0.

The difference equation for y2 is

an+2 = 2 n − λ

(n + 1)(n + 2)an, a0 = 0, a1 = 1,which has the solution

a2n= 0, a2n+1 = 2

nQn−1 k=0(2(n − k) − 1 − λ)

We equate coefficients of xn to obtain a recurrence relation.

n−2

Y

k=0 even k

k(k + 1) − α(α + 1)
, even n,

a1n!

n−2

Y

k=1 odd k







1n!

n−2

Y

k=0 even k







1n!

n−2

Y

k=1 odd k

k(k + 1) − α(α + 1)





xn

2 We determine the radius of convergence of the series solutions with the ratio test.

lim

n→∞

an+2xn+2

anxn

< 1

lim

n→∞

x2 < 1

x2 < 1Thus we see that the radius of convergence of the series is 1 We knew that the radius of convergence would be

at least one, because the nearest singularities of the coefficients of (23.3) occur at x = ±1, a distance of 1 fromthe origin This implies that the solutions of the equation are analytic in the unit circle about x = 0 The radius

of convergence of the Taylor series expansion of an analytic function is the distance to the nearest singularity

3 If α = 2n then a2n+2 = 0 in our first solution From the recurrence relation, we see that all subsequent coefficientsare also zero The solution becomes an even polynomial

y1 =

2n

X

m=0 even m







1m!

m−2

Y

k=0 even k







1m!

m−2

Y

k=1 odd k

k(k + 1) − α(α + 1)





xm

Figure 23.4: The First Four Legendre Polynomials

5 From our solutions above, the first four polynomials are

1x

6 We note that the first two terms in the Legendre equation form an exact derivative Thus the Legendre equationcan also be written as

We verify that for the first four polynomials the value of the integral is 2/(2n + 1) for n = m.

The indicial equation for this problem is

zi = ei log z = cos(log z) + i sin(log z), and z−i = e−i log z = cos(log z) − i sin(log z),

we can write a new set of solutions that are real-valued on the positive real axis as linear combinations of w1 and w2

u1 = 1

2(w1+ w2), u2 =

12i(w1− w2)

Solution 23.4

Consider the equation w00+ w0/(z − 1) + 2w = 0

We see that there is a regular singular point at z = 1 All other finite values of z are ordinary points of the equation

To examine the point at infinity we introduce the transformation z = 1/t, w(z) = u(t) Writing the derivatives withrespect to z in terms of t yields

t4u00+ 2t3u0− t

2u01/t − 1+ 2u = 0

0

+z − 18z2 w = 0

We see that z = 0 is a regular singular point of the equation The indicial equation is

 

α − 14

We multiply the differential equation by 8z2 to put it in a better form Substituting a Frobenius series into thedifferential equation,

[8(n + α)(n + α − 1) + 10(n + α) − 1] an = −an−1

8(n + α)2 + 2(n + α) − 1.The First Solution Setting α = 1/4 in the recurrence formula,

an(α1) = − an−1

8(n + 1/4)2+ 2(n + 1/4) − 1

an(α1) = − an−1

2n(4n + 3).Thus the first solution is

The Second Solution Setting α = −1/2 in the recurrence formula,

8(n − 1/2)2+ 2(n − 1/2) − 1

an= − an−1

2n(4n − 3)

Thus the second linearly independent solution is

Equating the coefficient of the zn term gives us

If {w1, w2} is the fundamental set of solutions, then the initial conditions demand that w1 = 1 + 0 · z + · · · and

w2 = 0 + z + · · · We see that w1 will have only even powers of z and w2 will have only odd powers of z

2n+1

Since the coefficient functions in the differential equation are entire, (analytic in the finite complex plane), the radius

of convergence of these series solutions is infinite

Solution 23.7

w00+ 12zw

Differentiating the series for the first solution,

(n + 1)an+1+ an= 0 → an+1= − an

(n + 1/2)(n + 1).

We can combine the above two equations for an

an+1 = − an

(n + 1/2)(n + 1), for n ≥ 0

Solving this difference equation for an,



−12



b0+12

dz = −1

t2dtd

dz = −t

2 ddt

The equation for u is then

α(α − 1) + 1 = 0

α = 1 ± i

√32Since the roots of the indicial equation are distinct and do not differ by an integer, a set of solutions has the form

(

t(1+i

√ 3)/2

Noting that

t(1+i

√ 3)/2

= t1/2exp i

√3

2 log t

!, and t(1−i

√ 3)/2

= t1/2exp −i

√3

2 log t

!

We can take the sum and difference of the above solutions to obtain the form

u1 = t1/2cos

√3

Putting the answer in terms of z, we have the form of the two Frobenius expansions about infinity.

w1 = z−1/2cos

√3

1 We write the equation in the standard form

y00+b − x

x y

0− a

xy = 0Since b−xx has no worse than a first order pole and ax has no worse than a second order pole at x = 0, that is aregular singular point Since the coefficient functions have no other singularities in the finite complex plane, allthe other points in the finite complex plane are regular points

Now to examine the point at infinity We make the change of variables u(ξ) = y(x), ξ = 1/x

y0 = dξdx



u = ξ4u00+ 2ξ3u0The differential equation becomes

xy00+ (b − x)y0− ay1

at infinity

2 The coefficient functions are

3 Since one of the roots of the indicial equation is zero, and the other root is not a negative integer, one of the

solutions of the differential equation is a Taylor series.

For c0 = 1, the recurrence relation has the solution

ck= (a)kx

k

(b)kk!.Thus one solution is

Now we see if the second solution has the Frobenius form There is no a1x term because y2 is only determined up to

an additive constant times y1

y2 = 1 + O(x2)

We substitute y2 into the differential equation and equate coefficients of powers of x.

xy00+ 2xy0+ 6 exy = 0O(x) + O(x) + 6(1 + O(x))(1 + O(x2)) = 0

6 = O(x)The substitution y2 = 1 + O(x) has yielded a contradiction Since the second solution is not of the Frobenius form, ithas the following form:

y2 = y1ln(x) + a0+ a2x2+ O(x3)The first three terms in the solution are

1 + 6a0 = 0

y2 = −1

6 + x ln x − 4x

2ln x + O(x2)

23.9 Quiz Solutions

Solution 23.1

The series P∞

n=1an converges if the sequence of partial sums, SN =PN

n=1an, converges That is,

indepen-|f (z) − SN(z)| =

< 

for all z in the domain

There is no relationship between absolute convergence and uniform convergence

an

... recurrence formula,

8(n − 1 /2) 2< /small>+ 2( n − 1 /2) −

an= − an−1

2n(4n − 3)

an(α1) = − an−1

8(n + 1 /4) 2< /small>+