**Chapter 5 Series solutions of ODEs, Special functions**

**1 Power series**

A *power series* is an infinite series of the form

where

- , , , are constants, the
*coefficients* of the series.
- is a also a constant, the
*center* of the series.

The *Taylor series* of a function is also a power series. It is of the form

A *Maclaurin series* is a Taylor series with center .

**Example 1.1.**

Maclaurin series

**Exercise 1.1.**

Find the Maclaurin series of .

**2 The power series method for solving ODEs**

Given

- First represent , and by power series in powers of (or if solution in powers of is wanted).
- Next, assume a solution in form of a power series of with unknown coefficients:

And substitute the power series from of , and into (1) to solve for the unknown coefficients , , , , by comparing coefficients of in the L.H.S. and R.H.S. of (1).

**Example 2.**

The power series method to solve .

**Solution.**

First let the solution by of the form

Hence,

Then substitute into the ODE,

Grouping terms of , we have

By comparing the coefficients of in the L.H.S. and R.H.S. we gives

Thus,

**Exercise 2.**

Solve the ODE by using the power series method.

**3 Theory of the power series method**

**3.1 Basic concepts**

Given a power series

- Assume is variable

- is the center
- The coefficients are real
- Its th partial sum is

where .

- The
*remainder* of is

For example, consider the geometric series

we have

- Thus, associated with , there is a sequence of partial sums
- If for some this sequence converges, say,

then is *convergent* at , the number is called the *value* or *sum* of at . And we have

- For any positive , (which depends on the value of ) s.t.

**3.2 Convergence interval and radius of convergence**

With respect to the convergence of the power series there are three cases.

**Case 1**. The series always converges at , because for all its terms are 0, perhaps except for the first one, . In exceptional cases may be the only for which converges. Such a series is of no practical interest.

**Case 2**. If there are further values of for which the series converges, these values form an interval, called the *convergence interval*. If this interval is finite, its midpoint is , so that it is of the form

and the series converges for all such that and diverges for all such that . This is called the *radius of convergence* and it can be found by the *ratio test* or the *root test*, as below.

provided these limits exist and are not zero.

**Case 3**. The convergence interval may sometimes be infinite, *i.e.*, the power series converges . For instance, if either of the limits above is 0 (in other words, if ), this case occurs.

**Example 3.2.1.**

For Case 1, consider a series

we have , and by the ratio test,

It only converges at . Such a series is useless.

**Example 3.2.2.**

For Case 2, for the geometic series we have

In fact, , and from the ratio test we obtain . That is, the geometric series converges and represents when .

Try the following MAXIMA program to see how the Taylor series converges within the given interval, and diverges elsewhere....

*load(draw);*

*tay(n, x) := block(*

* [ts : taylor(1/(1-x__), x__, 0, n)],*

* subst(x__=x, ts)*

*)$*

*with_slider(*

* /* first two arguments are the parameter and parameter values */*

* n, makelist(i, i, 1, 20),*

* /* the rest of arguments are for plot2d command */*

* [1/(1-x), '(tay(n, x))],*

* [x, -2, 2],*

* [y, -10, 10]*

*)$*

Next, try the following MAXIMA program to see how the series centred around 3 converges/diverges...

*load(draw);*

*tay(n, x) := block(*

* [ts : taylor(1/(1-x__), x__, 3, n)],*

* subst(x__=x, ts)*

*)$*

*with_slider(*

* /* first two arguments are the parameter and parameter values */*

* n, makelist(i, i, 1, 20),*

* /* the rest of arguments are for plot2d command */*

* [1/(1-x), '(tay(n, x))],*

* [x, 0, 6],*

* [y, -10, 10]*

*)$*

**Example 3.2.3.**

For Case 3, in the case of the series

we have . Hence by the ratio test,

so that the series converges for all .

**Example 3.2.4.**

*Hint for some problems*

Find the radius of convergence of the series

*Solution*.

This is a power series of with coefficients , so that in (3b),

Thus . Hence the series converges for , i.e. .

**Exercise 3.2.**

Determine the radius of convergence.

**3.3 Existence of power series solutions and Real analytic functions**

**3.3.1 Real analytic function**

A real function is called *analytic at a point* if it can be represented by a power series in powers of with radius of convergence .

**3.3.2 Existence of power series solutions**

Consider

If , and in (4) are analytic at , then every solution of the ODE is analytic at and can thus be represented by a power series in powers of with radius of convergence .

For an ODE

the same is true if and are all analytic at and .

**Exercise 3.3.**

Consider the ODE

Does it have a power series solution? If yes, find it in the power of .

**ADVANCED TOPICS**

**4 Legendre's equation**

This is Legendre's equation

where is a given constant. A typical application of Legendre's Equation is for problems with *spherical symmetry*.

**4.1 Solving Legendre's equation**

Dividing (5) by the coefficient of , we see

The coefficients and of the new equation are analytic at .

Hence Legendre's equation has power series solutions of the form

Substituting it into the ODE and comparing the coefficients of the powers of of both sides, we obtain the general formula

This is called a *recurrence relation* or *recursion formula*. It gives each coefficient in terms of the second one preceding it, except for and , which are left as arbitrary constants.

and so on. By inserting these expressions for the coefficients into (6) we obtain

where

These series converge for and linearly independent solutions. Hence it is a general solution of (5) on the interval .

**Exercise 4.1.**

Prove

(a) Legendre's ODE has a power series solution.

(b) The recursion formula above.

**4.2 Legendre polynomials **

In various applications,

- The power series solutions of ODEs reduce to polynomials, i.e. they terminate after finitely many terms.
- For Legendre's equation this happens when is a non-negative integer because then the right side of the recursion formula is zero for , so that , , , . Hence if is even, reduces to a polynomial of degree . If is odd, the same is true for .

These polynomials, multiplied by some constants, are called Legendre polynomials, . The standard choice of a constant is

(and if ).

Then solving for in terms of , that is,

**Note**: The choice of constants above makes for every . This motivates the choice.

For we then obtain

**Exercise 4.2.1.**

Try to write down , and . Find the general terms when .

In general,

The resulting solution of Legendre's differential equation is called the *Legendre polynomial of degree* and is denoted by .

From the above we obtain

where or , whichever is an integer. The first few of these functions are

Figure 1: Legendre polynomials

**Exercise 4.2.**

Obtain and .

**4.3 Applications of Legendre polynomials in physics**^{1} ^{1}http://en.wikipedia.org/wiki/Legendre_polynomials

Figure 2: Electric potenital

Consider the electric potential at point ,

Then, using the cosine law,

Since the Maclaurin series of

Let

For , let

Consider the Legendre polynomials and comparing to and , we have

**Exercise 4.3.**

Kreyszig 9th Edition Problem set 5.3 question 14(b)

Potential theory.

Let and be two points in space. Show that

This formula has applications in potential theory.

**5 Frobenius method**

Any ODE of the form

where and are analytic at , has AT LEAST ONE SOLUTION that can be represented in the form

where the exponent can be any real or complex number and is choose such that .

The equation also has a second solution s.t. the two solutions are linearly independent and the 2nd solution may be similar to the first one with a different and different coefficients, or may contain a logarithmic term.

**5.1 Indicial equation, Indicating the form of solutions**

Multiple the standard form by , we have

Expanding and in power series,

Differentiating this term by term

Substituting into the original ODE, and equating the sum of coefficients of each power of to zero, we have a system of equations with coefficient as unknowns. In particular, consider the coefficient of

Since , this gives

This is called the *indicial equation* of the ODE.

The Frobenius method yields a basis of solutions. One of the two solutions will always be of the from (12), where is a root of the indicial equation. The other solution will be of a form indicated by the indicial equation. There are three cases:

**Case 1**. Distinct roots and they do not differing by an integer. A basis is

and

**Case 2**. A double root . A basis is

and

**Case 3**. Distinct roots and differing by an integer. A basis is

and

**Example 5.1.1.**

For the Euler-Cauchy equation

substitution of gives the auxiliary equation

which is the indicial equation and is a very special form of (12). For different roots , we get a basis , , and for a double root we get a basis , .

Accordingly, for this simple ODE, Case 3 plays no extra role.

**Example 5.1.2.**

illustration of Case 2.

Solve the ODE

Solution. Write the ODE in standard form.

**Ex 5.1.1.**

What do and become?

By inserting (12) and its derivatives into it, we obtain

The smallest power is .

**Ex 5.1.2.**

Write down the indicial equation in this example and solve it.

By equating the sum of its coefficients to zero we have

Hence this indicial equation has the double root .

We insert this value into (15) and equate the sum of the coefficients of the power to zero, obtaining

thus . Hence , and by choosing we obtain the solution

To obtain the second independent solution we use the method of reduction of order. Or we can use the above discussion, let . Then using substitution to find .

We obtain

(Please see section 2.5 of chapter 2 and try youself.)

**Example 5.1.3.**

Illustration of Case 3.

Solve the ODE

**Solution**. Substituting the Mclaurin series of the solution and and its derivatives into it, we have

In the first series we set and in the second , thus . Then

Since in the second series, the lowest power is and the indicial equation is

The roots are and . They differ by an integer. This is Case 3.

From the recurrence relation above, with we have

This gives the recurrence relation

Hence , , successively. Taking , we get as a first solution

Again, we use the method of reduction of order and we obtain

Or we use the above discussion, let and use substitution to find .

**Exercise 5.1.**

Find a basis of solutions for the following ODE by the Frobenius method.

**6 Bessel's equation**

Bessel's equation

- is real and non-negative
- It can be solved by the Frobenius method
- Bessel's equation often relates to problems of potentials which show
*cylindrical symmetry*, e.g. heat conduction, electrical fields or membrane vibrations in cylinders.

Accordingly, we substitute the series

and its derivatives into the ODE. This gives

Take in the first, second and fourth series, and to in the third series. Then equating the sum of coefficients of to zero, we have

From this we obtain the indicial equation

The roots are and . For , (20b) reduces to . Hence since .

From (c) in the recurrence relation above, since , we have . So we only need to deal with even-numbered coefficients.

Let , we have

Solving for gives the recursion formula

Hence

and so on, and in general

**6.1 Bessel's function for integer **

For , the recurrence relationship becomes

For ease of further manipulation, we set

So,

With these coefficients and we get a particular solution of the ODE as

is called the *Bessel function of the 1st kind of order* and it converges for all .

**Example 6.1.**

For we obtain from above, the Bessel function of order

which looks similar to cosine.

For we obtain the Bessel function of order

which looks a little bit like a decaying sinusoidal function.

Figure 3: Bessel functions of the first kind and

**6.2 Bessel Function for Any **

By defining the Gamma function

By integration by parts we obtain

Now consider ,

Then consider , we obtain , , and in general

This shows that the gamma function does in fact generalize the fractorial function.

Now we take

Then (21) becomes

Since we have ,

so that

Hence

With these coefficients and we get a particular solution of the ODE as

is called the Bessel function of the 1st kind of order and it also converges for all .

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