THE METHOD OF FROBENIUS - Loyola University Chicago

[Pages:13]THE METHOD OF FROBENIUS

We have studied how to solve many differential equations via series solutions. In this section we learn how to extend series solutions to a class of differential equations that appear at first glance to diverge in our region of interest.

Let's consider the equation: 2x2 y + 7x(x + 1) y - 3y = 0 (1)

and we are interested in finding the series solution to this equation in the vicinity of x0=0.

Your first instinct might be divide each term in (1) to produce:

y + 7x(x + 1) y - 3 y = 0 (2)

2x2

2x2

In the vicinity of x0=0, it appears that this equation is undefined and will not yield meaningful solutions to the equation (1) near 0.

The necessary conditions for solving equations of the form of (2)

However, the method of Frobenius provides us with a method of adapting our series solutions techniques to solve equations like this if certain conditions hold. Let's consider now those conditions.

First, we will write our second order differential equation as: y + P(x) y + Q(x) y = 0 (3)

When transformed into this form, our original equation in (1)

has P(x) = 7x(x + 1) and Q(x) = - 3 .

2x2

2x2

The theorem of Frobenius shows that if both (x-x0)P(x) and (x-x0)2Q(x) have meaningful series solutions around x0, then a series solution to the differential equation can be found.

Let's apply this theorem to eq. (2) to see if the conditions of this theorem hold:

We want to find a series solution in the neighborhood of x0=0, so (x-x0) = x. Then, we construct the terms xP(x) and x2Q(x) and see if they have are well behaved at x= 0. It is easy to show that both xP(x) and x2Q(x) are well behaved at x0=0 and will have power

series that converge near 0. Thus, the conditions of the Theorem of Frobenius are met,

and we can find a power series solution for (1).

The Method of Frobenius

If the conditions described in the previous section are met, then we can find at least one solution to a second order differential equation by assuming a solution of the form:

y = x r an x n = an x n+r (4)

n=0

n=0

where r and an are constants to be determined, and n = 0, 1, 2, 3,...While n is always an integer, there is no such constraint on r. We will find that r may have negative or fractional values.

Eq. (4) looks very familiar to our earlier work with series solutions, and we should expect that we will have to derive recursion relations to find the exact values of an. The obvious difference is that we also have to determine the values of the exponent r; it is these factors of xr that allow us to find solutions to our differential equations. (Note that other authors and sources will use a staggeringly diverse array of symbols to represent the exponent that we have called r. There seems to be no standard for this; do not be confused, they are all dummy variables referring to the same concept. But we will use r throughout.)

The indicial equation and the values of r

The first step in using the method of Frobenius is to determine the values of r that allow us to solve the differential equation. We do this by exploiting the fact that this method produces a series where the first non zero term is a0.

We substitute our assumed form of the series solution (4) into our original equation and determine the proper values of r by setting n=0; in other words, we find an expression that will ensure the a0 term is non-zero. Let's apply this technique to our original equation (1). Substituting (4) into (1) gives us:

2x 2 (n + r)(n + r - 1)an x n+r-2 + 7x 2 (n + r)an x n+r-1 + 7x (n + r)an x n+r-1 - 3 an x n+r = 0

n=0

n=0

n=0

n=0

(5)

Let's look carefully at (5). Notice that unlike our earlier work with series solutions, we do not change the lower limit of summation when we differentiate our solution. This is because of the presence of the factor xr; as noted above, the value of r is chosen so that the first non zero term of the expansion is the a0 term, so we do not change the lower limit of summation upon differentiation as we did in our prior work with series solutions. Now, let's rewrite (5) by consolidating all powers of x inside the summation signs, and get:

2 (n + r)(n + r - 1)an x n+r + 7 (n + r)an x n+r+1 + 7 (n + r)an x n+r - 3 an x n+r = 0 (6)

n=0

n=0

n=0

n=0

We see immediately that three of the summations are to the power of n+r; and that only one summation (the second one in (6)) is to a different power. We know that we want to re-index this sum so that it is also to the power of n+r? and when we do this, eq (6) becomes:

2 (n + r)(n + r - 1)an x n+r + 7 (n + r - 1)an-1 x n+r + 7 (n + r)an x n+r - 3 an x n+r = 0

n=0

n =1

n=0

n=0

(7)

Now it is appropriate to change the lower limit of the second summation since we have to re-index each term that involves n.

We are now in a position to determine the values of r for this equation. Remember that r must be chosen so that the a0 term is non zero. We can "strip out" the n=0 term in (7) and solve for the values of r. Setting n= 0 in (7) gives us:

[2r(r - 1) + 7r - 3]a0 = 0 (2r 2 + 5r - 3)a0 = 0 (8)

Since we know that a0 is non-zero, eq. (8) implies: 2r 2 + 5r - 3 = 0 (9)

Equation (9) is called the indicial equation, and its solution gives us the values of r we use in finding our solution in the form of (4).

Solving (9) is easily done by factoring: (2r - 1)(r + 3) = 0 (10)

and our values of r are r= ? and r=-3; we will see that our solutions to the differential equation will involve factors of x1/2 and x ? 3.

In fact, we can derive the indicial equation even before re-indexing summations. We look for those summations that involve the lowest power of x, and use those terms with n=0 to derive the indicial equation. In equation (6) above, the first, third and fourth sums are the terms with the lowest power of x, and these are the terms that we use to derive the indicial equation.

Finding the recursion relation

The next step in this method is a familiar one to us, namely, finding the recursion relation that will enable us to determine the values of an. We begin by referring back to eq. (7):

2 (n + r)(n + r - 1)an x n+r + 7 (n + r - 1)an-1 x n+r + 7 (n + r)an x n+r - 3 an x n+r = 0

n=0

n =1

n=0

n=0

We know that since we have all our exponents to the same power, we can combine like terms to produce:

[2(n + r)(n + r -1) + 7(n + r) - 3]an + 7(n + r -1)an-1 = 0 (11)

or equivalently:

an

=

- 7(n + r - 1)an-1 [2(n + r) - 1][(n + r) + 3]

(12)

And this is the recursion relation for this particular differential equation.

We can make a couple of important points about recursion relations in the method of Frobenius. First, you will have to use this recursion relation twice; once to determine values of an when r = ? , and a second time to determine values of an when r = -3. Second, notice the denominator in (12). It is highly recommended that you factor the coefficient of an in writing your final recursion relation.

Review the steps in deriving the indicial equation (10). You set n = 0 in deriving (10). If you replace r with n+r , you will have determined the coefficient of an for all values of n (including non zero values). Look now at the denominator in (12); the factors in the denominator are those you would derive by replacing r with (n+r) in (10).

Of course, it is critical that you determine the indicial equation correctly, else you will propagate your errors all the way through the problem.

Finding the values of an

Now that we have a recursion relation, we can calculate the coefficients an. As noted above, we have to do this separately for each value of r.

It is standard to start with the greater value of r.

When r= ? :

an

=

- 7(n + r - 1)an-1 [2(n + r) -1][(n + r) + 3]

=

- 7(n - 1/ 2)an-1 [2(n + 1/ 2) -1][n + 7 / 2]

=

- 7(2n - 1)an-1 2n(2n + 7)

for n 1 (13)

This recursion relation gives us the following values:

a1

=

- 7a0 18

,

a2

=

- 21a1 44

=

147 792 a0

(14)

And we can write the first solution to the differential equation as:

y 1

=

a0 x1/ 2 (1 -

7 18

x

+

147 792

x2

+ ...)

(15)

When r = -3:

We go back to the original recursion relation (12) and begin as:

an

=

- 7(n + r - 1)an-1 [2(n + r) -1][(n + r) + 3]

=

- 7(n - 4)an-1 [2(n - 3) -1][n]

=

- 7(n - 4)an-1 n(2n - 7)

for n 1

(16)

This recursion relation yields the following coefficients:

a1

=

- 21 5

a0

,

a2

=

- 7a1 3

=

49 5

a0

,

a3

=

- 7a0 3

=

- 343 15

a0

(17)

And the solution corresponding to this value of r is then given by:

y2

=

a0 x -3 (1 -

21 x 5

+

49 5

x2

-

343 15

x 3 ...)

(18)

The general solution to the differential equation

We have in essence found the complete solution to our original equation in (1). We then write the general solution as:

y

=

c1 y1

+ c2 y2

=

c1

x1

/

2

(1

-

7 18

x + 147 792

x2

+ ...) + c2 x -3 (1 -

21 x + 5

49 5

x2

-

343 x3 ) 15

(19)

There is not much new in (19); the only point to make is to take explicit note of the coefficients. Above we calculated coefficients in the two cases (i.e., where r = ? and when r = -3) and called the coefficients in both cases an as is our custom. In writing the final solution (19), we just have to remember that y1 and y2 are independent solutions to the original differential equation, so we need to keep in mind that the coefficients c1 and c2 are likely to be different.

Summary of the technique

We can use the method of Frobenius whenever we have a differential equation that meets the conditions described above.

Once you have verified that the method will apply to the equation under scrutiny:

1) Assume a solution of the form y = x r an x n = an xn+r

n=0

n=0

2) Substitute into equation

3) Re-index as needed so that all summations contain xn+r

4) Derive the indicial equation by setting n = 0; this step tells you the needed values of r in order that the first non zero term be the a0 term.

5) Derive recursion relation for equation.

6) Sequentially substitute values of r into recursion relation to determine coefficients for each solution to the differential equation

7) Substitute appropriate values of r and an into y = xr an x n for each value of r. n=0

8) Write the general solution of the differential equation.

The Second Solution in Frobenius' Method--Three Cases

When we began discussion of Frobenius' method, we said that this technique provides at least one solution to a differential equation that meets the conditions that (x-x0) P(x) and (x-x0)2 Q(x) are well defined in the neighborhood of x0. In the section immediately above, we glossed over any nuance implied by this statement and found both solutions to the differential equation.

There are actually three different possible cases that arise in using Frobenius' method, and the nature of the second solution to the equation is dependent upon which case we encounter.

The method of Frobenius will always produce a solution for the larger value of r. The second solution depends on:

CASE I:

If the two roots of the indicial equation, r1 and r2 are such that the quantity (r1 ? r2) is not an integer, then the second solution to the differential equation is found in exactly the manner we have just used.

In short, we can find both solutions to the differential equation from:

y1 = x r1 an xn and y2 = x r2 bn x n (20)

where r1 and r2 are the solutions to the indicial equation, and I call the coefficients an and bn just to remind us that the sets of coefficients produced for the two summations are almost certain to be different from each other.

CASE II:

If the two roots of the indicial equation are equal, i.e., r1 = r2, then the method of Frobenius will only provide one solution to the differential equation.

The second solution will be of the form:

x

exp[- P(x)dx]

y2 = y1

( y12 )

dx (21)

Where P(x) is as defined earlier, y1 is the solution of the form of eq. (4), and y2 is the second solution to the differential equation.

We will show examples of how to use (21) to solve equations a little later in the document.

CASE III:

If the two roots of the indicial equation are different and are such that (r1 ? r2) is an integer, then the method of Frobenius may or may not produce the second solution to the differential equation. (Don't you love this case...)

The obvious question is how you can determine whether you the method of Frobenius will produce the second solution or not. The way to proceed is to find the roots of the indicial equation and the recursion relation. Use these data to find the solution for the larger root of the indicial equation.

Then, substitute the value of the lesser root of the indicial equation into the recursion formula and see if it produces a complete set of coefficients. If it does, then the method of Frobenius is valid for the second solution of the differential equation, and the second

solution comes simply from y = xr an xn where now r is the lesser root of the indicial n=0

equation.

However, if the recursion relation produces one or more infinite results when using the r2 root, then the method of Frobenius will not yield the second solution, and you have to apply eq. (21) to find that solution.

EXAMPLES

Below is an example showing of a CASE III equation which shows how to use eq. (21) to find the second solution.

Let's start with the differential equation:

x(x - 1) y + 3xy + y = 0 (22)

Check that this equation satisfies the conditions for using the Frobenius method. Here, P(x)=3/(x-1) and Q(x)=1/x(x-1).

Substituting y = x r an x n = an xn+r into (22) yields:

n=0

n=0

y = (n + r)(n + r -1)an x n+r - (n + r)(n + r -1)an x n+r-1 + 3 (n + r)an x n+r + an x n+r = 0

n=0

n=0

n=0

n=0

(23)

The indicial equation:

We use eq. (23) to find the indicial equation. As described before, the indicial equation provides us with the values of r that will solve the differential equation and will ensure that the first non zero term of the power series solution is the a0 term. We can find the indicial equation by setting n equal to zero in those terms of (23) bearing the lowest power of x. In this case, there is only one such term: the second one. (The exponent in the second term is n+r-1, in all other terms the exponent is n+r.)

Setting n=0 in the second term produces the indicial equation:

r(r - 1) = 0 (24)

for which the roots are easily seen to be r = 0, 1. This is an example of CASE III, where the difference in the roots of the indicial equation is an integer.

The recursion relation

We know that the method of Frobenius will produce at least one solution for the original equation (22). To produce the recursion relation we need to re-index only the second summation. Doing so in the typical way, we can proceed by writing the recursion relation:

an+1

=

(n + r (n +

+ 1)(n + r)(n + r

r + 1) + 1) an

=

(n + r + 1) (n + r)

an

(25)

For the greater root, we have that:

an+1

=

(n + r (n +

+ 1) r)

an

=

(n + 2) (n + 1)

an

(26)

The first solution

By now we are very adept at using recursion relations to find coefficients, and doing so allows us to write:

 ................
................

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