Frobenius Method

The Frobenius method is an approach to identify an infinite series solution to a second-order ordinary differential equation. Generally, the Frobenius method determines two independent solutions provided that an integer does not divide the indicial equation’s roots.

Consider the second-order ordinary differential equation given below:

Here, a(x), b(x), and c(x) are “suitable functions”.

Let us assume that these are the suitable rational functions that means polynomials divided by polynomials, such as in Bessel’s equation of order 1/2.

(d²y/dx²) + (1/x) (dy/dx) + [1 – (1/4x²)]y = 0

For this equation, we can illustrate the Frobenius method. The primary approach of Frobenius aims at solutions in the form of power series around some given point x₀ multiplied by (x − x₀) to a certain power. This can be expressed as:

Here, a_k and r are the constants that can be identified through the Frobenius method.

Learn:Ordinary Differential Equations

Let’s start with the Frobenius method to solve the second-order ordinary differential equation.

Step 1: Choose a suitable value for x₀. This can be done in two ways:

(i) If conditions are given for y(x) at some point, we can use that for x0.

(ii) If no conditions are given for y(x), we must choose x₀ as per our convenience. Generally, we prefer to choose x₀ = 0.

Step 2: If the given differential equation is of the form a(x) (d²y/dx²) + b(x) (dy/dx) + c(x) y = 0, then convert this, as mentioned above.

i.e., (d²y/dx²) + (1/x) (dy/dx) + [1 – (1/4x²)]y = 0 ………(1)

Then multiply the equation by 4x² so that we can avoid fractions to make the simplification easy.

4x²(d²y/dx²) + 4x (dy/dx) + (4x² – 1)y = 0

(or)

4x² y′′ + 4xy′ + (4x² – 1)y = 0……..(2)

Step 3: Let us assume that the solution is of the form

Step 3: Now, bring the factor (x – x₀)^r inside the summation. That means,

Step 4: From the assumed series of y, we need to calculate the respective modified power series for y′ and y′′ by differentiating “term-by-term”.

For this substitute x₀ = 0. Thus, we get;

Then, by differentiating y, we get;

Step 5: Substitute the expressions for y, y′ and y′′ in equation (2). On simplification of this we get;

Step 6: For each series obtained in the above equation, we need to change the index so that each series will be of the form:

Step 7: Convert the sum of series in the obtained equation into one big series. A few terms will likely have to be written separately, so we need to simplify to a possible extent.

Step 8: Now, the first term of the obtained series may be of the form;

a₀ [formula of r] (x – x₀)^something

Also, remember that each term of the series is 0.

From this, we can write the formula of r = 0

This yields a quadratic equation in r.

Solving this equation, we get two roots, r₁ and r₂.

Step 9: By substituting r1 in the last series equation, we get;

From this, we can get;

n th formula of ak’s = 0 for n0 ≤ n

Then, solve this for highest index = formula of n and lower indexed a_k’s

To simplify the terms at least a little, perform the change of indices so that the recursion formula will be derived and can be rewritten as;

a_k = formula of k and lower-indexed coefficients

Step 10: Using the recursion formula or any corresponding formulas to the lower-order terms, we need to find all the a_k ’s in terms of a₀ and, maybe, one other a_m.

Step 11: Using r = r₁ and the formulas, we have derived the coefficients. Now, write out the resultant series for y. Simplify it and factor out the arbitrary constant(s) if possible.

Step 12: Repeat steps 9 to 11 for substituting r = r₂.

Finally, the last step may yield y as an arbitrary linear combination of two distinct series. Therefore, this is referred to as the general solution to the given differential equation.

This can be further simplified as:

Frobenius Method Solved Example

Example:

Find the solution of 4xy′′ + 2y′ + y = 0 by Frobenius Method.

Solution:

Given differential equation is:

4xy′′ + 2y′ + y = 0….(1)

Let

So,

Substituting the expressions of y, y′ and y′′ in equation (1), we get;

This can be written as:

Dividing the above equation by x^r-1, we get;

Changing the indices of the bases, we get;

Now, we need to expand the summation to make the indices equal. This can be done as follows.

Now consider the term which is of the form a₀[term of r] = 0 .

i.e., a₀[4r(r – 1) + 2r] = 0

4r² – 4r + 2r = 0

4r² – 2r = 0

r² – (½)r = 0

r[r – (½)] = 0

Thus, r = ½, r = 0.

Now, by changing the indices of equation (2), we get;

For k ≥ 0, we get;

4(k + r + 1)(k + r)a_k+1 + 2(k + r + 1)a_k+1 + a_k = 0

From this, we can write a_k+1 as:

a_k+1 = -a_k/ [(2k + 2r + 2)(2k + 2r + 1)] ; k = 0, 1, 2, 3, etc., ………..(3)

Substitute r = r₁ = ½ in equ (3).

a_k+1 = -a_k/ [(2k + 3)(2k + 2)]

Substituting k = 0, 1, 2, 3, and so on, in the above equation, we get;

a₁ = -a₀/3.2 = -a₀/3!

a₂ = -a₁/5.4 = a₀/5!

a₃ = -a₂/7.6 = -a₀/7!

As we know, a₀ is an arbitrary constant. So, let a₀ = 1.

Therefore, a_k (r₁) = (-1)^k/ (2k + 1)!; k = 0, 1, 2, 3,….

Hence,

Similarly, by substituting r = r₂ = 0 in equ (3), we get;

Thus, we obtained the solutions of the given differential equation.

Practice Problems

1. Solve xy′′ + 2y′ + xy = 0 by Frobenius Method.
2. Solve: 5x² y′′ + x(1 + x)y′ − y = 0
3. Consider the differential equation xy” + y’ + 2xy = 0. Obtain the solution by Frobenius method.