The Method of Frobenius

The Method of Frobenius Department of Mathematics IIT Guwahati

If either p(x) or q(x) in y + p(x)y + q(x)y = 0 is not analytic near x 0, power series solutions valid near x 0 may or may not exist.

If either p(x) or q(x) in y + p(x)y + q(x)y = 0 is not analytic near x 0, power series solutions valid near x 0 may or may not exist. Example: Try to find a power series solution of about the point x 0 = 0. Assume that a solution exists. x 2 y y y = 0 (1) y(x) = a n x n

Substituting this series in (1), we obtain the recursion formula a n+1 = n2 n 1 a n. n + 1

Substituting this series in (1), we obtain the recursion formula a n+1 = n2 n 1 a n. n + 1 The ratio test shows that this power series converges only for x = 0. Thus, there is no power series solution valid in any open interval about x 0 = 0. This is because (1) has a singular point at x = 0.

Substituting this series in (1), we obtain the recursion formula a n+1 = n2 n 1 a n. n + 1 The ratio test shows that this power series converges only for x = 0. Thus, there is no power series solution valid in any open interval about x 0 = 0. This is because (1) has a singular point at x = 0. The method of Frobenius is a useful method to treat such equations.

Cauchy-Euler equations revisited Recall that a second order homogeneous Cauchy-Euler equation has the form ax 2 y (x) + bxy (x) + cy(x) = 0, x > 0, (2) where a( 0), b, c are real constants. Writing (2) in the standard form as y + p(x)y + q(x)y = 0, where p(x) = b ax, q(x) = c ax 2.

Cauchy-Euler equations revisited Recall that a second order homogeneous Cauchy-Euler equation has the form ax 2 y (x) + bxy (x) + cy(x) = 0, x > 0, (2) where a( 0), b, c are real constants. Writing (2) in the standard form as y + p(x)y + q(x)y = 0, where p(x) = b ax, q(x) = c ax 2. Note that x = 0 is a singular point for (2). We seek solutions of the form y(x) = x r and then try to determine the values for r.

Set L(y)(x) := ax 2 y (x) + bxy (x) + cy(x) and w(x) := x r. Now L(w)(x) = ax 2 r(r 1)x r 2 + bxrx r 1 + cx r = {ar 2 + (b a)r + c}x r.

Set L(y)(x) := ax 2 y (x) + bxy (x) + cy(x) and w(x) := x r. Now L(w)(x) = ax 2 r(r 1)x r 2 + bxrx r 1 + cx r = {ar 2 + (b a)r + c}x r. Thus, w = x r is a solution r satisfies ar 2 + (b a)r + c = 0. (3)

Set L(y)(x) := ax 2 y (x) + bxy (x) + cy(x) and w(x) := x r. Now L(w)(x) = ax 2 r(r 1)x r 2 + bxrx r 1 + cx r = {ar 2 + (b a)r + c}x r. Thus, w = x r is a solution r satisfies ar 2 + (b a)r + c = 0. (3) The equation (3) is known as the auxiliary or indicial equation for (2).

Case I: When (3) has two distinct roots r 1, r 2. Then L(w)(x) = a(r r 1 )(r r 2 )x r. The two linearly independent solutions are y 1 (x) = w 1 (x) = x r 1, y 2 (x) = w 2 (x) = x r 2 for x > 0.

Case I: When (3) has two distinct roots r 1, r 2. Then L(w)(x) = a(r r 1 )(r r 2 )x r. The two linearly independent solutions are y 1 (x) = w 1 (x) = x r 1, y 2 (x) = w 2 (x) = x r 2 for x > 0. Case II: When r 1 = α + iβ, r 2 = α iβ. Then x α+iβ = e (α+iβ) ln x = e α ln x cos(β ln x) + i e α ln x sin(β ln x) = x α cos(β ln x) + i x α sin(β ln x).

Case I: When (3) has two distinct roots r 1, r 2. Then L(w)(x) = a(r r 1 )(r r 2 )x r. The two linearly independent solutions are y 1 (x) = w 1 (x) = x r 1, y 2 (x) = w 2 (x) = x r 2 for x > 0. Case II: When r 1 = α + iβ, r 2 = α iβ. Then x α+iβ = e (α+iβ) ln x = e α ln x cos(β ln x) + i e α ln x sin(β ln x) = x α cos(β ln x) + i x α sin(β ln x). Thus (Exercise:), two linearly independent real-valued solutions are y 1 (x) = x α cos(β ln x), y 2 (x) = x α sin(β ln x).

Case III: When r 1 = r 2 = r 0 is a repeated roots. Then L(w)(x) = a(r r 0 ) 2 x r. Setting r = r 0 yields the solution y 1 (x) = w(x) = x r 0, x > 0. To find the second linearly independent solution, apply reduction of order method.

A second linearly independent solution is y 2 (x) = x r 0 ln x, x > 0.

A second linearly independent solution is y 2 (x) = x r 0 ln x, x > 0. Example: Find a general solution to Note that 4x 2 y (x) + y(x) = 0, x > 0. L(w)(x) = (4r 2 4r + 1)x r. The indicial equation has repeated roots r 0 = 1/2. Thus, the general solution is y(x) = c 1 x + c2 x ln x, x > 0.

The Method of Frobenius To motivate the procedure, recall the Cauchy-Euler equation in the standard form where y (x) + p(x)y (x) + q(x)y(x) = 0, (4) p(x) = p 0 x, q(x) = q 0 x 2 with p 0 = b/a q 0 = c/a.

The Method of Frobenius To motivate the procedure, recall the Cauchy-Euler equation in the standard form where y (x) + p(x)y (x) + q(x)y(x) = 0, (4) p(x) = p 0 x, q(x) = q 0 x 2 with p 0 = b/a q 0 = c/a. The indicial equation is of the form r(r 1) + p 0 r + q 0 = 0. (5) If r = r 1 is a root of (5), then w(x) = x r 1 is a solution to (4).

Observe that, x p(x) = p 0, a constant, and hence analytic at x = 0, x 2 q(x) = q 0, a constant, and hence analytic at x = 0

Observe that, x p(x) = p 0, a constant, and hence analytic at x = 0, x 2 q(x) = q 0, a constant, and hence analytic at x = 0 Now consider general second order variable coefficient ODE in the standard form y (x) + p(x)y (x) + q(x)y(x) = 0. (1)

Observe that, x p(x) = p 0, a constant, and hence analytic at x = 0, x 2 q(x) = q 0, a constant, and hence analytic at x = 0 Now consider general second order variable coefficient ODE in the standard form y (x) + p(x)y (x) + q(x)y(x) = 0. (1) Say x = 0 is a singular point of this ODE.

Observe that, x p(x) = p 0, a constant, and hence analytic at x = 0, x 2 q(x) = q 0, a constant, and hence analytic at x = 0 Now consider general second order variable coefficient ODE in the standard form y (x) + p(x)y (x) + q(x)y(x) = 0. (1) Say x = 0 is a singular point of this ODE. If we have xp(x) and x 2 q(x) are analytic near x = 0. i.e. xp(x) = p n x n, x 2 q(x) = q n x n.

Observe that, x p(x) = p 0, a constant, and hence analytic at x = 0, x 2 q(x) = q 0, a constant, and hence analytic at x = 0 Now consider general second order variable coefficient ODE in the standard form y (x) + p(x)y (x) + q(x)y(x) = 0. (1) Say x = 0 is a singular point of this ODE. If we have xp(x) and x 2 q(x) are analytic near x = 0. i.e. xp(x) = p n x n, x 2 q(x) = q n x n. Then near x = 0, lim xp(x) = p 0, a constant, x 0 lim x 0 x2 q(x) = q 0, a constant, Resembles Cauchy-Euler eqn.

Therefore a guess for the solution of ODE (??) is y(x) = x r a n x n.

Therefore a guess for the solution of ODE (??) is y(x) = x r Definition: A singular point x 0 of a n x n. y (x) + p(x)y (x) + q(x)y(x) = 0 is said to be a regular singular point if both (x x 0 )p(x) and (x x 0 ) 2 q(x) are analytic at x 0. Otherwise x 0 is called an irregular singular point.

Therefore a guess for the solution of ODE (??) is y(x) = x r Definition: A singular point x 0 of a n x n. y (x) + p(x)y (x) + q(x)y(x) = 0 is said to be a regular singular point if both (x x 0 )p(x) and (x x 0 ) 2 q(x) are analytic at x 0. Otherwise x 0 is called an irregular singular point. Example: Classify the singular points of the equation (x 2 1) 2 y (x) + (x + 1)y (x) y(x) = 0.

Therefore a guess for the solution of ODE (??) is y(x) = x r Definition: A singular point x 0 of a n x n. y (x) + p(x)y (x) + q(x)y(x) = 0 is said to be a regular singular point if both (x x 0 )p(x) and (x x 0 ) 2 q(x) are analytic at x 0. Otherwise x 0 is called an irregular singular point. Example: Classify the singular points of the equation (x 2 1) 2 y (x) + (x + 1)y (x) y(x) = 0. The singular points are 1 and 1. Note that x = 1 is an irregular singular point and x = 1 is a regular singular point.

Series solutions about a regular singular point Assume that x = 0 is a regular singular point for y (x) + p(x)y (x) + q(x)y(x) = 0 so that p(x) = p n x n 1, q(x) = q n x n 2.

Series solutions about a regular singular point Assume that x = 0 is a regular singular point for y (x) + p(x)y (x) + q(x)y(x) = 0 so that p(x) = p n x n 1, q(x) = q n x n 2. In the method of Frobenius, we seek solutions of the form w(x) = x r a n x n = a n x n+r, x > 0. Assume that a 0 0. We now determine r and a n, n 1.

Differentiating w(x) with respect to x, we have w (x) = (n + r)a n x n+r 1, w (x) = (n + r)(n + r 1)a n x n+r 2.

Differentiating w(x) with respect to x, we have w (x) = (n + r)a n x n+r 1, w (x) = (n + r)(n + r 1)a n x n+r 2. Substituting w, w, w, p(x) and q(x) into (4), we obtain (n + r)(n + r 1)a n x n+r 2 ( ) ( ) + p n x n 1 (n + r)a n x n+r 1 ( ) ( ) + q n x n 2 a n x n+r = 0.

Group like powers of x, starting with the lowest power, x r 2. We find that [r(r 1) + p 0 r + q 0 ]a 0 x r 2 + [(r + 1)ra 1 + (r + 1)p 0 a 1 +p 1 ra 0 + q 0 a 1 + q 1 a 0 ]x r 1 + = 0. Considering the first term, x r 2, we obtain {r(r 1) + p 0 r + q 0 }a 0 = 0. Since a 0 0, we obtain the indicial equation.

Group like powers of x, starting with the lowest power, x r 2. We find that [r(r 1) + p 0 r + q 0 ]a 0 x r 2 + [(r + 1)ra 1 + (r + 1)p 0 a 1 +p 1 ra 0 + q 0 a 1 + q 1 a 0 ]x r 1 + = 0. Considering the first term, x r 2, we obtain {r(r 1) + p 0 r + q 0 }a 0 = 0. Since a 0 0, we obtain the indicial equation. Definition: If x 0 is a regular singular point of y + p(x)y + q(x)y = 0, then the indicial equation for this point is r(r 1) + p 0 r + q 0 = 0, where p 0 := lim x x0 (x x 0 )p(x), q 0 := lim x x0 (x x 0 ) 2 q(x).

Example: Find the indicial equation at the the singularity x = 1 of (x 2 1) 2 y (x) + (x + 1)y (x) y(x) = 0.

Example: Find the indicial equation at the the singularity x = 1 of (x 2 1) 2 y (x) + (x + 1)y (x) y(x) = 0. Here x = 1 is a regular singular point. We find that p 0 = lim (x + 1)p(x) = lim (x x 1 x 1 1) 2 = 1 4, q 0 = lim x 1 (x + 1)2 q(x) = lim x 1 [ (x 1) 2 ] = 1 4. Thus, the indicial equation is given by r(r 1) + 1 4 r 1 4 = 0.

The method of Frobenius To derive a series solution about the singular point x 0 of a 2 (x)y (x) + a 1 (x)y (x) + a 0 (x)y(x) = 0, x > x 0. (8) Set p(x) = a 1 (x)/a 2 (x), q(x) = a 0 (x)/a 2 (x). If both (x x 0 )p(x) and (x x 0 ) 2 q(x) are analytic at x 0, then x 0 is a regular singular point and the following steps apply.

The method of Frobenius To derive a series solution about the singular point x 0 of a 2 (x)y (x) + a 1 (x)y (x) + a 0 (x)y(x) = 0, x > x 0. (8) Set p(x) = a 1 (x)/a 2 (x), q(x) = a 0 (x)/a 2 (x). If both (x x 0 )p(x) and (x x 0 ) 2 q(x) are analytic at x 0, then x 0 is a regular singular point and the following steps apply. Step 1: Seek solution of the form w(x) = a n (x x 0 ) n+r. Using term-wise differentiation and substitute w(x) into (8) to obtain an equation of the form A 0 (x x 0 ) r 2 + A 1 (x x 0 ) r 1 + = 0.

Step 2: Set A 0 = A 1 = A 2 = = 0. (A 0 = 0 will give the indicial equation f(r) = r(r 1) + p 0 r + q 0 = 0.)

Step 2: Set A 0 = A 1 = A 2 = = 0. (A 0 = 0 will give the indicial equation f(r) = r(r 1) + p 0 r + q 0 = 0.) Step 3: Use the system of equations A 0 = 0, A 1 = 0,..., A k = 0 to find a recurrence relation involving a k and a 0, a 1,..., a k 1.

Step 2: Set A 0 = A 1 = A 2 = = 0. (A 0 = 0 will give the indicial equation f(r) = r(r 1) + p 0 r + q 0 = 0.) Step 3: Use the system of equations A 0 = 0, A 1 = 0,..., A k = 0 to find a recurrence relation involving a k and a 0, a 1,..., a k 1. Step 4: Take r = r 1, the larger root of the indicial equation (We are assuming here that the indicial equation has two real roots. The case when the indicial equation has two complex conjugate roots is complicated and we do not go into it.), and use the relation obtained in Step 3 to determine a 1, a 2,... recursively in terms of a 0 and r 1.

Step 2: Set A 0 = A 1 = A 2 = = 0. (A 0 = 0 will give the indicial equation f(r) = r(r 1) + p 0 r + q 0 = 0.) Step 3: Use the system of equations A 0 = 0, A 1 = 0,..., A k = 0 to find a recurrence relation involving a k and a 0, a 1,..., a k 1. Step 4: Take r = r 1, the larger root of the indicial equation (We are assuming here that the indicial equation has two real roots. The case when the indicial equation has two complex conjugate roots is complicated and we do not go into it.), and use the relation obtained in Step 3 to determine a 1, a 2,... recursively in terms of a 0 and r 1. Step 5: A series expansion of a solution to (8) is w 1 (x) = (x x 0 ) r 1 a n (x x 0 ) n, x > x 0, where a 0 is arbitrary and a n s are defined in terms of a 0 and r 1.

Remark: The roots r 1 and r 2 of the indicial equation can be either real or imaginary, we discuss only the case when the roots are real. Let r 1 r 2. For the larger root r 1, we have already got one series solution (as mentioned earlier).

Remark: The roots r 1 and r 2 of the indicial equation can be either real or imaginary, we discuss only the case when the roots are real. Let r 1 r 2. For the larger root r 1, we have already got one series solution (as mentioned earlier). Qn. How to find the second linear independent solution of the give ODE?

Remark: The roots r 1 and r 2 of the indicial equation can be either real or imaginary, we discuss only the case when the roots are real. Let r 1 r 2. For the larger root r 1, we have already got one series solution (as mentioned earlier). Qn. How to find the second linear independent solution of the give ODE? Observation: We obtain the constants a n from the following equations: a 0f(r) = 0, a 1f(r + 1) + a 0(rp 1 + q 1) = 0, a 2f(r + 2) + a 0(rp 2 + q 2) + a 1[(r + 1)p 1 + q 1] = 0, a nf(r + n) + a 0(rp n + q n) + + a n 1[(r + n 1)p 1 + q 1] = 0.

Remark: The roots r 1 and r 2 of the indicial equation can be either real or imaginary, we discuss only the case when the roots are real. Let r 1 r 2. For the larger root r 1, we have already got one series solution (as mentioned earlier). Qn. How to find the second linear independent solution of the give ODE? Observation: We obtain the constants a n from the following equations: a 0f(r) = 0, a 1f(r + 1) + a 0(rp 1 + q 1) = 0, a 2f(r + 2) + a 0(rp 2 + q 2) + a 1[(r + 1)p 1 + q 1] = 0, a nf(r + n) + a 0(rp n + q n) + + a n 1[(r + n 1)p 1 + q 1] = 0. Note: f(r 1 ) = 0, f(r 2 ) = 0. If r 1 r 2 = N then f(r 2 + N) = f(r 1 ) = 0 cannot find a N of solution corresponds to r 2.

Theorem: Let x 0 be a regular singular point for y + p(x)y (x) + q(x)y(x) = 0. R be the radius of convergence of both (x x 0 )p(x)and (x x 0 ) 2 q(x) and let r 1 and r 2 be the roots of the associated indicial equation, where r 1 r 2.

Theorem: Let x 0 be a regular singular point for y + p(x)y (x) + q(x)y(x) = 0. R be the radius of convergence of both (x x 0 )p(x)and (x x 0 ) 2 q(x) and let r 1 and r 2 be the roots of the associated indicial equation, where r 1 r 2. Case a: If r 1 r 2 is not an integer, then there exist two linearly independent solutions of the form y 1 (x) = y 2 (x) = a n (x x 0 ) n+r 1 ; a 0 0, b n (x x 0 ) n+r 2, b 0 0, x 0 < x < x 0 + R x 0 < x < x 0 + R

Case b: If r 1 = r 2, then there exist two linearly independent solutions of the form y 1 (x) = a n (x x 0 ) n+r1 ; a 0 0, y 2 (x) = y 1 (x) ln(x x 0 ) + x 0 < x < x 0 + R b n (x x 0 ) n+r1, x 0 < x < x 0 + R. n=1

Case b: If r 1 = r 2, then there exist two linearly independent solutions of the form y 1 (x) = a n (x x 0 ) n+r1 ; a 0 0, y 2 (x) = y 1 (x) ln(x x 0 ) + x 0 < x < x 0 + R b n (x x 0 ) n+r1, x 0 < x < x 0 + R. n=1 Case c: If r 1 r 2 is a positive integer, then there exist two linearly independent solutions of the form y 1 (x) = a n (x x 0 ) n+r1 ; a 0 0, y 2 (x) = Cy 1 (x) ln(x x 0 ) + x 0 < x < x 0 + R b n (x x 0 ) n+r2, b 0 0, x 0 < x < x 0 + R where C is a constant that could be zero. *** End ***