the method of frobenius - iit guwahati

The Method of Frobenius

Department of MathematicsIIT Guwahati

SU/KSK MA-102 (2018)

If either p(x) or q(x) in

y′′ + p(x)y′ + q(x)y = 0

is not analytic near x0, power series solutions valid near x0may or may not exist.

Example: Try to find a power series solution of

x2y′′ − y′ − y = 0 (1)

about the point x0 = 0.

Assume that a solution

y(x) =∞∑n=0

exists.

SU/KSK MA-102 (2018)

If either p(x) or q(x) in

y′′ + p(x)y′ + q(x)y = 0

is not analytic near x0, power series solutions valid near x0may or may not exist.

Example: Try to find a power series solution of

x2y′′ − y′ − y = 0 (1)

about the point x0 = 0.

Assume that a solution

y(x) =∞∑n=0

exists.

SU/KSK MA-102 (2018)

Substituting this series in (1), we obtain the recursion formula

an+1 =n2 − n− 1

n+ 1an.

The ratio test shows that this power series converges only forx = 0. Thus, there is no power series solution valid in anyopen interval about x0 = 0. This is because (1) has a singularpoint at x = 0.

The method of Frobenius is a useful method to treat suchequations.

SU/KSK MA-102 (2018)

an+1 =n2 − n− 1

n+ 1an.

SU/KSK MA-102 (2018)

an+1 =n2 − n− 1

n+ 1an.

SU/KSK MA-102 (2018)

Cauchy-Euler equations revisited

Recall that a second order homogeneous Cauchy-Eulerequation has the form

ax2y′′(x) + bxy′(x) + cy(x) = 0, x > 0, (2)

where a(6= 0), b, c are real constants. Writing (2) in thestandard form as

y′′ + p(x)y′ + q(x)y = 0, where p(x) =b

ax, q(x) =

Note that x = 0 is a singular point for (2). We seek solutionsof the form

y(x) = xr

and then try to determine the values for r.

SU/KSK MA-102 (2018)

Cauchy-Euler equations revisited

Recall that a second order homogeneous Cauchy-Eulerequation has the form

ax2y′′(x) + bxy′(x) + cy(x) = 0, x > 0, (2)

where a(6= 0), b, c are real constants. Writing (2) in thestandard form as

y′′ + p(x)y′ + q(x)y = 0, where p(x) =b

ax, q(x) =

Note that x = 0 is a singular point for (2). We seek solutionsof the form

y(x) = xr

and then try to determine the values for r.

SU/KSK MA-102 (2018)

L(y)(x) := ax2y′′(x) + bxy′(x) + cy(x) and w(x) := xr.

L(w)(x) = ax2r(r − 1)xr−2 + bxrxr−1 + cxr

= {ar2 + (b− a)r + c}xr.

Thus,w = xr is a solution ⇐⇒ r satisfies

ar2 + (b− a)r + c = 0. (3)

The equation (3) is known as the auxiliary or indicial equationfor (2).

SU/KSK MA-102 (2018)

= {ar2 + (b− a)r + c}xr.

ar2 + (b− a)r + c = 0. (3)

SU/KSK MA-102 (2018)

= {ar2 + (b− a)r + c}xr.

ar2 + (b− a)r + c = 0. (3)

SU/KSK MA-102 (2018)

Case I: When (3) has two distinct roots r1, r2. Then

L(w)(x) = a(r − r1)(r − r2)xr.

The two linearly independent solutions are

y1(x) = w1(x) = xr1 , y2(x) = w2(x) = xr2 for x > 0.

Case II: When r1 = α + iβ, r2 = α− iβ. Then

xα+iβ = e(α+iβ) lnx = eα lnx cos(β lnx) + i eα lnx sin(β lnx)

= xα cos(β lnx) + i xα sin(β lnx).

Thus (Exercise:), two linearly independent real-valuedsolutions are

y1(x) = xα cos(β lnx), y2(x) = xα sin(β lnx).

SU/KSK MA-102 (2018)

L(w)(x) = a(r − r1)(r − r2)xr.

SU/KSK MA-102 (2018)

L(w)(x) = a(r − r1)(r − r2)xr.

SU/KSK MA-102 (2018)

Case III: When r1 = r2 = r0 is a repeated roots. Then

L(w)(x) = a(r − r0)2xr.

Setting r = r0 yields the solution

y1(x) = w(x) = xr0 , x > 0.

To find the second linearly independent solution, applyreduction of order method.

SU/KSK MA-102 (2018)

A second linearly independent solution is

y2(x) = xr0 lnx, x > 0.

Example: Find a general solution to

4x2y′′(x) + y(x) = 0, x > 0.

Note thatL(w)(x) = (4r2 − 4r + 1)xr.

The indicial equation has repeated roots r0 = 1/2. Thus, thegeneral solution is

y(x) = c1√x+ c2

√x lnx, x > 0.

SU/KSK MA-102 (2018)

A second linearly independent solution is

y2(x) = xr0 lnx, x > 0.

Example: Find a general solution to

4x2y′′(x) + y(x) = 0, x > 0.

Note thatL(w)(x) = (4r2 − 4r + 1)xr.

The indicial equation has repeated roots r0 = 1/2. Thus, thegeneral solution is

y(x) = c1√x+ c2

√x lnx, x > 0.

SU/KSK MA-102 (2018)

To motivate the procedure, recall the Cauchy-Euler equationin the standard form

y′′(x) + p̃(x)y′(x) + q̃(x)y(x) = 0, (4)

p̃(x) =p̃0x, q̃(x) =

q̃0x2

with p̃0 = b/a q̃0 = c/a.

The indicial equation is of the form

r(r − 1) + p̃0r + q̃0 = 0. (5)

If r = r1 is a root of (5), then w(x) = xr1 is a solution to (4).

SU/KSK MA-102 (2018)

To motivate the procedure, recall the Cauchy-Euler equationin the standard form

y′′(x) + p̃(x)y′(x) + q̃(x)y(x) = 0, (4)

p̃(x) =p̃0x, q̃(x) =

q̃0x2

with p̃0 = b/a q̃0 = c/a.

The indicial equation is of the form

r(r − 1) + p̃0r + q̃0 = 0. (5)

If r = r1 is a root of (5), then w(x) = xr1 is a solution to (4).

SU/KSK MA-102 (2018)

Observe that,

xp̃(x) = p̃0, a constant, and hence analytic at x = 0,

x2q̃(x) = q̃0, a constant, and hence analytic at x = 0

Now consider general second order variable coefficient ODE inthe 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 x2q(x) are analytic near x = 0. i.e.

xp(x) =∞∑n=0

pnxn, x2q(x) =

∞∑n=0

Then near x = 0, limx→0

xp(x) = p0, a constant,

limx→0

x2q(x) = q0, a constant,

Resembles Cauchy-Euler eqn.

SU/KSK MA-102 (2018)

Observe that,

y′′(x) + p(x)y′(x) + q(x)y(x) = 0. (1)

xp(x) =∞∑n=0

pnxn, x2q(x) =

∞∑n=0

limx→0

SU/KSK MA-102 (2018)

Observe that,

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 x2q(x) are analytic near x = 0. i.e.

xp(x) =∞∑n=0

pnxn, x2q(x) =

∞∑n=0

limx→0

SU/KSK MA-102 (2018)

Observe that,

y′′(x) + p(x)y′(x) + q(x)y(x) = 0. (1)

xp(x) =∞∑n=0

pnxn, x2q(x) =

∞∑n=0

limx→0

SU/KSK MA-102 (2018)

Observe that,

y′′(x) + p(x)y′(x) + q(x)y(x) = 0. (1)

xp(x) =∞∑n=0

pnxn, x2q(x) =

∞∑n=0

limx→0

SU/KSK MA-102 (2018)

Therefore a guess for the solution of ODE (??) is

y(x) = xr∞∑n=0

Definition: A singular point x0 of

y′′(x) + p(x)y′(x) + q(x)y(x) = 0

is said to be a regular singular point if both (x− x0)p(x) and(x− x0)2q(x) are analytic at x0. Otherwise x0 is called anirregular singular point.

Example: Classify the singular points of the equation

(x2 − 1)2y′′(x) + (x+ 1)y′(x)− y(x) = 0.

The singular points are 1 and −1. Note that x = 1 is anirregular singular point and x = −1 is a regular singular point.

SU/KSK MA-102 (2018)

y(x) = xr∞∑n=0

y′′(x) + p(x)y′(x) + q(x)y(x) = 0

(x2 − 1)2y′′(x) + (x+ 1)y′(x)− y(x) = 0.

SU/KSK MA-102 (2018)

y(x) = xr∞∑n=0

y′′(x) + p(x)y′(x) + q(x)y(x) = 0

(x2 − 1)2y′′(x) + (x+ 1)y′(x)− y(x) = 0.

SU/KSK MA-102 (2018)

y(x) = xr∞∑n=0

y′′(x) + p(x)y′(x) + q(x)y(x) = 0

(x2 − 1)2y′′(x) + (x+ 1)y′(x)− y(x) = 0.

SU/KSK MA-102 (2018)

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) =∞∑n=0

pnxn−1, q(x) =

∞∑n=0

qnxn−2.

In the method of Frobenius, we seek solutions of the form

w(x) = xr∞∑n=0

anxn =

∞∑n=0

anxn+r, x > 0.

Assume that a0 6= 0. We now determine r and an, n ≥ 1.

SU/KSK MA-102 (2018)

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) =∞∑n=0

pnxn−1, q(x) =

∞∑n=0

qnxn−2.

In the method of Frobenius, we seek solutions of the form

w(x) = xr∞∑n=0

anxn =

∞∑n=0

anxn+r, x > 0.

Assume that a0 6= 0. We now determine r and an, n ≥ 1.

SU/KSK MA-102 (2018)

Differentiating w(x) with respect to x, we have

w′(x) =∞∑n=0

(n+ r)anxn+r−1,

w′′(x) =∞∑n=0

(n+ r)(n+ r − 1)anxn+r−2.

Substituting w, w′, w′′, p(x) and q(x) into (4), we obtain

∞∑n=0

(n+ r)(n+ r − 1)anxn+r−2

(∞∑n=0

pnxn−1

)(∞∑n=0

(n+ r)anxn+r−1

(∞∑n=0

qnxn−2

)(∞∑n=0

anxn+r

SU/KSK MA-102 (2018)

Differentiating w(x) with respect to x, we have

w′(x) =∞∑n=0

(n+ r)anxn+r−1,

w′′(x) =∞∑n=0

(n+ r)(n+ r − 1)anxn+r−2.

Substituting w, w′, w′′, p(x) and q(x) into (4), we obtain

∞∑n=0

(n+ r)(n+ r − 1)anxn+r−2

(∞∑n=0

pnxn−1

)(∞∑n=0

(n+ r)anxn+r−1

(∞∑n=0

qnxn−2

)(∞∑n=0

anxn+r

SU/KSK MA-102 (2018)

Group like powers of x, starting with the lowest power, xr−2.We find that

[r(r − 1) + p0r + q0]a0xr−2 + [(r + 1)ra1 + (r + 1)p0a1

+p1ra0 + q0a1 + q1a0]xr−1 + · · · = 0.

Considering the first term, xr−2, we obtain

{r(r − 1) + p0r + q0}a0 = 0.

Since a0 6= 0, we obtain the indicial equation.

Definition: If x0 is a regular singular point ofy′′ + p(x)y′ + q(x)y = 0, then the indicial equation for thispoint is

r(r − 1) + p0r + q0 = 0,

p0 := limx→x0

(x− x0)p(x), q0 := limx→x0

(x− x0)2q(x).

SU/KSK MA-102 (2018)

Group like powers of x, starting with the lowest power, xr−2.We find that

[r(r − 1) + p0r + q0]a0xr−2 + [(r + 1)ra1 + (r + 1)p0a1

+p1ra0 + q0a1 + q1a0]xr−1 + · · · = 0.

Considering the first term, xr−2, we obtain

{r(r − 1) + p0r + q0}a0 = 0.

Since a0 6= 0, we obtain the indicial equation.

Definition: If x0 is a regular singular point ofy′′ + p(x)y′ + q(x)y = 0, then the indicial equation for thispoint is

r(r − 1) + p0r + q0 = 0,

p0 := limx→x0

(x− x0)p(x), q0 := limx→x0

(x− x0)2q(x).

SU/KSK MA-102 (2018)

Example: Find the indicial equation at the the singularityx = −1 of

(x2 − 1)2y′′(x) + (x+ 1)y′(x)− y(x) = 0.

Here x = −1 is a regular singular point. We find that

p0 = limx→−1

(x+ 1)p(x) = limx→−1

(x− 1)−2 =1

q0 = limx→−1

(x+ 1)2q(x) = limx→−1

[−(x− 1)−2] = −1

Thus, the indicial equation is given by

r(r − 1) +1

4r − 1

SU/KSK MA-102 (2018)

Example: Find the indicial equation at the the singularityx = −1 of

(x2 − 1)2y′′(x) + (x+ 1)y′(x)− y(x) = 0.

Here x = −1 is a regular singular point. We find that

p0 = limx→−1

(x+ 1)p(x) = limx→−1

(x− 1)−2 =1

q0 = limx→−1

(x+ 1)2q(x) = limx→−1

[−(x− 1)−2] = −1

Thus, the indicial equation is given by

r(r − 1) +1

4r − 1

SU/KSK MA-102 (2018)

The method of FrobeniusTo derive a series solution about the singular point x0 of

a2(x)y′′(x) + a1(x)y

′(x) + a0(x)y(x) = 0, x > x0. (8)

Set p(x) = a1(x)/a2(x), q(x) = a0(x)/a2(x).

If both (x− x0)p(x) and (x− x0)2q(x) are analytic at x0, thenx0 is a regular singular point and the following steps apply.

Step 1: Seek solution of the form

w(x) =∞∑n=0

an(x− x0)n+r.

Using term-wise differentiation and substitute w(x) into (8) toobtain an equation of the form

A0(x− x0)r−2 + A1(x− x0)r−1 + · · · = 0.

SU/KSK MA-102 (2018)

The method of FrobeniusTo derive a series solution about the singular point x0 of

a2(x)y′′(x) + a1(x)y

′(x) + a0(x)y(x) = 0, x > x0. (8)

Set p(x) = a1(x)/a2(x), q(x) = a0(x)/a2(x).

If both (x− x0)p(x) and (x− x0)2q(x) are analytic at x0, thenx0 is a regular singular point and the following steps apply.

Step 1: Seek solution of the form

w(x) =∞∑n=0

an(x− x0)n+r.

Using term-wise differentiation and substitute w(x) into (8) toobtain an equation of the form

A0(x− x0)r−2 + A1(x− x0)r−1 + · · · = 0.

SU/KSK MA-102 (2018)

Step 2: Set A0 = A1 = A2 = · · · = 0. (A0 = 0 will give theindicial equation f(r) = r(r − 1) + p0r + q0 = 0.)

Step 3: Use the system of equations

A0 = 0, A1 = 0, . . . , Ak = 0

to find a recurrence relation involving ak and a0, a1, . . ., ak−1.

Step 4: Take r = r1, the larger root of the indicial equation(We are assuming here that the indicial equation has two realroots. The case when the indicial equation has two complexconjugate roots is complicated and we do not go into it.), anduse the relation obtained in Step 3 to determine a1, a2, . . .recursively in terms of a0 and r1.

Step 5: A series expansion of a solution to (8) is

w1(x) = (x− x0)r1∞∑n=0

an(x− x0)n, x > x0,

where a0 is arbitrary and an’s are defined in terms of a0 and r1.

SU/KSK MA-102 (2018)

A0 = 0, A1 = 0, . . . , Ak = 0

w1(x) = (x− x0)r1∞∑n=0

an(x− x0)n, x > x0,

SU/KSK MA-102 (2018)

A0 = 0, A1 = 0, . . . , Ak = 0

w1(x) = (x− x0)r1∞∑n=0

an(x− x0)n, x > x0,

SU/KSK MA-102 (2018)

A0 = 0, A1 = 0, . . . , Ak = 0

w1(x) = (x− x0)r1∞∑n=0

an(x− x0)n, x > x0,

where a0 is arbitrary and an’s are defined in terms of a0 and r1.SU/KSK MA-102 (2018)

Remark: The roots r1 and r2 of the indicial equation can beeither real or imaginary, we discuss only the case when theroots are real. Let r1 ≥ r2. For the larger root r1, we havealready got one series solution (as mentioned earlier).

Qn. How to find the second linear independent solution of thegive ODE?Observation: We obtain the constants an from the followingequations:

a0f(r) = 0,

a1f(r + 1) + a0(rp1 + q1) = 0,

a2f(r + 2) + a0(rp2 + q2) + a1[(r + 1)p1 + q1] = 0,

· · ·anf(r + n) + a0(rpn + qn) + · · ·+ an−1[(r + n− 1)p1 + q1] = 0.

Note: f(r1) = 0, f(r2) = 0.If r1 − r2 = N then f(r2 +N) = f(r1) = 0 cannot find aN ofsolution corresponds to r2.

SU/KSK MA-102 (2018)

Remark: The roots r1 and r2 of the indicial equation can beeither real or imaginary, we discuss only the case when theroots are real. Let r1 ≥ r2. For the larger root r1, we havealready got one series solution (as mentioned earlier).Qn. How to find the second linear independent solution of thegive ODE?

Observation: We obtain the constants an from the followingequations:

a0f(r) = 0,

a1f(r + 1) + a0(rp1 + q1) = 0,

a2f(r + 2) + a0(rp2 + q2) + a1[(r + 1)p1 + q1] = 0,

SU/KSK MA-102 (2018)

Remark: The roots r1 and r2 of the indicial equation can beeither real or imaginary, we discuss only the case when theroots are real. Let r1 ≥ r2. For the larger root r1, we havealready got one series solution (as mentioned earlier).Qn. How to find the second linear independent solution of thegive ODE?Observation: We obtain the constants an from the followingequations:

a0f(r) = 0,

a1f(r + 1) + a0(rp1 + q1) = 0,

a2f(r + 2) + a0(rp2 + q2) + a1[(r + 1)p1 + q1] = 0,

SU/KSK MA-102 (2018)

Remark: The roots r1 and r2 of the indicial equation can beeither real or imaginary, we discuss only the case when theroots are real. Let r1 ≥ r2. For the larger root r1, we havealready got one series solution (as mentioned earlier).Qn. How to find the second linear independent solution of thegive ODE?Observation: We obtain the constants an from the followingequations:

a0f(r) = 0,

a1f(r + 1) + a0(rp1 + q1) = 0,

a2f(r + 2) + a0(rp2 + q2) + a1[(r + 1)p1 + q1] = 0,

SU/KSK MA-102 (2018)

Theorem: Let x0 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− x0)p(x)and(x− x0)2q(x) and let r1 and r2 be the roots of the associatedindicial equation, where r1 ≥ r2.

Case a: If r1 − r2 is not an integer, then there exist twolinearly independent solutions of the form

y1(x) =∞∑n=0

an(x− x0)n+r1 ; a0 6= 0, x0 < x < x0 +R

y2(x) =∞∑n=0

bn(x− x0)n+r2 , b0 6= 0, x0 < x < x0 +R

SU/KSK MA-102 (2018)

Theorem: Let x0 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− x0)p(x)and(x− x0)2q(x) and let r1 and r2 be the roots of the associatedindicial equation, where r1 ≥ r2.

Case a: If r1 − r2 is not an integer, then there exist twolinearly independent solutions of the form

y1(x) =∞∑n=0

an(x− x0)n+r1 ; a0 6= 0, x0 < x < x0 +R

y2(x) =∞∑n=0

bn(x− x0)n+r2 , b0 6= 0, x0 < x < x0 +R

SU/KSK MA-102 (2018)

Case b: If r1 = r2, then there exist two linearly independent solutions ofthe form

y1(x) =

∞∑n=0

an(x− x0)n+r1 ; a0 6= 0, x0 < x < x0 +R

y2(x) = y1(x) ln(x− x0) +

∞∑n=1

bn(x− x0)n+r1 , x0 < x < x0 +R.

Case c: If r1 − r2 is a positive integer, then there exist two linearlyindependent solutions of the form

y1(x) =∞∑

an(x− x0)n+r1 ; a0 6= 0, x0 < x < x0 +R

y2(x) = Cy1(x) ln(x− x0) +

∞∑n=0

bn(x− x0)n+r2 , b0 6= 0, x0 < x < x0 +R

where C is a constant that could be zero.

*** End ***

SU/KSK MA-102 (2018)

Case b: If r1 = r2, then there exist two linearly independent solutions ofthe form

y1(x) =

∞∑n=0

an(x− x0)n+r1 ; a0 6= 0, x0 < x < x0 +R

y2(x) = y1(x) ln(x− x0) +

∞∑n=1

bn(x− x0)n+r1 , x0 < x < x0 +R.

Case c: If r1 − r2 is a positive integer, then there exist two linearlyindependent solutions of the form

y1(x) =

∞∑n=0

an(x− x0)n+r1 ; a0 6= 0, x0 < x < x0 +R

y2(x) = Cy1(x) ln(x− x0) +

∞∑n=0

bn(x− x0)n+r2 , b0 6= 0, x0 < x < x0 +R

where C is a constant that could be zero.

*** End ***

SU/KSK MA-102 (2018)

the method of frobenius - iit guwahati

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