SET 6

MATH 543: FROBENIUS METHOD

References: Hildebrand and Sadri Hassan.

The following theorem summarizes the Frobenius method: (Proved in Class please see your lecture notes and also DK)

Theorem 1. Suppose that the differential equation Lu = 0, where L is a second order differential operator, has a regular singular point at z = z_{0} with the roots r_{1} and r_{2} of the indical equation. There are three possible cases: (Assuming Re(r_{1}) > Re(r_{2}))

1. r_{1} − r_{2} ≠ an integer

2. r_{1} − r_{2} = N(a non-negative integer) and recursion relation is consistent 3. r_{1}−r_{2} = N(a non-negative integer) and recursion relation is not consistent.

Then, in the first two cases, there exists a bases of {u_{1}, u_{2}} of solutions of Lu = 0. These solutions are of the form

u_{1}(z) = (z − z_{0})^{r_{1}} ∑_{k=0}^{∞} C_{k}^{1} (z − z_{0})^{k},  (1)
u_{2}(z) = (z − z_{0})^{r_{2}} ∑_{k=0}^{∞} C_{k}^{2} (z − z_{0})^{k}  (2)

and the third case , the bases {u_{1}, u_{2}} is of the following form

u_{1}(z) = (z − z_{0})^{r_{1}} ∑_{k=0}^{∞} C_{k} (z − z_{0})^{k},  (3)
u_{2}(z) = Cu_{1}(z) ln(z − z_{0}) + (z − z_{0})^{r_{2}} ∑_{k=1}^{∞} B_{k} (z − z_{0})^{k}  (4)

where the power series about (z − z_{0}) are convergent in a neighborhhod of z = z_{0} A method when r_{1} − r_{2} = N Let z = z_{0} be a regular singular point of differential equation

Lu = u′′ + p(z) u′ + q(z)u = 0,  (5)

where

A(z) = (z − z_{0}) p(z), B(z) = (z − z_{0})^{2} q(z)  (6)

are analytic in a neighborhood D of z = z_{0} hence A(z) and B(z) are analytic in D then

A(z) = ∑_{k=0}^{∞} a_{k} (z − z_{0})^{k}, B(z) = ∑_{k=0}^{∞} b_{k} (z − z_{0})^{k},  (7)

Using the anzats

u(z) = (z − z_{0})^{r} ∑_{k=0}^{∞} C_{k} (z − z_{0})^{k},  (8)

in the DE we get

λ_{0}(n + r) C_{n} + ∑_{m=0}^{n−1} [(m + r)a_{n−m} + b_{n−m}] C_{m},  (9)

This is the recursion relations mentioned in the above theorem 1. Here

λ_{0}(r) = r(r − 1) + r a_{0} + b_{0}  (10)

If the indical equation is not imposed λ_{0}(r) ≠ 0 and the remaining

C_{1}(r), C_{2}(r), ⋯, C_{n}(r)  (11)

are determined through the recursion relation. Since n = 0 term is not set to zero then the differential equation is not satisfied but becomes

Lu = C_{0} λ_{0}(r) (z − z_{0})^{r−2}  (12)

where C_{0} is the undetermined constant. The solution of this equation is

u(r, z) = (z − z_{0})^{r} ∑_{k=0}^{∞} C_{k}(r) (z − z_{0})^{k},  (13)

In the case 3 , when the difference of the indices is a nonegative number and recursion realtions are inconsistent the the usual anzats fails. To find the correct anzats and a method of solution we use (12) and (13).

r_{1} = r_{2} case Eq.(12) becomes

Lu = C_{0} (r − r_{1})^{2} (z − z_{0})^{r−2},  (14)

Taking the derivative of both sides wrt r and letting r = r_{1} we obtain the second solution as

u_{2}(z) = \frac{du(r, z)}{dr}|_{r=r_{1}},  (15)

Using the form of u(r, z) given in (13) we obtain

u_{2}(z) = u_{1}(z) ln(z − z_{0}) + (z − z_{0})^{r_{1}} ∑_{k=0}^{∞} C′_{k}(r_{1}) (z − z_{0})^{k}  (16)

where C′_{k}(r_{1}) = \frac{d}{dr} C_{k}(r) |_{r=r_{1}}

r_{1} − r_{2} = N a positive integer case Eq.(12) becomes

Lu = C_{0}(r − r_{1})(r − r_{2}) (z − z_{0})^{r−2}  (17)

Multiplying both sides by r − r_{2} and taking derivatives of both sides wrt r and taking limit as r goes to r_{2} we obtain

u_{2}(z) = \frac{d}{dr} [(r − r_{2}) u(r, z)] at r = r_{2}  (18)

Using the expression (13) for u(r, z) we obtain

u_{2}(z) = Cu_{1}(z) ln(z − z_{0}) + (z − z_{0})^{r_{2}} ∑_{k=0}^{∞} C̃_{k} (z − z_{0})^{k}  (19)

where

C̃_{k} = \frac{d}{dr}[(r − r_{2}) C_{k}(r)] |_{r=r_{2}}

Problems

1. An example for the case r_{1} − r_{2} = N with consistent recursion realtions. Solve z^{2}u′′ + (z^{2} + z)u′ − u = 0 about z = 0 (solved in Class)

2. An example for the case r_{1} = r_{2} with inconsistent recursion relarions. Solve z^{2}u′′ + zu′ + z^{2}u = 0 about z = 0. You can find the solution u_{1} by using the recursion relations. This solution is known as the Bessel function of order 0,

u_{1}(z) = J_{0}(z) = ∑_{k=0}^{∞} \frac{(−1)^{k}(z/2)^{2k}}{(k!)^{2}}

The second solution u_{2}(z) is found by using (16). Prove that it takes the form

u_{2}(z) = J_{0}(z) ln z + ∑_{k=0}^{∞} φ(k)\frac{(z/2)^{2k}}{(k!)^{2}}

where

φ(k) = \{ \begin{matrix}∑_{m=1}^{k} \frac{1}{m}, & for k = 1, 2, ⋯ \\ 0, & for (k = 0) \end{matrix}  (20)

3. An example for the case r_{1} −r_{2} = N with inconsistent recursion relations. Solve z u′′ − u = 0 about z = 0. Again u_{1}(z) will be found by the standard method

u_{1}(z) = ∑_{k=0}^{∞} \frac{z^{k+1}}{k!(k + 1)!}

And the second solution is found by using the formula (19)

u_{2}(z) = u_{1} ln z + 1 − ∑_{k=1}^{∞} \frac{φ(k) + φ(k − 1)}{k!(k − 1)!} z^{k}

4. Find the solutions of the Bessel’s equation z^{2}u′′ + zu′ + (z^{2} − p^{2})u = 0. Solutions of this equation are known as the Bessel’s functions of order p. Solve this equation for p = not integer and for p = integer.

5. Show that for the differential equation z u′′ + 3u′ + 4zu = 0 the condition u(0) = 1 determines a unique solution, and that u′(0) cannot also be prescribed. Determine this solution.

6. Use the method of Frobenius to obtain the general solution of each of the following differential equations, valid near z = 0:

(1) z^{2}u′′ − 2zu′ + (2 − z^{2})u = 0

(2) (z − 1)u′′ − zu′ + u = 0

(3) 2zu′′ + (1 − 2z)u′ − u = 0

(4) z^{2}u′′ + zu′ + (z^{2} − \frac{1}{4})u = 0

(5) zu′′ − u′ + 4z^{3}u = 0

(6) zu′′ + 2u′ + zu = 0

(7) z(1 − z)u′′ − 2u′ − 2 + 2u = 0

7. Determine the two values of the constant α for which it is true that all solutions of the equation zu′′ + (z − 1)u′ − αu = 0 are regular at z = 0 , and obtain the general solution in each of these cases

8. (a) Show that the equation zu′′ + u′ − u = 0 possesses equal indices r_{1} = r_{2} = 0 at z = 0

(b) Obtain the regular solution u_{1}(z)

(c) Assume the second solution of the form u_{2}(z) = Cu_{1}(z) ln z + v(z) where C ≠ 0. Find v(z).

9. (a) Show that the equation zu′′ − zu′ − u = 0 possesses indices r_{1} = 1, r_{2} = 0 at z = 0.

(b) Obtain the regular solution u_{1}(z)

(c) Assume the general form for the second solution as u_{2}(z) = Cu_{1} ln z+v(z), where C ≠ 0. Find v(z).