Method of Frobenius. General Considerations. L. Nielsen, Ph.D. Dierential Equations, Fall Department of Mathematics, Creighton University

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1 Method of Frobenius General Considerations L. Nielsen, Ph.D. Department of Mathematics, Creighton University Dierential Equations, Fall 2008

2 Outline 1 The Dierential Equation and Assumptions 2 3 Main Theorem How to Calculate Coecients in the Hard Cases

3 Outline The Dierential Equation and Assumptions 1 The Dierential Equation and Assumptions 2 3 Main Theorem How to Calculate Coecients in the Hard Cases

4 The DE The Dierential Equation and Assumptions Our DE is L[y] = x 2 d 2 y dx 2 + x 2 p(x) dy dx + x 2 q(x)y = 0. We will assume that the singular point x = 0 is a regular singular point. Since x = 0 is a R.S.P., we know that we can expand xp(x) and x 2 q(x) as convergent Taylor series about x = 0. We set y = a nx n+r, a 0 0.

5 The DE The Dierential Equation and Assumptions Our DE is L[y] = x 2 d 2 y dx 2 + x 2 p(x) dy dx + x 2 q(x)y = 0. We will assume that the singular point x = 0 is a regular singular point. Since x = 0 is a R.S.P., we know that we can expand xp(x) and x 2 q(x) as convergent Taylor series about x = 0. We set y = a nx n+r, a 0 0.

6 The DE The Dierential Equation and Assumptions Our DE is L[y] = x 2 d 2 y dx 2 + x 2 p(x) dy dx + x 2 q(x)y = 0. We will assume that the singular point x = 0 is a regular singular point. Since x = 0 is a R.S.P., we know that we can expand xp(x) and x 2 q(x) as convergent Taylor series about x = 0. We set y = a nx n+r, a 0 0.

7 The DE The Dierential Equation and Assumptions Our DE is L[y] = x 2 d 2 y dx 2 + x 2 p(x) dy dx + x 2 q(x)y = 0. We will assume that the singular point x = 0 is a regular singular point. Since x = 0 is a R.S.P., we know that we can expand xp(x) and x 2 q(x) as convergent Taylor series about x = 0. We set y = a nx n+r, a 0 0.

8 Outline 1 The Dierential Equation and Assumptions 2 3 Main Theorem How to Calculate Coecients in the Hard Cases

9 We compute the rst two derivatives of y = a nx n+r in the usual way. We then use the Taylor series for the coecients to obtain L[y] = [ ] + r)anx (n n+r (n + r)(n + r 1)a n x n+r + + =[r(r 1) + p 0 r + q 0 ]a 0 x r + ( m=0 pmx m ) ( )( ) q m x m a n x n+r m=0 {[(1 + r)r + p 0 (1 + r) + q 0 ]a 1 + (rp 1 + q 1 )a 0 }x r {[(n + r)(n + r 1) + p 0 (n + r) + q 0 ]a n + n 1 k=0 [(k + r)p n k + q n k ]a k } x n+r +

10 Simplication The expression for L[y] on the previous slide can be simplied by letting F (r) = r(r 1) + p 0 r + q 0. Then L[y] = a 0 F (r)tx r + [a 1 F (1 + r) + (rp 1 + q 1 )a 0 ] x 1+r + { n 1 } +a n F (n + r)x n+r + [(k + r)p n k + q n k ]a k x n+r k=0 + We now set the coecient of each power of x equal to zero. The results are...

11 Outline 1 The Dierential Equation and Assumptions 2 3 Main Theorem How to Calculate Coecients in the Hard Cases

12 Coecients We have, rst of all, the indicial equation. Also, we have n 1 F (n + r)a n = F (r) = r(r 1) + p 0 r + q 0 = 0, k=0 [(k + r)p n k + q n k ]a k, n 1 Note that the indicial equation is a quadratic equation in r. It's roots determine the values r 1 and r 2 for which there may be solutions. The second equation shows that, in general, a n depends on r and all of the preceding coecients.

13 Coecients We have, rst of all, the indicial equation. Also, we have n 1 F (n + r)a n = F (r) = r(r 1) + p 0 r + q 0 = 0, k=0 [(k + r)p n k + q n k ]a k, n 1 Note that the indicial equation is a quadratic equation in r. It's roots determine the values r 1 and r 2 for which there may be solutions. The second equation shows that, in general, a n depends on r and all of the preceding coecients.

14 Coecients We have, rst of all, the indicial equation. Also, we have n 1 F (n + r)a n = F (r) = r(r 1) + p 0 r + q 0 = 0, k=0 [(k + r)p n k + q n k ]a k, n 1 Note that the indicial equation is a quadratic equation in r. It's roots determine the values r 1 and r 2 for which there may be solutions. The second equation shows that, in general, a n depends on r and all of the preceding coecients.

15 Outline 1 The Dierential Equation and Assumptions 2 3 Main Theorem How to Calculate Coecients in the Hard Cases

16 The Recurrence Relation We can solve the recurrence relation F (n + r)a n = n 1 k=0 [(k + r)p n k + q n k ]a k for a n provided that F (1 + r), F (2 + r),...,f (n + r) are not zero. If F (n + r) vanishes for some positive integer n, then n + r must be a root of the indicial equation. Hence, if the indicial equation has two real roots r 1, r 2 with r 1 > r 2 and r 1 r 2 not equal to an integer, then our DE has two solutions; one corresponding to each value of r obtained from the indicial equation. It can be shown that these solutions converge wherever the series for xp(x) and x 2 q(x) converge. What happens when the roots to the indicial equation dier by a positive integer or are equal?

17 The Recurrence Relation We can solve the recurrence relation F (n + r)a n = n 1 k=0 [(k + r)p n k + q n k ]a k for a n provided that F (1 + r), F (2 + r),...,f (n + r) are not zero. If F (n + r) vanishes for some positive integer n, then n + r must be a root of the indicial equation. Hence, if the indicial equation has two real roots r 1, r 2 with r 1 > r 2 and r 1 r 2 not equal to an integer, then our DE has two solutions; one corresponding to each value of r obtained from the indicial equation. It can be shown that these solutions converge wherever the series for xp(x) and x 2 q(x) converge. What happens when the roots to the indicial equation dier by a positive integer or are equal?

18 Equal Roots When the roots to the indicial equation are equal, we have only one solution of the form y 1 (x) = x r 1 a n (r 1 )x n. One can show that there is a second linearly independent solution of the form y 2 (x) = y 1 (x)ln x + x r 1 b n x n where the coecients b n need to be calculated - a sometimes dicult problem. We remark, however, that the second solution is often rejected on the grounds that there is a logarithmic singularity at x = 0 - many physical applications don't want such a singularity.

19 Roots Diering by a Positive Integer Here we have r 1 = r 2 + N for some positive integer N. In this case we have a solution y 1 (x) = x r 1 a n(r 1 )x n, but it may not be possible to find a second solution of this form when r = r 2. Why? It is because F (r 2 + N) = F (r 1 ) = 0 and the recurrence relation becomes n 1 0 a N = k=0 [(k + r)p n k + q n k]a k when n = N. This equation cannot be satised for any choice of a N, if the right side of this equation is nonzero. In this case, the second solution has the form y 2 (x) = y 1 (x)ln x + x r 2 b n x n

20 Roots Diering by a Positive Integer, II What if n 1 k=0 [(k + r)p n k + q n k ]a k = 0? Then it's clear that a N is arbitrary and we can obtain a second solution of the form y = x r 2 b n x n. This can be seen via the example on the next slide(s).

21 Example: Roots Diering by a Positive Integer Example Use the method of Frobenius to solve x 2 d 2 y dx 2 x dy dx (x 2 + 5/4)y = 0. Solution: It is clear that x = 0 is a R.S.P. (Check!) We let y = a nx n+r and we obtain the following, after computing the derivatives and adjusting the series: Indicial Equation: r 2 2r 5 = 4 0, so that r 1 = 5 2 and r 2 = 1 2. These roots dier by a positive integer. We also obtain the condition [ ] (r + 1)r (r + 1) 5 a 4 1 = 0 and the recurrence relation [ (n + r)(n + r 1) (n + r) 5 ] a n a n 2 = 0. 4 We will always be able to nd a solution corresponding to the larger root and we proceed to do this next.

22 Example, Continued Let r = 5 2. Then 4a 1 = 0 and so a 1 = 0. The recurrence relation is a n = a n 2 n(n + 3), n 2. We obtain, via induction, that all of the odd indexed coecients vanish and that a 2n = Our solution is therefore y 1 (x) = a 0 x 5/2 [1 + a 0 2 n n![5 7 9 (2n + 3)], n 1. n=1 x 2n 2 n n![5 7 9 (2n + 3)] ].

23 Example, Continued We now consider r = 1. Let us assume (hopefully, but 2 without justication!) that there is a second solution of the standard form. Letting r = 1 2 results in 2a 1 = 0 and a n = a n 2, n 2, n 3. n(n 3) Some thought shows that a 3 is arbitrary! Also, we have a 2 = a 0 /2, and a 2n+1 = a 2n = a 0 2 n n![3 5 7 (2n 3)], n 3 a 3 2 n 1 (n 1)![5 7 9 (2n + 1)], n 2.

24 Example, Continued Our solution in this case is therefore { } y 2 = a 0 x 1/2 1 x 2 2 x x 2n + n=3 2 n n![3 5 7 (2n 3)] } a 3 x {x 1/2 3 + n=2 x 2n+1 2 n 1 (n 1)![5 7 9 (2n + 1)] It turns out that we are free to take a 3 = 0 (remember, it's arbitrary) and so we are left with only the rst series for the second solution. However, it can be seen that we can nd two linearly independent solutions to the DE by using the smaller root alone. A good rule of thumb is to work out the solution corresponding to the smaller root rst, in the hope that this smaller root by itself may lead directly to the general solution. Keep in mind, though, that this won't always work.

25 A Second Example Example We consider the DE t 2 d 2 y dt 2 + t dy dt + (t2 1/4)y = 0. Proceeding using the method of Frobenius we nd the following: [r(r 1) + r 1 4 ]a 0 = (r )a 0 = 0 - here we get r 1 = 1/2 and r 2 = 1/2. [(1 + r) ]a 1 = 0 [(n + r) ]a n = a n 2, n 2

26 A Second Example Example We consider the DE t 2 d 2 y dt 2 + t dy dt + (t2 1/4)y = 0. Proceeding using the method of Frobenius we nd the following: [r(r 1) + r 1 4 ]a 0 = (r )a 0 = 0 - here we get r 1 = 1/2 and r 2 = 1/2. [(1 + r) ]a 1 = 0 [(n + r) ]a n = a n 2, n 2

27 A Second Example Example We consider the DE t 2 d 2 y dt 2 + t dy dt + (t2 1/4)y = 0. Proceeding using the method of Frobenius we nd the following: [r(r 1) + r 1 4 ]a 0 = (r )a 0 = 0 - here we get r 1 = 1/2 and r 2 = 1/2. [(1 + r) ]a 1 = 0 [(n + r) ]a n = a n 2, n 2

28 Second Example, Larger Root We set r = 1/2 to obtain a 1 = 0 and nd that the recurrence relation is a n = a n 2 n(n + 1), n 2. This relation along with a 1 = 0, tells us that all of the odd coecients vanish and that a 2n = ( 1) n (2n)!(2n + 1). (We've set a 0 = 1.) We obtain from this recurrence relation that y 1 = sin t t.

29 Second Example, Smaller Root Setting r = 1/2, we see that, since 1 + ( 1/2) = 1/2, we cannot nd any value for a 1 - it is arbitrary. Set a 1 = 0, then. This causes all of the odd coecients to be zero and the recurrence relation, gives us that Our second solution is therefore a n = a n 2 n(n 1), n 2, a 2n = ( 1)n (2n)!. y 2 = cos t t.

30 Outline Main Theorem How to Calculate Coecients in the Hard Cases 1 The Dierential Equation and Assumptions 2 3 Main Theorem How to Calculate Coecients in the Hard Cases

31 Frobenius Theorem Main Theorem How to Calculate Coecients in the Hard Cases Let r 1 and r 2 be roots of the indicial equation with r 1 r 2 if they are real. Then the dierential equation has two linearly independent solutions on the interval (0,ρ) (ρ is determined by the radius of convergence of the series for tp(t) and t 2 q(t)) of the following form: If r 1 r 2 is not a positive integer, then y 1 (t) = t r 1 If r 1 = r 2, then y 1 (t) = t r 1 a n (r 1 )t n, y 2 (t) = t r 2 a n (r 2 )t n. a n (r 1 )t n, y 2 (t) = y 1 (t)ln t + t r 1 b n t n.

32 Last Conclusion Main Theorem How to Calculate Coecients in the Hard Cases If r 1 = r 2 + N, a positive integer, then for some constant a (possibly zero), y 1 (t) = t r 1 a n (r 1 )t n, y 2 (t) = ay 1 (t)ln t + t r 2 b n t n.

33 Outline Main Theorem How to Calculate Coecients in the Hard Cases 1 The Dierential Equation and Assumptions 2 3 Main Theorem How to Calculate Coecients in the Hard Cases

34 Main Theorem How to Calculate Coecients in the Hard Cases Calculation of Coecients with Equal Roots We run into trouble if the indicial equation has equal roots r 1 = r 2 because our DE then only has one solution in the form y = a nt n+r. The method of nding a second solution goes as follows. Let us rewrite the series solution as follows: y(t) = y(t, r) = a n (r)t n+r (1) to emphasize that the solution y(t) depends on our choice of r. Then, using notation developed above in our general discussion of the method, L[y](t, r) = a 0 F (r)t r + { n=1 a n (r)f (n + r) + n 1 k=0 [(k + r)p n k + q n k ]a k } t n+r.

35 Main Theorem How to Calculate Coecients in the Hard Cases Calculation of Coecients with Equal Roots We now think of r as a continuous variable and determine a n as a function of r by requiring that the coecient of t n+r be zero for n 1. Thus a n (r) = n 1 k=0 [(k + r)p n k + q n k ]a k. (2) F (n + r) With this choice of a n (r), we see that L[y](t, r) = a 0 F (r)t n+r. (3) In the case of equal roots, F (r) = (r r 1 ) 2, so that equation (3) can be written in the form L[y](t, r) = a 0 (r r 1 ) 2 t r.

36 Main Theorem How to Calculate Coecients in the Hard Cases Calculation of Coecients with Equal Roots Since L[y](t, r 1 ) = 0, we obtain one solution ( y 1 (t) = t r 1 a 0 + n=1 a n (r 1 )t n ). Observe now, that L[y](t, r) =L r [ y r ] (t, r) = r a 0(r r 1 ) 2 t r (4) also vanishes when r = r 1. = 2a 0 (r r 1 )t r + a 0 (r r 1 ) 2 (ln t)t r

37 Main Theorem How to Calculate Coecients in the Hard Cases Calculation of Coecients with Equal Roots Thus the second solution is y 2 (t) = r y 1(t, r) r=r1 [ an (r)t n+r ] = r = { a n (r 1 )t n+r 1 = y 1 (t)ln t + } r=r 1 ln t + da n dr (r 1)t n+r 1 da n dr (r 1)t n+r 1 (5)

38 Main Theorem How to Calculate Coecients in the Hard Cases Calculation of Coecients when Roots Dier by an Integer Here we suppose that the zeros of the indicial equation, r 1 > r 2, are such that r 1 r 2 = N N. It may be the case that there are two solutions of the form However, there is always a solution of the form x r a n x n. (6) y = x r 1 a n x n. (7) A second solution of this form for the other zero of the indicial equation will exist if the recursion relation is welldened for all n when using the second (smaller) zero r 2. If the recursion relation is not welldened for the second root, there is a second solution of the form

39 Main Theorem How to Calculate Coecients in the Hard Cases Calculation of Coecients when Roots Dier by an Integer y 2 (x) = ay 1 (x)ln x + x r 2 d dr [(r r 2)a n (r)] r=r2 x n. (8) The constant a may or may not be zero but we won't pursue how to calculate its value here.

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