Lecture 21 Power Series Method at Singular Points Frobenius Theory

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1 Lecture 1 Power Series Method at Singular Points Frobenius Theory 10/8/011 Review. The usual power series method, that is setting y = a n 0 ) n, breaks down if 0 is a singular point. Here breaks down means cannot find all solutions. It s possible to completely solve one class of DE a y + by + cy = 0 1) Euler equations at singular points. The solutions involve r r not 0,1,, ) and ln, which are not analytic at 0 = 0 and therefore cannot be represented as a n 0 ) n. Remark 1. Note that α+iβ = α cos β ln + sinβ ln ). It turns out, once we include these two new ingredients into our ansatz, we can solve equations at singular points as long as those singular points are regular singular : 0 is regular singular point if 0 is singular; The Method of Frobenius. 0 ) p) and 0 ) q) are analytic at 0. The method of Frobenius is a modification to the power series method guided by the above observation. This method is effective at regular singular points. The basic idea is to look for solutions of the form Consider the equation Let 0 be a regular singular point. That is p) 0 ) = 0 ) r a n 0 ) n. ) y + p) y + q) y = 0. 3) p n 0 ) n, q) 0 ) = q n 0 ) n 4) with certain radii of convergence. To make the following discussion easier to read, we assume 0 =0. Substitute the epansion y = r a n n 5) into the equation we get ) r a n n + p) ) r a n n + q) r a n n = 0. 6) 1

2 Lecture 1 Power Series Method at Singular Points Frobenius Theory Now compute r a n n ) ) = a n n+r = n +r) n + r 1) a n n+r. 7) p) ) r a n n q) r ) = p) a n n+r ) = p) n +r) a n n+r 1 ) = p) ) n + r)a n n+r ) ) = p n n n +r)a n n+r = { n p n m m + r)a m } n+r. 8) a n n = r q n n ) a n n ) Now the equation becomes = n q n m a m ] n+r. 9) { n + r)n + r 1)a n + n m + r) p n m + q n m ] a m } n+r = 0. 10) Or equivalently { n + r)n + r 1)+n + r) p 0 + q 0 ] a n + m +r) p n m + q n m ] a m } n+r = 0. 11) This leads to the following equations: n 1): n +r) n + r 1)+n +r) p 0 + q 0 ] a n + n =0): r r 1) + p 0 r + q 0 ] a 0 = 0, 1) m + r) p n m + q n m ] a m = 0. 13) The n = 0 equation is singled out because if we require a 0 0 which is natural as when a 0 = 0, we have y = r+1 b m m where b m = a m+1.), then it becomes a condition on r: r r 1) + p 0 r + q 0 = 0. 14) This is called the indicial equation and will provide us with two roots r 1, r Some complicated situation may arise, we will discuss them later). These two roots are called eponents of the regular singular point =0. After deciding r, the n 1 relations provide us with a way to determine a n one by one. It turns out that there are three cases: r 1 r with r 1 r not an integer; r 1 = r ; r 1 r is an integer. Before we discuss these cases in a bit more detail, let s state the following theorem which summarizes the method of Frobenius in its full glory.

3 10/8/011 3 Theorem. Consider the equation y + p) y + q) y = 0 15) at an regular singular point 0. Let ρ be no bigger than the radius of convergence of either 0 ) p or 0 ) q. Let r 1, r solve the indicial equation Then r r 1) + p 0 r + q 0 = 0. 16) 1. If r 1 r and r 1 r is not an integer, then the two linearly independent solutions are given by y 1 )= 0 r1 a n 0 ) n, y )= 0 r ā n 0 ) n. 17) The coefficients a n and ā n should be determined through the recursive relation n + r)n + r 1)+n + r) p 0 + q 0 ] a n + m + r) p n m + q n m ] a m =0. 18). If r 1 =r, then y 1 is given by the same formula as above, and y is of the form y ) = y 1 ) ln r1 d n 0 ) n. 19) 3. If r 1 r is an integer, then take r 1 to be the larger root More precisely, when r 1,r are both comple, take r 1 to be the one with larger real part, that is Rer 1 ) Rer )). Then y 1 is still the same, while Note that c may be 0. y )=cy 1 ) ln r e n 0 ) n. 0) All the solutions constructed above converge at least for 0 < 0 < ρ Remember that 0 is a singular point, so we cannot epect convergence there). Remark 3. Note that, although ρ is given by radii of convergence of 0 ) p and 0 ) q, in practice, it is the same as the distance from 0 to the nearest singular point of p and q no 0 ) factor needed. Remark 4. The proof of this theorem is through careful estimate of the size of a n using the recurrence relation. See R. P. Agarwal & D. O Regan Ordinary and Partial Differential Equations: With Special Functions, Fourier Series, and Boundary Value Problems Lecture 5. Remark 5. In fact the converse of this theorem is also true. That is if all solutions of the equation satisfies lim 0 0 r y)=0 1) for some r, then 0 ) p and 0 ) q are analytic at 0. This is called Fuchs Theorem. Its proof is a tour de force of comple analysis and can be found in K. Yosida Lectures on Differential and Integral Equations, pp Eamples. In this class we will only require solving equations with r 1 r not an integer. Eample 6. Solve at 0 = 0. y + 1 ) y + 1 y =0 )

4 4 Lecture 1 Power Series Method at Singular Points Frobenius Theory Solution. We first write it into the standard form y + 1/) y + 1 y =0. 3) Thus p)= 1/ and q)= 1. It is clear that p) and q) are analytic so 0 is a regular singular point, and the method of Frobenius applies. Now we write y = a n n+r. 4) Substitute into the equation, we have ) ) a n n+r + 1/ a n n+r + 1 a n n+r = 0. 5) As p and q are particularly simple, we write the equation as ) ) ) a n n+r + a n n+r 1 a n n+r + 1 a n n+r = 0. 6) Carrying out the differentiation, we reach n + r) n + r 1) a n n+r + n + r) a n n+r 1 1 n + r) a n n+r + 0. Shifting inde: Now the equation becomes r r 1) r + 1 ] a 0 r + { an n+r = n + r)a n n+r 1 = n + r 1)a n+r. 8) n + r) n + r 1) 1 n + r) + 1 7) ]a n + n + r 1) a } n+r = 0. The indicial equation is 9) r r 1) r + 1 = 0 r 1 =1, r = 1. 30) Their difference is not an integer. To find y 1 we set r = r 1 =1. The recurrence relation n + r)n +r 1) 1 n +r) + 1 ] a n + n + r 1)a = 0 31) becomes nn +1) 1 n +1)+ 1 ] a n + n a =0 3) which simplifies to This gives Setting a 0 =1 we obtain a n = 1) n y 1 )= a n = n +1 a. 33) n n +1)n 1) 3 a 0. 34) 1) n n n+1) n 1) 3 n. 35)

5 10/8/011 5 so To find y we set r = r =1/. The recurrence relation becomes Finall the general solution is y) =C 1 a n = 1 n a a n = 1) n 1 n! a 0 36) y ) = 1/ 1) n 1) n 1 n! n = 1/ e. 37) n n+1) n 1) 3 n + C 1/ e. 38) Remark 7. Of course, for anyone who can remember the formulas, there is no need to do all these differentiation and inde-shifting. Remark 8. After getting r 1,, one can also write y 1, y eplicitly and solve a n by substituting them into the equation. See homework 8 solution. Remark 9. For detailed discussion of the why and how of the other two cases, see my Math note at They are not required for Math