Modular Forms course lecture notes 12

Modular Forms course lecture notes 12 - Exceptional regular...

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( May 14, 2011 ) Exceptional regular singular points of second-order ODEs Paul Garrett [email protected] http: / / e garrett/ 1. Solving second-order ODEs 2. Examples 3. Convergence Frobenius’ method for solving u 00 + b ( x ) x u 0 + c ( x ) x 2 u = 0 (with b, c analytic near 0) is slightly more complicated when the indicial equation α ( α - 1) + b (0) α + c (0) = 0 has repeated roots or roots differing by an integer . An important example in which this occurs is the differential equation in radial coordinates for spherical functions on hyperbolic n -space: u 00 + ( n - 1) coth r · u 0 - λ u = 0 The point 0 is a regular singular point for this equation. The indical equation is α ( α - 1) + ( n - 1) α = 0 When n = 2, the indicial equation has a double root 0. When n 3, the roots are 0 and 2 - n , which differ by an integer. 1. Solving second-order ODEs [1.1] Frobenius’ method Let E = d 2 dx 2 + b ( x ) x d dx + c ( x ) x 2 (with b, c [[ x ]]) be the differential operator and let p ( α ) = α ( α - 1) + b (0) α + c (0) [ α ] Consider f ( x, α ) = x α · X n 0 a n x n (on x > 0, with a 0 6 = 0) with a n depending on α . We can solve Ef ( x, α ) = p ( α ) · a 0 · x α by recursively solving for the coefficients a n , as follows. Expand For example, see [Coddington 1961], chapter 4, section 6. 1

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Paul Garrett: Exceptional regular singular points of second-order ODEs (May 14, 2011) Ef ( x, α ) = p ( α ) · a 0 · x α + h p ( α + 1) · a 1 + (something involving a 0 ) i · x α +1 + h p ( α + 2) · a 2 + (something involving a 0 , a 1 ) i · x α +2 + . . . That is, the coefficient of x α + n - 2 in Ef is h ( α + n )( α + n - 1) + b (0)( α + n ) + c (0) i · a n + (linear combination of a ’s with ‘ < n ) The coefficient of x α is p ( α ) · a 0 . For n 1, as long as the coefficient p ( α + n ) of a n is non-zero, the condition that the coefficient of x α + n vanish determines a n in terms of a with ‘ < n . Thus, when the two roots of p ( α ) = 0 do not differ by an integer, the differential equation has a solution of the form x α n 0 a n x n with either root α , since the recursion for the coefficients succeeds. It is not difficult to prove that the series has positive radius of convergence: we prove this later. [1.1.1] Remark: In any case, a first solution can be obtained by taking α to be the solution of the indicial equation with larger real part, thus avoiding failure of the recursion. The problem is to obtain a second solution when the roots of the indicial equation are equal or differ by an integer. [1.2] Simpler exceptional case The simpler exceptional case is that p ( α ) = ( α - α o ) 2 . Then the recursion for the coefficients a n succeeds, but produces just one solution f ( x, α o ). To obtain a second solution, apply ∂/∂α to Ef ( x, α ) = p ( α ) · a 0 · x α and evaluate at α = α o : E ∂f ∂α ( x, α o ) = p 0 ( α o ) a 0 x α o + p ( α o ) a 0 x α o log x = 0 Given the expansion f ( x, α ) = x α n a n x n , the expansion of the second solution is ∂f ∂α ( x, α o ) = x α o X n ∂a n ∂α x n + log x · x α o X n a n x n = x α o X n ∂a n ∂α x n + log x · f ( x, α o ) Since a 0 does not depend on α , the differentiation in α annihilates this term. Thus, the leading term of the
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