Linear differential equation Expansion at irregular singular point

Transcription

1 Linear differential equation Epansion at irregular singular point Epansion around 0 for y'' 1 y 0. Eample in Bender & Orszag Section.4. Initial change of variable y e s If we replace y by e s : y[_]:=ep[s[]] the general second order equation in y[] (only LHS shown; RHS = 0) gets transformed into one in s[] eq1=y''[]+p[] y'[]+q[] y[] s q s p s s s s s Multiplying through by e s cleans this up a bit : eq=epand[eq1*ep[-s[]]] q p s s s In our particular case, p[] and q[] are defined as given below. The ODE for s[] then simplifies: p[_]:=0 ; q[_]:=-^(-) ; eq=eq 1 s s First approimation to s[] At this stage, the s''[]-term is usually asymptotically small compated to the other two terms. We assume this to be the case, and solve the simplified first order ODE this leads to DSolve 1 s 0, s, s C1, s C1 The second order equation we started with has two possible solutions. We choose to consider the one

2 0_ ODE epansion Irregular singular point.nb The second order equation we started with has two possible solutions. We choose to consider the one that grows as 0, i.e. choose s_ : To check that our assumption about s''[] small compared to S'[]^ and ^(-), we compare these s'', s'^, ^, 1 5, 1 As 0, the assumption was indeed OK. Refinement of approimation to s[] Let c[] be the first correction to s[], i.e. s_ : c and the equation for c[] (still, we work with the LHS only; RHS=0) becomes eq c c which simplifies to eq4=epand[%] c c c 5 We now know that c 1 s, implying c ', c'' 5. The last two terms in the equation are small compared to the first one. The dominant balance must occur between the first two terms. Hence we solve DSolve c 0, c, 5 c C1 Log 4 For net approimation to s[], we therefore try s_ : 4 Log d where d[] will have to satisfy

3 0_ ODE epansion Irregular singular point.nb Epand[eq] 16 d d d d This time, we know d Log[], implying d ' 1, d ''. Again, the dominant balance has to be between the first two terms, so DSolve d 16 d 0, d, 16 C1 At this point, the corrections go to zero as 0, i.e. we can no longer ignore the integration constant. We could continue in the same style, and get more corrections to the finction s[], but it is now often more practical to aim instead for a refinemt of the original function y[]. We have so far found that y[] Ep[ 4 Log+c[1] ] c[] 16 4 e (1 +...) 16 Net task is to generate some more terms in the last epansion shown. Refinement of approimation to y[] Substitution of y_ : into the the ODE for y[] gives e Simplify y'' p y' q y e 18 4 e 56 e or simpler eq5 Simplify e 18 4 e 56 e With e 1, e' 1, e'', the only term above that can reach the size of the first one is the third one : 18 4 e'. Hence, we solve DSolve e 0, e, e C1

4 4 0_ ODE epansion Irregular singular point.nb The epansion for y following the eponential factor is therefore It looks very 51 likely that it will continue as a series in 1. Calling this function w, it will have to cause y to satisfy y_ : 4 w i.e. the following epression needs to be zero y'' p y' q y w w w 14 4 w This can be cleaned up a bit eq7 Epand w 16 w w w We can now try to get the first n (say, n=6) coefficients through n 6 ; w_ 1 Sumak ^k, k, 1, n 1 a1 a a a4 5 a5 a6 Net, plug this into the equation for w[] Epandeq7 a a1 a 1 a a a 4 14 a a4 5 4 a a a a6 After getting rid of the leading factor (and throwing in an etra 1 so that we can put to zero in turn the coefficients for 1,, etc; Mathematica's routine Ccoefficient does not like to etract the coefficient for 0 eq6 Epand a1 5 a1 a 1 a 48 a 45 a 64 a a a5 117 a5 96 a a6 we get the successive coefficients through

5 0_ ODE epansion Irregular singular point.nb 5 Do ai ai. SolveCoefficient eq6, ^1, i 0, ai1; Printai, i, 1, n This epansion has apparently not encountered any problems. We can easily obtain any number of terms we wish. The algebra in this very last step (running out many coefficients) is trivial once we have the simple ODE that w[] satisfies. In Bender & Orszag, pages 85-87, it is shown how we can do the recusrions so that we get all the coefficients in closed form.