I have an ODE
V''[z] + ( (I z0^2 w)/(3 (z - z0)^2) + (2 - (4 I z0 w)/9)/(z - z0)) V'[z] +
(-(12/41) I z0 w + (23 z0^2 w^2)/369)/(z - z0)^2 V[z] == 0
Mathematica can solve this exactly. But I just need the asymptotic behavior of the solution near z = z0. I have tried Series near z = z0, but the solution still looks messy. Is there any way to directly obtain the asymptotic behavior in Mathematica?
The general solution:
ex = DSolveValue[
V''[z] + ((I z0^2 w)/(3 (z - z0)^2) + (2 - (4 I z0 w)/9)/(z -
z0)) V'[
z] + (-(12/41) I z0 w + (23 z0^2 w^2)/369)/(z - z0)^2 V[z] ==
0, V[z], z] /. C[1] -> 1 /. C[2] -> 1 // Simplify
involves some Hypergeometric1F1 functions. (Note that we set the two free constants C[1]=C[2]=1 for simplicity here. Set them as you need for your boundary conditions later.)
Trying to expand ex around z=z0 causes some messy Floor functions to appear, because Mathematica is careful about the symbolic parameters potentially being complex. Instead of using Mathematica, we can just look up the expansion - see the last equation here:
$$\, _1F_1(a;b;z)\propto \frac{e^z \Gamma (b) \left(O\left(\left(\frac{1}{z}\right)^1\right)+1\right) z^{a-b}}{\Gamma (a)}+\frac{(-z)^{-a} \Gamma (b) \left(O\left(\left(\frac{1}{z}\right)^1\right)+1\right)}{\Gamma (b-a)}$$
Which is valid for $\left| z\right| \to \infty $. Dropping the $O\left(\left(\frac{1}{z}\right)^1\right)$ terms we get:
ex2 = ex /.Hypergeometric1F1[a_, b_, c_] ->Gamma[b]/Gamma[b - a] (-c)^-a+Gamma[b]/Gamma[a] Exp[c] c^(a - b);
This new expression ex2 now features only rational functions in z-z0 to some powers dependent on the parameters w,z0 and constant factors dependent of the same parameters.
One can also check for some particular values of parameters (i.e. w=Pi, z0=1) that the asymptotics works correctly. Real part of the ratio of full solution and approximation properly goes to $1$ around z=z0=1:
Plot[{1, Re[ex/ex2]} /. w -> \[Pi] /. z0 -> 1, {z, 0, 2}]
and imaginary part of the ratio properly vanishes there:
Plot[{1, Im[ex/ex2]} /. w -> \[Pi] /. z0 -> 1, {z, 0, 2}]
z=z0. What is this equation modeling? What are some usual boundary conditions that may be applied to resolve constants of integration? These details may help further. $\endgroup$z - z0I get a complicated looking expression. I want to have only the leading (and/or sub-leading) singular behavior. I just wanted to know if there is any way to extract that from the original ODE. $\endgroup$DSolveValueor some such. $\endgroup$