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Consider an ODE eigensystem

$$ \begin{bmatrix} 0 & d_1-\mathrm id_2 \\ d_1+\mathrm id_2 & 0 \end{bmatrix} \begin{bmatrix} a(y) \\ b(y) \end{bmatrix} = \lambda \begin{bmatrix} a(y) \\ b(y) \end{bmatrix}, $$

where $$d_1=-\mathrm{i}(p+qy)\partial_y+ry+s$$ $$d_2=-\mathrm{i}(u+vy)\partial_y+wy+t,$$

$p,\,q,\,r\,,s\,,u\,,v\,,w,\,t$ are just real constants, and $\mathrm i$ is the imaginary unit. Is it solvable by Mathematica?

I tried the following to reduce it to a 2nd-order ODE of $b$ with coefficients quadratic in $y$. But DSolve only gives a useless DifferentialRoot form after a long wait.

variables = {a[y], b[y]};
Fop1[F_, pm_] := 
  (r y + s) F - I (p D[F, y] + q (y D[F, y])) + 
   pm (-I) ((w y + t) F - I (u D[F, y] + v (y D[F, y])));
lhs = {Fop1[b[y], 1], Fop1[a[y], -1]};
eqe = 
  FullSimplify[
    Eliminate[
      Thread[
        Flatten[{D[(lhs - λ variables) // First, y], 
        lhs - λ variables}] == 0], 
      {a[y], a'[y]}]]
DSolve[eqe, b[y], y]

However, when $u,\,v=0$ or $p,\,q=0$, it becomes solvable, although the coefficients are still quadratic polynomials of $y$. Therefore, I was wondering if the more general case could be tackled as well. But I don't know how to proceed.

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  • 1
    $\begingroup$ Looks very unlikely to get a general solution to the underlying ODE with quadratic coefficients. If you play around with general equations of the form it requires several coefficients to be zero to get a special function representation to be found. $\endgroup$ – KraZug Jan 21 at 10:47
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Think that

$$ (d_1-i d_2)b(y) = \lambda a(y)\\ (d_1+i d_2)a(y) = \lambda b(y) $$

can be handled as

$$ (d_1+i d_2)(d_1-i d_2)b(y) = \lambda (d_1+i d_2) a(y) = \lambda^2 b(y) $$

then

$$ (d_1^2+i(d_2d_1-d_1d_2)+d_2^2)b(y) = \lambda^2 b(y) $$

In this case $d_1d_2 \ne d_2d_1$ and

$$ d_1^2 +i(d_2d1-d_1d_2)+ d_2^2 = p_2(y)\partial_x^2+p_1(y)\partial_x+p_0(y) $$

with

$$ \left\{ \begin{array}{rcl} p_2(y) & = &-(p+q y)^2-(u+v y)^2\\ p_1(y) & = &-p (q+i (2 r y+2 s-v))-q^2 y-i q (2 y (r y+s)+u)-(u+v y) (2 i t+v+2 i w y)\\ p_0(y) & = & -i r (p+q y)-w (p+q y)+(r y+s)^2+r (u+v y)+(t+w y)^2-i w (u+v y) \end{array} \right. $$

and finally

$$ p_2(y)b''(y)+p_1(y)b'(y)+(p_0(y)-\lambda^2)b(y) = 0 $$

This DE can be solved with a series expansion solution, proposing

$$ b(y) = \sum_{k=0}^n\alpha_k y^k $$

Clear[p, q, u, v, r, s, w, t]
d1[f_, y_] := -I (p + q y) D[f, y] + (r y + s) f
d2[f_, y_] := -I (u + v y) D[f, y] + (w y + t) f
d11 = d1[d1[b[y], y], y]
d22 = d2[d2[b[y], y], y]
d12 = d2[d1[b[y], y], y] - d1[d2[b[y], y], y]
DE = d1[d1[b[y], y], y] + I d1d2 + d2[d2[b[y], y], y] - lambda^2 b[y]
c0 = D[DE, b[y]]
c1 = D[DE, b'[y]]
c2 = D[DE, b''[y]]
Operator = c2 D[#, {y, 2}] + c1 D[#, y] + c0 # &;
n = 4;
Sumb = Sum[Subscript[alpha, k] y^k, {k, 0, n}];
res = Operator[Sumb] /. {Subscript[alpha, 0] -> Subscript[b, 0], 
Subscript[alpha, 1] -> Subscript[b, 1]};
coefs = CoefficientList[res, y];
equs = Thread[coefs == 0];  

For[k = 1; Alphas = {}; equsk = equs[[1]]; subs = {}, 
    k <= Length[equs] - 1, k++, 
    solalphak = Solve[equsk, Subscript[alpha, k + 1]];
    AppendTo[Alphas, Subscript[alpha, k + 1] /. solalphak][[1]];
    AppendTo[subs, solalphak];
    equsk = equs[[k + 1]] /. Flatten[subs]
]
Alphas

So we can extract the $\alpha_k$ in Alphas

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  • $\begingroup$ Thanks. But it's wrong from the 4th line as you can easily check using my code. $\endgroup$ – xiaohuamao Jan 21 at 17:16
  • $\begingroup$ @xiaohuamao Yes. A forgotten # in the Operator definition. Now it is fixed. Thanks. $\endgroup$ – Cesareo Jan 21 at 18:12
  • $\begingroup$ I mean your Latex derivation is wrong from the 4th line. $\endgroup$ – xiaohuamao Jan 21 at 18:22
  • $\begingroup$ @xiaohuamao Yes. A wrong defined operator. Now I hope it is fixed. Thanks. $\endgroup$ – Cesareo Jan 21 at 19:31
  • $\begingroup$ The expansion is an undetermined Laurent series with unknown negative powers. Not as simple as you've assumed. And the 1st-order operators are non-Abelian. And I still find sign errors in your 2nd-order operator. $\endgroup$ – xiaohuamao Jan 21 at 19:48

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