first attempt
(apparently this isn't what the OP wanted)
I have edited the equation a bit because D
is a reserved symbol and because subscripts are inconvenient:
eq[l_,m_] = d λ (KroneckerDelta[-1, m] KroneckerDelta[1, l] - KroneckerDelta[1, l] KroneckerDelta[1, m]) + a V[l, m] +
m R V[l, m] +
1/2 b (-Sqrt[(((-1 + l + m) (l + m))/((-1 + 2 l) (1 + 2 l)))] V[l-1, m-1] +
Sqrt[((-1 + l - m) (l - m))/((-1 + 2 l) (1 + 2 l))] V[l-1, m+1] +
Sqrt[((1 + l - m) (2 + l - m))/((1 + 2 l) (3 + 2 l))] V[l+1, m-1] -
Sqrt[((1 + l + m) (2 + l + m))/((1 + 2 l) (3 + 2 l))] V[l+1, m+1]);
For every value of l
, we have 2l+1
equations that can be used to determine {V[l+1,-l-1],...,V[l+1,l+1]}
. There are 2l+3
unknowns, so we cannot determine all coefficients. Here I've chosen to solve for the {V[l+1,-l-1],...,V[l+1,l-1]}
and leave V[l+1,l]
and V[l+1,l+1]
as free parameters:
S[l_] := First@Solve[Table[eq[l, m] == 0, {m, -l, l}],
Table[V[l+1, m], {m, -l-1, l-1}]]
Try it out:
S[2]
(* {V[3, -3] -> ...,
V[3, -2] -> ...,
V[3, -1] -> ...,
V[3, 0] -> ...,
V[3, 1] -> ...} *)
We can stack these solutions in order to express the coefficients at a given value of l
in terms of all smaller l
-values:
Expand[S[2] /. S[1] /. S[0]]
(* complicated expression for V[3,-3]...V[3,1] in terms of
V[0, 0], V[1, 0], V[1, 1], V[2, 1], V[2, 2], V[3, 2], V[3, 3]} *)
programmatically:
Clear[F];
F[l_] := Module[{f = S[l]},
Do[f = f /. S[L], {L, l - 1, 0, -1}];
Expand[f]]
Now calling F[2]
gives the same result as the above Expand[S[2] /. S[1] /. S[0]]
.
second attempt
If you want to express $V_{0,0}$ in terms of $\{V_{L,-L},V_{L,-L+1},\ldots,V_{L,L}\}$ for a specific $L$, you can solve all the equations for $l<L$ for all the variables with $l<L$. For example, with $L=4$ we express $V_{0,0}$ in terms of $\{V_{4,-4},V_{4,-3},V_{4,-2},V_{4,-1},V_{4,0},V_{4,1},V_{4,2},V_{4,3},V_{4,4}\}$:
With[{L = 4},
FullSimplify[V[0, 0] /.
First@Solve[
Flatten[Table[eq[l, m] == 0, {l, 0, L - 1}, {m, -l, l}]],
Flatten[Table[V[l, m], {l, 0, L - 1}, {m, -l, l}]]]]]
-((b (-10290 Sqrt[7] a^7 d λ +
42 Sqrt[15]
a^2 b^3 R (35 Sqrt[7] R^2 (-V[4, -2] + V[4, 2]) +
2 b^2 (7 V[4, -4] - Sqrt[7] V[4, -2] + Sqrt[7] V[4, 2] -
7 V[4, 4])) -
294 Sqrt[15]
a^4 b^3 R (7 V[4, -4] - Sqrt[7] V[4, -2] + Sqrt[7] V[4, 2] -
7 V[4, 4]) +
6 Sqrt[15]
b^3 R (21 Sqrt[7] b^2 R^2 (V[4, -2] - V[4, 2]) +
49 R^4 (7 V[4, -4] + 4 Sqrt[7] V[4, -2] -
4 Sqrt[7] V[4, 2] - 7 V[4, 4]) +
b^4 (-7 V[4, -4] + Sqrt[7] V[4, -2] - Sqrt[7] V[4, 2] +
7 V[4, 4])) +
49 a^5 (174 Sqrt[7] b^2 d λ +
2940 Sqrt[7] d R^2 λ +
b^3 (-7 Sqrt[15] V[4, -4] + 2 Sqrt[105] V[4, -2] -
3 Sqrt[42] V[4, 0] + 2 Sqrt[105] V[4, 2] -
7 Sqrt[15] V[4, 4])) -
14 a^3 (153 Sqrt[7] b^4 d λ +
1680 Sqrt[7] b^2 d R^2 λ +
36015 Sqrt[7] d R^4 λ +
35 Sqrt[3]
b^3 R^2 (7 Sqrt[5] V[4, -4] + Sqrt[35] V[4, -2] -
3 Sqrt[14] V[4, 0] + Sqrt[35] V[4, 2] +
7 Sqrt[5] V[4, 4]) -
2 b^5 (7 Sqrt[15] V[4, -4] - 2 Sqrt[105] V[4, -2] +
3 Sqrt[42] V[4, 0] - 2 Sqrt[105] V[4, 2] +
7 Sqrt[15] V[4, 4])) +
a (162 Sqrt[7] b^6 d λ +
252 Sqrt[7] b^4 d R^2 λ +
50274 Sqrt[7] b^2 d R^4 λ +
370440 Sqrt[7] d R^6 λ +
14 Sqrt[3]
b^5 R^2 (28 Sqrt[5] V[4, -4] + Sqrt[35] V[4, -2] -
6 Sqrt[14] V[4, 0] + Sqrt[35] V[4, 2] +
28 Sqrt[5] V[4, 4]) +
49 Sqrt[3]
b^3 R^4 (77 Sqrt[5] V[4, -4] + 8 Sqrt[35] V[4, -2] -
27 Sqrt[14] V[4, 0] + 8 Sqrt[35] V[4, 2] +
77 Sqrt[5] V[4, 4]) -
3 b^7 (7 Sqrt[15] V[4, -4] - 2 Sqrt[105] V[4, -2] +
3 Sqrt[42] V[4, 0] - 2 Sqrt[105] V[4, 2] +
7 Sqrt[15] V[4, 4]))))/(3 Sqrt[
42] (1715 a^9 - 1225 a^7 (2 b^2 + 21 R^2) +
7 a^5 (156 b^4 + 2380 b^2 R^2 + 15435 R^4) -
a^3 (174 b^6 + 3045 b^4 R^2 + 31850 b^2 R^4 + 145775 R^6) +
3 a (3 b^8 + 58 b^6 R^2 + 651 b^4 R^4 + 5880 b^2 R^6 +
20580 R^8))))
You can do this for larger values of $L$, but it gets more difficult.
eqs
defined as the list of all equations explicitly given above, the number of distinct coefficientsV
isLength@Union@Cases[eqs, {_, _}, Infinity]
, which evaluates to18
, while the number of equations isLength@eqs
, which evaluates to10
. Hence, the coefficientsV
in these equations are underdetermined. Of course,V
in a larger set of equations might not be underdetermined, but it is impossible to say without knowing the general form of the equations. By the way use==
instead of=
in the equations to avoid syntax errors. $\endgroup$