# How to find the particular solution in symbolic form of an overdetermined system of equations

I want to use mathematica to reproduce a conclusion in a paper, which is solve an overdetermined system of equations, $$\begin{array}{l} c^{2}(n-1)^{2}\left(a_{0}+a_{1} Y+a_{2} Y^{2}\right)^{3}+a(n-1)^{2}\left(a_{0}+a_{1} Y+a_{2} Y^{2}\right)^{2}+b n(2 n-1)\left(1-Y^{2}\right)^{2}\left(a_{1}+2 a_{2} Y\right)^{2} \\ -b n(n-1)\left(a_{0}+a_{1} Y+a_{2} Y^{2}\right)\left(1-Y^{2}\right)\left(-2 Y\left(a_{1}+2 a_{2} Y\right)+\left(1-Y^{2}\right) 2 a_{2}\right)=0 . \end{array}$$

Expanding the above expression into a polynomial in terms of $$Y$$ and equating the coefficients of all powers of $$Y$$ to zero yields an overdetermined system of equations.

The equations is quite lengthy, so I will list it in the end, but it involves natural number $$n$$ greater than 1 and real constants $$a,b,c$$. The variables to solve for are $$a_0, a_1,a_2$$.

The conclusion is,

When $$\frac{a}{b}=-\frac{4n^{2}}{\left(n-1\right)^{2}}$$, we can get two sets of solutions, \begin{aligned} &1)\quad a_0=-\frac{a\left(n+1\right)}{2c^2n},\quad a_1=0,\quad a_2=\frac{a\left(n+1\right)}{2c^2n},\quad c=c,\\ &2)\quad a_0=\frac{2bn\left(n+1\right)}{c^2\left(n-1\right)^2},\quad a_1=0,\quad a_2=-\frac{2bn\left(n+1\right)}{c^2\left(n-1\right)^2},\quad c=c, \end{aligned}

Here is my attempt,

(* This part of the code is for deriving the equation, so you can ignore it. *)
eqn = c^2*(n - 1)^2*S^3 + a*(n - 1)^2*S^2 +
b*n*(2*n - 1)*(1 - Y^2)^2*(DS)^2 -
b*n*(n - 1)*S*(1 - Y^2)*(-2*Y*DS + (1 - Y^2)*D2S);

S[Y_] := a0 + a1*Y + a2*Y^2;

DS = D[S[Y], Y];
D2S = D[DS, Y];

(* Here, we can obtain the formula mentioned above *)
seqn = eqn /. {S -> a0 + a1*Y + a2*Y^2, DS -> DS, D2S -> D2S}

(* get the overdetermined system of equations *)
coeList = CoefficientList[Expand[seqn], Y];
equList

(* in this way it produces two sets of trivial solutions *)
solutions = Solve[equList, {a0, a1, a2}];
solutions

In fact, I attempted to manually derive the condition $$\frac{a}{b}=-\frac{4n^{2}}{\left(n-1\right)^{2}}$$ but couldn't find it. I also tried solving the system by using it as a condition(already known) and use Reduce but the answer solves $$n$$ directly, which is what I don’t want.

solutions =
ToRules[Reduce[
Append[equList, a/b == -4*n^2/(n - 1)^2], {a0, a1, a2}]];
solutions

(* Here is the output *)
Sequence[{n -> 0, c -> 0, a -> 0},
{n -> 0, c -> 0, a -> 0, a1 -> 0},
{n -> 0, c -> 0, a -> 0, a2 -> 0},
{n -> 0, c -> 0, a -> 0, a1 -> 0, a2 -> 0},
{a -> -((4 b n^2)/(-1 + n)^2), a0 -> 0, a1 -> 0, a2 -> 0},
{a -> -((4 b n^2)/(-1 + n)^2), a0 -> (-3 a^2 + 13 a b + a^2 n - 23 a b n + 52 b^2 n^2 + 12 b^2 n^3)/(2 c^2 (a + 4 b n^2 + 3 b n^3)), a1 -> 0,
a2 -> (3 a - a n - 4 b n)/(2 c^2)},
{c -> 0, a -> -((4 b n^2)/(-1 + n)^2), a0 -> 0, a1 -> 0, a2 -> 0}, {n -> -1,
a -> -b, a0 -> 0, a1 -> 0, a2 -> 0}, {n -> -1, a -> -b, a0 -> b/c^2,
a1 -> 0, a2 -> 0}, {n -> -1, c -> 0, a -> -b, a0 -> 0, a1 -> 0,
a2 -> 0}, {n -> -1, c -> 0, a -> -b, a1 -> 0, a2 -> -a0}, {n -> 0,
a -> 0, a0 -> 0, a1 -> 0, a2 -> 0}, {n -> 5/3, a -> -25 b,
a0 -> (20 b)/c^2, a1 -> 0,
a2 -> -a0}, {a -> -((4 b n^2)/(-1 + n)^2), a0 -> 0, a1 -> 0,
a2 -> 0}, {a -> -((4 b n^2)/(-1 + n)^2), a0 -> -(a/c^2), a1 -> 0,
a2 -> 0}]

So my question is,

1. could the condition $$\frac{a}{b}=-\frac{4n^{2}}{\left(n-1\right)^{2}}$$ be found in using Mathematica?
2. How to solve this overdetermined system of equations without solving $$n$$ directly.

Any help would be greatly appreciated!

The "big" overdetermined system of equations,by the way, the author of the paper use Maple to solve this but did not give the code. \begin{aligned} Y^{0}: & 2 b n^{2} a_{1}^{2}-b n a_{1}^{2}+c^{2} a_{0}^{3}+a a_{0}^{2}-2 n c^{2} a_{0}^{3}-2 n a a_{0}^{2}+n^{2} c^{2} a_{0}^{3} \\ & +a n^{2} a_{0}^{2}-2 b n^{2} a_{0} a_{2}+2 b n a_{0} a_{2}=0, \\ & 2 b n^{2} a_{0} a_{1}-4 a n a_{0} a_{1}-2 b n a_{1} a_{2}+3 c^{2} a_{0}^{2} a_{1}-6 c^{2} a_{0}^{2} a_{1} n+6 b n^{2} a_{1} a_{2} \\ Y^{1}: & 3 c^{2} a_{0}^{2} a_{1} n^{2}-2 b n a_{1} a_{0}+2 a n^{2} a_{0} a_{1}+2 a a_{0} a_{1}=0, \\ & a a_{1}^{2}-6 c^{2} n a_{0}^{2} a_{2}+3 c^{2} n^{2} a_{0} a_{1}^{2}+3 c^{2} a_{0} a_{1}^{2}-2 b n a_{2}^{2}-8 b n a_{0} a_{2}+2 a a_{0} a_{2}-4 a n a_{0} a_{2} \\ Y^{2}: & -2 a n a_{1}^{2}+8 b n^{2} a_{0} a_{2}+a n^{2} a_{1}^{2}-2 b n^{2} a_{1}^{2}-6 c^{2} n a_{0} a_{1}^{2}+6 b n^{2} a_{2}^{2}+3 c^{2} n^{2} a_{0}^{2} a_{2}+3 c^{2} a_{0}^{2} a_{2} \\ & +2 a n^{2} a_{0} a_{2}=0,\\ Y^{3}: & 6 c^{2} n^{2} a_{0} a_{1} a_{2}+2 a a_{1} a_{2}+2 b n a_{0} a_{1}+6 c^{2} a_{0} a_{1} a_{2}-2 b n a_{1} a_{2}-2 c^{2} n a_{1}^{3}-2 b n^{2} a_{0} a_{1} \\ & -6 b n^{2} a_{1} a_{2}-4 a n a_{1} a_{2}+c^{2} a_{1}^{3}+2 a n^{2} a_{1} a_{2}-12 c^{2} n a_{0} a_{1} a_{2}+c^{2} n^{2} a_{1}^{3}=0, \\ Y^{4}: & a n^{2} a_{2}^{2}+3 c^{2} a_{1}^{2} a_{2}+3 c^{2} n^{2} a_{2}^{2} a_{0}-8 b n^{2} a_{2}^{2}+6 b n a_{0} a_{2}+b n a_{1}^{2}-2 a a_{2}^{2} n+3 c^{2} a_{2}^{2} a_{0} \\ & -6 c^{2} a_{1}^{2} a_{2} n+a a_{2}^{2}+3 c^{2} a_{1}^{2} n^{2} a_{2}-6 b n^{2} a_{0} a_{2}-6 c^{2} a_{2}^{2} a_{0} n=0, \\ Y^{5}: & 4 b n a_{1} a_{2}+3 c^{2} n^{2} a_{1} a_{2}^{2}-6 c^{2} n a_{1} a_{2}^{2}+3 c^{2} a_{1} a_{2}^{2}=0, \\ Y^{6}: & 2 b n^{2} a_{2}^{2}+2 b n a_{2}^{2}+c^{2} a_{2}^{3}+c^{2} a_{2}^{3} n^{2}-2 c^{2} a_{2}^{3} n=0 . \end{aligned}

• Your questions are unclear to me. Is a/b==-4n^2/(n-1)^2an additional condition or an expected result? What means "...without solving n directly"? Commented Jan 11 at 9:38
• @UlrichNeumann a/b==-4n^2/(n-1)^2 is an additional condition to get the result; the solutions in the second code block give the exact value of n which is not expected, the result of the paper gives $a_0=\frac{2bn\left(n+1\right)}{c^2\left(n-1\right)^2}$. Commented Jan 11 at 10:19
• It looks like this additional condition isn't necessary, see my answer. Commented Jan 11 at 10:26

Try

eqn = c^2*(n - 1)^2*S[Y]^3 + a*(n - 1)^2*S[Y]^2 +
b*n*(2*n - 1)*(1 - Y^2)^2*(S'[Y])^2 -
b*n*(n - 1)*S[Y]*(1 - Y^2)*(-2*Y*S'[Y] + (1 - Y^2)*S''[Y]) /.
S -> Function[Y, a0 + a1*Y + a2*Y^2]

cond=Reduce[0 == CoefficientList[eqn, Y] , {a0, a1, a2 }] //Simplify[#, {Element[n, Integers], n > 1}] &
(*(a == 0 && c == 0 && ((a1 == 0 && a2 == 0) || b == 0)) ||
(a1 ==0 && ((c != 0 && ((a == -((4 b n^2)/(-1 + n)^2) &&
a0 == -(((1 + n) (a (-1 + n)^2 - 8 b n^2))/(6 c^2 (-1 + n)^2 n)) &&
a2 == (a0 (a + a0 c^2) (-1 + n))/(2 b n) && a0 != 0 &&b != 0)
|| (a0 == -(a/c^2) && a2 == 0))) || (a0 == 0 &&a2 == 0)))*)

Perhaps ToRules helps to create a more easily readable result

ToRules[cond]
(*Sequence[
{a -> 0, c -> 0, a1 -> 0, a2 -> 0},
{a -> 0, c -> 0,b -> 0},
{a1 -> 0, a -> -((4 b n^2)/(-1 + n)^2),a0 -> -(((1 + n) (a (-1 +n)^2 - 8 b n^2))/(6 c^2 (-1 + n)^2 n)), a2 -> (a0 (a + a0 c^2) (-1 + n))/(2 bn)},
{a1 -> 0, a0 -> -(a/c^2),a2 -> 0},
{a1 -> 0, a0 -> 0, a2 -> 0}]*)
• Thank your advice! I am new to Mathematica,in addition,I don’t want it to solve a, b, c, n. I just treat them as constants and express them in the solution as the paper did,I don't know if you understand. Commented Jan 11 at 10:27
• These are additional conditions to make solutions a0, a1,a2 valid Commented Jan 11 at 10:29
• Oh,I understand your comment, after tying Solve[equList, {a0, a1, a2, a, b, c, n}] // MatrixForm, I found a solution {a1 -> 0, a2 -> -a0, a -> -((2 a0 c^2 n)/(1 + n)), b -> (a0 c^2 (-1 + n)^2)/(2 n (1 + n))} in the output and it is the result of the paper! Commented Jan 11 at 12:05
• Aren't you looking for a solution a0,a1,a2? Probably case 3 in the last block ToRules[cond] of my answer is what you are looking for Commented Jan 11 at 12:42
• Yes,but case 3 a2 is expressed by a0 which is not expected,I want three of them to be represented independently by a, b, c, n Commented Jan 11 at 12:51