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I need to solve the following system of equations symbolically for a_i, b_i, and c_i:

$$\beta_1 (1-a_i-b_i-c_i) \sum_{j=1}^{n}A_{i,j}(a_j+c_j)-\delta_1 a_i+\delta_2 c_i - \varepsilon \beta_2 a_i \sum_{j=1}^{n}B_{i,j}(b_j+c_j)=0,\\ \beta_2 (1-a_i-b_i-c_i) \sum_{j=1}^{n}B_{i,j}(b_j+c_j)-\delta_2 b_i+\delta_1 c_i - \varepsilon \beta_1 b_i \sum_{j=1}^{n}A_{i,j}(a_j+c_j)=0,\\ \varepsilon \beta_2 a_i \sum_{j=1}^{n}B_{i,j}(b_j+c_j) + \varepsilon \beta_1 b_i \sum_{j=1}^{n}A_{i,j}(a_j+c_j)-(\delta_1+\delta_2)c_i=0. $$

Here, $a_i$, $b_i$ and $c_i$ are considered to be probabilities for some node $i$ while $A$ and $B$ are adjacency matrices such that $A_{i,j}$ denotes whether $i$ is connected to $j$ or not. I have attempted to solve the above equations using the following code, but failed:

Remove["Global`*"]
Solve[{Subscript[β, 
 1] (1 - Subscript[a, i] - Subscript[b, i] - Subscript[c, i]) Sum[
  Subscript[A, i, j] (Subscript[a, j] + Subscript[c, j]), {j, 1, 
   n}] - Subscript[δ, 1] Subscript[a, i] + 
Subscript[δ, 2] Subscript[c, 
 i] - ϵ Subscript[β, 2] Subscript[a, i]
  Sum[Subscript[B, i, j] (Subscript[b, j] + Subscript[c, j]), {j, 
   1, n}] == 0,
Subscript[β, 
 2] (1 - Subscript[a, i] - Subscript[b, i] - Subscript[c, i]) Sum[
  Subscript[B, i, j] (Subscript[b, j] + Subscript[c, j]), {j, 1, 
   n}] - Subscript[δ, 2] Subscript[b, i] + 
Subscript[δ, 1] Subscript[c, 
 i] - ϵ Subscript[β, 1] Subscript[b, i]
  Sum[Subscript[A, i, j] (Subscript[a, j] + Subscript[c, j]), {j, 
   1, n}] == 0,
ϵ Subscript[β, 2] Subscript[a, i]
  Sum[Subscript[B, i, j] (Subscript[b, j] + Subscript[c, j]), {j, 
   1, n}] + ϵ Subscript[β, 1] Subscript[b, i]
  Sum[Subscript[A, i, j] (Subscript[a, j] + Subscript[c, j]), {j, 
   1, n}] - (Subscript[δ, 1] + Subscript[δ, 
   2]) Subscript[c, i] == 0,}, {Subscript[a, i], Subscript[b, i], Subscript[c, i]}]

Could anyone kindly tell me how to obtain the symbolic solutions for this system. Thank you in advance!

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  • $\begingroup$ Make up yer mind, are these differential or difference equations? For the former, you use DSolve[]; for the latter, RSolve[]. $\endgroup$ Commented May 15, 2015 at 8:19
  • $\begingroup$ These are differential equations where the $t$ variable has been dropped as these equations are investigated in steady-state condition. So, using DSolve does not help. $\endgroup$
    – Adee_lib
    Commented May 15, 2015 at 8:23
  • $\begingroup$ It is a algebraic system which includes Schur product.I think you could send it to Mathematics Stackexchange.:) $\endgroup$
    – WateSoyan
    Commented May 15, 2015 at 10:30

1 Answer 1

3
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Not an answer but too long for comment:

If I modify slightly your input and choose 'n=2`

 n = 2;
Solve[Table[{Subscript[β, 
        1] (1 - Subscript[a, i] - Subscript[b, i] - 
         Subscript[c, i]) Sum[
        Subscript[A, i, j] (Subscript[a, j] + Subscript[c, j]), {j, 1,
          n}] - Subscript[δ, 1] Subscript[a, i] + 
      Subscript[δ, 2] Subscript[c, 
        i] - ϵ Subscript[β, 2] Subscript[a, i] Sum[
        Subscript[B, i, j] (Subscript[b, j] + Subscript[c, j]), {j, 1,
          n}] == 0, 
    Subscript[β, 
        2] (1 - Subscript[a, i] - Subscript[b, i] - 
         Subscript[c, i]) Sum[
        Subscript[B, i, j] (Subscript[b, j] + Subscript[c, j]), {j, 1,
          n}] - Subscript[δ, 2] Subscript[b, i] + 
      Subscript[δ, 1] Subscript[c, 
        i] - ϵ Subscript[β, 1] Subscript[b, i] Sum[
        Subscript[A, i, j] (Subscript[a, j] + Subscript[c, j]), {j, 1,
          n}] == 0, ϵ Subscript[β, 2] Subscript[a, 
        i] Sum[Subscript[B, i, 
          j] (Subscript[b, j] + Subscript[c, j]), {j, 1, 
         n}] + ϵ Subscript[β, 1] Subscript[b, i] Sum[
        Subscript[A, i, j] (Subscript[a, j] + Subscript[c, j]), {j, 1,
          n}] - (Subscript[δ, 1] + 
         Subscript[δ, 2]) Subscript[c, i] == 0}, {i, n}] // 
  Flatten, Table[{Subscript[a, i], Subscript[b, i], 
    Subscript[c, i]}, {i, n}] // Flatten]

I see that even in this case we have 6 quadratic equations

Mathematica graphics

in the variable Mathematica graphics

so it is unlikely there is an analytical solution of the general case.

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