1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Make up yer mind, are these differential or difference equations? For the former, you use DSolve[]; for the latter, RSolve[]. $\endgroup$ – J. M. will be back soon May 15 '15 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 May 15 '15 at 8:23
  • $\begingroup$ It is a algebraic system which includes Schur product.I think you could send it to Mathematics Stackexchange.:) $\endgroup$ – WateSoyan May 15 '15 at 10:30
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.