solving a system of nonlinear equations in terms of parameter (essentially rearrangement)
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
I'm having a trouble solving a system of equations with multiple parameters and unknowns
syms a b c k t p1 p2 v1 v2 x
eqn_x = x == a + ((p2-v2)-(p1-v1))/(2*t*(1-a-b)) + (1-a-b)/2 ;
eqn_p1 = p1 == c - k + (1/3)*(v1-v2) + t*(1-a-b)*(1+(a-b-4*v1-2*v2)/3) ;
eqn_p2 = p2 == c - k + (1/3)*(v2-v1) + t*(1-a-b)*(1+(a-b-4*v2-2*v1)/3) ;
eqn_v1 = v1 == (c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b)) ;
eqn_v2 = v2 == (c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b)) ;
eqns = [eqn_x, eqn_p1, eqn_p2, eqn_v1, eqn_v2] ;
S = solve(eqns, [p1 p2 v1 v2 x]) ;
This is my code, the goal is to get v1 and v2 term rearranged in terms of only a, b, c, k, & t terms.
there are 5 functions and 5 unknowns. though it is not a system of linear equation because v1 and v2 term depends on quadratic function of x, I guess there is a possible rearrangement.
But what I get is p1, p2, v1, v2, x being [0x1 sym]. How do I get those rearrangement?
0 commentaires
Réponses (1)
Torsten
le 26 Août 2019
Does this work ?
syms a b c k t x
eqn = x == a + (((c - k + (1/3)*(((c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b)))-((c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b)))) + t*(1-a-b)*(1+(a-b-4*((c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b)))-2*((c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b))))/3))-((c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b))))-((c - k + (1/3)*(((c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b)))-((c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b)))) + t*(1-a-b)*(1+(a-b-4*((c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b)))-2*((c - k + (1/3)*t*(1-a-b)*(3+b-a) + t*(2*(1-b-x)^2 - (x-a)^2))/(1+2*t*(1-a-b))))/3)) -((c - k + (1/3)*t*(1-a-b)*(3+a-b) + t*(2*(x-a)^2 - (1-b-x)^2))/(1+2*t*(1-a-b)))))/(2*t*(1-a-b)) + (1-a-b)/2 ;
S = solve(eqn,x)
2 commentaires
Torsten
le 26 Août 2019
Modifié(e) : Torsten
le 26 Août 2019
I inserted the expressions for v1 and v2 in p1 and p2 and then inserted v1, v2 and the revised p1 and p2 all in the first equation x == ...
I only inspected this equation superficially, but it seems it is a quadratic equation in x that MATLAB should be able to solve.
Voir également
Catégories
En savoir plus sur Calculus dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!