Symbolize but the solutions number of the equation changed

3 vues (au cours des 30 derniers jours)
Lingling Fan
Lingling Fan le 17 Fév 2019
Commenté : chirag rohit le 16 Mar 2019
Screen Shot 2019-02-16 at 11.57.29 PM.png
To solve the det A ==0 equation, if I symbolize eta_yy, eta_xy..., I get q_sol as 4*1. But if I take eta_yx, eta_xy... as a number, I get q_sol as 6*1. Why is it?
Thank you in advance

Réponses (1)

chirag rohit
chirag rohit le 16 Mar 2019
clc
clear all;
syms q
eta_yy = 2;
eta_yx = 3;
eta_xy = 4;
eta_xx = 2;
eta_zy = 4;
eta_zx = 5;
eta_yz = 6;
eta_xz = 3;
eta_zz = 5;
ky = 8;
kx = 7;
k0 = 2.5;
A2 = [eta_yy, -eta_yx; -eta_xy, eta_xx];
A1a = [-ky*eta_zy, ky*eta_zx; kx*eta_zy, -kx*eta_zx];
A1b = [-ky*eta_yz, ky*eta_yz; kx*eta_xz, -kx*eta_xz];
A1 = A1a + A1b;
A0a = [ky^2 * eta_zz, -kx * ky * eta_zz; -kx * ky * eta_zz, kx^2 * eta_zz];
A0b = [kx^2 * eta_yy - kx * ky * eta_yx, kx*ky*eta_yy - ky^2*eta_yx; ...
kx * ky * eta_xx - kx^2 * eta_xy, ky^2 * eta_xx - kx * ky * eta_xy];
A0 = A0a + A0b - k0^2*eye(2);
Am1 = [ky^2 * kx * eta_zx - kx^2 * ky * eta_zy, ky^3 * eta_zx - ky^2 * kx *eta_zy;...
kx^3 * eta_zy - kx^2 * ky * eta_zx, kx^2 * ky * eta_zy - ky^2 * kx * eta_zx];
A_mat = A2 * q^2 + A1 * q + A0 + Am1 * q^(-1);
% simplify(A_mat);
collect(det(A_mat),q);
q_sol = solve(det(A_mat)==0,q)
>> vpa(q_sol,4)
ans =
0.2184 - 10.71i
0.2184 + 10.71i
2.867
25.07
  1 commentaire
chirag rohit
chirag rohit le 16 Mar 2019
Respected ma'am,
Here i have taken random values of eta_xx,.... And tried to solve eqation. I get q_sol as 4*1. In both ways i get q_sol as 4*1. If you are choosing some specified values for variables please mention.

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