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Effacer les filtres

solving multiple equations with variable later

2 vues (au cours des 30 derniers jours)
Yokuna
Yokuna le 23 Sep 2021
Commenté : Walter Roberson le 25 Sep 2021
I want to solve the set of equations and plot x1 vs t and x2 vs t. The problem is due to sols(tidx,1). I want to calculate x1 as 140 +sols(tidx,1)^2 as given in equations. The value of sols(tidx,1) is updating below in the loop. How to find x1 by updating xL1 at each instant?
syms x1 x2 xL1 xL2 t
eqns = [
x1-140-xL1==0;
x2-160-xL2==0;
xL1==sols(tidx,1)^2;
xL2==sols(tidx,2)^2;
x1+x2==300+xL1+xL2;];
E2 = lhs(eqns)-rhs(eqns);
F = matlabFunction(E2, 'vars', {[x1,x2,xL1,xL2], t});
T = linspace(0,20,500);
nT = length(T);
x0 = [140+140*140, 160+160*160, 140*140, 160*160];
sols = zeros(nT, 4);
options = optimoptions(@fsolve, 'Algorithm', 'levenberg-marquardt', 'display', 'none');
for tidx = 1 : nT
[thissol, ~, exitflag, output] = fsolve(@(x) F(x,T(tidx)), x0, options);
p=thissol;
sols(tidx,:)=p;
x0=[sols(tidx,1),sols(tidx,2),sols(tidx,3),sols(tidx,4)]
end
figure(1)
plot(T, sols(:,1));
hold on
plot(T, sols(:,2));
hold on

Réponses (1)

Walter Roberson
Walter Roberson le 25 Sep 2021
syms x1 x2 xL1 xL2 t x3 x4
x1 = 140 + xL1;
x2 = 160 + xL2;
eqn1 = x1 + x2 + x3 + x4 == 300 + xL1 + xL2
eqn1 = 
lhs(eqn1) - rhs(eqn1) == 0
ans = 
but we do not know what x3 and x4 are.
If x3 and x4 happen to satisfy that x3 == -x4 then your equations are always satisfied for finite xL1, xL2; otherwise they cannot be satisfied.
  1 commentaire
Walter Roberson
Walter Roberson le 25 Sep 2021
You edited your code. With the new code, the fifth equation is the sum of the first two equations, and x1 and x2 can be directly calculated from sols. There is no point going through fsolve.

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