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fsolve result is not desirable even giving a close starting point

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I have three nonlinear equations with three unknowns and I used fsolve function.
I know that x=[0.3173;0.3173;3.6590] is a good enough solution for the three equations. However, using fsolve, I could not get to this solution even for a very close starting point as shown below. Instead the result is (upon plugging these results into the 3 equations, the values are not close to 0):
x =
0.999999999999936
1.000000000000067
1.000000000000036
How can this be solved? Thanks!
CODE:
x0=[0.31; 0.31; 3.5];
options=optimset('Display','iter');
[x,fval]=fsolve(@myfun1,x0,options)
function F=myfun(x)
F=[13*cos(x(3)) - 5*x(1)*x(3)^2 - 5*x(2)*x(3)^2 + 13*x(1)*x(3)*sin(x(3)) + 13*x(2)*x(3)*sin(x(3)) + 10*x(1)*x(2)*x(3)^2 - 13*x(1)*x(2)*x(3)^2*cos(x(3)) + 40;
5*x(3) - 13*sin(x(3)) + 30*x(1)*x(3) + 13*x(1)*x(3)*cos(x(3)) + 13*x(2)*x(3)*cos(x(3)) - 5*x(1)*x(2)*x(3)^3 + 13*x(1)*x(2)*x(3)^2*sin(x(3));
390*x(3)^2*sin(x(3)) + 300*x(1)*x(3)^3 - 25*x(1)*x(3)^5 + 25*x(2)*x(3)^5 - 150*x(3)^3 - 260*x(1)*x(3)^3*cos(x(3)) - 130*x(2)*x(3)^3*cos(x(3)) + 130*x(1)*x(3)^4*sin(x(3)) - 130*x(2)*x(3)^4*sin(x(3)) - 169*x(1)*x(3)^3*cos(x(3))^2 + 169*x(2)*x(3)^3*cos(x(3))^2 - 169*x(1)*x(3)^3*sin(x(3))^2 + 169*x(2)*x(3)^3*sin(x(3))^2
];
  2 Commenti
Matt J
Matt J il 31 Gen 2014
Modificato: Matt J il 31 Gen 2014
In future, please highlight your code and apply the
formatting button, as I have just done for you now.

Accedi per commentare.

Risposta accettata

Matt J
Matt J il 31 Gen 2014
Modificato: Matt J il 31 Gen 2014
Possibly because you're passing @myfun1 to fsolve instead of @myfun? When I make this change and run your code, I get the solution you expect.
  2 Commenti
Qingbin
Qingbin il 31 Gen 2014
Modificato: Qingbin il 31 Gen 2014
Thanks! Sorry for the mistake. One question to follow up is that how can I give a good estimate of the starting point. In the above problem, I want to solve for the max{x(3)} and min{x(3)}.
Matt J
Matt J il 31 Gen 2014
There is no science to the initial guess. If you want solutions with the max/min possible x(3) value, try an initial guess with parameters well above/below what x(3) would naturally be.

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Più risposte (1)

Alex Sha
Alex Sha il 3 Set 2023
For Qingbin's equations, although it is a problem that has passed a long time, it is worth and interesting to have a try, there are multi-solutions:
The first one:
x1: -12.7613758329431
x2: -12.7613758329431
x3: -0.755039591199301
The second one:
x1: 0.101047394240414
x2: -3.71598808815167
x3: -1.63192533266968
The third one:
x1: -0.282297628581163
x2: 0.369875309676247
x3: 3.09470113469319
The fourth one:
x1: 0.317262018077202
x2: 0.317262018077202
x3: 3.65895757009226

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