Solve nonlinear equation with different parameters

2019 2 14
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更新时间：2019 2 19
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0 个投票

Hello,
I would like to solve a (quite complicated) equation that looks like this:
pi/2*C*x + 0.5*(3 - gamma^2 - 2*gamma)*C*V^2 + 2*A*V + 2*x*B^2 - pi/2*F;
where gamma is also a function of x and it's defined as following:
gamma = gamma0*exp(-D*atan((B*x)./(C*Vdc)))*sqrt(1+((B*x)/(C*V)));
So, in the end, it can be considered as one (long) equation.
The unkowns are x and Vdc. I was thinking of defining the value of Vdc as a linsopace vector, for example, and see the value of x for each variation of Vdc (by making a loop). The question is: how can I solve the equation?
Also, would it be possible to solve the same equation by making varying other parameters of it (by making a loop on them)?
Thank you in advance for your help!

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Star Strider
Star Strider 2019-2-14

0 个投票

Do something like this (although with the correct values for the constants):
A = 13;
B = 5;
C = 7;
D = 3;
F = 11;
V = 42;
gamma0 = 31;
gamma = @(x,Vdc) gamma0*exp(-D*atan((B*x)./(C*Vdc)))*sqrt(1+((B*x)/(C*V)));
eqn = @(x,Vdc) pi/2*C*x + 0.5*(3 - gamma(x,Vdc).^2 - 2*gamma(x,Vdc))*C*V^2 + 2*A*V + 2*x*B^2 - pi/2*F;
Vdc = linspace(0, 10, 5);
for k1 = 1:numel(Vdc)
xs(k1) = fsolve(@(x) eqn(x,Vdc(k1)), 1);
end
figure
plot(Vdc, xs)
grid
Experiment to get the result you want.

12 个评论
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letoppina
letoppina 2019-2-15
When computing, I get the following error:
Objective function is returning undefined values at initial point. FSOLVE cannot continue.
How should I proceed?
Walter Roberson
Walter Roberson 2019-2-15
编辑：Walter Roberson 2019-2-18
don't start at 0. You divide by something times the current vdc value which generates a divide by 0 when the vdc is 0.
Star Strider
Star Strider 2019-2-15
@Walter — Thank you.
@letoppina — Note that the initial estimate I use here is 1.
letoppina
letoppina 2019-2-18
Thanks, the error was on the initial condition (although the ouput is still not the one I expected..).
Moreover, if i want to make varying another variable involved in the equations (let's say for example B), should I just make another loop outside the existing one?
Star Strider
Star Strider 2019-2-18
As always, my pleasure.
‘... should I just make another loop outside the existing one?’
That is likely the best approach to change one other variable.
letoppina
letoppina 2019-2-19
编辑：letoppina 2019-2-19
Hi,
sorry to bother again but I have another problem: when I compute, xs becomes a vector composed of values all equal to the initial condition (whatever I put in). Why?
Star Strider
Star Strider 2019-2-19
I have no idea.
What values are you using for the parameters?
Also, post your code.
letoppina
letoppina 2019-2-19
here is the code I am using now (I have modified a little the equation involved since I am making a test but the pricniple should be the same):
clc
clear all
close all
f0 = 170;
w0 = 2*pi*f0;
beta = 0.1;
L = 28e-3;
C0 = 50e-9;
alpha = 1e-3;
C = 0.0219;
wsw = 1/(sqrt(L*C0));
Rsw = 200;
M = 20e-4;
acc = 0.2;
Fm = M*acc;
gamma0 = exp(-(Rsw*pi)/(2*L*wsw));
Vdc = linspace(0.1e-12, 20, 100);
eqn = @(x,Vdc) pi/2*C*w0.*x^2 + 1/2*C0*(1 - gamma0^2).*Vdc.^2 + Vdc.*2*alpha.*x - (1-gamma0)*C0.*Vdc.^2 - pi/2*Fm.*x;
for ii=1:length(Vdc)
xs(ii) = fsolve(@(x) eqn(x,Vdc(ii)), 0.01e-13)
end
figure
plot(Vdc, xs)
grid
Star Strider
Star Strider 2019-2-19
Your initial estimate in your fsolve call is sufficiently close to 0 that your function is 0 there, satisfying the criteria fsolve uses as its convergence test.
You can change the convergence criterion using an options structure, however it’s easier to try a different starting value. You will get what appears to be an acceptable result with this call:
xs(ii) = fsolve(@(x) eqn(x,Vdc(ii)), 1);
letoppina
letoppina 2019-2-19
Thank you for your answer. The reason I put such a small value as initial condition is because the order of magnitude of my solution has to be anyway between 10^-6 and 10^-4. How is it possible to change the convergence criterion?
Star Strider
Star Strider 2019-2-19
‘the order of magnitude of my solution has to be anyway between 10^-6 and 10^-4’
True. However fsolve is a zero-finding function, so if you give it something very close to 0 initially, it is satisfied that it has solved the problem and stops iterating. The solution it arrives at is the value of the parameter of interest in your function that makes your function sufficiently close to 0.
You can change the various options with the optimoptions (link) function. However there is no need for you to do that, since fsolve is functioning normally, and delivers a reasonable result if you let it find the value of the parameter that results in your function being sufficiently close to 0.
Walter Roberson
Walter Roberson 2019-2-19
Note that instead of passing a scalar x0 you can pass a vector with two elements; fsolve will only search within the given range. This does require that the function has a different sign at the two endpoints.

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