how can I remove error in fzero?

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Parham Babakhani Dehkordi 2015-4-2

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评论： Torsten 2015-4-2

Hi everybody, I have some input parameters such as: D=0.040; A=pi*D^2/4; Jo=0.74; Jw=[0.35 0.25 0.19 0.12 0.08 .06 0.04]; ro_o=910; ro_w=1000; mu_w=0.001; mu_o=0.92;

for p=1:length(Jw)

fun=@(Hw,Jw) (((8*ro_o*((D-(2*sqrt(Hw*A/pi)))*(Jo/(1-Hw))*ro_o/mu_o)^-1.0)*(Jo/(1-Hw))^2)*(pi*(D-(2*sqrt(Hw*A/pi)))*(1/(1-Hw))))-((0.023*((Jw(p)*D*ro_w/mu_w)^-.2)*ro_w*(Jw(p)/Hw)^2*pi*D)) ;

Hw(p)=fzero(@(Hw) fun(Hw,Jw(p)),[0 0.4]);

end

if I run this, I would get an error as follows: Error using fzero (line 242) Function values at interval endpoints must be finite and real.

Error in by_using_fminbnd (line 21) Hw(p)=fzero(@(Hw) fun(Hw,Jw(p)),[0 0.4]);

Thus, I tried to use fsolve, it is good but the results have imaginary part which I do not want it, my result must be real positive values, I wrote the codes for fsolve:

for p=1:length(Jw)

Hw(p) = fsolve(@(Hw) (((8*ro_o*((D-(2*sqrt(Hw*A/pi)))*(Jo/(1-Hw))*ro_o/mu_o)^-1.0)*(Jo/(1-Hw))^2)*(pi*(D-(2*sqrt(Hw*A/pi)))*(1/(1-Hw))))-((0.023*((Jw(p)*D*ro_w/mu_w)^-.2)*ro_w*(Jw(p)/Hw)^2*pi*D)), 0.4)

end

is there anybody to help me?

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Answer 1

Parham Babakhani Dehkordi 2015-4-2

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/195776-how-can-i-remove-error-in-fzero#answer_173702

在 MATLAB Online 中打开

D=0.040; A=pi*D^2/4; Jo=0.74; Jw=[0.35 0.25 0.19 0.12 0.08 .06 0.04]; ro_o=910; ro_w=1000; mu_w=0.001; mu_o=0.92;

fun=@(Hw,Jw) (((8*ro_o*((D-(2*sqrt(Hw*A/pi)))*(Jo/(1-Hw))*ro_o/mu_o)^-1.0)*(Jo/(1-Hw))^2)*(pi*(D-(2*sqrt(Hw*A/pi)))*(1/(1-Hw))))-((0.023*((Jw*D*ro_w/mu_w)^-.2)*ro_w*(Jw/Hw)^2*pi*D)) ;

for p=1:length(Jw)

Hw(p)=fzero(@(Hw) fun(Hw,Jw(p)),[-0.4 0.4]);

end

your guess is right, this an error again:

Error using fzero (line 242) Function values at interval endpoints must be finite and real.

Error in by_using_fminbnd (line 21) Hw(p)=fzero(@(Hw) fun(Hw,Jw(p)),[-0.4 0.4]);

Do you have any other functions or approach that you can suggest me to avoid going to an infinite value?

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Parham Babakhani Dehkordi 2015-4-2

在 MATLAB Online 中打开

D=0.040; A=pi*D^2/4; Jo=0.74; Jw=[0.35 0.25 0.19 0.12 0.08 .06 0.04]; ro_o=910; ro_w=1000; mu_w=0.001; mu_o=0.92;

fun=@(Hw,Jw) (((8*ro_o*((D-(2*sqrt(Hw*A/pi)))*(Jo/(1-Hw))*ro_o/mu_o)^-1.0)*(Jo/(1-Hw))^2)*(pi*(D-(2*sqrt(Hw*A/pi)))*(1/(1-Hw))))-((0.023*((Jw*D*ro_w/mu_w)^-.2)*ro_w*(Jw/Hw)^2*pi*D)) ;

for p=1:length(Jw)

Hw(p)=fzero(@(Hw) fun(Hw,Jw(p)),[-0.4 0.4]);

end

your guess is right, this an error again:

Error using fzero (line 242) Function values at interval endpoints must be finite and real.

Error in by_using_fminbnd (line 21) Hw(p)=fzero(@(Hw) fun(Hw,Jw(p)),[-0.4 0.4]);

Do you have any other functions or approach that you can suggest me to avoid going to an infinite value?

Torsten 2015-4-2

You have sqrt(Hw*A/pi) in your formula for the function.

sqrt(-0.4*A/pi) gives a complex number ...

Best wishes

Torsten.

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Answer 2

Torsten 2015-4-2

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https://ww2.mathworks.cn/matlabcentral/answers/195776-how-can-i-remove-error-in-fzero#answer_173695

在 MATLAB Online 中打开

In your function definition, you divide by Hw. Thus your left interval endpoint is not allowed to be 0.

Further:

Did you test that your function produces results of different sign at the endpoints of the interval ? My guess is no since you did not recognize that f(0)=Infinity.

Further you should use:

D=0.040; A=pi*D^2/4; Jo=0.74; Jw=[0.35 0.25 0.19 0.12 0.08 .06 0.04]; ro_o=910; ro_w=1000; mu_w=0.001; mu_o=0.92;
fun=@(Hw,Jw) (((8*ro_o*((D-(2*sqrt(Hw*A/pi)))*(Jo/(1-Hw))*ro_o/mu_o)^-1.0)*(Jo/(1-Hw))^2)*(pi*(D-(2*sqrt(Hw*A/pi)))*(1/(1-Hw))))-((0.023*((Jw*D*ro_w/mu_w)^-.2)*ro_w*(Jw/Hw)^2*pi*D)) ;
for p=1:length(Jw)
   Hw(p)=fzero(@(Hw) fun(Hw,Jw(p)),[0 0.4]);
end