How to fix integral calculation error when the function handle become non-symbolic?

Question

rasoul rad 2016-12-25

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评论： rasoul rad 2016-12-27

Hi, I'm trying to optimize a problem using genetic algorithm. I write below code and use ga for optimization. It dose work and find x as answer:

SOC_new=100;
DOD_new=0;
t=1
x=sym('x');
SOC=@(x)(SOC_new);
DOD=@(x)(DOD_new);
Uc=@(x)(U0-g*DOD(x)+R_c*x./CN+R_c*M_c*x./CN*SOC(x)/(Cc-SOC(x)));
I_gas=@(x)(CN/100*I_gas0*exp(cu*(Uc(x)-U_gas0)+cT*(T-Tgas0)));
I_ch=@(x)((x-I_gas(x))/CN);
SOC=@(x)(SOC(x)+quadgk(I_ch(x),0,t)); %%or @(x)(SOC(x)+integral(I_ch(x),0,t)) or @(x)(SOC(x)+integral(I_ch(x),0,t,'ArrayValued',true)) or @(x)(SOC(x)+int(I_ch(x),x,0,t))==> "" all of these commands work""
DOD=@(x)(1-SOC(x));
P_c=@(x)(Uc(x)*x);

After optimization, when i want to see the value of SOC i use below because x is found:

>>SOC(x)

But i got the following error:

Error using quadgk (line 96)
First input argument must be a function handle.

or this:

Error using integral (line 96)
First input argument must be a function handle.

or this:

Undefined function 'int' for input arguments of type 'double'.

While i want to see the value of I_ch, i got no error:

>> I_ch(x)
ans =
    -4.2614e+06

How can i fix the integral error? I'll be grateful for any help.

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Answer 1

Walter Roberson 2016-12-25

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/318111-how-to-fix-integral-calculation-error-when-the-function-handle-become-non-symbolic#answer_248413

在 MATLAB Online 中打开

SOC=@(x) SOC_new * ones(size(x));
DOD=@(x) DOD_new * ones(size(x));

Note: you have

SOC=@(x)(SOC(x)+quadgk(I_ch(x),0,t));
DOD=@(x)(1-SOC(x));

Please use different variables than SOC and DOD when you define these. Although what you coded is legal, it confuses the readers, and it is likely to fail if you have a loop. Additional variable names are nearly free; brain cells trying to figure out self-referential anonymous functions are not.