Solving Equation Depending on Parameter

Question

onsagerian 2018-7-24

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/411800-solving-equation-depending-on-parameter

评论： Star Strider 2018-7-24

Hello,

An equation in a reference paper has been simply translated into a Matlab code (see below) for solving it, and it must have particular values of x depending on the parameter "gamma", but I got an error. Would you check the equation and answer me what I need to make a correction?

format long e
T=0.1;
k=1.0;        %k(1)       default value is 1.0
k_1=0.2;      %k(-1)      default value is 2.0   (modified: 1.0)
k_2=0.00001;      %k(-1*)      default value is 0.1 (Note: k_2=0.00001 higher error rate for n=1 compared to n>1)
%k_2=[0.1:0.1:1.0];
a_p=0.0001;     %k(p)     default value is 0.0001
k_p=a_p/10;    %k(-p)     Assuming this ratio is the constant (a_p/10)
a_q=0.00015;    %k(q)     default value is 0.00015
k_q=0.00002;    %k(-q)    default value is 0.00002
theta=0.01;              %default value is 0.01
m1=0.0000005;             %default value is 0.0000005
m2=0.0000002;             %default value is 0.0000002  
W=0.00002;                  %default value is 0.002
n=6;
syms x
gamma=(a_p*k*k_2)/(k_p*m1*k_1);
equation=gamma-[1+(theta-x)/(1-theta)*[(1-theta)/(x^(1/(n+1))-theta)]^(n+1)]==0;
sol=solve(equation,x);

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Aquatris 2018-7-24

编辑：Aquatris 2018-7-24

在 MATLAB Online 中打开

You should put your code in a nice format. I cannot say which parts are comment which parts are not. For instance, you cannot do

syms x gamma=(a_p*k*k_2)/(k_p*m1*k_1);

in a single line. It should be

syms x 
gamma = (a_p*k*k_2)/(k_p*m1*k_1);

Is it due to your copy paste or do you actually use it this way cannot be determined from your question.

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Answer 1

Star Strider 2018-7-24

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/411800-solving-equation-depending-on-parameter#answer_330045

在 MATLAB Online 中打开

There appears to be one valid value for ‘x’:

x =
0.2014042781988740154717685102689

The reason is that although there are 7 roots to the equation for ‘x’, it is the only one that meets the criteria set out in the ‘conditions’ result, that being:

-0.4487989505128276054946633404685 < angle(z2) & angle(z2) <= 0.4487989505128276054946633404685

This results from instructing solve to return the conditions as well:

sol = solve(equation,x, 'ReturnConditions',1);

that returns a polynomial in ‘z2’ with roots:

z2 =
                                             0.2014042781988740154717685102689
 - 0.16246739411681022588655095122335 + 0.083052533086836812066634941174803i
 - 0.16246739411681022588655095122335 - 0.083052533086836812066634941174804i
  - 0.032602888099564513824045864450354 + 0.1866174981726704349497552909463i
  - 0.032602888099564513824045864450354 - 0.1866174981726704349497552909463i
    0.12933518938326112554046904728674 - 0.14965579028283080032788468715005i
    0.12933518938326112554046904728674 + 0.14965579028283080032788468715005i

the angles of these roots being:

angle_z2 =
                                   0
   2.6690291408928938742331335149207
  -2.6690291408928938742331335149207
   1.7437551108769301908411899136693
  -1.7437551108769301908411899136693
 -0.85810582106481765317999373280298
  0.85810582106481765317999373280298

with the only value meeting the ‘angle’ requirement being the real root.

Also, please highlight your code, then press the [{}Code] button to format it properly. This makes it easier for everyone to read it and work with it. (I did that for you this time.)

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onsagerian 2018-7-24

I appreciate your help, and also thank you for letting me know how to make the format in a proper manner using the button "{} code". But,I still have a question. I used the format sol = solve(equation,x, 'ReturnConditions',1); to get proper solutions for the equation, but there was no return value. No error message, but not response. It is the only thing I made a correction. Do you know why this happened?