solving a system of equation using symbolic expressions returns empty solution

Question

Tae Lim 2022-5-6

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1713475-solving-a-system-of-equation-using-symbolic-expressions-returns-empty-solution

评论： Tae Lim 2022-5-9

采纳的回答： Walter Roberson

在 MATLAB Online 中打开

Hello,

I'm trying to solve a system of equation using symbolic expressions but it outputs empty value.

% values are predefined for the following variables: CG0, HC, R, T, K0, Cm0, beta
syms RH q_H2O CG K Cm
eq1 = q_H2O == Cm * CG*K*RH / ((1-K*RH)*(1+(CG-1)*K*RH));
eq2 = CG == CG0*exp(HC/R/T);
eq3 = K == K0*exp(HK/R/T);
eq4 = Cm == Cm0*exp(beta/T);
eqns = [eq1, eq2, eq3, eq4];
[~, ~, ~, q_H2O] = solve(eqns);

Running this code yields:

ans = 
  struct with fields:
       CG: [0×1 sym]
        K: [0×1 sym]
       RH: [0×1 sym]
    q_H2O: [0×1 sym]

I was expecting a result that is a function of 'RH' (q_H2O = f(RH))

But if I predefine 'RH' and run the code (e.g. RH=0.5), this code outputs a value as a result.

I would appreciate any comments!

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Answer 1

Walter Roberson 2022-5-7

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1713475-solving-a-system-of-equation-using-symbolic-expressions-returns-empty-solution#answer_958855

在 MATLAB Online 中打开

syms CG0  HC R T K0 Cm0 beta
syms RH q_H2O CG K Cm HK
eq1 = q_H2O == Cm * CG*K*RH / ((1-K*RH)*(1+(CG-1)*K*RH));
eq2 = CG == CG0*exp(HC/R/T);
eq3 = K == K0*exp(HK/R/T);
eq4 = Cm == Cm0*exp(beta/T);
eqns = [eq1, eq2, eq3, eq4];
sol = solve(eqns, [q_H2O, CG, K, Cm] )

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Tae Lim 2022-5-9

在 MATLAB Online 中打开

Hi Walter, Thank you again for your comments. I am still confused since I previously used the code to solve another system of equations without issues. Please see below:

ns0 = 2;
chi = 2;
T0 = 296;
T = 300;
b0 = 2e5;
H_CO2 = 60;
R = 8;
t0 = 1;
alpha = 1;
syms P_CO2 q ns b t
eq1 = q == ns*b*P_CO2 / (1+(b*P_CO2)^t)^(1/t);
eq2 = ns == ns0 * exp(chi * (1-T0/T));          % maximum adsorption capacity
eq3 = b == b0 * exp(H_CO2/R/T0 * (T0/T-1));    % adsorption affinity
eq4 = t == t0 + alpha * (1-T0/T);              % heterogeneity of the adsorption system
eqns = [eq1, eq2, eq3, eq4];
solve(eqns)
ans = 
  struct with fields:
     b: 3434813230887443/17179869184
    ns: 4625311226299041/2251799813685248
     q: (15887080197064170033712929842163*P_CO2)/(38685626227668133590597632*(((3434813230887443*P_CO2)/17179869184)^(76/75) + 1)^(75/76))
     t: 76/75

I do not really see the primary difference between this code and the code above. In essence, I'm just trying to generate an equation in the form y=f(x) so I can either evaluate 'y' in specific values of 'x' or generate plots of 'y' as a function of 'x'. I appreciate your help!

Torsten 2022-5-9

编辑：Torsten 2022-5-9

As Walter already said:

In your previous code, you specified to solve for one solution variable q_H2O with four equations. This is not possible with solve - the number of equations mustn't exceed the number of variables solved for.

In the code above, you didn't specify the variables to solve for. So MATLAB chose b, ns, q and t as sympathic and expressed q in terms of the last of the five symbolic variables, namely P_CO2. If you try harder, you might find the rule MATLAB applied to choose the variables it solved for (maybe alphabetic order or something similar).

Tae Lim 2022-5-9

Hi Walter and Torsten, thank you for your comments! I ended up simplifying my codes to evaluate just one function with one unknown since I can fix all the other variables into certain values (for now). I don't need to worry about a system of equations in this case. I appreciate all your help!

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