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Trouble using fsolve with two non-linear equations and 3 parameter

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Keith Contrera
Keith Contrera le 25 Avr 2020
Clôturé : MATLAB Answer Bot le 20 Août 2021
I am trying to solve for a and d in the equations
  • K_H2O = (((2*a)/(6-a))/(((4-2*a-d)/(6-a))*((4-2*a-d)/(6-a))^0.5)*(p^0.5))
  • K_OH = (((2*d)/(6-a))/(((2-a-d)/(6-a))^(0.5)*((4-2*a-d)/(6-a))^(0.5)))
Using fsolve. I also need to pass parameters K_H2O, K_OH and p to the function. So far I have
K_H2O = 20.999;
K_OH = 1.0478;
p = 50;
guess = [2;2];
[a,d] = fsolve(eql_coef(K_H2O, K_OH, p), guess)
and in my function "eql_coef.m" I have
function F = eql_coef(K_H2O, K_OH, p)
F(1) = (((2*a)/(6-a))/(((4-2*a-d)/(6-a))*((4-2*a-d)/(6-a))^0.5)*(p^0.5))-K_H2O;
F(2) = (((2*d)/(6-a))/(((2-a-d)/(6-a))^(0.5)*((4-2*a-d)/(6-a))^(0.5)))-K_OH;
end
It returns
Undefined function or variable 'a'.
Error in eql_coef (line 2)
F(1) = (((2*a)/(6-a))/(((4-2*a-d)/(6-a))*((4-2*a-d)/(6-a))^0.5)*(p^0.5))-K_H2O;
I am pretty sure I have provided all the information I need and this is a syntac issue. Please help.
  2 commentaires
Muhammad Usman
Muhammad Usman le 25 Avr 2020
your unknows are a & d and you are creating function for coefficients, that's seem wrong.
Keith Contrera
Keith Contrera le 25 Avr 2020
I thought you need to create a function to solve for variables a and d? How do you suggest I go about it?

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Réponses (2)

Muhammad Usman
Muhammad Usman le 25 Avr 2020
I think it will works like in this way:
function F = eql_coef(x)
% x(1) == a
% x(2) == d
K_H2O = 20.999;
K_OH = 1.0478;
p = 50;
F = [(((2*x(1))/(6-x(1)))/(((4-2*x(1)-x(2))/(6-x(1)))*((4-2*x(1)-x(2))/(6-x(1)))^0.5)*(p^0.5))-K_H2O;
(((2*x(2))/(6-x(1)))/(((2-x(1)-x(2))/(6-x(1)))^(0.5)*((4-2*x(1)-x(2))/(6-x(1)))^(0.5)))-K_OH];
end
and then:
guess = [2;2];
options = optimoptions('fsolve','Display','iter');
[x,fval] = fsolve(@eql_coef,guess,options);
It produced results like:
Norm of First-order Trust-region
Iteration Func-count f(x) step optimality radius
0 3 850.247 656 1
1 6 712.211 1 127 1
2 9 524.365 1 54.2 1
3 12 506.84 1 39.3 1
4 15 488.656 1 32.7 1
5 16 488.656 1 32.7 1
6 19 470.782 0.25 32.1 0.25
7 20 470.782 0.625 32.1 0.625
8 23 459.878 0.15625 31.6 0.156
9 24 459.878 0.390625 31.6 0.391
10 25 459.878 0.0976563 31.6 0.0977
11 28 458.193 0.0244141 31.5 0.0244
12 29 458.193 0.0610352 31.5 0.061
13 30 458.193 0.0152588 31.5 0.0153
14 33 457.931 0.0038147 31.5 0.00381
15 34 457.931 0.00953674 31.5 0.00954
16 37 457.766 0.00238419 31.5 0.00238
17 38 457.766 0.00596046 31.5 0.00596
18 41 457.664 0.00149012 31.5 0.00149
19 42 457.664 0.00372529 31.5 0.00373
20 43 457.664 0.000931323 31.5 0.000931
21 46 457.648 0.000232831 31.5 0.000233
22 47 457.648 0.000582077 31.5 0.000582
23 50 453.253 0.000145519 31.4 0.000146
24 51 453.253 0.000363798 31.4 0.000364
25 54 453.247 9.09495e-05 31.4 9.09e-05
26 55 453.247 0.000227374 31.4 0.000227
27 58 453.243 5.68434e-05 31.4 5.68e-05
28 59 453.243 0.000142109 31.4 0.000142
29 62 453.241 3.55271e-05 31.4 3.55e-05
30 65 453.235 8.88178e-05 31.4 8.88e-05
31 66 453.235 0.000222045 31.4 0.000222
32 67 453.235 5.55112e-05 31.4 5.55e-05
33 68 453.235 1.38778e-05 31.4 1.39e-05
34 69 453.235 3.46945e-06 31.4 3.47e-06
35 72 453.234 8.67362e-07 31.4 8.67e-07
36 73 453.234 2.1684e-06 31.4 2.17e-06
37 74 453.234 5.42101e-07 31.4 5.42e-07
38 75 453.234 1.35525e-07 31.4 1.36e-07
39 76 453.234 3.38813e-08 31.4 3.39e-08
40 79 453.234 8.47033e-09 31.4 8.47e-09
41 80 453.234 2.11758e-08 31.4 2.12e-08
42 83 453.234 5.29396e-09 31.4 5.29e-09
43 84 453.234 1.32349e-08 31.4 1.32e-08
44 87 453.234 3.30872e-09 31.4 3.31e-09
45 88 453.234 8.27181e-09 31.4 8.27e-09
46 91 453.234 2.06795e-09 31.4 2.07e-09
47 92 453.234 5.16988e-09 31.4 5.17e-09
48 95 453.234 1.29247e-09 31.4 1.29e-09
49 96 453.234 3.23117e-09 31.4 3.23e-09
50 99 453.234 8.07794e-10 31.4 8.08e-10
51 100 453.234 2.01948e-09 31.4 2.02e-09
52 103 453.234 5.04871e-10 31.4 5.05e-10
53 104 453.234 1.26218e-09 31.4 1.26e-09
54 107 453.234 3.15544e-10 31.4 3.16e-10
55 108 453.234 7.88861e-10 31.4 7.89e-10
56 109 453.234 1.97215e-10 31.4 1.97e-10
57 112 453.234 4.93038e-11 31.4 4.93e-11
58 113 453.234 1.2326e-10 31.4 1.23e-10
59 116 453.234 3.08149e-11 31.4 3.08e-11
60 117 453.234 7.70372e-11 31.4 7.7e-11
61 120 453.234 1.92593e-11 31.4 1.93e-11
62 121 453.234 4.81482e-11 31.4 4.81e-11
63 124 453.234 1.20371e-11 31.4 1.2e-11
64 125 453.234 3.00927e-11 31.4 3.01e-11
65 128 453.234 7.52316e-12 31.4 7.52e-12
66 129 453.234 1.88079e-11 31.4 1.88e-11
67 132 453.234 4.70198e-12 31.4 4.7e-12
No solution found.
fsolve stopped because the relative size of the current step is less than the
default value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the default value of the function tolerance.
<stopping criteria details>
Hope so it will helpful for you
Bjorn Gustavsson
Bjorn Gustavsson le 25 Avr 2020
You need to make the function eql_coef depend on the coefficients a and d, something like this:
function F = eql_coef(a_d,K_H2O, K_OH, p)
a = a_d(1);
d = a_d(2);
F(1) = (((2*a)/(6-a))/(((4-2*a-d)/(6-a))*((4-2*a-d)/(6-a))^0.5)*(p^0.5))-K_H2O;
F(2) = (((2*d)/(6-a))/(((2-a-d)/(6-a))^(0.5)*((4-2*a-d)/(6-a))^(0.5)))-K_OH;
end
Now you have a function that will depend on a and d, that fsolve can adjust. So you call fsolve like this:
a_d_guess = [2;2];
[a_d_solution] = fsolve(@(a_d) eql_coef(a_d,K_H2O, K_OH, p), a_d_guess)
HTH
  2 commentaires
Keith Contrera
Keith Contrera le 25 Avr 2020
This gives me imaginary solutions and I need real solutions. How would I fix this?
Alex Sha
Alex Sha le 26 Avr 2020
Hi, I get the solution below:
a: 1.09358949887241
d: 0.427017129161841

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