equation solver solve() returns empty result
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I tried using symbolic equation solver to get the results
syms V a b K
eq1 = pi^2*(a+b).*(b-a).^2 == V
eq2 = V == 8000
eq3 = a == K*b
eq4 = K == 0.2:0.1:0.7
a = solve([eq1,eq2,eq3,eq4],a)
And I got
a =
Empty sym: 0-by-1
It should have solutions while it returned nothing. Did I have something wrong?
3 个评论
madhan ravi
2019-2-2
why not solve the first three equations and substitute K values?
kairui wang
2019-2-2
Walter Roberson
2019-2-2
编辑：Walter Roberson
2019-2-2
eq4 = K == 0.2:0.1:0.7
defines a system of 6 equations,
[ K == 1/5, K == 3/10, K == 2/5, K == 1/2, K == 3/5, K == 7/10]
all of which have to be true simultaneously in order for solve() to find a solution.
You need to use symbolic K and substitute in the values later -- or you need to loop with specific numeric K values.
Also, you must solve() for as many variables as your provide equations. With four equations, you need to solve for four variables simultaneously. If you look at the solution I provided, I provide two equations and I solve for two variables.
采纳的回答
Walter Roberson
2019-2-2
V = 8000;
Kvals = 0.2:0.1:0.7;
syms a b K
eq1 = pi^2*(a+b).*(b-a).^2 == V
eq3 = a == K*b;
sol = solve([eq1, eq3], [a, b]);
a = subs(sol.a, K, Kvals)
You will find that the result is 3 (different solutions) per K value. Two of the solutions are complex valued and the third is real valued.
2 个评论
kairui wang
2019-2-2
Walter Roberson
2019-2-3
At the end add
a(all(imag(a)~=0,2), :) = [];
This will delete all rows in which all of the values are complex-valued. If there should happen to be any row in which some of the values are only real and some are complex then this code will not delete such a row. The code could be changed to create a cell array in which each entry had 1 to 3 real-valued solutions (since the number of real-valued could potentially change according to k value).
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