Solve non-linear system of equations with variable number of unknowns

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Ash Ash
Ash Ash 2020-12-8
评论： Ash Ash 2020-12-25
Hi everyone, I need some help. I have written a script to form system of nonlinear equations using the symbolic toolbox (because the equation forms are not fixed and dependant on certain parameters) and I need to solve them. I know how to solve a system of nonlinear equations with multiple unknowns, by parameterising the function as follows:
matfun=matlabFunction( symbolic_eqn );
funMiddleMan = @(f) matfun(f(1),f(2),f(3),f(4),f(5),f(6),f(7),f(8),f(9),f(10),...
f(11),f(12)); % the function has 12 unknowns
x0=zeros(1,12);
[soln_fs,fval_fs]=fsolve(funMiddleMan,x0);
However, for the case where I wish to vary the number of unknowns in "symbolic_eqn", what should I do do make this work? For now, the only idea I can think of is by writing an inelegant list of conditional statements of the form:
switch num_unknowns
case N
funMiddleMan = @(f) matfun(f(1),f(2),...,f(N)); %for N number of variables
[...]
end
Would you please guide me on how I can solve this problem?
Thank you for your help and time.
  1 个评论
Ash Ash
Ash Ash 2020-12-10
If it's not possible to do so and I need to use a list of conditional statements please do let me know. Thank you

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采纳的回答

Walter Roberson
Walter Roberson 2020-12-10
Ran in:
%{
matfun = matlabFunction( symbolic_eqn, 'vars', [list variables here]);
%}
for example
N = 5;
syms F [1 N]
expr = sum(randn(1,N) .* F);
f1 = matlabFunction( expr, 'vars', F)
f1 = function_handle with value:
@(F1,F2,F3,F4,F5)F1.*3.599379160893824e-1-F2.*1.002299459586992-F3.*1.613867501380812e-1-F4.*1.083541721318095+F5.*2.742004284674245e-1
But most of the time it is easier to instead
f2 = matlabFunction( expr, 'vars', {F} )
f2 = function_handle with value:
@(in1)in1(:,1).*3.599379160893824e-1-in1(:,2).*1.002299459586992-in1(:,3).*1.613867501380812e-1-in1(:,4).*1.083541721318095+in1(:,5).*2.742004284674245e-1
That second use of matlabFunction expects to be passed a row vector of N values, rather than N individual values.
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Ash Ash
Ash Ash 2020-12-25
aha! I get it now, I'm familiar with the usage of "sym". Thank you for responding . Merry Christmas to you!

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