Solving a system of function handles with fsolve

I have a problem using fsolve in order to solve a system of equations.

a1 to a6 are anonymous functions which are supposed to be equal to 0. T is a vector with six unknown variables T1 to T6. Each anonymous function consists of a term with at least two of those variables.

So with 6 equations and 6 unknowns the solver should be able to solve the system of equations.

I am calling the function like this:

T0 = [0 0 0 0 0 0];
Temp = fsolve(CalcTemps,T0)
function fun = CalcTemps(T)
   global a1 a2 a3 a4 a5 a6
   fun(1) = a1;
   fun(2) = a2;
   fun(3) = a3;
   fun(4) = a4;
   fun(5) = a5;
   fun(6) = a6;
end

In my understanding the code is not working as it is not possible to store function handles in fun. I also tried to create fun as a cell array so that the function handles can be stored but then I get an error as fun has to be a function handle and not a cell array.

So my question is: Is it possible to store multiple function handles (a1 to a6) in a single function handle (fun)?

Answer 1

[Stephan]

Hi,

use:

calc_temp
function calc__tenp

T = @(T1,T2,T3,T4,T5,T6) [T1 T2 T3 T4 T5 T6];   % unknown variable vector
T_Amb = 20;

a1 = @(T)5*T(2)-T(1);                           % anonymous functions
a2 = @(T)2*T(3)-4*T(2);
a3 = @(T)T(4)-T(3);
a4 = @(T)2*T(5)-3*T(4);
a5 = @(T)3*T(6)-4*T(5);
a6 = @(T)5*T_Amb-4*T(6);

T0 = [20 20 20 20 20 20];                       % initial guess
Temp = fsolve(@CalcTemps,T0)

function fun = CalcTemps(T)
  fun(1) = a1(T);       % ERROR is produced in this line
  fun(2) = a2(T);
  fun(3) = a3(T);
  fun(4) = a4(T);
  fun(5) = a5(T);
  fun(6) = a6(T);
end
end

Results:

Temp =

     31.2500    6.2500   12.5000   12.5000   18.7500   25.0000

Best regards

Stephan

[Lennart Vogt]

Yes it is working now. Thanks a lot!

[Stephan]

Please accept useful answers in order to help other people with similar Problems finding helpful Solutions in the future.

Answer 2

[Matt J]

You need to evaluate your function at T,

function fun = CalcTemps(T)
   global a1 a2 a3 a4 a5 a6
   fun(1) = a1(T);
   fun(2) = a2(T);
   fun(3) = a3(T);
   fun(4) = a4(T);
   fun(5) = a5(T);
   fun(6) = a6(T);
end

[Lennart Vogt]

Alright I have done so, thank you so far. I am getting the error "Not enough input arguments." for the line

fun(1) = a1(T);

Any idea where that might come from?

[Matt J]

You haven't shown us the definition of a1(), but apparently it expects more input arguments than just T.

[Lennart Vogt]

The anonymous function contains other variables and has the following structure:

k = 2;
v = [1 2 3];
a1 = @(T) v(k)*T;

Is there a way to do some kind of interim solution so that the anonymous function transforms from

a1 = @(T) v(k)*T;

to

a1 = @(T) 4*T;

?

I think that would make my code a lot easier to continue.

[Matt J]

Ah, sorry. The problem has nothing to do with a1. You have an error in your call to fsolve.

Temp = fsolve(@CalcTemps,T0); % was missing the '@'

[Lennart Vogt]

Yes that was definetly missing. Still I am getting an error "Index exceeds array bounds.".

My functions are full of other variables as I mentioned before. I think i am getting an error with those variables.

Do you know a way to compute them as I showed in my previous comment?

[Matt J]

You could define a1 like so

c=v(k);

a1=@(T) c*T;

but I don't think it will make a difference.

[Lennart Vogt]

So basically I can reduce my code to the folowing structure:

T = @(T1,T2,T3,T4,T5,T6) [T1 T2 T3 T4 T5 T6];   % unknown variable vector
T_Amb = 20;

a1 = @(T)5*T(2)-T(1);                           % anonymous functions
a2 = @(T)2*T(3)-4*T(2);
a3 = @(T)T(4)-T(3);
a4 = @(T)2*T(5)-3*T(4);
a5 = @(T)3*T(6)-4*T(5);
a6 = @(T)5*T_Amb-4*T(6);

T0 = [20 20 20 20 20 20];                       % initial guess
Temp = fsolve(@CalcTemps,T0)

function fun = CalcTemps(T)

   global a1 a2 a3 a4 a5 a6

   fun(1) = a1(T);       % ERROR is produced in this line
  fun(2) = a2(T);
  fun(3) = a3(T);
  fun(4) = a4(T);
  fun(5) = a5(T);
  fun(6) = a6(T);
end

Still I am getting an error "Index exceeds array bounds." inside of the function.