Solve equations in a loop with fsolve

Hi there,

I have two problems solving an equation:

Problem 1:

The equation below is to be solved component by component and the results are to be stored line by line in the vector F1. So far so good, how do I teach the loop to use the correct column for the calculations (e.g. f1 (f ,:) or f8 (f ,:) without integrating the function into fsolve?

tau = 0.1
f4 = [3; 2; 6; 8]
f8 = [2; 6; 7; 3]
 eq = @(s) s*tau-(0.1.*s^2+3.54.*s-9.53).*f4.^2-f8;
 for f = 1:1:length (f4)
     F1 (f,:) = fsolve (eq, 0)
end

Problem 2:

The eq described above actually consists of two equations:

eq1 = 0.01*s.^2+3.54.*s-y*9.53
eq2 = y.*f4.^2-f8-s.*tau

It would be desirable to be able to insert both equations separately. Here is the variable y, which disappears after summarizing. Is there a way to combine this with the "problem" above?

Thanks a lot!

[darova]

Shouldn't the function be dependent?

[Mepe]

s is the variable we are looking for. f4 and f8 are given by the vectors. Do these still have to be specified as you have declared?

Answer 1

[Matt J]

Your equations are quadratic and therefore generally have two solutions, s. Fsolve cannot find them both for you. Why aren't you using roots()? Regardless, here are the code changes pertaining to your question:

Problem 1

tau = 0.1
f4 = [3; 2; 6; 8]
f8 = [2; 6; 7; 3]
 
 for i = 1:1:length (f4)
     
     eq = @(s) s*tau-(0.1.*s^2+3.54.*s-9.53).*f4(i).^2-f8(i);
     
     F1 (i,:) = fsolve (eq, 0);
     
end

Problem 2

for i = 1:1:length (f4)
     
     eq1 = @(sy) [0.01*sy(1).^2+3.54.*sy(1)-sy(2)*9.53 ; ...
                   sy(2).*f4(i).^2-f8(i)-sy(1).*tau];
     
     F2 (i,:) = fsolve (eq, [0,0]);
     
end

[Mepe]

Many many thanks! that has taken me a lot further. I will rethink the alternative use of roots ().

I have two more questions about your solution.

tau = 0.1
f4 = [3; 2; 6; 8]
f8 = [2; 6; 7; 3]
for i = 1:1:length (f4)
     eq1 = @(sy) [0.01*sy(1).^2+3.54.*sy(1)-sy(2)*9.53 ; sy(2).*f4(i).^2-f8(i)-sy(1).*tau];
     F2 (i,:) = fsolve (eq1, [0,0]);   
end

What exactly do sy(1) or sy(2) mean in the equation definition?

in F2 the solution is filled with a 2x4 matrix. How does that come about?

[Matt J]

What exactly do sy(1) or sy(2) mean in the equation definition?

In Problem 2, you have two unknowns, s and y. Fsolve requires that they be bundled into a vector sy=[s,y].

in F2 the solution is filled with a 2x4 matrix. How does that come about?

Each row is a solution [s,y] corresponding to i=1,2,3,4.

[Mepe]

Now I see. Thanks a lot for your help!!!

[Mepe]

I still have a question.

Now if I needed both solutions to the quadratic equation, how exactly would that work with the roots () command? according to help, a vector with numerical values must be used. How can I apply this to my problem?

Answer 2

[darova]

This is the correct form

tau = 0.1
f4 = [3; 2; 6; 8]
f8 = [2; 6; 7; 3]
 eq = @(s,f4,f8) s*tau-(0.1.*s^2+3.54.*s-9.53).*f4.^2-f8;
 for f = 1:1:length (f4)
     F1 (f,:) = fsolve (@(s)eq(s,f4(f),f8(f), 0);
end

[Mepe]

Thank you very much for your support. I really like your idea of how the fsolve expression is built!

Unfortunately I get an error message (The input to FSOLVE should be either a structure with valid fields or consist of at least two arguments).

Can you help again here?

tau = 0.1
f4 = [3; 2; 6; 8]
f8 = [2; 6; 7; 3]
 eq = @(s,f4,f8) s*tau-(0.1.*s^2+3.54.*s-9.53).*f4.^2-f8;
 for f = 1:1:length (f4)
     F1 (f,:) = fsolve (@(s)eq(s,f4(f),f8(f), 0));
end

[darova]

I made a terrible mistake

[Mepe]

No problem. Thanks a lot for this solution!!!