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Hi all,

I have a system of equations and I would like to solve for e, τ, and k. However, I want to assume that the parameter R can take on several different values (e.g. 0.5, 1, 1.5). In particular, I hope to plot a graph of how the solution for any of my variables e, τ or k varies with R. I am really struggling with how this can be implemented in MATLAB. Any help will be much appreciated.

Ameer Hamza
on 9 Apr 2020

Edited: Ameer Hamza
on 9 Apr 2020

See fsolve(): https://www.mathworks.com/help/optim/ug/fsolve.html

Write your system of equation in the form

Where , and are the three equations in your queston. Then apply fsolve() function.

This is the code:

f_1 = @(e, tau, k, R) 1./(0.8*e.^0.7-e-tau-k) - 0.336*e.^-0.3./(tau+R.*k);

f_2 = @(e, tau, k, R) 1./(0.8*e.^0.7-e-tau-k) - 0.5./tau;

f_3 = @(e, tau, k, R) 1./(0.8*e.^0.7-e-tau-k) - (R-0.3)./(tau+R.*k);

f = @(e, tau, k, R) [f_1(e,tau,k,R); f_2(e,tau,k,R); f_3(e,tau,k,R)];

R = [0.5, 1, 1.5];

sol = zeros(numel(R), 3);

for i=1:numel(R)

sol(i,:) = fsolve(@(x) f(x(1),x(2),x(3), R(i)), rand(1,3));

end

plot(R, sol);

legend({'e', 'tau', 'k'});

Ameer Hamza
on 9 Apr 2020

Teck, your equation does seem to have multiple solutions, and the solution you mentioned for R=1 is correct. I tried different initial points, and values of R. The fsolve() is able to find a solution if R >= 0.8, but for smaller values, it cannot find a solution

f_1 = @(e, tau, k, R) 1./(0.8*e.^0.7-e-tau-k) - 0.336*e.^-0.3./(tau+R.*k);

f_2 = @(e, tau, k, R) 1./(0.8*e.^0.7-e-tau-k) - 0.5./tau;

f_3 = @(e, tau, k, R) 1./(0.8*e.^0.7-e-tau-k) - (R-0.3)./(tau+R.*k);

f = @(e, tau, k, R) [f_1(e,tau,k,R); f_2(e,tau,k,R); f_3(e,tau,k,R)];

R = [0.5, 0.8, 1, 1.5, 2];

sol = zeros(numel(R), 3);

for i=1:numel(R)

sol(i,:) = fsolve(@(x) f(x(1),x(2),x(3), R(i)), 1e-6.*rand(1,3));

f(sol(i,1), sol(i,2), sol(i,3), R(i)) % to check if the 3 equations are true.

end

plot(R, sol);

legend({'e', 'tau', 'k'});

Torsten
on 9 Apr 2020

Your equations can only be solved for one value of R, namely R = 0.3 + 0.366*exp(-0.3).

This can be seen by dividing equation 1 by equation 3.

Torsten
on 9 Apr 2020

ok, then at least we can explicitly solve for e depending on R.

Taking the reciprocals of all the 6 expressions and inserting the result for e from the first step, we can solve for tau and k since we have a linear system of equations.

Ameer Hamza
on 9 Apr 2020

Torsten's analysis is correct. This system does have a closed-form solution. I used a symbolic toolbox to solve this system of equation, and it gives more accurate results (as expected) at R=0.5 as compared to the numeric solver. Following code shows the solution with symbolic approach

syms e tau k R

eq1 = 1./(0.8*e.^0.7-e-tau-k) - 0.336*e.^-0.3./(tau+R.*k);

eq2 = 1./(0.8*e.^0.7-e-tau-k) - 0.5./tau;

eq3 = 1./(0.8*e.^0.7-e-tau-k) - (R-0.3)./(tau+R.*k);

r = [0.5 1 1.5];

sols = zeros(numel(r), 3);

for i=1:numel(r)

sol = solve([eq1==0 eq2==0 eq3==0], [e tau k]);

sols(i,1) = subs(sol.e(1), R, r(i));

sols(i,2) = subs(sol.tau(1), R, r(i));

sols(i,3) = subs(sol.k(1), R, r(i));

end

sols = double(sols);

plot(r, sols);

legend({'e', 'tau', 'k'});

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