# How to plot graph of value of variable when a parameter is varied

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Teck Lim 2020 年 4 月 9 日
コメント済み: Ameer Hamza 2020 年 4 月 9 日
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.

### 採用された回答

Ameer Hamza 2020 年 4 月 9 日

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'});

#### 5 件のコメント

Teck Lim 2020 年 4 月 9 日
For my pen and paper calculations, I assumed R=1 and I obtained:
τ=0.01698
e=0.0866
k=0.00679
Certainly, I am not ruling out multiple solutions. How would the algorithm respond if that was the case? Does that mean the solution will depend on the initial guess?
Ameer Hamza 2020 年 4 月 9 日
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'});
Teck Lim 2020 年 4 月 9 日

### その他の回答 (1 件)

Torsten 2020 年 4 月 9 日
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.

#### 4 件のコメント

Torsten 2020 年 4 月 9 日
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.
Teck Lim 2020 年 4 月 9 日
Thanks for the suggestion. I will try that out.
Ameer Hamza 2020 年 4 月 9 日
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'});