error in matlab code

I have been trying to solve an optimal problem in which a body of mass m have to [be] transfered to y(tf) = 0 from its initial position y(o) = 0 such that its final velocity is zero satisfying the performance index.

, where t = [to, tf].

But I am not getting the optimal trajectories using this code. Could anyone help me to find where the mistake [is]?

clear all
c0 = [0.1; 0.1; 0.1; 0.1; 5]; % Make a starting guess at the solution
[c, fval] = fsolve (@myfun1, c0) % Call optimizer

Solver stopped prematurely.

fsolve stopped because it exceeded the function evaluation limit,
options.MaxFunctionEvaluations = 5.000000e+02.

c = 5×1

0.0046 -0.0198 -0.0881 0.7725 12.1323

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fval = 5×1

-0.2275 0.0271 -0.0881 -0.0968 0.0024

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c1 = c(1); c2 = c(2); c3 = c(3); c4 = c(4); c5 = c(5);
t=0:0.01:c5;
x1op = -(c1*(t.^3))/6-(c2*(t.^2))/2+(c3.*t)+c4

x1op = 1×1214

0.7725 0.7716 0.7707 0.7698 0.7690 0.7681 0.7672 0.7663 0.7655 0.7646 0.7638 0.7629 0.7620 0.7612 0.7603 0.7595 0.7586 0.7578 0.7569 0.7561 0.7552 0.7544 0.7536 0.7527 0.7519 0.7511 0.7502 0.7494 0.7486 0.7477

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x2op =-(c1*(t.^2))/2+-(c2.*t)+c3

x2op = 1×1214

-0.0881 -0.0879 -0.0877 -0.0875 -0.0873 -0.0871 -0.0869 -0.0867 -0.0865 -0.0863 -0.0861 -0.0859 -0.0857 -0.0855 -0.0853 -0.0851 -0.0850 -0.0848 -0.0846 -0.0844 -0.0842 -0.0840 -0.0838 -0.0836 -0.0834 -0.0833 -0.0831 -0.0829 -0.0827 -0.0825

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uop = -(c1.*t)-c2

uop = 1×1214

0.0198 0.0198 0.0198 0.0197 0.0197 0.0196 0.0196 0.0195 0.0195 0.0194 0.0194 0.0193 0.0193 0.0193 0.0192 0.0192 0.0191 0.0191 0.0190 0.0190 0.0189 0.0189 0.0188 0.0188 0.0187 0.0187 0.0187 0.0186 0.0186 0.0185

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lam1= c1;
H = 1.5 * uop.^2 + x2op.* lam1;
figure (1)
plot(t, uop)

figure(2)
plot(t,x1op,t,x2op)

figure(3)
plot(t,H)

function F = myfun1(c)
    x10 = 1; 
    x20 = 0; 
    x1f = 0; 
    x2f = 0; 
    
    
    
    u=-(c(1)*c(5))-c(2);
    
    
    
    ud = c(1);
    
    
    
    % the conditions x1(t = 0), x1(t = tf), x2(t = 0), x2(t = tf)
    
    
    
    
    f1 = c(4)-x10;
    f2 =-(c(1)*(c(5)^3))/6 -(c(2)*(c(5)^2))/2 +(c(3)*c(5)) + 1 -x1f;
    f3 = c(3)-x20;
    f4 = -(c(1)*(c(5)^2))/2 -(c(2)*c(5))-x2f;
    f5 = 1.5 *u.^2 -f4.*ud ;% the condition H(tf) = 0
    
    
    
    F = [f1;f2;f3;f4;f5]; % the five unknown system F(C) = 0
end

[Walter Roberson]

clear all
c0 = [0.1; 0.1; 0.1; 0.1; 5]; % Make a starting guess at the solution
opts = optimoptions('fsolve', 'MaxIterations', 1e6, 'MaxFunEval', 1e7);
[c, fval] = fsolve (@myfun1, c0, opts) % Call optimizer

Equation solved, inaccuracy possible.

The vector of function values is near zero, as measured by the value
of the function tolerance. However, the last step was ineffective.

c = 5×1

0.0000 -0.0001 -0.0075 0.9828 210.0654

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fval = 5×1

-0.0172 0.0000 -0.0075 -0.0026 0.0000

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c1 = c(1); c2 = c(2); c3 = c(3); c4 = c(4); c5 = c(5);
t=0:0.01:c5;
x1op = -(c1*(t.^3))/6-(c2*(t.^2))/2+(c3.*t)+c4

x1op = 1×21007

0.9828 0.9827 0.9826 0.9826 0.9825 0.9824 0.9823 0.9823 0.9822 0.9821 0.9820 0.9820 0.9819 0.9818 0.9817 0.9817 0.9816 0.9815 0.9814 0.9814 0.9813 0.9812 0.9811 0.9811 0.9810 0.9809 0.9808 0.9808 0.9807 0.9806

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x2op =-(c1*(t.^2))/2+-(c2.*t)+c3

x2op = 1×21007

-0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075 -0.0075

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uop = -(c1.*t)-c2

uop = 1×21007

1.0e-03 * 0.1026 0.1026 0.1026 0.1026 0.1026 0.1026 0.1025 0.1025 0.1025 0.1025 0.1025 0.1025 0.1025 0.1025 0.1025 0.1024 0.1024 0.1024 0.1024 0.1024 0.1024 0.1024 0.1024 0.1024 0.1023 0.1023 0.1023 0.1023 0.1023 0.1023

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lam1= c1;
H = 1.5 * uop.^2 + x2op.* lam1;
figure (1)
plot(t, uop)

●

figure(2)
plot(t,x1op,t,x2op)

figure(3)
plot(t,H)

function F = myfun1(c)
    x10 = 1; 
    x20 = 0; 
    x1f = 0; 
    x2f = 0; 
    
    
    
    u=-(c(1)*c(5))-c(2);
    
    
    
    ud = c(1);
    
    
    
    % the conditions x1(t = 0), x1(t = tf), x2(t = 0), x2(t = tf)
    
    
    
    
    f1 = c(4)-x10;
    f2 =-(c(1)*(c(5)^3))/6 -(c(2)*(c(5)^2))/2 +(c(3)*c(5)) + 1 -x1f;
    f3 = c(3)-x20;
    f4 = -(c(1)*(c(5)^2))/2 -(c(2)*c(5))-x2f;
    f5 = 1.5 *u.^2 -f4.*ud ;% the condition H(tf) = 0
    
    
    
    F = [f1;f2;f3;f4;f5]; % the five unknown system F(C) = 0
end

Answer 1

[Divyanshu]

Hello @Devapriya,

I think the reason for the incorrect trajectories can be the error 'Solver stopped prematurely'. This MATLAB answer thread addresses a similar issue, just try to follow this and re-run your script.

Hopefully, after modifying the 'MaxIterations' and 'MaxFunctionEvaluations' parameters you would get correct results from the script.