F1(u,v)=0,F2(u,v)=0 with 4 parameters a,b,c,d, 3 of which (a,c,d) are given

조회 수: 2 (최근 30일)
I have two equations F1(u,v)=0,F2(u,v)=0 with 4 parameters a,b,c,d, 3 of which (a,c,d) are given. For the range b=0.1:0.1:9, I have to collect all positive real solutions of the system F1(u,v)=0,F2(u,v)=0. How to then plot each solutions obtained for b=0.1:0.1:9
How to do that. Note that the solutions (u,v) should be positive.
a = 1.5; c = 0.4; d = 0.9;
b=0.1:0.1:9;
syms u v
F1= u*(1 - u) - b*u*v/(c*u + 1);
F2= 0.5*b*u*v/(c*u + 1) + u*v/(u + 1) - d*v^2;
E=[F1==0,F2==0];
solve(E,u,v)
plot(u,v)

채택된 답변

Star Strider
Star Strider 2024년 6월 10일
a = 1.5; c = 0.4; d = 0.9;
b=0.1:0.1:9;
syms u v
F1 = @(u,v,b) u*(1 - u) - b*u*v/(c*u + 1);
F2 = @(u,v,b) 0.5*b*u*v/(c*u + 1) + u*v/(u + 1) - d*v^2;
% E=[F1==0,F2==0];
for k = 1:numel(b)
[uv(:,k),fv(k)] = fmincon(@(p)norm([F1(p(1),p(2),b(k)); F2(p(1),p(2),b(k))]), rand(2,1), [],[],[],[], zeros(2,1));
end
Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance. Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance. Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance. Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the value of the step size tolerance and constraints are satisfied to within the value of the constraint tolerance. Local minimum possible. 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format shortG
uv
uv = 2x90
1.9593e-06 0.91147 0.86145 0.8088 0.75463 0.70008 0.6463 0.59434 0.54505 0.49901 0.45659 0.4179 0.38287 0.35132 0.323 0.29762 0.27487 0.25447 0.23614 0.21965 0.20479 0.19135 0.17917 0.16811 0.99999 0.14884 0.14042 0.1327 0.1256 0.11906 0.0012415 0.60404 0.62098 0.63263 0.63888 0.63985 0.63591 0.62762 0.61571 0.60098 0.58423 0.56617 0.54742 0.52845 0.50964 0.49125 0.47345 0.45634 0.44001 0.42445 0.40969 0.3957 0.38246 0.36993 5.7414e-06 0.34686 0.33624 0.32619 0.31666 0.30763
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
fv
fv = 1x90
1.0e+00 * 2.399e-06 2.6284e-09 1.6137e-09 2.9782e-09 5.5332e-09 5.0974e-09 1.6383e-09 1.7269e-09 5.4861e-09 3.6968e-09 2.5587e-09 2.2643e-09 1.7091e-09 2.8115e-09 2.2959e-09 4.0353e-09 5.9438e-09 6.0736e-09 3.0128e-09 3.4053e-09 5.7217e-09 8.3134e-10 3.362e-09 5.8099e-09 7.9969e-06 4.2445e-09 2.3542e-09 6.0049e-09 3.598e-09 1.9788e-09
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
figure
plot(b, uv(1,:), 'DisplayName','u(b)')
hold on
plot(b,uv(2,:), 'DisplayName','v(b)')
hold off
grid
xlabel('b')
ylabel('Parameter Value')
legend('Location','best')
If you want to elimiinate the ‘spike’ discontinuities, use the ismissing function to detect the ‘outlier’ values (that appear to be either 0 or 1 in the ‘uv’ matrix, other functions may also be useful), set them to NaN, and then use the fillmissing function to interpolate them. I leave that to you.
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Atom
Atom 2024년 6월 10일
This is amazing... Thank you greatly.

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