Want to plot these coupling functions r(t) and p(t) (in image) on a graph. I tried using this code and ended up with a blank graph(without errors) . Kindly help. Please note t varies as 2 to 3 with stepping of 0.01 .
z1 = -1.0629;
z2 = 0.5948;
%Code for r(t)
r_int1 = @(s) (2/3 - s).^(-1/2) .* (6 + abs(r(s))+abs(p(s))) ./ (5.*exp(s+4).*(1+abs(r(s))+abs(p(s))));
r_int2 = @(s) (t-s).^(-1/2) .* (6+abs(r(s))+abs(p(s))) ./ (5*exp(s+4).*(1+abs(r(s))+abs(p(s))));
r = @(t) ( z1.*t.^(-1/3)/gamma(1/2) .* integral(r_int1, 0, 2/3) ) + ( (1/gamma(1/2)) .* integral(r_int2, 0, t) );
%Code for p(t)
p_int1 = @(s) (3/2 - s).^(-4/5) .* (1/exp(2*s)) .* (sin(r(s))+sin(p(s)));
p_int2 = @(s) (t - s).^(-4/5) .* (1/exp(2*s)) .* (sin(r(s))+sin(p(s)));
p = @(t) ( z2.*t.^(-3/2)/gamma(1/5) .* integral(p_int1, 0, 3/2) ) + ( (1/gamma(1/5)) .* integral(p_int2, 0, t) );
fp = fplot(r,p,[2 3]);
fp.Marker = '*';
Funtions to be plotted

2 commentaires

Cris LaPierre
Cris LaPierre le 8 Juin 2021
Without seeing your code, it's hard to say why your plot was empty.
Toshith Vats
Toshith Vats le 8 Juin 2021
Kindly check now I have upated

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Réponses (1)

If I don't use fplot, I do get a more descriptive error message. Based on the code you have shared, you are using your function handle r in you computation of r_int1 and r_int2 before it has been defined.
z1 = -1.0629;
z2 = 0.5948;
%Code for r(t)
r_int1 = @(s) (2/3 - s).^(-1/2) .* (6 + abs(r(s))+abs(p(s))) ./ (5.*exp(s+4).*(1+abs(r(s))+abs(p(s))));
r_int2 = @(s) (t-s).^(-1/2) .* (6+abs(r(s))+abs(p(s))) ./ (5*exp(s+4).*(1+abs(r(s))+abs(p(s))));
r = @(t) ( z1.*t.^(-1/3)/gamma(1/2) .* integral(r_int1, 0, 2/3) ) + ( (1/gamma(1/2)) .* integral(r_int2, 0, t) );
%Code for p(t)
p_int1 = @(s) (3/2 - s).^(-4/5) .* (1/exp(2*s)) .* (sin(r(s))+sin(p(s)));
p_int2 = @(s) (t - s).^(-4/5) .* (1/exp(2*s)) .* (sin(r(s))+sin(p(s)));
p = @(t) ( z2.*t.^(-3/2)/gamma(1/5) .* integral(p_int1, 0, 3/2) ) + ( (1/gamma(1/5)) .* integral(p_int2, 0, t) );
% fp = fplot(r,p,[2 3]);
% fp.Marker = '*';
plot(r(2:.01:3),p(2:0.01:3))
Unrecognized function or variable 'r'.

Error in solution (line 5)
r_int1 = @(s) (2/3 - s).^(-1/2) .* (6 + abs(r(s))+abs(p(s))) ./ (5.*exp(s+4).*(1+abs(r(s))+abs(p(s))));

Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);

Error in integralCalc/vadapt (line 132)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);

Error in integralCalc (line 75)
[q,errbnd] = vadapt(@AtoBInvTransform,interval);

Error in integral (line 87)
Q = integralCalc(fun,a,b,opstruct);

Error in solution (line 7)
r = @(t) ( z1.*t.^(-1/3)/gamma(1/2) .* integral(r_int1, 0, 2/3) ) + ( (1/gamma(1/2)) .* integral(r_int2, 0, t) );

6 commentaires

Toshith Vats
Toshith Vats le 8 Juin 2021
Thanks for the help. Possibly, rearranging the order of equations would not help (leading to similar error ), could you please help me out in solving this issue.
Toshith Vats
Toshith Vats le 9 Juin 2021
Any hint regarding the solution for this problem would be helpful.
Cris LaPierre
Cris LaPierre le 9 Juin 2021
The problem is you are using the function r inside your definition of r. It is possible to define recursive functions in MATLAB, but you need to have some sort of exit condition. Did you intend to define a recursive function, or are r(t) and r(s) in fact different functions?
Toshith Vats
Toshith Vats le 9 Juin 2021
They are recursive functions
Cris LaPierre
Cris LaPierre le 9 Juin 2021
I'm not sure I've seen recursive integration before. Perhaps someone else can provide insight on that.
Toshith Vats
Toshith Vats le 11 Juin 2021
Thabnks fot the help

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le 11 Juin 2021

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