symbolic code ends with an error
显示 更早的评论
Tr = 0.5; M = 1; Kp = 0.1; K = 02; L = 0.5; D = 3; Rd = 0.5; Pr = 2; S1 = 0.01;
Nb = 02; Nt = 1; Le = 1; Kc = 1; m = 0.5; t1 = 0.5 ; s1 = 0.5; Ec = 0.5;
p1 = -2.1501; p2 = -1.0774; p3 = 1.8615; p4 = -1.5575;
t = sym('t'); x= sym('x');
f = zeros(1,1,'sym'); w = zeros(1,1,'sym'); g = zeros(1,1,'sym'); h = zeros(1,1,'sym');
fa = zeros(1,3,'sym'); wa = zeros(1,3,'sym'); ga = zeros(1,3,'sym'); ha = zeros(1,3,'sym');
f(1)= (p1*x^2)/2 + x; w(1) = p2*x - m*p1;g(1) = p3*x - t1 + 1;h(1)= p4*x - s1 + 1;
for i=1:3
fa(i) = subs(f(i),x,t); dfa = diff(fa(i),t,1); d2fa = diff(dfa,t,1);
wa(i) = subs(w(i),x,t); ga(i) = subs(g(i),x,t); ha(i) = subs(h(i),x,t);
dwa = diff(wa(i),t,1); dga = diff(ga(i),t,1); dha = diff(ha(i),t,1);
If1 = int(((M+(1/Kp))*dfa + dfa ^2 - dfa * d2fa - K*wa(i) -L*(ga(i)+D*ha(i))/(1+K)),t,0,t); If2 = int(If1,t,0,t);
f(i+1) = int(If2,t,0,x);
Iw1 = int((dfa*wa(i)- fa(i)*dwa + K*(2*wa(i)+ d2fa)/(1+(K/2))),t,0,t);
w(i+1) = int(Iw1,t,0,x);
Ig1 = - int((3*Rd *(Tr-1)*(1+(Tr-1)*ga(i))^2* dga^2 + Pr*(fa(i)*dga + Nb*dga*dha + Nt*dga^2 + S1*ga(i))+ ...
(1+K)*Ec*Pr* d2fa^2)./(1+Rd*(1+(Tr-1)*ga(i))^3),t,0,t);
g(i+1) = int(Ig1,t,0,x);
Ih1 = int( Pr*Le*Kc*ha(i) - fa(i)*dha + (Nt/Nb)*((3*Rd *(Tr-1)*(1+(Tr-1)*ga(i))^2* dga^2 + Pr*(fa(i)*dga + Nb*dga*dha + Nt*dga^2 + S1*ga(i))+(1+K)*Ec*Pr* d2fa^2)./(1+Rd*(1+(Tr-1)*ga(i))^3)),t,0,t);
h(i+1) = int(Ih1,t,0,x);
% disp(vpa(f(i+1)))
end
disp(vpa(g(2)))
f = f(1)+f(2)+f(3); w= w(1)+w(2)+w(3); g = g(1)+g(2)+g(3); h = h(1)+h(2)+h(3);
11 个评论
Ameer Hamza
2020-4-22
What are you trying to do in this code?
MINATI PATRA
2020-4-23
Integration but in g nd h part one variable is divided that's why code gives NAN, some modification needed.
Ameer Hamza
2020-4-23
Can you show the mathematical equations for this integration?
madhan ravi
2020-4-23
dbstop if naninf
MINATI
2020-4-23
MINATI
2020-4-23
MINATI
2020-4-24
Walter Roberson
2020-4-24
Your code uses S, which is not defined. However, S1 is defined.
MINATI
2020-4-24
Walter Roberson
2020-4-24
On the first round, i=1, your integral for IG1 is undefined if t > (4000*2^(1/3))/3723 + 1000/1241 and is -inf when t is exactly that value. For reasons I am not clear on at the moment, MATLAB returns nan when it notices this.
Furthermore, MATLAB only returns this nan if you integrate the expression (which is in t) from a numeric constant less than (4000*2^(1/3))/3723 + 1000/1241 to t. If you create a new symbolic variable, B, and integrate the expression from 0 to B, then NaN is not returned: an unresolved integration is returned intead.
This suggests a work-around: integrate to a symbolic variable such as B. Since the resulting expression will have an unresolved integral, the result will be of the form int(expression,t,0,B), and since B is independent of t, then in theory the result will be in terms of B instead of in terms of t -- which will then be important in the next step because the next step assumes that t is present in the equation, so you would have to change that. Similar changes are needed for some of the other equations.
MINATI PATRA
2020-4-24
Ok I will try and let u know the proceed
采纳的回答
Walter Roberson
2020-4-24
Closer, but not completely debugged. It slows down a fair bit.
I use the change of variables trick to avoid the nan being immediately generated. However, if you just left it at that, you would continue to encounter nan as further rounds of integration were done. To avoid this, I piecewise() the formula, defining it as 0 outside the valid reason. This leads to bunches of nested formulas with conditional tests.
The whole thing is intended to construct f, g, w, h equations in terms of x, with x not having any defined restriction in range, but with x implicitly having a lower bound of 0. Because of the singularities, you cannot just define formulas without conditions: the formulas would be invalid outside of 0 <= x <= (4000*2^(1/3))/3723 + 1000/1241. So the equations get messy.
Part of the problem here is that MATLAB doesn't find a closed form formula for some of the integrals even restricted to the proper range. This is a weakness in MATLAB; some of them have closed forms, in terms of log and arctan and square roots, but MATLAB is not able to find them.
Tr = 0.5; M = 1; Kp = 0.1; K = 02; L = 0.5; D = 3; Rd = 0.5; Pr = 2; S1 = 0.01;
Nb = 02; Nt = 1; Le = 1; Kc = 1; m = 0.5; t1 = 0.5 ; s1 = 0.5; Ec = 0.5;
p1 = -2.1501; p2 = -1.0774; p3 = 1.8615; p4 = -1.5575;
syms x t
assume(x >= 0 & t>=0);
Iters = 3;
f = zeros(1,Iters+1,'sym');
w = zeros(1,Iters+1,'sym');
g = zeros(1,Iters+1,'sym');
h = zeros(1,Iters+1,'sym');
fa = zeros(1,Iters,'sym');
wa = zeros(1,Iters,'sym');
ga = zeros(1,Iters,'sym');
ha = zeros(1,Iters,'sym');
f(1)= (p1*x^2)/2 + x; w(1) = p2*x - m*p1;g(1) = p3*x - t1 + 1;h(1)= p4*x - s1 + 1;
syms A1 A2 B C E
assume(A1>=0 & A2>=0 & A2>=A1 & B>=0 & C>=0 & E>=0);
for i=1:Iters
fa(i) = subs(f(i),x,t);
dfa = diff(fa(i),t,1);
d2fa = diff(dfa,t,1);
wa(i) = subs(w(i),x,t);
ga(i) = subs(g(i),x,t);
ha(i) = subs(h(i),x,t);
dwa = diff(wa(i),t,1);
dga = diff(ga(i),t,1);
dha = diff(ha(i),t,1);
If1 = int(((M+(1/Kp))*dfa + dfa ^2 - dfa * d2fa - K*wa(i) -L*(ga(i)+D*ha(i))/(1+K)),t,0,A1);
If2 = int(If1,A1,0,A2);
f(i+1) = int(If2,A2,0,x);
Iw1 = int((dfa*wa(i)- fa(i)*dwa + K*(2*wa(i)+ d2fa)/(1+(K/2))),t,0,B);
w(i+1) = int(Iw1,B,0,x);
Ig1_temp1 = (3*Rd *(Tr-1)*(1+(Tr-1)*ga(i))^2* dga^2 + Pr*(fa(i)*dga + Nb*dga*dha + Nt*dga^2 + S1*ga(i))+ ...
(1+K)*Ec*Pr* d2fa^2)./(1+Rd*(1+(Tr-1)*ga(i))^3);
[N1, D1] = numden(Ig1_temp1);
D1_singular = solve(D1, t, 'MaxDegree', 4);
Ig1_temp2 = piecewise(t < D1_singular, Ig1_temp1, 0);
Ig1 = -int(Ig1_temp2, t, 0, C);
g(i+1) = int(Ig1,C,0,x);
Ih1_temp1 = Pr*Le*Kc*ha(i) - fa(i)*dha + (Nt/Nb)*((3*Rd *(Tr-1)*(1+(Tr-1)*ga(i))^2* dga^2 + Pr*(fa(i)*dga + Nb*dga*dha + Nt*dga^2 + S1*ga(i))+(1+K)*Ec*Pr* d2fa^2)./(1+Rd*(1+(Tr-1)*ga(i))^3));
[N2, D2] = numden(Ih1_temp1);
D2_singular = solve(D2, t, 'MaxDegree', 4);
Ih1_temp2 = piecewise(t < D2_singular, Ih1_temp1, 0);
Ih1 = int(Ih1_temp2, t, 0, E);
h(i+1) = int(Ih1,E,0,x);
% disp(vpa(f(i+1)))
end
disp(vpa(g(2)))
f = f(1)+f(2)+f(3); w= w(1)+w(2)+w(3); g = g(1)+g(2)+g(3); h = h(1)+h(2)+h(3);
3 个评论
Walter Roberson
2020-4-24
As I was concerned about, eventually the above will fail in try to determine the singularties of Ig1_temp1 (the equation being integrated to form Ig1). You have a problem that way...
MINATI
2020-4-24
Walter Roberson
2020-4-24
yes, you could use a different computer language. Maple is able to do at least one of the integrals that matlab is not able to do.
Possibly it might be possible to get further if you could put an upper bound on x, or if you could indicate what result you want from the integral if the singularity is crossed.
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Code Performance 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
