The symbolic code is not running

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MINATI
MINATI le 4 Nov 2019
Commenté : MINATI le 10 Nov 2019
syms t x a p q r a1 a2 A pr
f(1)=x+p*x^2/2;g(1)=a*x+q*x^2/2;h(1)=1+r*x;
for i=1:5 %(Can I take i=0:5)
fa(i) = subs(f(i),x,t);ga(i) = subs(g(i),x,t);ha(i) = subs(h(i),x,t);
f(i+1) =f(i)+a1*int(int(int((diff(fa(i),t,3)+(fa(i)+ga(i))*diff(fa(i),t,2)+ a1*diff(fa(i),t,1)*(diff(fa(i),t,1)+diff(ga(i),t,1))),t,0,x)));
g(i+1) =g(i)+a1*int(int(int((diff(ga(i),t,3)+(fa(i)+ga(i))*diff(ga(i),t,2)+ a1*diff(ga(i),t,1)*(diff(fa(i),t,1)+diff(ga(i),t,1))),t,0,x)));
h(i+1) =h(i)+pr*a2*int(int((diff(ha(i),t,2)+(fa(i)+ga(i))*diff(ha(i),t,1)+ A*ha(i)*(diff(fa(i),t,1)+diff(ga(i),t,1))),t,0,x));
end
f=f(1)+f(2)+f(3)+f(4)+f(5);
disp(f(i+1))
figure(1)
fplot(x,f) %% (for FIG. a1=1;a2=2;A=1;pr=1;)
  10 commentaires
Walter Roberson
Walter Roberson le 10 Nov 2019
You have triple nested integrals, but you only have bounds for one of the levels, which leads you open to issues about ending up with whatever constant of integration that the routines decide to throw in. Wouldn't it be better to use definite integrals for all of the calculations? At the very least you should be indicating the variable of integration.
MINATI
MINATI le 10 Nov 2019
ok
Thanks Walter
for your interest and time

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