# The symbolic code is not running

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MINATI on 4 Nov 2019
Commented: MINATI on 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;)

Walter Roberson on 10 Nov 2019
I had to stop calculating on the 5th iteration, as it was using 80 gigabytes of memory.
Walter Roberson on 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 on 10 Nov 2019
ok
Thanks Walter