The symbolic code is not running

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;)

[MINATI]

Please anyone have a look

Thanks in advance

[MINATI]

Dear Walter

if you have time to follow, please

[KSSV]

What do you mean by not running?

[Walter Roberson]

The code is running for me, doing everything it is documented to do.

It is slow code, but that is to be expected when each step involves a triple integral of everything that has gone before. And remember that when you compute the same expression multiple times in an expression, it is faster to compute it once and store into a variable and use the variable See for example the way you compute diff(ga(i),t,1) multiple times and remember that diff(ga(i),t,2) involves first computing diff(ga(i),t,1)

%(Can I take i=0:5)

Yes, just remember to add 1 to i in every place that you use i as an index, as it is not possible to index anything at 0. So you would assign to fa(i+1) and to f(i+1+1) and so on.

[MINATI]

Is there any way to speed it up

[Walter Roberson]

Find the expressions that are computed multiple times, such as diff(ga(i),t,1) and store them in variables and use the variables. For example,

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)));

can be changed to

dfa = diff(fa(i),t,1)));
d2fa = diff(dfa,t,1);
d3fa = diff(d2fa,t,1);
dga = diff(ga(i),t,1);
f(i+1) =f(i)+a1*int(int(int((d3fa+(fa(i)+ga(i))*d2fa + a1*dfa*(dfa+dga)),t,0,x)));

This will be more efficient.

However, most of the time will still be spent doing the triple integrals.

Are you trying to do something like a lagrange interpolating polynomial?

[MINATI]

@Walter

Are you trying to do something like a lagrange interpolating polynomial?

No, its Adomian decomposition method to solve coupled ODEs.

it takes so much of time but MATHEMATICA did it quickly.

[Walter Roberson]

I had to stop calculating on the 5th iteration, as it was using 80 gigabytes of memory.

[Walter Roberson]

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]

ok

Thanks Walter

for your interest and time