thank you

i hope that made sense. sorry if it doesnt

I am looking for exact solutions to the navier-stokes equations and in this particular part of my analysis I have had to write the similarity function F(eta) as a power series in R, so F(eta)=add(F[n]*R^n,n=0..nmax) then write F[n](eta) as a power series in alpha, F[n](eta)=add(F[n,m](eta)*alpha^m). So I have a double subscript. I will be looking at the differential equation written as a series and equating powers of alpha and R.

Is this what you meant?

> bcs:=n,m->seq(seq(F[n,m](1) = `if`(n=0 and m=0, 1,  0), n=0..nmax), m=0..nmax),
>           seq(seq(F[n,m](-1) = `if`(n=0 and m=0, -1,  0), n=0..nmax), m=0..nmax),
>           seq(seq(D(F[n,m])(1) = 0, n=0..nmax), m=0..nmax),
>           seq(seq(D(F[n,m])(-1) = 0, n=0..nmax), m=0..nmax);

thanks alec

greatly appreciated

thank you robert. greatly appreciated

thanks very much

scrap this! ive sorted this problem now. thank you if anyone took at look at it :)

thank you so much to both of you.

so i guess all i really need differentiating with respect to t is

Xi:=F(eta,t)/(a(t)^2);

where

eta=y/a(t)

thanks so much for all your help. sorry my description wasnt very good. I should've written F(eta,t) depends on eta and t but eta=y/a(t) so it also depends on t from the a(t) as well. x is unimportant. it can be considered as a constant as the differentiation is with respect to t.

thanks again

mj

Thanks thats a great help. Is there a way of making this more efficient. Here is my worksheet

> restart;
> with(plots);
> nmax:=5:
> Order:=nmax+1:

We will look at the symmetrical case of suction first. So at u=0, v=-v_w at y=a and u=0, v=v_w at y=-a.
Start with the Navier-Stokes Equations rewritten as a differential equation (de1)

> de1:=diff(F(eta,R),eta$4)-R*(diff(F(eta,R),eta)*diff(F(eta,R),eta$2)-F(eta,R)*diff(F(eta,R),eta$3)):

Write F in a series of powers of R

> S1:=series(eval(de1,F(eta,R)=add(F[k](eta)*R^k,k=0..nmax)),R):

Extract the individual equations for each power upto R^nmax

> des1:=[seq(coeff(S1,R,k),k=0..nmax)]:

Define the boundary conditions

> bcs:=k->(F[k](-1)=`if`(k=0,-1,0),F[k](1)=`if`(k=0,1,0),D(F[k])(-1)=0,D(F[k])(1)=0):

Solve the boundary value problem for each power

> for k from 1 to nmax+1 do
>       gsol:=unapply(rhs(dsolve(des1[k],useint)), eta):
>       F[k-1]:=subs(solve(eval({bcs(k-1)},F[k-1]=gsol)),eval(gsol));
> end do:

F''(1)+3

> c:=eval(diff(add(F[k](eta)*R^k,k=0..nmax),eta$2),eta=1)+3:
> AS:=Sum(a[n]*R^n,n=1..nmax):
> sol:=solve(series(AS,R,nmax)=A,R):
> for k from 1 to nmax do
>       a[k]:=coeff(c,R,k);
> end do:
> sol:=convert(sol,polynom);

i need it to work upto about nmax:=30 but anything past nmax:=8 takes too long to compute on my computer.

Thanks

i want to plot the outcome of this. i have all the coefficients a[n] but it wont plot becuase i have the order term in there O(A^4). how can i remove this?

thank you

thank you so much. i'm so greatful

mj

yes it is the same. i do apologise. all the answers were very helpful and greatly appreciated but i still couldnt get it to work past nmax=15. i was just getting very frustrated not being able to do anything about it. guess i cant. once again im very sorry.

mj