Loop and NMinimize
- To: mathgroup at smc.vnet.net
- Subject: [mg42914] Loop and NMinimize
- From: "Devesh Roy" <roy at econ.bsos.umd.edu>
- Date: Sat, 2 Aug 2003 04:12:40 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Hi everybody I am a new bie and need some urgent help. I have to finish
this paper and this is the last bit which is holding me back. The
question is related to NMinimize and then looping it.
The problem is as follows.
I have a non linear function F(x,y) =G(x,y)+H(x,y). I minimize this
function F using NMinimize for a given parameter t. Now I need to see
how the solution changes as t changes and that I need to do by writing
in a loop. I also need to get the values of G and H at the mimimum as I
move in the loop.
Once I obtain this all I need is to obtain the plots against the
parameter t.
Hope I am clear. Looking forward to the reply and thank you all.
Thanksd
Devesh
>>> jf alcover <jf.alcover at bdpme.fr> 08/01/03 1:25 AM >>>
Farkhanda <farkhanda_yusaf at yahoo.com> wrote in message
news:<bfrhop$qsr$1 at smc.vnet.net>...
> solve the heat equation using explicit difference approximation
> with the help of programming in mathematica
> let u= u(x,t)
> with the given boundary conditions u[0, t] =u[1, t] = 0 for 0 <=
t <= 1
> and the initial condition u(x,0)=x when 0<=x<=1/2
> u(x,0)=1-x when 1/2 <=x<=1
> The explicit difference scheme is given as
> [u(x,t+delta t)-u(x,t)]/(delta t) =[u(x+delta x,t)-2u(x,t)+u(x-delta
x,t)]/(delta x )^2
> with a condition
> (delta t) / (delta x )^2 <=(1/2)
Here is an example (not thoroughly tested,
not the best way, just my way ! )
of what could be done with Mathematica :
The subscripts used are k for x[k] and n for t[n].
ClearAll[u, x, t];
deltat = 0.01;
deltax = 0.2;
(* we check the stability condition : *)
(deltat) / (deltax )^2
0.25
kmax = Floor[1/deltax]
5
nmax = Floor[1/deltat]
100
(* boundary conditions : *)
u[0, n_Integer /; 0 <= n <= nmax] = 0;
u[kmax, n_Integer /; 0 <= n <= nmax] = 0;
u[k_Integer /; 0 <= k <= kmax/2, 0] := k deltax;
u[k_Integer /; kmax/2 <= k <= kmax, 0] := 1 - k deltax;
(* difference scheme : *)
eq = (u[k, n + 1] - u[k, n])/(deltat) == (u[k + 1, n] - 2u[k, n] +
u[k - 1, n])/(deltax )^2;
(* solve for u[k,n+1] : *)
sol = Solve[eq, u[k, n + 1]]
{{u[k, 1 + n] ->
0.01 (100. u[k, n] + 25. (u[-1 + k, n] - 2. u[k, n] + u[1 + k,
n]))}}
(* we need another subscript because on left hand side n+1 is not
allowed : *)
sol2 = sol[[1]] /. n -> m - 1
{u[k, m] ->
0.01 (100. u[k, -1 + m] +
25. (u[-1 + k, -1 + m] - 2. u[k, -1 + m] + u[1 + k, -1 + m]))}
sol2[[1, 2]] // Simplify
0.25 u[-1 + k, -1 + m] + 0.5 u[k, -1 + m] + 0.25 u[1 + k, -1 + m]
(* we plug this formula into u[k,m] : *)
u[k_Integer, m_Integer /; m < 0] = 0;
u[k_Integer /; k < 0 || k > kmax] = 0;
u[k_Integer /; -kmax <= k <= 2kmax, m_Integer /; m > 0] :=
(* cache is useful for speed *)
u[k, m] = 0.25 u[-1 + k, -1 + m] + 0.50 u[k, -1 + m] + 0.25 u[1 + k, -1
+ m];
(* results for the 5 first layers : *)
Table[u[k, n], {k, 0, kmax}, {n, 0, 5}]
{{0., 0., 0., 0., 0., 0.},
{0.2, 0.2, 0.1875, 0.171875, 0.15625, 0.141602},
{0.4, 0.35, 0.3125, 0.28125, 0.253906, 0.229492},
{0.4, 0.35, 0.3125, 0.28125, 0.253906, 0.229492},
{0.2, 0.2, 0.1875, 0.171875, 0.15625, 0.141602},
{0., 0., 0., 0., 0., 0.}}
I hope this will help you to build your own program.