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Re: Avoiding divide by zero error in NDSolve
*To*: mathgroup at smc.vnet.net
*Subject*: [mg62463] Re: Avoiding divide by zero error in NDSolve
*From*: Peter Pein <petsie at dordos.net>
*Date*: Fri, 25 Nov 2005 02:25:02 -0500 (EST)
*References*: <dm49jj$sul$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
pradeep suresh schrieb:
> Hi all,
> I am having this problem of mathematica not letting me specify the exact
> intial values to a differential equation set. a simple example follws
>
> Simple Mass Balance Equation set in Chemical Systems.
>
> Nsp = 5;
> k1 = 100;
> For[i = 1, i < (Nsp + 1), i++, k2[i] = k1/Nsp];
> eqn1 = {D[M[t], t] == k1, M[0] == 0};
> eqn2 = Table[{D[M[t]y[i][t], t] == k2[i],y[i][0] ==k2[i]/k1},{i,Nsp}];
> var1 = Table[y[i][t], {i, Nsp}];
> var2 = Join[{M}, var1];
> sol=NDSolve[{eqn1,eqn2},var2,{t,0,10}]
>
> Mathematica refuses to solve this system showing error messages of
> encountering 1/0. If i change the initial condition of M to some
> arbitrary nonzero value, it is able to solve this system although now
> the set of equations are practically meaningless for me. Is there any
> way i can retain the meaning of my equations and still get them solved
> by mathematica? Plz advice!
>
> Thanks & Happy thanksgiving,
> Pradeep
>
Hi Pradeep,
eqn1 determines m as m[t]=k1 t. Substituting this into eqn2 leaves you
with _very_ simple equations:
In[1]:=
nsp = 5; k1 = 100;
Do[k2[i] = k1/nsp, {i, nsp}];
msol = DSolve[{m'[t] == k1, m[0] == 0}, m, t][[1]]
Out[3]=
{m -> Function[{t}, 100*t]}
In[4]:=
eqn2 = Simplify[Flatten[
{m'[t]*y[#][t]==k2[#], y[#][0]==k2[#]/k1}& /@ Range[nsp]] /. msol]
Out[4]=
{5*y[1][t] == 1, 5*y[1][0] == 1,
5*y[2][t] == 1, 5*y[2][0] == 1,
5*y[3][t] == 1, 5*y[3][0] == 1,
5*y[4][t] == 1, 5*y[4][0] == 1,
5*y[5][t] == 1, 5*y[5][0] == 1}
so y[i][t]==1/5, i=1..5
Regards,
Peter
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