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