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Re: DSolve Issues

  • To: mathgroup at smc.vnet.net
  • Subject: [mg88357] Re: [mg88335] DSolve Issues
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sat, 3 May 2008 06:13:58 -0400 (EDT)
  • Reply-to: hanlonr at cox.net

Use exact numbers

Clear[Phi];

eqn1 = Phi''[r] + 2*r^(-1) Phi'[r] + 104479/1000000*Phi[r] == 0;

soln1 = DSolve[{eqn1, Phi[0] == K}, Phi[r], r][[1]] //
  FullSimplify

{Phi[r] -> (1000*K*Sin[(Sqrt[104479]*r)/
              1000])/(Sqrt[104479]*r)}

eqn1 /. NestList[D[#, r] &, soln1[[1]], 2] // Simplify

True

Limit[soln1[[1, 2]], r -> 0]

K

bc1 = soln1[[1, 2]] /. r -> 3 R;

bc2 = D[soln1[[1, 2]], r] /. r -> 3 R // Simplify;

eqn2 = Phi''[r] + 2*Phi'[r]/r - 2831/100*Phi[r] == 0;

soln2 = DSolve[{eqn2, Phi[3 R] == bc1, Phi'[3 R] == bc2}, Phi[r], r][[1]] // 
   ExpToTrig // FullSimplify

{Phi[r] -> 
     (1000*K*Cosh[(1/10)*Sqrt[2831]*
                (r - 3*R)]*
            Sin[(3*Sqrt[104479]*R)/1000])/
         (Sqrt[104479]*r) + 
       (10*K*Cos[(3*Sqrt[104479]*R)/1000]*
            Sinh[(1/10)*Sqrt[2831]*
                (r - 3*R)])/(Sqrt[2831]*r)}

eqn2 /. NestList[D[#, r] &, soln2[[1]], 2] // Simplify

True

(soln2[[1, 2]] /. r -> 3 R) == bc1

True

(D[soln2[[1, 2]], r] /. r -> 3 R) == bc2 // Simplify

True

solnR = Simplify[Reduce[bc1 == 0, R], {K != 0, R > 0, C[1] == 1}][[1]] // 
  ToRules

{R -> (2000*Pi)/(3*Sqrt[104479])}

soln2[[1, 2]] /. solnR

(10*K*Sinh[(1/10)*Sqrt[2831]*
          (r - (2000*Pi)/Sqrt[104479])])/
   (Sqrt[2831]*r)


Bob Hanlon

---- donkorgi12 <ringtailinblacklw02 at gmail.com> wrote: 
> I am solving the following Differential Equation
> 
> Phi''[r]+2*r^(-1)Phi'[r]+0.104479*Phi[r]==0 and Phi[0]==K (some
> constant)   ; kinda has a cos/sin solution
> 
> 2.71828^(-0.323232 \[ImaginaryI] r) ((0.+
>      0. \[ImaginaryI]) + (0.+ 0. \[ImaginaryI]) 2.71828^(
>     0.646465 \[ImaginaryI] r) + (0.+ 1.54687 \[ImaginaryI]) K - (0.+
>       1.54687 \[ImaginaryI]) 2.71828^(0.646465 \[ImaginaryI] r) K)
> 
> all divided by r.
> 
> My problem is that Mathematica is not treating those "zeros".... as
> well zeros. Thus, the solution cannot really be used. In fact, if I
> manually reproduce the solution and remove those "zeros", then the
> solution is fine.
> 
> I have another similar ODE Anyone have any ideas.
> 
> Phi''[r]+2*r^(-1)Phi'[r]- 287.31*Phi[r]==0 and Phi[3R]==0(some
> constant)  ; kinda has a cosh/sinh solution
> 
> As you might have notice I need to match these two solutions and their
> derivatives at some point.  The ratio of which gives me what I desire
> the value of R.
> 
> Yet, Mathematica treats those "zeros" as something else.
> 
> 
> 



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