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Re: How to simplify?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg96224] Re: [mg96161] How to simplify?
*From*: Bob Hanlon <hanlonr at cox.net>
*Date*: Mon, 9 Feb 2009 05:38:36 -0500 (EST)
*Reply-to*: hanlonr at cox.net
S[r_] := 1/2 r BesselJ[1, r];
S2[r_] = y[r] /. DSolve[
y''[r] + 1/r y'[r] + S[r] == BesselJ[0, r], y[r], r][[1]] //
FullSimplify
C[1]*Log[r] + C[2] + (r*BesselJ[1, r])/2 - 1
You can get the desired result by setting the arbitrary constants to the obvious values
S2[r] /. {C[1] -> 0, C[2] -> 1}
(r*BesselJ[1, r])/2
Or more generally,
Reduce[S[r] == S2[r]]
(Log[r] == 0 && C[2] == 1) || (Log[r] != 0 && C[1] == (1 - C[2])/Log[r])
S2[r] /. C[1] -> (1 - C[2])/Log[r]
(r*BesselJ[1, r])/2
Or use boundary conditions that are consistent with the desired results
DSolve[{y''[r] + 1/r y'[r] + S[r] == BesselJ[0, r],
y[1] == (BesselJ[1, 1])/2, y'[1] == BesselJ[0, 1]/2},
y[r], r][[1]] // FullSimplify
{y[r] -> (r*BesselJ[1, r])/2}
Bob Hanlon
---- Aaron Fude <aaronfude at gmail.com> wrote:
=============
Hi,
I'm sorry for totally belaboring this point, but I am having a hard
time getting Mathematica be useful for me in this one respect. The
following code shows that the linear ODE that I am trying to solve has
1/2 r BesselJ[1, r] as the particular solution. DSolve, however,
returns an answer that I'm sure is correct. I tested it - numerically,
the particular part is exactly 1/2 r BesselJ[1, r].
But for someone who is looking for analytical insight, the answer is
not useful. What can be done to simplify the expression so that it
appears as 1/2 r BesselJ[1, r]
S[r_] := 1/2 r BesselJ[1, r];
D[S[r], {r, 2}] + 1/r D[S[r], r] + S[r] // FullSimplify
DSolve[y''[r] + 1/r y'[r] + S[r] == BesselJ[0, r], y[r],
r] // FullSimplify
I will gladly accept the answer "Nothing" and move on to looking for
alternative solutions. Also note, this is just a model problem for me
in preparation for more complicated ones. Last time I got several
responses that said - "if you already know the solution to this
problem, why are you trying to solve it?"
Many thanks in advance,
Aaron
--
Bob Hanlon
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