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Re: I have an operator, How do I DSolve it?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg96213] Re: I have an operator, How do I DSolve it?
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Mon, 9 Feb 2009 05:36:35 -0500 (EST)
  • Organization: The Open University, Milton Keynes, UK
  • References: <gmndhs$s04$1@smc.vnet.net>

In article <gmndhs$s04$1 at smc.vnet.net>,
 Aaron Fude <aaronfude at gmail.com> wrote:

> I have defined a differential operator, but what's the syntax for
> using it in DSolve?
> 
> I have the Bessel operator
> 
> B[m_, f_] := (D[f[#], {#, 2}] +
>     1/# D[f[#], {#, 1}] + (1 - m^2/#^2) f[#]) &
> 
> It's the Bessel operator for now, but I will soon replace it with a
> different one. This is for testing puposes only.
> 
> Now, how do I use it in DSolve?


    DSolve[B[m, S][r] == 0, S[r], r]


> So far I use
> 
> DSolve[S''[r] + 1/r S'[r] + (1 - m^2/r^2) S[r] == 0, S[r], r]
> 
> but that's not a good solution because I have to different expressions
> for the same thing. I want to used B only.

In[1]:= B[m_, 
  f_] := (D[f[#], {#, 2}] + 
    1/# D[f[#], {#, 1}] + (1 - m^2/#^2) f[#]) &

In[2]:= DSolve[B[m, S][r] == 0, S[r], r]

Out[2]= {{S[r] -> BesselJ[m, r] C[1] + BesselY[m, r] C[2]}}

In[3]:= DSolve[S''[r] + 1/r S'[r] + (1 - m^2/r^2) S[r] == 0, S[r], r]

Out[3]= {{S[r] -> BesselJ[m, r] C[1] + BesselY[m, r] C[2]}}

In[4]:= %% === %

Out[4]= True

Regards,
--Jean-Marc


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