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