Re: Questionable solution from DSolve
- To: mathgroup at smc.vnet.net
- Subject: [mg63646] Re: Questionable solution from DSolve
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Mon, 9 Jan 2006 04:48:22 -0500 (EST)
- Organization: The Open University, Milton Keynes, UK
- References: <dpqjr1$2hq$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
dkjk at bigpond.net.au wrote:
> Hi group,
>
> Consider the differential equation
>
> D[f[r,z], {r, 2}] + (1/r)D[f[r, z], {r, 1}] - (kz^2 + 1^2)f[r, z] ==
> g[r,z]
>
> With the RHS set to 0, the solution is simply
>
> BesselJ[0, I*r*Sqrt[kz^2+1]]C[1][z] + BesselY[0,
> -I*r*Sqrt[kz^2+1]]C[2][z]
>
> as expected. But if I put g[r,z] = - Exp[-a * r^2 - b * z^2]
>
> I obtain
>
> BesselJ[0, I*r*Sqrt[kz^2+1]]C[1][z] + BesselY[0,
> -I*r*Sqrt[kz^2+1]]C[2][z]
>
> plus some integral with 1 in the lower terminal and r in the upper
> terminal. This is very odd, since the solution is not even
> dimensionally correct! The variable r can be considered to have units
> of metres whereas unity is dimensionless. You might claim that the
> original differential equation is not dimensionally corect to begin
> with, but multiply any of the terms by arbitrary constants and you
> still end up with unity in the lower terminal.
>
> Is this solution rubbish, or have I overlooked something ?
>
> Thanks in advance.
>
> James
>
> DSolve[D[f[r,
> z], {r, 2}] + (1/r)D[f[r, z], {r, 1}] - (
> kz^2 + 1^2)f[r, z] == -Exp[-a*r^2 - b*z^2], f[r, z], {r, z}]
>
Hi James,
The integral you can see with with variable names of the form K$nnn...
are integrals that cannot be carried out: "Solutions given by DSolve
sometimes include integrals that cannot be carried out explicitly by
Integrate. Dummy variables with local names are used in such integrals
(Online Help -> Built-in Functions -> DSolve)." For instance, check out
the example of an Abel-type equation in the example section.
In[1]:=
$Version
Out[1]=
"5.2 for Microsoft Windows (June 20, 2005)"
In[2]:=
eqn=D[f[r,z],{r,2}]+(1/r)D[f[r,z],{r,1}]-(k z^2+1^2)f[r,z]\[Equal]g[r,z]
Out[2]=
(1,0)
2 f [r, z] (2,0)
-(1 + k z ) f[r, z] + ------------ + f [r, z] ==
r
g[r, z]
In[3]:=
DSolve[eqn /. g[r, z] -> 0, f, {r, z}]
Out[3]=
{{f -> Function[{r, z},
2
BesselJ[0, I r Sqrt[1 + k z ]] C[1][z] +
2
BesselY[0, -I r Sqrt[1 + k z ]] C[2][z]]}}
In[4]:=
DSolve[eqn /. g[r, z] -> -Exp[(-a)*r^2 - b*z^2], f,
{r, z}]
Out[4]=
{{f -> Function[{r, z},
2
BesselY[0, -I r Sqrt[1 + k z ]]
2 2
1 -a K$6369 - b z
Integrate[-(-) E K$6369 Pi
2
2
BesselJ[0, I K$6369 Sqrt[1 + k z ]],
{K$6369, 1, r}] +
2
BesselJ[0, I r Sqrt[1 + k z ]]
2 2
1 -a K$222 - b z
Integrate[- E K$222 Pi
2
2
BesselY[0, -I K$222 Sqrt[1 + k z ]],
{K$222, 1, r}] +
2
BesselJ[0, I r Sqrt[1 + k z ]] C[1][z] +
2
BesselY[0, -I r Sqrt[1 + k z ]] C[2][z]]}}
Best regards,
/J.M.