       Re: shooting method

• To: mathgroup at smc.vnet.net
• Subject: [mg92024] Re: shooting method
• From: dh <dh at metrohm.ch>
• Date: Wed, 17 Sep 2008 04:30:18 -0400 (EDT)
• References: <gapfd4\$o7q\$1@smc.vnet.net>

```
Hi Luca,

there lurk some numerical problems.

First note that the equation is homogeneous, the function is only

determined up to a constant factor. We may therefore choose the starting

slope as we please.

As the Potential goes to infinity near zero the function becomes very

small there. It is therefore numerical difficult to start the

integration there. It is much less problematic to start the integration

at the node and integrate towards zero. We therefore set fun[B]==0 and

arbitrarily fun'[B]==1. And we take care not to set A too far inside the

classical forbidden region, because this is calling for troubles. Here

is an example:

A=3/4;B=3/2;

slope=-1;

fun0[E0_]:=Module[{y,x,eq},

eq={-y''[x]+400((1/x)^12-(1/x)^6) y[x]-E0*y[x] ==0,

y[B]==0,y' [B]==slope};

y/.NDSolve[eq,y,{x,A,B}][]]

fun[E0_?NumericQ]:=fun0[E0][A];

res=FindRoot[fun[E0],{E0,-50}]

fun[E0]/.res

Plot[fun0[E0/.res][x],{x,A,B}]

hope this helps, Daniel

Luca Petrone wrote:

> Dear All,

>

> Using a shooting method, I would like to find the smallest eigenvalue of

>

> -y''[x] + 400((1/x)^12 - (1/x)^6) y[x] ==  E0*y[x]

>

> (the quantum-mechanical steady state energy of an anharmonic oscillator with Lennard-Jones potential)

> with the boundary conditions :

> y[A]==0

> y[B]==0

> with A<<1 and B >> 1

> Can anyone help about how to implement it ?

>

> Thank you

>

> Luca

>

--

Daniel Huber

Metrohm Ltd.

Oberdorfstr. 68

CH-9100 Herisau

Tel. +41 71 353 8585, Fax +41 71 353 8907

E-Mail:<mailto:dh at metrohm.com>

Internet:<http://www.metrohm.com>

```

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