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Re: Imposing boundary condition at infinity


The basic idea is to apply the BC at some finite but large r.  For the
purpose of this discussion let us write your ODE  as

Ï?''[r] + 2  Ï?[r]/r == α Ï?[r] + β Exp[-γ r^2]

where α, β γ are constants . For large r we can approximate  the
above ODE as

Ï?''[r] == α Ï?[r]

The general solution of the above ODE  that satisfies the BC at
infinity is

Ï?[r] = C[1] Exp[-α^1/2 r]

where I have assumed that α^1/2 is real and postive.  Differentiating
this equation gives

Ï?'[r] = -C[1]α^1/2 Exp[-α^1/2 r].

Thus for large r the following must hold

Ï?'[r] = -α^1/2 Ï?[r]

Hence at some finite but large r say r=r1 the original ODE must satisfy

Ï?'[r1] = -α^1/2 Ï?[r1]

Now you are in a position to solve your original BVP but posed on a
finite domain. Select a valuer for r1 and solve your BVP using the
above as a BC at r=r1.  Then check that the solution does not change
when r1 is increased.



dkjk at wrote:
> Hi all,
> I need to solve a differential equation whose solution has the
> contraint that it tends to zero in the infinite limit. Mathematica will
> not, however, allow \[Phi][Infinity]==0 as one of the contraints. Does
> mathematica have another notation for dealing with this constraint?
> Thanks.
> James
> \!\(DSolve[{\(1\/r\^2\) D[r\^2*D[Ï?[r], r],
>             r] == \(1\/ϵ\) \((Ï?[r] \((\(2*ec^2*n0\)\/\(k*T\))\) +
> \(27*
>             q\)\/\(2*Pi*
>             a\^3*\@\(2*Pi\)\)*
>                 Exp[\(-\(\(9  r^2\)\/\(2  a^2\)\)\)])\), Ï?[0] ==
> \(\(3\ \@2\ \
> q\ λ\)\/Ï?\^\(3/2\) - \(\[ExponentialE]\^\(a\^2\/\(18\ λ\^2\)\)\ q\ \
> \@\(a\^2\/λ\^2\)\ λ\)\/Ï? + \(a\ \[ExponentialE]\^\(a\^2\/\(18\
> λ\^2\)\)\
>                     q\ Erf[a\/\(3\ \@2\ λ\)]\)\/
>         Ï?\)\/\(4\ a\ ϵ\ λ\), Ï?[Infinity] == 0}, Ï?, r]\)

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