Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: DSolve

  • To: mathgroup at smc.vnet.net
  • Subject: [mg83175] Re: DSolve
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Wed, 14 Nov 2007 04:41:05 -0500 (EST)
  • Organization: Uni Leipzig
  • References: <fhc3n2$4vv$1@smc.vnet.net>
  • Reply-to: kuska at informatik.uni-leipzig.de

Hi,

from

eqn = D[u[x], {x, 2}] + \[Lambda]^2 (1/4 - x^2) u[x] == 0;
sol = DSolve[eqn, u[x], x]

you see that Mathematica find a solution

ins = eqn /. Flatten[{#, D[#, x], D[#, {x, 2}]} & /@  sol[[1]]] //
   FullSimplify

gives not zero, but

ins /. x -> 2. /. \[Lambda] -> -1.

show you that Mahematica has found a solution.


But it find no
solution for the boundary conditions you gave can't be satisfyed
because from:

s1 = 0 == (u[x] /. sol[[1]]) /. x -> -1/2;
s2 = 0 == (u[x] /. sol[[1]]) /. x -> 1/2;

{s1, s2} /. \[Lambda] -> -2 // N

you see that C[1] must be zero becaus the ParabolicCylinderD[] function
is complex and this gives that also C[2] must be zero too

Regards
   Jens

Raj wrote:
> eqn = D[u[x], {x, 2}] + \[Lambda]^2 (0.25 - x^2) u[x] == 0
> DSolve[{eqn, u[-1/2] == 0, u[1/2] == 0}, u[x], x]
> 
> This returns {{u[x] -> 0}} while another CAS system returns a solution
> in terms of WhittakerW function.
> 
> Am I doing something wrong or is Mathematica not able to solve this
> equation symbolically?
> 
> Thanks,
> 
> Raj
> 
> 


  • Prev by Date: Re: DSolve
  • Next by Date: Choosing preferred functions for Trig Simplification?
  • Previous by thread: Re: DSolve
  • Next by thread: Re: DSolve