       Re: differential equation with buondary conditions

• To: mathgroup at smc.vnet.net
• Subject: [mg30323] Re: differential equation with buondary conditions
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Sat, 11 Aug 2001 03:39:46 -0400 (EDT)
• References: <9kqk0f\$4g6\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Gustavo,

We have,

DSolve[{y''[x]+k y[x]==0, y==0,y[a]==0},y,x]

{{y -> Function[{x}, 0]}}

This is the generic solution, the only other possibility is for special
values of k and/or a (see below)

Leave out the boundary conditions

sol=DSolve[{y''[x]+k y[x]==0},y,x]

{{y -> Function[{x}, C*Cos[Sqrt[k]*x] +
C*Sin[Sqrt[k]*x]]}}

Get the function

ys = y/.sol[]

Function[{x}, C*Cos[Sqrt[k]*x] + C*Sin[Sqrt[k]*x]]

The first boundary condition gives

ys==0

C == 0

Using this on the second boundary condition we get

ys[a]==0/.C->0

C*Sin[a*Sqrt[k]] == 0

This entails C =0 unless Sin[a*Sqrt[k]] =0.
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"Gustavo Seabra" <gseabra at swbell.net> wrote in message
news:9kqk0f\$4g6\$1 at smc.vnet.net...
> Hello,
>
>     I'm trying to make Mathematica solve the following:
>
>                             y''[x] + k y[x] == 0
>
> subject to the boundary conditions:
>         y[x<0] = 0
>         y[x>a] = 0
> so that y[x] != 0 only if 0 < x < a.
> (yes, it's the "particle in a 1-d box problem.)
>
> If I just do: DSolve[{y''[x] + k y[x]  ==  0}, y[x], x]
> it works fine, giving:
> {{y[x] -> C Cos[Sqrt[k] x] + C Sin[Sqrt[k] x]}},
> which is perfectly ok.
>
> But if I include the boundary conditions y == y[a] == 0,
> it doesn't work.
>
>     Any ideas?
> --
> -----------------------------------------------------------------
> Gustavo Seabra - Graduate Student
> Chemistry Department
> Kansas State University
> -----------------------------------------------------------------
>
>

```

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