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Re: boundary value problem
*To*: mathgroup at smc.vnet.net
*Subject*: [mg30273] Re: boundary value problem
*From*: "Allan Hayes" <hay at haystack.demon.co.uk>
*Date*: Sat, 4 Aug 2001 20:02:02 -0400 (EDT)
*References*: <9kg1bb$ihd$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Richard,
Sometimes DSolve will work
DSolve[{f''[x] == -f[x],f[0]== a,f'[L]== b},f,x]
{{f -> Function[{x}, a*Cos[x] + b*Sec[L]*Sin[x] +
a*Sin[x]*Tan[L]]}}
NDSolve may work in a totally numerical situation (and techniques like the
shooting method may be used)
NDSolve[{f''[x]== -f[x],f[0]==1,f'[2]==3},f,{x,-3,3}]
{{f -> InterpolatingFunction[]}}
A series solution can be tried ( I'll send you my package separately).
<< "haypacks`Calculus`SeriesDSolve`"
n=4;
SeriesDSolve[{f''[x] == -f[x],f[0]==a,f'[L]==b},f,{x,0,n}]
{{f -> Function[x, a - ((6*b + 6*a*L - a*L^3)*x)/
(3*(-2 + L^2)) - (a*x^2)/2 -
((-6*b - 6*a*L + a*L^3)*x^3)/(18*(-2 + L^2)) +
(a*x^4)/24]}}
ff = %[[1,1,2]] ;
ff[0]
a
ff'[L]//Simplify
b
Simplify[f''[x]+f[x]/.f->ff]
(x^3*(24*b + a*(24*L - 4*L^3 - 6*x + 3*L^2*x)))/
(72*(-2 + L^2))
Usually, we will need n to be much bigger, and we may find problems with
singularities.
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565
"Richard S Hsu" <rhsu at U.Arizona.EDU> wrote in message
news:9kg1bb$ihd$1 at smc.vnet.net...
> Hi,
>
> Where I could find a mathematica program to solve
> a boundary value problem :
>
> f''[ x ] == g[x], f[0] = a, f'[L] = b ,
>
> where g[x] is a known function, a,b and L are constants.
>
> Thanks.
>
>
>
>
>
>
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