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. > > > > > >