From: spellucci@mathematik.tu-darmstadt.de (Peter Spellucci)
Subject: Re: error analysis to series soultion of a PDE
Date: 12 Jan 2001 18:36:28 GMT
Newsgroups: sci.math.num-analysis
Summary: Estimating errors in series solutions of a PDE

In article <mjcakm5471zq@forum.mathforum.com>,
 stevechou@home.com (steve) writes:
|> Does anyone know if a series solution to a given PDE(heat equation) is
|> given, how to find the numbers of term such that an error is given.

you must consider a bounded domain, formally select  truncation indices for the
series (a double series, I guess) , express the rest of the series in terms of 
these truncation indices, estimating the terms occuring on the domain from above
and finally solve for the indices, given a precision level delta. 
this has unlike for a single series not an unique solution, you get a 
region of possible indices. In some simple cases you get a series solution
which is expressed with the help of the series expansion of the initial
value. then anything is much simpler. for example, if 
you have 
  u_t=u_xx, 0<x<1 
  u(0,t)=u(1,t)=0
  u(x,0)=u0(x)= \sum_k  gamma_k *sin(k*pi*x)

with the solution 
  u(x,t)= \sum_k exp(-k^2*pi^2*t)*gamma_k*sin(k*pi*x)
then , because t>=0 and |sin(.)|<=1
 theerror in truncating the series at k0 is simply bounded by 
  \sum_{k0+1 to infinity} |gamma_k|

hope this helps
peter
==============================================================================

From: spellucci@mathematik.tu-darmstadt.de (Peter Spellucci)
Subject: Re: Error Analysis
Date: 15 Jan 2001 11:57:50 GMT
Newsgroups: sci.math.num-analysis

In article <zge6udmp3nj6@forum.mathforum.com>,
 z74142000@hotmail.com (Stock Well) writes:
|> Suppose U(x,t)=4/pi*Summation of(1/n
|> *e^[-(n^2)((pi)^2)-1)t]}*sin(n*pi*x) with respect to n from 1, 3 ,
|> 5....to infinity is a solution to a heat equation.  I would like to
|> know how many terms in the series in order to get an accuratcy to
|> 0.001 for all x in [0, 1] and t=0.1

using |sin(...)| <=1 , 1/n<=1/2 for n>=2, n^2>=2n for n>=2 , pi/4>=0.75 
and exp(-(pi^2-1)*0.1)=.41190586443964450917=c 
we get as a sufficient condition    
   n0>=2 and 
   0.00075*2  > =  \sum_{n=n0 to infinity}  (c^2)^n=(c^2)^n0/(1-(c^2))
solving for n0 (as a real first) we get
   n0=3.777.....
hence n0=4.
done
hope that helps
peter