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MathGroup Archive 1997

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Re: Boundary Value Problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg7059] Re: Boundary Value Problem
  • From: Troy.D.Goodson at jpl.nasa.gov
  • Date: Sat, 3 May 1997 22:04:51 -0400 (EDT)
  • Organization: Reference.Com Posting Service
  • Sender: owner-wri-mathgroup at wolfram.com

On 24 Apr 1997 05:04:15 -0400, "Rick A. Sprague" <sprague at egr.msu.edu> wrote:
> Hello,
> 
> Would anybody know how to solve this problem?  As a mathcad user for about
> 1 1/2 years, I have been trying to make the transition to MMA by doing
> every problem out of the mathcad manual to learn MMA equivalents.  I have
> come to a problem, though, that seems to have stumped MMA.  Can the
> following problem be solved with out writing a lengthy MMA program?
> 
> y'''''[x]+y[x]==0
> 
> Conditions
> y[0]==0
> y'[0]==7
> 
> y[1]==1
> y'[1]==10
> y''[1]=5

This is a little late, and maybe someone else has solved it, but here is a solution from Matlab

The solution I get is
y(0)=        0
y'(0)=       7.0000
y''(0)=     12.2378
y'''(0)=     9.3279
y''''(0)=  -39.5937



The trick is that this is a linear problem and be transformed to:
y1 = y
y2 = y'
y3 = y''
y4 = y'''
y5 = y''''

y5' = -y

so if x = Transpose[y1 y2 y3 y4 y5], then
dx/dt = Ax, where...

>> A=[0 1 0 0 0;
      0 0 1 0 0;
      0 0 0 1 0;
      0 0 0 0 1;
     -1 0 0 0 0] 

>> A1=A(:,1:2)

A1 =

     0     1
     0     0
     0     0
     0     0
    -1     0

>> B=exp(A);
% The solution to the system is the matrix exponential
% B = e^(A*t)
%
% The TPBVP can be solved if we break up B:
%
%  B = [ B1   B2   ]     (B1 is 5x2)
%      [      . . .]     (B2 is 3x3)
%      [      . . .]
%
% we know y1(0), y2(0) but we need y3(0), y4(0), & y5(0)
% so...
% evaluate multiply the
% leftmost 5x2 portion of B with the 2x1 vector of known
% initial conditions
>> B1=B(:,1:2);
>> c=B1*[0; 7];
>> c1=c(1:3,:)

c1 =

   19.0280
    7.0000
    7.0000

>> d1=[1; 10; 5];    
>> e1=d1-c1 

e1 =

  -18.0280
    3.0000
   -2.0000
%
%  now we can solve e1=B2*yo, where
%  yo = Transpose[y3(0) y4(0) y5(0)]
%
>> B2=B(1:3,3:5)

B2 =

    1.0000    1.0000    1.0000
    2.7183    1.0000    1.0000
    1.0000    2.7183    1.0000

>> y2=inv(B2)*e1;
>> y0=[0; 7; y2]

y0 =

         0
    7.0000
   12.2378
    9.3279
  -39.5937

>> B*y0

%
%  this shows that the solution matches the t=1 
%  boundary conditions
%
ans =

    1.0000
   10.0000
    5.0000
  -79.0611
  -11.0280


Troy
http://www.csun.edu/~kg46825/TGoodson.html


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