This preview shows page 1. Sign up to view the full content.
Unformatted text preview: y(x) 1 eRx (1 ; ) eRx(1 ; ) c + ;10
) e (1 +
2 where
1. (If = e
determine c as ; and the solution as 2R x>0 then the above relation is exact at x = 1.) We would then
11
c R2 ;1
1
y(x) 2eR2 ;1
Rx x > 0: With R large, there is the possibility of having catastrophic growth in the solution even
when is small. In order to demonstrate this, we solved this problem using the MATLAB
RungeKutta procedure ode45. While an explicit RungeKutta code is not the most
e cient solution procedure for this problem, e ciency was not our main concern. The
maximum errors in the solution component y1 with R = 1 10 are reported in Table 7.1.1.
The error has grown by six decades for a tenfold increase in R. Indeed, the procedure
failed to nd a solution of the BVP with R = 100. R ke1k =ky1k
1 2:445 10 8
10 2:940 10 2
Table 7.1.1: Maximum errors in y1 for Example 7.1.1
1 1 ;
; Let's pursue this di culty with a bit more formality. The exact solution of the linear
4 BVP (7.1.1) can be written as (Problem 1...
View
Full
Document
This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.
 Spring '14
 JosephE.Flaherty

Click to edit the document details