{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 11bc give lya qc lva l ryb rvb r 713a the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lv(a) = l : Ry(b) Rv(b) r (7.1.3a) The solution exists since LY L Q = RY((a) = RY(b) (7.1.3b) b) has full rank m for a well-posed problem. Once c has been determined, the solution of the BVP is determined by (7.1.2a). The procedure requires the solution of the m + 1 vector IVPs (7.1.2b,c). Although this technique seems straight forward, we recall that the stability of BVPs is quite di erent from that for IVPs. An IVP for (7.1.1a) is stable when the eigenvalues of A have non-positive real parts. This is generally not the case for BVPs where the eigenvalues of A for stable problems can have positive and negative real parts. Let's explore a simple example. Example 7.1.1 ( 1], Section 4.2). Consider the second-order linear BVP y ; R2 y = 1 y(0) = y(1) = 0 00 and write this as a rst-order system by letting y2 = y : R 0 y1 = y This leads to the symmetric system 0 0 y = R R y + 1=R 0 0 with the initial and terminal conditions 1 0]y(0) = 0 1 0]y(1) = 0: 2 We'll let the particular solution satisfy =R2 v(0) = ;10 : 0 0 v = R R v + 1=R 0 0 Thus, =R2 v(x) = ;10 : The set of fundamental solutions satis es 0 Y= R R Y 0 Y(0) = I: 0 The matrix 0 A= R R 0 has the eigenvalues = R thus, an IVP would be unstable. We'll have to investigate the stability of the BVP...
View Full Document

{[ snackBarMessage ]}