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 wellposed 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 nonpositive 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 secondorder linear BVP y ; R2 y = 1 y(0) = y(1) = 0 00 and write this as a rstorder 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...
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 Spring '14
 JosephE.Flaherty
 Boundary value problem, BVPs

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