The Computable Differential Equation Lecture ... - Bruce E. Shapiro
The Computable Differential Equation Lecture ... - Bruce E. Shapiro The Computable Differential Equation Lecture ... - Bruce E. Shapiro
158 CHAPTER 8. BOUNDARY VALUE PROBLEMS and hence, since q = 0, ⎛ ⎜ y(t) = c 1 ⎝ where the parameter vector c = ( c 1 equation 8.16. Substituting 8.16 into 8.28, If we define the matrix then t 2 + 1 2t t 2 − 1 2t 2 ⎞ ⎛ ⎟ ⎜ ⎠ + c 2 ⎝ t 2 − 1 2t t 2 + 1 2t 2 ⎞ ⎟ ⎠ (8.30) c 2 ) T is determined by the boundary conditions, y(a) = Y (a)c (8.31) y(b) = Y (b)c − Y (b) ∫ b a Y −1 (s)q(s)ds (8.32) d = B a y(a) + B b y(b) (8.33) = (B a Y (a) + B b Y (b))c − B b Y (b) ∫ b a Y −1 (s)q(s)ds (8.34) Q = B a Y (a) + B b Y (b) (8.35) c = Q −1 [d − B b Y (b) ∫ b a ] Y −1 (s)q(s)ds (8.36) Example 8.3. Solve the boundary value problem considered in section 8.1 using this formalism. Solution. We have a = 1 and b = 2, so ( ) 1 0 Y (a) = ; Y (b) = 0 1 ( 5/4 ) 3/4 3/8 5/8 (8.37) and hence (see example 8.1), ( ) ( ) 1 0 1 0 Q = + 0 0 0 1 Hence Q −1 = ( 0 0 1 0 ) ( ) 5/4 3/4 = 3/8 5/8 ( 1 0 ) −5/3 4/3 and therefore (since q = 0 for this problem) ( ) c = Q −1 1 0 d = = −5/3 4/3 ( ) 1 −1/3 ( ) 1 0 5/4 3/4 (8.38) (8.39) (8.40) The solution of the differential equation is then (we only state the first component since that is all we are really interested in, as it originated in a 2nd order equation): u(t) = t2 + 1 2t − 1 t 2 − 1 = t2 + 2 3 2t 3t (8.41) Math 582B, Spring 2007 California State University Northridge c○2007, B.E.Shapiro Last revised: May 23, 2007
CHAPTER 8. BOUNDARY VALUE PROBLEMS 159 Theorem 8.1. If the matrix Q = B a Y (a) + B b Y (b) is invertible then the solution to y ′ = A(t)y + q(t) (8.42) B a y(a) + B b y(b) = d (8.43) exists and is given by [ y(t) = Y (t) c + ∫ t c = Q −1 [d − B b Y (b) a ] Y −1 (s)q(s)ds ∫ b a ] Y −1 (s)q(s)ds (8.44) (8.45) If we define the matrix Φ(t) = Y (t)Q −1 then y(t) = Φ(t) + ∫ b a G(t, s)(q)(s)ds (8.46) where G(t, s) is the Green’s function of the differential equation { Φ(t)B a φ(a)Φ −1 (s), s ≤ t G(t, s) = −Φ(t)B b Φ(b)Φ −1 (s), s > t (8.47) 8.3 Shooting Revisited We return to solving the general boundary value problem y ′ = y(t, y) (8.48) g(y(a; c), y(b; c)) = 0 (8.49) where we have used the notation y(t, c) to indicate the solution at t that satisfies the initial condition y(a; c) = c. The we define the function H(c) = g(c, y(b, c)) = 0 (8.50) To solve this equation for c we can iterate using Newton’s method c i+1 = c i − ( ) ∂H −1 H(c i ) (8.51) ∂c Using the notation of 8.17, where g = g(u, v), by the chain rule ∂H ∂c = ∂H ∂u ∂u ∂c + ∂H ∂v ∂v ∂c (8.52) = B a Y (a) + B b Y (b) (8.53) = Q (8.54) c○2007, B.E.Shapiro Last revised: May 23, 2007 Math 582B, Spring 2007 California State University Northridge
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CHAPTER 8. BOUNDARY VALUE PROBLEMS 159<br />
<strong>The</strong>orem 8.1. If the matrix Q = B a Y (a) + B b Y (b) is invertible then the solution<br />
to<br />
y ′ = A(t)y + q(t) (8.42)<br />
B a y(a) + B b y(b) = d (8.43)<br />
exists and is given by<br />
[<br />
y(t) = Y (t) c +<br />
∫ t<br />
c = Q −1 [d − B b Y (b)<br />
a<br />
]<br />
Y −1 (s)q(s)ds<br />
∫ b<br />
a<br />
]<br />
Y −1 (s)q(s)ds<br />
(8.44)<br />
(8.45)<br />
If we define the matrix Φ(t) = Y (t)Q −1 then<br />
y(t) = Φ(t) +<br />
∫ b<br />
a<br />
G(t, s)(q)(s)ds (8.46)<br />
where G(t, s) is the Green’s function of the differential equation<br />
{<br />
Φ(t)B a φ(a)Φ −1 (s), s ≤ t<br />
G(t, s) =<br />
−Φ(t)B b Φ(b)Φ −1 (s), s > t<br />
(8.47)<br />
8.3 Shooting Revisited<br />
We return to solving the general boundary value problem<br />
y ′ = y(t, y) (8.48)<br />
g(y(a; c), y(b; c)) = 0 (8.49)<br />
where we have used the notation y(t, c) to indicate the solution at t that satisfies<br />
the initial condition y(a; c) = c. <strong>The</strong> we define the function<br />
H(c) = g(c, y(b, c)) = 0 (8.50)<br />
To solve this equation for c we can iterate using Newton’s method<br />
c i+1 = c i −<br />
( ) ∂H −1<br />
H(c i ) (8.51)<br />
∂c<br />
Using the notation of 8.17, where g = g(u, v), by the chain rule<br />
∂H<br />
∂c = ∂H ∂u<br />
∂u ∂c + ∂H ∂v<br />
∂v ∂c<br />
(8.52)<br />
= B a Y (a) + B b Y (b) (8.53)<br />
= Q (8.54)<br />
c○2007, B.E.<strong>Shapiro</strong><br />
Last revised: May 23, 2007<br />
Math 582B, Spring 2007<br />
California State University Northridge