The Computable Differential Equation Lecture ... - Bruce E. Shapiro
The Computable Differential Equation Lecture ... - Bruce E. Shapiro The Computable Differential Equation Lecture ... - Bruce E. Shapiro
150 CHAPTER 7. DELAY DIFFERENTIAL EQUATIONS y, y ′ , y ′′ , . . . , y (p+1) ; however, we may not even know where all the discontinuities are. For a system with time-dependent delay, Discontinuities can arise because y ′ = f(t, y(t), y(t − τ(t, y))) (7.87) y(t) = g(t), t ≤ t 0 (7.88) • The initial function g(t) may have discontinuities. • The initial function g(t) may not link smoothly to the solution at t = t 0 . This will lead to a discontinuity in y ′′ at the next step, y ′′′ at the following step, and so on. • Discontinuities in the derivatives of f, τ, and g. In particular, one can construct pathological examples where the distance between discontinuities gets smaller and smaller, approaching zero as t → ξ for some number ξ. This vanishing delay can lead one to require infinitely small grid spacing, which is not possible. The subject of discontinuity is subtle and non-trivial. We will not consider the details here (see [3] for details) except to point out that when the delay is not constant they can arise in unexpected places. When the delay is constant, the functions f and g are smooth and the solution matches g at t = 0 then the only discontinuities correspond to multiples of the delay. Definition 7.3. Let y n+1 = k∑ α n,i y n+1−i + h n+1 φ(y n , . . . , y n−k+1 ) (7.89) i=1 be a general multistep method for the initial value problem y ′ = f(t, y), y(t 0 ) = y 0 . Then a continuous extension of the method is a piecewise polynomial interpolant η(t) η(t n + θh n+1 ) = j n+i ∑ n+1 i=1 β n,i (θ)y n+jn−i+1 + h n+1 φ(y n+jn , . . . , y n−in ) (7.90) η(t n ) = y n (7.91) η(t n+1 ) = y n+1 (7.92) that is computed on an interval I = [t n−in , t n+jn+1], where [t n , t n+1 ] ⊂ I. A continuous Runge-Kutta method, for example, is y(t n + θh) = y n + h Y ni = y n + h 2∑ b i (θ)f(t n , Y ni (7.93) i=1 2∑ a ij Y nj (7.94) j=1 Math 582B, Spring 2007 California State University Northridge c○2007, B.E.Shapiro Last revised: May 23, 2007
CHAPTER 7. DELAY DIFFERENTIAL EQUATIONS 151 The order conditions for the continuous extensions of the Runge-Kutta 4-stage method, for example, become (compare with 5.128 through 5.131). For first order: ∑ b i (θ) = θ (7.95) For second order, For third order, i ∑ b i (θ)c i = 1 2 θ2 (7.96) i ∑ b i (θ)c 2 i = 1 3 θ3 (7.97) i ∑ b i (θ)a ij c j = 1 6 θ3 (7.98) For fourth order, the following conditions must also be satisfied: i,j ∑ b i (θ)c 3 i = 1 4 θ4 (7.99) i ∑ b i (θ)c i a ij c j = 1 8 θ4 (7.100) i,j ∑ i,j b i (θ)a ij c 2 j = 1 12 θ4 (7.101) ∑ b i (θ)a ij a jk c k = 1 24 θ4 (7.102) i,j,k A second order interpolant for the traditional Runge-Kutta 4-stage method is given by ( b 1 (θ) = − 1 2 θ + 2 ) θ (7.103) 3 b 2 (θ) = 1 3 θ (7.104) b 3 (θ) = 1 3 θ (7.105) ( 1 b 4 (θ) = 2 θ − 1 θ (7.106) 3) A third order interpolant is given by ( 2 b 1 (θ) = 3 θ2 − 3 ) 2 θ + 1 θ (7.107) b 2 (θ) = (− 2 3 θ + 1 ) θ 2 (7.108) c○2007, B.E.Shapiro Last revised: May 23, 2007 Math 582B, Spring 2007 California State University Northridge
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CHAPTER 7. DELAY DIFFERENTIAL EQUATIONS 151<br />
<strong>The</strong> order conditions for the continuous extensions of the Runge-Kutta 4-stage<br />
method, for example, become (compare with 5.128 through 5.131). For first order:<br />
∑<br />
b i (θ) = θ (7.95)<br />
For second order,<br />
For third order,<br />
i<br />
∑<br />
b i (θ)c i = 1 2 θ2 (7.96)<br />
i<br />
∑<br />
b i (θ)c 2 i = 1 3 θ3 (7.97)<br />
i<br />
∑<br />
b i (θ)a ij c j = 1 6 θ3 (7.98)<br />
For fourth order, the following conditions must also be satisfied:<br />
i,j<br />
∑<br />
b i (θ)c 3 i = 1 4 θ4 (7.99)<br />
i<br />
∑<br />
b i (θ)c i a ij c j = 1 8 θ4 (7.100)<br />
i,j<br />
∑<br />
i,j<br />
b i (θ)a ij c 2 j = 1<br />
12 θ4 (7.101)<br />
∑<br />
b i (θ)a ij a jk c k = 1 24 θ4 (7.102)<br />
i,j,k<br />
A second order interpolant for the traditional Runge-Kutta 4-stage method is given<br />
by<br />
(<br />
b 1 (θ) = − 1 2 θ + 2 )<br />
θ (7.103)<br />
3<br />
b 2 (θ) = 1 3 θ (7.104)<br />
b 3 (θ) = 1 3 θ (7.105)<br />
( 1<br />
b 4 (θ) =<br />
2 θ − 1 θ (7.106)<br />
3)<br />
A third order interpolant is given by<br />
( 2<br />
b 1 (θ) =<br />
3 θ2 − 3 )<br />
2 θ + 1 θ (7.107)<br />
b 2 (θ) =<br />
(− 2 3 θ + 1 )<br />
θ 2 (7.108)<br />
c○2007, B.E.<strong>Shapiro</strong><br />
Last revised: May 23, 2007<br />
Math 582B, Spring 2007<br />
California State University Northridge
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