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復合微差分方程演算法 - MTDDE Algorithm

參考文獻:
ABELL, VLECK  - COMPUTATION OF  MIXED TYPE FUNCTIONAL DIFFERENTIAL BOUNDARY VALUE PROBLEM. AMS 65L10, 65L20, 35K57, 74N99



keys: ODE, BVP, RK

DDE Algorithm For d-dimension DDE, y = f(x,y,y ,y ) with delay and advance part: m p y (x) = y(x s) m y (x) = y(x s) p We consider this linear case y (x) = A(x)y(x)+C(x)y (x)+D(x)y (x)+q(x) m p d where A(x),C(x),D(x) are d×d matrices and q : R → R . Given a mesh π = {x < x < ... < x }, consider on a subelement [x ,x ] in π. By 1 2 N+1 i i+1 implicit Runge-Kutta scheme, choose a set of points {0 ? ρ < ρ < ··· < 1 2 ρ ? 1} we have collocation point x = x +h ρ in [x ,x ] and coe?cients k ij i i j i i+1 α , β with j, = 1,2,...,k such that j j f = f(x ,y ) ij ij ij  k y = y +h α f ij i i j i=1 k y = y +h β f i+1 i i j ij j=1 Now the collocation technic witch interpolate the solution y by x x x x i i y (x) = y (x )L ( ) = f L ( ) ij j ij j π h h i i j j x ξ x k i )dξ withLagrangebasis{L } oforderk. Theny (x) = y(x )+ f L ( j π i ij j j=1 j h x i i Due to fix the delay and advance part, we note t = (x x )/h and x i i t x ψ (t ) = L (t)dt ,j = 1,2,...,k j x j 0 thus y (x) = y(x )+h f ψ (t ) π i i ij j x j and than for each x there is a integer w between 1 and N, x ? [x ,x ] ij ij w w+1 such that y (x ) = y(x s) ? y +h f ψ (t ) m ij ij w w wr r x s ij r    Similarly y (x ) ? y +h f ψ (t ). Now note y , y be preceding p ij v v vr r x +s mij pij ij r approximations, respectively. Thus we have following equations: f = A(x )y +C(x )y +D(x )y +q(x ) i = 1,2,...,N;j = 1,2,...k ij ij ij ij mij ij pij ij y = y +h α f ij i i j i y = y +h f ψ (t ) mij w w wr r x s ij r y = y +h f ψ (t ) pij v v vr r x +s ij r y = y +h β f i = 1,2,...,N i+1 i i j ij j B y +B y = β 1 a a b b Total 4NK +N +1 matched the unknows: f i = 1,2,...,N;j = 1,2,...k ij y ij y mij y pij y i = 1,2,...,N +1 i Total 4NK +N +1 Remark. The w,v are depended by i,j, be careful! For solving the system by change the first equation be q(x ) = A(x )y +C(x )y +D(x )y ij ij i ij w ij v = (h α A(x ),··· ,h α A(x ))f f i j1 ij i jk ij i ij = (h ψ (t )C(x ),··· ,h ψ (t )C(x ))f w 1 x s ij w k x s ij w ij ij = (h ψ (t )D(x ),··· ,h ψ (t )D(x ))f v 1 x +s ij v k x +s ij v ij ij


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