Implicit interval multistep methods for solving the initial value problem

18 M. Jankowska, A. Marciniak where fn = f (tn, yn). Let us note that in general to apply (2.4) we need to solve the system of nonlinear equations with respect to the unknowns y n. After replacing the unknown values y(tn-k), y(tn-k+1), ..., y(tn-1) approximations yn-k, yn-k+1, ..., yn-1 obtained by applying another method (for example by a Runge-Kutta method) and neglecting the error term h ̄γk+1ψ(η, y(η)) we are given the following formula known as the k-step implicit Adams-Moulton method and where (2.3) (2.4) (2.2) Let us choose a positive integer m and select the mesh points t 0 , t 1 , . . . , tm where tn nh for each n = 0,1, . . . , m and h = ξ / m. Assume that an integer k =1,2, . . . will state how many approximations yn_k, yn_k+1,..., yn-1 of the exact solution at the previous k mesh points have to be known to determine the approximation yn at the point tn. As shown in [7] the exact solution of (1) considered on the interval [tn-1, tn] has the form and there exists a constant L > 0 such that for each and all we have fulfilled if the function f is determined and continuous in the set exists and is unique. The theory of ordinary differential equations states that these conditions are where We will assume that the solution of (2.1) and (2.1) subject to an initial condition Implicit Interval Multistep Methods for Solving the Initial Value Problem 19 Let us denote: Δt and Δy sets in which the function f(t, y) is defined, i.e. for n k, k +1,..., m, where (3.1) terval one-step method, for example an interval method of Runge-Kutta type (see [3], [10] or [11]). Then, the implicit interval method of Adams-Moulton type we define as follows F(T ,Y) an interval extension of f(t, y) (for a definition of interval extension see e.g. [4], [13] or [16]), • the function F(T,Y) is defined and continues for all and • the function F(T,Y) is monotonic with respect to inclusion, i.e. • for each and for each there exists a constant L > 0 such that where d(A) denotes the width of A (if A = ( A1, A2, ..., AN) T , then the number d(A) is • the function and • the function Now, after making the above assumptions, we are able to construct the implicit multistep interval method of Adams-Moulton type. First, let us assume that and the intervals Yi such as for i = 1, 2 , . . . , k -1 are known. We can obtain such Yiby applying an indefined by d(A) = is defined for all is monotonic with respect to inclusion. an interval extension of Let us assume that 3. IMPLICIT INTERVAL MULTISTEP METHODS OF ADAMS-MOULTON TYPE 20 M. Jankowska, A. Marciniak where Let us note that (3.2) is a nonlinear interval equation with respect to Yn (n = k, k +1,..., m). It follows that in each step of the method we have to solve an interval equation of the form (3.3) (3.2) for n = k, k +1,..., m, where The formula (3.1) can be written in the equivalent form In particular for a given k we get the following methods: Implicit Interval Multistep Methods for Solving the Initial Value Problem 21 where If we assume that the function G is a contraction mapping, then the well-known fixed-point theorem (see e.g. [9] or [14]) implies that the iteration process (3.4) is convergent to Y*, i.e. for an arbitrary choice of Let us recall that a function G is called a contraction mapping if where p is a metric, and α < 1 denotes a constant. Let us note that using the fact that the equation (3.1) can be written in the form (3.5) (3.6) For the equation (3.5) the iteration process (3.4) is of the form and we usually choose Yn (0) = Yn_1. 4. THE E X A C T SOLUTION VS. I N T E R V A L SOLUTIONS For the method (3.2) we can prove that the exact solution of the initial value problem (2.1) belongs to the intervals obtained by this method. Before that it is convenient to present the lemma. 22 M. Jankowska, A. Marciniak where Because and ti = ih for i = 0,1, . . . , m, we get where follows that From the assumption we have and from the Lemma 1 it (4.5) (4.4) for n = k, k+1,..., m, where Yn = Y ( t n ) are obtained from the method (3.2). Proof. Let us consider the formula (2.2) for n = k. We get Theorem 1. If and for i = 1, 2 , . . . , k-1, then for the exact solution y(t) of the initial value problem (2.1) we have From (2.3), (4.2) and (4.3) the inclusion (4.1) follows immediately. (4.3) But This implies that Proof. Since F(T, Y) is an interval extension of f(t, y), then for each for i = n-k, n k + 1 , ...,n-1, and hence we get the inclusion as follows Moreover, and for each This fact implies that where (4.2) (4.1) for any j = 0,1, . . . , k 1 , k we have where Yi = Y(ti), than Lemma 1. Applying Taylor's formula we have where the constants A, B and C are independent of h. Proof. Substituting (3.3) and (3.6) into (3.2) we have Implicit Internal Multistep Methods for Solving the Initial Value Problem 23 (4.6) Moreover, y'(t) = f(t, y(t)). Since then (5.1) and Yn for n = k, k +1,..., m are obtained from (3.2), then where Theorem 2. If the intervals Yn for n = 0,1, . . . , k-1 are known, i = 0,1, . . . , m, but according to the formula (3.2) this is the interval Yk. This conclusion ends the proof for n = k. In a similar way we show the thesis of this theorem for n = k + 1, k + 2,..., m. • 5. WIDTHS OF I N T E R V A L SOLUTIONS Thus, we have shown that y (tk) belongs to an interval As we assumed, Ψ is an interval extension of ψ . Thus, applying (4.6) and (4.7), we have (4.7) In addition, F(T,Y) is an interval extension of f (t, y), and hence for each and For these reasons we can state that where . Taking into account the above considerations, from the formula (4.5) we get 24 M. Jankowska, A. Marciniak