Reference no: EM132296307

Question - (Implicit - explicit Euler method.) We write an initial value problem in the form

(5)

i.e., we decompose the right-hand side of the ODE into two parts. With the usual notation, consider the following method for problem (5)

(6) y_n+1 = y_n + hf(t_n+1, y_n+1) hg(t_n, y_n), n = 0, . . . , N - 1,

with y₀ := y⁰. Obviously, method (6) is a combination of the implicit and the explicit Euler methods, and it reduces to them, when g = 0 and f = 0, respectively. Prove that the order of accuracy of the new method is one, equal to the order of the methods we combined to construct it. Assume now that f satisfies the one-sided Lipschitz condition

∀t ∈ [a, b] ∀z, w ∈ R (f(t, z) - f(t, w))(z - w) ≤ 0,

and g satisfies the global Lipschitz condition with constant L. that is

∃L ≥ 0 ∀t ∈ [a, b] ∀z, w ∈ R |g(t, z) - g(t, w)| ≤ L|z - w|.

Prove d convergence (error estimate) of the method.

[Hint: Use the ODE to check that

y(t_n+1) - y(t_n) - hf(t_n+1, y(t_n+1)) - hg(t_n, y(t_n))

= y(t_n+1) - y(t_n) - hy'(t_n+1) - h[G(t_n+1) - G(t_n)]

with G(t) := g(t, y(t)).]

[Comment: In some cases, when the functions f and g exhibit different behaviour, method (6) combines the advantages of both methods, from which it was constructed, without inheriting their drawbacks. For instance, if we use only the explicit Euler method, the constant in the error estimate necessarily depends also on f. On the other hand, if f is, e.g., linear, the computation of y_n+1 in (6) is very easy, while if we use only the implicit Euler method and g is nonlinear, then to advance in time we need to solve a nonlinear equation at every time level.]