Math 660 Lecture 4: FDM for evolutionary equations: ODE solvers

Transcription

1 Math 660 Lecture 4: FDM for evolutionary equations: ODE solvers Consider the ODE u (t) = f(t, u(t)), u(0) = u 0, where u could be a vector valued function. Any ODE can be reduced to a first order system, so this is general enough. The ODE solvers are all approximations to tn+1 u(t n+1 ) = u(t n ) + f(s, u(s))ds. t n tn+1 t n f(s, u(s)) will be approximated by data u 0, u 1,..., u n+1. 1 Examples (Sections in Leveque s book.) If we approximate f(s, u(s)) f(t n, u n ), then we have the forward Euler: u n+1 = u n + kf(t n, u n ) f(s, u(s)) f(t n+1, u n+1 ), we have the backward Euler: u n+1 = u n + kf(t n+1, u n+1 ) f(s, u(s)) 1 2 (f(t n, u n )+f(t n+1, u n+1 )), then we have the trapezoidal method: u n+1 = u n + k 2 (f(t n, u n ) + f(t n+1, u n+1 )) Runga-Kutta methods (multi-stage, one-step methods) The general r-stage Runga-Kutta method is Y i = u n + k a ij f(t n + c j k, Y j ), i = 1, 2,..., r u n+1 = u n + k b j f(t n + c j k, Y j ) 1

2 Clearly, Y i is an approximation of the value at t n + c i k. Hence, r a ij = c i. For other coefficients, we can apply the method to f(t, u) = λu. Then, Y i = u n + k a ij λy j, e λk u n = u n + k b j λy j Solving out Y j and expanding both sides, we can determine the coefficients. The most frequently used schemes are RK2, RK3, RK4. RK2 is not unique. Above methods are all one-step method. The consistency is again measured by the local truncation error (LTE) where u n is replaced by u(t n ),: LT E = 1 (LHS RHS) k note that we have divided k here because un+1 u n k is in the same order as the derivative. This is different from the so-called one-step error. The order of the method is the order of the LTE. Direct Taylor expansion shows that the two Euler methods are first order while the trapezoidal method is a second order method. Linear Multistep Methods (LMM) are another class of ODE solvers. The solvers involve the values at several steps. The most frequently used are the Adams methods u n+r = u n+r 1 + k β j f(t n+j, u n+j ). If β r = 0, we have the Adams-Bashforth methods. The local truncation error is defined similarly LT E = 1 k (LHS RHS) with un replaced by true values. 2 Consistency, zero stability and convergence (Chap. 6 in Leveque.) The ODE solvers are said to be consistent if the local truncation error goes to zero as k 0. An ODE solver is convergent if for a problem u = f(t, u) where f is continuous and Lipschitz continuous in u, we have u n u(t ) 0 as k 0 with nk = T, where T is in the largest interval of existence. 2

3 For one step solvers, u n+1 = u n + kψ(u n, t n, k), as long as Ψ is continuous and Lipschitz continuous in u, the solver is stable and then consistent. Here, stable means that the global error introduced by the m-th step error will not be amplified too much. Zero stability of LMMs For linear multi-step methods, we need the notion of zero-stability to ensure convergence. Consider the LMM α j u n+j = k β j f(u n+j, t n+j ). The zero stability considers the stability for k 0. Hence, we can explore the behavior of the scheme for f = 0. Then, we have α j u n+j = 0, which is a linear difference equation. The characteristic equation is ρ = α j ζ j, and the general solution is determined by the roots. If the solution u n < C (f = 0) as n, then it is zero stable. The condition (called the root condition) is Theorem 1. Suppose ζ j are the roots of the characteristic equation. The LMM is zero stable if ζ j 1 and ζ j < 1 if it is repeated. It can be shown that for LMMs: Zero stability + consistency convergence. For one-step LMMs such as forward/backward Euler and trapezoidal method, there is only one root ζ j = 1. They are zero-stable. 3

4 3 Stability region (Chap. 7 in Leveque.) We have seen that zero stability can ensure the convergence. However, the convergence is in the k 0 limit. For some problems, we must choose very small k to get convergence though it is zero stable. To figure out the restrictions on the step size, we need the notion of stability region. Apply the method on the test equation u = λu and define z = kλ. Usually, the method yields for one step method and u n+1 = R(z)u n (α j zβ j )u n+j = 0. for LMMs. The stability region is the set of complex z-values for which the solutions u n is guaranteed to be bounded: u n < C. For one step method, we require R(z) 1. For LMM, we require, to satisfy the root-condition. π = ρ(ζ) zσ(ζ) = (α j zβ j )ζ j Corollary 1. An LMM is zero stable if and only if z = 0 is in the stability region. It is clear that now we should choose k so that kλ falls into the stability region for any eigenvalue with Re(λ) < 0. The method is then stable whenever z falls into the region of stability. Remark 1. The above analysis is reasonable since λ can be understood as the Jacobian at t = t n, and it is general enough. Examples: 4

5 Note that the midpoint method (leapfrog method) u n+1 = u n 1 + 2kf(t n, u n ) is unstable for any finite k but it is zero-stable. u n+1 = u n 1 + 2zu n ζ 2 2zζ 1 = 0. For ζ 1, both roots must have magnitude 1 since their product is 1. ζ 1 = e iθ and ζ 2 = e iθ. z = 1 2 (ζ 1 ζ ) = i sin θ if ζ = eiθ. θ ±π/2 since ζ 1 ζ 2. Hence, the stability region is the open interval from i to i. The stability region of the backward Euler is {z : z 1 1}. u n+1 = u n + zu n+1 u n+1 = 1 1 z un. Hence, ζ = 1 1 z and 1 1 z 1. For the fourth order Runge-Kutta: Y 1 = u n Y 2 = u n zy 1 = ( z)un Y 3 = u n zy 2 = ( z( z))un Y 4 = u n + zy 3 = u n (1 + z z z3 ) u n+1 = u n + z 6 (Y 1 + 2Y 2 + 2Y 3 + Y 4 ) = (1 + z z ! z ! z4 )u n Hence, the stability region is determined by 4 Stiff problems 1 + z z ! z ! z4 1 In the so-called stiff problems, we care about a slowly varying solution while solutions nearby are rapidly varying with much smaller time scales. In some typical physical applications, the transition to the equilibrium solution is fast but the equilibrium solution itself changes slowly. We care the equilibrium solution instead of the fast transition. For stiff problems, designing numerical schemes is challenging since the fast transition corresponds to negative eigenvalues with large absolute value. 5

6 For the method to be stable, we need kλ to fall into the stability region. However, for explicit methods, the intersection of the stability region and negative real axis usually has a finite length. The explicit schemes requires that k to be very small for stiff problems. The issue is that we don t care the fast transition, i.e. we only care the smaller eigenvalues but the eigenvalues for fast transition put restrictions. We hope k 1/ λ slow instead of 1/ λ fast. A scheme is said to be A-stable if its stability region contains the whole left half plane. A scheme is said to be A(α)-stable if the region π α arg(z) π + α lies in the stability region. Clearly, if we use A-stable schemes, we won t face instability even if our k is large. Sometimes, this is not enough since we hope the modes for the fast transitions to damp instead of just being stable. Then, we require a scheme to be L-stable. Consider that a method applied to u = λu and we have u n+1 = R(z)u n. The method is said to be L-stable if it s A-stable and lim z R(z) = 0. Example: The trapezoidal method is A-stable but not L-stable. The backward Euler is L-stable. Actually, for trapezodial, we have u n+1 u n k = 1 2 λ(un + u n+1 ) R(z) = 1 + z/2 1 z/2. Similarly, for backward Euler, we have R(z) = 1 1 z. 6