Numerical Differential Equations. An Undergraduate Course. Ronald H.W. Hoppe

Transcription

1 Numerical Differential Equations An Undergraduate Course Ronald H.W. Hoppe

2 Contents 1. Ordinary Differential Equations Initial-value Problems: Theoretical foundations and preliminaries Basic definitions Existence and uniqueness Continuous Dependence on the initial data Elementary examples of numerical integrators Stiff and non-stiff systems One-step methods for initial-value problems Definitions Convergence, consistency, and stability Explicit Runge-Kutta methods Asymptotic expansion of the global discretization error Explicit extrapolation methods Adaptive step size control for explicit one-step methods Multi-step methods Basic definitions Convergence, consistency, and stability Extrapolatory and interpolatory multi-step methods Predictor-corrector methods BDF (Backward Difference Formulas) 51

3 1.4 Numerical integration of stiff systems Linear stability theory A-stability, L-stability, and a(α)-stability Implicit and semi-implicit Runge-Kutta methods Linear multi-step methods and Dahlquist s second barrier A(α)-stability of implicit BDF Numerical solution of differential-algebraic systems Numerical solution of boundary-value problems Theoretical foundations and preliminaries Shooting methods Finite difference approximations Galerkin approximations 77 Literature 80

4 Partial Differential Equations Theoretical foundations and preliminaries Linear second order elliptic boundary value problems Preliminaries Finite difference approximations Sobolev spaces Finite element approximations The heat equation Preliminaries Method of lines and Rothe s method Finite difference approximations Finite element approximations The wave equation Preliminaries Finite difference approximations Finite element approximations 147 Literature 152 Exercises 154

5 Numerical Differential Equations 1 1. Ordinary Differential Equations 1.1 Initial-value Problems: Theoretical foundations and preliminaries Basic definitions An ordinary differential equation is an equation which contains an unknown function in one independent variable along with certain derivatives of that function. The order of the differential equation corresponds to the highest derivative that occurs in the equation. Definition 1.1. (First order system) (i) Assume that g : I D 1 D 2 R d,g = (g 1,...,g d ) T, where I := [a,b] R,D i R d,1 i 2, and y : I R d,y = (y 1,...,y d ) T with y = (y 1,...,y d )T,y i := dy i dx i,1 i d. Assume further that (y(x),y (x)) D 1 D 2,x I. Then (1.1a) g(x,y(x),y (x)) = 0, x I, is called a nonlinear system of ordinary differential equations of first order in implicit form. Afunctiony C 1 (I) d satisfying (1.1a)iscalledasolutionofthesystem of ordinary differential equations of first order. (ii) Let α R d,α = (α 1,...,α d ) T, and assume that y C 1 (D) satisfies (1.1a) and (1.1b) y(a) = α. Then, (1.1a),(1.1b) is called an initial value problem for the system of ordinary differential equations of first order. (iii) Assume that B : I D R d d,b = (B ij ) d i,j=1,d R d, and f : I D R d,f = (f 1,...,f d ) T. Assume further that y : I R d as in (i) with y(x) D,x I. Then (1.2) B(x,y(x)) y (x) = f(x,y(x)), x I, is called a quasilinear system of ordinary differential equations of first order in implicit form. In case B = B(y),f = f(y), the system (1.2) is said to be autonomous. (iv) In the special case B = I we obtain (1.3) y (x) = f(x,y(x)), x I.

6 2 Ronald H.W. Hoppe The system (1.3) is referred to as a system of ordinary differential equations of first order in explicit form. (v) Assume that A,B : I R d d,a = (A ij ) d i,j=1,b = (B ij ) d i,j=1 and f : I R d,f = (f 1,...,f d ) T. Let further y be as in in (iii). Then (1.4a) (1.4b) B(x) y (x) = A(x) y(x)+f(x),x I, y (x) = A(x) y(x)+f(x), x I, are called linear systems of ordinary differential equations of first order in implicit resp. explicit form. Definition 1.2. (n-th order system) (i) Assume that g : I D R d, I := [a,b] R,D R (n+1)d,n N, and y : I R d with y (i) (x) := d i y(x)/dx i,1 i n. Assume further that (y(x),y (1) (x),...,y (n) (x)) T D,x I. Then (1.5a) g(x,y(x),y (1) (x),...,y (n) (x)) = 0, x I, is called a system of ordinary differential equations of n-th order in implicit form. (ii) Assume that α R nd,α = (α 0,,α n 1 ) T,α i R d,0 i n 1. Moreover, suppose that (1.5b) y (i) (a) = α i, 0 i n 1. Then, (1.5a),(1.5b) is called an initial value problem for the system of ordinary differential equations of n-th order. (iii) The definitions of quasilinear resp. linear systems of n-th order and of systems of n-th order in explicit form can be given analogously. In the sequel we will restrict ourselves to systems of first order, since higher order equations can be equivalently written as a first order system as the following result shows. Lemma 1.3. (Equivalence of n-th order equation and first order system) Any ordinary differential equation of n-th order is equivalent to a system of ordinary differential equations of first order. Proof. Without restriction of generality we assume that d = 1 and that the ordinary differential equation of n-th order is explicit, i.e., y (n) (x) = f(x,y(x),y (x),...,y (n 1) (x)), x I.

7 Numerical Differential Equations 3 Setting y i (x) := y (i 1) (x),x I,1 i n, there holds y 1 (x) = y 2(x) y 2(x) = y 3 (x) =, y n 1 (x) = y n(x) y n (x) = f(x,y 1(x),...,y n 1 (x)) i.e., (y 1,...,y n ) satisfies a system of first order Existence and uniqueness We consider an initial-value problem for a system of first order ordinary differential equations in explicit form. The continuity of the right-hand side of the ordinary differential equation is sufficient for the existence of a solution. Theorem 1.4. (Peano s existence theorem) Assume that I := [a,b] R,D R d,f C(I D), and α D. Then, the initial value problem (1.6a) (1.6b) y (x) = f(x,y(x)), x I, y(a) = α has a solution y C 1 (D). However, the continuity of the right-hand side does not guarantee uniqueness as the following example shows: Example 1.5. (Non-uniqueness) We consider the initial-value problem y (x) = y(x), x R, y(2) = 1. If y = y(x),x R is a solution, so is z(x) := y( x), i.e., it suffices to consider positive solutions. Separation of variables results in dy = dx = 2 y = x+c = y(x;c) = (x+c)2. y 4

8 4 Ronald H.W. Hoppe The condition y(2) = 1 yields (C + 2) 2 = 4 = C = 0. Multiple solutions: x 2 /4, x > 0 y(x;a) = 0, a x 0 (a 0) (x a) 2 /4, x < a { x y(x) = 2 /4, x > 0 0, x 0. If the right-hand side additionally satisfies a Lipschitz-condition in its second argument, we do have uniqueness. Theorem 1.6. (Existence and uniqueness theorem of Picard- Lindelöf) Assume that f : I D R d,i := [a,b] R,D R d, and α D. Assume further that (1.7a) f C(I D), and that there exists a constant L > 0 such that for all (x,y i ) I D,1 i 2, it holds (1.7b) f(x,y 1 ) f(x,y 2 ) L y 1 y 2. Then, the initial value problem (1.6a),(1.6b) has one and only one solution Continuous Dependence on the initial data We study the continuous dependence of the solution on the initial data. Lemma 1.7. (Gronwall s lemma) Let Φ : I R,I := [a,b] R,Φ C(I), and assume that Then, it holds Φ(x) α+β x Proof. Let ε > 0. We consider a Φ(s) ds, x I, α 0,β > 0. Φ(x) α exp(β (x a)), x I. Ψ(x) := (α+ε) exp(β (x a)), x I = Ψ (x) = β ψ(x) = ψ(x) = ψ(a)+β x a Ψ(s) ds.

9 We show Obviously, we have Numerical Differential Equations 5 Φ(x) < Ψ(x), x I. Φ(a) α α+ε = Ψ(a). In contradiction to the assertion we assume the existence of x (a,b] such that Φ(x) Ψ(x). Set x 0 := min { x (a,b] Φ(x) = Ψ(x) }. We have Φ(x) Ψ(x),x [a,x 0 ], whence Φ(x 0 ) α+β x 0 Φ(s) ds < α+ε+β x 0 Ψ(s) ds = Ψ(x 0 ) a contradicting the assumption Φ(x 0 ) = Ψ(x 0 ). a Theorem 1.8. (Continuous dependence on the initial data) Under the assumptions of the Theoremof Picard-Lindelöflet y i,1 i 2, two solutions of the system y (x) = f(x,y(x)),x I, with respect to the initial conditions y i (a) = α i. Then, it holds y 1 (x) y 2 (x) α 1 α 2 exp(l(x a)), x I. Proof. In view of we have y i (x) = α i + x y 1 (x) y 2 (x) α }{{} 1 α 2 + =: Φ(x) α 1 α 2 + }{{}}{{} L =: α =: β x a a f(s,y i (s)) ds, 1 i 2, x a y 1 (s) y 2 (s) }{{} =: Φ(s) Gronwall s lemma allows to conclude. f(s,y 1 (s)) f(s,y 2 (s)) ds ds.

10 6 Ronald H.W. Hoppe Elementary examples of numerical integrators We consider an initial-value problem for a scalar ordinary differential equation of first order in explicit form y (x) = f(x,y(x)), x [a,b], y(a) = α. We introduce a partition of the interval I according to I h := {a =: x 0 < x 1 < < x N := b}, N N, with step sizes h k := x k+1 x k,0 k N 1. The integration of the differential equation over the subinterval [x k,x k+1 ] yields (1.8) y(x k+1 ) y(x k ) = x k+1 x k f(x,y(x)) dx. The idea is to approximate the integral on the right-hand side by a quadrature formula. Example 1.9. (Explicit and implicit Euler method) The integral on the right-hand side of (1.8) describes the area enclosed by the graph of the function f(,y( )) and the interval [x k,x k+1 ]. In the explicit Euler method it is approximated by the area of the rectangle [x k,x k+1 ] [0,f(x k,y(x k ))] (cf. Figure 1 left). In the implicit Euler method we approximate the integral by the area of the rectangle [x k,x k+1 ] [0,f(x k+1,y(x k+1 ))] (cf. Figure 1 right). y y f(x,y(x)) f(x,y(x)) x k x k+1 x x k x k+1 x Figure 1. The explicit (left) and the implicit (right) Euler method. We replace the unknown function values y(x k ) and y(x k+1 ) by some approximations y k and y k+1 and thus obtain:

11 Numerical Differential Equations 7 (i) Explicit Euler method: For k = 0,1,,N 1 compute (1.9a) (1.9b) y k+1 = y k +h k f(x k,y k ), y 0 = α. We note that y k+1 can be explicitly computed by the evaluation of the right-hand side in (1.9a). (ii) Implicit Euler method: For k = 0,1,,N 1 compute (1.10a) (1.10b) y k+1 = y k +h k f(x k+1,y k+1 ), y 0 = α. We note that (1.10a) requires the solution of a nonlinear equation, i.e., y k+1 is implicitly given. Example (Implicit trapezoidal rule) In the implicit trapezoidal rule the integral on the right-hand side of (1.8) is approximated by the area of the trapeze as depicted in Figure 2: x k+1 x k f(x,y(x)) dx h k ( ) f(x k,y(x k ))+f(x k+1,y(x k+1 )). y x k f(x,y(x)) x k+1 x Figure 2. The implicit trapezoidal rule. Implicit trapezoidal rule: For k = 0,1,,N 1 compute ) (1.11a) y k+1 = y k +h k (f(x k,y k )+f(x k+1,y k+1 ), (1.11b) y 0 = α.

12 8 Ronald H.W. Hoppe Example (Explicit and implicit midpoint rule) Intheexplicitmidpointrule,thedifferentialequationy (x) = f(x,y(x)) is integrated over the interval [x k,x k+2 ] resulting in y(x k+2 ) y(x k ) = x k+2 x k f(x,y(x)) dx. In the explicit midpoint rule we use a quadrature formula which approximates the integral by the area of the rectangle [x k,x k+2 ] [0,f(x k+1, y(x k+1 ))] (cf. Figure 3 left): x k+2 x k f(x,y(x)) dx (h k +h k+1 ) f(x k+1,y(x k+1 )). Intheimplicitmidpointrule, thedifferentialequationy (x) = f(x,y(x)) is integrated over the interval [x k,x k+1 ] and a quadrature formula is used which approximates the integral by the area of the rectangle [x k,x k+1 ] [0,f(x k +h k /2,y(x k +h k /2))] (cf. Figure 3 right): x k+1 x k f(x,y(x)) dx h k f(x k +h k /2,y(x k +h k /2)). Since we do not want to evaluate the function f at intermediate points, the function value f(x k +h k /2,y(x k +h k /2)) is approximated by the arithmetic mean of the function values with respect to x k and x k+1 : f(x k +h k /2,y(x k +h k /2)) 1 ( ) f(x k,y xk )+f(x k+1,y(x k+1 )). 2 y y f(x,y(x)) f(x,y(x)) x k x k+1 x k+2 x x x k x k+1 Figure 3. The explicit (left) and the implicit (right) midpoint rule.

13 Numerical Differential Equations 9 Again, we replace the unknown function values y(x k ),y(x k+1 ), and y(x k+2 ) by some approximations y k,y k+1, and y k+2. We note that the explicitmidpointrulerequirestwostartvaluesy 0 andy 1. Giveny 0 = α, y 1 canbecomputed, e.g., bytheexpliciteulermethod. Wethusobtain: (i) Explicit midpoint rule: For k = 0,1,,N 2 compute (1.12a) (1.12b) y k+2 = y k +(h k +h k+1 ) f(x k,y k ), y 0 = α, y 1 = α+h 0 f(a,α). We note that y k+2 can be explicitly computed by the evaluation of the right-hand side in (1.12a). (ii) Implicit midpoint rule: For k = 0,1,,N 1 compute (1.13a) y k+1 = y k + h ( ) k f(x k,y k )+f(x k+1,y k+1 ), 2 (1.13b) y 0 = α. We note that y k+1 is implicitly given, i.e., the computation requires the solution of a nonlinear equation. Remark (Comparison of explicit and implicit methods) Since explicit methods only require function evaluations, whereas implicit methods require the solution of nonlinear equations and are thus computationally more expensive, the question is why one should use implicit methods instead of explicit methods. The answer is that some systems of ordinary differential equations require the solution by implicit methods as will be illustrated in the next subsection Stiff and non-stiff systems A system of ordinary differential equations is called stiff, if the solution has components with extremely different growth behavior. Otherwise, it is said to be non-stiff. Example (Stiff system) We consider the following initial-value problem for a linear system of first order ( y 1 (x) ) y 2 (x) = 1 ( ) }{{} =: A ) ( ) 2 =. 0 ( y1 (0) y 2 (0) ( y1 (x) y 2 (x) ), x 0,

14 10 Ronald H.W. Hoppe We diagonalize the matrix A by a similarity transformation TAT, where T := (e 1 e 2 ) with e i,1 i 2, being the orthonormal eigenvectors associated with the eigenvalues λ 1,λ 2 of A: Computation of the eigenvalues of A as the zeroes of the characteristic polynomial det(λi A) = λ 2 1λ+100 = λ 1 = 100, λ 2 = 1. The orthonormal eigenvectors are given by ( 2 ) 1 e 1 = 1, 2 e 2 = Using the transformation matrix ( 2 1 ) 1 2 T := ( we obtain the transformed linear system ( ( ) ) ( ) y1 T = TAT y y1 T 2 y 2 ( ) ( ỹ 1 (x) ỹ 2(x) = 0 1 ( ) ( ) ỹ1 (0) = 2. ỹ 2 (0) 2 It has the solution, ). ( ỹ1 ) = T ỹ 2 = ) ) ( ỹ1 (x) ỹ 2 (x) ( y1 y 2 ),, x 0, ỹ 1 (x) = 2 exp( 100x), ỹ 2 (x) = 2 exp( x). The back transformation ( ) ( ) y1 ỹ1 = T y 2 ỹ 2 results in y 1 (x) = exp( 100x)+exp( x), y 2 (x) = exp( 100x) exp( x). The linear system is stiff, since the solution component exp( 100x) decays much faster than the solution component exp( x). We now consider the numerical solution of the initial-value problem from the example above with respect to an equidistant grid with grid

15 Numerical Differential Equations 11 points x k = kh,k N,h > 0, using the explicit Euler method. The application of this method to the transformed linear system results in ỹ 1,k = ỹ 1,k 1 100hỹ 1,k 1 = (1 100h)ỹ 1,k 1 = (1 100h) k ỹ 1,0, ỹ 2,k = ỹ 2,k 1 hỹ 2,k 1 = The back transformation yields (1 h)ỹ 2,k 1 = (1 h) k ỹ 2,0. y 1,k = (1 100h) k +(1 h) k, y 2,k = (1 100h) k (1 h) k. We want that the approximate solution decays as the solution of the continuous problem, i.e., y 1,k,y 2,k 0 for k. Hence, we must have 1 100h < 1 h < On the other hand, we consider the numerical solution by the implicit Euler method. The application of this method to the transformed linear system gives ỹ 1,k = ỹ 1,k hỹ 1,k = ỹ 1,k = (1+100h) 1 ỹ 1,k 1 = (1+100h) k ỹ 1,0, ỹ 2,k = ỹ 2,k 1 +hỹ 2,k = ỹ 2,k = (1+h) 1 ỹ 2,k 1 = (1+h) k ỹ 2,0. Transforming back, we obtain y 1,k = (1+100h) k +(1+h) k, y 2,k = (1+100h) k (1 h) k. Obviously, we have y 1,k,y 2,k 0 for k without any restriction on the step size h. Remark (Implicit methods for stiff systems) The application of the explicit and implicit Euler method to the stiff system from Example 1.13 shows that the explicit Euler method is much less suited than the implicit Euler method, since it requires the choice of a very small step size to guarantee that the approximate solution behaves in the same way as the exact solution. This does not only yield a significant higher amount of computational work but also may lead to rounding errors. The next sections will show indeed that

16 12 Ronald H.W. Hoppe explicit methods are well suited for non-stiff problems, whereas stiff problems require the application of implicit schemes.

17 Numerical Differential Equations One-step methods for initial-value problems Definitions We consider the initial-value problem (1.14a) (1.14b) y (x) = f(x,y(x)), x I := [a,b], y(a) = α. We assume f C(I D),D R d,α D, and suppose that there exists L 0 such that (1.15) f(x,y 1 ) f(x,y 2 ) L y 1 y 2 for all (x,y i ) I D,1 i 2. Definition (Explicit and implicit one-step methods) Let I h := {x k = a + kh 0 k N},N N, be an equidistant partition of I = [a,b] of step size h = (b a)/n and Φ : I h I h lr d lr d R + R d. Then, the scheme (1.16a) (1.16b) y k+1 = y k +h Φ(x k,x k+1,y k,y k+1,h), 0 k N 1, y 0 = α is called a one-step method with increment function Φ. The one-step method is said to be explicit, if Φ does not depend on y k+1, and is called implicit otherwise. Remark (Interpretation as a difference equation) Setting I h := I h\{x N } and introducing the grid function y h : I h R d, the one-step method (1.16a),(1.16b) can be equivalent written as the difference equation of first order (1.17a) y h (x + h) = y h (x)+h Φ(x,x+h,y h (x),y h (x+h),h), x I h, (1.17b) y h (a) = α Convergence, consistency, and stability Definition (Convergence and order of convergence) The grid function (1.18) e h (x) := y h (x) y(x), x I h, is called the global discretization error. The one-step method (1.16a), (1.16b) is said to be convergent, if it holds (1.19) max e h (x) 0 (h 0). x I h

18 14 Ronald H.W. Hoppe It is said to be convergent of order p, if (1.20) max x I h e h (x) = O(h p ) (h 0). The convergence can be characterized by consistency and stability. Consistency guarantees that the one-step method (1.16a),(1.16b) is a meaningful approximation of the initial-value problem (1.14a),(1.14b). Wecannot expect that theexact solution y of theinitial-value problem satisfies the one-step method, but if we insert the exact solution into the equivalent difference equation, we should have y(x + h) y(x) = h Φ(x,x+h,y(x),y(x+h),h), h h 0 h 0 y (x) f(x,y(x)). Definition (Consistency, order of consistency) The grid function τ h (x) := h 1 (y(x+h) y(x)) Φ(x,x+h,y(x),y(x+h),h), is called the local discretization error. The one-step method (1.16a), (1.16b) is said to consistent with the initial-value problem (1.14a), (1.14b), if (1.21) h τ h (x) 0 (h 0). x I h It is said to be consistent of order p, if (1.22) h x I h τ h (x) = O(h p ) (h 0). Lemma (Sufficient condition for consistency) Assume that (1.23) max τ h (x) 0 (h 0). x I h Then, the one-step method(1.16a),(1.16b) is consistent with the initialvalue problem (1.14a),(1.14b). The condition (1.23) holds true if and only if (1.24) as h 0. max Φ(x,x+h,y(x),y(x+h),h) f(x,y(x)) 0 x I h Proof. The proof is left as an exercise.

19 Numerical Differential Equations 15 Example (Consistency order: explicit Euler method) The explicit Euler method reads y h (x+h) = y h (x) + h f(x,y h (x)), y h (a) = α. Hence, the local discretization error is given by Taylor expansion leads It follows that τ h (x) = h 1 (y(x+h) y(x)) f(x,y(x)). y(x+h) = y(x) + h y (x)+ 1 2 h2 y (x)+o(h 3 ) = h 1 (y(x+h) y(x)) = y (x)+ 1 2 h y (x)+o(h 2 ) = f(x,y(x))+ 1 2 h y (x)+o(h 2 ). τ h (x) = 1 2 h y (x)+o(h 2 ). Hence, if y C 2 (I), the explicit Euler method is consistent of order p = 1. Example (Consistency order: implicit trapezoidal rule) The implicit trapezoidal rule reads y h (x+h) = y h (x)+ h 2 [f(x,y h(x))+f(x+h,y h (x+h))], y h (a) = α. By Taylor expansion we obtain y(x+h) = y(x)+h y (x)+ 1 2 h2 y (x)+ 1 6 h3 y (x)+o(h 4 ) = h 1 (y(x+h) y(x)) = y (x)+ 1 2 h y (x)+ 1 6 h2 y (x)+o(h 3 ) = h 1 (y(x+h) y(x)) = f(x,y(x)) h (f x(x,y(x))+f y (x,y(x)) f(x,y(x)))+ 1 6 h2 y (x)+o(h 3 ), and f(x+h,y(x+h)) = f(x,y(x))+h f x (x,y(x)) + h f y (x,y(x)) y (x)+o(h 2 ) = 1 (f(x,y(x))+f(x+h,y(x+h))) = f(x,y(x)) h (f x(x,y(x))+f y (x,y(x)) f(x,y(x)))+o(h 2 ).

20 16 Ronald H.W. Hoppe It follows that τ h (x) = O(h 2 ), i.e., for y C 3 (I) the implicit trapezoidal rule if of consistency order p = 2. In contrast to the consistency which related the one-step method and the initial-value problem, stability is alone a property of the one-step method. It states the continuous dependence of the solution of the one-step method on its data. We restrict ourselves to the explicit one-step method (1.25a) (1.25b) y h (x+h) = y h (x)+h Φ(x,y h (x),h), y h (a) = α. Definition (Asymptotic stability) Consider the perturbed one-step method ( ) (1.26a) z h (x+h) = z h (x)+h Φ(x,z h (x),h)+σ h (x), (1.26b) z h (a) = α+β h. The explicit one-step method (1.25a),(1.25b) is called asymptotically stable, if there exist h max > 0 and for each ε > 0 a number δ > 0 such that for all perturbations σ h,β h with β h +h x I h (x) σ h (x) < δ for h < h max the solutions of the perturbed one-step method (1.26a), (1.26b) satisfy max (y h z h )(x) +h (D h y h D h z h )(x) < ε, x I h x I h where (D h y h )(x) := h 1 (y h (x+h) y h (x)), x I h. A significant tool in the proof of the continuous dependence of the exact solution on the initial data was Gronwall s lemma (cf. Lemma 1.7). A discrete analog of Gronwall s lemma will play a prominent role in the proof of the asymptotic stability of the one-step method (1.25a),(1.25b).

21 Numerical Differential Equations 17 Lemma (Discrete Gronwall s lemma) Consider a grid function v h : I h R d with v j := v h (x j ),0 j N, and assume that there exist δ 0 and η 0 such that (1.27a) (1.27b) v j+1 δ h v 0 η, hold true. Then we have (1.28) j v k +η, 0 j N 1, k=0 v j η exp(δ (x j a)), 0 j N. Proof. Suppose that v j M,0 j N. By induction we prove that (1.29) v j M δm (x j a) m m! +η m 1 l=0 δ l (x j a) l. l! (i) Induction basis: Obviously, the assertion (1.28) holds true for j = 0,m N, and m = 0,0 j N. (ii) Induction assumption: The assertion (1.28) holds true for some m N and 0 j N. (iii) Induction conclusion: We have v j+1 δ h j (M δm (x k a) m k=0 Using the elementary inequality h j (x k a) l k=0 for m+1 and j +1 we obtain x j+1 a m! v j+1 M δm+1 (x j+1 a) m+1 (m+1)! m 1 +η l=0 δ l (x k a) l ) +η. l! (x a) l dx = (x j+1 a) l+1, l+1 +η m δ l (x j+1 a) l. l! l=0 Theorem (Sufficient condition for asymptotic stability) Assume that the increment function Φ satisfies the following Lipschitz condition: There exist a neighborhood U I D of the graph {(x,y(x)) x I} of the exact solution y of the initial-value problem

22 18 Ronald H.W. Hoppe and numbers h max > 0 and L Φ 0 such that for all 0 < h < h max and all (x,y i ) U,1 i 2, it holds (1.30) Φ(x,y 1,h) Φ(x,y 2,h) L Φ y 1 y 2. Then, for all perturbations σ h,β h with β h +h x I h σ h (x) δ and x I h resp. x I h the solution y h of the unperturbed one-step method (1.25a),(1.25b) and the solution z h of the perturbed one-step method (1.26a),(1.26b) satisfy ( (y h z h )(x) β h +h ) (1.31a) σ h (x) exp(l Φ (x a)), x I h (1.31b) (D h y h D h z h )(x) L Φ (y h z h )(x) + σ h (x). In particular, this implies asymptotic stability of the one-step method (1.25a),(1.25b). Proof. Without restriction of generality let U = I D. We set w h (x) := z h (x) y h (x), x I h. With w j := w h (x j ) it follows that ( w j+1 = w j +h Φ(x j,z j,h)+σ j Φ(x j,y j,h) w j+1 = w 0 +h j k=0 The Lipschitz condition (1.30) yields ( ) σ k +Φ(x k,z k,h) Φ(x k,y k,h). w j+1 β h +h N σ k +L Φ h k=0 ), w 0 = β h = j w k. Then, the discrete Gronwall s lemma implies (1.31a). The assertion (1.31b) is a direct consequence of (1.31a) and k=0 (D h w h )(x) = Φ(x,z h (x),h)+σ h (x) Φ(x,y h (x),h). An important result is that consistency and asymptotic stability of the one-step method (1.25a),(1.25b) implies convergence with the order of convergence being the order of consistency.

23 Numerical Differential Equations 19 Theorem (Consistency and stability imply convergence) Assume that the one-step method (1.25a),(1.25b) is consistent with the initial-value problem (1.14a),(1.14b) and asymptotically stable. Then it holds (1.32a) (1.32b) max (y h y)(x) 0 (h 0), x I h (D h y h D h y)(x) 0 (h 0). x I h Proof. We set z h (x) := y(x),x I h. It follows that σ h (x) = τ h (x) and β h = 0. In view of the consistency, for each ε > 0 there exist h max (ε) > 0 and δ(ε) > 0 such that for 0 < h < h max (ε) we have h x I h τ h (x) < δ(ε). The asymptotic stability implies (1.32a),(1.32b). Theorem (Characterization of convergence) Assume that the one-step method (1.25a),(1.25b) satisfies the Lipschitz condition (1.30). Then, the following two statements are equivalent: (i) The one-step method (1.25a),(1.25b) is convergent and (1.32a), (1.32b) hold true. (ii) The one-step method (1.25a),(1.25b) is consistent with the initialvalue problem (1.14a),(1.14b). For consistent one-step methods there exists h max > 0 such that for all 0 < h < h max the following a priori error estimate holds true: (1.33) (y h y)(x) (h x I h τ h (x) ) exp(l Φ (x a)). If additionally max τ h (x) 0 (h 0) is satisfied, then we have x I h (1.34) max (D h y h D h y)(x) 0 (h 0). x I h Proof. We first show that (ii) implies (i). Theorem 1.24 gives the asymptotic stability of the one-step method and Theorem 1.25 implies convergence in the sense of (1.32a),(1.32b). The a priori estimate (2.90) and (1.34) follow from Theorem 1.24 with z h (x) := y(x) so that σ h (x) = τ h (x) and β h = 0.

24 20 Ronald H.W. Hoppe Next, we show that (i) implies (ii). In view of the convergence, for sufficiently small h the Lipschitz condition (1.30) can be applied to whence Φ(x,y h (x),h) Φ(x,y(x),h), τ h (x) = D h y(x) Φ(x,y(x),h) = (D h )y(x) Φ(x,y(x),h) D h y h (x)+φ(x,y h (x),h) (D h y h D h y)(x) +L Φ (y h y)(x). Multiplication by h and summation over x I h allows to conclude. Remark (Implicit one-step methods) The definition of asymptotic stability can be generalized to implicit one-step methods. The corresponding stability and convergence results hold true as well. Remark (Lipschitz condition for the increment function) If the increment function Φ only involves the right-hand side f of the initial-value problem, the Lipschitz condition (1.30) follows from the corresponding Lipschitz condition for f.

25 1.2.3 Explicit Runge-Kutta methods Numerical Differential Equations 21 TheexplicitEulermethod(1.9a),(1.9b)usesapproximateslopesf(x k,y k ) f(x k,y(x k )) of the graph of the exact solution in x k (cf. Figure 4). y y 2 y 1 α y(x) x 0 x 1 x 2 x Figure 4. Geometric interpretation of the explicit Euler method. In order to increase the accuracy, the idea is to use a suitable linear combination of slopes f(x k,y(x k )) and f(x k +a 21 h,y(x k +a 21 h)),a 21 > 0. This gives rise to the explicit one-step method (1.35a) (1.35b) y k+1 = y k +h [ b 1 f(x k,y k ) + y 0 = α, b 2 f(x k +a 21 h,y k +a 21 h f(x k,y k ) ], k 0, where b i R,1 i 2. The goal is to determine a 21,b 1,b 2 such that the one-step method (1.35a),(1.35b) has the order of consistency p = 2. Assuming f to be sufficiently smooth, Taylor expansion yields (1.36a) (1.36b) y(x+h) y(x) = y (x)+ h h 2 y (x)+o(h 2 ) = f(x,y(x))+ h ( ) f x (x,y(x))+f y (x,y(x))f(x,y(x)) +O(h 2 ), 2 b 1 f(x,y(x))+b 2 f(x+a 21 h,y(x)+a 21 hf(x,y(x))) = (b 1 +b 2 )f(x,y(x)) + ( ) h a 21 b 2 f x (x,y(x))+a 21 b 2 f y (x,y(x))f(x,y(x)) +O(h 2 ). A comparison of the coefficients in (1.36a) and (1.36b) results in b 1 + b 2 = 1, a 21 b 2 = 1 2.

26 22 Ronald H.W. Hoppe Choosing β := b 2 gives b 1 = 1 β, a 21 = 1 2β. Definition (Method of Runge and Heun) The explicit one-step method (1.37a) (1.37b) y k+1 = y k +(1 β) h f(x k,y k ) + β h f(x k + 1 2β h,y k + h 2β f(x k,y k )), k 0, y 0 = α, is called the method of Runge. In particular, for the special choice β = 1 it is called the method of Heun. 2 We now study the question whether by a suitable choice of β we can achieve an order of consistency p = 3. To this end we determine a further term in the Taylor expansion: y(x+h) y(x) h f + h 2 [f x +f y f]+ h2 6 (1 β)f(x,y(x))+βf(x+ h f + h 2 ( f x +f y f )+ h2 8β = y (x)+ h 2 y (x)+ h2 6 y (x)+o(h 3 ) = ( ) f xx +2ff xy +f 2 f yy +f x f y +ffy 2 +O(h 3 ), Again, a comparison of coefficients yields τ h (x) = h2 6 2β,y(x)+ h 2β f(x,y(x))) = ) (f xx +2ff xy +f 2 f yy +O(h 3 ). ( (1 3 ) 4β ) (f xx +2ff xy +f 2 f yy )+f x f y +ffy 2 +O(h 3 ). Hence, the consistency order p = 3 can not be achieved by any choice of β. However, if we choose β = 3, the sum of the coefficients in front 4 of the leading error term of τ h is minimized. The construction of the method of Runge can be generalized and gives rise to explicit Runge-Kutta methods of higher order.

27 Numerical Differential Equations 23 Definition (Explicit s-stage Runge-Kutta method) The explicit one-step method (1.38a) (1.38b) y k+1 = y k + y 0 = α, s b j k j, k 0, j=1 with k 1 := h f(x k,y k ), k 2 := h f(x k +a 21 h,y k +a 21 k 1 ), k 3 := h f(x k +a 31 h+a 32 h,y k +a 31 k 1 +a 32 k 2 ), i 1 i 1 k i := h f(x k +c i h,y k + a ij k j ), c i := a ij,1 i s, j=1 is called an explicit s-stage Runge-Kutta method. It is uniquely determined by the s(s+1)/2 parameters a ij,2 i s,1 j i 1, and b i,1 i s. j=1 An explicit s-stage Runge-Kutta method can be described by the socalled Butcher scheme c 1 = 0 a 21 c 2 a 31 a 32 c s a s1 a s2 a s,s 1 b 1 b 2 b s 1 b s Lemma (Consistency of s-stage Runge-Kutta methods) Under the assumption (1.39) s b i = 1 i=1 the explicit s-stage Runge-Kutta method (1.38a),(1.38b) is consistent with the initial-value problem (1.14a),(1.14b).

28 24 Ronald H.W. Hoppe Proof. For the local discretization error we obtain τ h (x) = y(x+h) y(x) h s i 1 b i f(x+c i h,y(x)+ a ij k j ), i=1 where k j is given as in (1.38a) with y k replaced by y(x). For h 0 it follows that s τ h (x) y (x) b i f(x,y(x)) = y (x) f(x,y(x)) = 0. i=1 The goal is to determine the parameters a ij and b i in such a way that the highest possible consistency order p can be achieved. This can be realized by graph-theoretical concepts (Butcher trees). For details we refer to the textbooks by Butcher and Hairer/Norsett/Wanner. The following table contains the number N(s) of parameters to be determined and the number M(s) of equations to be satisfied as well as the maximal achievable consistency order p(s): j=1 s N s M s p s If M(s) < N(s), the free parameters are chosen such that the leading term in the local discretization error gets minimized. Example (Kutta s third order method) The 4 equations to be satisfied by the 6 unknown parameters are given by b 1 + b 2 + b 3 = 1, b 2 a 21 + b 3 (a 31 + a 32 ) = 1 2, b 2 a b 3 (a a 31 a 32 + a 2 32) = 1 3, b 3 a 32 a 21 = 1 6. The associated Butcher scheme reads

29 Numerical Differential Equations Example (Classical fourth order Runge-Kutta method) The Butcher scheme for the classical fourth order Runge-Kutta method is given by

30 26 Ronald H.W. Hoppe Asymptotic expansion of the global discretization error An asymptotic expansion of the global discretization error is a prerequisite for extrapolation methods that will be treated in the subsequent subsection. We assume that the explicit one-step method (1.25a),(1.25b) has consistency order p. Then, the local discretization error admits the asymptotic expansion (1.40) y(x+h) y(x) h Φ(x,y(x),h) = d p+1 (x) h p+1 +d p+2 (x) h p+2 +. A natural question is whether the global discretization error has a similar asymptotic expansion. Theorem (Theorem of Gragg) Let f C p+k (I D) and let Φ C p+k (I D lr + ),k 1, be the increment function of an explicit one-step method with Φ(x,y,0) = f(x,y),(x,y) I D. Moreover, assume that there exist functions d p+i C k i (I),1 i k, such that the local discretization error satisfies k y(x+h) y(x) h Φ(x,y(x),h) = d p+i (x) h p+i +O(h p+k+1 ), where x I h. Then, the global discretization error admits the asymptotic expansion i=1 k 1 e h (x) = e p+i (x) h p+i +E p+k (x;h) h p+k, x I h. i=0 Here, the coefficient functions e p+i,0 i k 1, are solutions of the linear initial-value problems e p+i(x) = f y (x,y(x)) e p+i (x) d p+i+1 (x), x I, e p+i (a) = 0. Further, there exist h max > 0 and a constant C > 0, independent of h, such that E p+k (x;h) C, x I h, 0 < h < h max. Remark (Implicit one-step methods) Gragg s theorem can be generalized to implicit one-step methods. However, the iterative solution of the associated nonlinear system of equations may perturb the asymptotic expansion.

31 Numerical Differential Equations 27 Example (Asymptotic expansions of the explicit/implicit Euler method) The explicit and the implicit Euler methods (1.9a), (1.9b), and (1.10a),(1.10b) admit asymptotic expansions of the global discretization error in h. Particular interest is on one-step methods that admit an asymptotic expansion of the global discretization error in h 2. Theorem (Theorem of Stetter) Assume that the one-step method y h (x+h) = y h (x)+h Φ(x,x+h,y h (x),y h (x+h),h), x I h, y h (a) = α, is symmetric, i.e., Φ(x,x+h,y h (x),y h (x+h),h) = Φ(x+h,x,y h (x+h),y h (x), h). Under the assumptions of Theorem 1.34 it holds e 2m+1 (x) = 0, x I, i.e., there exists an asymptotic expansion of the global discretization error in h 2. Example (Explicit midpoint rule) Although the explicit midpoint rule (cf. Example 1.11) formally is a two-step method, it can be equivalently written as a system of two one-step methods to which the assumptions of Theorem 1.37 apply. Hence, the explicit midpoint rule admits an asymptotic expansion of the global discretization error in h 2.

32 28 Ronald H.W. Hoppe Explicit extrapolation methods We consider an explicit one-step method (1.25a),(1.25b) which admits an asymptotic expansion of the global discretization error in h γ. Example (Extrapolation in case of an asymptotic expansion in h) In Theorem 1.34 we assume p = 1,γ = 1. For the step sizes h and h/2 we obtain (1.41a) (1.41b) y h (x) y(x) = e 1 (x) h+e 2 (x;h) h 2, y h/2 (x) y(x) = 1 2 e 1(x) h+ 1 4 E 2(x;h) h 2. Multiplication of (1.41b) with 2 and subtraction of (1.41a) yield (2 y h/2 (x) y h (x)) y(x) = O(h 2 ). }{{} =:ŷ h (x) Now, consider the polynomial p 1 P 1 (R) with p 1 (h i ) = y hi (x),1 i 2, where h 1 := h and h 2 := h/2. We obtain p 1 (t) = 2 h (y h(x) y h/2 (x)) (t h 2 )+y h/2(x) = p 1 (0) = 2 y h/2 (x) y h (x) = ŷ h (x). Hence, the construction of the approximation ŷ h (x) of order h 2 corresponds to an extrapolation to the step size h = 0. Example (Extrapolation in case of an asymptotic expansion in h) In Theorem 1.34 we assume p = 2,γ = 2. For the step sizes h and h/2 we obtain (1.42a) (1.42b) y h (x) y(x) = e 2 (x) h 2 +E 4 (x;h) h 4, y h/2 (x) y(x) = 1 4 e 2(x) h E 4(x;h) h 4. Multiplication of (1.42b) with 4 and subtraction of (1.42a) implies 4 y h/2 (x) 3 y h (x) = O(h 4 ) = 4 y h/2 (x) 3 y h (x) y(x) = O(h 4 ). }{{ 3 } =:ŷ h (x)

33 Numerical Differential Equations 29 We consider the polynomial p 1 P 1 (R) in t 2 with p 1 (h i ) = y hi (x),1 i 2, where h 1 := h and h 2 := h/2. It follows that p 1 (t 2 ) = 4 3h 2 (y h(x) y h/2 (x)) (t 2 h2 4 )+y h/2(x) = p 1 (0) = 4 3 y h/2(x) 1 3 y h(x) = ŷ h (x). Again, the construction of the approximation ŷ h (x) of order h 4 corresponds to an extrapolation to the step size h = 0. In practice, more than two step sizes are used for extrapolation: Given a basic step size H > 0, we choose a step size sequence (1.43) characterized by the sequence and compute the approximations h i := H/n i, n i N, i = 1,2,..., F := {n 1,n 2,...}, y(h;h i ) := y hi, i = 1,2,... The associated sequence T i,1 = y(h;h i ),i = 1,2,..., forms the first column of an extrapolation tableau. In case of polynomial extrapolation of Aitken-Neville type we determine a polynomial in h γ such that T ik (h) := a 0 +a 1 h γ +...+a k 1 h (k 1)γ, T ik (h j ) := y(h;h j ), j = i,i 1,...,i k +1, andextrapolatetoh = 0: T ik := T ik (0). Thisgivesrisetotherecursion T ik = T i,k 1 + T i,k 1 T i 1,k 1 n ( i, i k, k 2. n i k+1 ) γ 1 The extrapolation tableau looks as follows: y(h;h 1 ) : T 11 y(h;h 2 ) : T 21 T 22 y(h;h 3 ) : T 31 T 32 T 33 : : y(h;h i ) : T i1 T i2 T i3 T ii

34 30 Ronald H.W. Hoppe Common choices of the step size sequence are given by: (i) Harmonic sequence (ii) Romberg sequence n i = i, i N, F H = {1,2,3,4,5,...}. n i = 2 i, i N, F R = {2,4,8,16,32,...}. (iii) Bulirsch sequence { 3 2 n i = (i 2)/2, i even 2 (i+1)/2, i odd, F B = {2,3,4,6,8,12,16,...}.

35 Numerical Differential Equations Adaptive step size control for explicit one-step methods Lety k+1 beanapproximationcomputedbyanexplicit one-stepmethod of order p: y k+1 = y k +h Φ(x k,y k,h). We are looking for an easily computable a posteriori estimator for the global discretization error e k+1 := y k+1 y(x k+1 ). Lemma (Error estimator) Assume that ŷ k+1 is a more accurate approximation of y(x k+1 ) according to Then ŷ k+1 y(x k+1 ) q y k+1 y(x k+1 ), 0 q < 1. ˆε k+1 := ŷ k+1 y k+1 is an estimator of e k+1 in the sense that 1 1+q ˆε k+1 e k q ˆε k+1. Proof. The assertions can be easily deduced by an application of the right- and left-hand side of the triangle inequality. We now assume that ŷ k+1 is an approximation of y at x k+1 of order p+1, i.e., ŷ k+1 y(x k+1 ) C h p+1. Given a prespecified accuracy ε > 0, for a reasonable new step size ĥ := x k+2 x k+1 we should have (1.44) C ĥp+1. = ρ ε. where 0 < ρ < 1 represents a safety factor. If y k+1 satisfies ˆε k+1 := ŷ k+1 y k+1. = ε. = C h p+1, then C. = ˆε k+1 /h p+1. Inserting into (1.44) and solving for ĥ yields ĥ = h ( ρ ε ˆε k+1 ) 1/(p+1). There are two strategies to determine ŷ k+1 : (i) extrapolation, (ii) embedded Runge-Kutta methods of higher order.

36 32 Ronald H.W. Hoppe We first consider extrapolation: Let y k+1 and ȳ k+1 be the approximations associated with the step sizes h and h/2, i.e., (1.45a) y k+1 y(x k+1 ). = C h p, (1.45b) ȳ k+1 y(x k+1 ). = C ( h 2 )p = 2 p C h p. Subtraction of (1.45b) from (1.45a) yields and hence, y k+1 ȳ k+1. = C h p (1 2 p ) = C h p 2p 1 2 p, C. = 2p 1 2 p h p (y k+1 ȳ k+1 ). If we insert C into (1.45b) and solve for y(x k+1 ), we obtain ŷ k+1 = ȳ k+1 + ȳk+1 y k+1. 2 p 1 Embedded Runge-Kutta methods of higher order are based on Runge- Kutta-Fehlberg methods: Let y k+1 be an approximation of y(x k+1 ) obtained by an explicit s-stage Runge-Kutta method of order p: s y k+1 = y k + b i k i, i=1 i 1 k i = h f(x k +c i h,y k + a ij k j ), c i = i 1 a ij. j=1 We compute ŷ k+1 as the solution of an explicit (s+t)-stage embedded Runge-Kutta method of order p+1: s+t ŷ k+1 = y k + ˆbi k i, i=1 j=1 i 1 k i = h f(x k +c i h,y k + a ij k j ), c i = i 1 a ij. j=1 The Butcher scheme of the embedded(s+t)-stage Runge-Kutta method reads j=1

37 Numerical Differential Equations 33 c 2 a 21 c s a s1 a s,s 1 c s+t a s+t,1 a s+t,s 1 a s+t,s+t 1 y k+1 b 1 b s ŷ k+1 ˆb1 ˆbs ˆbs+t A drawback is that the computation is continued with y k instead of ŷ k. Therefore, in practice one considers Dormand-Prince type embedded Runge-Kutta methods: We proceed as above but continue with ŷ k, i.e., y k+1 = ŷ k + s b i k i, i=1 i 1 k i = h f(x k +c i h,ŷ k + a ij k j ), and use the so-called Fehlberg trick for the reduction of the number of free parameters. For t = 1 the Fehlberg trick works as follows: In the following step x k+1 x k+2 use as In view of ŷ k+1 = ŷ k + s+1 k s+1 = h f(x k +c s+1 h,ŷ k + j=1 j=1 s a s+1,j k j ) j=1 k 1 = h f(x k+1,ŷ k+1 ). ˆbj k j this gives rise to c s+1 = 1, ˆbs+1 = 0, ˆbi = a s+1,i, 1 i s. Example (Dormand/Prince embedded 7-stage Runge- Kutta-Fehlberg method) The Butcher scheme for the Dormand/Prince embedded 7-stage Runge- Kutta-Fehlberg method is as follows:

38 34 Ronald H.W. Hoppe

39 Numerical Differential Equations Multi-step methods Basic definitions In the sequel we will use the grid-point sets I h := { x j = a+j h 0 j N}, h := (b a)/n, I h := { x j I h m j N }, m 2. Definition (Multi-step method) Givenrealnumbersα 0,α 1,,α m,α m 0,afunctionΦ : I h R d(m+1) R + R d and vectors α (0),,α (m 1) R d, the scheme (1.46a) (1.46b) 1 h m α k y j+k = Φ(x j+m,y j,...,y j+m ;h), 0 j N m, k=0 y j = α (j), 0 j m 1, is called a multi-step method. The multi-step method is said to be explicit, if Φ(x,z 0,...,z m ;h) does not depend on z m, and implicit otherwise. Using the grid function y h : I h R d, the multi-step method (1.46a), (1.46b) can be written as a difference equation of m-th order: (1.47a) (1.47b) 1 h m α k y h (x+kh) = Φ(x+mh,y h (x),...,y h (x+mh);h), k=0 y h (a+jh) = α (j), 0 j m 1. Example (Milne-Simpson method) For the scalar first order ordinary differential equation y (x) = f(x,y(x)), x [a,b], integration over [x k,x k+2 ] [a,b] yields y(x k+2 ) y(x k ) = x k+2 x k f(x,y(x)) dx. Approximating the right-hand side by the Simpson rule, we obtain the Milne-Simpson rule y k+2 = y k + h ( ) f(x k,y k )+4 f(x k+1,y k+1 )+f(x k+2,y k+2 ). 3

40 36 Ronald H.W. Hoppe Definition (Linear multi-step methods) If there exist real numbers β 0,β 1,,β m, such that the function Φ is given by m Φ(x,z 0,,z m ;h) = β k f(x k,z k ), x m I h, k=0 the multi-step method (1.46a),(1.46b) is said to be a linear multi-step method. Remark (Start ramp) For j > 1 the determination of the initial vectors α (j),0 j m 1, can be done by a multi-step method of a lower step number. This is called the start ramp. Definition (Characteristic polynomials) For a multi-step method (1.46a),(1.46b) the polynomial m ρ(z) := α k z k k=0 is called the characteristic polynomial of the multi-step method. In case of a linear multi-step method, we can assign another characteristic polynomial associated with the increment function Φ: m σ(z) := β k z k. k=0

41 Numerical Differential Equations Convergence, consistency, and stability Definition (Convergence and convergence order) If y C 1 ([a,b]) is the exact solution of the initial-value problem (1.6a),(1.6b) and y h : I h R d is the approximation obtained by the multi-step method (1.46a),(1.46b), again we denote by e h (x) := y h (x) y(x),x I h, the global discretization error. The solution y h of the multi-step method (1.46a),(1.46b) is said to convergence to the solution y of the initial-value problem, if max x I h e h (x) 0 (h 0). The multi-step method (1.46a),(1.46b) is said to convergence of order p > 0, if there exists a constant C > 0, independent of h, such that (1.48) max e h (x) C h p (h 0). x I h As in the case of one-step methods, the convergence of multi-step methods can be characterized by consistency and stability. Definition (Consistency and order of consistency) Let y C 1 ([a,b]) be the exact solution of the initial-value problem (1.6a),(1.6b). Then, the grid function τ h (x) := 1 m α k y(x (m k)h) Φ(x,y(x mh),,y(x);h), h k=0 τ h (x j ) := y(x j ) α (j), 0 j m 1, is called the local discretization error of the multi-step method (1.46a), (1.46b). The multi-step method is said to be consistent with the initial-value problem, if max τ h(x j ) +h τ h (x) 0 (h 0). 0 j m 1 x I h The multi-step method has the order of consistency p > 0, if there exists a constant C > 0, independent of h, such that (1.49) max τ h(x j ) +h τ h (x) C h p (h 0). 0 j m 1 x I h

42 38 Ronald H.W. Hoppe Theorem (Consistency of multi-step methods) The multi-step method (1.46a),(1.46b) is consistent with the given initial value problem, if (1.50a) (1.50b) (1.50c) α (j) α (h 0), 0 j m 1, ρ(1)y = 0, Proof. The equivalence h Φ(x,y(x mh),,y(x);h) x I h ρ (1)f(x,y(x)) 0 (h 0). α (j) α (h 0) τ h (x j ) 0 (h 0), 0 j m 1, is obvious. Moreover, by Taylor expansion we obtain y(x mh+kh) = y(x mh)+khy (x mh)+o(h 2 ) = τ h (x) = 1 m α k y(x mh) + h k=0 m kα k y (x mh) Φ(x,y(x mh),,y(x);h)+o(h). k=1 Observing y(x mh) = y(x)+o(h), y (x mh) = y (x)+o(h) = f(x,y(x))+o(h), we see that (1.50b) and (1.50c) are sufficient for h x I τ h h (x) 0 (h 0). For linear multi-step methods the following result holds true: Theorem (Order of consistency of linear multi-step methods) Let f C p (I D) and assume τ h (x j = O(h p ), 0 j m 1. Then, for a linear multi-step method each of the following conditions is equivalent to the method having an order of consistency p. C 1 : For each l = 0,1,,p it holds m (k l α k l k l 1 β k ) = 0, k=0

43 Numerical Differential Equations 39 C 2 : z = 0 is a zero of order p+1 of the entire function C 3 : The initial-value problem has the order of consistency p. ϕ(z) := ρ(exp(z)) z σ(exp(z)). y (x) = y(x), x I, y(a) = 1, C 4 : The order of consistency is p for a class of initial-value problems whose solutions span the linear space of polynomials of degree p. Proof. We first prove the equivalence of C 1 and C 4. We note that f C p (I D) implies y C p+1 (I). Taylor expansion yields h τ h (x+mh) = m ( k=0 y(x+kh) = y(x)+ y (x+kh) = p l=1 p 1 l=0 α k y(x+kh) hβ k f(x+kh,y(x+kh)) }{{} = y (x+kh) p l=1 1 l! kl h l y (l) (x)+o(h p+1 ), 1 l! kl h l y (l+1) (x)+o(h p ) = 1 (l 1)! kl 1 h l 1 y (l) (x)+o(h p ) = (1.51) h τ h (x+mh) = m ( α k y(x)+ k=0 It follows that p l=1 ( α k l! k l β ) k (l 1)! kl 1 ) h l y (l) (x) +O(h p ). C 1 = τ h (x+mh) = O(h p ). Conversely, the consistency order p implies C 1, if in (1.51) we use the functions y = 1, y = x,, y = x p. The equivalence of C 4 can be shown similarly. We next show the equivalence of C 3. If the linear multi-step method is consistent of order p, then C 3 obviously holds true. Conversely, assume that C 3 is satisfied. Inserting y(x) = exp(x a) into (1.51) and ),

44 40 Ronald H.W. Hoppe summing over I h yields h τ h (x+mh) = }{{} = O(h p ) p 1 ( m ) (k l α k l k l 1 β k ) l! l=0 k=0 h l 1 h exp(x a) +O(h p ) x+mh I } h {{} b exp(x a)dx (h 0) a }{{} = = O(h p ) m (k l α k l k l 1 β k ) = 0, 0 l p. k=0 Finally, we prove the equivalence of C 2. Taylor expansion of ϕ(z) around z = 0 results in p ϕ(z) = ρ(exp(z)) z σ(exp(z)) = ϕ (l) (0) z l +O(h p ) = ϕ (l) (0) = 0 l=0 m (k l α k l l 1 β k ) = 0, 0 l p. k=0 For the stability of the multi-step method (1.47a),(1.47b) we consider the linear space C(I h ) of grid functions on I h equipped with the norm z h := m 1 j=0 z h (x j ) +h x I h The multi-step method define a mapping A h : C(I h ) C(I h ) A h z h (x j ) := 1 h z h (x). z j α (j), 0 j m 1, m α k z j m+k Φ(x j,z j m,,z j,h), m j N k=0. Definition (Local Lipschitz stability of multi-step methods) The multi-step method (1.47a),(1.47b) is said to be locally Lipschitz

45 Numerical Differential Equations 41 stable in z h C(I h ), if there exist positive numbers h max,δ, and η such that for all 0 < h < h max and all w h C(I h ) with it holds (1.52) A h z h A h w h δ z h (x) w h (x) η A h z h A h w h, x I h. The number η is called the stability barrier, whereas the number δ is referred to as the stability threshold. We suppose that the increment function Φ satisfies the following Lipschitz condition: There exist a neighborhood U I R d of the graph (x,y(x)), x I, of the solution y C 1 (I) of the initial-value problem (1.6a),(1.6b) and numbers h max > 0 and L > 0 such that for all (x,y k ),(x,y k ) U (I h R d ),0 k m, and for all 0 h < h max it holds (1.53) m Φ(x,y 0,,y m,h) Φ(x,y 0,,y m,h) L y k y k. Theorem (Characterization of local Lipschitz stability) Assume that (1.53) holds true. Then, the multi-step method (1.47a), (1.47b) is locally Lipschitz stable in r h y := y Ih if and only if the multistep method with Φ 0 is locally Lipschitz stable. Proof. Without restriction of generality let U = I R d. The local Lipschitz stability in r h y isequivalent tothefact thatforv h := r h y w h and h < h max it holds (1.54) ( v h (x) η t I h,t x (A h r h y A h w h )(t) + m 1 j=0,t j x k=0 ) v h (t j ). Now, let Ψ : I h R d(m+1) R + R d another increment function that satisfies (1.53) and let Ãh the mapping associated with the increment function Φ+Ψ, i.e., (1.55) A h z h (x j ) = Ãhz h (x j )+Ψ(x j,z j m,,z j,h). If we insert (1.55) into (1.56) ( v h (x) η t I h,t x (A h r h y A h w h )(t) + m 1 j=0,t j x ) v h (t j ),