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A New Robust Numerical Scheme for Nonlinear ODE
Systems
Léo Agélas, Thibaut-Hugues Gallois, Thierry Soriano
To cite this version:
A New Robust Numerical Scheme for Nonlinear ODE
Systems
L. Ag´elasa,∗, T.-H. Galloisa,∗, T. Sorianob,∗
aDepartment of Applied Mathematics, IFP Energies Nouvelles, 1-4, avenue de Bois-Pr´eau,
F-92852 Rueil-Malmaison, France
bLaboratoire d’Ing´enierie des Syst`emes M´ecaniques et des Mat´eriaux, SUPMECA-Institut
Sup´erieur de m´ecanique, Univ. de Toulon USTV BP 20132, F-83957 La Garde, France
Abstract
To solve ODEs systems, implicit numerical schemes are often used because of their good stability. Among the most widely used implicit methods for stiff prob-lem are Backward Differentiation Formulas (BDF). However, solving implicit time stepping requires the use of Newton’s algorithm which can be very con-suming CPU-time. Then, in this paper, we propose a linearly-implicit method built from BDF scheme and which keeps a good stability. The scheme is based on a linearization of the BDF implicit term and the use of an interpolation of right order. The method obtained is linear for each time step and thus faster that the Newton method. Numerical tests on a challenging real-world test problem reveal that the method proposed is a promising alternative to well-established approaches based on Newton algorithm.
Keywords: non linear ode systems, implicit numerical schemes, stability.
Introduction
In this paper, we study time integration methods for initial value problems involving large stiff systems of nonlinear Ordinary Differential Equations ODEs in autonomous form: t ∈ [0, T ], d¯y dt = f (¯y), ¯ y(0) = y0, (1) where ¯y ∈ Rd, f ∈ Cp(Rd)d.
Many problems in physics, engineering, chemistry, biology and other yield initial
∗Authors listed alphabetically
Email addresses: leo.agelas@ifpen.fr (L. Ag´elas ),
value problems involving systems of ODEs and many of these problems are known as stiff ODEs. Some of these physical systems, such as dynamics of
5
fluids within a porous medium, the atmosphere and oceans, or the behavior of materials, arise from spatial discretisations of nonlinear time-dependent Partial Differential Equations (PDEs), where the entries of the vector ¯y are the discrete solution values and the stiffness of the system is characterised by the presence of one or more eigenvalues of the Jacobian matrix J =∂f∂ ¯y with large negative real
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parts. The existence of fast and slow dynamics poses considerable challenges to the solution of the semi-discrete PDEs (1) by explicit time stepping methods. Specifically, due to the Courant-Friedrichs-Lewy (CFL) stability condition, the largest allowable step sizes are bounded above by the shortest (fastest) time scale in the system. To avoid stability restrictions on the step size, implicit time
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integration methods are becoming widely used in the simulation of large-scale evolutionary PDEs (1). Perhaps the most widely used implicit methods for stiff problems are the Backward Differentiation Formulas (BDF) (see [7, 18, 28]). The BDF method is an implicit multipoint method that uses known solution data determined at several previous time points as well as the unknown solution at
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the current point. The order of the error term can be easily adjusted by choosing a different number of previous data points. This method is known to provide good solution stability in solving stiff differential equations for which an implicit solution is essential. Implicit time stepping as BDF requires the solution of large nonlinear systems of equations, coupling all variables in the model, at each time
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step. However, solving nonlinear systems requires the computations of Jacobian matrices for each time stepping. For many systems the exact Jacobian J can be both costly and difficult to obtain, e.g., due to the size of the application and the use of complex spatial discretization schemes.
Then, in this paper, we propose a new class of integration methods that we call
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Linearized Interpolation Backward Differentiation Formulas (LIBDF). These are linearly-implicit methods which enjoy the benefit of requiring only solving one linear system per iteration in time, as opposed to solve one nonlinear system in the case of BDF. Moreover, in the linear case (f is linear), the scheme LIBDF degenerates to the BDF method and as a consequence, satisfies all the criteria
35
of a good numerical scheme (consistency, order, zero-stability and absolute-stability, convergence). Of particular interest in this work is the use of the equilibrium point of the system (1). Indeed, the construction of the LIBDF lies on the existence of equilibrium points of the system and we have shown in the case of stable equilibrium that the bound of the global error does not depend
40
on the time even in the nonlinear case. Moreover, this bound does not blow up as the stiffness of the problem becomes large.
The paper is organized as follows: In section 1, we introduce the LIBDF method. In section 2, we deal with the zero-stability and the absolute stability of the scheme and in Lemmata 2.1,2.2, we give some bounds on the eigenvalues in
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estimate.
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Let us now introduce the context and the following considerations of our problem :
Since f ∈ Cp(Rd)d, then thanks to Cauchy-Lipschitz Theorem (see the first chapter of [9]), there exists T0 > 0 such that the IVP (1) admits on [0, T0] an unique solution ¯y ∈ Cp+1([0, T
0])d. Then, we take T such that 0 < T ≤ T∗,
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where T∗ is the maximal time of existence of the solution y.
We assume that there exists a constant C0> 0 which does not depend on T such that,
max
0≤k≤p+1t∈[0,T ]sup |¯y
(k)(t)| ≤ C
0. (2)
For any function g ∈ Ck([0, T ]), we denote by Mk(g) the value Mk(g) := sup
t∈[0,T ]
|g(k)(t)|.
1. The Linearized Interpolated Backward Differentiation Formulae For N ∈ N∗ and a non-decreasing sequence {ti}0≤i≤N subdivision of the interval [0, T ] such that t0 = 0, tN = T and hi = ti+1− ti, 0 ≤ i ≤ N − 1 the step-size, the BDF methods of order p take the form (see the introduction of [31] and [20]) : for all p − 1 ≤ n ≤ N − 1,
yn+1= p−1 X i=0
αi,nyn−i+ βnhnf (yn+1), (3)
where the variable coefficients αi,n and βn are functions of the step-size ratios
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ωi = hi/hi−1 for i = n, ..., n + 2 − p (see [] to infer the expression of the coefficients).
For any sequence u = {ui}i∈J0,N Kand for any p − 1 ≤ n ≤ N − 1, we denote by Pp,n(t, u) the Lagrange interpolation polynomial (see Definition 6.1 [33]) of degree p − 1 for the set of points {(tn−i, un−i) : i = 0, 1, .., p − 1}. We denote by Pp,n(u) the value Pp,n(tn+1, u).
For a function g ∈ Cp([0, T ]) and for u = {g(ti)}
i∈J0,N K, thanks to Theorem 6.2 of [33], we get, |Pp,n(u) − g(tn+1)| ≤Mp(g) p! p−1 Y i=0 (tn+1− tn−i), (4)
By virtue of (4), for any sequence u = {ui}i∈J0,N Kand for any p−1 ≤ n ≤ N −1, we say that Pp,n(u) is an approximation of un+1of order p.
1.1. Expression of LIBDF scheme
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The idea of the construction of our scheme p-LIBDF lies on an approximation of order p of the implicit term f (yn+1) as follows:
where y = {yi}i∈J0,N K, c is a zero of the function f . Then, similarly as (3), we obtain our new scheme, which reads:
yn+1= p−1 X i=0
αi,nyn−i+ βnhn(f (Pp,n(y)) + f0(c)(yn+1− Pp,n(y))).
As the implicit term is now linearized, it is possible to inverse the scheme and make it explicit :
Definition 1.1.1. The p-LIBDF scheme is the following: ∀p ∈ N∗,
yn+1= (I − βnhnf0(c))−1 p−1 X i=0
αi,nyn−i+ βnhn(f (Pp,n(y)) − f0(c)Pp,n(y)) !
. (6) For the sake of simplicity, we take for the dimension of space d = 1 and our
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analysis will focus on the BDF methods with uniform steps-size. In the case of uniform steps-size, we have for all 0 ≤ i ≤ N , ti = ih, where h = NT is the unique step-size. We observe that the variable coefficients αi,n and βn do not depend no longer on the subscript n, then the definition 6 becomes in the case of uniform steps-size:
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Definition 1.1.2. The p-LIBDF scheme is the following: ∀p ∈ N∗,
yn+1= (1 − βhf0(c))−1 p−1
X i=0
αiyn−i+ βh(f (Pp,n(y)) − f0(c)Pp,n(y)) !
, (7)
where β and {αi}i∈J0,p−1K are the coefficients defining the usual p-BDF scheme (3) in the case of uniform steps-size. From Lemma 2.3 of [19], we infer
that these coefficients are given by β = p X `=1 1 ` !−1
and for all 0 ≤ i ≤ p − 1,
αi= β(−1)i p X `=i+1 C`i+1 ` .
Furthermore, for any sequence u = {ui}i∈
J0,N K and for any p − 1 ≤ n ≤ N − 1, we observe that if we apply the formula given by Lagrange in [23],
Pp,n(u) = p−1 X k=0
(−1)kCpk+1un−k. (8)
From (4), we deduce that for a function g ∈ Cp([0, T ]) and for u = {g(ih)}i∈ J0,N K, we get,
2. Stability of the p-LIBDF
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2.1. Zero-Stability
We recall the definition of zero-stability of a general numerical method F : Rm+1 7−→ R`, ` ≤ m by generalizing the formulation of S¨uli in [33]. The solution y = {yi}i∈
J0,mKof F verifies the equation F (y, h) = 0. We consider now u a solution of F computed with a round-off error ξ = {ξj}j∈
J0,`K: F (y, h) = ξ
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with ξ small enough. For m ∈ N∗, we denote by k · k
m the euclidian norm in Rm.
Definition 2.1.1. Consider the following IVP :
y0= 0; y(0) = y0, (10) y and u are its numerical solutions defined as described in the previous para-graph, the numerical method is called zero-stable if :
ky − ukm≤ Ckξk` (11) where C may depend on T the length of the simulation interval but is indepen-dant of h.
Considering this definition, we have the following proposition
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Proposition 2.1. The p-LIBDF scheme is zero-stable for 1 ≤ p ≤ 6.
Proof. The proof follows immediately from the fact that for f = 0, the p-LIBDF scheme corresponds exactly to the p-BDF scheme and the latter is zero-stable for 1 ≤ p ≤ 6 (see [16], see also [14]).
2.2. Absolute Stability
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We recall the definition of the absolute stability according to Dahlquist in [10]:
Definition 2.2.1. A numerical method is absolute stable for a given fixed λh, if all its solutions tend to zero as n → ∞ when the method is applied to a differential equation of the form,
y0= λy (12)
where λ is a complex constant with negative real part.
Considering this definition, we have the following proposition :
Proposition 2.2. The region of Absolute Stability of the p-LIBDF scheme is
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the same as the p-BDF scheme’s one.
In what follows, we assume that 1 ≤ p ≤ 6.
We recall that the region of absolutely stability is the set of points λh in the complex plane for which the method is absolutely stable. Then, indifferently the p-LIBDF or p-BDF is absolutely stable for a given value of λh if each root zr= zr(λh) of the associated stability polynomial π(z; (λh)) := ρ(z) − (λh)σ(z) with σ(z) = βzp, satisfies |z r(λh)| < 1, where ρ(z) = zp− p−1 X i=0 αizp−1−i (see section 12.11 in [33]). For µ > 0, the stability polynomial π(·, −µβ) can be seen also as the characteristic polynomial of the companion matrix Ap(µ) of size p defined by (see Theorem C.3.5 of [22]):
Ap(µ) = α0 1+µ α1 1+µ ... αp−2 1+µ αp−1 1+µ 1 0 ... 0 0 ... ... ... ... ... 0 0 ... 0 0 0 0 ... 1 0 . (13)
Therefore, the roots zr(−µβ) of the stability polynomial π(·; −µβ) associated to the method p-BDF are the eigenvalues of the matrix Ap(µ) and it is well-known (see [3] and [26]) that these roots satisfy |zr(−µβ)| < 1 provided that µ ∈ R and µ > 0. For µ = 0, we know that Ap(µ) is diagonalisable, then we introduce the non-empty set Dp= {µ ∈ R+; Ap(µ) is diagonalisable }. Since the function Fp : µ 7−→ |det(Ap(µ))| is continuous over R+, then we deduce that Dp which can be rewritten as {µ ∈ R+; Fp(µ) > 0} = F−1
p (]0, +∞[) is an open set of R+. For any µ ∈ Dp, µ > 0, we have Ap(µ) = Pp(µ)Dp(µ)Pp(µ)−1, with Pp(µ) is an invertible matrix of size p where each of its column is an eigenvector of Ap(µ) and Dp(µ) is the diagonal matrix of size p composed of the eigenvalues of Ap(µ). Then, we define the following norm on Rp: for all x ∈ Rp,
kxkµ:= kPp(µ)−1xk∞. (14) We define also the induced matrix norm as follows: for all B matrix of size p,
k|Bk|µ:= sup x∈Rp,x6=0
kBxkµ kxkµ
= kPp(µ)−1BPp(µ)k∞. (15)
We observe that k|Ap(µ)k|µ = k|Dp(µ)k|∞ = ρv(Ap(µ)) < 1 for µ > 0, the
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largest eigenvalue in module of Ap(µ). For the proof of our Theorem 4.2, we need Lemma 2.2. Lemma 2.2 is obtained thanks to the following Lemma. Lemma 2.1. For all µ0 > 0, there exists θp > 0 depending only on p and µ0 such that for all µ ∈ Dp, µ ≥ µ0, we have,
ρv(Ap(µ)) ≤ 1 (1 + θpµ)1p
Proof. Let µ ≥ µ0 and z ∈ C an eigenvalue of Ap(µ), therefore z satisfies the 110 following equation: (1 + µ)zp− p−1 X i=0 αizp−1−i= 0, which implies, (1 + µ)zp= p−1 X i=0 αizp−1−i. Since |z| ≤ ρv(Ap(µ)) ≤ 1, we deduce (1 + µ)|z|p≤ p−1 X i=0 |αi||z|p−1−i≤ p−1 X i=0 |αi|. (16) Then, we obtain |z| ≤ γp 1 + µ with γp = p−1 X i=0 |αi| ≥ 1 as p−1 X i=0 αi = 1. Since γp 1 + µ = 1 1 +γ1 p(µ − (γp− 1))
, then for all µ ≥ 2(γp−1), we have µ ≥ µ2+(γp−1)
and therefore we deduce that if µ ≥ 2(γp− 1) then γp 1 + µ ≤ 1 1 +2γµ p . We set 115 µp= max(2(γp− 1), µ0).
Then, using (16), we infer that for all µ ≥ µp,
|z| ≤ 1 1 + 2γµ p 1p . (17)
The function g : µ 7−→ ρv(Ap(µ)) is continuous and then reach its maximum
over [µ0, µp] that we denote ρp,max < 1. By setting αp = 1 ρpp,max − 1 µp , we have 1 (1 + αpµp) 1 p
= ρp,max. Therefore, we deduce that for all µ ∈ [µ0, µp],
|z| ≤ ρv(Ap(µ)) ≤ ρp,max= 1 (1 + αpµp)1p ≤ 1 (1 + αpµ)1p . (18) By taking νp= min(αp,2γ1
p) and using (17),(18), we deduce that for all µ ≥ µ0,
|z| ≤ 1 (1 + νpµ)1p
,
Now, we use Lemma 2.1 to obtain the following Lemma,
Lemma 2.2. There exists αp> 0 depending only on p such that for all µ ∈ Dp,
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µ > 0, we have,
ρv(Ap(µ)) ≤ 1 (1 + αpµ)1p
.
Proof. Thanks to Theorem 2.1 in [19], we infer, ρ(z)
σ(z) log(z) = 1 + O(|z − 1|
p), z → 1, (19)
where σ(z) = βzp. Equation (19) implies, ρ(z)
σ(z) log(z) → 1 as z → 1. (20) Since µ ∈ Dp, then the companion matrix Ap(µ) (13) is diagonalizable. Thanks to Corollary C.3.6 of [22] (see also subsection 10.3 of [1]), we deduce that the eigenvalues of Ap(µ) (which are also the roots of the polynomial π(·, −µβ)) are all distincts. Then, let z0(µ) ∈ C be the unique eigenvalue of Ap(µ) such that
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ρv(Ap(µ)) = |z0(µ)|, notice also that z0(µ) is a root of π(·, −µβ). For µ = 0, we deduce:
• the number 1 is an eigenvalue of Ap(µ). • the number 1 is a simple root of π(·, −µ
β), thanks to the zero-stability of p-BDF for 1 ≤ p ≤ 6(see [16] and Theorem 12.4 of [33]),
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• the number 1 is the only eigenvalue of Ap(µ) in module equal to one, thanks to the strong stability of the p-BDF for 1 ≤ p ≤ 6 (see Theorem 6 in [21] used with l = 1 since Brown’s (k,1) method is the k-BDF method). Therefore, we get also z0(µ) = 1 for µ = 0.
Then, as also the polynomial function µ 7−→ π(·, −µβ) is continuous with real
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coefficients and the roots of this latter are all distincts, we deduce that z0(µ) → 1 as µ → 0+ and there exists µ
1 > 0 such that for all µ ∈ [0, µ1], z0(µ) is a real number. Since z0(µ) is an eigenvalue of Ap(µ), then we get 0 = π(z0(µ)), −µβ) = ρ(z0(µ)) +µβσ((z0(µ)), then we deduce that
ρ(z0(µ)) σ(z0(µ)) = −
µ
β. Therefore, thanks to (20) combined with the fact that z0(µ) → 1 as µ → 0+, we infer that,
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− µ
β log z0(µ)→ 1 as µ → 0 +.
Therefore, we deduce that there exists 0 < µ2≤ µ1such that for all 0 < µ ≤ µ2, we get z0(µ) real, 0 < z0(µ) ≤ 1 and,
− µ
β log z0(µ) ≤3
which implies, since 0 < z0(µ) ≤ 1,
0 ≤ z0(µ) ≤ e −2µ
3β . (21)
Since for all x ≥ 0, ex ≥ 1 + x, then for p ≥ 1, epx ≥ 1 + x which implies e−x≤ 1
(1+x)1p
. Therefore from (21), we have for all 0 < µ ≤ µ2
0 ≤ z0(µ) ≤ 1 (1 +2µ3β)1p
, (22)
which means that for all 0 < µ ≤ µ2
ρv(Ap(µ)) ≤ 1 (1 +2µ3β)1p
. (23)
And thanks to Lemma 2.1 used with µ0= µ2, we conclude the proof.
3. Consistency error and order of the p-LIBDF scheme
We recall the definition of consistency and order of a numerical scheme (see
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Definitions 2.5 and 2.6 in [25]) :
Definition 3.0.2. Suppose that the numerical scheme can be written as yn+1= Ψ(tn+1, yn+1−p, yn−p, . . . , yn, h) for all p − 1 ≤ n ≤ N − 1. For ¯y ∈ Cp+1([0, T ]) solution of (1), we define the local truncation error committed by one step of the method : τn= y(t¯ n+1)−Ψ(tn+1,¯y(tn+1−p),¯y(tn−p),...,¯y(tn),h)
h , then the method is said to be consistent if : τ = sup
p−1≤n≤N −1
|τn| is such that
lim
h→0τ = 0. (24)
The method has order of accuracy p if :
τ = O(hp). (25)
Considering the definition of consistency and order, we have Proposition 3.1. To prove Proposition 3.1, we need to use a constant Lipschitz of f . Since f ∈ Cp
(R), then if we take as constant Lipschitz the value sup z∈R
|f0(z)|, this value may be equal to infinity. To avoid this situation, thanks to (2), we introduce
150
the interval Jp defined by Jp := [m0(¯y) − Mp(¯y)hp, m1(¯y) + Mp(¯y)hp], where m0(¯y) = inf
t∈[0,T ]y(t) and m1(¯¯ y) = supt∈[0,T ]y(t). Then, we introduce Lp¯ the con-stant Lipschitz of f on Jpdefined by Lp = sup
z∈Jp
|f0(z)|. This constant Lipschitz Lp of f is then sufficient to obtain the proof of our Proposition 3.1.
It is important to notice, thanks to (2), that there exists an interval J which do
155
not depend on T such that Jp ⊂ J , and hence Lp admit a upper bound which does not depend on T , for instance L = sup
Proposition 3.1. The p-LIBDF scheme has order of accuracy p and its local truncation error τn satisfies for all p − 1 ≤ n ≤ N − 1,
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|τn| ≤ (CpMp+1(¯y) + Mp(¯y)(Lp+ |f0(c)|)) hp, where Cp> 0 is a constant depending only on p.
Proof. The p-LIBDF scheme reads:
yn+1= p−1 X i=0
αiyn−i+ βh(f (Pp,n(y)) + f0(c)(yn+1− Pp,n(y))).
To study the consistency and the order of p-LIBDF scheme, we consider the local truncation error:
τn= 1 h y(t¯ n+1) − p−1 X i=0
αiy(t¯ n−i) + βh(f (Pp,n( ¯y)) + f0(c)(¯y(tn+1) − Pp,n( ¯y))) !!
,
where ¯y is the sequence defined by ¯y = {¯y(ti)}i∈J0N K. The local truncation error τn can be rewritten as:
τn= ˜τn+ β(f (¯y(tn+1)) − f (Pp,n( ¯y)) − f0(c)(¯y(tn+1) − Pp,n( ¯y))), (26) where ˜τn is the local truncation error of the p-BDF scheme given by,
˜ τn= 1 h y(tn+1) −¯ p−1 X i=0
αiy(tn−i) + βhf (¯¯ y(tn+1)) !!
,
It is well-known that the p-BDF scheme has order of accuracy p (see [33]) and hence there exists a constant Cp > 0 such that for all p − 1 ≤ n ≤ N − 1, |˜τn| ≤ CpMp+1(¯y) hp. Furthermore, thanks to (9), we have Pp,n( ¯y) ∈ Jp and since f is a Lipschitz function on Jp of constant Lp, we deduce that,
|f (¯y(tn+1)) − f (Pp,n( ¯y))| ≤ Lp|¯y(tn+1) − Pp,n( ¯y)|. (27)
Then, we infer,
|f (¯y(tn+1)) − f (Pp,n( ¯y)) − f0(c)(¯y(tn+1) − Pp,n( ¯y))|
≤ (Lp+ |f0(c)|) |¯y(tn+1) − Pp,n( ¯y))| ≤ Mp(¯y)(Lp+ |f0(c)|) hp,
where we have used again(9) for the last inequality. Therefore, from (26), we obtain for all p − 1 ≤ n ≤ N − 1, |τn| ≤ C1hp, with C1 = (CpMp+1(¯y) +
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4. Convergence of order p for the p-LIBDF
Convergence of order p means that for sufficiently accurate starting approx-imations y0, y1, ..., yp−1, the global error satisfies max
0≤i≤N|yi − ¯y(ti)| = O(h p).
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Thanks to the Propositions 2.1 and 3.1, we get the convergence of order p for the p-LIBDF by borrowing the arguments used for the proof of the Dahlquist’s Equivalence Theorem (see theorem 6.3.4 of [11] or Theorem 5.10 of [17]). In most of existing proofs, the constant symbolized by the O(hp) notation is proportional to eLT, where T is the length of the simulation interval and L a
175
Lipschitz constant of the function f (see [33] p. 318, [6] p. 271 and [2] p. 41). However, this constant can be greatly improved in some situations, indeed in this section, under the assumption of existence of a stable equilibrium point, through Theorem 4.2, we obtain a constant independent of T and which does not blow up as the best Lipschitz constant or the stiffness of the problem get
180
more and more large.
To prove our Theorem 4.2, we need to introduce αp,µ > 0 the real defined by, αp,µ= sup kxk∞ke1kµ kxkµ ; x ∈ R p, x 6= 0 , (28)
the norm k · k∞,µdefined by k · k∞,µ:= k · kµ
ke1kµ and to use a serie of Lemmata. These Lemmata are obtained under the assumption that there exists a stable equilibrium point of the system (1). In Lemma 4.2, we show that all the numer-ical sequence obtained from p-LIBDF scheme is contained in a ball whose its radius Rµ may be very large and may depend on µ = βh|f0(c)|. In Proposition 4.1, under the assumption that sup
λ>0
αp,λ< +∞, we show in fact that the radius Rµadmits a lower bound independent of µ and hence all the numerical sequence lies inside a ball with its radius independent of µ. In Lemma 4.3, we show that the continuous solution of (1) is also contained in a ball.
Thanks to (28), we can notice that for all x ∈ Rp,
kxk∞≤ αp,µkxk∞,µ. (29) We denote by R(µ) := ρv(Ap(µ)). Then, we begin with the following Lemma. Lemma 4.1. The sequence {un}n∈N∗ defined by un+1 ≤ anun+ b, u0 = 0 is
majored for n ≥ 1 by : un ≤ n X k=1 n−1 Y l=k al ! b. (30)
Proof. The proof follows easily by using an induction argument.
then we get for all p ≤ n ≤ N , kDnk∞,µ≤ R, where, Dn= yn− c yn−1− c ... yn+1−p− c . (31)
The real R > 0 satisfies, R ≥ Rµ where,
Rµ= ( inf J with J :=nx ∈ R+; g(x) ≥ 2|f0(c)| νp,µ (1 + µ) 1−R(µ) µ o if J 6= ∅ +∞ elsewhere (32) and the function g is defined for all x ≥ 0, by:
g(x) = x sup z∈Ix |f00(z)|, where Ix=c − x√νp,µ, c + x√νp,µ and νp,µ= (2p− 1)2α2 p,µ. 185
Proof. The p-LIBDF scheme is defined for all p − 1 ≤ n ≤ N − 1 by
yn+1= p−1 X k=0
αkyn−k+ hβ(f (Pp,n(y)) + f0(c)(yn+1− Pp,n(y))). (33)
Since f (c) = 0 and p−1 X k=0
αk= 1, then from (33), we deduce,
yn+1−c = p−1 X k=0
αk(yn−k−c)+hβ(f (Pp,n(y))−f (c)+f0(c)((yn+1−c)−(Pp,n(y))−c)). (34) By using a Taylor expansion of order two, we deduce that there exists θn ∈ (c, Pp,n(y)) such that,
f (Pp,n(y)) − f (c) + f0(c)((yn+1− c) − (Pp,n(y)) − c) = f 00(θn)
2 (Pp,n(y)) − c) 2. (35) We introduce c = {ci}i∈J0,N K with ci= c. We can notice that
p−1 X k=0 (−1)kCpk+1= 1, indeed p−1 X k=0 (−1)kCpk+1= p X `=1 (−1)`−1Cp`= − p X `=0 (−1)`Cp`+ 1 = −(1 + (−1)) p + 1 = 1. Then thanks to (8), we deduce that
Hence, after setting fn= f00(θn) 2 Pp,n(d) 2 and µ = βh|f0(c)|, we obtain, Dn+1= Ap(µ)Dn+ Fn(µ), (37) where, Dn= dn dn−1 ... dn+1−p , Fn= βhfn 1+µ 0 ... 0 . (38)
We take the norm k · kµ of equation (37) to obtain,
kDn+1kµ≤ k|Ap(µ)k|µkDnkµ+ kFn(µ)kµ. (39)
We set R(µ) = k|Ap(µ)k|µand we notice kFn(µ)kµ= βh|fn|
1 + µke1kµ. We re-write (39) as follows,
kDn+1kµ≤ R(µ) kDnkµ+βh|fn|
1 + µke1kµ. (40) Let us give an estimate of fn. We notice thanks to (8) that |Pp,n(d)| ≤ γpkDnk∞ and thanks to (29), we obtain,
|Pp,n(d)| ≤ γpαp,µkDnk∞,µ. (41) Therefore, thanks to (41) we get |fn| ≤
νp,µ
2 |f 00(θ
n)| kDnk2∞,µ with νp,µ = γ2
pαp,µ2 . Then from (40), we get,
kDn+1k∞,µ≤ R(µ) kDnk∞,µ+
νp,µβh 2(1 + µ)|f
00(θn)| kDnk2
∞,µ. (42) Since c = Pp,n(c), then Pp,n(y) = Pp,n(y) − c + c = Pp,n(y − c) + c = Pp,n(d) + c and thanks again to (41) this give us that θn ∈ (c, c + Pp,n(d)) ⊂ In := c −√νp,µkDnk∞,µ, c +√νp,µkDnk∞,µ. Therefore, we deduce that |f00(θ
n)| ≤ sup
z∈In
|f00(z)|. Hence, from (42) we get,
kDn+1k∞,µ≤ R(µ) kDnk∞,µ+
νp,µβh 2(1 + µ)z∈Isupn
|f00(z)| kDnk2
∞,µ. (43) Let R > 0 such that R ≥ kDp−1k∞,µ.
We prove by induction the following proposition P(n): for all p − 1 ≤ n ≤ N , kDnk∞,µ≤ R.
The proposition P(n) is true for n = p − 1. Let us assume that for n ∈ Jp − 1, N − 1K, the Proposition P(n) is true, then from (43), it is inferred,
kDn+1k∞,µ≤ R(µ) R + νp,µβh 2(1 + µ)z∈Isupn
Then to obtain
kDn+1k∞,µ≤ R, (45)
thanks to (44) it suffices to get
190 R(µ) R + βhνp,µ 2(1 + µ)R 2 sup z∈IR |f00(z) ≤ R, where IR=c − √ νp,µR, c + √
νp,µR, which means also, βhνp,µ
2(1 + µ)R supz∈IR
|f00(z) ≤ 1 − R(µ). (46)
Since µ = βh|f0(c)|, then we get βh = |f0µ(c)|, henceforth, inequality (46) can be
re-written as follows, R sup z∈IR |f00(z)| ≤ 2|f 0(c)| νp,µ (1 + µ) 1 − R(µ) µ . (47)
Therefore if (47) holds then we get (45). We consider the function g defined for all x ≥ 0, by:
g(x) = x sup z∈Ix
|f00(z)|,
where Ix=c − x√νp,µ, c + x√νp,µ. Since g is continuous and g(0) = 0, then the value Rµ defined by Rµ= infnx ∈ R+; g(x) ≥ 2|f0(c)|
νp,µ (1 + µ)
1−R(µ) µ
o is strictly positive and can be eventually equal to +∞ in the case where the set is
195
empty.
By requiring that the initial datas y0, y1, ..., yp−1are such that kDp−1k∞,µ≤ Rµ, then by taking R = Rµ, inequality (47) holds and hence (45) holds also which means that the proposition P(n + 1) is true. Therefore, we deduce that for all for all p − 1 ≤ n ≤ N ,
200
kDnk∞,µ≤ R, which concludes the proof.
Proposition 4.1. Let c be a zero of f such that f0(c) < 0. Let µ = βh|f0(c)| ∈ Dp, µ > 0. Under the assumption that there exists κp > 0 depending only on p such that sup
λ>0
αp,λ ≤ κp (see 28 for αp,λ) then there exists ηp > 0 depending only on p such that the real Rµ defined in (32) satisfies Rµ≥ Rp where Rp> 0 is the real defined by,
Rp=
inf ˜J with ˜J := {x ∈ R+; g(x) ≥ ηp|f0(c)|} if ˜J 6= ∅
+∞ elsewhere (48)
and the function ˜g is defined for all x ≥ 0, by: ˜
g(x) = x sup z∈Qx
where Qx= [c − xχp, c + xχp] and χp= (2p− 1)κp.
Proof. We recall that Rµ is defined by:
Rµ= ( inf J with J :=nx ∈ R+; g(x) ≥ 2|f0(c)| νp,µ (1 + µ) 1−R(µ) µ o if J 6= ∅ +∞ elsewhere (49) and the function g is defined for all x ≥ 0, by:
205 g(x) = x sup z∈Ix |f00(z)|, where Ix=c − x√νp,µ, c + x√νp,µ and νp,µ= (2p− 1)2α2 p,µ.
Since αp,µ≤ κp, we have νp,µ≤ %p:= (2p− 1)κp, then we get for all x ≥ 0, Ix⊂ Qx:=c − x√%p, c + x√%p . (50) Furthermore, thanks to Lemma 2.2, there exists ςp > 0 depending only on p such that for all µ ∈ Dp, µ > 0, we have,
R(µ) ≤ 1 (1 + ςpµ)1p
,
which give us combined with the fact that αp,µ≤ κp,
2|f0(c)| νp,µ (1 + µ) 1 − R(µ) µ ≥ 2|f 0(c)| (2p− 1)2κ2 p 1 + µ µ 1 − 1 (1 + ςpµ)1p ! . (51)
We introduce the positive continuous function G on ]0, +∞[ defined for all µ > 0 by: Gp(µ) := 1 + µ µ 1 − 1 (1 + ςpµ)p1 ! . We notice that Gp(µ) → 1 pςp as µ → 0 and G(µ) → 1 as µ → +∞, then we deduce that Υp := inf
µ>0Gp(µ) > 0. Then, from (51), we infer, 2|f0(c)| νp,µ (1 + µ) 1 − R(µ) µ ≥ 2|f 0(c)| (2p− 1)2κ2 p Υp. (52)
Therefore, thanks to (50) and (52), from (49) we obtain that Rµ ≥ Rp, where Rp> 0 is a real depending only on p, defined by:
and the function ˜g is defined for all x ≥ 0, by: 210 ˜ g(x) = x sup z∈Qx |f00(z)|,
which concludes the proof.
Lemma 4.3. Let c be a zero of f such that f0(c) < 0 and 0 < ϑ < 1. Then there exists a constant Rϑ> 0 such that if |y0− c| < Rϑ, then the solution ¯y of (1) satisfies for all t ∈ [0, T ], |y(t) − c| ≤ |y0− c|eϑf0(c)t.
A possible value for Rϑ is:
215
Rϑ= sup{R > 0; ∀z ∈ R, |z − c| ≤ R ⇒ f0(z) ≤ ϑf0(c)}. Proof. Since f (c) = 0, then from (1), we get for all t ∈ [0, T ],
d(¯y(t) − c)
dt = f (¯y(t)) − f (c). (54) Multiplying equation (54) with ¯y(t) − c yields to,
1 2
d|¯y(t) − c|2
dt = (f (¯y(t)) − f (c))(¯y(t) − c). (55) Thanks to Taylor expansion, for any t ∈ [0, T ], there exists θ(t) ∈ (c, y(t)) such that,
f (¯y(t)) − f (c) = f0(θ(t))(¯y(t) − c). Then, from (55), we obtain,
1 2
d|¯y(t) − c|2 dt = f
0(θ(t))|¯y(t) − c|2. (56)
Since f0is continuous and f0(c) < 0, then for any 0 < ϑ < 1 there exists Rϑ> 0 such that for all z ∈ R such that |z − c| ≤ Rϑ, we have f0(z) ≤ ϑf0(c) < 0. We require that y0be such that |y0− c| < Rϑ.
Let us show that for all t ∈ [0, T ], |y(t) − c| ≤ Rϑ. If there exists t0∈ [0, T ] such that |y(t0) − c| > Rϑ, then, the value t∗= inf{t ∈ [0, T ]; |y(t) − c| > Rϑ} is well defined. Due to the definition of t∗ and the fact that y is continuous, we have
|y(t∗) − c| = Rϑ (57) and then t∗6= 0.
Furthermore, thanks again to the definition of t∗, we get that for all t ∈ [0, t∗], |y(t) − c| ≤ Rϑ (58) and since θ(t) ∈ (c, y(t)) then we get also |θ(t) − c| ≤ Rϑ which implies that f0(θ(t)) ≤ ϑf0(c) < 0. Therefore, from (56), we deduce that for all t ∈ [0, t∗],
1 2
d|¯y(t) − c|2 dt ≤ ϑf
We notice 1 2 d dt |¯y(t) − c|2e−2ϑf0(c)t= e−2ϑf0(c)t 1 2 d|¯y(t) − c|2 dt − ϑf 0(c)|¯y(t) − c|2 .
Therefore inequality (59) can be rewritten as follows, 1 2 d dt |¯y(t) − c|2e−2ϑf0(c)t≤ 0. (60) Then, from (60), we deduce that for all t ∈ [0, t∗],
|¯y(t) − c| ≤ |y0− c|eϑf
0(c)t
. (61)
Since |y0− c| < Rϑ and f0(c) < 0, then we obtain that |¯y(t∗) − c| < Rϑ which leads to a contradiction with (57). Therefore, we have that for all t ∈ [0, T ],
220
|y(t) − c| ≤ Rϑ. From (58), by applying to T the same arguments used for t∗ , we infer that for all t ∈ [0, T ],
|¯y(t) − c| ≤ |y0− c|eϑf
0(c)t
, which concludes the proof.
Now, we turn to the proof of our Theorem. Thanks to Proposition 3.1 and (2), the third condition in Theorem 4.2 (where τ is defined in Definition
225
3.0.2) can be traduced in terms of a condition on h independent of the time of simulation T . For simplicity, we assume that there is no error on the initial datas (which implies Ep−1= 0)
Theorem 4.2. Let c be a zero of f such that f0(c) < 0. Let µ = β|f0(c)|h ∈ Dp and 0 < ϑ < 1. Under the assumption that there exists κp > 0 depending only
230
on p such that sup λ>0
αp,λ≤ κp (see 28 for αp,λ), there exists Rp > 0 depending only on p, Rϑ> 0 depending only on ϑ, Θp,ϑ> 0, Φp,ϑ> 0 and Ψp,ϑ depending only on p, ϑ such that if
kDp−1k∞,µ ≤ Rp |y0− c| ≤ Rϑ τ ≤ Θp,ϑ e −2|f 0 (c)|Ψp,ϑ |f0(c)|2 Ep−1 = 0.
Proof. The p-LIBDF scheme is the following, for all p − 1 ≤ n ≤ N − 1,
yn+1= p−1 X k=0
αkyn−k+ hβ(f (Pp,n(y)) + f0(c)(yn+1− Pp,n(y))). (63)
The definition of the local truncation error of the p-LIBDF yields to,
¯
y(tn+1) = p−1 X k=0
αky(tn−k) + βh(f (Pp,n( ¯¯ y)) + f0(c)(¯y(tn+1) − Pp,n( ¯y)))) + hτn, (64) where ¯y = {¯y(ti)}i∈
J0,N K.
235
By substracting (63) and (64), and using the fact that the application Pp,n(·) is linear, we deduce: n+1= p−1 X k=0 αkn−k+ βh(f (Pp,n(y)) − f (Pp,n( ¯y)) + f0(c)(n+1− Pp,n())) − hτn, (65) where = {i}i∈J0,N Kwith i = yi− ¯y(ti). By using a Taylor expansion of order two, we deduce that there exists θn∈ (Pp,n(y), Pp,n( ¯y)) ⊂ (0, T ) such that,
f (Pp,n(y)) = f (Pp,n( ¯y)) + f0(Pp,n( ¯y))Pp,n() +f 00(θn)
2 Pp,n()
2. (66) By plugging the expression (66) in (65), it comes:
(1 − βhf0(c))n+1 = p−1 X k=0 αkn−k+ βh(f0(Pp,n( ¯y)) − f0(c))Pp,n() +βhf 00(θn) 2 Pp,n() 2− hτn. (67)
By using a Taylor expansion of order one, we deduce that there exists βn ∈ (c, Pp,n( ¯y)) ⊂ (0, T ) such that,
Then, we get, (1 − hβf0(c))n+1= p−1 X k=0 αkn−k+ φn− hτn, which implies, n+1= p−1 X k=0 αk 1 − hβf0(c)n−k+ φn− hτn 1 − hβf0(c). Hence, by setting µ = βh|f0(c)|, we obtain,
En+1= Ap(µ)En+ Bn, (71) where, En= n n−1 ... n+1−p , Bn = φn−hτn 1+µ 0 ... 0 (72)
We take the norm k · kµ of equation (71) to obtain,
kEn+1kµ ≤ k|Ap(µ)k|µkEnkµ+ kBnkµ. (73)
From (70), we deduce, 240 |φn| ≤ βhKn |Pp,n( ¯y − c)| |Pp,n()| + 1 2Pp,n() 2 ,
where Kn = max(|f00(θn)|, |f00(βn)|). Thanks to (8) with γp= 2p−1, we deduce, |Pp,n()| ≤ γpkEnk∞. (74) Then, we obtain,
|φn| ≤ βhKn(γp|Pp,n( ¯y − c)|kEnk∞+ 0.5γp2kEnk 2 ∞). Thanks to (29), after setting νp:= γpκp≥ γpαp,µ, we deduce,
|φn| ≤ βhKn(νp|Pp,n( ¯y − c)|kEnk∞,µ+ 0.5νp2kEnk 2 ∞,µ). Furthermore, we notice that kBnk = |φn−hτn|
1+µ ke1kµ. We set En= kEnk∞,µand R(µ) = k|Ap(µ)k|µ, then from (73), we deduce,
where τ = max n∈Jp−1,N −1K
|τn|. Thanks to Lemma 4.2 and Proposition 4.1, there exists Rp> 0 depending only on p such that
kDp−1k∞,µ≤ Rp (76)
implies for all p ≤ n ≤ N
kDnk∞,µ≤ Rp (77)
where for all p − 1 ≤ n ≤ N ,
Dn= yn− c yn−1− c ... yn+1−p− c . (78) Thanks to (29), we get kDnk∞≤ αp,µkDnk∞,µ≤ κpkDnk∞,µ. (79) Furthermore, thanks to Lemma 4.3, there exists Rϑ > 0 depending only on ϑ such that
|y0− c| ≤ Rϑ (80)
implies for all t ∈ [0, T ]
|¯y(t) − c| ≤ |y0− c|eϑf0(c)t. (81) In what follows, we assume that (76) and (80) hold. Then (77) and (81) hold
245
and thanks also to (79), we infer that there exists Rp,ϑ > 0 depending only on p, ϑ such that for all p − 1 ≤ n ≤ N − 1,
Kn ≤ Kp,ϑ:= sup z∈[c−Rp,ϑ,c+Rp,ϑ]
|f00(z)|.
Thanks again to (81), we get |Pp,n( ¯y − c)| ≤ zp(tn) := ζpexp(−ϑ|f0(c)|tn−p) with ζp= (2p− 1)|y0− c|. Inequality (75) becomes, En+1≤ R(µ) +βhKp,ϑνp 1 + µ zp(tn) En+βhKp,ϑν 2 p 2(1 + µ) E 2 n+ hτ 1 + µ. (82) We set 0 = hτ
1 + µ. We prove by induction the following proposition : for all p − 1 ≤ n ≤ N ,
En≤ θ0, (83)
with θ > 1 satisfying (86). We require that Ep−1≤ θτ , then the proposition is true for n = p − 1.
We suppose that the proposition is true until the rank n and we show that
250
By using the induction hypothesis, we get : En+1≤ R(µ) +βhKp,ϑνp 1 + µ zp(tn) En+βhKp,ϑν 2 p 2(1 + µ) θ 22 0+ 0. (84) We set b = 01 + βhKp,ϑνp2 2(1+µ) θ 20. We use Lemma 4.1 to obtain:
En+1≤ n+1 X k=1 n Y l=k R(µ) + βhKp,ϑνp 1 + µ zp(tl) ! b
Thanks to Lemma 2.2, there exists %p> 0 depending only on p such that,
R(µ) ≤ G(µ) := 1 (1 + %pµ)1p . (85) Then, we get, En+1≤ n+1 X k=1 n Y l=k G(µ) +βhKp,ϑνp 1 + µ zp(tl) ! b ≤ n+1 X k=1 G(µ)n−k+1 n Y l=k 1 + βhKp,ϑνp (1 + µ)G(µ)zp(tl) ! b ≤ n+1 X k=1 G(µ)n−k+1 n Y l=k e βhKp,ϑνp (1+µ)G(µ)zp(tl) ! b = n+1 X k=1 G(µ)n−k+1ePn l=k βhKp,ϑνp (1+µ)G(µ)zp(tl) ! b
Since G(µ) := 1 (1+%pµ) 1 p , then we have (1 + µ)G(µ) ≥ 1 υp(1 + µ) 1−1 p where
υp= max(1, %p)1p. Since the second derivative of h 7−→ e−ϑ|f0(c)|his decreasing,
then we get that for all h > 0, h
1 − e−ϑ|f0(c)|h ≤ 1 ϑ|f0(c)| Therefore, we get, 260 ωp,µ,ϑ≤ ψp,ϑ:= exp Kp,ϑζpνpυp β ϑ|f0(c)| ,
and we obtain En+1≤ b ψp,ϑ
1 − G(µ). In what follows, we require that,
θ = 3 2
ψp,θ
1 − G(µ). (86)
After using the value of b, we deduce that to get En+1≤ θ0, it suffices to have
1 +βhKp,ϑν 2 p 2(1 + µ) θ 20 ! ψp,ϑ 1 − G(µ) ≤ θ. (87)
By using the value of 0, inequality (87) becomes,
1 + βKp,ϑν 2 p 2 θ 2 h 1 + µ 2 τ ! ψp,ϑ 1 − G(µ) ≤ θ. (88)
Inequality (88) can be rewritten as,
τ ≤ 2 1 + µ h 2 1−G(µ) ψp,ϑ θ − 1 βKp,ϑν2 pθ2 = 4 9 (1 + µ)2 βKp,ϑν2 pψp,ϑ2 1 − G(µ) h 2 , (89)
where we have used the value of θ (86). Since h = β|fµ0(c)| and G(µ) :=
1 (1+%pµ)
1 p
, then inequality (89) becomes,
τ ≤ 4 9 β|f0(c)|2 Kp,ϑν2 pψ2p,ϑ (1 + µ) 1 − 1 (1+%pµ) 1 p µ 2 . (90)
We introduce the positive function w on ]0, +∞[ defined for all x > 0 by:
1 as x → +∞, hence we deduce that Γp= inf x>0w(x) > 0. Then, inequality (90) holds if we have, τ ≤ 4 9 β|f0(c)|2 Kp,ϑν2 pψ2p,ϑ Γ2p. (91)
Then by requiring that inequality (91) holds, we obtain that En+1 ≤ θ0 and the proposition at the rank n + 1 is true. Therefore, we deduce that for all p − 1 ≤ n ≤ N ,
En≤ θ0,
which gives us with the values of θ and 0,
En ≤ 3 2 ψp,ϑ (1 − G(µ)) hτ 1 + µ = 3 2β|f0(c)| ψp,µ,h (1 − G(µ)) µ 1 + µτ = 3ψp,ϑ 2β|f0(c)| τ w(µ) ≤ 3ψp,ϑ 2β|f0(c)| τ Γp .
which allows us to conclude the proof.
265
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