Rigorous numerics for nonlinear operators with tridiagonal dominant linear parts

Rigorous numerics for nonlinear operators with tridiagonal dominant linear parts Maxime Breden Laurent Desvillettes Jean-Philippe Lessard Abstract We propose a method to compute solutions of infinite dimensional nonlinear operators fx = 0 with tridiagonal dominant linear parts We recast the operator equation into an equivalent Newton-like equation x = T x = x Afx, where A is an approximate inverse of the derivative Df x at an approximate solution x We develop rigorous computer-assisted bounds to show that T is a contraction near x, which yields existence of a solution Since Df x does not have an asymptotically diagonal dominant structure, the computation of A is not straightforward This paper provides a method to obtain A proposes a new rigorous computational method to prove existence of solutions of nonlinear operators with tridiagonal dominant linear parts Keywords Tridiagonal operator Contraction mapping Rigorous numerics Fourier series Mathematics Subject Classification 200 47H0 42A0 65L0 34B08 ntroduction Tridiagonal operators arise naturally in the theory of orthogonal polynomials, ordinary differential equations ODEs, continued fractions, numerical analysis of partial differential equations PDEs, integrable systems, quantum mechanics solid state physics Some differential operators can be represented by infinite tridiagonal matrices acting in sequence spaces, as it is the case for instance for differentiation in frequency space of the Hermite functions Other examples come from the study of ODEs like the Mathieu equation, the spheroidal wave equation, the Whittaker-Hill equation the Lamé equation While there exist many well developed methods efficient algorithms in the literature for solving linear tridiagonal matrix equations computing their inverses, our proposed method has a different flavour We aim at developing a computational method to prove, in a mathematically rigorous constructive sense, existence of solutions of infinite dimensional nonlinear operators of the form fx = Lx + Nx = 0, CMLA, ENS Cachan & CNRS, 6 avenue du Président Wilson, 94235 Cachan, France mbreden@ens-cachanfr CMLA, ENS Cachan & CNRS, 6 avenue du Président Wilson, 94235 Cachan, France desville@cmlaens-cachanfr Département de Mathématiques et de Statistique, Université Laval, 045 avenue de la Médecine, Québec, QC, GV0A6, Canada jean-philippelessard@matulavalca

where L is a tridiagonal linear operator N is a nonlinear operator The domain of the operator f is the space of algebraically decaying sequences { } Ω s def def = x = x k k 0 : x s = sup{ x k k s } < 2 The assumptions on the linear nonlinear parts of are that L : Ω s Ω s s L N : Ω s Ω s s N, for some s L > s N ntuitively, this means that the linear part dominates the nonlinear part Since Ω s Ω s2 for s > s 2, one has that f : Ω s Ω s s L General nonlinear operators fx = 0 defined on Ω s arise in the study of bounded solutions of finite infinite dimensional dynamical systems For instance, x = x k k 0 may be the infinite sequence of Fourier coefficients of a periodic solution of an ODE, a periodic solution of a delay differential equation DDE or an equilibrium solution of a PDE with Dirichlet, periodic or Neumann boundary conditions The unknown x may also be the infinite sequence of Chebyshev coefficients of a solution of a boundary value problem BVP, the Hermite coefficients of a solution of an ODE defined on an unbounded domain or the Taylor coefficients of the solution of a Cauchy problem n case the differential equation is smooth, the decay rate of the coefficients of x will be algebraic or even exponential [] n this paper, we chose to solve in the weighed l Banach space Ω s which corresponds to C k solutions n order to exploit the analyticity of the solutions, we could follow the idea of [2] solve in weighed l Banach spaces This choice of space is not considered in the present paper n recent years, there has been several successful attempts in solving fx = 0 in Ω s using the field of rigorous numerics This field aims at constructing algorithms that provide an approximate solution to the problem together with precise bounds within which the exact solution is guaranteed to exist in the mathematically rigorous sense Equilibria of PDEs [3, 4, 5], periodic solutions of DDEs [6], fixed points of infinite dimensional maps [7] periodic solutions of ODEs [8, 9] have been computed using such methods One popular idea in rigorous numerics is to recast the problem fx = 0 as a problem of looking for the fixed points of a Newton-like equation of the form T x = x Afx, where A is an approximate inverse of Df x, where x is a numerical approximations obtained by computing on a finite dimensional projection of f n [3, 4, 6, 7, 9, 5], the nonlinear equations under study have asymptotically diagonal or block-diagonal dominant linear parts which facilitates the computation of approximate inverses n contrast, the present work considers problems with tridiagonal dominant linear parts To the best of our knowledge, this is first attempt to compute rigorously solutions of such problems While our proposed approach is designed for the moment for a specific class of operators see assumptions 4 5, we believe it is a first step toward solving rigorously more complicated nonlinear operators with tridiagonal dominant linear parts The paper is organized as follows n Section 2, we present the method to compute, with the help of the computer, pseudo-inverses of tridiagonal operators of a certain class n Section 3, we recast the problem fx = 0 as a fixed point problem T x = x Afx where A is a pseudo-inverse, we present the rigorous computational method to prove existence of fixed points of T n Section 4, we present an application finally in Section 5, we conclude by presenting some interesting future directions 2 Computing pseudo-inverses of tridiagonal operators Given three sequences λ k k 0, µ k k 0, β k k 0 x Ω s, define formally the tridiagonal linear operator Lx = L k x k 0 of by k 0 2

L k x = λ k x k + µ k x k + β k x k+, k 3 L 0 x = µ 0 x 0 + β 0 x Assume that there exist real numbers s L >, 0 < C C 2 an integer k 0 such that k, λ k, µ k, β k C 2 k k 0, C µ k 4 k s L k s L Assume further the existence of δ k s L 0, k 0 0 such that 2 k s L k k 0, λ k µ k, β k δ 5 µ k Therefore under assumptions 4 5, L defined by 3 is a tridiagonal operator such that L : Ω s Ω s s L ndeed, for x Ω s, then Lx s sl = sup{ L k x k s s L } k 0 C 2 sup { x k k s } + sup k 0 k 0 { x k k s } + sup k 0 { x k+ k s } < From now on, assume for sake of simplicity that s N = 0, that is the nonlinear part N of satisfies N : Ω s Ω s Note that any combination of convolutions satisfies this property by the algebra property of Ω s for s > [5, 0] Assume that using a finite dimensional projection f m : R m R m of, we computed a numerical approximation x, that is f m x 0 We identify x R m x = x, 0, 0, 0, 0, Ω s The idea is to construct a ball B x r = x + B 0 r = x + {x Ω s : x s r} = {x Ω s : x x s r} centered at x containing a unique solution of by showing that a certain Newton-like operator T x = x Afx is a contraction on B x r This requires constructing A an approximate inverse of Df x = L x + DN x n order to do so, the structures of L x DN x need to be understood From 3 4, L x is a tridiagonal operator with entries growing to infinity at the rate k s L Moreover, since DN x : Ω s Ω s, then it is a bounded linear operator As mentioned above, the expectation is that the coefficients of x decay fast to zero Therefore, a reasonable approximation A for Df x is given by A = D 0 β m λ m µ m β m 0 λ m+ µ m+ β m+, 6 with D def = Df m x for m large enough Again we assume that m k 0 We wish to find its inverse in terms of D, β k k m, µ k k m λ k k m We assume therefore that A x = y, 7 3

where x y are the infinite vectors x 0 x x =, y = y 0 y The infinite part of 7 writes µ m β m 0 0 λ m+ µ m+ β m+ 0 0 λ m+2 µ m+2 β m+2 x m x m+ = y m λ m x m y m+ 8 We introduce the notations of the book of PG Ciarlet from Theorem 43-2 on page 42 in [] Note that only the δ n are really useful: a 2 = λ m+, a 3 = λ m+2,, b = µ m, b 2 = µ m+,, c = β m, c 2 = β m+, δ n n N defined by the induction formula δ 0 =, δ = b, δ n = b n δ n a n c n δ n 2, for n 2 Let define the tridiagonal operator T by T = b c 0 0 a 2 b 2 c 2 0 0 a 3 b 3 c 3 For any infinite vector x = x 0,, x k, T, we introduce the notation x F = x 0,, x m T x = x m,, x m+k, T Using the notation e =, 0, 0, 0, 0, T, the system 8 becomes T x = y λ m x m e 9 From Theorem 43-2 in [], we compute an LU-decomposition of the above tridiagonal operator in 9 as T = L U, where δ 0 0 δ 0 c 0 δ L = a 0 2 δ 0 δ 0 a 3 δ 2 U δ = 0 2 δ c 2 δ 0 0 3 δ 2 0 Hence, the system 8 becomes L z = y λ m x m e combined with U x = z, that is 0 0 z m y m λ m x m δ a 0 2 δ 0 z m+ δ 0 a 3 δ 2 = y m+, 4

combined with δ δ 0 c 0 δ 0 2 δ c 2 δ 0 0 3 δ 2 x m x m+ = Both infinite systems 2 can be solved explicitly System leads to z m = y m λ m x m, for any k z m+k = y m+k + z m z m+ 2 l δ k l a k l+2 a k+ y m+k l + k+ δ 0 a 2 a k+ λ m x m δ k δ k which we rewrite with infinite matrix/vectors notations as where z = z m z m+ z m+2 z = L [y λ m x m e ] = L y λ m x m v, 3, y = y m y m+ y m+2, v = L e = δ a 0 2 δ a 3 a 2 δ 0 δ 2 a 4 a 3 a 2 δ 0 δ 3 The second system 2 leads to the infinite sum for any k 0 x m+k = δ k δ k+ z m+k + l δ k δ k+l+ c k+ c k+l z m+k+l, which we also rewrite with infinite matrix/vector notations as Coupling 3 4, we end up with x = U z 4 x = U z = U [L y λ m x m v ] = U L y λ m x m w, 5 where w = U v Denoting U L l0 the first line of the infinite matrix U L remember that w 0 denotes the first element of w, we can rewrite the first line of 5 as x m = U L l0 y λ m x m w 0 6 Then, we investigate the finite part of the linear system 7 t writes x 0 x D + =, x m 2 x m 0 0 0 β m x m y 0 y y m 2 y m 5

or, according to 6, D x 0 x x m 2 x m + β m 0 0 0 U L l0 y λ m x m w 0 = y 0 y y m 2 y m Defining 0 0 0 0 0 0 0 0 K = D β m λ m, 0 0 0 0 0 0 0 w 0 together with its inverse K resp last column K c m, last line K l m, last south-east element K m,m of this inverse, we end up with x F = K { y F β m U L y l0 } K c m { } = K y F β m K c m { U L l0 with the tensor product notation The last line of this identity reads } y, 7 x m = K l m y F β m { U L l0 y } K m,m 8 Coming back to 5 using 8 we see that x = U L y λ m x m w = U L y λ m [K { l m y F β m U L y l0 { } = U L y λ m w K l m y F } K m,m ] w { U +β m λ m K m,m w L l0 y { } } = λ m w {K l m y F 9 } { U + U L + β m λ m K m,m {w L } y l0 Putting together 7 9, we end up with A = λ m { w } { K β m } {K l m } } { U K c m U L + Λ L l0 } 6

where { } { U Λ = β m λ m K m,m w L } l0 Now to get an approximate pseudo inverse of A we would like to get a numerical approximation of K However the definition of K involves w 0 which cannot be computed explicitly By definition w = U L e so using again the computations made in this section we get w 0 = U v 0 = δ 0 δ v m + = δ 0 δ + l δ 0 c c l v m+l δ l+ So given a computational parameter L, we define δ 2 0 δ l δ l+ c c l a 2 a l+ w = δ L 0 δ0 2 + c c l a 2 a l+, 20 δ δ l δ l+ 0 0 0 0 0 0 0 0 K = D β m λ m 0 0 0 0 0 0 0 w Now we can consider A m a numerically computed inverse of K then define the approximate pseudo inverse of A as { } { U A m β m A m c m L } l0 A = { } } λ m w {A m l m U L, + Λ where { } { U Λ = β m λ m A m m,m w L } l0 Lemma 2 Assuming m k 0 δ < 2, U : Ω s Ω s+s L Proof Let z Ω s x = U z Using 4 the formula above, we get x m+k δ k δ k+ z m+k + δ k δ k+ z m+k + δ k δ k+l+ c k+ c k+l z m+k+l 2 δ l δ k δ k+l+ b k+ b k+l z m+k+l 22 7

Now remember that for all k 2, δ k = b k δ k a k c k δ k 2, so We introduce u k = δ k δ k b k δ k δ k b k a k c k δ k 2 b k δ k δ2 b k δ k 2 δ k which then satisfies u = u k δ2, k 2 u k The study of the inductive sequence defined as above in the equality case yields that for any k, γ u k, where γ = 2 + 4 δ2 is the largest root of x = δ2 x see figure 08 08 06 06 04 04 02 02 0 0 0 02 04 06 08 a δ = 02 0 02 04 06 08 b δ = 05 Figure : The iterations of u n+ = δ 2 /u n with u = We can then rewrite 22 to get x m+k δ k δ k+ z m+k + δ l δ k δ k+l δ k+ δ k+l+ b k+ b k+l z m+k+l δ k δ k+ z m+k + δ l δ k+l u k+ u k+l δ k+l+ z m+k+l l δ δ k+l γ δ k+l+ z m+k+l l δ γ γ b k+l+ z m+k+l z l s δ C γ γ k + l + s L m + k + l s 8

Finally, since δ < 2 < γ, x Ω s+s L x m+k m + k s+s L z s C γ δ γ Lemma 22 Assuming m k 0 δ < 2, L : Ω s Ω s m + k s+s L k + s L m + k s Proof Let y Ω s z = L y Using 3 the formula above without the last term since we do not consider here L y λ m x m e, we get z m+k y m+k + y m+k + y m+k + y m+k + δ k l δ k δ l δ k l δ k a k l+2 a k+ y m+k l b k l+2 b k+ y m+k l δ l δ k l δ k b k l+ b k l+2 b k+ δ k l+ δ k b k l+ δ l b k+ u k l+ u k b k l+ y m+k l, where we use the sequence u k introduced in the previous proof So we get z m+k z m+k m + k s C 2 y s C l δ b k+ γ b k l+ y m+k l, y m+k l l sl s δ k + m + k, γ k + l m + k l k q k it is enough to show that given 0 < θ < q 0, θ l is bounded k l uniformly in k to prove that z Ω s ndeed, k q [ k 2 k ] q k θ l θ l + k l k l k l=[ k 2 ]+ [ k 2 ] 2 q θ l + θ [ k 2 ]+ k 2 kq 2q θ + θ[ k 2 ]+ k q+ 2, k θ l k l which is bounded uniformly in k since the last term goes to 0 when k goes to +, the proof is complete q 9

Proposition 23 Assuming m k 0 δ < 2, A : Ωs Ω s+s L Proof Consider y = y F, y T Ω s Let x = x F, x T = Ay Then, by definition of the operator A in 2, { } { U x F = A m y F β m A m c m L } y l0 = A m y F β m { U L l0 y } Am c m By the previous lemmas, U L y Ω s+sl, in particular U L l0 y = U L y 0 is well defined so is x F Again using 2, { } } x = λ m w {A m l m y F + U L y + Λy Remember that w = U L e, therefore w Ω s for any s, so according to the previous lemmas the definition of Λ see 2, x Ω s+s L 3 Computations of fixed points of the operator T Our main motivation for computing approximate inverses is to prove existence, in a mathematically rigorous sense, of a fixed point of the Newton-like operators T in a set centered at a numerical approximation x The Newton-like operator has the form T x = x Afx, 23 where A is the approximate inverse 2 of Df x computed using the theory of Section 2 By Proposition 23, A : Ω s Ω s+s L Since f : Ω s Ω s s L, then T : Ω s Ω s Before proceeding further, we endow Ω s with the operation of discrete convolution More precisely, given x = x k k 0, y = y k k 0 Ω s, extend x, y symmetrically by x = x k k Z, ỹ = y k k Z where x k = x k, ỹ k = y k, for k The discrete convolution of x y is denoted by x y defined by x y k = x k ỹ k2 k +k 2 =k k,k 2 Z t is known that for s >, Ω s, is an algebra under discrete convolution eg see [0], that is, given x, y Ω s, x y Ω s This fact will be especially useful when computing in practice the Z bounds as defined in the following results Theorem 3 For fixed s >, consider T : Ω s Ω s with T = T k k 0, T k R Assume that there exists a point x Ω s vectors Y = {Y k } k 0 Z = {Z k r} k 0 with Y k, Z k r R satisfying T x x k Y k 24 sup b,b 2 B 0r [ DT x + b b 2 Z ]k k r, k 0 25 f there exists r > 0 such that Y + Zr s < r, then the operator T is a contraction in B x r there exists a unique ˆx B x r that satisfies T ˆx = ˆx 0

Proof We perform the proof in two parts: i T x x s < r for all x B x r This implies that T B x r B x r; ii T is a contraction, that is there exists κ 0, such that for every x, y B x r, one has that T x T y s κ x y s For all k 0 for any x, y B x r, the Mean Value Theorem implies that T k x T k y = DT k zx y for some z = zk {tx + ty : t [0, ]} B x r Since r x y x y s B 0 r then from 25 T k x T k y = DT rx y kz x y s r x y s Z kr x y s 26 r The triangular inequality applied component-wise using y = x above gives T k x x k T k x T k x + T k x x k Z k r + Y k hence Therefore for any x B x r T k x x k Y k + Z k r T x x s = sup{ T k x x k ωk} s sup{ Y k + Z k r ωk} s = Y + Zr s < r k 0 k 0 This proves that T B x r B x r From 26, we obtain that for any x, y B x r, T k x T k y Z kr r x y s thus Since Zr s Y + Zr s < r, then T x T y s Zr s x y s 27 r κ def = Zr s <, r we can conclude that T : B x r B x r is a contraction An application of the Contraction Mapping Theorem on the complete metric space B x r gives the existence unicity of a solution ˆx of the equation T x = x in B x r Now we are going to see how to get such bounds Y Section 32 Zr Section 33 as well as how to find in an efficient way a r > 0 such that Y + Zr s < r Section 34 We start by computing some estimate to control the action of U L 3 Some preliminary computations Suppose y = y m, y m+, T is a given infinite vector We want to bound component wise x = U L y Let θ = δ γ Again introducing z = L y using the computations made in the proof of lemma 2 lemma 22, we get x m+k γ + θ l b k+l+ z m+k+l

Putting the two together, z m+k θ k l b k+ b l+ y m+l with + k+l x m+k θ k+2l j y m+j γ b j=0 j+ = y m+j + + θ k+2l j y m+j + + γ b j=0 j+ b j+ j=k+ l=j k = y m+j θ k j + γ b j=0 j+ θ 2 + y m+j θ j k b j+ θ 2 j=k+ = γ θ 2 θ k j y m+j + b j=0 j+ + θ j k y m+j b j+ j=k+ = η θ k j y m+j + µ m+j + θ j k y m+j, µ m+j j=0 η = j=k+ γ θ 2 n particular, for w = w m, w m+, T def = U L θ k+2l j e we have for all k 0 w m+k ηθ k µ m 28 More generally, if y is such that y m+k = 0 for any k K, then k K 2, x m+k η θ k l y K m+l µ m+l + l=k+ θ l k y m+l µ m+l 29 k K, K x m+k ηθ k y m+l θ l µ m+l 30 We will also need a bound of x m+k m + k s+s L that is uniform in k for k large enough x m+k m + k s+s L η y s C η y s C s+sl m + k θ k l + m + l + l=k+ s+sl m + k θ k l + θ m + l θ s+sl m + k θ l k m + l 2

We assume here that m 2 which will always be the case in practice, fix a computational parameter M N such that M max m ln θ s s L m ln θ + s + s L + 2 4m ln θ 2 ln 4, θ ln θ 2, m, say that for all k < M, x m+k m + k s+s L η y s C Then for k M, we split the remaining sum s+sl [ k 2 ] m + k θ k l = m + l θ k 2 θ M 2 k 2 s+sl k [ k] m + k θ k l + m + l s+sl m + k + θ k k m s+sl m + M + θ M 2 m 3 s+sl m + k θ k l + θ 32 m + l θ l=[ k 2 ] 2 2s+s L + θ s+sl m + k θ k l + m + l M M 2 2s+s L + θ m + k l=k [ k] s+sl m + k k m + M m + M M 33 Let us detail a bit why the last inequality holds First consider, for x > 0, ϕ x = θ x 2 xm + x s+s L, whose derivative is ϕ x = θ x ln θ xm + x s+s L + m + x s+s L + s + s L x m + x s+s L = m + x s+s L θ x ln θ m + xx + m + x + s + s L x = m + x s+s L θ x ln θ x 2 + m ln θ + s + s L + x + m Notice that since 0 < θ <, the discriminant of ln θx 2 + m ln θ + s + s L + x + m = m ln θ + s + s L + 2 4m ln θ, is positive but still, by the definition of M in 3, for any x M ϕ x 0 so ϕ k ϕ M for all k M Then consider ϕ 2 x = θ x x ϕ 2x = θ x ln θ 2 x x + so for x = θ x 2 x ln θ + 2, 4 ln θ 2, ϕ 2x 0 so ϕ 2 k ϕ 2 M for all k M Finally we consider s+sl m + k θ k l m + l s+sl 3

m + x ϕ 3 x = m + x then, x m + x x m + x ϕ 2 x 3x = m + x x 2 x + 2 x m = 2 x m + x x 2 so for x m, ϕ 3x 0 ϕ 3 k ϕ 3 M for all k M Putting all this together we get 33 So we can define s+sl χ = χθ, m, M, s, s L def = θ M M m + M 2 +θ M M 2 m 2 2s+s L + s+sl m + M θ m + M, M 34 state that for all k M x m+k m + k s+s L η y s χ + θ 35 C θ Finally, we need to bound the error made by using w instead of w 0 for the definition 2 of A Using 5 together with the sequence u l introduced in the proof of Lemma 2, we get w 0 w l=l δ 0 δ µ m = δ 0 2 δ l δ l+ c c l a 2 a l+ l=l l=l δ 2l u u l u 2 θ 2l 32 Computation of the Y bounds u l+ θ 2L µ m θ 2 36 Now for the rest of this paper we assume for the sake of clarity that the nonlinearity N of f in is a polynomial of degree two The generalization to a polynomial nonlinearity of higher degree requires only the use of the estimates developed in [5] to bound terms like x x p n where x,, x p B 0 r Moreover, as long as one is interested in problems with nonlinearities built from elementary functions of mathematical physics powers, exponential, trigonometric functions, rational, Bessel, elliptic integrals, etc, our method is applicable ndeed, since these nonlinearities are themselves solutions of first or second order linear ODEs, they can be appended to the original problem of interest in order to obtain a strictly polynomial nonlinearity, albeit in a higher number of variables This stard trick is explained in more details in [2] The first step is to bound T x x = Af x, 4

where here in the following, applied to vectors or matrices must be understood component-wise Note that since we suppose that f is at most quadratic x is constructed such that x k = 0 for all k m, we have that f x m+k = 0 for all k m According to 2, Af x F A m f x F + β m U L f x 0 so using 29 with K = m, we can set Using 2 again, Y F = A m f x F + β m η m 2 θ l f x m+l µ m+l Am Af x λ m f x βm l m F + A m m,m U so using 28, 29 30 again with K = m, we can set Am Y m+k = f x l m F + η θ k l f x m+l µ m+l Am Ỹ m+k = f x l m F m 2 + ηθ k m 2 βm + A m m,m η + η m 2 l=k+ Am c m, A m c m f x w 0 + U L f x, L θ l f x m+l µ m+l θ l k f x m+l, 0 k m 3 µ m+l m 2 βm + A m m,m η f x m+l θ l µ m+l, k m 2 We then take an integer M such that This yields that Therefore we can set M max k M, θ l f x m+l µ m+l ηθ k λ m µ m ηθ k λ m µ m m 2, s ln θ m 37 Ỹ m+k Ỹm+M ωm+m s ωm+k s Y m+k = Ỹm+k, ωm+m s m 2 k M Y m+k = Y m+m ωm+k s, k > M You will see in Section 34 the rationale behind this choice 33 Computation of the Z bounds For y, z B 0 r, we need to bound DT x + y z = ADf x + y z = AA z A Df x + y A z 5

33 Estimates for AA z According to 6 2, AA z 0 F = A m Dz F + β m U β m z m L T z + λ m z m e 0 A m c m = A m Dz F + β m z m A m c m β m z m + λ m z m w 0 A m c m = A m KzF + β m λ m w w 0 z m A m c m, so AA z F = A m K z F + β m λ m w w 0 z m A m c m Again using 6 2, we get AA z 0 = λ m A m l m Dz F + w + U β m z m L + Λ T z + λ m z m e = z + λ m w A m l m Dz F β m A m m,m z m + z m + β m A m m,m z + λ m z m w 0 = z + λ m A m l m Dz F + z m + β m λ m A m m,m z m w 0 w = z + λ m z m A m l m KzF + β m λ m A m m,m z m w w 0 = z + λ m A m K z F + β m λ m A m m,m z m w w 0 l m w w so AA z = λ m A m K l m z F + β m λ m A m m,m z m w w 0 w T We introduce W s = ω0 s,, ωk s, For z B 0 r we have, using 36, AA z F Am K WF s + β m λ m θ 2L µ m ωm s θ2 A m c m r, using also 28, AA z m+k Am K WF s + l m As in Section 32, we then assume 37 define Z by β m λ m θ 2L µ m ωm s θ2 A m m,m ZF = Am K WF s + β m λ m θ 2L µ m ωm s θ2 A m c m r, ηθ k λ m r, k 0 µ m 6

Zm+k = Am K WF s + l m β m λ m θ 2L µ m ωm s θ2 A m m,m Z m+k = Z m+m ωm+m s ωm+k s, k > M By the definition of M, AA z Z, for all z B 0 r 332 Estimates for A Df x + y A z ηθ k λ m r, 0 k M, µ m We assumed that the nonlinear part N was polynomial of degree 2 so Df x + y can be written as a finite Taylor expansion: Df x + y = Df x + D 2 f x y Df x + y A z = Df x A z + D 2 f x y, z f we denote by σ the coefficient of degree 2 of f, we have that D 2 f x y, z = 2σy z We then bound the convolution product using Lemma 32 Let s 2, x, y Ω s, K 6 L computational parameters k 0, x y k αkk s x s y s ωk s, where + 2 L L αkk s = 2 + 2 L 2 + 2 l s + 2 s L s, k = 0 k l s + 2 s L s + l s + 2 K s L s + 2 K k s l s k l s, k < K s 4 lnk 2 + + π2 6 2 K 3 K + s, k K 2 Proof See [3] for a proof of this bound [0] for a similar bound for < s < 2 t is important to notice here that αk sk = αs K K for all k K For the rest of this paper we assume that m is taken 6 which will allow us to use Lemma 32 with K = m For y, z B 0 r, we get D 2 f x y, z 2 σ α s mw s r 2, We define C 2 = 2 σ α s mw s so that for all y, z B 0 r D 2 f x y, z C 2 r 2 7

Now we focus on the order one term According to the definition 6 of A, we have 0 Df x A z 0 = Df xz F F Df m xz F 0 β m z m = 2σ x z F x z F F, where in the convolution product, z F must be understood as the infinite vector z F, 0,, 0, T So Df x A z 0 = 0 for all z B 0 r, Df x A z k 2 σ r m ω s l=m k k+l x l, k m Then, remembering that Df x = L + DN x 6, we get that Df x A z = DN xz = 2σ x z, so using Lemma 32, for all z B 0 r, Putting both together we define Df x A z m+k 2 σ αs m+k m x s ωm+k s r, k 0 C 0 x = 0, C k x = 2 σ m ω s l=m k k+l x l, k m, to get that for all z B 0 r Cm+k x = 2 σ αs mm x s ωm+k s, k 0, Df x A z C xr Finally, A Df x + y A z A C xr + C 2 r 2, we are left to bound A C x A C 2 According to 2, A C x F A m CF x + β m U L C x 0 A m c m, using 32 U L C x 0 η C x s C θω s+s L m 2η σ αs mm x s C θω s+s L m 8

so we set DF x = A m CF x + 2 β m η σ αmm x s s C θω s+s L m Still according to 2, A C x λ m A m l m C F x w + U λ m L A m c m C x + λm β m A m U m,m A m l m C F x + 2 β m A m m,m η σ α s mm x s C θω s+s L m Now we take some M as in 3 use 28, 32 35 to set Dm+k x = A m l m CF x + 2 β m A m m,m η σ αmm x s s C θω s+s η λ m L µ m θk D m+m x = so that Similarly, we set D 2 m+m = so that D 2 m+k = Finally we can set + 2η σ αs mm x s C ω s+s L m+k θ k l m + k m + l m s+sl + θ θ A m l m CF x + 2 β m A m m,m η σ αmm x s s C θω s+s L + 2η σ αs mm x s C ω s+s L m+m χ + θ, θ Dm+k x = Dm+M x ωs m+m ωm+k s, k > M A C x D x D 2 F = A m C 2 F + 2 β m η σ α s mm C θω s+s L m m A m c m,, 0 k < M, A m l m CF 2 + 2 β m A m m,m η σ αmm s C θω s+s η λ m L µ m θk + 2η σ αs mm C ω s+s L m+k θ k l m + k m + l m s+sl + θ θ η λ m µ m θm, 0 k < M, A m l m CF 2 + 2 β m A m m,m η σ αmm s C θω s+s η λ m L m µ m θm + 2η σ αs mm C ω s+s L D 2 m+k = D 2 m+m ωm+m s ωm+k s, k > M A C 2 D 2 Z 2 = D xr + D 2 r 2 Z = Z + Z 2 is such that, for all y, z B 0 r DT x + y z Z L C x 0 w w + U L C x m+m χ + θ, θ 9

34 Radii polynomials interval arithmetic We are now left to find a radius r > 0 such that for every k 0, the radii polynomials {P k r} k satisfy P k r def = Y k + Z k r r ωk s < 0 Note that since we constructed Y Z so that for every k M Y m+k = Y m+m ω s m+m ω s m+k ωm+m s Z m+k = Z m+m ωm+k s, it is enough to find a r > 0 such that for all 0 k m + M, P k r < 0 To do so, we compute numerically for each 0 k m + M def k = {r > 0 P k r < 0}, m+m def = k k=0 f is empty then the proof fails, we should try again with some larger parameters m M f is non empty, we pick an r check rigorously, using the interval arithmetic package NTLAB [4], that for all 0 k m + M, P k r < 0 which according to Theorem 3 proves that T defined in 23 is a contraction on B s x, r, yielding the existence of a unique solution of fx = 0 in B s x, r 4 An example application Equations of the following form 2 + cos ξu ξ + uξ = σuξ 2 + gξ 38 u 0 = u π = 0, where g is a 2π-periodic even smooth function, fall into the framework developed in Section 2 ndeed consider the cosine Fourier expansions of u g uξ = k Z x k coskξ, gξ = k Z g k coskξ Then 38 can be rewritten as fx = 0, where for all k f 0 x def = x 0 + x + σ x x 0 g 0 f k x def = 2 k 2 x k + + 2k 2 x k + 2 k + 2 x k+ + σ x x k g k 39 Then we do have that the linear part of 39 is as in 3, given by L k x = λ k x k + µ k x k + β k x k+, with def def µ 0 =, β 0 =, 20

for all k Let s fix some m 2 With we have together with def λ k = 2 k 2 def, µ k = + 2k 2 def β k = 2 k + 2 C = 2, C 2 = 3 δ = 4 k m, k, λ k k 2 C µ k k 2 We now focus on the example where, µ k, k 2 β k k 2 m + 2 m 2 +, 2 C 2, λ k µ k gξ def = 2 + 3 cosξ + 2 cos2ξ,, β k δ so that uξ = cosξ is a trivial solution for σ = 0 n the next section we are going to use rigorous computation to prove the existence of solutions for σ 0 compute those solutions 4 Results Starting from σ = 0 we first use stard pseudo-arclength continuation techniques to numerically get some non trivial approximate solutions for σ 0 We computed 250 different solutions half for σ > 0 the other half for σ < 0 See Figure 2 for a diagram summing up those computations, where each point represent a solution of 38 µ k 7 6 5 kxk 4 3 2 0 08 06 04 02 0 02 04 06 Figure 2: Branch of solutions of 38 Then we use the rigorous computation method described in this paper to prove, for each numerical solution, the existence of a true solution in a small neighbourhood of the numerical 2

approximation We keep m = 20 Fourier coefficient for the numerical computation use M = 20 the decay rate s = 2 for the proof The bounds of Lemma 32 as well as the error on ω 36 are computed with L = 00 For each numerical solution the proof succeeds More precisely, the defined in Section 34 on which all radii polynomials should be negative always contains [4 0, 0 4 ] we prove rigorously using interval arithmetic that indeed they all are negative for r = 0 0 Hence the hypotheses of Theorem 3 hold we have that within a ball of radius r = 0 0 in Ω s centered on the numerical approximation lays a unique solution to 38 Therefore the existence of the solutions represented Figure 2 is rigorously proven, within a margin of error that is to small to be depicted The codes to perform the proofs can be found at [7] Notice that existence of solutions of 38 could certainly have been obtained in different more classical ways, for example using perturbative methods when σ is close to 0, or using a variational approach that is, considering 38 as the Euler-Lagrange equation related to the critical points of a functional, or even using topological tools such as the Leray-Schauder theory The advantage of our method is that it gives us more quantitative information than those approaches: indeed it enables to provide more than one solution for some values of σ,, maybe more importantly, it gives a very precise localization of this or these solutions in terms of Fourier coefficients something that looks very hard to obtain with qualitative PDEs methods 5 Conclusion A first interesting future direction of research would be to adapt our proposed approach to rigorously compute connecting orbits of ODEs using spectral methods For instance, we would like to investigate the possibility of combining Hermite spectral methods with our approach to compute homoclinic orbits eg see [5, 6] Since the differential operator in frequency space of the Hermite functions is tridiagonal, adapting our approach to this class of operator could lead to a new rigorous numerical method for connecting orbits t would also be interesting to adapt our method to the case of looking for solutions in the sequence space l def ν = {x = x k k 0 : x ν = x k ν k < } k 0 for some ν With this choice of Banach space, we could exploit the fact that l ν is naturally a Banach algebra under discrete convolutions This could greatly simplify the nonlinear analysis Note that assumption 5 requires the tridiagonal operator to have symmetric ratios between the diagonal terms the upper lower diagonal terms This is a restriction that we hope could be relaxed Since many interesting problems involve tridiagonal operators with non symmetric ratios as in the case of differentiation in frequency space of the Hermite functions, we believe that this is a promising route to follow Finally, generalizing our approach to problems with block-tridiagonal structures could also be a valuable project Acknowledgement The research leading to this paper was partially funded by the french ANR blanche project Kibord: ANR-3-BS0-0004 22

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