Lecture 4: High oscillations

Transcription

1 Geometric Numerical Integration TU München Ernst Hairer January February Lecture 4: High oscillations Table of contents A Fermi Pasta Ulam type problem Application of classical integrators 4 3 Exponential trigonometric integrators 5 4 Modulated Fourier expansion 9 5 Invariants of the modulated Fourier expansion 6 Long-time energy conservation 5 7 Behavior of the Störmer Verlet discretization 6 8 Exercises 9 This lecture deals with numerical approaches for second order Hamiltonian systems with highly oscillatory solutions. We focus on the situation where the product of time step size and highest frequency in the system is not small. A Fermi Pasta Ulam type problem The problem of Fermi, Pasta & Ulam see also the recent lecture notes 3 is a simple model for simulations in statistical mechanics which revealed highly unexpected dynamical behaviour. We consider a modification consisting of a chain of m mass points, connected with alternating soft nonlinear and stiff linear springs, and fixed at the end points see Galgani, Giorgilli, Martinoli & Vanzini 4 and Figure. The variables q,...,q m q = q m+ = stand for the displacements Large parts are taken from Geometric Numerical Integration by Hairer, Lubich & Wanner. E. Fermi, J. Pasta & S. Ulam, Studies of non linear problems. Los Alamos Report No. LA , later published in E. Fermi: Collected Papers Chicago 965, and Lect. Appl. Math. 5, G. Gallavotti ed., The Fermi Pasta Ulam problem. Lecture Notes in Physics, vol. 78, Springer, Berlin, 8. A status report. 4 L. Galgani, A. Giorgilli, A. Martinoli & S. Vanzini, On the problem of energy equipartition for large systems of the Fermi Pasta Ulam type: analytical and numerical estimates, Physica D 59 99,

2 q q q m q m stiff harmonic soft nonlinear Figure : Chain with alternating soft nonlinear and stiff linear springs of the mass points, and p i = q i for their velocities. The motion is described by a Hamiltonian system with total energy m Hp, q = p i + pi ω m m + q i q i + q i+ q i 4, 4 i= where ω is assumed to be large. With the symplectic change of coordinates i= x,i = q i + q i /, x,i = q i q i /, ẋ,i = p i + p i /, ẋ,i = p i p i /, where x,i is a scaled displacement of the ith stiff spring, x,i a scaled expansion compression of the ith stiff spring, we get a Hamiltonian system with Hx, ẋ = m ẋ,i + ẋ,i i= m + i= m i= + ω x,i + x, x, i= 4 x,i+ x,i+ x,i x,i + x,m + x,m 4 Besides the fact that the total energy is exactly conserved, the system has a further interesting feature. We consider the oscillatory energy I = I I m, I j = ẋ,j + ω x,j, 3 where I j denotes the harmonic energy of the jth stiff spring, and the kinetic energies corresponding to the slow and fast motions m m T = ẋ,i, T = ẋ,i. 4 i= For an illustration we choose m = 3 as in Figure, ω = 5, x, =, ẋ, =, x, = ω, ẋ, =, and zero for the remaining initial values. Figure shows the different time scales that are present in the evolution of the system. i=.

3 H I H I T T T.4.8 H I h = /ω H I I I I Figure : Different time scales in the FPU type problem ω = 5. Time Scale ω. The vibration of the stiff linear springs is nearly harmonic with almost-period π/ω. This is illustrated by the plot of T in the first picture. Time Scale ω. This is the time scale of the motion of the soft nonlinear springs, as is exemplified by the plot of T in the second picture of Figure. Time Scale ω. A slow energy exchange among the stiff I springs takes place on the scale ω. In the third picture see also the zoom to the right, the initially excited first.4 I stiff spring passes energy to the second one, and then also the third stiff spring begins to vibrate. The picture also illustrates that the problem is very sensitive to perturbations of the initial data: the grey curves of each of I I, I, I 3 correspond to initial data where 5 has been added to x,, ẋ, and ẋ,. The displayed solutions of the first three pictures have been computed very accurately by an adaptive integrator. Time Scale ω N, N. The oscillatory energy I has only Oω deviations from the initial value over very long time intervals. The fourth picture of Figure shows the total energy H and the oscillatory energy I as computed by an exponential integrator see Section 3 with step size h = /ω =.4, which is nearly as large as the time interval of the first picture. No drift is seen for H or I. 3

4 Application of classical integrators Which of the methods, discussed in Lecture, produce qualitatively correct approximations when the product of the step size h with the high frequency ω is relatively large? Linear stability analysis. To get an idea of the maximum admissible step size, we neglect the nonlinear term in the differential equation, so that it splits into the two-dimensional problems ẏ,i =, ẋ,i = y,i and ẏ,i = ω x,i, ẋ,i = y,i. 5 Omitting the subscripts, the solution of 5 is yt cosωt sin ωt y =. ω xt sin ωt cosωt ω x The numerical solution of a one-step method applied to 5 yields yn+ yn = Mhω, 6 ω x n+ ω x n and the eigenvalues λ i of Mhω determine the long-time behaviour of the numerical solution. Stability i.e., boundedness of the solution of 6 requires the eigenvalues to be less than or equal to one in modulus. For the explicit Euler method we have λ, = ± ihω, so that the energy I n = yn + ω x n/ increases as + h ω n/. For the implicit Euler method we have λ, = ± ihω, and the energy decreases as + h ω n/. For the implicit midpoint rule and for all symplectic Runge Kutta methods, the matrix Mhω is orthogonal and therefore I n is exactly preserved for all h and n. Finally, for the symplectic Euler method and for the Störmer Verlet scheme we have Mhω = hω hω h ω, Mhω = h ω hω h ω 4 hω h ω For both matrices, the characteristic polynomial is λ h ω λ +, so that the eigenvalues are of modulus one if and only if hω. Numerical experiments. We apply several methods to the FPU type problem, with ω = 5 and initial data as in Figure. Figure 3 presents the numerical results for H.8, I, I, I, I 3 the last picture only for I and I obtained with the implicit midpoint rule, the classical Runge Kutta method of order 4, and the Störmer Verlet scheme. For the small step size h =. all methods give satisfactory results, although the energy exchange is not reproduced accurately 4.

5 implicit mid-point h =. Runge Kutta, order 4 h =. Störmer Verlet h =. h =.5 h =.8 h =.5 Figure 3: Numerical solution for the FPU problem with data as in Figure. over long times compare with exact solution in Figure. The explicit Runge Kutta method gives completely wrong solutions for larger step sizes or for small step sizes and larger time intervals. The values of H and I are still bounded over very long time intervals for the Störmer Verlet method, but the relative error is very large for step sizes close to the stability limit h =.5 corresponds to hω = 5. These phenomena call for an explanation, and for numerical methods with an improved behaviour. 3 Exponential trigonometric integrators The Störmer Verlet scheme for ẍ = gx is obtained by replacing the second derivative with the difference h x n+ x n +x n. For a differential equation ẍ + Ω x = gx, gx = Ux, Ω = 7 ωi we replace the linear part by h x n+ coshω x n + x n, which reproduces the exact solution for ẍ + Ω x =. The derivative of the exact solution for this linear problem satisfies with sincξ = sin ξ/ξ h sinchω ẋt = xt + h xt h, which leads to an approximation of the derivative. 5

6 Two-step formulation. We consider numerical integrators of the form x n+ coshω x n + x n = h ΨgΦx n h sinchω ẋ n = x n+ x n. 8 Here Ψ = ψhω and Φ = φhω, where the filter functions ψ and φ are bounded, even, real-valued functions with ψ = φ =. In our numerical experiments we will consider the following choices of ψ and φ A ψξ = sinc ξ φξ = Gautschi B ψξ = sincξ φξ = Deuflhard C ψξ = sincξ φξ φξ = sincξ García-Archilla & al. 3 D ψξ = sinc ξ φξ of 4 Hochbruck & Lubich 4 E ψξ = sinc ξ φξ = Hairer & Lubich 5 One-step formulation. Eliminating x n from the two relations in 8, we obtain the following equations x n+ = coshωx n + Ω sin hωẋ n + h Ψ g n 9 ẋ n+ = Ω sin hωx n + cos hωẋ n + h Ψ g n + Ψ g n+ where g n = gφx n and Ψ = ψ hω, Ψ = ψ hω with even functions ψ, ψ defined by ψξ = sincξ ψ ξ, ψ ξ = cosξ ψ ξ. Method 9- is of order for h, and it is symmetric whenever is satisfied. The method is symplectic if in addition see Exercise ψξ = sincξ φξ. W. Gautschi, Numerical integration of ordinary differential equations based on trigonometric polynomials, Numer. Math P. Deuflhard, A study of extrapolation methods based on multistep schemes without parasitic solutions, Z. angew. Math. Phys B. García-Archilla, J.M. Sanz-Serna & R.D. Skeel, Long-time-step methods for oscillatory differential equations, SIAM J. Sci. Comput M. Hochbruck & Ch. Lubich, A Gautschi-type method for oscillatory second-order differential equations, Numer. Math a E. Hairer & Ch. Lubich, Long-time energy conservation of numerical methods for oscillatory differential equations, SIAM J. Numer. Anal

7 Energy exchange between stiff components. Figure 4 shows the energy exchange of the six methods A-F applied to the FPU problem with the same data as in Figure. The figures show again the oscillatory energies I, I, I 3 of the stiff springs, their sum I = I +I +I 3 and the total energy H.8 as functions of time on the interval t. Only the methods B, D and F give a good approximation of the energy exchange between the stiff springs. By the use of mod- A B C D E F Figure 4: Energy exchange between stiff springs for methods A-F h =.35, ω = 5. Method F is not considered in these notes. A B C π π 3π 4π π π 3π 4π π π 3π 4π D E F π π 3π 4π π π 3π 4π π π 3π 4π Figure 5: Maximum error of the total energy on the interval [, ] for methods A - F as a function of hω step size h =.. 7

8 ulated Fourier expansions see below it can be shown that a necessary condition for a correct approximation of the energy exchange is ψhωφhω = sinchω, which is satisfied for method B. The good behaviour of method D comes from the fact that here ψhωφhω.95 sinchω for hω =.5. Near-conservation of total and oscillatory energies. Figure 5 shows the maximum error of the total energy H as a function of the scaled frequency hω step size h =.. We consider the long time interval [, ]. The pictures for the different methods show that in general the total energy is well conserved. Exceptions are near integral multiples of π. Certain methods show a bad energy conservation close to odd multiples of π, other methods close to even multiples of π. Only method E shows a uniformly good behaviour for all frequencies. In B C F.97π π.3π.97π π.3π.97π π.3π Figure 6: Zoom close to π or π of the maximum error of the total energy on the interval [, ] for three methods as a function of hω step size h =.. A B C π π 3π 4π π π 3π 4π π π 3π 4π D E F π π 3π 4π π π 3π 4π π π 3π 4π Figure 7: Maximum deviation of the oscillatory energy on the interval [, ] for methods A - F as a function of hω step size h =.. 8

9 Figure 6 we show in more detail what happens close to such integral multiples of π. If there is a difficulty close to π, it is typically in an entire neighbourhood. Close to π, the picture is different. Method C has good energy conservation for values of hω that are very close to π, but there are small intervals to the left and to the right, where the error in the total energy is large. Unlike the other methods shown, method B has poor energy conservation in rather large intervals around even multiples of π. Methods A and D conserve the total energy particularly well, for hω away from integral multiples of π. Figure 7 shows similar pictures where the total energy H is replaced by the oscillatory energy I. For the exact solution we have It = Const + Oω. It is therefore not surprising that this quantity is not well conserved for small values of ω. None of the considered methods conserves both quantities H and I uniformly for all values of hω. 4 Modulated Fourier expansion Let us consider second order ordinary differential equations ẍ + Ω x = gx, gx = Ux, Ω =. 3 ωi To motivate a suitable ansatz for the solution, we notice that the general solution of ẍ + ω x = is xt = c e iωt + c e iωt, that of ẍ + ω x = x is xt = e iωt z t+e iωt z t with z ± t = c ± e ±iαt and α = ω + ω = Oω. If gx contains quadratic terms, also products of e ±iωt z ± t will be involved. Modulated Fourier expansion of the exact solution. We aim in writing the solution of 3 as xt = z e ikωt z k t, z k k t = t z k k Z t, 4 where the coefficient functions z k t are partitioned according to the partitioning of Ω in 3. This expansion is called modulated Fourier expansion 5 of the solution. It is essential that the coefficient functions z k t do not contain high oscillations. More precisely, we search for functions such that the derivatives of z k t up to a certain order are bounded uniformly when ω. 5 E. Hairer & Ch. Lubich, Long-time energy conservation of numerical methods for oscillatory differential equations, SIAM J. Numer. Anal

10 Inserting the expansion 4 into the differential equation 3, and comparing the coefficients of e ikωt yields for ω =, ω = ω z k j + ikωżk j + ω j kω z k j = sα=k m! gm j z z α,...,z αm, 5 where the sum ranges over all m and all multi-indices α = α,..., α m with α j, having a given sum sα = m j= α j. Among the solutions of these second order differential equations we have to select those, whose derivatives are uniformly bounded in ω. To achieve this, we determine the dominant term in the left-hand side of 5 for ω, put the other terms to the right-hand side, and eliminate higher derivatives by iteration until a sufficiently high order. In this way we get a second order differential equation for z t, first order differential equations for z ± t, algebraic equations for all other zj kt. This construction can best be understood by first studying the smooth solution of problems like z + ω z = gt or z + iωż = gt. Initial values for the differential equations are obtained from 4 and its derivative taken at t = : x = z k, ẋ = ikωz k + ż k. k Z k Z Using the above algebraic relations for zj k t and the differential equations for z ± t, these equations constitute a nonlinear system that defines uniquely z, ż, z± as functions of x and ẋ. This construction yields a unique formal expansion of the form 4 for the initial value problem 3. Because of the non-convergence of the series, they have to be truncated by neglecting terms of size Oω N. The above construction shows that under the bounded-energy condition for the initial values, ẋ + Ωx E, 6 the coefficient functions of 4 satisfy on a finite time interval t T, z = O, z ± = Oω 3, z k = Oω k z = Oω, z ± = Oω, z k = Oω k. 7

11 Modulated Fourier expansion of the numerical solution. To get insight into the numerical solution of x n+ coshωx n + x n = h ΨgΦx n 8 we aim in writing the numerical solution as x n = x h nh with x h t = k Z e ikωt zh k t. 9 As we proceeded for the analytic solution, we insert the ansatz 9 into the numerical method 8 and compare the coefficients of e ikωt. Using the relation e ikωt coskωh zht k + h! zk ht +... x h t + h + x h t h = k Z + i sinkωh hż k h h3... t + z k 3! h t and the abbreviations c k = cos h kω, sk j = sinc h ω j kω we obtain i 3 sk kωh... z k j + ck z j k + isk kωżk j + sk j s k j ωj kω zj k = m! Ψ jg m j Φz Φz α,...,φz αm, sα=k which reduces to 5 in the limit h. Again we have to determine the dominant terms in the right-hand side this time asymptotically for h and hω c >, so that ω = Oh. Under the numerical non-resonance condition sin kωh c h for k =,..., N we get a system of differential and algebraic equations for zj k. As before, we get a second order differential equation for z t, first order differential equations for z ± t, and algebraic relations for the other zk j t. Also initial values are obtained in the same way as for the analytic solution. To avoid the difficulty with the nonconvergence of the arising series, we truncate them by neglecting terms of size Oh N+. Under suitable assumptions on the filter functions, the coefficients of the modulated Fourier expansion are bounded as follows: z = O, z± = Oω, z k = Oω k z = Oω, z ± = Oω, z k = Oω k. 6 Components of the functions z k t of 9 are written without the subscript h. Notice, nevertheless, that they are different from those of 4. We want to avoid a further subscript, and hope that there will not arise any confusion.

12 An expansion for the derivative ẋ n = x h nh can be immediately obtained by inserting 9 into the second equation of 8. Notice that x h t is not the time derivative of x h t. 5 Invariants of the modulated Fourier expansion The equation 3 is a Hamiltonian system with the Hamiltonian Hx, ẋ = ẋtẋ + x T Ω x + Ux, and we have seen in the numerical experiments that the quantity plays an important role. Ix, ẋ = ẋ + ω x 3 Invariants for the exact solution. In the modulated Fourier expansion of the analytic solution xt, denote y k t = e ikωt z k t for all k, and collect them in By 5 these functions satisfy y =...,y, y, y, y, y,.... ÿ k + Ω y k = sα=k m! Um+ y y α,...,y αm. 4 The important and somewhat surprising observation is that with the expression Uy = Uy + m! Um y y α,...,y αm 5 sα= the differential equation for yt obtains the Hamiltonian structure ÿ k + Ω y k = y k Uy. 6 This system does not only have the Hamiltonian as a first integrals, but it has the following two formal first integrals: Hy, ẏ = ẏ k T ẏ k + y k T Ω y k + Uy k Z Iy, ẏ = i ω 7 k y k T ẏ k. k Z

13 Theorem For a fixed N, and with a suitable truncation of the series in 5 and 7, we have Hyt, ẏt = Hy, ẏ + Oω N 8 Hyt, ẏt = Hxt, ẋt + Oω, 9 where the constants symbolized by O are independent of ω and t with t T, but depend on E of 6, N and T. Similar estimates are obtained for the second invariant Iyt, ẏt = Iy, ẏ + Oω N 3 Iyt, ẏt = Ixt, ẋt + Oω. 3 Proof. Taking the scalar product of 6 with ẏ k and summing over all k shows that both sides become total derivatives. This leads to the explicit formula for Hy, ẏ. The proof of 3 is somewhat more tricky. We note that with the vector yλ, whose components are e ikλ y k, the expression Uyλ is independent of λ. Its derivative with respect to λ thus yields = d dλ Uyλ λ= = i k y k T k Uy, 3 k Z for all y. Using this relation, a multiplication of 6 with k y k leads to the explicit formula for Iy, ẏ which proves 3 after suitable truncation. By the bounds 7, we have for t T Hy, ẏ = ẏ + ẏ + ω y + Uy + Oω. 33 On the other hand, we have from 3 and 4 Hx, ẋ = ẏ + ẏ + ẏ + ω y + y + Uy + Oω. 34 Using y = eiωt z and ẏ = eiωt ż + iωz from ż = Oω that ẏ + ẏ = iωy y together with y = y, it follows + Oω and ẏ = ω y + Oω. Inserted into 33 and 34, this yields 9. Property 3 is obtained in the same way as for the Hamiltonian. Invariants for the numerical solution. We introduce the differential operator LhD := e hd coshω + e hd = cosihd coshω = 4 sin hω + h D + h4 D 4... so that the function x h t of 9 formally satisfies the difference scheme 35 LhDx h t = h Ψg Φx h t. 36 We insert the modulated Fourier expansion x h t = k eikωt z k h t = k yk h t 3

14 with yh kt = eikωt zh k t, expand the right-hand side into a Taylor series around Φyh t, and compare the coefficients containing the factor eikωt. This yields, for gx = Ux, the following formal equations for the functions yh kt, LhDyh k = h Ψ m! Um+ Φyh Φy α h,...,φyαm h, 37 sα=k which are the numerical analogue of 4. With the extended potential U h y h = UΦyh + m! Um Φyh Φy α h,...,φyαm h, 38 sα= this system can be formally written as Ψ Φh LhD y k h = k U h y h. 39 The essential difference to the situation in 6 is the appearance of higher even derivatives in the left-hand expression. To find a first invariant of the differential equation 39, we take its scalar product with ẏ k h = yh k, so that the right-hand side becomes the total derivative d Uy dt h. Also the left-hand sides can be written as a total derivative, which is a consequence of relations of the type Re ż T z l = Re d dt żt z l... z l T z l+ ± zl T z l. To find a second invariant, we take the scalar product with k y k h, and use an identity similar to that of 3, so that the right-hand side vanishes. To write the left-hand side as a total derivative we use Imz T z l+ = Im d z T z l+ ż T z l +... ± z l T z l+. dt A careful elaboration of these ideas see Section XIII.6 of our monograph on Geometric Numerical Integration we obtain the following result. Theorem Under suitable assumptions on the filter functions and on the step size, which will be stated below in more detail, there exist functions H h y and I h y such that the following holds for t = nh T : H h yt = Hh y + Oth N, I h yt = Ih y + Oth N H h yt = Hxn, ẋ n + Oh, I h yt = Ixn, ẋ n + Oh. The constants symbolized by O depend on the energy E, on the truncation index N and on T, but not on ω. 4

15 6 Long-time energy conservation The statements of the preceding section are valid on intervals [, T], where T is a fixed value independent of ω. We show here how the results on energy conservation can be extended to much longer time intervals. Conservation of the oscillatory energy for the analytic solution. Whereas the conservation of the total energy Hx, ẋ along the analytic solution is obvious, the near-conservation of the oscillatory energy is a non-trivial, but typical feature of highly oscillatory problems. Theorem 3 If the solution xt of 3 stays in a compact set for t ω N, then Ixt, ẋt = Ix, ẋ + Oω + Otω N. The constants symbolized by O are independent of ω and t with t ω N, but depend on the initial energy E and on the truncation index N. Proof. With a fixed T >, let y j denote the vector of the modulated Fourier expansion terms that correspond to starting values xjt, ẋjt on the exact solution. For t = n + θt with θ <, we have by Theorem, We note Ixt, ẋt Ix, ẋ = Iy n θt, ẏ n θt + Oω Iy, ẏ + Oω = Iy n θt, ẏ n θt Iy n, ẏ n + n Iy j+, ẏ j+ Iy j, ẏ j + Oω. j= Iy j+, ẏ j+ Iy j, ẏ j = Oω N, because, by the quasi-uniqueness of the coefficient functions, we have for the truncated modulated Fourier expansion that y j+ = y j T + Oω N and ẏ j+ = ẏ j T + Oω N. This yields the result. Energy conservation for the numerical solution. We are now able to prove one of the main results of this lecture the near-conservation of the total energy H and the oscillatory energy I over long time intervals. In the previous section we emphasized the ideas without giving details on the assumptions for rigorous error estimates. We state them here without proof. the energy bound: ẋ + Ωx E ; 5

16 the condition on the numerical solution: the values Φx n stay in a compact subset of a domain on which the potential U is smooth; the conditions on the filter functions: ψ and φ are even, real-analytic, and have no real zeros other than integral multiples of π; furthermore, they satisfy ψ = φ = and ψhω C sinc hω, φhω C sinc hω, ψhωφhω C 3 sinchω ; the condition hω c > ; the non-resonance condition : for some N, 4 sin khω c h for k =,..., N. Theorem 4 Under the above conditions, the numerical solution of 3 obtained by the method 9 with satisfies, for nh h N+, Hx n, ẋ n = Hx, ẋ + Oh Ix n, ẋ n = Ix, ẋ + Oh. The constants symbolized by O are independent of n, h, ω satisfying the above conditions, but depend on N and the constants in the conditions. Proof. These long-time error bounds can be obtained as in the proof of Theorem 3. One only has to use the estimates of Theorem instead of those of Theorem. 7 Behavior of the Störmer Verlet discretization In applications, the Störmer Verlet method is often used with step sizes h for which the product with the highest frequency ω is not small, so that backward error analysis does not provide any insight into the long-time energy preservation. For example, in spatially discretized wave equations, hω is known as the CFL number, which is typically kept near. Values of hω around.5 are often used in molecular dynamics simulations. Consider now applying the Störmer Verlet method to the nonlinear model problem 3, x n+ x n + x n = h Ω x n h Ux n h ẋ n = x n+ x n 4 with hω < for linear stability. The method is made accessible to the analysis 6

17 of the preceding sections by rewriting it as an exponential integrator x n+ cosh Ω x n + x n = h Ux n Ω =. 4 h sinch Ω x n = x n+ x ωi n with ψξ = φξ =, and with modified frequency ω, defined by hω h ω = cosh ω or, equivalently, sin The velocity approximation x n is related to ẋ n by = hω. ẋ n = sinch Ω x n or ẋ n, = x n, ẋ n, = sinch ω x n,. 43 Theorem 5 Let the Störmer Verlet method be applied to 3 with initial values satisfying 6, and with a step size h for which < c hω c < and sin kh ω c h for k =,..., N for some N and c >. Suppose further that the numerical solution values x n stay in a region on which all derivatives of U are bounded. Then, we have for nh h N+ Hx n, ẋ n + γ ẋ n, = Const + Oh Ix n, ẋ n + γ ẋ n, = Const + Oh along the numerical solution, where γ γ = hω/ hω/ The constants symbolized by O are independent of n, h, ω with the above conditions. Proof. The condition < c hω c < implies sin kh ω c > for k =,, and hence conditions 4 are trivially satisfied with h ω instead of hω. We are thus in the position to apply Theorem 4 to 4, which yields Hx n, x n = Hx, x + Oh Ĩx n, x n = Ĩx, x + Oh for nh h N+, 44 where H and Ĩ are defined in the same way as H and I, but with ω in place of ω. 7 hω

18 With the relations 43 we get Ĩx n, x n = x n, + ω x n, ω = Ix n, ẋ n ω ẋ n, + x n, ω = Ix n, ẋ n + ω γ ẋ n,. 45 Similarly, we get for the Hamiltonian Hx n, x n = ẋ n, + x n, + ω x n, + Ux n and hence 44 yields the result. = Hx n, ẋ n Ix n, ẋ n + Ĩx n, x n = Hx n, ẋ n + γ ẋ n, + ω Ĩxn, x ω n, 46 For fixed hω c > and h, the maximum deviation in the energy does not tend to, due to the highly oscillatory term γ ẋ n,. This term is bounded because of ẋ n, x n, Ĩx n, x n Const. It is possible to prove that the average over a fixed sufficiently large number of consecutive values ẋ n, is constant. Numerical experiments. We consider the FPU-type problem with ω = 5 discussed in the beginning of this lecture. Figure 8 shows the error in the total energy of the Störmer Verlet method with four different step sizes. For the largest Figure 8: Error in the total energy of the FPU-type problem ω = 5 obtained for the Störmer Verlet method applied with four different step sizes: hω =.95,,.5, 5. 8

19 step size where hω =.95 the error is very large and of the size of γ. Nevertheless, there is no drift in the energy. Halving the step size decreases the error by a factor close to 6. This is in good agreement with the form of the function γ = γhω. Our second experiment Figure 9 shows again the energy error for various choices of ω and h. It illustrates the fact that the term γ ẋ n, dominates the error for large step sizes and depends essentially on the product hω. We can also see that the error oscillates around a constant which is different from the theoretical value of the Hamiltonian..3.3 hω =.5 ω = 5..9 hω = ω = Hx, ẋ = hω =.5 ω =..9 hω = ω = Hy, ẏ = 3 4 Figure 9: Total energy of the FPU-type problem along numerical solutions of the Störmer Verlet method. 8 Exercises. Interpret the exponential integrator 9 as a splitting method: a halfstep of a symplectic Euler type method for ẍ = gx, a step of the exact solution for ẍ + Ω x =, and finally the adjoint of the first half-step.. Show that a method 9 satisfying is symplectic if and only if ψξ = sincξ φξ for ξ = hω. 9

20 3. The change of coordinates z n = χhωx n transforms 9 into a method of identical form with φ, ψ, ψ, ψ replaced by χφ, χ ψ, χ ψ, χ ψ. Prove that, for hω satisfying sinchωφhω/ψhω >, it is possible to find χhω such that the transformed method is symplectic. 4. Prove that for infinitely differentiable functions gt the solution of the differential equation ẍ + ω x = x + gt can be written as xt = yt + cosωt ut + sinωt vt, where yt, ut, vt are given by asymptotic expansions in powers of ω. Hint. Use the variation-of-constants formula and apply repeated partial integration. 5. Consider a Hamiltonian Hp R, p I, q R, q I and let Hp, q = Hp R, p I, q R, q I for p = p R + ip I, q = q R + iq I. Prove that in the new variables p, q the Hamiltonian system becomes ṗ = H p, q, q H q = p, q. p