The Fourier spectral method (Amath Bretherton)

Transcription

1 The Fourier spectral method (Amath Bretherton) 1 Introduction The Fourier spectral method (or more precisely, pseudospectral method) is a very accurate way to solve BVPs with smooth solutions on periodic domains. It can be extended to handle homogeneous Dirichlet and eumann BCs. It is a function-space method that approximates an arbitrary periodic function on an interval [, L] by a finite sum of (assumed even) complex exponentials: u(x) u () (x) = 1 û m exp(ik m x) (1) with wavenumbers k m = 2πM m /L. (2) The first /2 exponentials correspond to zero and positive harmonics M m, and the remaining /2 exponentials correspond to the negative harmonics: { m 1 1 m /2 M m = m 1 /2 < m (3) The û m are called the (discrete) Fourier coefficients. We need pairs of positive and negative harmonics to make up cosines and sines and create a real-valued function. The apparently unpaired /2 harmonic is exactly equal to the /2 harmonic at all the grid points (an example of aliasing), so with some care we can treat both these harmonics with a single term. With harmonics, we can represent the exact value of the function u(x) at grid points, which we take to be uniformly spaced across the interval, such that. u j = u(x j ), x j = (j 1)h, h = L/, j = 1,...,. (4) The next grid point x +1 = L would be redundant, since it is periodically equivalent to the first grid point x 1 =. 1

2 2 The Discrete Fourier Transform Given the gridpoint values u j, we calculate the corresponding Fourier coefficients û m from the definition (1): u j 1 û m exp(ik m x j ), j = 1,..., By manipulating the exponential, we find that for 1 m /2, while for /2 < m, exp(ik m x j ) = exp 2πi L M m(j 1) L = exp 2πi (j 1)(m 1) exp(ik m x j ) = exp 2πi L M m(j 1) L = exp 2πi (j 1)(m 1) = exp 2πi 2πi (j 1)(m 1) exp (j 1)( ) = exp 2πi (j 1)(m 1) exp[ 2πi(j 1)] = exp 2πi (j 1)(m 1) Thus u j 1 û m exp[2πi(j 1)(m 1)/] j = 1,..., (5) This is a system of equations in the unknowns û m. Luckily, it has an elegant solution in terms of the discrete Fourier transform (DFT). Let u j, j = 1,..., be a sequence of possibly complex values. Its DFT is defined as the sequence û m, m = 1,...,, where û m = u j exp[ 2πi(m 1)(j 1)/] (6) j=1 As this notation implies, these û m can be shown to be exactly the desired coefficients that solve (5). In fact, the transformation (5) from the vector û of Fourier coefficients û m to the vector u of gridpoint values u j is called the inverse DFT or IDFT. 2

3 2.1 Proof that the IDFT is the inverse of the DFT It suffices to show that applying the IDFT to the DFT of a sequence recovers the original sequence. By direct substitution of (6) into (5): u j = 1 exp[2πi(j 1)(m 1)/] u n exp[ 2πi(m 1)(n 1)/] = A jn u n n=1 A jn = 1 exp[2πi(j 1)(m 1)/] exp[ 2πi(m 1)(n 1)/] n=1 = 1 exp[2πi(j n)(m 1)/] (8) If j = n, all the terms in the series are 1, so A jj = 1. If j n, we are summing a geometrical series with ratio exp[2πi(j n)(m 1)/]. This is computed to be We conclude that A jn = exp[2πi(j n)( + 1 1)/] exp[2πi(j n)(1 1)/] exp[2πi(j n)(m 1)/] 1 A jn u n = u j, i. e. that substituting the DFT into the IDFT returns the original sequence, as claimed. 2.2 The FFT - a fast algorithm to compute the DFT if = 2 p n=1 = (7) Direct use of these formulas would require O( 2 ) flops to obtain the sequence û m from the sequence u n, or vice versa. However, when is a power of 2, a very efficient algorithm called the fast Fourier transform (FFT) can calculate the DFT in O( log ) flops. The FFT is what makes the DFT a practical tool for large problems, and is among the most influential of all numerical algorithms. The FFT can be generalized (with a little loss of efficiency) to all which can be factored into products of small primes (e. g. 2, 3, and 5). Because software packages use the FFT algorithm whenever possible, the word FFT is often (loosely) used in place of DFT. In MATLAB, uhat = fft(u) calculates the DFT of the sequence u, and u = ifft(uhat) calculates the inverse DFT of uhat. MATLAB uses the fast FFT algorithm only if is a power of 2. 3

4 2.3 Matrix representation of the DFT and IDFT If u is the vectors of gridpoint values u i, then the DFT and inverse DFT can be written as matrix operators û = DF T (u) = 1/2 F u u = IDF T (y) = 1/2 F û (9) whereû is the DFT of u, and F mj = 1/2 exp[ 2πi(m 1)(j 1)] (1) and F is the conjugate transpose of F. The matrix form of the statement that the IDFT is the inverse of the DFT is F 1 = F This is equivalent to saying that F is a unitary matrix (i. e. F F = F F = I). 3 Use of DFT for differentiation We can approximate du/dx for all x using the DFT interpolating function: In particular, at the grid points u (x) d dx 1 û m exp(ik m x) 1 ik m û m exp(ik m x) u (x j ) 1 ik m û m exp(ik m x j ) (11) The RHS is just the IDFT of the sequence ik m û m. That is, just like with Fourier transforms, differentiation in physical space is just like multiplication by ik in wavenumber space. 3.1 The unbalanced harmonic M m = /2 One minor complication is the unbalanced harmonic M m = /2 corresponding to index m = / Assuming u j is a real-valued function, it is easy to show that û m will also be real for this harmonic. We d 4

5 like to interpret the component of the interpolant u () (x) due to this harmonic as 1 û m (e ix/2 + e ix/2 )/2 At the gridpoints e ixj/2 = e ixj/2 = ( 1) j 1, so there it does not matter whether we interpret the interpolant using the negative or positive harmonic. However, between the gridpoints, it does matter, so this will affect the derivative of the interpolant even at the gridpoints. We can calculate the derivative that we want (with equal weighting of positive and negative harmonics) as At the gridpoints, this equals d dx 1 û m (e ix/2 + e ix/2 )/2 = (i/4)(e ix/2 e ix/2 )û m (i/4)û m (e ixj/2 e ixj/2 ) = Thus, we want to zero the term in the derivative corresponding to this harmonic. On the other hand, the formula we are using so far will give a contribution from this term of d dx 1 û m e ix/2 xj = (i/2)( 1) j 1 û m Since û m is real for this harmonic, the spurious contribution to the derivative will always be imaginary, and can be removed simply by taking the real part of the IDFT. The contributions to the derivative from all the other matched pairs of harmonics will be real, so they won t be affected by this. 3.2 Implementation of the derivative using DFT/IDFT in Matlab Thus, to use the DFT to calculate a derivative of a periodic function sampled on a uniform grid: 1. Calculate the DFT û of the sampled function u 2. Multiply element m of the DFT by ik m 3. Take the real part of the IDFT of the resulting vector to get the derivative at the grid points. The following MATLAB algorithm can be used for this purpose (assuming L and have already been defined); 5

6 uhat = fft(u) M = [:(/2-1) (1-/2):(-1)]; k = 2*pi*M/L; dudx = real(ifft(1i*k.*uhat)) 3.3 Higher derivatives Higher derivatives are easily calculated using the same approach as above; to take a p th derivative, the Fourier coefficients are just multiplied by (ik) p. 4 Relation to complex Fourier analysis The DFT interpolant formula (1) is a truncated version of the complex Fourier series for u(x), which has the form where the coefficients u(x) = c M exp(2πimx/l) c M = 1 L M= L u(x) exp( 2πiM x/l)dx, (12) Indeed, the discrete Fourier coefficients give approximations to the c M that are accurate when M m : c Mm û m /, m = 1,..., (13) In fact, the DFT formula for û m can be obtained by approximating the integral in c Mm as a Riemann sum centered on the gridpoints x j. 5 Accuracy of DFT-based interpolation and differentiation 5.1 Smoothness matters! The accuracy of Fourier interpolation and differentiation depends on the relative size of the Fourier coefficients of harmonics are that must be truncated if only harmonics can be retained. One can show that for an infinitely-differentiable L-periodic function u(x), the Fourier coefficients û m corresponding to high harmonics M m tend to zero faster than any inverse power of M m. This also implies that the error of Fourier 6

7 interpolation and differentiation will decrease faster than any negative power of the smallest neglected harmonic, i..e. faster than any inverse power of. SInce h = L/, this also implies that for C L-periodic functions, the error of Fourier interpolation and differentiation goes to zero faster than any power of the grid spacing h. This is sometimes referred to as spectral accuracy. On the other hand, if u(x) is less smooth, the Fourier coefficients die off less rapidly at large harmonics, and the error of Fourier differentation decreases less rapidly with h. If u(x) is R times differentiable over the entire interval [, L] then its discrete Fourier coefficients are O( M m (R+1) ), and the Fourier coefficients of its p the derivative are O( M m (R p+1) ). Thus a periodic step function (which is not continuous at the steps) has Fourier coefficients which are O( M 1 ), and can t accurately be differentiated with the DFT, while a sawtooth (which is continuous, but whose derivative is not continuous at the teeth) has Fourier coefficients which are O( M 2 ), and its derivative (the periodic step function) has rather slowly convergent Fourier coefficients of O( M 1 ). Hence, for problems with non-smooth terms or solutions, Fourier spectral methods lose most of their accuracy advantages. 5.2 The DFT power spectrum A good approximation to u(x) can be obtained from the DFT interpolant as long as û m 2 is sufficiently small for the highest retained harmonics (in practice, 1 6 of the largest Fourier mode squared amplitude is usually OK). For smooth functions u(x), the required can be small (1 or less). For a less smooth function, the required may be very large. It is useful to make a semilogy plot of the power spectrum of the squared amplitude of the Fourier modes û m 2 / 2 vs. harmonic M m to check whether the highest harmonics do in fact have small amplitude. 5.3 Examples of DFT interpolants of smooth and nonsmooth functions The solid lines in the left panels of Figure 1 shows a step, sawtooth, and C function periodic over [, 1]. In the right panel are plotted the DFT power spectrum and for comparison, the analytically derived squared amplitudes of the exact Fourier coefficients. The analytical form of the Fourier coefficients in the 7

8 Step function, = 16 y(x) y i y (x) Squared amplitude Fourier power spectrum yhat DFT x.5 Sawtooth, = 16 y(x) y i y (x) Squared amplitude M Fourier power spectrum yhat DFT x /(1.6 cos x), = 16 y(x) y i y (x) Squared amplitude M Fourier power spectrum yhat DFT x M Figure 1: Left panels: Function u(x) (solid), grid points u i, and FFT-based interpolating function u () (x). Right panels: Power spectrum of exact complex Fourier components and of the DFT. 8

9 three cases are: Step: u(x) = sgn(x.5) c M = 2i/(πM), M odd, Sawtooth: u(x) = 2x 1.5 c M = 2/(πM) 2, M odd, Swell: u(x) = (1.6 cos(x)) 1 c M = M. For the swell function, the squared amplitudes of all Fourier coefficients for M > 6 are less that 1 6 of the leading Fourier coefficient, so a very accurate representation of the swell can be obtained with only a few Fourier modes. For the small harmonics M, there is extremely good agreement between the squared DFT Fourier coefficient amplitudes and those from the exact full complex Fourier series. For larger M, the agreement is less good, especially when u(x) is not very smooth. The chain-dashed line on the left panel of Figure 1 is the DFT-based approximation u (x) to the original function using = 16 grid points. For the C swell function, it is extremely good. For the step function, ringing near the edges of the step is evident. 5.4 Accuracy of DFT-based differentiation on smooth vs. nonsmooth functions DFT-based differentiation is extremely accurate for smooth functions u(x) well resolved by the grid, but is less accurate for functions with derivative discontinuities. This can be seen in Figure 2. The derivative of a sawtooth function is the periodic step function shown at left. The DFT estimates of its derivative oscillate around the step function. In contrast, for the C swell function at right, the DFT gives an excellent approximation to du/dx at the gridpoints despite the small number of gridpoints used. 6 Fourier spectral interpolation The DFT-based approximation u (x) to the continuous function u(x) can be used for interpolation. The matlab function u =interpft(un,intp) evaluates an interpolating function based on the -vector of values un (assumed equally spaced through a wavelength) at intp grid points (also equally spaced through a wavelength). This is done by taking the DFT of un, filling out the middle of the resulting vector with zeros to bring it to length intp, renormalizing by multiplying by intp/, then taking an inverse DFT. 9

10 3 Sawtooth, = /(1.6 cos(2πx)), = Exact DFT 5 dy/dx dy/dx x 5 Exact DFT x Figure 2: Exact and DFT-computed derivative of [, 1] periodic sawtooth and swell functions. 7 Use of DFT for localized functions on unbounded domains Frequently, the DFT is used for approximating localized functions on the unbounded domain (, ). The method is to restrict u(x) to a finite domain [X 1, X 2 ] which is chosen large enough so that u(x) is very small (1 3 or less of its maximum value) outside this domain. Then u(x) is treated as periodic on this domain. For example, we can use a DFT to numerically differentiate exp( x 2 /2) in this manner (Figure 3). If we restrict the domain to [-4, 4] and take = 16 grid points, du/dx has only small errors at the grid points (left) compared to the exact value x exp( x 2 /2). The maximum error,.13, occurs near x = -4, suggesting that it is primarily due to the domain truncation. The power spectrum of the DFT (right) falls off rapidly with M up through M = 6, then levels off. The leveling is due to the derivative discontinuity in u(x) when interpreted as a periodic function on the truncated domain. In this example, enlarging the domain somewhat would make u(x) so small at the boundary that the domain truncation effects could be made totally negligible. To avoid degrading accuracy by increasing the grid spacing, correspondingly more grid points (and hence a larger FFT) would be required. In this problem, the computational effort is minute, but in other problems, an optimal choice of domain truncation may involve a tradeoff (for a given allowable computational effort) between wanting a large domain and wanting a sufficiently fine grid spacing. 1

11 1 Gaussian over [ 4, 4], = DFT Power spectrum.5 1 dy/dx Y m 2.5 Exact DFT x M m Figure 3: Left: Exact derivative of u(x) = exp( x 2 /2), and DFT-computed derivative with = 16 grid points, truncating domain to [-4, 4]. Right: Power spectrum of the DFT of u(x) 8 Application to 1D BVPs The main application of Fourier spectral methods is to multidimensional elliptic equations in simple rectangularbox geometries, e. g. commonly used in simulation of fluid flows. However, 1D examples illustrates the concepts well. 1D periodic ODE example Consider the ODE u xx u = f(x) = 1/(1.6 cos(x)) (14) with periodic BCs on [, 2π]. The forcing function is C and periodic on [, 2π], so we expect a Fourier method to achieve spectral accuracy. The Fourier spectral method with gridpoints for this problem can be written: ˆf = DFT(f), û m = ˆf m /(k 2 m + 1), u = Re[IDFT(û)] Matlab script FS 1DBVP.m implements this method for = 2 p and a range of p (Fig. 4). The left panel shows f(x); the middle panel shows the solution for various, interpolated between grid points using interpft 11

12 2.5 f(x) -.8 Fourier spectral u(x) 1 Max-norm h/2-h soln diff f u =2 =4 =8 =16 =32 δ x/l x/l h Figure 4: RHS f(x), FS solution u(x) at grid points and between, and error convergence plot..ote h = 2π/. The right panel assesses solution convergence from the max-norm difference δ between the p + 1 and p solutions at their common gridpoints. Spectral accuracy is achieved : δ decreases faster than any power of h. 9 Homogeneous Dirichlet and eumann BCs For an ODE with only even derivatives, the Fourier spectral method is easily extensible to homogeneous Dirichlet and eumann BCs at x =, L/2. The Dirichlet case is handled using odd extension. We assume the RHS and solution are specified as f H (x), u H (x) on the half-interval < x < L/2 and we extend the domain rightward to x = L. We let f(x), u(x) be the odd extension of f H (x), u H (x) to the extended domain, f H (x), < x < L/2 f(x) =, x =, L/2, L f H (L x), L/2 < x < L and similarly for u(x). ote that if f H is nonzero at either domain endpoint, its odd extension will have discontinuities there, and this will greatly degrade the error convergence even if even if f H (x) is smooth in the domain interior. By construction, f and its derivative are the same at x = and x = L, so a periodic solution can be found to the extended problem, allowing application of the Fourier spectral method. One can easily show by substitution that an ODE with only second derivatives holds over the extended domain < x < L, except perhaps at the reflection point L/2. At that point, the ODE is automatically 12

13 f Odd extension of f H (x) x/l u Fourier spectral u(x) -.2 =2 -.4 =4 =8 -.6 =16 = x/l δ 1 Max-norm h/2-h soln diff O(h -2 ) Figure 5: RHS, FS solution, and error convergence plot for odd extension of Dirichlet problem on the half-interval < x < L/2 1 h satisfied, because the RHS and all even derivatives of u are zero, e. g. d 2 u u(l/2 + ɛ) 2u(L/2) + u(l/2 ɛ) (L/2) = lim dx2 ɛ ɛ 2, u H (L/2 ɛ) 2() + u H (L/2 ɛ) = lim ɛ ɛ 2, =. FS 1DBVP.m implements odd extension on the ODE (14) on the half-interval < x < L/2 = π with homogeneous Dirichlet BCs, giving Fig. 5. The odd extension of the RHS (left panel) has discontinuities at and L/2. o solution is plotted for = 2 since in this case the only Fourier grid points are at the half-interval boundaries and L/2, where the solution is zero by construction. With larger, one can see the solution satisfies the Dirichlet BCs at x/l =,.5 and it looks like a periodic odd extension outside this interval. The RHS endpoint discontinuities keep the error convergence slow (O(h 2 )), such that the spectral method has no accuracy advantage over finite differences for this problem. The eumann case is handled using even extension. We extend the domain rightward to x = L. We let f(x) be the even extension of f H (x) to [, L]: { fh (x), x L/2 f(x) = f H (L x), L/2 < x L and similarly for u(x). If this is implemented on the the half-interval < x < L/2 = π for the ODE (14), the 13

14 even extension of f H is exactly the same as the periodic f that we already defined, so the Fourier spectral solution will be identical to the periodic problem. 14