( )e ikx. ( ) = ˆq k

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1 ! Revised December 15, :57 PM! 1 Chapter 13: Spectral Methods Copyright 2015, David A. Randall Introduction Assume that q( x,t) is real and integrable. If the domain is periodic, with period L, we can express q( x,t) exactly by a Fourier series expansion: ( ) = ˆq k t q x,t ( )e ikx. k= (1) The complex coefficients ˆq k ( t) can be evaluated using ˆq k ( t) = 1 L x+l/2 q( x,t)e ikx dx. Recall that the proof of (1) and (2) involves use of the orthogonality condition x L/2 (2) where 1 L x+ L /2 e ikx e ilx dx = δ k,l, x L /2 (3) δ k,l 1, k = l 0, k l is the Kronecker delta. Eqs. (1) and (2) are a transform pair. They can be used to go back and forth between ordinary space and wave-number space. From (1), we see that the x -derivative of q satisfies (4)

2 ! Revised December 15, :57 PM! 2 q x,t ( )e ikx. ( ) = ik ˆq k t k= (5) Inspection of (5) shows that q should be clear. does not receive a contribution from ˆq 0 ; the reason for this A spectral model uses equations similar to (1), (2), and (5), but with a finite set of wave numbers, and with x defined on a finite mesh: n ( ) ˆq k t q x j,t ( )e ikx j, k= n ˆq k ( t) = 1 M M q ( x,t j )e ikx j, n k n, j=1 (6) (7) n ( ) ik ˆq k t q x j,t ( )e ikx j. k= n The sums in (6) and (8) are truncated in that they do not include wave numbers outside the range ±n. The value of n is chosen by the modeler. Note that we have used approximately equal signs in (6) and (8), but not in (7). The sum that appears in (7) is over a grid with M points. The relationship between M and n is discussed below. Apart from the effects of round-off error, the transform (6) is exactly reversible by the inverse transform (7), provided that M is large enough. Substitution of (6) into (7) gives ˆq k ( t) = 1 M n ˆq l ( t)e ilx j M l= n e ikx j for n k n. j=1 This is, of course, a rather circular substitution, but the result serves to clarify some basic ideas. If expanded, each term on the right-hand side of (9) involves the product of two wave numbers, l and k, each of which lies in the range n to n. The range for wave number l is explicitly spelled out in the inner sum on the right-hand side of (9); the range for wave number k is understood because, as indicated, we wish to evaluate the left-hand side of (9) for k in the range n to n. Because each term on the right-hand side of (9) involves the product of two Fourier (8) (9)

3 R k = R k (13)! Revised December 15, :57 PM! 3 modes with wave numbers in the range n to n, each term includes wave numbers up to ±2n. We therefore need 2n +1 complex coefficients, i.e., 2n +1 values of the q! k ( t). Because q is real, it must be true that ˆq k = ˆq k *, where the * denotes the conjugate. To see why this is so, consider the contributions of the +k and k coefficients to the sum in (6): (10) Τ k ( x j ) ˆq k t ( )e ikx j + ˆq k ( t)e ikx j R k e iθ e ikx j + R k e iµ e ikx j = R k e i ( θ+kx j ) + R k e i ( µ kx j ). (11) Here R k e iθ ˆq k ( t) and R k e iµ ˆq k t independence, our assumption that q x j,t Τ k ( x j ) must be zero, for all x j. It follows that ( ), and R k and R k are real and non-negative. By linear ( ) is real for all x j implies that the imaginary part of The only way to satisfy (12) for all x j is to set, R k sin ( θ + kx j ) + R k sin ( µ kx j ) = 0 for all x j. (12) and From (14), it follows that θ = µ. θ + kx j = ( µ kx j ) = µ + kx j. (14) Eq. (10) follows from (13) and (14). It implies that ˆq k and ˆq k together involve only two distinct real numbers, so that if you know the value of ˆq k you can immediately write down ˆq k. In addition, it follows from (10) that ˆq 0 is real. Therefore, the 2n +1 complex values of ˆq k embody the equivalent of only 2n +1 distinct real numbers. The Fourier representation up to

4 ! Revised December 15, :57 PM! 4 wave number n is thus equivalent to representing the real function q( x,t) on 2n +1 equally spaced grid points, in the sense that the information content is the same. We conclude that, in order to use a grid of M points to represent the amplitudes and phases of all waves up to k = ±n, we need M 2n + 1 ; we can use a grid with more than 2n +1 points, but not fewer. As a very simple example, a highly truncated Fourier representation of q, including just wave numbers zero and one, is equivalent to a grid-point representation of q using 3 grid points. The real values of q assigned at the three grid points suffice to compute the coefficient of wave number zero (i.e., the mean value of q ) and the phase and amplitude (or sine and cosine coefficients ) of wave number one. Substituting (7) into (8) gives q x l,t M q ( x,t j )e kx j e ikx l. k= n j=1 ( ) n M ik Reversing the order of summation leads to (15) q x l,t M ( ) α j ( ) l q x j,t, j=1 (16) where α j l i M n ke ik ( x l x j ). k= n The point of this little exercise is that (16) can be interpreted as a finite-difference approximation. It is a member of the family of approximations discussed many times in this course, but it is special in that it involves all grid points in the domain. From this point of view, spectral models can be regarded as a class of finite-difference models. Now consider the one-dimensional advection equation with a constant current, c: q t (17) = c q. (18) Substituting (6) and (8) into (18) gives

5 ! Revised December 15, :57 PM! 5 By linear independence, we obtain n n d ˆq k dt eikx = c ik ˆq k e ikx. k= n k= n (19) d ˆq k dt = ikc ˆq k for n k n. (20) Note that d ˆq 0 will be equal to zero; the interpretation of this should be clear. We can use (20) to dt predict ˆq k ( t). When we need to know q x j,t d ˆq k dt ( ), we can get it from (6). Compare (20) with ( ) kδx sin kδx = ikc ˆq k, which, as we have seen, is obtained by using centered second-order space differencing. The spectral method gives the exact advection speed for each Fourier mode, while the finite difference method gives a slower value. Similarly, spectral methods give the exact phase speeds for linear waves propagating through a uniform medium, while finite-difference methods generally underestimate the phase speeds. Keep in mind, however, that the spectral solution is not really exact, because only a finite number of modes are kept. In addition, the spectral method does not give the exact answer, even for individual Fourier modes, when the advection speed (or the phase speed) is spatially variable. Another strength of spectral methods is that they make it very easy to solve boundary value problems. As an example, consider (21) 2 q = f ( x, y), (22) as a problem to determine q for given f ( x, y). In one dimension, (22) becomes d 2 q dx = f ( x). 2 We assume periodic boundary conditions and compute the Fourier coefficients of both q and f, using (7). Then (23) can be written as (23)

6 ! Revised December 15, :57 PM! 6 n n ( k 2 ) ˆq k e ikx = ˆfk e ikx, k= n k= n (24) Invoking linear independence to equate coefficients of e ikx, we find that ˆq k = ˆf k k 2 for n k n (unless k = 0 ). (25) Eq. (25) can be used to obtain ˆq k, for k = 1,n. Then q( x) can be constructed using (6). This completes the solution of (23), apart from the application of an additional boundary condition to determine ˆq 0. The solution is exact for the modes that are included; it is approximate because not all modes are included. One issue with spectral models involves the representation of topography. In many models, e.g., those that use the sigma coordinate, it is necessary to take horizontal derivatives of the terrain height in order to evaluate the horizontal pressure gradient force. The terrain heights have to be expanded to compute spectral coefficients, and of course the expansion is truncated at some finite wave number. With bumpy continents and flat oceans, as schematically shown in Fig. 13.1, Fig. 13.1: The Earth is bumpy. truncation leads to bumpy oceans. Various approaches have been suggested to alleviate this problem (Hoskins, 1980; Navarra, et al., 1994; Boutelou, 1995; Holzer, 1996; Lindberg and Broccoli, 1996). Now consider a nonlinear problem, such as momentum advection, i.e., u t = u u, (26) again with a periodic domain. Fourier expansion gives

7 ! Revised December 15, :57 PM! 7 n n dû k = û l e ilx k= n dt eikx l= n n m= n imû m e imx. (27) Our goal is to predict û k ( t) for k in the range n to n. Wave numbers outside that range are excluded by the definition of our chosen truncation. The right-hand-side of (27) involves products of the form e ilx e imx, where l and m are in the range n to n. These products can generate new wave numbers, some of which lie outside the range n to n. Those that lie outside this range are simply neglected, i.e., they are not included when we evaluate the left-hand side of (27). For a given Fourier mode, (27) implies that dû k dt α α = im û l û m e i l+m l= α m= α ( )x e ikx, for n k n. Here the quantity in curly braces is a sum over wave numbers and is therefore defined at grid points. The summations on the right-hand side of (29) are over the range ±α. We must choose α large enough so that we pick up all possible combinations of l and m that lie in the range n to n. See Fig The circled X s in the figure denote excluded triangular regions. The number of points in each triangular region is n 1 The number of points retained is therefore given by ( ) ( ) = n( n 1) ( 2n +1) 2 2 n n 1 = 4n 2 + 4n ( ) ( n 2 n) = 3n 2 + 5n +1. This is the number of terms that must be evaluated in (27). The number of terms in the product of sums on the right-hand-side of (27) is thus of order n 2, i.e., it grows very rapidly as n increases. The amount of computation therefore grows rapidly as n increases, and of course the problem is twice as hard in two dimensions. At first, this poor scaling with problem size appeared to make. (28) (29) (30) (31)

8 ! Revised December 15, :57 PM! 8 spectral methods prohibitively expensive for nonlinear (i.e., realistic) problems. Note that the same issue would arise in a linear problem with spatially variable coefficients. Fig. 13.2: Cartoon table of l + m, showing which ( l,m) pairs can contribute to wave numbers k in the range n to n in Eq. (29). The pairs in the triangular regions marked by X s do not contribute. A way around this practical difficulty was proposed by Orszag, and independently by Eliassen et al., both in They suggested a transform method in which (6) and (8) are used to evaluate both u and u transform the result back into wavenumber space. on a grid. We then compute the produce u u on the grid, and We can obtain an exact result exact up to wave number n by making the number of grid points used large enough to allow the exact representation, for wave numbers in the range n to n, of quadratic nonlinearities like u u. Of course, here exact means exact up to wave number n. Because the solution is exact for wave numbers up to n, there is no error for those wave numbers, and in particular, there is no aliasing error. Therefore, a model of this type is not subject to aliasing instability arising from quadratic terms like u u. Aliasing can still arise, however, from cubic or higher-order nonlinearities. To investigate the transform method, we proceed as follows. By analogy with (7), we can write

9 ! Revised December 15, :57 PM! 9 u u k ( ) u ( ) = 1 M u x j M x j e ikx j, n k n. j=1 (32) Here we are computing the spectral coefficients of the product u u u x j, by starting from ( ) u ( ), which is the grid-point representation of the same quantity. Using (6) and (8), we x j can express u x j ( ) and u ( ) in terms of Fourier series: u u k x j = 1 M n û l e ilx n j M l = n imû m e mx j m= n e ikx j, n k n. j =1 It is important to note that, in (33), the derivative, i.e., u (33), has been computed using the spectral method, rather than a grid-point method. The grid is being used to allow efficient implementation of the spectral method. Eq. (33) is analogous to (9). When expanded, each term on the right-hand side of (33) involves the product of three Fourier modes (k, l, and m), and therefore includes zonal wave numbers up to ±3n. We need 3n +1 complex coefficients to encompass wave numbers up to ±3n. Because u u is real, those 3n +1 complex coefficients actually correspond to 3n +1 independent real numbers. Therefore, we need grid points to represent u u M 3n + 1 (34) exactly, up to wave number n. This is about 50% more than the 2n +1 grid points needed to represent u itself exactly up to wave number n. This larger grid is sometimes called a non-aliasing grid. In practice, the transform method to solve (26) works as follows: 1. Initialize the spectral coefficients û k, for n k n. This would normally be done by using measurements of u to initialize u on a grid, and then using a transform to obtain the spectral coefficients.

10 ! Revised December 15, :57 PM! Evaluate both u and u on a grid with M points, where M 3n + 1. Here u computed using the spectral method, i.e., Eq. (8), and the result obtained is used to compute grid-point values of u. is 3. Form the product u u on the grid. 4. Using (33), transform u u! coefficients u u. k back into wave-number space, for n k n. This gives the 5. Predict new values of the û k, using dû k dt! = u u k. 6. Return to Step 2, and repeat this cycle as many times as desired. Note that the grid-point representation of u contains more information ( 3n +1 real values) than the spectral representation ( 2n +1 real values). For this one-dimensional example the ratio is approximately 3/2. The additional information embodied in the grid-point representation is thrown away in Step 4 above, when we transform from the grid back into wave-number space. Therefore, the additional information is not remembered from one time step to the next. In effect, we throw away about 1/3 of the information that is represented on the grid. This is the price that we pay to avoid errors (for wave numbers up to ±n ) in the evaluation of quadratic nonlinearities. As illustrated above, the transform method uses a grid to evaluate quadratic nonlinearities. The same grid is used to implement complicated and often highly nonlinear physical parameterizations. The transform method was revolutionary; it made spectral models a practical possibility. Spectral methods on the sphere Spectral methods on the sphere were first advocated by Silberman (1954). A function F that is defined on the sphere can be represented by where the ( ) = F m n Y m n ( λ,ϕ) F λ,ϕ, m= n= m (35)

11 ! Revised December 15, :57 PM! 11 Y n m ( λ,ϕ) = e imλ P m n sinϕ ( ) (36) are spherical harmonics, and the P m n ( sinϕ ) are the associated Legendre functions of the first kind, which satisfy P n m ( 2n)! ( x) = 2 n n! ( n m)! + n m ( 1 x2 ) m 2 x n m n m ( )( n m 1) 2( 2n 1) ( )( n m 1) ( n m 2) ( n m 3) 2 4( 2n 1) ( 2n 3) x n m 2 x n m 4. (37) Here m is the zonal wave number and n m is the meridional nodal number. As discussed in m the Appendix on spherical harmonics, it has to be true that n m. The spherical harmonics Y n are the eigenfunctions of the Laplacian on the sphere: ( ) a 2 n n +1 2 Y m n = Y m n. Here a is the radius of the sphere. See the Appendix for further explanation. (38) We can approximate F by a truncated sum: F = M N ( m) F m m n Y n. m= M n= m (39) Here the overbar indicates that F is an approximation to F. Recall that m is the zonal wave number. In (39), the sum over m from M to M ensures that F is real. The choice of N m ( ) is discussed below. For smooth F, F converges to F very quickly, in the sense that the rootmean-square error decreases quickly towards zero. Why should we expand our variables in terms of the eigenfunctions of the Laplacian on the sphere? The Fourier representation discussed earlier is also based on the eigenfunctions of the Laplacian, in just one dimension, i.e., sines and cosines. There are infinitely many differential operators. What is so special about the Laplacian? A justification is that: The Laplacian consumes scalars and returns scalars, unlike, for example, the gradient, the curl, or the divergence; The Laplacian can be defined without reference to any coordinate system;

12 ! Revised December 15, :57 PM! 12 The Laplacian is isotropic, i.e., it does not favor any particular direction on the sphere; The Laplacian is simple. How should we choose N ( m)? This is called the problem of truncation. The two bestknown possibilities are rhomboidal truncation and triangular truncation. In both cases, we can choose the value of M, i.e., the highest zonal wave number to be included in the model, and the value of N follows. Rhomboidal truncation: N = M + m, and Triangular truncation: N = M. (40) (41) Fig. 13.3: Rhomboidal and triangular truncation, both illustrated for the case M = 4. From Jarraud and Simmons (1983). In both cases, n m. For a given value of M, rhomboidal truncation includes more spherical harmonics than triangular truncation. Rhomboidal and triangular truncation are illustrated in Fig With rhomboidal truncation, the value of N (the maximum value of n to be included in the sum) increases with the value of m, in such a way that the highest meridional nodal number is the same for all values of m. In the case of triangular truncation, the two-dimensional wave number N is a the same for all of the spherical harmonics that are included in the model, so the highest meridional nodal number is smaller for the larger values of m. The figure also shows that, for a given value of M, rhomboidal truncation includes more spectral coefficients than triangular truncation. The numbers of complex coefficients needed are

13 ! Revised December 15, :57 PM! 13 and ( M +1) 2 + M 2 + M for rhomboidal truncation, (42) ( M +1) 2 for triangular truncation. You can count the x s in Fig to persuade yourself that these formulas are correct. (43) Finally, the figure shows that, with both types of truncation, choosing the value of M is enough to determine which spectral coefficients are included. It is conventional to designate the resolution of a spectral model by designating the type of truncation by R or T, and appending the value of M, as in T106. The equivalent grid-point resolution is discussed later in this chapter. Triangular truncation has the beautiful property that it is not tied to a coordinate system, in the following sense: In order to actually perform a spherical harmonic transform, it is necessary to adopt a spherical coordinate system ( λ,ϕ). There are, of course, infinitely many such systems, which differ in the orientations of their poles. There is no reason, in principle, that the coordinates have to be chosen in the conventional way, in which the poles of the coordinate system coincide with the Earth s poles of rotation. The choice of a particular spherical coordinate system is, therefore, somewhat arbitrary. Suppose that we are given an analytical function on the sphere. We choose two different spherical coordinate systems (tilted with respect to one another in an arbitrary way), perform a triangularly truncated expansion in both, and then transform the results back to physical space. It can be shown that the two results will be identical, i.e., F( λ 1,ϕ 1 ) = F( λ 2,ϕ 2 ), where the subscripts indicate alternative spherical coordinate systems. This means that the arbitrary orientations of the spherical coordinate systems used have no effect whatsoever on the results obtained. The coordinate system used disappears at the end. Triangular truncation is very widely used today, in part because of this nice property, which is not shared by rhomboidal truncation.. As shown in Fig. 13.4, triangular truncation represents the observed kinetic energy spectrum more efficiently than does rhomboidal truncation (Baer, 1972). The thick lines in the figure show the observed kinetic energy percentage that comes from each component. The thin, straight, diagonal lines show the modes kept with triangular truncation. With rhomboidal truncation the thin lines would be horizontal, and more modes would be kept, but they would not add much useful information. (44)

14 ! Revised December 15, :57 PM! 14 In order to use (39), we need a spherical harmonic transform, analogous to a Fourier transform. From (36), we see that a spherical harmonic transform is equivalent to the combination of a Fourier transform and a Legendre transform. The Legendre transform is formulated using a classical method called Gaussian quadrature. The idea is as follows. Suppose that we are given a function f ( x) defined on the interval 1 x 1, and we wish to evaluate 1 I = f ( x)dx, 1 by a numerical method. If f ( x) is defined at a finite number of points, denoted by x j, then (45) where the w i are weights. N I f ( x i )w i, i=1 (46) Fig. 13.4: Percentage of total kinetic energy in each spectral component. From Jarraud and Simmons (1983), based on Baer (1972). Consider the special case in which f ( x) is itself a weighted sum of Legendre polynomials, as in a Legendre transform. We want the transform to be exact, within the round-off error of

15 ! Revised December 15, :57 PM! 15 the machine, so that we can recover f ( x) without error. Gauss showed that for such a case (44) gives the exact value of I, provided that the x i are chosen to be the roots of the highest Legendre polynomial used. In other words, we can use (44) to evaluate the integral (43) exactly, provided that we choose the latitudes so that they are the roots of the highest Legendre polynomial used. These latitudes can be found by a variety of iterative methods, and of course this only has to be done once, before the model is run. The Gaussian quadrature algorithm is used to perform the Legendre transform. With the transform method described earlier, the number of grid points needed to avoid errors in the evalution of quadratic nonlinearities exceeds the number of degrees of freedom in the spectral representation. The number of grid points around a latitude circle must be 3M + 1. ( ) The number of latitude circles must be 3M + 1 for triangular truncation, and so the total 2 number of grid points needed is ( 3M + 1)2. Referring back to (46), we see that, for large M, 2 the grid representation uses about 2.25 times as many equivalent real numbers as the triangularly truncated spectral representation. A similar conclusion holds for rhomboidal truncation. Computing the physics on a nonaliasing grid is standard procedure, but wasteful. In summary, the spectral transform method as it is applied to global models works as follows: First, we choose a spectral truncation, e.g., T42. Then we identify the number of grid points needed in the longitudinal and latitudinal directions, perhaps with a view to avoiding aliasing due to quadratic nonlinearities. Next, we identify the highest degree Legendre polynomial needed with the chosen spectral truncation, and find the latitudes where the roots of that polynomial occur. These are called the Gaussian latitudes. At this point, we can set up our Gaussian grid. The horizontal derivatives are evaluated in the spectral domain, essentially through multiplication by wave number. When we transform from the spectral domain to the grid, we combine an inverse fast Fourier transform with an inverse Legendre transform. The nonlinear terms and the model physics are computed on the grid. Then we use the Legendre and Fourier transforms to return to the spectral domain. The basic logic of this procedure is the very similar to that described earlier for the simple one-dimensional case. We have a fast Fourier transform, but no one has yet discovered a fast Legendre transform, although some recent work points towards one. Lacking a fast Legendre transform, the operation count for a spectral model is of O N 3 harmonics used. Finite difference methods are, in effect, of O N 2 ( ), where N is the number of spherical ( ). This means that spectral

16 ! Revised December 15, :57 PM! 16 models become increasingly expensive, relative to grid-point models, at high resolution. Further comments are given later in this chapter. The equivalent grid resolution of spectral models Laprise (1992) distinguishes four possible ways to answer the following very natural question: What is the equivalent grid-spacing of a spectral model? 1. One might argue that the effective grid spacing of a spectral model is the average distance between latitudes on the Gaussian grid. With triangular truncation, this is the same as the spacing between longitudes at the Equator, which is L 1 = 2πa. Given the radius of the 3M + 1 Earth, and using units of thousands of kilometers, this is equivalent to 13.5 / M. For a T31 model (with M = 31 ), we get L km. An objection to this measure is that, as discussed above, much of the information on the Gaussian grid is thrown away when we transform back into spectral space. 2. A second possible measure of resolution is half the wavelength of the shortest resolved zonal wave at the Equator, which is L 2 = πa, or about 20 / M in units of thousands of M kilometers. For a T31 model, L km. 3. A third method is based on the idea that the spectral coefficients, which are the prognostic variables of the spectral model, can be thought of as a certain number of real variables per unit area, distributed over the Earth. A triangularly truncated model has the equivalent of ( M + 1) 2 real coefficients. The corresponding resolution is then L 3 = 4πa 2 ( M + 1) = 2 π a, which works out to about 725 km for a T31 model. 2 M A fourth measure of resolution is based on the equivalent total wave number associated with the Laplacian operator, for the highest mode. The square of this total wave number is Κ 2 = M ( M + 1). Suppose that we equate this to the square of the equivalent total wave number 2a 2 on a square grid, i.e. Κ 2 = k x 2 + k y 2 corresponding wavelength is L 4 = π k =, and let k x = k y = k for simplicity. One half of the 2πa M thousands of kilometers. For a T31 model this gives about 900 km., which is equivalent to 28.3 / M in units of

17 ! Revised December 15, :57 PM! 17 These four measures of spectral resolution range over more than a factor of two. The measure that makes a spectral model look good is L 1, and so it is not surprising that spectral modelers almost always use it when specifying the equivalent grid spacing of their models. Semi-implicit time differencing As we have already discussed in Chapters 5 and 8, gravity waves limit the time step that can be used in a primitive-equation (or shallow water) model. A way to get around this is to use semi-implicit time differencing, in which the gravity wave terms of the equations are treated implicitly, while the other terms are treated explicitly. This can be accomplished much more easily in a spectral model than in a finite-difference model. A detailed discussion of this approach will not be given here, but the basic ideas are as follows. The relevant terms are the pressure-gradient terms of the horizontal equations of motion, and the mass convergence term of the continuity equation. These are the same terms that we focused on in the discussion of the pole problem, in Chapter 11. The terms involve horizontal derivatives of the height field and the winds, respectively. Typically the Coriolis terms are also included, so that the waves in question are inertia-gravity waves. Consider a finite-difference model. If we implicitly difference the gravity-wave terms, the resulting equations will involve the n +1 time-level values of the heights and the winds at multiple grid points in the horizontal. This means that we must solve simultaneously for the new values of the heights and winds. Such problems can be solved, of course, but they can be computationally expensive. For this reason, most finite-difference models do not use semiimplicit time differencing. In spectral models, on the other hand, we prognose the spectral coefficients of the heights and winds, and so we can apply the gradient and divergence operators simply by multiplying by wave number (roughly speaking). This is a local operation in wave-number space, so it is not necessary to solve a system of simultaneous equations. The use of semi-implicit time differencing allows spectral models to take time steps several times longer than those of (explicit) grid-point models. This is a major advantage in terms of computational speed, which compensates, to some extent, for the expense of the spectral transform. Conservation properties and computational stability Because the spectral transform method prevents aliasing for quadratic nonlinearities, but not cubic nonlinearities, spectral models are formulated so that the highest nonlinearities that appear in the equations (other than in the physical parameterizations) are quadratic. This means that the equations must be written in advective form, rather than flux form. As a result, spectral models do not exactly conserve anything - not even mass - for a general, divergent flow.

18 ! Revised December 15, :57 PM! 18 It can be shown, however, that in the limit of two-dimensional non-divergent flow, spectral models do conserve kinetic energy and enstrophy. Because of this property, they are well behaved computationally. Nevertheless, all spectral models need some artificial diffusive damping to avoid computational instability. In contrast, it is possible to formulate finitedifference models that are very highly conservative and can run indefinitely with no artificial damping at all. Moisture advection The mixing ratio of water vapor is non-negative. We have already discussed the possibility of spurious negative mixing ratios caused by dispersion errors in finite-difference schemes, and we have also discussed the families of finite-difference advection schemes that are signpreserving and do not suffer from this problem. Spectral models have a very strong tendency to produce negative water vapor mixing ratios (e.g., Williamson and Rasch, 1994). In the global mean, the rate at which negative water is produced can be a significant fraction of the globally averaged precipitation rate. Negative water vapor mixing ratios occur not only locally on individual time steps, but even in zonal averages that have been time-averaged over a month. Because of this disastrous situation, many spectral models are now using non-spectral methods for advection (e.g. Williamson and Olson, 1994). This means that they are only partly spectral. Linear grids When non-spectral methods are used to evaluate the nonlinear advection terms, the motivation for using the high-resolution, non-aliasing grid disappears. Such models can then use a coarser linear grid, with the same number of grid points as the number of independent real coefficients in the spectral representation. This leads to a major savings in computational cost. Physical parameterizations Because most physical parameterizations are highly nonlinear, spectral models evaluate such things as convective heating rates, turbulent exchanges with the Earth s surface, and radiative transfer on their Gaussian grids. The tendencies due to these parameterizations are then applied to the prognostic variables, which are promptly transformed into wave-number space. Recall that when this transform is done, the spectral representation contains less information than is present on the grid, due to the spectral truncation used to avoid aliasing due to quadratic nonlinearities. This means that if the fields were immediately transformed back onto the grid (without any changes due, e.g., to advection), the physics would not see the fields that it had just finished with. Instead, it would see spectrally truncated versions of these fields.

19 ! Revised December 15, :57 PM! 19 For example, suppose that the physics package includes a convective adjustment that is supposed to adjust convectively unstable columns so as to remove the instability. Suppose that on a certain time step this parameterization has done its work, removing all instability as seen on the Gaussian grid. After spectral truncation, some convective instability may re-appear, even though physically nothing has happened! In effect, the spectral truncation that is inserted between the grid domain and the spectral domain prevents the physical parameterizations from doing their work properly. This is a problem for all spectral models. It is not an issue when the physics is evaluated on a linear grid that has the same number of degrees of freedom as the spectral representation. Reduced linear grids The cost of the Legendre transforms can be reduced by decreasing the numbers of grid points around latitude circles, near the poles (Hortal and Simmons, 1990). Recall that a similar idea was tried with grid point methods. The approach works with spectral methods because with high resolution the spectral coefficients corresponding to the largest values of m are very close to zero near on the poles. Summary In summary, the spectral method has both strengths and weaknesses: Strengths: Especially with triangular truncation, it eliminates the pole problem. It gives the exact phase speeds for linear waves and advection by a constant current such as solid-body rotation. It converges very rapidly, and gives good results with just a few modes. Semi-implicit time-differencing schemes are easily implemented in spectral models. Weaknesses: Spectral models do not exactly conserve anything - not even mass. Partly because of failure to conserve the mass-weighted total energy, artificial damping is needed to maintain computational stability. Spectral models have bumpy oceans. Because of truncation in the transform method, physical parameterizations do not always have the intended effect. Moisture advection does not work well in the spectral domain.

20 ! Revised December 15, :57 PM! 20 At high resolution, spectral methods are computationally expensive compared to grid point models.

21 ! Revised December 15, :57 PM! 21 Problems 1. Write subroutines to compute Fourier transforms and inverse transforms, for arbitrary complex q x j ( ). The number of waves to be included in the transform, and the number of grid points to be used in the inverse transform, should be set through the argument lists of subroutines. Let where q( x j ) = 14 cos( k 0 x j ) + 6i cos( k 1 x j ) + 5, k 0 = 2π L 0 and L 0 = X 4, (47) k 1 = 2π L 1 and L 1 = X 8. (48) Here X is the size of the periodic domain. Compute the Fourier coefficients starting from values of x j on a grid of M points, for M = 3, M = 9, M = 17, and M = 101. Discuss your results. 2. Consider a periodic step function, defined by H ( x) = 1 when the integer part of x is odd, H ( x) = +1 when the integer part of x is even. With this definition, H ( x) is discontinuous at integer points on the real line, and infinitely differentiable elsewhere. Sample H ( x) on the domain 1 x 1, using M evenly spaced points, for M = 11, M = 101, and M = 1001, and feel free to go to larger values if you like. Plot the results for 1 x 1, for each value of M. Discuss in terms of the Gibbs phenomenon. 3. For the transform method with a one-dimensional problem, many grid points would be needed to give the exact answer up to wave number n for the case of cubic nonlinearities?