Math 660-Lecture 17: Spectral methods: Fourier pseduo-spectral method Spectral accuracy of Fourier pseudo-spectral method The spatial accuracy of Fourier pseudo-spectral method is usually O(hm) 8m > 0 for smooth functions and O(cN ) for analytic functions. This is known as the spectral accuracy. We aim to understand why we have such spectral accuracy 1 Aliasing formula We now consider the effect of sampling. Suppose we discretize the space with step h and xj = jh. Let's recall the semi-discrete Fourier Transform of v, given by 1 X −iξxj v^(ξ) = h vje : j=−∞ Note that this is different from the semi-discrete transform we have used in the von-Neumann analysis. The only difference is that we don't have the extra p1 here. This extra constant doesn't 2π matter. 2 Theorem 1. Suppose u 2 L (R) and has a first derivative with bounded variation. Let vj = u(xj). v^(ξ) is the semi-discrete Fourier transform of v and u^(ξ) is the Fourier Transform of u. Then, for any ξ 2 [−π=h; π=h), we have 1 X v^(ξ) = u^(ξ + 2πj=h): j=−∞ Comment: This aliasing formula is related to the Nyquist sampling the- orem in signaling processing. We now consider the aliasing formula for functions defined on the torus. Recall that the DFT is defined by N X −ikxn v^k = vne n=1 Theorem 2. Suppose u is a good periodic function on [0; 2π) such that its Fourier series converges to u. Let vj = u(xj). Then, the DFT of v and 1 Fouerier series of u are related by 1 X v^k = N u^k+mN ; k = −N=2 + 1; : : : ; N=2: m=−∞ Note that here we used a different scaling for DFT compared with semi-discrete Fourier Transform and the Fourier series for functions on the torus. That's why we have the extra N = 2π=h here. For the functions defined on R, we have Corollary 1. With the same notations, we have the following claims for any ξ 2 [−π=h; π=h): • If u has p−1 continuous derivatives and are in L2 for some p ≥ 1 and a pth derivative of bounded variation, then ju^(ξ) − v^(ξ)j = O(hp+1): • If u has infinitely many continuous derivatives in L2, then ju^(ξ) − v^(ξ)j = O(hm); 8m > 0: • If u can be analytically extended to jIm(z)j < a such that there exits a constant C independent of y 2 (−a; a), ku(· + iy)kL2 ≤ C, then ju^(ξ) − v^(ξ)j = O(e−π(a−)=h); 8 > 0: Proof. We can estimate that X h Z ju^(ξ) − v^(ξ)j ≤ ju^(ξ + 2πj=h)j ∼ ju^(ξ)jdξ 2π j6=0 jξj>π=h R 1 p+1 In the first case, we have Ch jξj>π=h jξjp+1 dξ = O(h ). The other cases follow similarly. For the periodic case, we then have the following corollary: (p−1) (j) (j) Corollary 2. Suppose u 2 C ([0; 2π]; C), u (0) = u (2π) for j = (p) 0; 1; : : : ; p − 1 and that u is piecewise continuous. Let v^k be the DFT of its sampling and u^k be its Fourier series coefficients. Then, 1 ju^(k) − v^ j = O(N −p) = O(hp): N k 2 2 The spectral accuracy Using the aliasing formula, it's not hard to find the spectral accuracy results. For the semi-discrete Fourier transform, supposew ^(ξ) = (ik)νv^(ξ) for ξ 2 [−π=h; π=h) and vj = u(xj). Then, we have the following theorem: Theorem 3. For any ξ 2 [−π=h; π=h): • If u has p − 1 continuous derivatives and are in L2 for some p ≥ ν + 1 and a pth derivative of bounded variation, then (ν) p−ν jwj − u (xj)j = O(h ): • If u has infinitely many continuous derivatives in L2, then (ν) m jwj − u (xj)j = O(h ); 8m > 0: • If u can be analytically extended to jIm(z)j < a such that there exits a constant C independent of y 2 (−a; a), ku(· + iy)kL2 ≤ C, then (ν) −π(a−)=h jwj − u (xj)j = O(e ); 8 > 0: Similarly, we have (p−1) (j) (j) Theorem 4. Suppose u 2 C ([0; 2π]; C), u (0) = u (2π) for j = (p) 0; 1; : : : ; p − 1 and that u is piecewise continuous. Let v^k be the DFT of ν its sampling and u^k be the Fourier series coefficients. Suppose w^ = (ik) v^ and ν ≤ p − 1. Then, (ν) p−ν jwj − u (xj)j = O(h ): The proof follows easily from the aliasing formulas. We would like to omit here. 3 Finite Element method for 1D second order el- liptic equations (In Sec. 1.1 of Johnson). Consider the elliptic equation −(a(x)u0)0 + c(x)u(x) = f(x); x 2 (0; 1); u(0) = u(1) = 0: (D) where a(x) ≥ a > 0 and c(x) ≥ 0. • In the book, the problem is a special case, −u00 = f. Since there's no big difference from that equation, let's use this more general form. 3 3.1 Minimization of Energy and variational form (weak for- mulation) The equation corresponds to an energy Z 1 1 Z 1 1 Z 1 F (u) = a(x)u0(x)2dx + cu2dx − f(x)u(x)dx: 0 2 0 2 0 It's easy to show using the variational principle or principle of virture work that the elliptical equation is the problem min F (u); u(0) = u(1) = 0 (M) u From this energy form, we take the variation and have Z 1 Z 1 Z 1 a(u; v) = a(x)u0(x)v0(x)dx + cuvdx = f(x)v(x)dx; 0 0 0 where v = δu. This is the weak formulation of the differential equation. The variational problem can be derived easily formally: we multiply any test function v with v(0) = v(1) = 0 and integrate. The variational problem is convenient since it can be used for equations like −(au0)0 + bu0 + cu = f, yielding Z 1 Z 1 Z 1 Z 1 a(x)u0(x)v0(x)dx + bu0vdx + cuvdx − f(x)v(x)dx = 0 0 0 0 0 1 0 where c− 2 b ≥ 0 is required. Note that the second term can also be changed R 1 0 R 1 0 to − 0 buv dx − 0 b uvdx which doesn't really matter. 1 0 Assume that a; b; c are continuous and a(x) ≥ a > 0; c − 2 b ≥ 0. The weak form (or variation formulation) of −(au0)0 + bu0 + cu = f is: 1 Find u 2 H0 such that Z 1 Z 1 Z 1 Z 1 a(x)u0(x)v0(x)dx + bu0vdx + cuvdx − f(x)v(x)dx = 0 (V ) 0 0 0 0 1 for all v 2 H0 , where Z 1 1 n 0 2 2 o H0 = (v ) + v dx < 1; v(0) = v(1) = 0 0 is the space in which the weak formulation make sense. 1 In Mathematics, H0 is a kind of Sobolev spaces (Read Sec. 1.5 for more details). The H1 norm is defined to be s Z 1 0 2 2 kvkH1 = (v ) + v dx: 0 4 Remark 1. If the solutions are nice enough, then (D), (M) and (V ) are equivalent. If the data are not smooth enough, (D) may not have classical solutions but (V ) or (M) may still have solutions. These solutions will then be called the weak solutions of the PDE. 5.