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Epiphany AMV Notes 3 Analysis in Many Variables 2019-20 Epiphany Term Epiphany AMV Notes 3 Contents 12 Convergence of Fourier series 61 12.1 Proving convergence . 61 12.2 Functions with discontinuities . 64 12.3 The Gibbs phenomenon . 68 13 Generalised functions 71 14 Proofs of the three big theorems 78 14.1 Green's theorem in the plane . 78 14.2 Stokes' theorem . 79 14.3 The divergence theorem . 82 14.4 Further examples . 85 12 Convergence of Fourier series So far we have used Fourier series to represent various functions f(x). Depending on the boundary P1 nπx conditions, we've seen the sine Fourier series f(x) = n=1 Bn sin( L ), the cosine Fourier series f(x) = P1 nπx P1 2πinx=L n=0 An cos( L ), and the exponential Fourier series f(x) = n=−∞ Ane , all on 0 ≤ x ≤ L, with the boundary conditions being Dirichlet, Neumann and periodic respectively. In each case we used a version of the `Fourier Trick' to find the coefficients An or Bn in these series, under the assumption that they could converge to the right answers. Now we must consider whether these series representations truly give us the function f(x): can the Fourier series converge to f(x) at all, and if so, for what types of functions f(x) does this work? Note, the material in this chapter is not examinable - though it is interesting! It also provides some motivation for the following chapter, on generalised functions, which is examinable. 12.1 Proving convergence For simplicity we'll restrict attention to the exponential Fourier series, and set L = 2π. Then the basic claim is that for at least some functions f we can write 1 X iny f(y) = Ane n=−∞ on 0 ≤ y ≤ 2π. (I've used y instead of x here as x is being saved for the variable in the Fourier integral.) As in last year's analysis course, the precise meaning of the right hand side of this equation is the limit of a sequence of partial sums, and so we need to examine the truncated series SN (y), where N X 1 Z 2π S (y) = A einy;A = f(x)e−inx dx : (12.1) N n n 2π n=−N 0 and N 2 Z+. The hope is to show that lim SN (y) = f(y) N!1 which will be enough to establish pointwise convergence of the partial Fourier sums to the desired answer. To start with we will suppose that f is periodic on [0; 2π], and continuously differentiable there. We manipulate SN (y) as follows: N X iny SN (y) = An e substitute for An n=−N N X 1 Z 2π = f(x) e−inx dx einy now swap the sum and integral 2π n=−N 0 N Z 2π 1 X = f(x) e−in(x−y) dx : 2π 0 n=−N 61 (Note, there is no problem swapping the sum and the integral here since the sum is finite.) Now N 2N 1 X 1 X e−inx = eiNx e−inx pulling out a factor and changing the limits 2π 2π n=−N n=0 −i(2N+1)x 1 (1 − e ) n+1 = eiNx using the geometric sum formula Pn rk = 1−r 2π (1 − e−ix) k=0 1−r 1 (ei(N+1=2)x − e−i(N+1=2)x) e−i(N+1=2)x = eiNx 2π (eix=2 − e−ix=2) e−ix=2 1 1 sin((N + 2 )x) 1 iα −iα = 1 : using sin α = (e − e ), twice 2π sin( 2 x) 2i Thus Z 2π 1 1 sin((N + 2 )(x − y)) SN (y) = f(x) 1 dx (12.2) 0 2π sin( 2 (x − y)) Z 2π 1 (x − y) sin((N + 2 )(x − y)) = f(x) 1 dx (12.3) 0 2 sin( 2 (x − y)) π(x − y) Z 2π = fe(y)(x) hN (x − y) dx (12.4) 0 where sin((N + 1 )x) h (x) = 2 (12.5) N πx and, for x 6= y, (x − y) fe(y)(x) = f(x) 1 : 2 sin( 2 (x − y)) Note that limx!y fe(y)(x) = f(y). Hence if set fe(y)(y) = f(y), then fe(y)(x) is a continuous function of x with the same value at x = y as f has there. Thus, if we could show that Z 2π lim fe(y)(x)hN (x − y) dx = fe(y)(y) N!1 0 then we would be done, given that fe(y)(y) = f(y). From now on we'll set y = 0 and write fe(0)(x) as fe(x) to save ink, but the arguments work exactly the same for any non-zero value of y, after a simple change of variables. Our aim has become to show that, for any continuously differentiable function fe(x), Z 2π lim fe(x)hN (x) dx = fe(0) : (12.6) N!1 0 To this end, let's investigate the function hN (x), as defined by equation (12.5)), and its behaviour as N ! 1. The left side of figure 30 shows hN (x) for two different values of N: the top plot showing h10(x) and the bottom one h100(x). Note that the vertical scales are not the same! Comparing these two plots, there are a couple of notable features: • For x 6= 0, hN (x) oscillates with increasing frequency as N increases. N+1=2 • At x = 0, hN (0) = limx!0 hN (x) = π , and grows with N. Thus limN!1 hN (x) is not a well-defined function. However, all we really care about is the integral of hN (x) multiplied by some other function, and this might be better behaved as N ! 1. Let's pick some a0 < 0 and define Z x 0 0 gN (x) = hN (x ) dx (12.7) a0 62 Figure 30: Plots of hN (x) (left) and gN (x) (right), for N = 10 (top) and N = 100 (bottom). (For animated versions of these plots, see http://maths.dur.ac.uk/users/P.E.Dorey/AMV/hNandgN.html.) Plots of g10(x) and g100(x) are shown on the right side of figure 30, for a0 = −2. We see that gN (x) does not diverge as N ! 1, but rather appears to converge to a step function: 1 x > 0 ; g (x) ! N 0 x < 0 : (Though, there does seem to be something slightly odd going on near to x = 0; this is related to the Gibbs phenomenon, which will be discussed at the end of this chapter.) Later on we will prove that this is indeed the case, by showing that fhN (x)g is an example of a δ-convergent sequence, defined as follows: Definition: A sequence fhN (x)g is called a δ-convergent sequence if: b 1 a < 0 < b (I) lim R h (x) dx = N!1 a N 0 a < b < 0 or 0 < a < b R b (II) for any M > 0 and jaj ≤ M, jbj ≤ M, j a hN (x) dxj < γ, where γ is some constant independent of a, b and N. This next step is the key to the proof of the convergence of the truncated Fourier sums. It follows from the fact that hN (x) is a δ-convergent sequence. Let's take a < 0 < b, and use our assumption that f(x), the function whose Fourier series we are looking at, is differentiable in the interval [a; b] (soon we will 0 relax this a bit, but let's make life easy to begin with). Since hN (x) = gN (x), we can integrate by parts to find: Z b Z b Z b 0 b 0 hN (x)f(x) dx = gN (x)f(x) dx = gN (x)f(x) a − gN (x)f (x) dx : a a a Taking the limit N ! 1 allows us to replace gN (x) by 0 for x < 0 and by 1 for x > 0 (part II of the definition of δ-convergence allows us to take this limit inside the integral sign), so in the limit the 63 right-hand side becomes Z b 0 b 1:f(b) − 0:f(a) − f (x)dx = f(b) − [f(x)]0 = f(0) : 0 Hence Z b lim hN (x)f(x)dx = f(0) when a < 0 < b (12.8) N!1 a which is exactly what we wanted to show. R b We'll write the LHS of (12.8) as a δ(x)f(x)dx, thereby defining δ(x), the Dirac delta function. It is not a function in the usual sense, but is rather defined by its value when multiplied by another (better- behaved) function and integrated. Shifting the integration variable from x to x − c its key property can be summed up as follows: Definition: Equation (12.8) defines the Dirac delta function δ(x), and can be symbolised in a neat, general form as: Z b δ(x − c)f(x)dx = f(c) when a < c < b : (12.9) a If we apply this definition (and convergence result) to our original goal, equation (12.6), we see that the Fourier series representation of the function f(x) does indeed converge: Z 2π Z 2π lim f(y)hN (y − x)dy = δ(y − x)f(y)dy = f(x) : N!1 0 0 The Dirac delta function is not a traditional function since it is not defined like a normal function { it does not have numerical values for each value of x.
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