The Fourier series for a 2π-periodic function

Transcription

1 The Fourier series for a 2π-periodic function Let f : ( π, π] R be a bounded piecewise continuous function which we continue to be a 2π-periodic function defined on R, i.e. f (x + 2π) = f (x), x R. The Fourier series of this function is written as where a n = 1 π π π f (x) a (a n cos(nx) + b n sin(nx)) f (x) cos(nx) dx and b n = 1 π π π f (x) sin(nx) dx. When certain sufficient conditions about f hold we have that f (x) = a (a n cos(nx) + b n sin(nx)) at all the points of continuity of f (x). MA2715, 2017/8 Week 30, Page 1 of 20

2 MA2715, 2017/8 Week 30, Page 2 of 20 Sufficient conditions for pointwise convergence Let f (x) be piecewise continuous on ( π, π] with the left and right limits f (x ) and f (x+) existing at all points. Suppose also that for some δ > 0 f (x) is differentiable in (x δ, x) and in (x, x + δ) with a left and right derivative at x. These are sufficient conditions for a m 0 lim m 2 + (a n cos(nx) + b n sin(nx)) = At points of continuity f (x ) + f (x+). 2 a m 0 lim m 2 + (a n cos(nx) + b n sin(nx)) = f (x). There are no known conditions which are both necessary and sufficient for the pointwise convergence of Fourier series.

3 MA2715, 2017/8 Week 30, Page 3 of 20 Example: Q1 of exercise sheet 1, if π x < π 3, f (x) = 0, if π 3 x π 3, 1, if π 3 < x < π. Part of the question asks you to show that the Fourier series is as follows. S(x) = 2 ( 1) n cos( nπ 3 ) sin(nx). π n

4 Sketch of the piecewise constant function S(x) = 2 π ( 1) n cos( nπ 3 ) n sin(nx) π 7π/3 5π/3 π π/3 π/3 π 5π/3 7π/3 3π 0 The function is an odd function which is why there are only sine terms in the series. f (x) = S(x) at the points of continuity. f (x) S(x) at the points of discontinuity, i.e. π, π/3, π/3 etc. MA2715, 2017/8 Week 30, Page 4 of 20

5 MA2715, 2017/8 Week 30, Page 5 of May 2017 question with terms up to sin(12x)

6 MA2715, 2017/8 Week 30, Page 6 of May 2017 question with terms up to sin(48x)

7 MA2715, 2017/8 Week 30, Page 7 of 20 Integrating a Fourier series This is valid but take care with the constant term. At points of continuity we have f (t) = a (a n cos(nt) + b n sin(nt)). For all x ( π, π) we can write φ(x) = x 0 f (t) dt = a 0x 2 + ( an When a 0 0 we complete the details by using x 2 = ( 1) n+1 which is valid for x < π. n sin(nx) = sin(x) sin(2x) 2 n sin(nx) b ) n (cos(nx) 1). n + sin(3x) 3

8 MA2715, 2017/8 Week 30, Page 8 of 20 Differentiating a Fourier series Suppose that the 2π-periodic extension of f (x) is continuous so that we have f (x) = a (a n cos(nx) + b n sin(nx)). In particular this means that f ( π+) = f (π ). In this case we can differentiate term-by-term to give f (x) ( na n sin(nx) + nb n cos(nx)). We have equality at all points of continuity of f (x).

9 MA2715, 2017/8 Week 30, Page 9 of 20 On ( π, π] let and f (x) = g(x) = Example { 0, π < x < 0, x(π x), 0 x π, { 0, π < x < 0, π 2x, 0 x π. Note that g(x) = f (x) in ( π, 0) and (0, π). g(x) is discontinuous at x = 0 and at ±π. The 2π periodic extension of f (x) is continuous.

10 Sketch of f (x) and g(x) in ( π, π) MA2715, 2017/8 Week 30, Page 10 of 20 π π π π π 0 π/2 π

11 MA2715, 2017/8 Week 30, Page 11 of 20 Techniques to get the coefficients of f (x) and g(x) In (0, π) we have polynomials in x and integration by parts is needed. At any stage we want to differentiate the polynomial term. We can get the series for g(x) and then use term-by-term integration to get the series for f (x). Alternatively we can first get the series for f (x) and then differentiate to get the series for g(x).

12 MA2715, 2017/8 Week 30, Page 12 of 20 Fourier series for f (x) and g(x) in the example Which ever route its taken the series obtained are as follows. ( 2 g(x) n 2 π (1 + ( 1)n+1 ) cos(nx) + 1 ) n (1 + ( 1)n ) sin(nx), ( f (x) = π n 3 π (1 + ( 1)n+1 ) sin(nx) 1 ) n 2 (1 + ( 1)n ) cos(nx). Note that if we consider either function on (0, π) and consider the odd extension of g(x) or the even extension of f (x) then the functions created have period π.

13 MA2715, 2017/8 Week 30, Page 13 of 20 The heat equation where Fourier series can be used Let u = u(x, t) denote temperature in a one-dimensional body 0 x L for time t 0. The heat equation is u t = u κ 2 x 2 and this governs how u changes in space and in time. Here κ > 0 is known as the thermal diffusivity of the body. Consider the following specific set-up in which we know the initial temperature distribution u(x, 0) = u 0 (x) and for all time t 0 we have boundary conditions u(0, t) = 0 and u(l, t) = 0. This was the type of problem that Fourier was considering when he created what we now know as Fourier series.

14 MA2715, 2017/8 Week 30, Page 14 of 20 The space time domain 0 x L, 0 t t Boundary Condition u(0, t) = 0 PDE u t = κ 2 u x 2 Boundary Condition u(l, t) = 0 Initial condition: u(x, 0) = u 0 (x) x The problem is find u(x, t) in 0 < x < L, t > 0.

15 MA2715, 2017/8 Week 30, Page 15 of 20 The separation of variables technique In this case it first involves finding a solution of the PDE of the form u(x, t) = X (x)t (t). If we substitute into the PDE then we get XT = κx T giving X X = T κt. In the last version as the left hand side only depends on x and the right side only depends on t the only way it can be true for all x and t is if they are constant. To get a solution which remains bounded as t requires that the constant must be negative and if we write this as m 2 then we have T = m 2 κt, X = m 2 X.

16 Expressions for X (x) and T (t) MA2715, 2017/8 Week 30, Page 16 of 20 T = m 2 κt, X = m 2 X. The general solution to these have the form where B and C are constants. T (t) = (const) exp( m 2 κt), X (x) = B sin(mx) + C cos(mx), If we add to the requirements the boundary conditions then C = 0 and sin(ml) = 0 and this requires that m = nπ L, n = 1, 2,....

17 The Fourier series solution X (x)t (t) is hence of the form sin ( nπx ) exp ( n2 π 2 ) L L 2 κt, n = 1, 2, As linear combinations will also satisfy the PDE and the boundary conditions a candiddate general solution is of the form u(x, t) = ( nπx ) b n sin exp ( n2 π 2 ) L L 2 κt. b n, n = 1, 2,... are obtained from the initial conditions as u 0 (x) = ( nπx ) b n sin. L Hence we can satisfy the PDE, the boundary conditions and the initial conditions. MA2715, 2017/8 Week 30, Page 17 of 20

18 Convergence in the L 2 ( π, π) norm and steps towards Parseval s identity For real valued functions defined on ( π, π) we can define the following inner product and we can define the L 2 norm by (f, g) = π π f 2 = (f, f ). f (x)g(x) dx, The Fourier coefficients can we written using this notation. (f, 1) = πa 0, (f, cos(nx)) = πa n, (f, sin(nx)) = πb n. Let S N denotes the truncated series S N (x) = a N m=1 For n N we similarly have (a m cos(mx) + b m sin(mx)). (S N, 1) = πa 0, (S N, cos(nx)) = πa n, (S N, sin(nx)) = πb n. MA2715, 2017/8 Week 30, Page 18 of 20

19 Bessel s inequality From this it follows that Hence (f S N, S N ) = 0 and thus (f, S N ) = (S N, S N ). 0 f S N 2 = (f S N, f S N ) = (f S N, f ) = (f, f ) (f, S N ) = (f, f ) (S N, S N ) = f 2 S N 2. In terms of the Fourier coefficients (S N, S N ) = a N 0 2 (S N, 1) + (a m (S N, cos(mx)) + b m (S N, sin(mx))) m=1 ( a 2 N ) 0 = π 2 + (am 2 + bm) 2. is Bessel s inequality. m=1 lim S N 2 f 2 N MA2715, 2017/8 Week 30, Page 19 of 20

20 MA2715, 2017/8 Week 30, Page 20 of 20 Convergence and Parseval s identity For the functions we deal with f S N 2 = f 2 S N 2 0 as N and Parsevels s identity is a (am 2 + bm) 2 = 1 π m=1 π π f (x) 2 dx.