A proof for the full Fourier series on [ π, π] is given here.

Transcription

1 niform convergence of Fourier series A smooth function on an interval [a, b] may be represented by a full, sine, or cosine Fourier series, and pointwise convergence can be achieved, except possibly at the boundary points. If the smooth function satisfies the same boundary conditions as the eigenfunctions used for the expansion, uniform convergence can be obtained. Theorem: Let A n X n represents the full Fourier series, or Fourier sine series, or Fourier cosine series. The series A n X n converges to f(x) absolutely and uniformly on [a, b] provided that: (i) f(x) and f (x) are continuous on [a, b], and (ii) f(x) satisfies the same boundary conditions as those of X n. A proof for the full Fourier series on [ π, π] is given here. Proof: Let A n, B n denote the Fourier coefficients of f, and A n, B n denote the Fourier coefficients of f. se integration by parts to get π A n = f(x) cos nx dx π π = 1 nπ f(x) sin nx π π π π f (x) sin nx dx nπ = 1 n B n 1

2 Similarly, π B n = f(x) sin nx dx π π = 1 π nπ f(x) cos nx π π + f (x) cos nx dx π nπ = 1 n A n in which the periodicity of f(x) is used. Bessel s inequality for the f series gives 1 2 A ( A n 2 + B n 2 ) 1 π f 2 dx <. π π Then A n = ( ) 1/2 ( 1 n 2 1 n B n B n 2 ) 1/2 < continuity of f used in which the Schwarz s inequality has been applied. This means that the series A n converges absolutely. A similar statement can be made on B n. Therefore, the Fourier series converges absolutely. The sum of the Fourier series converges pointwise to f(x) everywhere as f is continuous and f( π) = f(π) (no jump anywhere). Then, 2

3 max f(x) S N (x) { } max ( A n cosnx + B n sin nx ) n=n+1 n=n+1 ( A n + B n ) 0 as N because it is the tail of a convergent series of numbers. Therefore, S N (x) f(x) uniformly. 3

4 Inhomogeneous problems in a finite interval Consider the inhomogeneous problem with time-dependent boundary values: u tt c 2 u xx = f(x, t) u(x, 0) = φ(x) u(0, t) = h(t) 0 < x < L u t (x, 0) = ψ(x) u(l, t) = k(t). The boundary conditions of this problem generally cannot be satisfied by the full, sine, or cosine Fourier series. Yet, the function u(x, t) can be represented pointwise in (0, L) by one of these series with time-dependent coefficients. Since the boundary conditions are of Dirichlet type, we use the Fourier sine series: u(x, t) = u n (t) sin nπx L. The equations for computing the coefficients of this series cannot be obtained by simply plugging the series into the pde. The spatial differential operator 2 cannot be moved inside x 2 the summation as the convergence is not uniform. It is therefore necessary to use a modified approach. Expansion method The technique is to expand u, u tt, and u tt into separate Fourier series with time-dependent coefficients. Multiplying both sides of the pde by sin nπx and integrating from 0 to L 4

5 L, one can get the following set of equations that connect the coefficients of the different Fourier series v n (t) c 2 w n (t) = f n (t). v n (t), w n (t), and f n (t) are coefficients of the Fourier sine series of u tt and u xx, and f(x, t), respectively. v n and w n can be related to the Fourier coefficient u n as following. v n (t) = 2 L w n (t) = 2 L L 0 L 0 2 u nπx sin t2 L dx = d2 u n dt, 2 2 u nπx sin x2 L dx = λ n u n (t) + 2nπL 2 [h(t) ( 1) n k(t)] where λ n = (nπ/l) 2. Therefore, u n satisfies the ode: integration by parts d 2 u n dt 2 + c2 λ n u n (t) = c 2 2nπL 2 [h(t) ( 1) n k(t)] + f n (t). The initial conditions for this ode are u n (0) = φ n and du n dt (0) = ψ n where φ n and ψ n are Fourier sine coefficients for φ(x) and ψ(x), respectively. Subtraction method The boundary conditions can be made homogeneous by subtracting any known function that satisfies them. We use the function 5

6 ( (x, t) = 1 x ) h(t) + x L L k(t) which satisfies the BCs. With the subtraction ũ(x, t) = u(x, t) (x, t), the problem can be converted to the following problem for ũ: ũ tt c 2 ũ xx = f(x, t) tt ũ(0, t) = 0 ũ(l, t) = 0 0 < x < L ũ(x, 0) = φ(x) (x, 0) ũ t (x, 0) = ψ(x) t (x, 0). Since the boundary conditions for ũ is now Dirichlet, we are assured of the uniform convergence of the Fourier sine series expansion of ũ(x, t) on [0, L] at any t. In particular, the Dirichlet boundary conditions are satisfied all the time. The Fourier sine series of the continuous function converges pointwise to on (0, L), but generally not at the boundary points. As u n (t) = ũ n (t) + n (t), the Fourier sine series of u(x, t) tends to h(t) and k(t) as x gets close to 0 and L, respectively. Note, however, the Gibb s phenomenon generally occurs. The equation for the Fourier sine series coefficient ũ n is d 2 ũ n dt + 2 c2 λ n ũ n (t) = f n (t) d2 n dt. 2 Through the substitution ũ n = u n n, this equation can be rewritten as d 2 u n dt 2 + c2 λ n [u n (t) n ] 6

7 = d2 u n dt 2 + c2 λ n u n (t) c 2 2nπL 2 [h(t) ( 1) n k(t)] = f n (t). The term c 2 λ n n re-introduces the boundary conditions h(t) and k(t) as two terms in the ode. This equation is the same as the one derived by the expansion method. The initial conditions are also the same. The wave equation in higher dimensions The solution of the wave equation in 3D can be expressed as a surface integral. The Kirchhoff-Poisson solution The Cauchy problem (CP) is: u tt = c 2 u x R 3, t 0 u( x, 0) = φ( x), u t ( x, 0) = ψ( x). where = 2 xx + 2 yy + 2 zz. This problem can be split into two parts according to the nature of the initial conditions: (1) WCP (2) VCP w tt = c 2 w w( x, 0) = φ( x), w t ( x, 0) = 0 7

8 v tt = c 2 v v( x, 0) = 0, v t ( x, 0) = ψ( x) Clearly, u = w + v is the solution of the original CP. If one can solve VCP, then one can also solve WCP. Lemma: If v is a solution of VCP with v t ( x, 0) = ψ( x), then w = v t is a solution of WCP with w( x, 0) = ψ( x). Theorem: (Kirchhoff s formula) Let ψ be continuous with continuous first and second partial derivatives for all x. Then for t > 0 and x R 3, the solution of VCP is v( x, t) = 1 ψ( y)dσ c 2 t S( x;t) in which S( x; t) = S t is the sphere with radius ct centered at x and dσ represents a surface element on S t. Proof: Let y = x + ct ξ, the integral formula can be written as v( x, t) = 1 c 2 t S t ψ( y)dσ = t where is the unit sphere in the ξ-space and dτ represents a surface element on. ψ( x + ct ξ)dτ (1) 8

9 Then v = t At the same time v t = 1 ψ( x + ct ξ)dτ = 1 c 2 t ψ( x + ct ξ)dτ + ct which can further be written as = 1 t v + 1 ψ ns dσ = 1 ct S t t v + 1 ct S t ψdσ. ψ n dτ (2) B t ψdv. n is the outward pointing unit normal on, n S is the outward pointing unit normal on S t, and B t is the solid ball with radius ct centered at x. Let I = ψdv, then v t = 1 B t t v + I ct. v tt = 1 t 2v + 1 t v t 1 ct 2I + I t ct = I t ct. Since I t = c ψdσ, S t v tt = 1 ψdσ = c 2 v. t S t The equation is satisfied. Furthermore, Eq.(1) gives 9

10 v( x, 0) = 0 and Eq.(2) gives v t ( x, 0) = 1 = ψ( x). ψ( x + 0 ζ)dσ = 0, ψ( x + c 0 ξ)dτ + c 0 The initial conditions are satisfied. ψ n dτ As a result, the solution of the CP is u( x, t) = t [ 1 c 2 t ] φ( y)dσ S t + 1 ψ( y)dσ. c 2 t S t This expression is known as Poisson s formula for the solution of the CP problem for the wave equation in 3D. Example: Consider the simple cases: (i) φ = 1, ψ = 0 and (ii) φ = 1, ψ = 1. For a given point x and time t, u( x, t) depends only on the initial data given on the sphere S t of radius ct about the point x. S t is the domain of influence of the initial conditions at time t. This was noted by Huygens in the seventeenth century (Huygens Principle). 10

11 Hadamard s method of descent The idea is to use Poisson s 3D formula to solve the 2D problem. Suppose that the 3D functions all depend only on the coordinates x and y. The initial conditions become u(x, y, z, 0) = φ(x, y), u t (x, y, z, 0) = ψ(x, y). The solution u(x, y, z, t) also depends only on x, y, and t. [ 1 c 2 t ] + 1 c 2 t u(x, y, t) = φ(ξ, η)dσ t S t (ξ, η, ζ) represents (x, y, z) in the integral. The surface integral can be projected onto the xy-plane as dξdη = n k dσ = ζ ct dσ = 1 (ct)2 (ξ x) ct 2 (η y) 2 dσ. Then, as S t projects two times on D t, ψ(ξ, η)dσ = S t ψ(ξ, η) 2ct D t (ct)2 (ξ x) 2 (η y) 2dξdη where D t is the disk with radius ct centered at (x, y). The formula for u(x, y, t) becomes S t ψ(ξ, η)dσ. 11

12 [ ] u(x, y, t) = 1 φ(ξ, η) 2πc t D t (ct)2 (ξ x) 2 (η y) 2dξdη + 1 ψ(ξ, η) 2πc D t (ct)2 (ξ x) 2 (η y) 2dξdη. The domain of influence is now the whole of D t. A disturbance at another point will eventually be felt at (x, y) and will then be felt at all later times. Huygens principle does not hold in the plane. plot example Problem Set 7 12