A Motivation for Fourier Analysis in Physics

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1 A Motivation for Fourier Analysis in Physics PHYS Southern Illinois University November 8, 2016 PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

2 The Vibrating String Consider a horizontally stretched string of length L = π that is fixed at its endpoints but is allowed to vibrate freely in the vertical direction. Let u(x, t) denote the vertical displacement as a function of position x and time t. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

3 The Vibrating String Consider a horizontally stretched string of length L = π that is fixed at its endpoints but is allowed to vibrate freely in the vertical direction. Let u(x, t) denote the vertical displacement as a function of position x and time t. The boundary conditions are u(0, t) = u(π, t) = 0 and initial conditions u(x, 0) = f (x) and u t (x, 0) = g(x). PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

4 In the previous lecture we showed that for x (0, π), the general motion of the vibrating string satisfies the wave equation 1 c 2 u tt = u xx. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

5 In the previous lecture we showed that for x (0, π), the general motion of the vibrating string satisfies the wave equation 1 c 2 u tt = u xx. Method 1: Express as a Superposition of Traveling Waves Idea: 1 Find the general solution u(x, t) for the wave equation for all (x, t) R 2 (i.e. with no boundary conditions). 2 Extend the solution to the vibrating string problem w/ boundary conditions to a solution for the wave equation on all (x, t) R 2. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

6 Recall Last lecture we showed that the general solution to the wave equation is a superposition of traveling waves: u(x, t) = F (x + ct) + G(x ct), where F and G are each functions defined on R. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

7 Recall Last lecture we showed that the general solution to the wave equation is a superposition of traveling waves: u(x, t) = F (x + ct) + G(x ct), where F and G are each functions defined on R. We now return to the original problem of solving the vibrating string problem on [0, π] with conditions u(0, t) = u(π, t) = 0, u(x, 0) = f (x) and (x, 0) = g(x). u t PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

8 Recall Last lecture we showed that the general solution to the wave equation is a superposition of traveling waves: u(x, t) = F (x + ct) + G(x ct), where F and G are each functions defined on R. We now return to the original problem of solving the vibrating string problem on [0, π] with conditions u(0, t) = u(π, t) = 0, u(x, 0) = f (x) and (x, 0) = g(x). u t The initial conditions u(x, 0) = f (x) and u t (x, 0) = g(x) are defined for x [0, π]. We need to extend these to functions ˆf and ĝ defined on all R. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

9 Odd Extension First extend f to an odd function on [ π, π]: { f (x) x [0, π] ˆf (x) := f ( x) x [ π, 0]. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

10 Odd Extension First extend f to an odd function on [ π, π]: { f (x) x [0, π] ˆf (x) := f ( x) x [ π, 0]. Periodic Extension For any y R, there exists an integer n such that y = x + 2πn for x [ π, π]. Then define ˆf (y) = f (x + 2πn). PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

11 Odd Extension First extend f to an odd function on [ π, π]: { f (x) x [0, π] ˆf (x) := f ( x) x [ π, 0]. Periodic Extension For any y R, there exists an integer n such that y = x + 2πn for x [ π, π]. Then define ˆf (y) = f (x + 2πn). Let ĝ(x) be the odd-periodic extension of g. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

12 Then if u(x, t) is any solution to the vibrating string problem, its odd-periodic extension û(x, t) solves the wave equation for x R with the conditions: 1 û(nπ, t) = 0 n Z, 2 û(x, 0) = ˆf (x), 3 û t (x, 0) = ĝ(x). PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

13 Then if u(x, t) is any solution to the vibrating string problem, its odd-periodic extension û(x, t) solves the wave equation for x R with the conditions: 1 û(nπ, t) = 0 n Z, 2 û(x, 0) = ˆf (x), 3 û t (x, 0) = ĝ(x). Remark We take an odd extension of u(x, t) so that its twice differentiable for all R. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

14 Then if u(x, t) is any solution to the vibrating string problem, its odd-periodic extension û(x, t) solves the wave equation for x R with the conditions: 1 û(nπ, t) = 0 n Z, 2 û(x, 0) = ˆf (x), 3 û t (x, 0) = ĝ(x). Remark We take an odd extension of u(x, t) so that its twice differentiable for all R. Since û(x, t) solves the wave equation for x R, we must have û(x, t) = F (x + ct) + G(x ct) (x, t) R 2. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

15 To determine the functions F and G, we impose the initial conditions: f (x) = F (x) + G(x), g(x) = cf (x) cg (x). PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

16 To determine the functions F and G, we impose the initial conditions: f (x) = F (x) + G(x), g(x) = cf (x) cg (x). Differentiate f (x) and add/subtract the two equalities to obtain: F (x) = 1 ( f (x) + 1 x ) g(x )dx + C 1 2 c 0 G(x) = 1 ( f (x) 1 x ) g(x )dx + C 2. 2 c 0 PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

17 To determine the functions F and G, we impose the initial conditions: f (x) = F (x) + G(x), g(x) = cf (x) cg (x). Differentiate f (x) and add/subtract the two equalities to obtain: F (x) = 1 ( f (x) + 1 x ) g(x )dx + C 1 2 c 0 G(x) = 1 ( f (x) 1 x ) g(x )dx + C 2. 2 c Since F (x) + G(x) = f (x), we have C 1 + C 2 = 0. 0 PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

18 Therefore, the general solution is: u(x, t) = 1 ( f (x + ct) + f (x ct) + 1 x+ct ) g(x )dx. 2 c x ct This is known as d Alembert s Formula. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

19 Therefore, the general solution is: u(x, t) = 1 ( f (x + ct) + f (x ct) + 1 x+ct ) g(x )dx. 2 c x ct This is known as d Alembert s Formula. The solution has two parts: 1 A superposition of the initial wave f (x) propagating left and right with velocity c. 2 A wave generated by the initial impulse g(x) whose wave fronts are propagating in both directions with velocity c. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

20 Method 2: Superposition of Standing Waves We attempt a separation of variables by attempting a solution of the form u(x, t) = φ(x)ψ(t). Recall we interpret this as a standing wave! PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

21 Method 2: Superposition of Standing Waves We attempt a separation of variables by attempting a solution of the form u(x, t) = φ(x)ψ(t). Recall we interpret this as a standing wave! This leads to the system of homogeneous ODEs: φ (x) λφ(x) = 0 ψ (t) λc 2 ψ(t) = 0. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

22 Method 2: Superposition of Standing Waves We attempt a separation of variables by attempting a solution of the form u(x, t) = φ(x)ψ(t). Recall we interpret this as a standing wave! This leads to the system of homogeneous ODEs: φ (x) λφ(x) = 0 ψ (t) λc 2 ψ(t) = 0. These described simple harmonic motion when λ < 0. Thus, let λ = m 2 : φ(x) = A cos(mx) + B sin(mx) ψ(t) = C cos(mct) + D sin(mct). PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

23 Impose boundary conditions u(0, t) = u(π, t) = 0 to obtain a solution u m (x, t) = (A m cos(mct) + B m sin(mct)) sin(mx) for every positive integer m = 1, 2,. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

24 Impose boundary conditions u(0, t) = u(π, t) = 0 to obtain a solution u m (x, t) = (A m cos(mct) + B m sin(mct)) sin(mx) for every positive integer m = 1, 2,. By linearity, an arbitrary linear combination u(x, t) = (A m cos(mct) + B m sin(mct)) sin(mx) m=1 should also solve the wave equation, provided the series converges. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

25 If the series converges and it is the most general solution to the wave equation, then the initial conditions demand: f (x) = A m sin(mx), g(x) = m=1 B m mc sin(mx). m=1 PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

26 If the series converges and it is the most general solution to the wave equation, then the initial conditions demand: f (x) = A m sin(mx), g(x) = m=1 B m mc sin(mx). m=1 The A m and B m are called Fourier coefficients. Use π 0 sin(mx) sin(nx)dx = δ mnπ/2 to obtain A m = 2 π π B m = 2 mcπ 0 π f (x) sin(mx)dx 0 g(x) sin(mx)dx. PHYS Southern Illinois University A Motivation for Fourier Analysis in Physics November 8, / 11

MA 201: Partial Differential Equations D Alembert s Solution Lecture - 7 MA 201 (2016), PDE 1 / 20

MA 201: Partial Differential Equations D Alembert s Solution Lecture - 7 MA 201 (2016), PDE 1 / 20 MA 201 (2016), PDE 2 / 20 Vibrating string and the wave equation Consider a stretched string of length