Periodic functions: simple harmonic oscillator

Transcription

1 Periodic functions: simple harmonic oscillator Recall the simple harmonic oscillator (e.g. mass-spring system) d 2 y dt 2 + ω2 0y = 0 Solution can be written in various ways: y(t) = Ae iω 0t y(t) = A cos ω 0 t + B sin ω 0 t The constants of integration A and B depend on the initial conditions (t = 0). The y(t) is periodic with period (or periodicity) τ 0 = ω 0

2 Periodic functions: wave motion If I take a snapshot of a wave (e.g. a water wave), I find y(x) = A sin x λ = A sin kx where A is the amplitude and λ is the wavelength and k is the wave number. The period (or periodicity) in this case is λ = k If we consider time-dependence with angular frequency ω, y(x, t) = A sin(kx ωt) If v is a constant (i.e. independent of λ), then v = ω k and y(x, t) = A sin λ (x vt)

3 More on periodic functions Consider again y(x, t) = A sin λ (x vt) At fixed t, y(x) is a periodic function of x with periodicity λ What about more general cases: Might have periodic y(x) with periodicity L

4 More on periodic functions Consider again y(x, t) = A sin λ (x vt) At fixed t, y(x) is a periodic function of x with periodicity λ At fixed x, y(t) is a periodic function of t with periodicity ω = λ v What about more general cases: Might have periodic y(x) with periodicity L Might have more complex pattern, not described by y(x) = sin x L or y(x) = cos x L

5 More on periodic functions Consider again y(x, t) = A sin λ (x vt) At fixed t, y(x) is a periodic function of x with periodicity λ At fixed x, y(t) is a periodic function of t with periodicity ω = λ v What about more general cases: Might have periodic y(x) with periodicity L Might have more complex pattern, not described by y(x) = sin x L or y(x) = cos x L Generally possible to write y(x) as a summation of periodic functions (e.g. sines and cosines)

6 Average value of a function We can define the average value of f (x) on the interval a < x < b b a f (x)dx b a An important example is the average of sin 2 nx or cos 2 nx where n is an integer (n 0) over the interval from < x < π First note the relationship sin 2 nx + cos 2 nx = 1, and then notice that sin2 nxdx = cos2 nxdx 1 π [ sin 2 nx + cos 2 nx ] dx = 1 sin 2 nxdx = 1 dx = 1 π

7 Average value of a function continued From these we finally see that 1 sin 2 nxdx = 1 cos 2 nxdx = 1 2 Another important example is f (x) = sin mx cos nx where m and n are integers (m 0 and n 0), again averaged over the interval < x < π 1 π sin mx cos nxdx = 0 ( ) ( ) To prove this, use sin mx cos nx = ie imx ie imx e inx +e inx 2 2

8 Average value of sin mx cos nx continued Consider the case m n, then we have integrals of the type, with N = m n, N = m + n, N = n m, or N = n m, and hence N is an integer N 0 e inx dx = 1 in einx π = 0 Next consider separately the case m = n, which gives sin nx cos nxdx = i 4 ( e inx e inx) ( e inx + e inx) dx = 0 In the homework, we will also show that for m n, 1 1 sin mx sin nxdx = 0 and cos mx cos nxdx = 0

9 Summary The important integrals we performed were averages over < x < π and can be summarized, with m and n integers and m 0 and n sin mx sin nxdx = 1 2 δ m,n cos mx cos nxdx = 1 2 δ m,n sin mx cos nxdx = 0 For second integral, we can add the case m = n = 0, which is a special case, and 1 π dx = 1

10 Fourier series, Fourier coefficients Let s imagine we can make a series of sin nx, cos nx functions, with n = 1, 2, 3,... f (x) = 1 2 a 0 + a n cos nx + b n sin nx The a n (including n = 0) and b n are the Fourier coefficients We do not yet know what kind of functions f (x) can be realized with this series Integrals can be solved to find appropriate values of the a n and b n to generate the function f (x)

11 Properties of the function f (x) We first note that the cos nx and sin nx functions are periodic with periodicity For integer k, sin n(x + 2kπ) = sin nx cos 2kπ = sin nx cos n(x + 2kπ) = cos nx cos 2kπ = cos nx This means that the Fourier series in the previous slide results in f (x) with periodicity described by f (x + 2kπ) = f (x) There are a few other restrictions but these will be dealt with later; periodicity is the most important consideration now

12 Expressions for a n and b n We can develop expressions for the coefficients a n and b n, and separately a 0 f (x) = 1 2 a 0 + a n cos nx + b n sin nx Multiply the expansion on the left- and right-hand sides by cos mx and integrate over one period On the left-hand side we get f (x) cos mxdx, which is equal to the right-hand side a 0 2 cos mxdx+ a n cos nx cos mxdx+ b n sin nx cos mxdx We did these integrals three slides earlier, and we use the results!

13 Expressions for a n and b n, continued a n = 1 π b n = 1 π For the a 0 case we obtain, a 0 = 1 π f (x) cos nxdx f (x) sin nxdx f (x)dx Note also that f (x) can be complex, so that the coefficients can be complex

14 Complex Fourier series Recalling the Euler formula, we can write cos nx = einx + e inx 2 sin nx = einx e inx 2i This suggest that instead of using cos nx and sin nx, we might use e inx f (x) = n= c n e inx To obtain the coefficients, multiply left and right sides by e imx and integrate over one period c n = 1 f (x)e inx dx

15 1 l Chapter7: nπx Fourier series Other intervals For a function with periodicity, we used functions with periodicity for the expansion, namely cos x, sin x, or e inx Consider now a function with periodicity 2l, so that f (x + 2l) = f (x) We can see that sin nπx l as does e inπx l We then have the relations and cos nπx l have the desired periodicity, f (x) = a a n cos nπx l Or we might use the complex series, + a n sin nπx l f (x) = n= The coefficients then are found from c n e nπx l