Fourier transforms We can imagine our periodic function having periodicity taken to the limits ± In this case, the function f (x) is not necessarily periodic, but we can still use Fourier transforms (related to Fourier series) Consider the complex Fourier series, periodic with periodicity 2l f (x) = n= c n e inπx l Write same thing in an equivalent form, using n = 1, f (x) = l π n= c n e inπx l ( ) nπ l
Fourier transforms continued Next we take the limit l, and the summation becomes an integral f (x) = g(k)e ikx dx Here k = nπl, dk = nπ/l, and g(k) = c n l/π We say that f (x) is the inverse Fourier transform of g(k) To get g(k) given f (x)(wesay g(k)isthefouriertransformof f(x)), we again start with the version for periodic functions, c n = 1 2l l l f (x)e inπx l dx Again use g(k) = c n l/π, k = nπ/l, and take the limits of integration from to g(k) = 1 2π f (x)e ikx dx
Example: Problem 21 Find the Fourier transform of f (x) = e x2 /2σ 2 g(k) = 1 e x2 /2σ 2 e ikx dx 2π Notice that for the exponent, we can write ( x x 2 /2σ 2 ikx = 2 1/2 σ + iσk ) 2 2 1/2 σ2 k 2 2 Then the integral can be written g(k) = 1 2π e σ 2 k 2 2 e h 1 2σ 2 (x+iσ 2 k) 2i dx Change variables to y = x + iσ 2 k, then dx = dy, and we get g(k) = 1 2π e σ 2 k 2 2 e y2 2σ 2 dx = σ e σ2 k 2 2 2π
Fourier sine and cosine transforms Less commonly used are the Fourier sine/cosine transforms 2 f s (x) = g s (k) sin kxdk π 0 2 g s (k) = f s (x) sin kxdx π 0 2 f c (x) = g c (k) cos kxdk π 0 2 g c (k) = f c (x) cos kxdx π 0 Here, it is assumed that f s (x) and g s (k) are odd functions of x and k respectively, and f c (x) and g c (k) are even functions of x and k respectively
Chapter 8: Ordinary differential equations: Introduction and goals By the end of the chapter you should be able to Solve separable differential equations Note that we will skip Laplace transforms in this chapter, only because of our time limit
Chapter 8: Ordinary differential equations: Introduction and goals By the end of the chapter you should be able to Solve separable differential equations Solve linear first-order equations Note that we will skip Laplace transforms in this chapter, only because of our time limit
Chapter 8: Ordinary differential equations: Introduction and goals By the end of the chapter you should be able to Solve separable differential equations Solve linear first-order equations Solve linear second-order equations with constant coefficients Note that we will skip Laplace transforms in this chapter, only because of our time limit
Chapter 8: Ordinary differential equations: Introduction and goals By the end of the chapter you should be able to Solve separable differential equations Solve linear first-order equations Solve linear second-order equations with constant coefficients Use Fourier transforms to solve differential equations Note that we will skip Laplace transforms in this chapter, only because of our time limit
Chapter 8: Ordinary differential equations: Introduction and goals By the end of the chapter you should be able to Solve separable differential equations Solve linear first-order equations Solve linear second-order equations with constant coefficients Use Fourier transforms to solve differential equations Use the Dirac delta function Note that we will skip Laplace transforms in this chapter, only because of our time limit
Chapter 8: Ordinary differential equations: Introduction and goals By the end of the chapter you should be able to Solve separable differential equations Solve linear first-order equations Solve linear second-order equations with constant coefficients Use Fourier transforms to solve differential equations Use the Dirac delta function Begin to apply Green functions for solving differential equations Note that we will skip Laplace transforms in this chapter, only because of our time limit
Separable equations If we have a differential equation with two variables, we can directly integrate if the equation is separable Consider the decay/growth equation dn dt = ±λn We can rewrite and get only N on the left side, and t on the right side dn N = λdt Integrate then, and we get up to a constant ln N = ±λt + C N = N 0 e ±λt
Example: Problem 1 Simple example, problem 1, solve xy = y if y = 1 when x = 2 Separate the equation dy y = dx x Solution is y = Cx, so for the conditions given C = 1/2, and y = (1/2)x Check it! dy dx = 1/2, so xy = (1/2)x = y
Linear first-order equations We consider equations here of the form y + Py = Q P and Q are functions of x (cases where coefficients are constants are dealt with later) First consider Q = 0, then equation is separable dy y We can solve as y = Ae R Pdx Write I = Pdx, the ye I = A = Pdx d dx (yei ) = e I y + e I yp = e I (y + Py) = e I Q
Linear first-order equations continued We can integrate the equation on the previous slide ye I = e I Qdx + c We can multiply by e I to obtain y = e I e I Qdx + ce I
Example: First order linear equation Example: Section 3, problem 11, solve y + y cos x = sin 2x Here we see P(x) = cos x and Q(x) = sin 2x We see that I = Pdx = cos xdx = sin x The solution we can write as, y = e sin x sin 2xe sin x dx + ce sin x In the integral, we can make the substitution u = sin x, and du = cos xdx, and also sin 2x = 2 sin x cos x, so that we get sin 2xe sin x dx = 2 ue u dx = 2(sin x 1)e sin x Finally we get the solution, y = 2(sin x 1) + ce sin x
Example, section 3 problem 11 Test the solution! (make sure it is correct) y + Py = 2 cos x c cos xe sin x + 2 cos x(sin x 1) + c cos xe sin x Finally the is equal to 2 cos x sin x = sin 2x
Second-order linear equations with constant coefficients We will consider first homogeneous equations, with a 2, a 1, and a 0 constants (independent of x and y) d 2 y a 2 dx 2 + a dy 1 dx + a 0y = 0 Example: Simple harmonic oscillator, a 2 = 1, a 1 = 0, a 0 = ω 2 0 We solved this before, d 2 y dt 2 + ω2 0y = 0 y = A cos ω 0 t + B sin ω 0 t = Ce iω 0t Let s try to systematically solve differential equations of this type
Characteristic equation Define the linear operator, D = d dx, and the D2 = d2 dx 2, then (a 2 D 2 + a 1 D + a 0 )y = 0 We can in general factor this equation (D a)(d b)y = 0 We can solve then if we take that if y = c 1 e ax + c 2 e bx, since D(c 1 e ax + c 2 e bx ) = c 1 ae ax + c 2 be bx
Characteristic equation continued (a 2 D 2 + a 1 D + a 0 )y = 0 We get from this the characteristic equation, treating D now as a number a 2 D 2 + a 1 D + a 0 = 0 A quadratic equation! D 1,2 = a 1 ± a1 2 4a 2a 0 2a 2 Then the solution is y = c 1 e D 1x + c 2 e D 2x (in the last slide, D 1 = a, D 2 = b) Also note that if 4a 2 a 0 > a 1 then D 1 and D 2 are complex!
A special case: Equal roots for characteristic equation We might end up with equal roots for the characteristic equation, for example (D a)(d a)y = 0 Here the operator D = d dx Apparently only gives solution y = c 1 e ax, but we will see there is one more solution Take u = (D a)y, then the u satisfies (D a)u = 0 Therefore we find u = Ae ax = (D a)y, which we then solve for y This is a first-order linear equation and we get y = (Ax + B)e ax
Example: Simple harmonic oscillator As an example, consider the simple harmonic oscillator, with D = d dt and D2 = d2 dt 2 d 2 y dt 2 + ω2 0y = (D 2 + ω 2 0)y = (D + iω 0 )(D iω 0 )y = 0 The roots of the characteristic equation then are ±iω 0, so finally y = c 1 e iω 0t + c 2 e iω 0t As we saw before, we can also use the Euler relation and write this in terms of sines and cosines, y = A cos ω 0 t + sin ω 0 t
Example: Damped simple harmonic oscillator When there is damping, we need to consider the differential equation d 2 y dy + 2b dt2 dt + ω2 0y = 0 We have from this, with D = d dt the characteristic equation D 2 + 2bD + ω0 2 = 0 The roots of the characteristic equation we find to be D 1,2 = b ± b 2 ω0 2 There three distinct regimes: b < ω 0 (underdamped), b > ω 0 (overdamped), and b = ω 0 (critically damped)
Damped simple-harmonic oscillator: Underdamped regime In the underdamped regime, b < ω 0 and the system shows damped oscillatory behavior y(t) = e bt ( c 1 e iβt + c 2 e iβt) = e bt (A sin βt + B cos βt) Here β = ω 2 0 b2 is the effective oscillation frequency
Damped simple-harmonic oscillator: Overdamped regime In the overdamped regime b > ω 0 and the system shows no oscillatory behavior The roots of the characteristic equation are real and equal to D 1,2 = b ± b 2 ω0 2 The equation of motion is given by h h y = Ae D1t + Be D2t b+ b = Ae 2 ω0 it i 2 b b + Be 2 ω0 2 t If we give the system a kick at t = 0, the displacement will exponentially decay to zero in a time faster than one period of oscillation
Damped simple-harmonic oscillator: Critical regime In the critical regime b = ω 0, the system is on the edge between the over and under-damped cases We get equal roots for the characteristic equation D 1,2 = b From what we saw in the case of equal roots, we find y = (At + B)e bt