FOURIER SERIES, GENERALIZED FUNCTIONS, LAPLACE TRANSFORM. 1. A (generalized) function f(t) of period 2L has a Fourier series of the form

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1 FOURIER SERIES, GENERAIZED FUNCTIONS, APACE TRANSFORM 1 Fourier Series 1.1 General facts 1. A (generalized) function f(t) of period has a Fourier series of the form or f(t) a + a 1 cos πt + a cos πt + + b 1 sin πt + b sin πt + f(t) a + a n cos nπt + b n sin nπt. The coefficients a n and b n can be found by the following formulas. a n = 1 f(t) cos nπt dt b n = 1 f(t) sin nπt dt 3. The Fourier series it convergent at every point t for which both the right limit f(t+) and left limit f(t ) exist and are finite. The sum of the Fourier series at t is the average of these two limits. 4. The terms in the Fourier series of a function f(t) must have the same symmetries as f(t) itself. For instance, an odd function will only have a sine functions in its Fourier series (no constant); an even function will only have a cosine functions in its Fourier series and the constant term; the Fourier series of a function that is odd about / will only have cos (n+1)πt the Fourier series of a function that is even about / will only have cos nπt It might be useful to recall that even even = even even odd = odd odd odd = even and sin nπt ; (n+1)πt and sin. 5. We differentiate Fourier series term by term. So given f(t) a + a n cos nπt + b n sin nπt, we get f nπ nπt (t) b n cos + nπ nπt a n sin 6. Integration is also done term by term, but the result is a Fourier series only of the series we integrated does not have a non-zero constant term. 1

2 1. How do we compute Fourier series? Directly from the definition using the formulas for a n and b n. Reducing to a known Fourier series (or to a Fourier series given on the exam) by differentiating, or integrating, or translating, or taking a dilation, etc... and maybe employing some trig identities. Trig identities are especially useful for computing Fourier series of, say, sin t. Try to remember that sin A = 1 cos A cos A = 1 + cos A sin(a ± B) = sin A cos B ± cos A sin B; cos(a ± B) = cos A cos B sin A sin B; sin A sin B = and the special values 1.3 Use in ODE cos(a B) cos(a + B) ; cos A cos B = sin sin A cos B = (n + 1)π sin(a + B) sin(a B) ; = ( 1) n ; cos nπ = ( 1) n ; cos(a B) + cos(a + B) ; sin π 4 = cos π 4 = 1 ; sin π 6 = cos π 3 = 1 ; sin π 3 = cos π 6 = 3. Consider the linear ODE with constant coefficients c n y (n) c 1 y + c y = f(t), c n, f(t) periodic with period. We can make use of Fourier series to find a particular solution y p of this ODE, as follows. (But the general solution is still y = y p + y h.) 1. First write down the Fourier series of f(t).. Replace f(t) by its Fourier series in the RHS of the ODE. It becomes something of the form c n y (n) c 1 y + c y = a + a n cos nπt + b n sin nπt. 3. For each term that appears in the Fourier series find a particular solution of the corresponding equation (use ERF or ERF or ERF etc... ). c n y (n) c 1 y + c y = a c n y (n) c 1 y + c y = a n cos nπt c n y (n) c 1 y + c y = b n sin nπt get y get y 1,n get y,n

3 4. A particular solution to the original ODE is the sum of all these pieces y p = y + y 1,n + y,n. Step and Delta.1 Definitions The step function u(t) = { 1 t > t < u a (t) = u(t a) = { 1 t > a t < a δ(t) = u (t) is for all t and has a spike of size 1 at t =. Properties 1. u = δ. u = u 3. δ = δ 4. f(t)u a (t) = δ a (t) = δ(t a) = u a(t) is for all t a and has a spike of size 1 at t = a { t < a f(t) t > a t < a 5. f(t)(u a (t) u b (t)) = f(t) a < t < b t > b 6. f(t)δ a (t) = f(a)δ a (t) is for all t a and has a spike of size f(a) at t = a 7. c b a < b f(t)δ a (t)dt = f(a) b < a < c c < a 3 Unit response and weight function p(d) = a n D n + + a 1 D + a I The weight function or unit impulse response of the operator p(d) is the unique function w(t) with the property that p(d)w = δ(t). 1. Solve the homogeneous equation p(d)x = for t > with the initial conditions (given by matching singularities) x (n 1) = 1/a n and x(+) = x (+) = x (n ) (+) =.. Take the solution you found and multiply by the step function u(t). This is your w(t). 3

4 The unit step response of the operator p(d) is the unique function v(t) with the property that Properties p(d)v = u(t). 1. Solve the inhomogeneous equation p(d)x = 1 for t > with the initial conditions (given by matching singularities) x(+) = x (+) = x (n 1) (+) =.. Take the solution you found and multiply by the step function u(t). This is your v(t). 1. v (t) = w(t). The solution to p(d)x = δ(t a) = δ a (t) is w(t a). 3. The solution to p(d)x = u(t a) = u a (t) is v(t a). 4 Convolution and Green s Formula The convolution of two functions f and g is a function (f g) given by Properties (f g)(x) = x f(u)g(x u) du. 1. f g = g fj. f (ag 1 + bg ) = af g 1 + bf g 3. f (g h) = (f g) h Green s formula gives a way of finding the solution y p to the IVP a n y (n) a 1 y + a y = f(t), y() = y () =... = y (n 1) () = in terms of the convolution of the forcing function f(t) with weight function w(t) associated to the operator p(d) = a n D n + + a 1 D + a I. Namely it tell us that the solution to the IVP is y p (t) = (f w)(t) = t f(t τ)w(τ)dτ = t f(τ)w(t τ)dτ. Note: This also gives a nice way of finding a particular solution y p to the equation a n y (n) a 1 y + a y = f(t). As always, the general solution is y = y p + y + h. If you have an IVP with any other initial conditions than rest, you can 1. use Green s formula to find y p ;. solve the associated homogeneous equation to find y h ; 3. write down the general solution y = y p + y + h; 4. use the initial condition to solve for the undetermined constants. 4

5 5 aplace Transform If the function f(t) is continuous at t =, there is no ambiguity about the way we define its aplace transform: (f)(s) = e st f(t)dt = F (s). When we have to deal with discontinuities or generalized functions, it makes sense to consider + (f)(s) = + e st f(t)dt that ignores the (possible) discontinuity at of f(t); (f)(s) = 5.1 General facts 1. af(t) + bg(t). f(t) g(t) 3. e at f(t) 4. tf(t) 5. t n f(t) e st f(t)dt that detects if f(t) has a discontinuity at. af (s) + bg(s) F (s)g(s) F (s a) F (s) ( 1) n F (n) (s) 6. u(t a)f(t a) 7. f (t) 8. f (t) 9. f (n) (t) e as F (s) for a sf (s) f() (use + for + ) s F (s) sf() f () (use + for + ) s n F (s) s n 1 f()... f (n 1) () (use + for + ) 1. (f (n) ) = s n F (s) 11. Basic transforms (u) = (1) = 1 s (t n ) = n! s n+1 (e at ) = 1 s a (cos at) = s a s + a (sin at) = s + a (δ(t)) = 1 (δ a (t)) = e as (a > ) 5

6 5. Use in ODE Finding solutions of ODEs: take he aplace transforms of both sides of an ODE, one obtains an (algeraic) equation for the aplace transform. Solve that and take the inverse aplace transform. That gives you a solution of the original ODE. Finding the weight function of an operator p(d): (w(t)) = 1 p(s) Pole diagram: the rightmost pole(s) of the aplace transform F (s) of a function f(t) tells us how f(t) behaves for very large t (as t ). If the right most pole is at s = a, then, for large t, f(t) e at If the right most poles are at s = a ± ib, then, for large t, f(t) e at cos bt In particular, f(t) as t if and only if the poles of F (s) have real part negative. More precisely, if F (s) has simple poles at a, and α ± iβ, i.e. if F (s) = then F (s) will be written as a sum of simple fractions Therefore, F (s) = A s a + B s α iβ + (s a)((s α) + β ), C s α + iβ = A s a + B s + C (s α) + β. or, equivalently, f(t) = Ae at + Be (α+iβ)t + Ce (α iβ)t f(t) = A e at + B e αt cos βt + C e αt sin βt. The dominant term as t gets very large will be the one with the largest coefficient on the exponential, which is to say the one corresponding to the rightmost pole. 6 Transfer function of p(d) W (s) = 1 p(s) = (w(t)) It has the property that a particular solution to p(d)x = e rt is given by x p = W (r)e rt (if p(r).) Note that this is an exponential solution. 6