MIT OpenCourseWarehttp://ocw.mit.edu 18.034 Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � � LECTURE 22. CONVOLUTION Motivation: buildup of a pollutant in a lake. Let’s say we have a lake and a pollutant is being dumped into it at the variable rate f(t). The pollutant degrades over time exponentially. If the lake begins at t =0with no pollutant, how much is in the lake at time t> 0? The small drip of pollutant added to the lake between t1 and t1 +Δt, where Δt small, is f(t1)Δt. Later t>t1, the drip reduces to e−a(t−t1)f(t1)Δt, where a> 0 is the decay constant. Adding them up, starting at the initial time t1 =0, we obtain that the amount is t (22.1) e −a(t−t1)f(t1) dt1. 0 Integral of this kind is called a convolution. We can solve this problem by setting up a differential equation. Let y(t) be the amount of pollutant in the lake at time t. Then, the amount of the chemical in the lake at time t +Δt is the amount at time t, minus the fraction that decayed plus the amount newly added: y(t +Δt)=y(t) − ay(t)Δt + f(t)Δt. Taking the limit as Δt → 0 we obtain y + ay = f(t), y(0)=0. It is straightforward that (22.1) gives the solution of the above initial value problem. The convolution integral. The convolution of f and g is defined as t (22.2) (f ∗ g)(t)= f(t1)g(t − t1) dt1. 0 It gives the response at the present time t as a weighted superposition over the input at times t1 <t. The weight g(t − t1) characterizes the system and f(t1) characterizes the history of the input. To ensure the existence of the integral, in what follows, we assume that f,g ∈ A. Example 22.1. Let f (t)=eB1t and g(t)=eB2t, where B1 = B2 are constants. Then, t eB1t − eB2t (f ∗ g)(t)= e B1t e B2(t−t1) dt1 = . 0 B1 − B2 Since Leat =1/(s − a) for any constant a, we have 1 � 1 1 � 1 1 L(f ∗ g)= − = =(Lf)(Lg). B1 = B2 s − B1 s − B2 s − B1 s − B2 It is not a coincidence, but rather, it is a property of convolution, as discussed below. The convolution operator acts like ordinary multiplication in that, if f,g,h are admissible then (i) (distibutive) f ∗ (g + h)=f ∗ g + f ∗ h, (ii) (commutative) f ∗ g = g ∗ f, (iii) (associative) f ∗ (g ∗ h)=(f ∗ g) ∗ h. 1� � � � � � � � However, the convolution operator differs from the multiplication operator. For example, f ∗1 = f, f ∗ f = f2 in general. Exercise. Show that t2 ∗ 1= t3/3 and cos t ∗cos t = 12 (t cos t + sin t). Nevertheless, convolution in the t-domain does corresponds to multiplication in the s-domain. Theorem 22.2 (Convolution Theorem). If f,g ∈ E, then f ∗ g ∈ E and L(f ∗ g)=(Lf)(Lg). Proof. If f,g ∈ E, then f ∗ g is continuous for all t ∈ [0, ∞). Since |f(t)| � A1eB1t and |g(t)| � A2eB2t, then t eB1t − eB2t |(f ∗ g)(t)| � A1e B1tA2e B2(t−t1) dt1 = A1A2 . 0 B1 − B2 Therefore, f ∗ g ∈ E. Better yet, |f|∗|g|∈E. In other words, L(f ∗ g) converges absolutely for large s. For simplicity, let f(t)=0 and g(t)=0 for t< 0, so that � ∞ � ∞ � ∞ Lf = e −stf(t) dt, Lg = e −st g(t) dt, and (f ∗ g)(t)= f(t1)g(t −t1) dt1. −∞ ∞ −∞ Indeed, the lower limit of the integral of f ∗ g can be replaced by −∞ since f (t1)=0 for t1 < 0 and the upper limit can be replaced by ∞ since g(t −t1)=0 for t − t1 < 0. Then, � ∞ � ∞ L(f ∗ g)(s)= e −st f(t1)g(t −t1) dt1 dt −∞ −∞ � ∞ ∞ = e −st g(t −t1) dt f(t1) dt1 −∞ −∞ � ∞ � ∞ = e −s(t1+t2)g(t2) dt2 f(t1) dt1 −∞ −∞ � ∞ = e −st1 Lg(s)f(t1) dt1 =(Lf)(Lg). −∞ Convolution allows an easy passage from the s-domain to the t-domain and leads to explicit solutions for a general inhomogeneous term f(t). Example 22.3. Solve the initial value problem y + ω2 y = ω2f(t), y(0) = y (0)=0, where ω2 is constant and f ∈ E. SOLUTION. Taking the transform, ω2 (s 2 + ω2)Ly = ω2Lf, Ly = Lf . s2 + ω2 ωSince L(sin ωt)= , it follows that Ly =(Lf)(Lω sin ωt). By the convolution theorem s2 + ω2 and by uniqueness, then t y(t)= f ∗ (ω sin ωt)= ωf(t1) sin ω(t − t1) dt1. 0 Note that we have a formula for the rest solution corresponding to the arbitrary function f . Lecture 22 2 18.034 Spring 2009� � � � � � The tautochrone. Starting from the rest state, suppose that a particle of mass m slides down a frictionless curve under gravity as shown in the figure below. The aim is to determine the shape of the curve so that the y time of descent is independent of the starting point. A curve of this kind is called a tautochrone. It comes from two Greek words, meaning ”same” and ”time.” The problem was solved by the Dutch mathematician Christian Huygens in 1673 as part v of his theory of pendulum clocks. Let y be the height at which the particle starts and let v be the speed of the particle at the height z. The change in the kinetic energy (1/2)mv2 at the height z is the change in the potential Figure 22.1. The tautochrone. energy, which leads to z 1 2 √ (22.3) mv = mg(y − z), v = 2g y − z, 2where g is the gravitational acceleration. Let σ = σ(y) be the arc from the rest state to the lowest point. The time of descent is σ(y) dσ y 1 dσ y 1 = dz ≡ φ(z) dz. 0 v 0 v dz 0 v Here, φ(y)= dσ/dy so that φ(z)= dσ/dy|y=z. Since the time of decent is constant, in view of (22.3), the problem reduces to y φ(z)(y − z)−1/2 dz = c1, 0 where c1 is constant. The left side is the convolution of φ and y−1/2. Taking the transform, then L(φ ∗ y −1/2)=(Lφ)(Ly −1/2)= Lc1 = c1 . s The first equality uses the convolution theorem. √Since∗ L[t−1/2]= πs−1/2, we have Lφ = c2s −1/2 , φ(y)= cy −1/2 , where c, c2 are constants. Since φ(y)= dσ/dy, the equation of the curve reduces to ��2dx c2 c2 1+ = or dx = − 1 dy. dy y y Exercise. By parameterizing y = c2 sin2 θ, show that the curve is a cycloid 2 2ccx = (2θ + sin 2θ), y = (1 − cos 2θ).2 2 When a particle slides down a frictionless curve, a related question is to find a path which minimizes the time of descent. …
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