Convolution integral and the Convolution Theorem.

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> 0is 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 → 0we 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