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Convolution: Mathematical Operation and Application in Differential Equations, Study notes of Differential Equations

Institute of Mathematics and ApplicationsDifferential Equations

The concept of convolution, a mathematical operation used to find the response at the present time as a weighted superposition of the input at previous times. The document also covers the properties of convolution and its relationship to the laplace transform. The lecture includes examples and the solution of an initial value problem using convolution.

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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>t
1, 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 <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.
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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 = 0 with no pollutant, how much is in the lake at time t > 0? The small drip of pollutant added to the lake between t 1 and t 1 +Δt, where Δt small, is f (t 1 )Δt. Later t > t 1 , the drip reduces to e−a(t−t^1 )f (t 1 )Δt, where a > 0 is the decay constant. Adding them up, starting at the initial time t 1 = 0, we obtain that the amount is t (22.1) e−a(t−t^1 )f (t 1 ) dt 1. 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 (t 1 )g(t − t 1 ) dt 1. 0 It gives the response at the present time t as a weighted superposition over the input at times t 1 < t. The weight g(t − t 1 ) characterizes the system and f (t 1 ) 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) = eB^1 t^ and g(t) = eB^2 t, where B 1 = B 2 are constants. Then, t (^) eB 1 t (^) − eB 2 t (f ∗ g)(t) = eB^1 t^ eB^2 (t−t^1 )^ dt 1 =. 0 B 1 −^ B 2 Since Leat^ = 1/(s − a) for any constant a, we have 1

L(f ∗ g) = − = = (Lf )(Lg). B 1 = B 2 s − B 1 s − B 2 s − B 1 s − B 2 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 = f 2 in general. Exercise. Show that t^2 ∗ 1 = t^3 / 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)| � A 1 eB^1 t^ and |g(t)| � A 2 eB^2 t, then t (^) eB 1 t (^) − eB 2 t |(f ∗ g)(t)| � A 1 eB^1 tA 2 eB^2 (t−t^1 )^ dt 1 = A 1 A 2. 0 B^1 −^ B^2 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 (t 1 )g(t − t 1 ) dt 1. −∞ ∞ −∞ Indeed, the lower limit of the integral of f ∗ g can be replaced by −∞ since f (t 1 ) = 0 for t 1 < 0 and the upper limit can be replaced by ∞ since g(t − t 1 ) = 0 for t − t 1 < 0. Then, � (^) ∞ � (^) ∞ L(f ∗ g)(s) = e−st^ f (t 1 )g(t − t 1 ) dt 1 dt −∞ −∞ � (^) ∞ ∞ = e−st^ g(t − t 1 ) dt f (t 1 ) dt 1 −∞ −∞ � (^) ∞ � (^) ∞ = e−s(t^1 +t^2 )g(t 2 ) dt 2 f (t 1 ) dt 1 �^ −∞ −∞ ∞ = e−st^1 Lg(s)f (t 1 ) dt 1 = (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 ′′^ + ω^2 y = ω^2 f (t), y(0) = y ′(0) = 0, where ω^2 is constant and f ∈ E. S OLUTION. Taking the transform, ω^2 (s^2 + ω^2 )Ly = ω^2 Lf, Ly = Lf. s^2 + ω^2 ω Since L(sin ωt) = , it follows that Ly = (Lf )(Lω sin ωt). By the convolution theorem s^2 + ω^2 and by uniqueness, then t y(t) = f ∗ (ω sin ωt) = ω f (t 1 ) sin ω(t − t 1 ) dt 1. 0 Note that we have a formula for the rest solution corresponding to the arbitrary function f. Lecture 22 2 18.034 Spring 2009