University:
Georgia Southern University
Course:
MATH 3230 | Ordinary Differential Equations
Academic year:

2024
Views:

111

Pages:

15
Author:

Jalen W.

Definition Let f (t) and g (t) be integrable on closed intervals in [0, ∞). The convolution of f (t) and g (t), denoted f (t) ∗ g (t) or (f ∗ g )(t), is defined by Z t (f ∗ g )(t) = f (v )g (t − v ) dv . 0 Interpretation of Convolution You may be wondering why we would even want to do such an operation. Let’s try to approach it with a rolling ball analogy using discrete time steps. I Each step in time, the ball is rolled an integer number of units to the right of its previous position with probability P(k) to roll k units. I Note that P(k|t = t0 ) = P(k), i.e. each ball roll is independent of the previous roll. I After two steps in time, the probability that the position of the ball will be n units to the right of the starting position is P(d = n | t = 2) = P(0) · P(n) + P(1) · P(n − 1) +P(2) · P(n − 2) + · · · + P(n)P(0) n X P(d = n | t = 2) = P(k)P(n − k). k=0 Interpretation of Convolution You may be wondering why we would even want to do such an operation. Let’s try to approach it with a rolling ball analogy using discrete time steps. I Now, if instead of integer steps to the right we allow continuous steps, then... n X P(d = n | t = 2) = P(k)P(n − k) k=0 BECOMES... Z P(d = n | t = 2) = n P(k)P(n − k) dk 0 Properties of Convolution Let f (t), g (t), and h(t) be integrable on closed intervals in [0, ∞). I Associative: (f ∗ (g ∗ h))(t) = ((f ∗ g ) ∗ h)(t) I Commutative: (f ∗ g )(t) = (g ∗ f )(t) Example Evaluate (f ∗ g )(t) given the following. f (t) = t 3 g (t) = t 2 Example Evaluate (f ∗ g )(t) given the following. f (t) = e a1 t g (t) = e a2 t Example Evaluate (f ∗ g )(t) given the following. f (t) = t g (t) = e at Brief Table of Convolutions f (t) g (t) (f ∗ g )(t) 1 g (t) Rt t t 1 3 6t t t2 1 4 12 t t2 t2 1 5 30 t 0 g (v ) dv Brief Table of Convolutions f (t) g (t) t e αt t2 e αt e α1 t e α2 t (f ∗ g )(t) 1 αt e α2 2 αt e α3 − α1 t − − α1 t 2 − 1 α2 2 t α2 e α1 t − e α2 t α1 − α2 − 2 α3 Brief Table of Convolutions (f ∗ g )(t) f (t) g (t) t sin(βt) 1 βt − 1 β2 sin(βt) t cos(βt) 1 β2 − 1 β2 cos(βt) e αt sin(βt) βe αt − α sin(βt) − β cos(βt) α2 + β 2 e αt cos(βt) αe αt + β sin(βt) − α cos(βt) α2 + β 2 Brief Table of Convolutions f (t) g (t) (f ∗ g )(t) sin(β1 t) sin(β2 t) β1 sin(β2 t)−β2 sin(β1 t) β12 −β22 cos(β1 t) cos(β2 t) β1 sin(β1 t)−β2 sin(β2 t) β12 −β22 sin(β1 t) cos(β2 t) β1 (cos(β2 t)−cos(β1 t)) β12 −β22 δ(t − c) g (t) u(t − c)g (t − c) Convolution Theorem Let f (t) and g (t) be functions with a Laplace Transforms, F (s) and G (s), for s > s0 . Then, L{(f ∗ g )(t)}(s) = F (s)G (s), or equivalently, L−1 {F (s)G (s)}(t) = (f ∗ g )(t). Example Solve the IVP using Laplace Transforms. y 00 + 4y 0 + 4y = 16e 2t , y (0) = 1, y 0 (0) = 1 Example Solve the IVP using Laplace Transforms. y 00 + 4y = 2u(t − 2), y (0) = 1, y 0 (0) = 1 Example Solve the IVP using Laplace Transforms. y 00 + 4y 0 = 16u(t − 2), y (0) = 1, y 0 (0) = −4

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