Class 11
We will address the following topics Convolution of functions
Consider the following question: Suppose that u(t) has Laplace transform U(s), v(t) has Laplace transform V(s), what is the inverse Laplace transform of U(s)V(s)? It turns out that the answer is given by (u v)(t) = t u(τ)v(t τ) dτ. We call this function the convolution of u and v.
Example We calculate the convolution 1 t n, where n. 1 t n = t 1 (t τ) n dτ We now make a change of variables x = t τ, hence Hence dx = dτ, τ = x = t, τ = t x =. 1 t n = t x n dx = t x n dx = x n+1 n+1 t = tn+1 n+1.
Theorem If u(t) and v(t) are continuous functions on the interval [, ), then we have Here is the proof (u v)(t) = (v u)(t), t [, ). (u v)(t) = t u(τ)v(t τ) dτ. We change variables by letting = t τ. Hence we have Hence (u v)(t) = d = dτ, τ = = t, τ = t =. t u(t )v( ) d = t v( )u(t ) d = (v u)(t).
Example Use the convolution property to find the general solution of using Laplace transforms. u = au + q(t) Applying the Laplace transform to the above equation, we obtain su(s) u() = au(s)+q(s), where Q(s) is the Laplace transform of q(t). Hence (s a)u(s) = u()+q(s). and U(s) = u() s a + 1 s a Q(s)
Example (continued) Hence, applying properties of convolution, u(t) = u()e at + q(t) e at = u()e at + t q(τ)e a(t τ) dτ.
Example Sketch the function f(t) = 2h 3 (t) 2h 4 (t) and find its Laplace transform. 2 1.5 1.5 1 2 3 4 5 6 t L[f(t)] = 2 e 3s s 2 e 4s s = 2 s (e 3s e 4s)
Example Compute the Laplace transform of f(t) = t 2 h 3 (t). L[f(t)] = = = = + = 3 τ= τ= τ= t 2 h 3 (t)e st dt t 2 e st dt (τ + 3) 2 e s(τ+3) dτ τ 2 e s(τ+3) dτ + 9e s(τ+3) dτ ( 2 s 3 + 6 s 2 + 9 ) e 3s. s τ= 6τe s(τ+3) dτ
More generally, we have Theorem L[h a (t)u(t a)] = U(s)e as Proof: L[h a (t)u(t a)] = = = = t= t=a τ= τ= = U(s) e as. h a (t)u(t a)e st dt u(t a)e st dt u(τ)e s(τ+a) dτ u(τ)e sτ dτ e as
Example Consider the initial value problem u + 4u = [h (t) h 2π (t)] cos(2t), u() =, u () =. The forcing term has the graph shown below 1.5 6 4 2 2 4 6 8 1 12 t.5 1
Example (continued) Applying the Laplace transform to the differential equation we obtain s 2 U(s)+4U(s) = s s 2 + 4 s s 2 + 4 e 2πs Solving for U(s) we obtain U(s) = = s s 2 + 4 [ 1 e 2πs]. s [ (s 2 + 4) 2 1 e 2πs].
Example We have [ ] [ ] L 1 s 1 (s 2 + 4) 2 = L 1 2 s 2(s 2 + 4) (s 2 + 4) = 1 [ [ ] 2 ] L 2 L 1 1 s (s 2 + 4) (s 2 + 4) = 1 sin 2t cos 2t 2 = 1 2 t sin 2τ cos 2(t τ) dτ Now we make the substitution = 2τ, hence d = 2dτ, and obtain
Example (continued) [ ] L 1 s (s 2 + 4) 2 = 1 2t sin cos(2t ) d 4 = 1 2t sin 2t 4 2 = 1 t sin 2t. 4 By the previous theorem, we also have [ ] L 1 s (s 2 + 4) 2 e 2πs = 1 4 (t 2π)h 2π(t) sin 2t. Hence u(t) = 1 4 [ ] t (t 2π)h 2π (t) sin 2t.
1.5 1.5.5 5 1 15 2 25 t 1 1.5
Many mechanical, electrical and biological systems have forcing terms which (essentially) act at a single instant of time: A shock absorber in a car is modeled by a damped spring mass system; it can be given an instantaneous" impulse by hitting a bump on the road. A serge of voltage in an electric circuit. Concentration of a substance in the blood stream after injection of medicine.
We can model the action of a bump on the road by the following family of functions. Fix an a > and let ǫ > (it represents a small" number). 1 for a ǫ b a,ǫ (t) = ǫ 2 t < a+ ǫ 2, otherwise. We can also express these functions via the formula b a,ǫ (t) = 1 ǫ ( h a ǫ 2 (t) h a+ ǫ 2 (t) ), where h stands for the Heaviside function.
Each of these functions has the property that (for ǫ/2 a) b a,ǫ (t) dt = 1. So the function represents a unit impulse" which occurs over length of time equal to ǫ. A couple of these functions are sketched below.
.5.4.3.2.1 1 2 3 4 5 t
1.8.6.4.2 1 2 3 4 5 t
2 1.5 1.5 1 2 3 4 5 t
4 3 2 1 1 2 3 4 5 t
8 6 4 2 1 2 3 4 5 t
We define" the delta function as the limit δ a (t) = lim ǫ b a,ǫ (t). Of course, this is not a function in the usual mathematical sense. (It would have value at t = a and everywhere else). None the less this informally defined object can be given a precise mathematical description - it is a generalized function. Surprisingly, one can make effective calculations with the δ function which give useful predictions.
has the sifting property For example, we have δ a (t)φ(t) dt = φ(a), where a. Clearly, therefore L[δ a (t)] = δ 4 (t)e 2(t 3)2 dt = e 2. δ a (t)e st dt = e as.
Example Consider the initial value problem { u = δ 4 (t), u() =. Applying the Laplace transform to the equation, we obtain Hence, su(s) = e 4s. U(s) = e 4s s. It follows that the answer is given by the Heaviside function u(t) = h 4 (t).
1.8.6.4.2 1 2 3 4 5 6 7 8 t
Example Consider the initial value problem u = δ 4 (t), u() =, u () =. Applying the Laplace transform to the equation, we obtain s 2 U(s) = e 4s. Hence, U(s) = e 4s s 2. The answer again involves the Heaviside function u(t) = h 4 (t)(t 4).
4 3 2 1 1 2 3 4 5 6 7 8 t
Example Consider the initial value problem u + u = δ 2 (t), u() =, u () =. Applying the Laplace transform to the equation, we obtain s 2 U(s)+sU(s) = e 2s. (s 2 + s)u(s) = e 2s. And finally, U(s) = e 2s s 2 + s.
Example Note that Therefore u(t) = h 2 (t) 1 s 2 + s = 1 s(s + 1) = 1 s 1 s+1. ( 1 e (t 2)) ( = h 2 (t) 1 e 2 t).
1.8.6.4.2 2 4 6 8 1 12 14 t
Example Consider the initial value problem u + u = δ 2 (t), u() =, u () =. Applying the Laplace transform to the equation, we obtain s 2 U(s)+U(s) = e 2s. (s 2 + 1)U(s) = e 2s. And finally, U(s) = e 2s s 2 + 1.
Example Therefore u(t) = h 2 (t) sin(t 2).
1.5 2 4 6 8 1 12 14 t.5 1