Generalized sources (Sect. 6.5). The Dirac delta generalized function. Definition Consider the sequence of functions for n, d n, t < δ n (t) = n, t 3 d3 d n, t > n. d t The Dirac delta generalized function is given by Remarks: lim δ n(t) = δ(t), t R. (a) There exist infinitely many sequences δ n that define the same generalized function δ. (b) For example, compare with the sequence δ n in the textbook.
The Dirac delta generalized function. d n delta R { } 3 d3 d d t t Remarks: (a) The Dirac δ is a function on the domain R {}, and δ(t) = for t R {}. (b) δ at t = is not defined, since δ() = lim n = +. (c) δ is not a function on R. Generalized sources (Sect. 6.5).
Properties of Dirac s delta. Remark: The Dirac δ is not a function. We define operations on Dirac s δ as limits n of the operation on the sequence elements δ n. Definition δ(t c) = lim δ n(t c), a δ(t) + b δ(t) = lim a δn (t) + b δ n (t) ], f (t) δ(t) = lim f (t) δn (t) ], b b δ(t) dt = lim δ n (t) dt, a a Lδ] = lim Lδ n]. Properties of Dirac s delta. Theorem Proof: δ(t) dt =, a >. /n δ(t) dt = lim δ n (t) dt = lim n dt ( δ(t) dt = lim n t /n )] = lim n ]. n We conclude: δ(t) dt =.
Properties of Dirac s delta. Theorem If f : R R is continuous, t R and a >, then t +a t δ(t t ) f (t) dt = f (t ). Proof: Introduce the change of variable τ = t t, I = t +a t δ(t t ) f (t) dt = δ(τ) f (τ + t ) dτ, /n I = lim δ n (τ) f (τ + t ) dτ = lim n f (τ + t ) dτ /n Therefore, I = lim n F (τ + t ) dτ, where we introduced the primitive F (t) = f (t) dt, that is, f (t) = F (t). Properties of Dirac s delta. Theorem If f : R R is continuous, t R and a >, then t +a t δ(t t ) f (t) dt = f (t ). /n Proof: So, I = lim n F (τ + t ) dτ, with f (t) = F (t). ] I = lim n We conclude: I = lim F (τ + t ) t +a t /n ( ) ] F t + n F (t ) n ( = lim n F t + ) ] F (t ). n δ(t t ) f (t) dt = f (t ). = F (t ) = f (t ).
Generalized sources (Sect. 6.5). Relation between deltas and steps. Theorem The sequence of functions for n, u n (t) =, t < nt, t n, t > n. u 3 u u /3 / t satisfies, for t (, ) (, /n) (/n, ), both equations, Remark: u n(t) = δ n (t), lim u n(t) = u(t), t R. If we generalize the notion of derivative as u (t) = lim δ n(t), then holds u (t) = δ(t). Dirac s delta is a generalized derivative of the step function.
Generalized sources (Sect. 6.5). Dirac s delta in Physics. Remarks: (a) Dirac s delta generalized function is useful to describe impulsive forces in mechanical systems. (b) An impulsive force transmits a finite momentum in an infinitely short time. (c) For example: The momentum transmitted to a pendulum when hit by a hammer. Newton s law of motion says, m v (t) = F (t), with F (t) = F δ(t t ). The momentum transfer is: I = lim t mv(t) t + t t t = lim t That is, I = F. t + t t t F (t) dt = F.
Generalized sources (Sect. 6.5). The Laplace Transform of Dirac s delta. Recall: The Laplace Transform can be generalized from functions to δ, as follows, Lδ(t c)] = lim Lδ n(t c)]. Theorem Proof: Lδ(t c)] = lim Lδ n(t c)], Lδ(t c)] = e cs. δ n (t) = n ( Lδ(t c)] = lim n Lu(t c)] L ( e cs Lδ(t c)] = lim n s e (c+ n )s s This is a singular limit,. Use l Hôpital rule. ) u(t) u ( t n ) ]. u ( t c ) ]) n = e cs ( e s n ) lim ( s. n)
The Laplace Transform of Dirac s delta. Proof: Recall: Lδ(t c)] = e cs ( e s n ) ( s e s lim ( n s = lim n) n ) ( ) s n ( e s n ) lim ( s. n) We therefore conclude that Lδ(t c)] = e cs. Remarks: (a) This result is consistent with a previous result: t +a t (b) Lδ(t c)] = δ(t t ) f (t) dt = f (t ). (c) Lδ(t c) f (t)] = δ(t c) e st dt = e cs. = lim e s n =. δ(t c) e st f (t) dt = e cs f (c). Generalized sources (Sect. 6.5).
Differential equations with Dirac s delta sources. Example Find the solution y to the initial value problem y y = δ(t 3), y() =, y () =. Solution: Compute: Ly ] Ly] = Lδ(t 3)]. Ly ] = s Ly] s y() y () (s ) Ly] s = e 3s, We arrive to the equation Ly] = s (s ) e 3s (s ), Ly] = Lcosh(t)] Lu(t 3) sinh(t 3)], We conclude: y(t) = cosh(t) u(t 3) sinh(t 3). Differential equations with Dirac s delta sources. Example Find the solution to the initial value problem y + 4y = δ(t π) δ(t π), y() =, y () =. Solution: Compute: Ly ] + 4 Ly] = Lδ(t π)] Lδ(t π)], (s + 4) Ly] = e πs e πs Ly] = e πs (s + 4) e πs (s + 4), that is, Ly] = e πs (s + 4) e πs (s + 4). Recall: e cs Lf (t)] = Lu(t c) f (t c)]. Therefore, Ly] = L u(t π) sin (t π) ]] L u(t π) sin (t π) ]].
Differential equations with Dirac s delta sources. Example Find the solution to the initial value problem y + 4y = δ(t π) δ(t π), y() =, y () =. Solution: Recall: Ly] = L u(t π) sin (t π) ]] L u(t π) sin (t π) ]]. This implies that, y(t) = u(t π) sin (t π) ] u(t π) sin (t π) ], We conclude: y(t) = u(t π) u(t π) ] sin(t).