LECTURE 12: LAPLACE TRANSFORM

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1 LECTURE 12: LAPLACE TRANSFORM 1. Definition and Quetion The definition of the Laplace tranform could hardly be impler: For an appropriate function f(t), the Laplace tranform of f(t) i a function F () which i equal to F () = L{f(t)} = f(t)e t dt. Naturally, jut by looking at thi expreion, one might ak: What are the domain and range of thi tranformation? I the tranformation one-to-one? That i, doe L{f(t)} = L{g(t)} imply f(t)=g(t)? I the tranformation linear? What doe mean? How can thi tranformation be applied? So, what are the domain and range of the Laplace tranformation? To be more explicit, we are really aking, for which function f(t), and for which value of i the integral f(t)e t dt well-defined? In fact, thi improper integral i defined a the limit lim f(t)e t dt. Thu we are aking, for which function f(t) doe thi limit exit? Before a characterization of uch function, let ee ome example of the Laplace tranform. 2. Example 2.1. L{1}. By the definition of the Laplace tranform, L{1} = lim 1 e t dt ( = lim 1 N ) e t t= ( = lim 1 e N + 1 ). Thi limit converge to 1 for >. Hence, L{1} = 1, >. Date: 6/1/15. 1

2 2 LECTURE 12: LAPLACE TRANSFORM 2.2. L{t}. By the definition, L{t} = lim = lim 1 te t dt [te t N t= = 1 lim (Ne N ) + 1 L{1} ] e t dt For >, the limit exit and we have L{t} = 1 2, > L{t n }. Again, L{t n } = lim = lim 1 t n e t dt [t n e t N t= = 1 lim (N n e N ) + n L{tn 1 }. ] nt n 1 e t dt For >, by L Hôpital rule, we have lim (N n e N ) =. Therefore, By induction, 2.4. L{co t}. By definition, L{t n } = n L{tn 1 }, >. L{t n } = n! n! L{1} =, >. n n+1 L{co t} = lim co t e t dt Now we need to evaluate the integral co t e t dt. By a trick introduced before, we could calculate the derivative of co t e t, in t e t with repect to t: (1) (2) So, (2) (1) give Therefore, (co t e t ) = in t e t co t e t, (in t e t ) = co t e t in t e t. (in t e t co t e t ) = (1 + 2 ) co t e t. co t e t dt = (in t co t)e t N t= = e N (in N co N)

3 For >, taking N give LECTURE 12: LAPLACE TRANSFORM 3 L{co t} = 1 + 2, >. Remark. We will ee a impler way to find the Laplace tranform of co at and in at once we have the firt hift theorem L{u c (t)}. Here {, t [, c) u c (t) = 1, t [c, ) i called the Heaviide tep function, where c i a non-negative contant. Again, we are facing the integral u c (t)e t dt. For N begin ufficiently large, thi integral conit of two part: integration on [, c) where u c (t) = and integration on [c, N] where u c (t) = 1. Therefore, L{u c (t)} = lim u c (t)e t dt = lim c e t dt = 1 e c, >. 3. The Appropriate function To which function can one apply the Laplace tranform? Firt of all, we need the integral f(t)dt to make ene, i.e., f(t) i integrable on the interval [, N]. The function f(t) being continuou certainly guarantee thi. However, we could alo allow a finitely number of dicontinuitie, ay < x 1 < x 2 <... < x k < N, o the integral i the um of the integral of f(t) on each of the interval [, x 1 ), [x 1, x 2 ),..., [x k, N] where f(t) i continuou. Such function are called piecewie continuou on the interval [, N]. A function i aid to be piecewie continuou on [, ) if it i piecewie continuou on any [, N], N >. Secondly, we need the limit lim f(t)e t dt to exit. Intuitively, fixing, thi i aying that the igned area of the region encloed by the graph of f(t)e t and the x-axi i finite. One implication i that f(t) hould not increae too fat. For intance, for = 1 and f(t) = e t, we have f(t)e t = 1 and the limit of the integral doe not exit. Alo note that for the ame f(t), the limit of the integral exit for > 1. More generally, if f(t) = Me αt for ome M, α >, then the limit lim Me αt e t dt = lim Me (α )t M dt = lim α (e(α )N 1) exit for all > α. What i exciting about thi fact i that it actually bring a lot appropriate function into our ight, baed on the comparion theorem below. Theorem. If f(x) and g(x) are piecewie continuou function defined on [, ) and f(x) g(x), then the convergence of g(x)dx implie the convergence of f(x)dx. The following corollary of the theorem tell u which function are appropriate to be put in the Laplace tranform.

4 4 LECTURE 12: LAPLACE TRANSFORM Corollary. Baed on the theorem, we can ay that if (a) f(t) i piecewie continuou on [, N] for any N > and (b) it i of the exponential order, then f(t)e t dt converge on the interval (α, ) for ome α >. Note: A function f(t) i aid to be of the exponential order if there exit contant M, α > uch that f(t) Me αt on [, ). From now on, we confine ourelve to the category of function which atify the condition in the corollary above and call thee function appropriate. A an exercie, you can check that all the f(t) in the previou ection are appropriate. 4. Propertie of the Laplace Tranform 4.1. Linearity. The Laplace tranform i linear. In fact, for f(t), g(t) appropriate function and L{f(t)} = F (), L{g(t)} = G(), both defined for (α, ), and any contant a, b, we have that L{af(t) + bg(t)} = = a (af(t) + bg(t))e t dt f(t)e t dt + b = al{f(t)} + bl{g(t)}. i defined for (α, ). Thi prove the linearity. g(t)e t dt 4.2. Firt hift theorem. If f(t) i appropriate and F () = L{f(t)} i defined for (α, ), then L{e at f(t)} = F ( a) i defined for (α + a, ). Proof. By definition, L{e at f(t)} = = = F ( a). e at f(t)e t dt f(t)e ( a)t dt Since we know that F () i defined for (α, ), F ( a) i defined for (α + a, ). Remark. The firt hift theorem work for complex-valued a. For intance, letting a = i, we have L{e at 1} = 1 a = + i On the other hand, by linearity, L{e it } = L{co t} + il{in t}. Hence, comparing the real and imaginary part, we have L{co t} = 1, L{in t} =

5 LECTURE 12: LAPLACE TRANSFORM Second hift theorem. If f(t) i appropriate and F () = L{f(t)} i defined for (α, ), then L{u c (t)f(t c)} = e c L{f(t)} = e c F (), > α. Proof. Firt, you may want to convince yourelf that the function u c (t)f(t c) i indeed the reult of hifting f(t) to the right by c. By definition, L{u c (t)f(t c)} = 4.4. Relation to the derivative. = = c u c (t)f(t c)e t dt f(t c)e t dt f(ξ)e (ξ+c)dξ (ξ = t c) = e c f(ξ)e ξ dξ = e c F (), > α L{f (t)}. Before tating the condition on f(t), let u calculate formally the Laplace tranform of f (t): L{f (t)} = f (t)e t dt. Now, uing integration by part (formally), we obtain L{f (t)} = f(t)e t t= ( ) = f() + L{f(t)}. f(t)e t dt Back to the quetion: what are the condition on the function f(t)? Firt, the reult above involve the Laplace tranform of f(t) and f (t), o it i reaonable to have both f(t), f (t) a being appropriate, i.e., piecewie continuou and of the exponential order. Second, by the fundamental theorem of calculu, the integration by part only applie to function that are continuou. You may wonder, in t f(t) continuou if f (t) exit? Remember, in our context, finitely many dicontinuitie can exit and f (t) may only exit piece-wiely. So, we enforce another condition, that i, f(t) i continuou. To ummarize, we tate thi a a propoition: Propoition. If f(t) i continuou and f(t), f (t) are appropriate, i.e., piecewie continuou and of the exponential order, then L{f (t)} = f() + L{f(t)}, > α for ome α >. Remark. In fact, α can be choen uing the fact that f(t), f (t) are of the exponential order. Do you ee why?

6 6 LECTURE 12: LAPLACE TRANSFORM L{f (n) (t)}. The previou derivative property can be applied ucceively and the reult generalized to higher derivative. Propoition. If f(t), f (t),..., f (n 1) (t) are continuou and f(t),..., f (n) (t) are appropriate, then L{f (n) (t)} = n L{f(t)} n 1 f() n 2 f ()... f (n 1) (), for ome α >. Note, in particular, that when n = 2, thi i jut L{f (t)} = 2 L{f(t)} f() f (), > α L{tf(t)}. Let u calculate L{tf(t)} formally by the definition L{tf(t)} = = d d tf(t)e t dt = d d L{f(t)} = d d F (). f(t)e t dt > α Here we have implicitly interchanged the order of integral and derivative. Particularly, auming the integral to be abolutely convergent enure thi. Of coure, the reult above can be extended to L{t n f(t)} = ( 1) n dn d n L{f(t)}. A an exercie, you can prove thi formula, then ue it to find L{t n }.

Chapter 4. The Laplace Transform Method

Chapter 4. The Laplace Transform Method Chapter 4. The Laplace Tranform Method The Laplace Tranform i a tranformation, meaning that it change a function into a new function. Actually, it i a linear tranformation, becaue it convert a linear combination