Lecture notes on Laplace transform

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The Laplace Transform: Theory and Applications Joel L. Schiff Springer. . . . . . . . . . . . . . . . . . Basic . . . . . . . . . . . . . 1 . . Principles . . . . . . . . . CHAPTER . Ordinary and partial differential equations describe the way certain quantities vary with time, such as the current in an electrical circuit, the oscillations of a vibrating membrane, or the flow of heat through an insulated conductor. These equations are generally coupled with initial conditions that describe the state of the system at time t  0. A very powerful technique for solving these problems is that of the Laplace transform, which literally transforms the original differ- ential equation into an elementary algebraic expression. This latter can then simply be transformed once again, into the solution of the original problem. This technique is known as the “Laplace transform method.” It will be treated extensively in Chapter 2. In the present chapter we lay down the foundations of the theory and the basic properties of the Laplace transform. 1.1 The Laplace Transform Suppose that f is a real- or complex-valued function of the (time) variablet 0 and s is a real or complex parameter. We define the 11. Basic Principles 2 Laplace transform of f as  ∞   −st F(s)  L f (t)  e f (t) dt 0  τ −st  lim e f (t) dt (1.1) τ→∞ 0 whenever the limit exists (as a finite number). When it does, the integral (1.1) is said toconverge. If the limit does not exist, the integral is said to diverge and there is no Laplace transform defined for f . The notation L(f ) will also be used to denote the Laplace transform of f , and the integral is the ordinary Riemann (improper) integral (see Appendix). The parameter s belongs to some domain on the real line or in the complex plane. We will choose s appropriately so as to ensure the convergence of the Laplace integral (1.1). In a mathematical and technical sense, the domain of s is quite important. However, in a practical sense, when differential equations are solved, the domain of s is routinely ignored. When s is complex, we will always use the notation s  x + iy. The symbol L is the Laplace transformation, which acts on   functions f  f (t) and generates a new function, F(s)  L f (t) . Example 1.1. If f (t) ≡ 1for t ≥ 0, then  ∞   −st L f (t)  e 1 dt 0    τ −st  e   lim  τ→∞ −s 0   −sτ e 1  lim + (1.2) τ→∞ −s s 1  s provided of course thats 0 (if s is real). Thus we have 1 L(1)  (s 0). (1.3) s1.1. The Laplace Transform 3 If s ≤ 0, then the integral would diverge and there would be no re- sulting Laplace transform. If we had taken s to be a complex variable, the same calculation, with Re(s) 0, would have given L(1)  1/s. In fact, let us just verify that in the above calculation the integral can be treated in the same way even if s is a complex variable. We require the well-known Euler formula (see Chapter 3) iθ e  cos θ + i sin θ, θ real, (1.4) iθ and the fact that e  1. The claim is that (ignoring the minus sign as well as the limits of integration to simplify the calculation)  st e st e dt  , (1.5) s for s  x + iy any complex number  0. To see this observe that   st (x+iy)t e dt  e dt   xt xt  e cos yt dt + i e sin yt dt by Euler’s formula. Performing a double integration by parts on both these integrals gives   xt e st e dt  (x cos yt + y sin yt) + i(x sin yt − y cos yt) . 2 2 x + y Now the right-hand side of (1.5) can be expressed as st (x+iy)t e e  s x + iy xt e (cos yt + i sin yt)(x − iy)  2 2 x + y xt  e  (x cos yt + y sin yt) + i(x sin yt − y cos yt) , 2 2 x + y which equals the left-hand side, and (1.5) follows. Furthermore, we obtain the result of (1.3) for s complex if we take Re(s) x 0, since then −sτ −xτ lim e  lim e  0, τ→∞ τ→∞1. Basic Principles 4 killing off the limit term in (1.3). Let us use the preceding to calculate L(cos ωt) and L(sin ωt) (ω real). Example 1.2. We begin with  ∞ iωt −st iωt L(e )  e e dt 0  τ (iω−s)t  e   lim  τ→∞ iω − s 0 1  , s − iω iωτ −sτ −xτ since lim e e  lim e  0, provided x  Re(s) τ→∞ τ→∞ −iωt 0. Similarly, L(e )  1/(s + iω). Therefore, using the linearity property of L, which follows from the fact that integrals are linear operators (discussed in Section 1.6),   iωt −iωt iωt −iωt L(e ) +L(e ) e + e  L  L(cos ωt), 2 2 and consequently,   1 1 1 s L(cos ωt)  +  . (1.6) 2 2 2 s − iω s + iω s + ω Similarly,     1 1 1 ω L(sin ωt)  −  Re(s) 0 . 2 2 2i s − iω s + iω s + ω (1.7) The Laplace transform of functions defined in a piecewise fashion is readily handled as follows. Example 1.3. Let (Figure 1.1) t 0 ≤ t ≤ 1 f (t)  1 t 1.Exercises 1.1 5 ft Ot FIGURE 1.1 From the definition,  ∞   −st L f (t)  e f (t) dt 0   1 τ −st −st  te dt + lim e dt τ→∞ 0 1    1 τ −st 1 −st   te 1 e −st    + e dt + lim   τ→∞ −s s −s 0 0 1 −s 1 − e    Re(s) 0 . 2 s Exercises 1.1   1. From the definition of the Laplace transform, compute L f (t) for 2t (a) f (t)  4t (b) f (t)  e (c) f (t)  2 cos 3t (d) f (t)  1 − cos ωt 2t t (e) f (t)  te (f) f (t)  e sin t  π  sin ωt 0t 1 t ≥ a ω (g) f (t)  (h) f (t)  π  0 ta 0 ≤ t ω1. Basic Principles 6 2 t ≤ 1 (i) f (t)  t e t 1. 2. Compute the Laplace transform of the function f (t) whose graph is given in the figures below. ftft a b OtOt FIGURE E.1 FIGURE E.2 1.2 Convergence Although the Laplace operator can be applied to a great many functions, there are some for which the integral (1.1) does not converge. 2 (t ) Example 1.4. For the function f (t)  e ,   τ τ 2 2 −st t t −st lim e e dt  lim e dt ∞ τ→∞ τ→∞ 0 0 for any choice of the variable s, since the integrand grows without bound as τ →∞. In order to go beyond the superficial aspects of the Laplace trans- form, we need to distinguish two special modes of convergence of the Laplace integral. The integral (1.1) is said to be absolutely convergent if  τ −st lim e f (t) dt τ→∞ 0   exists. If L f (t) does converge absolutely, then        τ  τ   −st −st e f (t) dt ≤ e f (t)dt → 0     τ τExercises 1.2 7    as τ →∞, for all τ τ. This then implies thatL f (t) also converges ∗ in the ordinary sense of (1.1). There is another form of convergence that is of the utmost im- portance from a mathematical perspective. The integral (1.1) is said to converge uniformly for s in some domain  in the complex plane if for anyε 0, there exists some number τ such that if τ ≥ τ , then 0 0    ∞   −st   e f (t) dt ε   τ for all s in . The point here is that τ can be chosen sufficiently 0 large in order to make the “tail” of the integral arbitrarily small, independent of s. Exercises 1.2 1. Suppose that f is a continuous function on 0, ∞) and f (t)≤ M ∞ for 0 ≤t ∞.   (a) Show that the Laplace transform F(s)  L f (t) con- verges absolutely (and hence converges) for any s satisfying Re(s) 0.   (b) Show that L f (t) converges uniformly if Re(s) ≥ x 0. 0   (c) Show that F(s)  L f (t) → 0as Re(s)→∞. t 2. Let f (t)  e on 0, ∞). t (a) Show that F(s)  L(e ) converges for Re(s) 1. t (b) Show that L(e ) converges uniformly if Re(s) ≥ x 1. 0 ∗ Convergence of an integral  ∞ ϕ(t)dt 0 is equivalent to the Cauchy criterion:   τ  ϕ(t)dt →0as τ →∞,τ τ. τ1. Basic Principles 8 t (c) Show that F(s)  L(e ) → 0as Re(s)→∞. 3. Show that the Laplace transform of the function f (t)  1/t,t 0 does not exist for any value of s. 1.3 Continuity Requirements Since we can compute the Laplace transform for some functions and 2 (t ) not others, such as e , we would like to know that there is a large class of functions that do have a Laplace tranform. There is such a class once we make a few restrictions on the functions we wish to consider. Definition 1.5. A function f has ajump discontinuity at a point t if both the limits 0 − + lim f (t)  f (t ) and lim f (t)  f (t ) 0 0 − + t→t t→t 0 0 − + − + exist (as finite numbers) and f (t )  f (t ). Here, t → t and t → t 0 0 0 0 mean that t → t from the left and right, respectively (Figure 1.2). 0 Example 1.6. The function (Figure 1.3) 1 f (t)  t − 3 ft ft ft Ott FIGURE 1.21.3. Continuity Requirements 9 ft Ot FIGURE 1.3 ft Ot FIGURE 1.4 has a discontinuity at t  3, but it is not a jump discontinuity since − + neither lim f (t) nor lim f (t) exists. t→3 t→3 Example 1.7. The function (Figure 1.4) 2 t − 2 e t 0 f (t)  0 t 0 has a jump discontinuity at t  0 and is continuous elsewhere. Example 1.8. The function (Figure 1.5) 0 t 0 f (t)  1 cos t 0 t is discontinuous at t  0, but lim + f (t) fails to exist, so f does not t→0 have a jump discontinuity at t  0.1. Basic Principles 10 ft Ot FIGURE 1.5 ft Obt FIGURE 1.6 The class of functions for which we consider the Laplace transform defined will have the following property. Definition 1.9. A function f is piecewise continuous on the in- + + terval 0, ∞) if (i) lim f (t)  f (0 ) exists and (ii) f is continuous t→0 on every finite interval (0,b) except possibly at a finite number of points τ ,τ ,...,τ in (0,b) at which f has a jump discontinuity 1 2 n (Figure 1.6). The function in Example 1.6 is not piecewise continuous on 0, ∞). Nor is the function in Example 1.8. However, the function in Example 1.7 is piecewise continuous on 0, ∞). An important consequence of piecewise continuity is that on each subinterval the function f is also bounded. That is to say, f (t)≤ M,τ tτ,i  1, 2,...,n − 1, i i i+1 for finite constants M . iExercises 1.3 11 In order to integrate piecewise continuous functions from 0 to b, one simply integrates f over each of the subintervals and takes the sum of these integrals, that is,     b τ τ b 1 2 f (t) dt  f (t) dt + f (t) dt +···+ f (t) dt. 0 0 τ τ 1 n This can be done since the function f is both continuous and bounded on each subinterval and thus on each has a well-defined (Riemann) integral. Exercises 1.3 Discuss the continuity of each of the following functions and locate any jump discontinuities. 1 1. f (t)  1 + t 1 2. g(t)  t sin (t  0) t  tt ≤ 1  3. h(t)  1  t 1 2 1 + t  sinh t  t  0 4. i(t)  t  1 t  0 1 1 5. j(t)  sinh (t  0) t t  −t 1 − e  t  0 6. k(t)  t  0 t  0 12na ≤t (2n + 1)a 7. l(t)  a 0, n  0, 1, 2, ... −1(2n + 1)a ≤t (2n + 2)a   t 8. m(t)  +1, for t ≥ 0,a 0, where x  greatest integer ≤ x. a1. Basic Principles 12 1.4 Exponential Order The second consideration of our class of functions possessing a well- defined Laplace transform has to do with the growth rate of the functions. In the definition  ∞   −st L f (t)  e f (t) dt, 0   when we takes 0 orRe(s) 0 , the integral will converge as long as f does not grow too rapidly. We have already seen by Example 1.4 2 t that f (t)  e does grow too rapidly for our purposes. A suitable rate of growth can be made explicit. Definition 1.10. A function f has exponential order α if there exist constantsM 0 and α such that for some t ≥ 0, 0 αt f (t)≤ Me,t ≥ t . 0 at Clearly the exponential function e has exponential order α  a, n whereas t has exponential order α for anyα 0 and any n ∈ N (Exercises 1.4, Question 2), and bounded functions like sin t, cos t, −1 −t tan t have exponential order 0, whereas e has order −1. How- 2 t ever, e does not have exponential order. Note that ifβα, then αt βt exponential order α implies exponential order β, since e ≤ e , t ≥ 0. We customarily state the order as the smallest value of α that works, and if the value itself is not significant it may be suppressed altogether. Exercises 1.4 1. If f and f are piecewise continuous functions of orders α and 1 2 β, respectively, on 0, ∞), what can be said about the continuity and order of the functions (i) c f + c f , c ,c constants, 1 1 2 2 1 2 (ii) f · g? n 2. Show that f (t)  t has exponential order α for anyα 0, n ∈ N. 2 t 3. Prove that the function g(t)  e does not have exponential order.1.5. The ClassL 13 1.5 The Class L We now show that a large class of functions possesses a Laplace transform. Theorem 1.11. If f is piecewise continuous on 0, ∞) and of exponen- tial order α, then the Laplace transform L(f ) exists for Re(s)α and converges absolutely. Proof. First, αt f (t)≤ M e,t ≥ t , 1 0 for some real α. Also, f is piecewise continuous on 0,t and hence 0 bounded there (the bound being just the largest bound over all the subintervals), say f (t)≤ M , 0tt . 2 0 αt Since e has a positive minimum on 0,t , a constant M can be 0 chosen sufficiently large so that αt f (t)≤ Me,t 0. Therefore,   τ τ −st −(x−α)t e f (t)dt ≤ M e dt 0 0  τ −(x−α)t  Me    −(x − α) 0 −(x−α)τ M Me  − . x − α x − α Letting τ →∞ and noting that Re(s) xα yield  ∞ M −st e f (t)dt ≤ . (1.8) x − α 0 Thus the Laplace integral converges absolutely in this instance (and hence converges) for Re(s)α. 1. Basic Principles 14 at Example 1.12. Let f (t)  e , a real. This function is continuous on 0, ∞) and of exponential order a. Then  ∞ at −st at L(e )  e e dt 0  ∞ −(s−a)t  e dt 0  ∞ −(s−a)t    e 1    Re(s)a .  −(s − a) s − a 0 The same calculation holds for a complex and Re(s) Re(a). Example 1.13. Applying integration by parts to the function f (t)  t (t ≥ 0), which is continuous and of exponential order, gives  ∞ −st L(t)  te dt 0   ∞ −st ∞  −te 1 −st   + e dt  s s 0 0 1    L(1) provided Re(s) 0 s 1  . 2 s Performing integration by parts twice as above, we find that  ∞ 2 −st 2 L(t )  e t dt 0   2  Re(s) 0 . 3 s By induction, one can show that in general,   n n L(t )  Re(s) 0 (1.9) n+1 s for n  1, 2, 3, ... . Indeed, this formula holds even for n  0, since 0  1, and will be shown to hold even for non-integer values of n in Section 2.1. Let us define the class L as the set of those real- or complex- valued functions defined on the open interval (0, ∞) for which theExercises 1.5 15 Laplace transform (defined in terms of the Riemann integral) exists   for some value of s. It is known that whenever F(s)  L f (t) exists for some value s , then F(s) exists for all s with Re(s) Re(s ), that 0 0 is, the Laplace transform exists for all s in some right half-plane (cf. Doetsch 2, Theorem 3.4). By Theorem 1.11, piecewise continuous functions on 0, ∞) having exponential order belong to L. However, there certainly are functions in L that do not satisfy one or both of these conditions. Example 1.14. Consider 2 2 t t f (t)  2te cos(e ). Then f (t) is continuous on 0, ∞) but not of exponential order. However, the Laplace transform of f (t),  ∞   2 2 −st t t L f (t)  e 2te cos(e )dt, 0 exists, since integration by parts yields   ∞   ∞ 2 2  −st t −st t L f (t)  e sin(e ) + s e sin(e ) dt  0 0    2 t − sin(1) + sL sin(e ) Re(s) 0 . and the latter Laplace transform exists by Theorem 1.11. Thus we have a continuous function that is not of exponential order yet nevertheless possesses a Laplace transform. See also Remark 2.8. Another example is the function 1 f (t)  √ . (1.10) t We will compute its actual Laplace transform in Section 2.1 in the context of the gamma function. While (1.10) has exponential order   α  0 f (t)≤ 1, t ≥ 1 , it is not piecewise continuous on 0, ∞) + since f (t)→∞ as t → 0 , that is, t  0 is not a jump discontinuity. Exercises 1.5 2 2 t t 1. Consider the function g(t)  te sin(e ).1. Basic Principles 16 (a) Is g continuous on 0, ∞)? Does g have exponential order? (b) Show that the Laplace transform F(s) exists for Re(s) 0. (c) Show that g is the derivative of some function having exponential order. 2. Without actually determining it, show that the following func- tions possess a Laplace transform. sin t 1 − cos t (a) (b) t t 2 (c) t sinh t 3. Without determining it, show that the function f , whose graph is given in Figure E.3, possesses a Laplace transform. (See Question 3(a), Exercises 1.7.) ft Oaaaat FIGURE E.3 1.6 Basic Properties of the Laplace Transform Linearity. One of the most basic and useful properties of the Laplace operatorL is that of linearity, namely, if f ∈ L forRe(s)α, 1 f ∈ L for Re(s)β, then f + f ∈ L for Re(s) maxα, β, and 2 1 2 L(c f + c f )  c L(f ) + c L(f ) (1.11) 1 1 2 2 1 1 2 21.6. Basic Properties of the Laplace Transform 17 for arbitrary constants c , c . 1 2 This follows from the fact that integration is a linear process, to wit,  ∞   −st e c f (t) + c f (t) dt 1 1 2 2 0   ∞ ∞ −st −st  c e f (t) dt + c e f (t) dt (f ,f ∈ L). 1 1 2 2 1 2 0 0 Example 1.15. The hyperbolic cosine function ωt −ωt e + e cosh ωt  2 describes the curve of a hanging cable between two supports. By linearity 1 ωt −ωt L(cosh ωt)  L(e ) +L(e ) 2   1 1 1  + 2 s − ω s + ω s  . 2 2 s − ω Similarly, ω L(sinh ωt)  . 2 2 s − ω n Example 1.16. If f (t)  a + a t + ··· + a t is a polynomial of 0 1 n degree n, then n    k L f (t)  a L(t ) k k0 n  a k k  k+1 s k0 by (1.9) and (1.11).  ∞ n Infinite Series. For an infinite series, a t , in general it is not n n0 possible to obtain the Laplace transform of the series by taking the transform term by term.1. Basic Principles 18 Example 1.17. ∞  n 2n (−1) t 2 −t f (t)  e  , −∞t ∞. n n0 Taking the Laplace transform term by term gives ∞ ∞ n n   (−1) (−1) (2n) 2n L(t )  2n+1 n n s n0 n0 ∞ n  1 (−1) (2n)··· (n + 2)(n + 1)  . 2n s s n0 Applying the ratio test,     u 2(2n + 1) n+1   lim  lim ∞,   2 n→∞ n→∞ u s n and so the series diverges for all values of s. 2 2 −t −t However,L(e ) does exist since e is continuous and bounded on 0, ∞). So when can we guarantee obtaining the Laplace transform of an infinite series by term-by-term computation? Theorem 1.18. If ∞  n f (t)  a t n n0 converges for t ≥ 0, with n Kα a ≤ , n n for all n sufficiently large andα 0,K 0, then ∞ ∞       a n n n L f (t)  a L(t )  Re(s)α . n n+1 s n0 n0 Proof. Since f (t) is represented by a convergent power series, it is continuous on 0, ∞). We desire to show that the difference        N   N          n n L f (t) − a L(t )  L f (t) − a t  n   n      n0 n01.6. Basic Properties of the Laplace Transform 19    N     n ≤ L f (t) − a t  x n   n0    ∞ −xt converges to zero as N →∞, where L h(t)  e h(t) dt, x  x 0 Re(s). To this end,      N   ∞        n n f (t) − a t    a t  n n     n0 nN+1 ∞ n  (αt) ≤ K n nN+1   N n  (αt) αt  K e − n n0  ∞ x n since e  x /n. As h ≤ g implies L (h) ≤ L (g) when the x x n0 transforms exist,      N  N n   (αt)   n αt L f (t) − a t  ≤ KL e − x n x   n n0 n0   N n  1 α  K − n+1 x − α x n0   N    n 1 1 α  K − x − α x x n0   → 0 Re(s) xα as N →∞. We have used the fact that the geometric series has the sum ∞  1 n z  , z 1. 1 − z n0 Therefore, N    n L f (t)  lim a L(t ) n N→∞ n0

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