Chapter 31. The Laplace Transform The Laplace Transform. The Laplace transform of the function f(t) is defined. e st f(t) dt, L[f(t)] =

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1 Chapter 3 The Laplace Transform 3. The Laplace Transform The Laplace transform of the function f(t) is defined L[f(t)] = e st f(t) dt, for all values of s for which the integral exists. The Laplace transform of f(t) is a function of s which we will denote ˆf(s). A function f(t) is of exponential order α if there exist constants t and M such that f(t) < M e αt, for all t > t. If t f(t) dt exists and f(t) is of exponential order α then the Laplace transform ˆf(s) exists for R(s) > α. Here are a few examples of these concepts. sin t is of exponential order. Denoting the Laplace transform of f(t) as F(s) is also common. 475

2 t e 2t is of exponential order α for any α > 2. e t2 is not of exponential order α for any α. t n is of exponential order α for any α >. t 2 does not have a Laplace transform as the integral diverges. Example 3.. Consider the Laplace transform of f(t) =. Since f(t) = is of exponential order α for any α >, the Laplace transform integral converges for R(s) >. ˆf(s) = e st dt [ = ] s e st = s Example 3..2 The function f(t) = t e t is of exponential order α for any α >. We compute the Laplace transform of this function. ˆf(s) = e st t e t dt = t e ( s)t dt [ ] = s t e( s)t [ ] = ( s) 2 e( s)t = for R(s) >. ( s) 2 s e( s)t dt 476

3 Example 3..3 Consider the Laplace transform of the Heaviside function, { for t < c H(t c) = for t > c, where c >. L[H(t c)] = Example 3..4 Next consider H(t c)f(t c). L[H(t c)f(t c)] = = e st dt c [ ] e st = s = e cs s = = c e st H(t c) dt c = e cs ˆf(s) 3.2 The Inverse Laplace Transform for R(s) > e st H(t c)f(t c) dt e st f(t c) dt e s(t+c) f(t) dt The inverse Laplace transform in denoted f(t) = L [ ˆf(s)]. 477

4 We compute the inverse Laplace transform with the Mellin inversion formula. f(t) = α+ı α ı e st ˆf(s) ds Here α is a real constant that is to the right of the singularities of ˆf(s). To see why the Mellin inversion formula is correct, we take the Laplace transform of it. Assume that f(t) is of exponential order α. Then α will be to the right of the singularities of ˆf(s). [ L[L [ ˆf(s)]] α+ı ] = L e zt ˆf(z) dz We interchange the order of integration. = Since R(z) = α, the integral in t exists for R(s) > α. α ı α+ı e st = α+ı α ı ˆf(z) α ı = α+ı ˆf(z) α ı s z dz e zt ˆf(z) dz dt e (z s)t dt dz We would like to evaluate this integral by closing the path of integration with a semi-circle of radius R in the right half plane and applying the residue theorem. However, in order for the integral along the semi-circle to vanish as R, ˆf(z) must vanish as z. If ˆf(z) vanishes we can use the maximum modulus bound to show that the integral along the semi-circle vanishes. This we assume that ˆf(z) vanishes at infinity. Consider the integral, C ˆf(z) s z dz, 478

5 Im(z) α+ir s Re(z) α-ir Figure 3.: The Laplace Transform Pair Contour. where C is the contour that starts at α ır, goes straight up to α + ır, and then follows a semi-circle back down to α ır. This contour is shown in Figure 3.. If s is inside the contour then C ˆf(z) dz = ˆf(s). s z Note that the contour is traversed in the negative direction. Since ˆf(z) decays as z, the semicircular contribution to the integral will vanish as R. Thus α+ı ˆf(z) dz = ˆf(s). α ı s z Therefore, we have shown than f(t) and ˆf(s) are known as Laplace transform pairs. L[L [ ˆf(s)]] = ˆf(s). 479

6 3.2. ˆf(s) with Poles Example 3.2. Consider the inverse Laplace transform of /s 2. s = is to the right of the singularity of /s 2. L [ s 2 ] = +ı ı e st s 2 ds Let B R be the contour starting at ır and following a straight line to + ır; let C R be the contour starting at + ır and following a semicircular path down to ır. Let C be the combination of B R and C R. This contour is shown in Figure 3.2. Im(s) α+ir C R B R Re(s) α-ir Figure 3.2: The Path of Integration for the Inverse Laplace Transform. 48

7 Consider the line integral on C for R >. e st ( C s ds = Res e st ) 2 s 2, = d ds est s= If t, the integral along C R vanishes as R. We parameterize s. = t s = + R e ıθ π, 2 θ 3π 2 e st = e t(+r eıθ ) = e t e tr cos θ e t Thus the inverse Laplace transform of /s 2 is e st C R s ds 2 C R est ds s 2 πr e t (R ) 2 as R L [ s 2 ] = t, for t. Let ˆf(s) be analytic except for isolated poles at s, s 2,...,s N and let α be to the right of these poles. Also, let ˆf(s) as s. Define B R to be the straight line from α ır to α + ır and C R to be the semicircular path 48

8 from α + ır to α ır. If R is large enough to enclose all the poles, then e st ˆf(s) ds = B R B R +C R e st ˆf(s) ds = N n= N Res(e st ˆf(s), sn ) n= Res(e st ˆf(s), sn ) C R e st ˆf(s) ds. Now let s examine the integral along C R. Let the maximum of ˆf(s) on C R be M R. We can parameterize the contour with s = α + R e ıθ, π/2 < θ < 3π/2. e st ˆf(s) ds = C R If t we can use Jordan s Lemma to obtain, 3π/2 π/2 3π/2 π/2 e t(α+r eiθ ) ˆf(α + R e ıθ )Rı e ıθ dθ e αt e tr cos θ RM R dθ π = RM R e αt e tr sin θ dθ < RM R e αt π tr. = M R e αt π t We use that M R as R. as R 482

9 Thus we have an expression for the inverse Laplace transform of ˆf(s). α+ı α ı N e st ˆf(s) ds = Res(e st ˆf(s), sn ) L [ ˆf(s)] = n= N Res(e st ˆf(s), sn ) n= Result 3.2. If ˆf(s) is analytic except for poles at s, s 2,...,s N and ˆf(s) as s then the inverse Laplace transform of ˆf(s) is f(t) = L [ ˆf(s)] = N Res(e st ˆf(s), sn ), for t >. n= Example Consider the inverse Laplace transform of. s 3 s 2 First we factor the denominator. s 3 s = 2 s 2 s. Taking the inverse Laplace transform, [ ] L s 3 s 3 ( = Res e st ) s 2 s, = d e st ds s + e t s= = ( ) + t 2 + et ( + Res e st ) s 2 s, 483

10 Thus we have that [ ] L = e t t, for t >. s 3 s 2 Example Consider the inverse Laplace transform of We factor the denominator. Then we take the inverse Laplace transform. [ ] L s 2 + s = Res s 3 2s 2 + s 2 Thus we have s 2 + s s 3 2s 2 + s 2. s 2 + s (s 2)(s ı)(s + ı). ( e st s 2 + s ) ( (s 2)(s ı)(s + ı), 2 + Res ( e st s 2 + s (s 2)(s ı)(s + ı), ı + Res = e 2t + e ıt ı2 + e ıt ı2 [ ] L s 2 + s = sint + e 2t, for t >. s 3 2s 2 + s 2 ) e st s 2 + s (s 2)(s ı)(s + ı), ı ) ˆf(s) with Branch Points Example Consider the inverse Laplace transform of s. s denotes the principal branch of s /2. There is a branch cut from s = to s = and s = e ıθ/2 r, for π < θ < π. 484

11 Let α be any positive number. The inverse Laplace transform of s is f(t) = α+ı e st α ı s ds. We will evaluate the integral by deforming it to wrap around the branch cut. Consider the integral on the contour shown in Figure 3.3. C + R and C R are circular arcs of radius R. B is the vertical line at R(s) = α joining the two arcs. C ǫ is a semi-circle in the right half plane joining ıǫ and ıǫ. L + and L are lines joining the circular arcs at I(s) = ±ǫ. C R + B L + L - C ε π/2 δ π/2+δ - C R Figure 3.3: Path of Integration for / s Since there are no residues inside the contour, we have ( + + B C + R L + + We will evaluate the inverse Laplace transform for t >. C ǫ + L + C R ) e st s ds =. 485

12 First we will show that the integral along C + R vanishes as R. π/2 ds = dθ + C + R π/2 δ π π/2 dθ. The first integral vanishes by the maximum modulus bound. Note that the length of the path of integration is less than 2α. ( π/2 ) dθ est s (2α) π/2 δ max s C + R = e αt R (2α) as R The second integral vanishes by Jordan s Lemma. A parameterization of C + R is s = R eıθ. π e R eıθ t π dθ R e ıθ er eıθ t R e ıθ dθ π/2 π/2 R π π/2 e R cos(θ)t dθ π/2 e Rt sin(φ) dφ R < R π 2Rt as R We could show that the integral along C R vanishes by the same method. Now we have ( ) + e st s ds =. B L + + C ǫ L

13 We can show that the integral along C ǫ vanishes as ǫ with the maximum modulus bound. ) s ds (max est s (πǫ) C ǫ e st s C ǫ < e ǫt ǫ πǫ as ǫ Now we can express the inverse Laplace transform in terms of the integrals along L + and L. f(t) α+ı e st s ds = e st s ds e st s ds L + L α ı On L +, s = r e ıπ, ds = e ıπ dr = dr; on L, s = r e ıπ, ds = e ıπ dr = dr. We can combine the integrals along the top and bottom of the branch cut. f(t) = = We make the change of variables x = rt. = π t We recognize this integral as Γ(/2). = π t Γ(/2) = πt rt ı e ( ) dr r e rt ı2 r dr e x x dx e rt ı r ( ) dr 487

14 Thus the inverse Laplace transform of s is f(t) = πt, for t > Asymptotic Behavior of ˆf(s) Consider the behavior of ˆf(s) = e st f(t) dt as s +. Assume that f(t) is analytic in a neighborhood of t =. Only the behavior of the integrand near t = will make a significant contribution to the value of the integral. As you move away from t =, the e st term dominates. Thus we could approximate the value of ˆf(s) by replacing f(t) with the first few terms in its Taylor series expansion about the origin. ] ˆf(s) e [f() st + tf () + t2 2 f () + dt as s + Using we obtain ˆf(s) f() s L [t n ] = n! s n+ + f () s 2 + f () s 3 + as s +. Example The Taylor series expansion of sin t about the origin is Thus the Laplace transform of sin t has the behavior sin t = t t3 6 + O(t5 ). L[sin t] s 2 s 4 + O(s 6 ) as s

15 We corroborate this by expanding L[sin t]. L[sin t] = s 2 + = s 2 + s 2 = s 2 ( ) n s 2n n= = s 2 s 4 + O(s 6 ) 3.3 Properties of the Laplace Transform In this section we will list several useful properties of the Laplace transform. If a result is not derived, it is shown in the Problems section. Unless otherwise stated, assume that f(t) and g(t) are piecewise continuous and of exponential order α. L[af(t) + bg(t)] = al[f(t)] + bl[g(t)] L[e ct f(t)] = ˆf(s c) for s > c + α L[t n f(t)] = ( ) n dn ds n [ ˆf(s)] for n =, 2,... If β f(t) t dt exists for positive β then [ ] t L f(τ) dτ = ˆf(s) s [ ] f(t) L = t s ˆf(σ) dσ. 489

16 L [ d f(t)] = s ˆf(s) f() dt [ ] d L 2 f(t) = s 2 ˆf(s) dt sf() f () 2 To derive these formulas, [ ] d L dt f(t) = e st f (t) dt = [ e st f(t) ] s e st f(t) dt = f() + s ˆf(s) [ ] d 2 L dt 2f(t) = sl[f (t)] f () = s 2 ˆf(s) sf() f () Let f(t) and g(t) be continuous. The convolution of f(t) and g(t) is defined h(t) = (f g) = The convolution theorem states t f(τ)g(t τ) dτ = t ĥ(s) = ˆf(s)ĝ(s). f(t τ)g(τ) dτ 49

17 To show this, ĥ(s) = = = = t e st f(τ)g(t τ) dτ dt τ e sτ f(τ) e st f(τ)g(t τ) dt dτ e sτ f(τ) dτ τ = ˆf(s)ĝ(s) e s(t τ) g(t τ) dt dτ e sη g(η) dη If f(t) is periodic with period T then L[f(t)] = T e st f(t) dt e st. Example 3.3. Consider the inverse Laplace transform of. First we factor the denominator. s 3 s 2 s 3 s = 2 s 2 s We know the inverse Laplace transforms of each term. [ ] [ ] L = t, L = e t s 2 s We apply the convolution theorem. [ ] L s 2 s = t τ e t τ dτ = e [ t τ e τ] t t et e τ dτ = t + e t 49

18 [ ] L = e t t. s 2 s Example We can find the inverse Laplace transform of s 2 + s s 3 2s 2 + s 2 with the aid of a table of Laplace transform pairs. We factor the denominator. s 2 + s (s 2)(s ı)(s + ı) We expand the function in partial fractions and then invert each term. s 2 + s (s 2)(s ı)(s + ı) = s 2 ı/2 s ı + ı/2 s + ı s 2 + s (s 2)(s ı)(s + ı) = s 2 + s 2 + L [ s 2 + s 2 + ] = e 2t + sin t 3.4 Constant Coefficient Differential Equations Example 3.4. Consider the differential equation y + y = cost, for t >, y() =. 492

19 We take the Laplace transform of this equation. Now we invert ŷ(s). sŷ(s) y() + ŷ(s) = s s 2 + s ŷ(s) = (s + )(s 2 + ) + s + ŷ(s) = /2 s + + s + 2 s 2 + y(t) = 2 e t + 2 cost + sin t, for t > 2 Notice that the initial condition was included when we took the Laplace transform. One can see from this example that taking the Laplace transform of a constant coefficient differential equation reduces the differential equation for y(t) to an algebraic equation for ŷ(s). Example Consider the differential equation We take the Laplace transform of this equation. From the table of Laplace transform pairs we know [ L s s 2 + y + y = cos(2t), for t >, y() =, y () =. s 2 ŷ(s) sy() y () + ŷ(s) = s s s ŷ(s) = (s 2 + )(s 2 + 4) + s s 2 + ] = cost, L [ 493 ] s = 2 sin(2t).

20 We use the convolution theorem to find the inverse Laplace transform of ŷ(s). y(t) = t t sin(2τ) cos(t τ) dτ + cost 2 = sin(t + τ) + sin(3τ t) dτ + cost 4 = [ cos(t + τ) ] t 4 3 cos(3τ t) + cost = ( cos(2t) + cost 3 4 cos(2t) + 3 ) cos(t) + cost = 3 cos(2t) cos(t) Alternatively, we can find the inverse Laplace transform of ŷ(s) by first finding its partial fraction expansion. Example Consider the initial value problem ŷ(s) = s/3 s 2 + s/3 s Without taking a Laplace transform, we know that since the Laplace transform has the behavior = s/3 s s/3 s 2 + s s 2 + y(t) = 3 cos(2t) cos(t) y + 5y + 2y =, y() =, y () = 2. y(t) = + 2t + O(t 2 ) ŷ(s) s + 2 s 2 + O(s 3 ), as s

21 3.5 Systems of Constant Coefficient Differential Equations The Laplace transform can be used to transform a system of constant coefficient differential equations into a system of algebraic equations. This should not be surprising, as a system of differential equations can be written as a single differential equation, and vice versa. Example 3.5. Consider the set of differential equations y = y 2 y 2 = y 3 y 3 = y 3 y 2 y + t 3 with the initial conditions y () = y 2 () = y 3 () =. We take the Laplace transform of this system. sŷ y () = ŷ 2 sŷ 2 y 2 () = ŷ 3 sŷ 3 y 3 () = ŷ 3 ŷ 2 ŷ + 6 s 4 The first two equations can be written as ŷ = ŷ3 s 2 ŷ 2 = ŷ3 s. 495

22 We substitute this into the third equation. sŷ 3 = ŷ 3 ŷ3 s ŷ3 s s 4 (s 3 + s 2 + s + )ŷ 3 = 6 s 2 6 ŷ 3 = s 2 (s 3 + s 2 + s + ). We solve for ŷ. We then take the inverse Laplace transform of ŷ. 6 ŷ = s 4 (s 3 + s 2 + s + ) ŷ = s 4 s + 3 2(s + ) + s 2(s 2 + ) y = t3 6 t e t + 2 sin t 2 cost. We can find y 2 and y 3 by differentiating the expression for y. y 2 = t2 2 t 2 e t + 2 cost + 2 sin t y 3 = t + 2 e t 2 sin t + 2 cost 496