We use the convolution theorem to find the inverse Laplace transform of ˆy(s). y(t) =  t 0 1 2 sin(2τ) cos(t − τ) dτ + cos t = 1 4  t 0 sin(t + τ) + sin(3τ − t) dτ + cos t = 1 4  −cos(t + τ) − 1 3 cos(3τ − t)  t 0 + cos t = 1 4  −cos(2t) + cos t − 1 3 cos(2t) + 1 3 cos(t)  + cos t = − 1 3 cos(2t) + 4 3 cos(t) Alternatively, we can find the inverse Laplace transform of ˆy(s) by first finding its partial fraction expansion. ˆy(s) = s/3 s 2 + 1 − s/3 s 2 + 4 + s s 2 + 1 = − s/3 s 2 + 4 + 4s/3 s 2 + 1 y(t) = − 1 3 cos(2t) + 4 3 cos(t) Example 31.4.3 Consider the initial value problem y  + 5y  + 2y = 0, y(0) = 1, y  (0) = 2. Without taking a Laplace transform, we know that since y(t) = 1 + 2t + O(t 2 ) the Laplace transform has the behavior ˆy(s) ∼ 1 s + 2 s 2 + O(s −3 ), as s → +∞. 1494 31.5 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 e quations. This should not be surprising, as a system of differential equations can be written as a single differential equation, and vice versa. Example 31.5.1 Consider the set of differential equations y  1 = y 2 y  2 = y 3 y  3 = −y 3 − y 2 − y 1 + t 3 with the initial conditions y 1 (0) = y 2 (0) = y 3 (0) = 0. We take the Laplace transform of this system. sˆy 1 − y 1 (0) = ˆy 2 sˆy 2 − y 2 (0) = ˆy 3 sˆy 3 − y 3 (0) = −ˆy 3 − ˆy 2 − ˆy 1 + 6 s 4 The first two equations can be written as ˆy 1 = ˆy 3 s 2 ˆy 2 = ˆy 3 s . 1495 We substitute this into the third equation. sˆy 3 = −ˆy 3 − ˆy 3 s − ˆy 3 s 2 + 6 s 4 (s 3 + s 2 + s + 1)ˆy 3 = 6 s 2 ˆy 3 = 6 s 2 (s 3 + s 2 + s + 1) . We solve for ˆy 1 . ˆy 1 = 6 s 4 (s 3 + s 2 + s + 1) ˆy 1 = 1 s 4 − 1 s 3 + 1 2(s + 1) + 1 − s 2(s 2 + 1) We then take the inverse Laplace transform of ˆy 1 . y 1 = t 3 6 − t 2 2 + 1 2 e −t + 1 2 sin t − 1 2 cos t. We can find y 2 and y 3 by differentiating the expression for y 1 . y 2 = t 2 2 − t − 1 2 e −t + 1 2 cos t + 1 2 sin t y 3 = t − 1 + 1 2 e −t − 1 2 sin t + 1 2 cos t 1496 31.6 Exercises Exercise 31.1 Find the Laplace transform of the following functions: 1. f(t) = e at 2. f(t) = sin(at) 3. f(t) = cos(at) 4. f(t) = sinh(at) 5. f(t) = cosh(at) 6. f(t) = sin(at) t 7. f(t) =  t 0 sin(au) u du 8. f(t) =  1, 0 ≤ t < π 0, π ≤ t < 2π and f(t + 2π) = f(t) for t > 0. That is, f(t) is periodic for t > 0. Hint, Solution Exercise 31.2 Show that L[af(t) + bg(t)] = aL[f(t)] + bL[g(t)]. Hint, Solution Exercise 31.3 Show that if f(t) is of exponential order α, L[ e ct f(t)] = ˆ f(s − c) for s > c + α. 1497 Hint, Solution Exercise 31.4 Show that L[t n f(t)] = (−1) n d n ds n [ ˆ f(s)] for n = 1, 2, . . . Hint, Solution Exercise 31.5 Show that if  β 0 f(t) t dt exists for positive β then L  f(t) t  =  ∞ s ˆ f(σ) dσ. Hint, Solution Exercise 31.6 Show that L   t 0 f(τ) dτ  = ˆ f(s) s . Hint, Solution Exercise 31.7 Show that if f(t) is periodic with period T then L[f(t)] =  T 0 e −st f(t) dt 1 − e −sT . Hint, Solution 1498 Exercise 31.8 The function f(t) t ≥ 0, is periodic with period 2T ; i.e. f(t + 2T) ≡ f(t), and is also odd with period T ; i.e. f(t + T ) = −f(t). Further,  T 0 f(t) e −st dt = ˆg(s). Show that the Laplace transform of f(t) is ˆ f(s) = ˆg(s)/(1 + e −sT ). Find f(t) such that ˆ f(s) = s −1 tanh(sT/2). Hint, Solution Exercise 31.9 Find the Laplace transform of t ν , ν > −1 by two methods. 1. Assume that s is complex-valued. Make the change of variables z = st and use integration in the complex plane. 2. Show that the Laplace transform of t ν is an analytic function for (s) > 0. Assume that s is real-valued. Make the change of variables x = st and evaluate the integral. Then use analytic continuation to extend the result to complex-valued s. Hint, Solution Exercise 31.10 (mathematica/ode/laplace/laplace.nb) Show that the Laplace transform of f(t) = ln t is ˆ f(s) = − Log s s − γ s , where γ = −  ∞ 0 e −t ln t dt. [ γ = 0.5772 . . . is known as Euler’s constant.] Hint, Solution Exercise 31.11 Find the Laplace transform of t ν ln t. Write the answer in terms of the digamma function, ψ(ν) = Γ  (ν)/Γ(ν). What is the answer for ν = 0? Hint, Solution 1499 Exercise 31.12 Find the inverse Laplace transform of ˆ f(s) = 1 s 3 − 2s 2 + s − 2 with the following methods. 1. Expand ˆ f(s) using partial fractions and then use the table of Laplace transforms. 2. Factor the denominator into (s − 2)(s 2 + 1) and then use the convolution theorem. 3. Use Result 31.2.1. Hint, Solution Exercise 31.13 Solve the differential equation y  + y  + y = sin t, y(0) = y  (0) = 0, 0 <   1 using the Laplace transform. This equation represents a weakly damped, driven, line ar oscillator. Hint, Solution Exercise 31.14 Solve the problem, y  − ty  + y = 0, y(0) = 0, y  (0) = 1, with the Laplace transform. Hint, Solution Exercise 31.15 Prove the following relation between the inverse Laplace transform and the inverse Fourier transform, L −1 [ ˆ f(s)] = 1 2π e ct F −1 [ ˆ f(c + ıω)], 1500 where c is to the right of the singularities of ˆ f(s). Hint, Solution Exercise 31.16 (mathematica/ode/laplace/laplace.nb) Show by evaluating the Laplace inversion integral that if ˆ f(s) =  π s  1/2 e −2(as) 1/2 , s 1/2 = √ s for s > 0, then f (t) = e −a/t / √ t. Hint: cut the s-plane along the negative real axis and deform the contour onto the cut. Remember that  ∞ 0 e −ax 2 cos(bx) dx =  π/4a e −b 2 /4a . Hint, Solution Exercise 31.17 (mathematica/ode/laplace/laplace.nb) Use Laplace transforms to solve the initial value problem d 4 y dt 4 − y = t, y(0) = y  (0) = y  (0) = y  (0) = 0. Hint, Solution Exercise 31.18 (mathematica/ode/laplace/laplace.nb) Solve, by Laplace transforms, dy dt = sin t +  t 0 y(τ) cos(t − τ) dτ, y(0) = 0. Hint, Solution Exercise 31.19 (mathematica/ode/laplace/laplace.nb) Suppose u(t) satisfies the difference-differential equation du dt + u(t) − u(t − 1) = 0, t ≥ 0, 1501 and the ‘initial condi tion’ u(t) = u 0 (t), −1 ≤ t ≤ 0, where u 0 (t) is given. Show that the Laplace transform ˆu(s) of u(t) satisfies ˆu(s) = u 0 (0) 1 + s − e −s + e −s 1 + s − e −s  0 −1 e −st u 0 (t) dt. Find u(t), t ≥ 0, when u 0 (t) = 1. Check the result. Hint, Solution Exercise 31.20 Let the function f(t) be defined by f(t) =  1 0 ≤ t < π 0 π ≤ t < 2π, and for all positive values of t so that f(t + 2π) = f(t). That is, f (t) is periodic with period 2π. Find the solution of the intial value problem d 2 y dt 2 − y = f(t); y(0) = 1, y  (0) = 0. Examine the continuity of the solution at t = nπ, where n is a positive integer, and verify that the solution is continuous and has a continuous derivative at these points. Hint, Solution Exercise 31.21 Use Laplace transforms to solve dy dt +  t 0 y(τ) dτ = e −t , y(0) = 1. Hint, Solution 1502 [...]... transform methods and show that i1 (0) = i2 (0) = i1 = where α= q(0) = 0 E0 −αt E0 e sin(ωt) + 2R 2ωL R 2L and ω 2 = 2 − α2 LC Hint, Solution Exercise 31.23 Solve the initial value problem, y + 4y + 4y = 4 e−t , y(0) = 2, y (0) = −3 Hint, Solution 1503 31.7 Hints Hint 31.1 Use the differentiation and integration properties of the Laplace transform where appropriate Hint 31.2 Hint 31.3 Hint 31 .4 If... = f (s − c) for s > c + α Solution 31 .4 First consider the Laplace transform of t0 f (t) ˆ L[t0 f (t)] = f (s) 1510 Now consider the Laplace transform of tn f (t) for n ≥ 1 ∞ e−st tn f (t) dt n L[t f (t)] = 0 d ∞ −st n−1 e t f (t) dt ds 0 d = − L[tn−1 f (t)] ds =− Thus we have a difference equation for the Laplace transform of tn f (t) with the solution dn L[t f (t)] = (−1) L[t0 f (t)] for n ∈ Z0+ ,... + e−sT for (s) > 0 ˆ Consider f (s) = s−1 tanh(sT /2) esT /2 − e−sT /2 esT /2 + e−sT /2 1 − e−sT = s−1 1 + e−sT s−1 tanh(sT /2) = s−1 We have T f (t) e−st dt = g (s) ≡ ˆ 0 15 14 1 − e−st s n By inspection we see that this is satisfied for f (t) = 1 for 0 < t < T We conclude: 1 for t ∈ [2nT (2n + 1)T ), −1 for t ∈ [(2n + 1)T (2n + 2)T ), f (t) = where n ∈ Z Solution 31 .9 The Laplace transform of... ln t] = e−st tν ln t dt 0 ˆ This integral defines f (s) for (s) > 0 Note that the integral converges uniformly for we can interchange differentiation and integration ∞ ˆ f (s) = 0 ˆ Since f (s) also exists for ∂ −st ν e t ln t dt = − ∂s ˆ (s) > 0, f (s) is analytic in that domain 15 19 ∞ t e−st tν Log t dt 0 (s) ≥ c > 0 On this domain Let σ be real and positive We make the change of variables x = σt ˆ... Laplace transform for all s in the right half plane L[tν ln t] = 1 sν+1 Γ(ν + 1) (ψ(ν + 1) − ln s) 1520 for (s) > 0 For the case ν = 0, we have 1 Γ(1) (ψ(1) − ln s) s1 −γ − ln s L[ln t] = , s L[ln t] = where γ is Euler’s constant ∞ e−x ln x dx = 0.57721566 29 γ= 0 Solution 31.12 Method 1 We factor the denominator ˆ f (s) = 1 1 = 2 + 1) (s − 2)(s (s − 2)(s − ı)(s + ı) We expand the function in partial... L[sin(t)] (s + 2 )2 + 1 − 4 We use a table of Laplace transforms to find the inverse Laplace transform of the first term L−1 1 (s + 2 )2 + 1 − 2 = 1 1− 4 2 e− t/2 sin 1− 2 4 We define 2 α= 1− 4 to get rid of some clutter Now we apply the convolution theorem to invert t y(t) = 0 y(t) = e− t/2 y s ˆ 1 − τ /2 e sin (ατ ) sin(t − τ ) dτ α 1 cos (αt) + The solution is plotted in Figure 31.5 for = 0.05 2 2 Evaluate... a2 1507 for (s) > 0 (s) > 0 3 d sin(at) dt a sin(at) = sL − sin(0) a L[cos(at)] = L L[cos(at)] = s2 s + a2 for (s) > 0 4 ∞ e−st sinh(at) dt L[sinh(at)] = 0 1 ∞ (−s+a)t e = − e(−s−a)t dt 2 0 ∞ 1 − e(−s+a)t e(−s−a)t = + for 2 s−a s+a 0 1 1 1 − = 2 s−a s+a L[sinh(at)] = s2 a − a2 for (s) > | (a)| (s) > | (a)| 5 d sinh(at) dt a sinh(at) = sL − sinh(0) a L[cosh(at)] = L L[cosh(at)] = s2 s − a2 1508 for (s)... 31.2 Hint 31.3 Hint 31 .4 If the integral is uniformly convergent and ∂g ∂s d ds is continuous then b b g(s, t) dt = a a ∂ g(s, t) dt ∂s Hint 31.5 ∞ e−tx dt = s 1 −sx e x Hint 31.6 Use integration by parts Hint 31.7 ∞ (n+1)T ∞ e 0 −st e−st f (t) dt f (t) dt = n=0 15 04 nT The sum can be put in the form of a geometric series ∞ αn = n=0 1 , 1−α Hint 31.8 Hint 31 .9 Write the answer in terms of the Gamma function... Hint 31.13 Hint 31. 14 Hint 31.15 Hint 31.16 1505 for |α| < 1 Hint 31.17 Hint 31.18 Hint 31. 19 Hint 31.20 Hint 31.21 Hint 31.22 Hint 31.23 1506 31.8 Solutions Solution 31.1 1 ∞ e−st eat dt L eat = 0 ∞ e−(s−a)t dt = 0 e−(s−a)t = − s−a L eat = 1 s−a ∞ for (s) > (a) 0 for (s) > (a) 2 ∞ e−st sin(at) dt L[sin(at)] = 0 1 ∞ (−s+ıa)t e = − e(−s−ıa)t dt ı2 0 ∞ 1 − e(−s+ıa)t e(−s−ıa)t + , for = ı2 s − ıa s +... sin (αt) − cos t 2α 4 t 15 10 5 20 40 60 80 100 -5 -10 -15 Figure 31.5: The Weakly Damped, Driven Oscillator Solution 31. 14 We consider the solutions of y − ty + y = 0, y(0) = 0, y (0) = 1 which are of exponential order α for any α > 0 We take the Laplace transform of the differential equation d (sˆ) + y = 0 y ˆ ds 2 1 y + s+ ˆ y= ˆ s s 2 /2 e−s 1 y (s) = 2 + c 2 ˆ s s s2 y − 1 + ˆ 15 24 We use that y (s) . answer for ν = 0? Hint, Solution 1 49 9 Exercise 31.12 Find the inverse Laplace transform of ˆ f(s) = 1 s 3 − 2s 2 + s − 2 with the following methods. 1. Expand ˆ f(s) using partial fractions and. transform methods and show that i 1 = E 0 2R + E 0 2ωL e −αt sin(ωt) where α = R 2L and ω 2 = 2 LC − α 2 . Hint, Solution Exercise 31.23 Solve the initial value problem, y  + 4y  + 4y = 4 e −t ,. c) for s > c + α. 1 49 7 Hint, Solution Exercise 31 .4 Show that L[t n f(t)] = (−1) n d n ds n [ ˆ f(s)] for n = 1, 2, . . . Hint, Solution Exercise 31.5 Show that if  β 0 f(t) t dt exists for