Advanced Engineering Mathematics

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1 Advanced Engineering Mathematics Note 6 Laplace Transforms CHUNG, CHIH-CHUNG

2 Outline Introduction & Partial Fractions Laplace Transform. Linearity. First Shifting Theorem (s-shifting) Transforms of Derivatives and Integrals. ODEs (Ch6.2) (Ch6.1) Unit Step Function (Heaviside Function). Second Shifting Theorem (t-shifting) (Ch6.3) Short Impulses. Dirac s Delta Function. Partial Fractions (Ch6.4) 2

3 Outline Convolution. Integral Equations (Ch6.5) Differentiation and Integration of Transforms. ODEs with Variable Coefficients (Ch6.6) Systems of ODEs (Ch6.7) Laplace Transform:General Formulas (Ch6.8) 3

4 L {e at (f)}=f(s-a) L (f )=s L (f) f(0) L (f )=s 2 L (f) sf(0)-f (0) L { f(t-a) u(t-a) } = e -as F(s)

5 Introduction The process of solving an ODE using the Laplace transform method consists of three steps, shown schematically Introduction 5

6 Introduction y(t) Y(s) y(t) Y(s) Introduction 6

7 Introduction key motivation for learning about Laplace transforms Solving an ODE is simplified to an algebraic problem Problems are solved more directly Use of the unit step function (Heaviside function in Sec. 6.3) and Dirac s delta (in Sec. 6.4) make the method particularly powerful for problems with inputs (driving forces) that have discontinuities or represent short impulses or complicated periodic functions. Introduction 7

8 Partial Fractions A/B A and B are polynomials The degree of A is less than the degree of B In case of the degree of A is less than the degree of B Partial Fractions 8

9 Partial Fractions 分母 B 的組合類別 : unrepeated real factor s a i unrepeated complex conjugate factor [s (α+β i )][s (α β i )] = s 2 2αs + (α 2 +β 2 ) repeated real factor (s a i ) 2, (s a i ) 3, etc. repeated complex conjugate [s 2 2αs + (α 2 +β 2 )] 2, [s 2 2αs + (α 2 +β 2 )] 3, etc. Partial Fractions 9

10 Partial Fractions unrepeated real factor s a i 同乘左式分母 1 = A 1 (s+1)(s+2) + A 2 s(s+2) + A 3 s(s+1) s = 0 1 = A 1 (0+1)(0+2) A 1 = ½ = 0.5 s = 1 1 = A 2 ( 1)( 1+2) A 2 = 1 s = 2 1 = A 3 ( 2)( 2+1) A 3 = ½ = 0.5 Partial Fractions 10

11 Partial Fractions unrepeated complex conjugate factor s 2 +4 =(s 2i)(s+2i), s 2 +2s+2=[s+(1 i)][s+(1+ i)] 同乘左式分母 1 = 20 = (As+B)(s 2 +2s+2) + (Ms+N)(s 2 +4) s 3 : 0 = A + M, s 2 : 0 = 2A + B + N s 1 : 0 = 2A + 2B + 4M, s 0 : 20 = 2B + 4N A = 2, B = 2, M = 2, N = 6 Partial Fractions 11

12 Partial Fractions repeated real factor 同乘左式分母 s 3 4s = A 2 (s 2)(s 1) + A 1 s (s 2)(s 1) + Bs 2 (s 1) + Cs 2 (s 2) unrepeated factor s= = B 4 1, 4 = 4B B = 1 s= = C 1 ( 1), 1 = C C = 1 Partial Fractions 12

13 unrepeated factor s 3 4s = A 2 (s 2)(s 1) + A 1 s (s 2)(s 1) + Bs 2 (s 1) + Cs 2 (s 2) Let s= = A2 ( 2) ( 1), 4 = 2A2 A2 = 2 A1 Solution 1 : 將上式微分 3s 2 8s = A 2 [(s 1)+(s 2)]+ A 1 [(s 2)(s 1)+s(s 1)+s(s 2)] + B[2s (s 1)+s 2 ] + C[2s(s 2) +s 2 ] s= 0 0 = A 2 ( 3) + A 1 2 A 1 = 3 Partial Fractions 13

14 A1 Solution 2 : 將上式同乘 s Let s 1 = A 1 + B + C = A 1 + ( 1) + ( 1) A 1 = 3 Partial Fractions 14

15 s 3 4s = A 2 (s 2)(s 1) + A 1 s (s 2)(s 1) + Bs 2 (s 1) + Cs 2 (s 2) A1 Solution 3 : 找一適當數字 ( 0,1,2), 例如 s = = A A B 18 + C 9 A 1 = 3 Partial Fractions 15

16 Partial Fractions repeated complex conjugate Ex. [s 2 2αs + (α 2 +β 2 )] 2 Needs partial fractions Ex. [s 2 2αs + (α 2 +β 2 )] 3 Needs partial fractions Partial Fractions 16

17 Laplace Transform. Linearity. First Shifting Theorem (s-shifting) Let f(t) be a given function that is defined for all t 0 Laplace transform: F(s) = L (f)= 0 e st f () t dt. Inverse transform: Note: ƒ(t) is such that the integral exists f(t) = L 1 (F) Ch6.1 Laplace Transform 17

18 Ex1:f(t)=1 for t 0, find F(s) F(s) = L (f)= e f ( t) dt = 0 e 1dt 1 e s st st st 1 0 = 0 = s (s>0) Ex2:f(t)= e at for t 0, where a is a constant F(s) = L (f)= st e f ( t) dt = 0 0 e st e at dt = 1 a s e ( s a) t 0 = s 1 a (s-a>0) Ch6.1 Laplace Transform 18

19 Ch6.1 Laplace Transform 19

20 Theorem 1 Linearity of the Laplace Transform L {af(t) + bg(t)}= al {f(t)} + bl {g(t)} Proof: L -1 {af(t) + bg(t)}= al -1 {f(t)} + bl -1 {g(t)} Ch6.1 Laplace Transform 20

21 Ex. Find the transforms of cosh(at) and sinh(at) cosh(at) = (e at + e at )/2, sinh(at) = (e at e at )/2 Ch6.1 Laplace Transform 21

22 Ex. Transform of t a (a>0) F(s) = L (f)= st = st a e f ( t) dt 0 e t dt 0 let st = x a x x dx 1 = x a e + e x dx = 0 a 1 0 s s s 1 s a+ 1 0 e x x ( a+ 1) 1 dx 1 = Γ a + s a ( 1) 1 + (Note: a!= Γ( a +1) ) Ch6.1 Laplace Transform 22

23 Ex. Transform of t n 已知 F(s) = L (f)= ( a 1) st a Γ + e t dt = 0 a+ 1 for integer a = n 0 L (t n )= Induction hypothesis( 歸納法 ) n=0 L (t n )= L (1)=0!/s 0+1 OK. s n! n+ s 1 n=n L (t n )=n!/s n+1 也成立 for n+1 L (t n+1 )= 0 e st t n + 1 = s n st n+ 1 1 dt = e t ( n + 1) s 0 s ( n + 1)! L (t n )= n+ 2 也成立 s 0 e st t n dt 23 Ch6.1 Laplace Transform 22

24 Theorem 2 - First Shifting Theorem Replacing s by s a in the Transform If ƒ(t) has the transform F(s) (where s > k for some k), then e at ƒ(t) has the transform F(s a) proof L {e at (f)}=f(s-a) e at (f) = L -1 {F(s-a)} Ch6.1 Laplace Transform 24

25 Ex. Find (f) Ch6.1 Laplace Transform

26 Existence and Uniqueness of Transforms Growth restriction A function f(t) has a Laplace transform if it does not grow too fast, say, if for all t 0 and some constant M and k, it satisfies the Growth restriction f(t) Me kt sometime called Growth of exponential order ƒ(t) need not be continuous, but it should not be too bad. The technical term is piecewise continuity ƒ(t) is piecewise continuous on a finite interval a t b Ch6.1 Laplace Transform 26

27 This then gives finite jumps as in Fig. 115 as the only possible discontinuities, but this suffices in most applications, and so does the following theorem. Ch6.1 Laplace Transform 27

28 Theorem 3: Existence Theorem for Laplace Transforms If f(t) is defined and piecewise continuous on every finite interval on the semi-axis t 0 and satisfies growth restriction for all t 0 and some constants M and k, f(t) Me kt then the Laplace transform L (f) exists for all s > k. Ch6.1 Laplace Transform 28

29 Uniqueness If the Laplace transform of a given function exists, it is unique. The inverse of a given transform is unique too. If the two continuous functions have the same transform, they are completely identical. If the two functions have the same transform, these functions cannot differ over an interval of positive length, although they may differ at isolated points. Ch6.1 Laplace Transform 29

30 Transforms of Derivatives and Integrals Theorem 1 The transforms of the first and second derivatives of ƒ(t) satisfy L (f )=s L (f) f(0) L (f )=s 2 L (f) sf(0)-f (0) (1) (2) f(t) Me kt (1) holds if ƒ(t) is continuous for all t 0 and satisfies the growth restriction (2) in Sec. 6.1 and ƒ'(t) is piecewise continuous on every finite interval on the semi-axis t 0. (2) holds if ƒ and ƒ' are continuous for all t 0 and satisfy the growth restriction and ƒ" is piecewise continuous on every finite interval on the semi-axis t 0. Ch 6.2 Transforms of Derivatives and Integrals 30

31 Proof = -f(0)+s L (f) s>k Ch 6.2 Transforms of Derivatives and Integrals 31

32 Theorem 2 Laplace Transform of the Derivative ƒ (n) of any order Let ƒ, ƒ',.,ƒ (n-1) be continuous for all t 0 and satisfy the growth restriction (2) in Sec Furthermore, let ƒ (n) be piecewise continuous on every finite interval on the semi-axis t 0. Then the transform of ƒ (n) satisfies L (f (n) )=s n L (f) s n-1 f(0)-s n-2 f (0)- - f (n-1) (0) Ch 6.2 Transforms of Derivatives and Integrals 32

33 Ex1. ƒ(t) = cosωt, ƒ(0) = 1 ƒ'(t) = - ω sinωt, ƒ'(0) = 0 ƒ"(t) = ω 2 cosωt L (f )=s 2 L (f) sf(0)- f (0) -ω 2 L (cosωt) = s 2 L (cosωt) - s Ch 6.2 Transforms of Derivatives and Integrals 33

34 Ex2. ƒ(t) = sinωt, ƒ(0) = 0 ƒ'(t) = ω cosωt, ƒ'(0) = ω ƒ"(t) = ω 2 sinωt L (f )=s 2 L (f) sf(0)- f (0) -ω 2 L (sinωt) = s 2 L (sinωt) - ω Ch 6.2 Transforms of Derivatives and Integrals 34

35 Ex3. ƒ(t) = t sinωt, ƒ(0) = 0 ƒ'(t) = sinωt + ωtcosωt, ƒ'(0) = 0 ƒ"(t) = 2ωcosωt ω 2 tsinωt L (f )=s 2 L (f) sf(0)- f (0) 2ω L (cosωt) ω 2 L (tsinωt) = s 2 L (f) Ch 6.2 Transforms of Derivatives and Integrals 35

36 Theorem 3 Laplace Transform of Integral Let F(s) denote the transform of a function ƒ(t) which is piecewise continuous for t 0 and satisfies a growth restriction (2), Sec Then, for s > 0, s > k, and t > 0, Ch 6.2 Transforms of Derivatives and Integrals 36

37 Proof set g(t) =, when k>0 Growth restriction f(t) Me kt g(t) satisfies the growth restriction Ch 6.2 Transforms of Derivatives and Integrals 37

38 also g'(t) = f(t), except at points which f(t) is discontinuous, hence g'(t) is piecewise continuous Ch 6.2 Transforms of Derivatives and Integrals 38

39 Ex. find the inverse of Ch 6.2 Transforms of Derivatives and Integrals 39

40 Solve Differential Equations y" + ay' + by = r(t), y(0) = K 0, y'(0) = K 1 Step 1. Setting up the subsidiary equation L ( y" + ay' + by = r ) L (y") + a L (y') + b L (y)= L (r) [s 2 L (y) sy(0) y'(0)] + a[sl (y) y(0)] + b L (y) = L (r) [s 2 Y(s) sy(0) y'(0)] + a[sy(s) y(0)] + by(s) = R(s) (s 2 + as + b)y(s) = (s + a)y(0) + y'(0) + R(s) Ch 6.2 Transforms of Derivatives and Integrals 40

41 Step 2. Solution of the subsidiary equation by algebra =Q(s) Y(s) = [(s + a)y(0) + y'(0) + R(s)] Y(s) = [(s + a)y(0) + y'(0)]q(s) + R(s)Q(s) where Q(s) is called transfer function and Q(s) = if initial conditions y(0) = y'(0) = 0 Y(s) = R(s)Q(s) Q=Y(s)/R(s) Note that Q depends neither on r(t) nor on the initial conditions (but only on a and b). Step 3. Inversion of Y to obtain y(t) = L 1 {Y(s)} 41 Ch 6.2 Transforms of Derivatives and Integrals

42 Ex. y" y = t, y(0) = 1, y'(0) = 1 Step.1 L ( y" y = t ) = L (y") L (y)= L (t) [s 2 L (y) sy(0) y'(0)] L (y) = L (t) [s 2 Y(s) s 1] Y(s) = 1/ s 2 Step.2 Y(s) = Step.3 (s 2 1) Y(s) = s /s 2 s s 1 s ( s 1) = s s 1 s s Ch 6.2 Transforms of Derivatives and Integrals 42

43 Ex. y" + y' + 9y = 0, y(0) = 0.16, y'(0) = 0 Step.1 L ( y" + y' + 9y = 0 ) = L (y") + L (y') + 9L (y) = L (0) [s 2 L (y) sy(0) y'(0)]+[s L (y) y(0)]+9 L (y) = 0 [s 2 Y(s) 0.16s] + [sy(s) 0.16] + 9Y(s) = 0 (s 2 +s +9) Y(s) = 0.16(s+1) Step.2 Ch 6.2 Transforms of Derivatives and Integrals

44 Step.3 Advantages of Laplace transform? Ch 6.2 Transforms of Derivatives and Integrals 44

45 Ex. y" + y = 2t, y(π/4)= π/2, y (π/4)= 2-2 initial conditions given at some t = t 0 > 0 instead of t=0 set t = ṫ+ t 0, so that t = t 0 gives ṫ = 0 and the Laplace transform can be applied set t = ṫ + π /4, so that t = t 0 = π /4 gives ṫ = 0 y" + y = 2t y (ṫ) + y(ṫ) = 2(ṫ + π /4), y(0)= π/2, y (0)= 2-2 L {y (ṫ) + y(ṫ) = 2(ṫ + π /4)} Partial fraction Ch 6.2 Transforms of Derivatives and Integrals 45

46 ṫ =t - π /4 sin(ṫ) = sin(t - π /4) = sint cos(π /4)-cost sin(π /4) = (1/ 2)(sint -cost) y = 2t - sint + cost Ch 6.2 Transforms of Derivatives and Integrals 46

47 Unit Step Function (Heaviside Function). Second Shifting Theorem (t-shifting) This section and the next one are extremely important because we shall now reach the point where the Laplace transform method shows its real power in applications and superiority over the classical approach of Chap. 2. We shall introduce two auxiliary functions, the unit step function or Heaviside function u(t a) (in Sec. 6.3) and Dirac s delta δ(t a) (in Sec. 6.4). Ch 6.3 Unit Step Function 47

48 Unit Step Function 0, if t < a ( t a < 0) u( t a) = ( a 1, if t > a ( t a > 0) 0) Ch 6.3 Unit Step Function 48

49 Laplace Transform of u(t a), u(t) L {u(t a)} = e -as /s, s>0 L {u(t )}= L {u(t 0)} = e -0s /s = 1/s = L (1), s>0 Proof: Ch 6.3 Unit Step Function 49

50 Ch 6.3 Unit Step Function 50

51 Ch 6.3 Unit Step Function

52 Ch 6.3 Unit Step Function 52

53 Theorem 1:Second Shifting Theorem; Time Shifting(t-shifting) If ƒ(t) has the transform F(s), then the shifted function ~ f ( t) = f ( t a) u( t a) = 0 f ( t a) if if t t < a > a ( t a < 0) ( t a > 0) ~ L {f(t)} = L { f(t-a) u(t-a) } = e -as F(s) = e -as L { f(t)} f(t-a) u(t-a) = L -1 {e -as F(s)} ~ Or L {f(t)} = L { f(t) u(t-a) } = e -as L { f(t+a)} Ch 6.3 Unit Step Function 53

54 Proof Set τ + a = t τ = t a, d τ =dt Lower limit τ = 0 t = a Ch 6.3 Unit Step Function 54

55 Ex. L {5sin(t 2)u(t 2)} known L { f(t-a) u(t-a) } = e -as L { f(t)} L {5sin(t 2)u(t 2)} = 5e -2s L { sin(t) } = Ch 6.3 Unit Step Function 55

56 Ex. f ( t) = t 2 2 / 2 cost if 0 < t < 1 if 1< t < π / 2 if t > π / 2 2*[u(t) - u(t-1)] t 2 /2*[u(t-1) - u(t-π/2)] cost*[u(t-π/2)] Ch 6.3 Unit Step Function 56

57 Step 1: f(t) = 2[u(t) u(t 1)] + t 2 /2*[u(t-1) - u(t-π/2)] + cost*[u(t-π/2)] Step 2: apply t-shifting theorem L {2[u(t) u(t 1)]} = L { f(t-a) u(t-a) } = e -as L { f(t)} L {t 2 /2*u(t-1)} = = e -s L (t 2 /2 + t + ½) = e -s ( ) = Ch 6.3 Unit Step Function 57

58 = e -πs/2 L (t 2 /2 + π t/2 + π 2 /8) Ch 6.3 Unit Step Function 58

59 = e -πs/2 L (-sint) Note: cos(t)=cos[(t- π/2)+ π/2] = cos (t- π/2) cos (π/2) - sin (t- π/2) sin (π/2) = - sin(t - π/2) Ch 6.3 Unit Step Function 59

60 Solution 2 Step 1: f(t) = 2[u(t) u(t 1)] + t 2 /2*[u(t-1) - u(t-π/2)] + cost*[u(t-π/2)] Step 2: apply t-shifting theorem L { f(t) u(t-a) } = e -as L { f(t+a)} L {t 2 /2*u(t-1)} = = Ch 6.3 Unit Step Function 60

61 Solution 2 apply t-shifting theorem L { f(t) u(t-a) } = e -as L { f(t+a)} Ch 6.3 Unit Step Function 61

62 Ex. (1) apply t-shifting theorem L { f(t-a) u(t-a) } = e -as L { f(t)} Ch 6.3 Unit Step Function 62

63 (2) apply t-shifting theorem L { f(t-a) u(t-a) } = e -as L { f(t)} Ch 6.3 Unit Step Function 63

64 (3) apply s-shifting theorem L {e at (f)}=f(s-a) apply t-shifting theorem L { f(t-a) u(t-a) } = e -as L { f(t)} Ch 6.3 Unit Step Function 64

65 Note: sin[π(t 1)] = sinπt cosπ cosπt sinπ = sinπt sin[π(t 2)] = sinπt cos2π cosπt sin2π = sinπt Ch 6.3 Unit Step Function 65

66 Short Impulses. Dirac s Delta Function Impulse function f k (t a) = (1/k) [ u(t a) u(t (a+k)) ] k Ch 6.4 Short Impulses Dirac s Delta Function 66

67 Laplace transform of function f k (t a) L {f k (t a)} apply t-shifting theorem L { f(t-a) u(t-a) } = e -as L { f(a)} = L { (1/k) [u(t-a)-u(t-(a+k)] } = Ch 6.4 Short Impulses Dirac s Delta Function 67

68 Dirac delta function δ(t a) (unit impulse function) 68 = = = a a g dt a t t g dt a t t g dt a t ) ( ) ( ) ( ) ( ) ( 1 ) ( δ δ δ Continuous function Ch 6.4 Short Impulses Dirac s Delta Function

69 Laplace transform of Dirac delta function δ(t a) 因 k 0 所以先進行微分 Ch 6.4 Short Impulses Dirac s Delta Function 69

70 Relation of u(t a) and δ(t a) Ch 6.4 Short Impulses Dirac s Delta Function 70

71 L {δ(t a)} = e -as L {u(t a)} = e -as /s = L {δ(t a)} /s By Theorem 3 in Ch. 6.2 Ch 6.4 Short Impulses Dirac s Delta Function 71

72 Ex. y" + 3y' + 2y = r(t) = u(t 1) u(t 2), y(0) = 0, y'(0) = 0 L { y" + 3y' + 2y = u(t 1) u(t 2) } [s 2 Y-sy(0)-y'(0)] + 3[sY-y(0)] + 2Y = (s 2 + 3s + 2)Y = (e -s -e -2s )/s =F(s) (e -s -e -2s ) Ch 6.4 Short Impulses Dirac s Delta Function 72

73 f(t) =L -1 { F(s)} = f(t a) u(t a) = L 1 { e as F(s) } (t-shifting) Y(s) = F(s) (e s e 2s )= e s F(s) e 2s F(s) y(t) = f(t 1) u(t 1) f(t 2) u(t 2) Ch 6.4 Short Impulses Dirac s Delta Function 73

74 Ex. y" + 3y' + 2y=δ(t 1), y(0)=0, y'(0)=0 L { y" + 3y' + 2y=δ(t 1) } [s 2 Y-sy(0)-y'(0)] + 3[sY-y(0)] + 2Y = e s (s 2 + 3s + 2)Y = e s Y ( s) e ( s + 1)( s + 2) 1 s s + 2 s = s = e = F( s ) e s f(t)= L 1 {F(s)} = e t e 2t Ch 6.4 Short Impulses Dirac s Delta Function 74

75 f(t a) u(t a) = L 1 { e as F(s) } (t-shifting) Y(s) = F(s) e s y(t) = f(t 1) u(t 1) Ch 6.4 Short Impulses Dirac s Delta Function 75

76 Ex. y" + 2y' + 2y = r(t),y(0) = 1, y'(0) = 5 r(t)=10sin2t [u(t) u(t π)] L {f(t) u(t a)} = e as L {f(t+a)} (t-shifting) Ch 6.4 Short Impulses Dirac s Delta Function 76

77 L { y" + 2y' + 2y = 10sin2t [u(t) u(t π)] } [s 2 Y-sy(0)-y'(0)]+2[sY-y(0)]+2Y = (1- e πs ) (s 2 + 2s + 2)Y (s 3) = (1- e πs ) Ch 6.4 Short Impulses Dirac s Delta Function 77

78 1st = L 1 {F(s)} = f(t) Ch 6.4 Short Impulses Dirac s Delta Function 78

79 2nd = F(s)e πs Ch 6.4 Short Impulses Dirac s Delta Function 79

80 3rd 80

81 Theorem 1:Laplace Transform of period function ƒ(t):a piecewise continuous function with period p Proof: Ch 6.4 Short Impulses Dirac s Delta Function 81

82 Second Integral set t =τ+ p, so dt = dτ, range p~2p 0~p f(τ + p ) = f(τ) Third Integral set t =τ+ 2p, so dt = dτ, range 2p~3p 0~p f(τ + 2p ) = f(τ) Ch 6.4 Short Impulses Dirac s Delta Function 82

83 Ch 6.4 Short Impulses Dirac s Delta Function 83

84 Ex. Find the Laplace transform of f(t) f(t) = (k/p) t, 0 < t < p f(t+p) = f(t), period = p Ch 6.4 Short Impulses Dirac s Delta Function 84

85 Convolution. Integral Equations (ch 6.5) L {f(t)g(t)} L {f(t)} L {g(t)} ex: f(t) = e t, g(t) = 1, f(t)g(t)= e t Convolution of f(t) and g(t) denoted by f*g( 用 f*g 表示 ) Ch 6.5 Convolution. Integral Equations 85

86 Theorem 1 Convolution Theorem If two functions ƒ(t) and g(t) satisfy the assumption in the existence theorem in Sec. 6.1, so that their transforms F(s) and G(s) exist, the product H(s) = F(s)G(s) is the transform of h(t) given by (1). L {f(t)} = F(s), L {g(t)} = G(s), and H(s)= F(s)G(s) Ch 6.5 Convolution. Integral Equations 86

87 Proof From definition Set t = p + τ p = t τ, dt = dp as p: 0, t: τ Ch 6.5 Convolution. Integral Equations 87

88 and H(s)=F(s)G(s) Ch 6.5 Convolution. Integral Equations 88

89 Ch 6.5 Convolution. Integral Equations 89

90 Properties of Convolution f*1 f Ch 6.5 Convolution. Integral Equations 90

91 Ex. H(s)=1/[s(s a)], h(t)=? H(s)=1/[s(s a)] = 1/(s a) 1/s = F(s) G(s) = L {h(t)} f(t) = L 1 {F(s)} = L 1 {1/(s a)} = e at, g(t) = L 1 {G(s)} = L 1 {1/s} = 1 h(t) = (f*g)(t) = e at *1 = Ch 6.5 Convolution. Integral Equations 91

92 Ex. H(s)=1/(s 2 +ω 2 ) 2, h(t)=? H(s)=1/(s 2 +ω 2 ) 1/(s 2 +ω 2 ) = F(s) G(s) f(t) = g(t) = L 1 {1/(s 2 +ω 2 )} = sinωt / ω h(t) = (f*g)(t) = = = = Ch 6.5 Convolution. Integral Equations 92

93 Ex. y" + ω 02 y = K sin ω 0 t, y(0) = 0, y'(0) = 0 Y(s) = L {y(t)} = L {y" + ω 02 y = K sin ω 0 t } s 2 Y sy(0) y'(0) + ω 2 0 Y = kω 0 /(s 2 + ω 02 ) (s 2 + ω 02 ) Y = kω 0 /(s 2 + ω 02 ) Y = kω 0 /(s 2 + ω 02 ) 2 Ch 6.5 Convolution. Integral Equations 93

94 Nonhomogeneous Linear ODEs : y" + ay' + by = r(t) from Ch6.2 L ( y" + ay' + by = r ) L (y") + a L (y') + b L (y)= L (r) [s 2 L (y) sy(0) y'(0)] + a[sl (y) y(0)] + b L (y) = L (r) [s 2 Y(s) sy(0) y'(0)] + a[sy(s) y(0)] + by(s) = R(s) 1/Q(s) (s 2 + as + b)y(s) = (s + a)y(0) + y'(0) + R(s) as y(0) = y'(0) = 0 Y(s)= R(s)Q(s) r(t) = L 1 {R(s)}, q(t) = L 1 {Q(s)} Ch 6.5 Convolution. Integral Equations 94

95 Ex. y" + 3y' + 2y = r(t) (Ch 6.4) r(t) = 1 if 1 < t < 2 and 0 otherwise, y(0) = y'(0) = 0 L { y" + 3y' + 2y = r(t)} s 2 Y sy(0) y'(0) + 3[sY y(0)] + 2Y = R(s) Y(s) = R(s) / (s 2 +3s+2) = R(s) Q(s) r(t) = u(t 1) u(t 2) = L 1 {R(s)} q(t) = e t e 2t Ch 6.5 Convolution. Integral Equations 95

96 y(t) = (r*q)(t) = = 0 < t <1 Ch 6.5 Convolution. Integral Equations 96

97 1< t < 2 y(t) = Ch 6.5 Convolution. Integral Equations 97

98 t > 2 y(t) = Ch 6.5 Convolution. Integral Equations 98

99 Ex. Volterra integral equation of the second kind =L (y) L (sint) = Y(s) Ch 6.5 Convolution. Integral Equations 99

100 Ex. Volterra integral equation = L (1+t) L (y) = Ch 6.5 Convolution. Integral Equations 100

101 Differentiation and Integration of Transforms (ch 6.6) Differentiation of Transforms L {f(t)} = F(s), or L 1 {F(s)} = f(t) then L {t f(t)} = F'(s), or L 1 {F'(s)} = t f(t) Ch 6.6 Differentiation and Integration of Transforms 101

102 Ex. Ch 6.6 Differentiation and Integration of Transforms 102

103 Sol. L {sinβt} = β/(s 2 +β 2 ) L {t sinβt} = [β/(s 2 +β 2 )]' = 2βs/(s 2 +β 2 ) 2 L {t sinβt/2β} = s/(s 2 +β 2 ) 2 L {cosβt} = s/(s 2 +β 2 ) L {t cosβt} = [s/(s 2 +β 2 )] = Ch 6.6 Differentiation and Integration of Transforms 103

104 Ch 6.6 Differentiation and Integration of Transforms 104

105 Ch 6.6 Differentiation and Integration of Transforms 105

106 Integration of Transforms = = If L 1 {F(s)} = f(t), then Ch 6.6 Differentiation and Integration of Transforms 106

107 Proof. = e -st /t Ch 6.6 Differentiation and Integration of Transforms 107

108 Ex. Find the inverse transform of Solution 1(by differentiation of Transforms ) F(s) = = ln (s 2 +ω 2 ) lns 2 L {t f(t)} = F'(s), or L 1 {F'(s)} = t f(t) L -1 {F'(s)} = 2cosωt 2 = t f(t) f(t) = 2 (cosωt 1) / t Ch 6.6 Differentiation and Integration of Transforms 108

109 Solution 2(by integration of transform) F(s) = f(t) = L -1 {F'(s)} G(s)=F (s) = = g(t) = L -1 {F'(s)} = 2cosωt 2 = 2(cosωt 1) Ch 6.6 Differentiation and Integration of Transforms 109

110 Ch 6.6 Differentiation and Integration of Transforms 110

111 Special Linear ODEs with Variable Coefficients L {y(t)} = Y(s) L {y'} = sy(s) y(0) L {t y'(t)} = = Y(s) s [sy(s) y(0)] L {y''} = s 2 Y(s) sy(0) y'(0) L {t y''(t)} = [s 2 Y(s) sy(0) y'(0)] = 2sY s 2 + y(0) Ch 6.6 Differentiation and Integration of Transforms 111

112 Ex. t 2 y' +y =0, y(0)=0 L {y(t)} = Y(s) L {y'} = sy y(0) L {t y'(t)} = [sy y(0)] = Y s L {t 2 y'(t)} = L {t ty'(t)} = [ Y s ] = 2 + s L { t 2 y' +y =0 } s Y = 0 no advantageous Ch 6.6 Differentiation and Integration of Transforms 112

113 If an ODE has coefficients at 2 + bt + c, then we would get a second- order ODE for Y(s). (no advantageous ) If an ODE has coefficients such as at + b, the subsidiary equation is a first-order ODE for Y(s), which is sometimes simpler than the given second-order ODE. The above shows that Laplace Transform method works well only for rather special ODEs with variable coefficients. (Laplace Transform 僅較適用於某些係數為變數之微分方程式 ) Ch 6.6 Differentiation and Integration of Transforms 113

114 Ex. ty" + (1 t)y' + ny = 0 (Laguerre s ODE) n =1,2,3,... L { ty" + (1 t)y' + ny = ty" + y' ty' + ny = 0 } [ 2sY s 2 + y(0)] + [sy y(0)] ( Y s ) + ny = 0 (s s 2 ) + (n + 1 s)y = 0 Ch 6.6 Differentiation and Integration of Transforms 114

115 兩邊積分 (let const. c=0) lny = n ln(s 1) (n+1) lns = ln(s 1) n + lns (n+1) = Y n n=0, Y = Y0 = 1/s, y0(t) =1 y 0 (t)=1,, y n (t) = n =1,2,3,... Rodrigues s formula Ch 6.6 Differentiation and Integration of Transforms 115

116 s-shifting : L {g(t)}= G(s) L {e at g(t)}= G(s a) Proof: =Y n y n (t) = n =1,2,3,... Thus, L {g(t)} = L {t n } = = G(s) f(t) = e t g(t) = e t t n L {f(t)} = L {e t t n } = G(s+1) = S-shifting Ch 6.6 Differentiation and Integration of Transforms 116

117 f(t) = e t t n f(0) = 0 f'(t) = e t t n + ne t t (n 1) =h(t n, t n-1 ) f'(0) = 0... f (n 1) (t) = h(t n, t n-1, t n-2,.,t) f (n 1)( 0) = 0 Ch 6.6 Differentiation and Integration of Transforms 117

118 L { } = L { } = =R(s) L { } =R(s-1) = = Y(s) S-shifting L 1 {Y(s)} = y n (t) = n =1,2,3,... Ch 6.6 Differentiation and Integration of Transforms 118

119 119

120

121 121

122 e iw =cosw-isinw definition + linearity S-shifting 122

123 Systems of ODEs The Laplace transform method may also be used for solving systems of ODEs a first-order linear system with constant coefficients y' 1 = a 11 y 1 + a 12 y 2 + g 1 (t) y' 2 = a 21 y 1 + a 22 y 2 + g 2 (t)..(1) Ch 6.7 Systems of ODEs 123

124 Y 1 (s) = L {y 1 (t)} = L (y 1 ), Y 2 (s) = L {y 2 (t)} = L (y 2 ), G 1 (s) = L {g 1 (t)} = L (g 1 ), G 2 (s) = L {g 2 (t)} = L (g 2 ) L ( y' 1 = a 11 y 1 + a 12 y 2 + g 1 ) L ( y' 2 = a 21 y 1 + a 22 y 2 + g 2 ) sy 1 y 1 (0) = a 11 Y 1 + a 12 Y 2 + G 1 sy 2 y 2 (0) = a 21 Y 1 + a 22 Y 2 + G 2 (a 11 s)y 1 + a 12 Y 2 = y 1 (0) G 1..(2) a 21 Y 1 + (a 22 s)y 2 = y 2 (0) G 2 Y 1 (s), Y 2 (s) and y 1 (t) = L 1 (Y 1 ), y 2 (t) = L 1 (Y 2 ) Ch 6.7 Systems of ODEs 124

125 By solving this system algebraically for Y 1 (s), Y 2 (s) and taking the inverse transform we obtain the solution y 1 (t) = L 1 (Y 1 ), y 2 (t) = L 1 (Y 2 ) of the given system (1). Note that (1) and (2) may be written in vector form (and similarly for the systems in the examples); thus, setting y(t) = [y1 y2] T, A = [ajk], g(t) = [g1 g2] T, Y(s) = [Y1 Y2] T, G(s) = [G1 G2] T y'(t) = Ay(t) + g(t) and (A si)y(s) = y(0) G(s) Ch 6.7 Systems of ODEs 125

126 y'(t) = Ay(t) + g(t)..(1) Ch 6.7 Systems of ODEs 126

127 (A si)y(s) = y(0) G(s)..(2) Ch 6.7 Systems of ODEs 127

128 Ex. y1(0) = 1, y'1(0) = sqrt(3k) y 2 (0) = 1, y' 2 (0) = - sqrt(3k) y" 1 = ky 1 + k(y 2 y 1 ) y" 2 = k(y 2 y 1 ) ky 2 s 2 Y 1 s sqrt(3k) = KY 1 + K (Y 2 Y 1 ) s 2 Y 2 s (- sqrt(3k))= K(Y 2 Y 1 ) KY 2 Ch 6.7 Systems of ODEs 128

129 (s 2 + 2K)Y 1 KY 2 = s + sqrt(3k) KY 1 + (s 2 + 2K)Y 2 = s sqrt(3k) (s 2 +2k+k)(s 2 +2k-k)= (s 2 +3k)(s 2 +k) Ch 6.7 Systems of ODEs 129

130 Ch 6.7 Systems of ODEs 130

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