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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