Laplace Transform Integration (ACM 40520)

Transcription

1 Laplace Transform Integration (ACM 40520) Peter Lynch School of Mathematical Sciences

2 Outline Basic Theory Residue Theorem Ordinary Differential Equations General Vector NWP Equation Lagrangian Formulation

3 Outline Basic Theory Residue Theorem Ordinary Differential Equations General Vector NWP Equation Lagrangian Formulation

4 The Laplace Transform: Definition For a function of time f (t), the LT is defined as ˆf (s) = e st f (t) dt. 0 Here, s is complex and ˆf (s) is a complex function of s.

5 The Laplace Transform: Definition For a function of time f (t), the LT is defined as ˆf (s) = e st f (t) dt. 0 Here, s is complex and ˆf (s) is a complex function of s. The inversion from ˆf (s) back to f (t) is f (t) = 1 e st ˆf (s) ds. 2πi C 1 where C 1 is a contour in the s-plane, parallel to the imaginary axis, to the right of all singularities of ˆf (s).

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7 Integral Transforms The LT is one of a large family of integral transforms They can be defined as F(s) = R K (s, t) f (t) dt where R is the range of f (t) and K (s, t) is called the kernel of the transform.

8 Integral Transforms The LT is one of a large family of integral transforms They can be defined as F(s) = R K (s, t) f (t) dt where R is the range of f (t) and K (s, t) is called the kernel of the transform. For example, the Fourier transform is f (ω) = e iωt f (t) dt f (t) = 1 2π e +iωt f (ω) dω

9 Integral Transforms The LT is one of a large family of integral transforms They can be defined as F(s) = R K (s, t) f (t) dt where R is the range of f (t) and K (s, t) is called the kernel of the transform. For example, the Fourier transform is f (ω) = e iωt f (t) dt f (t) = 1 2π e +iωt f (ω) dω The Hilbert transform is another... and many more.

10 The LT is a linear operator Therefore L{αf (t)} = L{f (t)} = ˆf (s) e st αf (t) dt = α e st f (t) dt. e st f (t) dt = αl{f (t)}.

11 The LT is a linear operator Therefore L{αf (t)} = Also L{f (t)} = ˆf (s) L{f (t)+g(t)} = e st αf (t) dt = α 0 e st f (t) dt. e st f (t) dt = αl{f (t)}. e st [f (t)+g(t)] dt = L{f (t)}+l{g(t)}.

12 The LT is a linear operator Therefore L{αf (t)} = Also L{f (t)} = ˆf (s) L{f (t)+g(t)} = Therefore e st αf (t) dt = α 0 e st f (t) dt. e st f (t) dt = αl{f (t)}. e st [f (t)+g(t)] dt = L{f (t)}+l{g(t)}. L{αf (t) + βg(t)} = αl{f (t)} + βl{g(t)}.

13 Basic Properties of the LT L{a} = a/s L{exp(at)} = 1/(s a)

14 Basic Properties of the LT L{a} = a/s L{exp(at)} = 1/(s a) L{exp(iωt)} = 1/(s iω)

15 Basic Properties of the LT L{a} = a/s L{exp(at)} = 1/(s a) L{exp(iωt)} = 1/(s iω) L{sin at} = a/(s 2 + a 2 ) L{cos at} = s/(s 2 + a 2 )

16 Basic Properties of the LT L{a} = a/s L{exp(at)} = 1/(s a) L{exp(iωt)} = 1/(s iω) L{sin at} = a/(s 2 + a 2 ) L{cos at} = s/(s 2 + a 2 ) L{df /dt} = sˆf (s) f (0)

17 Basic Properties of the LT L{a} = a/s L{exp(at)} = 1/(s a) L{exp(iωt)} = 1/(s iω) L{sin at} = a/(s 2 + a 2 ) L{cos at} = s/(s 2 + a 2 ) L{df /dt} = sˆf (s) f (0) All these results can be demonstrated immediately by using the definition of the Laplace transform L{f (t)}.

18 Outline Basic Theory Residue Theorem Ordinary Differential Equations General Vector NWP Equation Lagrangian Formulation

19 A Simple Oscillation Let f (t) have a single harmonic component f (t) = α exp(iωt)

20 A Simple Oscillation Let f (t) have a single harmonic component f (t) = α exp(iωt) Exercise: Show that the LT of f (t) is ˆf (s) = α s iω, a holomorphic function with a simple pole at s = iω.

21 A Simple Oscillation Let f (t) have a single harmonic component f (t) = α exp(iωt) Exercise: Show that the LT of f (t) is ˆf (s) = α s iω, a holomorphic function with a simple pole at s = iω. A pure (monochrome) oscillation in time transforms to a function with a single pole.

22 A Simple Oscillation Let f (t) have a single harmonic component f (t) = α exp(iωt) Exercise: Show that the LT of f (t) is ˆf (s) = α s iω, a holomorphic function with a simple pole at s = iω. A pure (monochrome) oscillation in time transforms to a function with a single pole. The position of the pole is determined by the frequency of the oscillation.

23 Again ˆf (s) = L{α exp(iωt)} = α s iω.

24 Again ˆf (s) = L{α exp(iωt)} = α s iω. The inverse transform of ˆf (s) is f (t) = 1 e st 1 ˆf (s) ds = 2πi C 1 2πi α exp(st) C 1 s iω ds.

25 Again ˆf (s) = L{α exp(iωt)} = α s iω. The inverse transform of ˆf (s) is f (t) = 1 e st 1 ˆf (s) ds = 2πi C 1 2πi α exp(st) C 1 s iω ds. We augment C 1 by a semi-circular arc C 2 in the left half-plane. Denote the resulting closed contour by C 0 = C 1 C 2.

26 Again ˆf (s) = L{α exp(iωt)} = α s iω. The inverse transform of ˆf (s) is f (t) = 1 e st 1 ˆf (s) ds = 2πi C 1 2πi α exp(st) C 1 s iω ds. We augment C 1 by a semi-circular arc C 2 in the left half-plane. Denote the resulting closed contour by C 0 = C 1 C 2. If it can be shown that this leaves the value of the integral unchanged, the inverse is an integral around a closed contour.

27 Again ˆf (s) = L{α exp(iωt)} = α s iω. The inverse transform of ˆf (s) is f (t) = 1 e st 1 ˆf (s) ds = 2πi C 1 2πi α exp(st) C 1 s iω ds. We augment C 1 by a semi-circular arc C 2 in the left half-plane. Denote the resulting closed contour by C 0 = C 1 C 2. If it can be shown that this leaves the value of the integral unchanged, the inverse is an integral around a closed contour. See textbook of Doetsch

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29 For an integral around a closed contour, f (t) = 1 α exp(st) 2πi C 0 s iω ds, we can apply the residue theorem:

30 For an integral around a closed contour, f (t) = 1 α exp(st) 2πi C 0 s iω ds, we can apply the residue theorem: f (t) = C 0 [Residues of ( )] α exp(st) s iω so f (t) is the sum of the residues of the integrand within C 0.

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32 Again f (t) = C 0 [Residues of ( )] α exp(st) s iω

33 Again f (t) = C 0 [Residues of ( )] α exp(st) s iω There is just one pole, at s = iω. The residue is ( ) α exp(st) lim (s iω) = α exp(iωt) s iω s iω

34 Again f (t) = C 0 [Residues of ( )] α exp(st) s iω There is just one pole, at s = iω. The residue is ( ) α exp(st) lim (s iω) = α exp(iωt) s iω s iω So we recover the input function: f (t) = α exp(iωt)

35 A Two-Component Oscillation Let f (t) have two harmonic components f (t) = a exp(iωt) + A exp(iωt) ω Ω

36 A Two-Component Oscillation Let f (t) have two harmonic components f (t) = a exp(iωt) + A exp(iωt) ω Ω The LT is a linear operator, so the transform of f (t) is ˆf (s) = a s iω + A s iω, which has two simple poles, at s = iω and s = iω.

37 A Two-Component Oscillation Let f (t) have two harmonic components f (t) = a exp(iωt) + A exp(iωt) ω Ω The LT is a linear operator, so the transform of f (t) is ˆf (s) = a s iω + A s iω, which has two simple poles, at s = iω and s = iω. The LF pole, at s = iω, is close to the origin. The HF pole, at s = iω, is far from the origin.

38 Again ˆf (s) = a s iω + A s iω.

39 Again ˆf (s) = a s iω + A s iω. The inverse transform of ˆf (s) is f (t) = 1 a exp(st) 2πi C 0 s iω ds + 1 A exp(st) 2πi C 0 s iω ds = a exp(iωt) + A exp(iωt).

40 Again ˆf (s) = a s iω + A s iω. The inverse transform of ˆf (s) is f (t) = 1 a exp(st) 2πi C 0 s iω ds + 1 A exp(st) 2πi C 0 s iω ds = a exp(iωt) + A exp(iωt). We now replace C 0 by a circular contour C centred at the origin, with radius γ such that ω < γ < Ω.

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42 Again: We replace C 0 by C with ω < γ < Ω.

43 Again: We replace C 0 by C with ω < γ < Ω. We denote the modified operator by L.

44 Again: We replace C 0 by C with ω < γ < Ω. We denote the modified operator by L. Since the pole s = iω falls within the contour C, it contributes to the integral. Since the pole s = iω falls outside the contour C, it makes no contribution.

45 Again: We replace C 0 by C with ω < γ < Ω. We denote the modified operator by L. Since the pole s = iω falls within the contour C, it contributes to the integral. Since the pole s = iω falls outside the contour C, it makes no contribution. Therefore, f (t) L {ˆf (s)} = 1 2πi a exp(st) C s iω ds = a exp(iωt).

46 Again: We replace C 0 by C with ω < γ < Ω. We denote the modified operator by L. Since the pole s = iω falls within the contour C, it contributes to the integral. Since the pole s = iω falls outside the contour C, it makes no contribution. Therefore, f (t) L {ˆf (s)} = 1 2πi a exp(st) C s iω ds = a exp(iωt). We have filtered f (t): the function f (t) is the LF component of f (t). The HF component is gone.

47 Exercise Consider the test function f (t) = α 1 cos(ω 1 t ψ 1 ) + α 2 cos(ω 2 t ψ 2 ) ω 1 < ω 2

48 Exercise Consider the test function f (t) = α 1 cos(ω 1 t ψ 1 ) + α 2 cos(ω 2 t ψ 2 ) ω 1 < ω 2 Show that the LT is [ α 1 e iψ 1 ] ˆf (s) = + eiψ1 2 s iω 1 s + iω 1 + α 2 2 [ e iψ 2 s iω 2 + eiψ2 s + iω 2 ]

49 Exercise Consider the test function f (t) = α 1 cos(ω 1 t ψ 1 ) + α 2 cos(ω 2 t ψ 2 ) ω 1 < ω 2 Show that the LT is [ α 1 e iψ 1 ] ˆf (s) = + eiψ1 2 s iω 1 s + iω 1 + α 2 2 [ e iψ 2 s iω 2 + eiψ2 s + iω 2 ] Show how, by choosing C with ω 1 < γ < ω 2, the HF component can be eliminated.

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51 Outline Basic Theory Residue Theorem Ordinary Differential Equations General Vector NWP Equation Lagrangian Formulation

52 Applying LT to an ODE We consider a nonlinear ordinary differential equation dw dt + iωw + n(w) = 0 w(0) = w 0

53 Applying LT to an ODE We consider a nonlinear ordinary differential equation dw dt + iωw + n(w) = 0 w(0) = w 0 The LT of the equation is (sŵ w 0 ) + iωŵ + n 0 s = 0. We have frozen n(w) at its initial value n 0 = n(w 0 ).

54 Applying LT to an ODE We consider a nonlinear ordinary differential equation dw dt + iωw + n(w) = 0 w(0) = w 0 The LT of the equation is (sŵ w 0 ) + iωŵ + n 0 s = 0. We have frozen n(w) at its initial value n 0 = n(w 0 ). We can immediately solve for the transform solution: ŵ(s) = 1 [ w 0 n ] 0 s + iω s

55 Using partial fractions, we write the transform as ( ) w0 ŵ(s) = + n ( 0 1 s + iω iω s 1 ) s + iω There are two poles, at s = iω and at s = 0.

56 Using partial fractions, we write the transform as ( ) w0 ŵ(s) = + n ( 0 1 s + iω iω s 1 ) s + iω There are two poles, at s = iω and at s = 0. The pole at s = 0 always falls within the contour C. The pole at s = iω may or may not fall within C.

57 Using partial fractions, we write the transform as ( ) w0 ŵ(s) = + n ( 0 1 s + iω iω s 1 ) s + iω There are two poles, at s = iω and at s = 0. The pole at s = 0 always falls within the contour C. The pole at s = iω may or may not fall within C. Thus, the solution is ( w 0 n ) 0 w (t) = iω exp( iωt) + n 0 iω : ω < γ n 0 iω : ω > γ

58 Again, ( w 0 n 0 w (t) = iω ) exp( iωt) + n 0 iω : ω < γ n 0 : ω > γ iω

59 Again, ( w 0 n 0 w (t) = iω ) exp( iωt) + n 0 iω : ω < γ n 0 : ω > γ iω So we see that, for a LF oscillation ( ω < γ), the solution w (t) is the same as the full solution w(t) of the ODE.

60 Again, ( w 0 n 0 w (t) = iω ) exp( iωt) + n 0 iω : ω < γ n 0 : ω > γ iω So we see that, for a LF oscillation ( ω < γ), the solution w (t) is the same as the full solution w(t) of the ODE. For a HF oscillation ( ω > γ), the solution contains only a constant term.

61 Again, ( w 0 n 0 w (t) = iω ) exp( iωt) + n 0 iω : ω < γ n 0 : ω > γ iω So we see that, for a LF oscillation ( ω < γ), the solution w (t) is the same as the full solution w(t) of the ODE. For a HF oscillation ( ω > γ), the solution contains only a constant term. Thus, high frequencies are filtered out.

62 Outline Basic Theory Residue Theorem Ordinary Differential Equations General Vector NWP Equation Lagrangian Formulation

63 A General Vector Equation We write the general NWP equations symbolically as dx dt + i LX + N(X) = 0 where X(t) is the state vector at time t.

64 A General Vector Equation We write the general NWP equations symbolically as dx dt + i LX + N(X) = 0 where X(t) is the state vector at time t. We apply the Laplace transform to get (s ˆX X 0 ) + i LˆX + 1 s N 0 = 0 where X 0 is the initial value of X and N 0 = N(X 0 ) is held constant at its initial value.

65 A General Vector Equation We write the general NWP equations symbolically as dx dt + i LX + N(X) = 0 where X(t) is the state vector at time t. We apply the Laplace transform to get (s ˆX X 0 ) + i LˆX + 1 s N 0 = 0 where X 0 is the initial value of X and N 0 = N(X 0 ) is held constant at its initial value. The frequencies are entangled. How do we proceed?

66 Eigenanalysis Ẋ + i LX + N(X) = 0

67 Eigenanalysis Ẋ + i LX + N(X) = 0 Assume the eigenanalysis of L is LE = EΛ where Λ = diag(λ 1,..., λ N ) and E = (e 1,..., e N ).

68 Eigenanalysis Ẋ + i LX + N(X) = 0 Assume the eigenanalysis of L is LE = EΛ where Λ = diag(λ 1,..., λ N ) and E = (e 1,..., e N ). More explicitly, assume that the eigenfrequencies split in two: [ ] ΛY 0 Λ = 0 Λ Z Λ Y Λ Z : Frequencies of rotational modes (LF) : Frequencies of gravity-inertia modes (HF)

69 We define a new set of variables: W = E 1 X.

70 We define a new set of variables: W = E 1 X. Multiplying the equation by E 1 we get E 1 Ẋ + i E 1 L(EE 1 )X + E 1 N(X) = 0

71 We define a new set of variables: W = E 1 X. Multiplying the equation by E 1 we get E 1 Ẋ + i E 1 L(EE 1 )X + E 1 N(X) = 0 This is just Ẇ + i ΛW + E 1 N(X) = 0

72 We define a new set of variables: W = E 1 X. Multiplying the equation by E 1 we get E 1 Ẋ + i E 1 L(EE 1 )X + E 1 N(X) = 0 This is just Ẇ + i ΛW + E 1 N(X) = 0 This equation separates into two sub-systems: Ẏ + i Λ Y Y + N Y (Y, Z) = 0 Ż + i Λ Z Z + N Z (Y, Z) = 0 where W = (Y, Z) T.

73 We define a new set of variables: W = E 1 X. Multiplying the equation by E 1 we get E 1 Ẋ + i E 1 L(EE 1 )X + E 1 N(X) = 0 This is just Ẇ + i ΛW + E 1 N(X) = 0 This equation separates into two sub-systems: where W = (Y, Z) T. Ẏ + i Λ Y Y + N Y (Y, Z) = 0 Ż + i Λ Z Z + N Z (Y, Z) = 0 The variables Y and Z are all coupled through the nonlinear terms N Y (Y, Z) and N Z (Y, Z).

74 We first consider a single component w: ẇ + iλw + n(w) = 0 Note that all other components may occur in the nonlinear term.

75 We first consider a single component w: ẇ + iλw + n(w) = 0 Note that all other components may occur in the nonlinear term. Holding the nonlinear term constant, we get or (sŵ w 0 ) + iλŵ + n 0 s = 0 ŵ(s) = 1 [ w 0 n ] 0 s + iλ s

76 We first consider a single component w: ẇ + iλw + n(w) = 0 Note that all other components may occur in the nonlinear term. Holding the nonlinear term constant, we get or (sŵ w 0 ) + iλŵ + n 0 s = 0 ŵ(s) = 1 [ w 0 n ] 0 s + iλ s This has two poles, at s = 0 and s = iλ: ŵ(s) = w 0 + n 0 /iλ s + iλ n 0/iλ s

77 Again, ŵ(s) = w 0 + n 0 /iλ s + iλ n 0/iλ s

78 Again, ŵ(s) = w 0 + n 0 /iλ s + iλ n 0/iλ s If λ is small, both poles are within C, so ( w (t) = L {ŵ} = w 0 + n ) ( 0 e iλt n0 ) iλ iλ an oscillation with frequency λ.,

79 Again, ŵ(s) = w 0 + n 0 /iλ s + iλ n 0/iλ s If λ is small, both poles are within C, so ( w (t) = L {ŵ} = w 0 + n ) ( 0 e iλt n0 ) iλ iλ an oscillation with frequency λ. If λ is large, the pole at s = iλ falls outside C, and ( w (t) = L n0 ) {ŵ} =. iλ,

80 Again, ŵ(s) = w 0 + n 0 /iλ s + iλ n 0/iλ s If λ is small, both poles are within C, so ( w (t) = L {ŵ} = w 0 + n ) ( 0 e iλt n0 ) iλ iλ an oscillation with frequency λ. If λ is large, the pole at s = iλ falls outside C, and ( w (t) = L n0 ) {ŵ} =. iλ This corresponds to putting ẇ = 0 in the equation: iλw + n 0 = 0,

81 General Solution Method We recall that the Laplace transform of the equation is (s ˆX X 0 ) + i LˆX + 1 s N 0 = 0 where X 0 is the initial value of X and N 0 = N(X 0 ) is held constant at its initial value.

82 General Solution Method We recall that the Laplace transform of the equation is (s ˆX X 0 ) + i LˆX + 1 s N 0 = 0 where X 0 is the initial value of X and N 0 = N(X 0 ) is held constant at its initial value. But now we take n t to be the initial time: (s ˆX X n ) + i LˆX + 1 s Nn = 0

83 General Solution Method We recall that the Laplace transform of the equation is (s ˆX X 0 ) + i LˆX + 1 s N 0 = 0 where X 0 is the initial value of X and N 0 = N(X 0 ) is held constant at its initial value. But now we take n t to be the initial time: (s ˆX X n ) + i LˆX + 1 s Nn = 0 The solution can be written formally: ˆX(s) = (s I + i L) 1 [ X n 1 s Nn ]

84 Again, ˆX(s) = (s I + i L) 1 [ X n 1 s Nn ]

85 Again, ˆX(s) = (s I + i L) 1 [ X n 1 s Nn ] We recover the filtered solution by applying L at time (n + 1) t. X = L {ˆX(s)} t= t

86 Again, ˆX(s) = (s I + i L) 1 [ X n 1 s Nn ] We recover the filtered solution by applying L at time (n + 1) t. X = L {ˆX(s)} t= t The procedure may now be iterated to produce a forecast of any length.

87 Again, ˆX(s) = (s I + i L) 1 [ X n 1 s Nn ] We recover the filtered solution by applying L at time (n + 1) t. X = L {ˆX(s)} t= t The procedure may now be iterated to produce a forecast of any length. Further details are given in Clancy and Lynch, 2011a,b

88 Outline Basic Theory Residue Theorem Ordinary Differential Equations General Vector NWP Equation Lagrangian Formulation

89 Lagrangian Formulation We now consider how to combine the Laplace transform approach with Lagrangian advection.

90 Lagrangian Formulation We now consider how to combine the Laplace transform approach with Lagrangian advection. The general form of the equation is DX Dt + i LX + N(X) = 0 where advection is now included in the time derivative.

91 Lagrangian Formulation We now consider how to combine the Laplace transform approach with Lagrangian advection. The general form of the equation is DX Dt + i LX + N(X) = 0 where advection is now included in the time derivative. We re-define the Laplace transform to be the integral in time along the trajectory of a fluid parcel: ˆX(s) e st X(t) dt T

92 We consider parcels that arrive at the gridpoints at time (n + 1) t. They originate at locations not corresponding to gridpoints at time n t.

93 We consider parcels that arrive at the gridpoints at time (n + 1) t. They originate at locations not corresponding to gridpoints at time n t. We denote the value at an arrival point by X n+1 A. The value at the departure point is X n D.

94 We consider parcels that arrive at the gridpoints at time (n + 1) t. They originate at locations not corresponding to gridpoints at time n t. We denote the value at an arrival point by X n+1 A. The value at the departure point is X n D. The initial values when transforming the Lagrangian time derivatives are X n D.

95 We consider parcels that arrive at the gridpoints at time (n + 1) t. They originate at locations not corresponding to gridpoints at time n t. We denote the value at an arrival point by X n+1 A. The value at the departure point is X n D. The initial values when transforming the Lagrangian time derivatives are X n D. The equations thus transform to (s ˆX X n D) + i LˆX + 1 s Nn+ 1 2 M = 0 where we evaluate nonlinear terms at a mid-point, interpolated in space and extrapolated in time.

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97 The solution can be written formally: ˆX(s) = (s I + i L) 1 [X n D 1 s Nn+ 1 2 M ]

98 The solution can be written formally: ˆX(s) = (s I + i L) 1 [X n D 1 s Nn+ 1 2 M ] The values at the departure point and mid-point are computed by interpolation.

99 The solution can be written formally: ˆX(s) = (s I + i L) 1 [X n D 1 s Nn+ 1 2 M ] The values at the departure point and mid-point are computed by interpolation. We recover the filtered solution by applying L at time (n + 1) t, or t after the initial time. X = L {ˆX(s)} t= t

100 The solution can be written formally: ˆX(s) = (s I + i L) 1 [X n D 1 s Nn+ 1 2 M ] The values at the departure point and mid-point are computed by interpolation. We recover the filtered solution by applying L at time (n + 1) t, or t after the initial time. X = L {ˆX(s)} t= t The procedure may now be iterated to produce a forecast of any length.

101 The solution can be written formally: ˆX(s) = (s I + i L) 1 [X n D 1 s Nn+ 1 2 M ] The values at the departure point and mid-point are computed by interpolation. We recover the filtered solution by applying L at time (n + 1) t, or t after the initial time. X = L {ˆX(s)} t= t The procedure may now be iterated to produce a forecast of any length. Further details are given in Clancy and Lynch, 2011a,b