Hankel Transforms - Lecture 10

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1 1 Introduction Hankel Transforms - Lecture 1 The Fourier transform was used in Cartesian coordinates. If we have problems with cylindrical geometry we will need to use cylindtical coordinates. Thus suppose the Fourier transform of a function f(x, y) which depends on ρ = (x 2 + y 2 ) 1/2. This is; F(α, β) = 1 2π dx dy f(ρ) e i(αx+βy) Change this to cylindrical coordinates using x = ρ cos(θ) and y = ρ sin(θ). F(α, β) = 1 2π dρ ρ f(ρ) 2π dθ e iρα cos(θ θ ) However the second integral is a representation of the cylindrical Bessel fuction. 2π dθ e iρα cos(θ θ ) = 2πT (αρ) F(α) = dρ ρ f(ρ) J (αρ) There is an inverse which is demonstrated by the completeness to the Bessel functions which can be used to expand the delta function. Thus f(ρ) = dα α F(α) J (αρ) This represents the transform for the th order Bessel function. Higher orders acn be obtained by increasing the dimension of the Fourier transformations. However, generalize this result by looking at the integral equation; f(α) = F(ρ) = ρ dρf(ρ) J n (αρ) α dα F(ρ) J n (αρ) Substitute F(ρ) into the first equation above, interchange the order of integration and use orthogonality to develope the delta function. f(α) = ρ dρ α dα J n (α ρ) J n (αρ) This also allows a similar relation to the Parseval s theorem for the Fourier transformation. 1

2 These are the Hankel transformations and are used for cylindrical geometry problems. α dα g(α)f(α) = ρ dρ G(ρ)f(ρ) 2 Hankel transformations of the derivative Suppose that; F(α) = f(ρ) = Then write; I = ρ dρ f(ρ) J n (αρ) ρ dρ f(ρ) J n (αρ) α dαf(α) J n (αρ) ρ dρ f(ρ) dρ J n(αρ) Integrate by parts. This results in ; I = [ρ df(ρ)j n (αρ)] Use the recurrence relations to write; dρ f(ρ) ρ J n dρ dρ J n dρ = J n + αρj n 1 nj n Assume that [ρ f(ρ)] =. The integral then is; I = (n 1) dρ f J n α ρ dρ f J n 1 For the 2 nd derivative one has provided the surface terms vanish; L = ρ dρ d2 f(ρ) dρ 2 J n (αρ) = dρ df(ρ) dρ ρ J n dρ However, since the Bessel function satisfies the Bessel ode; d dρ [ρ J (αρ)] = (α 2 (n/ρ) 2 )ρ J n (αρ) 2

3 Vo L z Vo y x L Figure 1: The geometry to find the potential outside a cylinder held at potential ±V Integrating each term by parts, Bessel s equation is obtained for f so the result is just, α 2 F n (α). 3 Example We are to find the electric potential outside a cylinder held at a potential V (z, ρ, θ) for L < z < L, see Figure 1. Laplace s equation in cylindrical coordinates is; (1/ρ) ρ [ρ V ρ ] + (1/ρ2 ) 2 V θ V z 2 = Represent the aximuthal dependence by a harmonic series. V V n (ρ, z) sin(nθ) V n = (1/π) 2π V sin(nθ) Choose to represent [ the potential] when ρ = a as; V < θ < π V = V π < θ < 2π 3

4 Then n must be odd and V n = V for L < x < L. This results in a pde for V n. (1/ρ) ρ [ρ V n ρ ] (n/ρ)2 V n + 2 V n z 2 = Apply the Hankel transformation by multiplying by J n and integrating. ρ dρ[ ρ [ρ V n ρ ] (n/ρ)2 V n ]J n (αρ) + 2 V z 2 = V = ρ dρ V n J n The solution is symmetric about z = so only look at the solution for z >. In this case, V n = α dα A(α) e αz J n noi This leads to the dual integral equations; V n = V n ρ = α dα A(α) e αz J n α 2 dα A(α) e αz J n < z < L z > L 4 Example Hydrodynamic Problem For any closed surface in an fluid having no sources of sinks (divergences), then the increase (decrease) in mass is due to the mass flowing into (out of) the volume. Let ρ be the density of the fluid. The total mass is M = ρ dτ. The rate of mass flow is given by; M t = surface ( V ˆn)ρ dσ In the above, V is the fluid velocity, and ˆn is the surface normal. By Gauss law (ρ V )dτ = surface (ρ V )dτ = t ρ dτ ( V ˆn)ρ dσ This gives the equation of continuity. ρ t + (ρ V ) 4

5 Suppose the rate of change of ρ due to convection on the fluid. A element of fluid at r and time t will move to the point, r + V δt at a time t + δt. Thus; δρ = ρ(r + vδt, t + δt) ρ(r, t) Use this to re-write; ρ t + ρ V + V ρ = dρ dt + ρ v = For a perfect fluid v = and the density is constant. Choose a velocity potential, V = φ. Thus Laplace s equation is obtained. 2 φ = Suppose the flow of a perfect fluid through a circular aperature ( a screen). The flow satisfies Laplace s equation and has the boundary conditions; V = g(r) r < a and z = V z = r > a and z = Apply the Hankel transform J (αr) Φ(α) = φ(r) = r dr φ J (αr) α dα ΦJ (αr) Using the form of the transformation for the derivatives, note the boundary conditions must be satisfied; d 2 Φ dz 2 α 2 Φ = This has solution; Φ = A e αz + B e αz Choose B = which results in the dual integral equations. Let a ρ = r and α = u/a. F(u) = ua(u/a) G(ρ) a 2 g(r) 5

6 G(ρ) = du F(u) e αz J (αr) < ρ < 1 = u du F(u) e αz J (αr) ρ > 1 5 Dual integral equations Dual integral equations occur when different boundary conditions need to be applied over different regions of the boundaries. We have seen how they are developed in the above examples. A general form for these equations is ; dy y α f(y) J ν (xy) = g(x) < x < 1 dy f(y) J ν (xy) = x > 1 These equations can be solved in particular cases by change of variable, particularly taking advantage of any symmetries in the problem. More generally, thay can be solved by a Mellin transform, which takes the form; F(s) = f(s) = 1 2π1 dxf(x) x s 1 c+i c i ds F(s) x s The development is complicated, requiring the result to be found by integration in the complex plane. Because of the detail required, and the scope of this class, this is not pursued further here. However, if your need to solve such equations your know where to begin. 6 Laplace Transform It was pointed oout that the Fourier transforms provides a solution for the steady-state problem. However, in many cases we look for transient solutions, so we need a different technique. In addition, the Fourier transformation of f(x) requires that dx f(x) converge. Thus, suppose that we have the condition that f(x) = for x <, and use; [ ] e f(x) γx f(x) x > x < 6

7 In the above γ >. Then look at the Fourier transformation. F(α) = 1 2π f(x) = 1 2π Use these equations to write; Define; f(x) = 1 2π dxe γx f(x) e iαx dα e γx F(α) e iαx dα e γx e iαx p = γ + iα dp = idα φ(p) = f(x) = 1 dx f(x ) e px dp φ(p) e px dx e γx f(x ) e iαx The function φ(p) is the Laplace transform of the function f(x). In almost cases the transform will now converge. Problems have been pushed into finding the inverse transform which requires integration in the complex plane. The inverse transform can be handled in many cases by the convolution theorem. Thus supppose the integrand in the inverse transform is a product of 2 known Laplace transforms. Or; 1 dp φ(p) Ψ(p)e px = 1 dp φ(p) e px g(y) e py 1 dp φ(p) Ψ(p)e px = 1 g(y) Since f(x y) = if (x y) <, the above is written; dp φ(p) e p(x y) 1 dp φ(p) Ψ(p)e px = dy g(y) f(x y) 7 Transformation of a derivative Consider the transformation; 7

8 φ n (p) = dx dr f dx r e px Integration by parts yields; [ dr 1 f dx r 1 p e px ] + p dx dr 1 f dx r 1 e px φ (r) (p) = r 1 p n f r n 1 () + p r φ(p) n= 8 Laplace transform of the δ function Suppose; dxδ(x ǫ) e px = e pǫ The inverse transformation is; δ(x ǫ) = 1 dp e (ǫ x)p Let iα = p γ and idα = dp. Substitution; δ(x ǫ) = 1 2π e (ǫ x)γ dα e i(ǫ x)α δ(x ǫ) = e (ǫ x)γ δ(x ǫ) = δ(x ǫ) 9 Transformation of the step function Suppose the function; lim t sin(tx) t = 2πδ(x) The above is the derivative of the step function, so its integral should be the step function. F(t) = dx sin(xt) x This is an improper integral since it does not converge. However, consider the Laplace transform; 8

9 Im p o Pole γ Re p ^ Figure 2: The contour to evaluate the inverse transformation of the step function dt e pt F(t) = dt e pt dx sin(xt) x Interchange the order of integration to obtain, π 2p. Now apply the inverse transformation. f(t) = 1 dp (π/2) ep t p Figure 2 shows the contour for the integration. The value of γ is moved so that the contour includes the singulatities, in this case p =. For t > close the contour in the negavive p plane as shown. The calculus of residues is used to obtain; f(t) = [(π/2) 1 ] = π/2 t > For t < replace p by p. The result is π/2 for t <. Note that; df(t) dt = 2πδ(t) = π dx cos(tx) 1 Resistive transmission line Suppose the i th increment of a transmission line appears as in Figure 3. Use Kirchoff s circuit laws to write the equations for the voltage and currents which flow and then let the 9

10 V i R L V (i+1) I i Ii C C Figure 3: An incremental element of a resistive transmission line length of the element δx. Define a capacitance per unit length, inductance per unit length, and resistance per unit lenght as, c = C/δx, l = L/δx, r = R/δx, respectively. The voltage equations are; (V i+1 V i ) = L i di i dt + I ir V i = Q i /C The current equations are; dq i dt = I i I i+1 = I i+1 + I i Combine these, use the expressions for the components per unit length, c, l, r, and let the length of the element δx. This results in the pde for the wave equation previously obtained. Let J = I in what follows. 2 J x 2 = cl 2 J t 2 + cr J t Then apply the Laplace transform; J (x, p) = dt e pt J(x, t) The initial conditions are set so that J(x, ) =. This removes the surface terms in the transform of the derivative. d 2 J dt 2 the solution is; = cpl p 2 J + cr p J 1

11 J = Ae αx + Be αx Use the step function for the initial conditions; J(, t) = ds sin(st) π/2 t > s = t = π/2 t < As we have seen the Laplace transform of this initial condition is; J (, p) = π/2p Thus from above the transform of the solution takes the form; J (x, p) = (π/2p)e αx α 2 = cl p 2 + cr p Now let cl = v the signal velocity, and re-arrange the terms in the solution. J (x, p) = (π/2p)e [ (p 2 +(r/l)p)](x/v) The inverse transform has the form; J(x, t) = 1 dpj (x, p)e pt The inverse transformation is accomplished using the convolution theorem. Let p = u r/2l and break the integrand into 3 components with inverse as given in the equations below. < t < x/v u [e 2 (r/2l) 2x/v e u x/v ] A(t) = (r/2l)(x/v) t2 (x/v) I 1((r/2l) t 2 (x/v) 2 ) t > x/v [ ] 2 e u(x/v) < t < x/v u (r/2l) 1 t > x/v 1 u (r/2l) e(r/2l)t In the above, I 1 is the modified bessel function of order 1. Then combining using the convolution theorem. J(x, t) = (π/2) (π/2) e (r/2l)(x/v) Θ(t x/v) t dτ e (r/2l)(t τ) A(τ) + Θ vanishes for negative argument and equals 1 for postitive argument. The above clearly 11

12 obeys causality being zero until the leading edge of the signal travels to the point x in time t at velocity v 12