Chapter IX The Integral Transform Methods IX.4 The Hankel Transform November 24, 2020 803 IX.4 The Hankel Transform IX.4.1 The Hankel Transform – Definition and Operational Properties 804 IX.4.2 Instantaneous Heat Source 807 IX.4.3 Heat Equation with Instantaneous Line Source 808 IX.4.4 Solution of PDE with Application of Two Integral Transforms 816 (Fourier and Hankel Transforms) – Diffusion of Light in Semi-Infinite Space IX.4.5 Finite Hankel Transform IX.4.5.1 Circular Domain (FHT-1) 820 IX.4.5.2 Development of Velocity Profile in a Pipe under a Pressure Gradient IX.4.5.3 Annular Domain (FHT-2) 827 IX.4.5.4 Viscous Fluid Flow between Two Cylinders IX.4.7.8 Some additional notes on Hankel Transform 842 804 Chapter IX The Integral Transform Methods IX.4 The Hankel Transform November 24, 2020 IX.4.1 HANKEL TRANSFORM The physical domain where we are supposed to apply the Hankel transform is the infinite 2-dimensional plane with polar coordinates (r,θ ). The Hankel transform which we consider in this section will be applied for transformation of the differential operator 1u∂∂ν 2 Lu ≡ ru− (1) ∂∂ 2 Hermann Hankel rr r r (1839-1873) where ν ∈ is a parameter and domain is 0r≤ <∞. The particular case of this operator with ν = 0 is the radial term of Laplacian in cylindrical coordinates. We assume that the function ur( ) is bounded for all r and that the function u and its derivative have radial decaying at infinity: ∂u <∞ = = ur( ) , ru r→∞ 0 , r 0 (2) ∂r r→∞ The bounded solution of the singular Sturm-Liouville problem (eigenvalue problem): Lu= µ u has non-trivial solution when the parameter is non-positive, let us rename it as µλ= − 2 . That yields the equation Jν (λ r) is a Bessel function 2 1u∂∂ν 2 of the 1st kind of order ν . r−=− uuλ (3) rr∂∂ r r 2 Jν (λ r) is the bounded solution which is the Bessel equation of order ν , the bounded solutions st of the Bessel Equation of which are the Bessel functions of the 1 kind 2 22 2 r y′′++ ry ′ (λν r −) y = 0 Jrν (λ ) for r∈∞[ 0, ) The self-adjoint form of the operator (1) is 1u∂∂ν 2 Lu ≡ ru− (4) rr∂∂ r r Jr0 ( ) Therefore, the weight function is pr( ) = r, which is used for Jr( ) 1 definition of the weighted inner product. Jr2 ( ) Formally, the Hankel transform can be constructed from the two- dimensional Fourier transform with transition to polar coordinates and application of the integral representation of the Bessel functions [Debnah]. Then the Hankel transform and the inverse Hankel transform are defined as ∞ Hankel Transform of order ν Hν { fr( )} = fνν(λλ) = ∫ f( r) J( r) rdr (5) 0 ∞ −1 Inverse Hankel transform of order ν Hfνν{ (λ )} = fr( ) = ∫ fνν(λ) J( λ r) λλ d (6) 0 Chapter IX The Integral Transform Methods IX.4 The Hankel Transform November 24, 2020 805 Apply the Hankel transform (5) of order ν to δµ(r − ) , µ > 0 ∞ Hrν {δµ( − )} =∫δµλ(r − ) Jν ( r) rdr 0 = Jν (λµ) µ Then δµ(r − ) the inverse Hankel transform (6) of Jν (λµ) µ −1 δµ(r − ) = HJνν{ (λµ) µ} = ∞ = ∫ Jνν(λµ) µ J( λ rd) λ λ 0 ∞ = µ∫ Jνν( λ rJ) ( λµ) λ d λ 0 from which follows the orthogonality of Bessel functions: ∞ δµ(r − ) λµ λ λ λ = µ > Orthogonality of Bessel functions ∫ Jνν( ) J( rd) , 0 (7) 0 µ Then we can formulate the Hankel Integral Theorem: ∞∞ = µ λµ µ µ λ λ λ The Hankel Integral Theorem fr( ) ∫∫ f( ) Jνν( ) d J( r) d (8) λµ=00 = Proof using the orthogonality relationship (7): ∞∞ µ λµ µ µ λ λ λ ∫∫f( ) Jνν( ) dJ( rd) λµ=00 = ∞∞ = µ λµ λ λ λ µ µ ∫∫f( ) Jνν( ) J( rd) d µλ=00= ∞ λµ( − r) = ∫ f (µ ) µ dµ p0= µ = fr( ) ■ The Hankel integral theorem equation (8) can be used for definition of the Hankel transform pair (5,6). However, it should be noted, that equation (8) in this notes is derived by application of equations (5,6) assuming that they are true. We can state there is no circular reasoning (derivation of the result starting with assumption that it is true), if equations (5,6) are originated from some external derivation, for example, by the change of variables in Fourier transform to polar coordinates. 806 Chapter IX The Integral Transform Methods IX.4 The Hankel Transform November 24, 2020 Properties of Hankel transform Both, the Hankel transform and its inverse are linear operators. 1 λ Hν { f( ar)} = fν scaling a2 a δλ( ) H1{ } = 0 λ e−ar H0 r δ (rc− ) H0 r H0 { Ha( − r)} −λ2 2 1 He−cr = e 4c 0 { } 2c −ar λ He1 = { } 3 (a22+ λ ) e−ar H1 r H0 { rf′( r) } =−−2f00(λλ) f′( λ) H00{ fr( ) } = f (λ ) reciprocity Have to be checked! Chapter IX The Integral Transform Methods IX.4 The Hankel Transform November 24, 2020 807 Operational Property Consider application of the Hankel trnsform of order ν to differential operator (1), in which the additional term defines the order of the applied Hankel transform: 1u∂∂ν 2 ∞ 1 ∂ ∂u ν 2 − = λ − Hrν 2 u ∫ rJν ( r) r 2 udr rr∂∂ r r 0 r ∂r ∂r r ∞ ∂ ∂u ∞ ν 2 = Jν (λr) r dr − rJν (λ r) u dr ∫ ∂r ∂r ∫ 2 0 0 r ∞ ∂u ∞ ν 2 = λ − λ ∫ Jν ( r)d r ∫ rJν ( r) 2 u dr 0 ∂r 0 r ∞ ∂u ∞ ∂u ∞ ν 2 = r Jν (λr) − r d[Jν (λr)] − rJν (λ r) u dr ∂ ∫ ∂ ∫ 2 r 0 0 r 0 r ∞ ∂u ′ ∞ ν 2 = − λ − λ ∫ r [Jν ( r)] dr ∫ rJν ( r) 2 u dr 0 ∂r 0 r ∞ ′ ∞ ν 2 = − λ − λ From self-adjoint form of ∫ r[Jν ( r)] du ∫ rJν ( r) 2 u dr 0 0 r the Bessel Equation: ∞ ∞ ∞ 2 ν 2 ′′ν ′ 2 = ruJr(λλ) + ud rJr ( ) − rJ(λ r) u dr (ry′) +− +λ r y = 0 { νν} ∫ { } ∫ ν 2 r 0 0 0 r write ∞ ′ ∞ 2 2 ′ ν ′ ν 2 = λ − λ ′ =−+ λ ∫ u r Jν ( r) dr rJν ( r) 2 u dr (ry) r 2 y { } ∫ r 0 0 r Then for y= J(λ r) ν ∞ ν 2 ∞ ν 2 = − λ2 λ − λ ∫ur 2 Jν ( r)dr ∫ rJν ( r) 2 u dr 0 r 0 r ν 2 ′ λλ′ =2 − λ rJrrνν( ) 2 Jr( ) r ∞ ∞ ν 2 ∞ ν 2 = −λ2 (λ ) + λ − λ ∫ rJν r udr ∫ rJν ( r) 2 udr ∫ rJν ( r) 2 u dr 0 0 r 0 r ∞ 2 = −λ ∫ rJν (λr)udr 0 2 = −λλuν ( ) 2 ∞ 2 2 ν 1u∂∂ν 2 Hν ∇− ur( ) = rJν (λ r) r− u dr = −λλuν ( ) (9) 2 ∫ ∂∂2 r 0 rr r r 0th order and 1st order Hankel Transforms 1u∂∂ ∞ 1u∂∂ ∞ = λ = −λλ2 λ λ th H0 r ∫ rJ0 ( r) r dr u 0 ( ) u0 ( ) = ∫ rJ0 ( r) u( r) dr 0 order rr∂∂ r 0 rr∂∂ r 0 1∂∂ uu ∞ 1∂∂ uu ∞ −= λ − = −λλ2 λ λ st Hr1 2 ∫ rJ1 ( r) r2 dr u1 ( ) u1 ( ) = ∫ rJ1 ( r) u( r) dr 1 order rr∂∂ r r 0 rr∂∂ r r 0 We will use mostly the 0th order Hankel transform (in this case, index 0 in the transformed function is usually omitted) 808 Chapter IX The Integral Transform Methods IX.4 The Hankel Transform November 24, 2020 EXAMPLE TRANSIENT HEAT TRANSFER Cooling of the bell-shaped temperature profile Consider the axisymmetric case of the Heat Equation in cylindrical coordinates 2 1∂∂ u 1u ∂ ∇≡ur = r∈∞[ 0, ) t > 0 rr∂∂ rα ∂ t −br2 −br2 ur0 ( ) = e u( r,0) = u0 ( r) = e (initial condition) There are no boundary conditions, but we assume that the function u and its derivative has radial decaying at infinity: ∂u infinite metal = = ru r→∞ 0 r 0 plate ∂r r→∞ Physically it can be treated as a heated to bell shaped temperature profile infinite flat plate thermoinsulated from both sides. Heat is conducted along the plate. 1) Transformed equation Apply 0th order Hankel transform ∞ λ = λ ut( , ) ∫ rJ0 ( r) u( r,t) dr 0 Then equation is transformed to 1u∂ −=λ 2u α ∂t With transformed initial condition (see p.806) λ2 − ∞∞ 4b 2 e λ λλ= −br = u( ,0) = ∫∫rJ00( r) u dr rJ 0( r) e dr 00 2b That is the 1st order linear ODE ∂u +=αλ 2u0 ∂t The solution is 1 −+αλ2 2 1 t u(λ ,t) = u(λ ,0) e−αλ t = e 4b 2b 2) Inverse transform Solution of the problem (use table p.806) 1 2 −+αλt −1 1 4b u(r,t) = He0 2b λ2 − 4 1 + α 1 − 4t 1 = He1 b denote c = 0 1 2b + 4tα b r2 − λ2 1 − +4tα 2c− 1 c − 2 1 Hankel example p.808.mws u( r,t ) = He1 4c ` = e cr ` = e b 0 + α 2b 2c b (1 4 bt) r2 − 1 1 +4tα u(r,t) = e b (1+ 4α bt) Chapter IX The Integral Transform Methods IX.4 The Hankel Transform November 24, 2020 809 IX.4.2 INSTANTENEOUS ENERGY SOURCE We consider the volumetric heat source S(r,t) which releases its entire energy spontaneously at the time t = t0 at the points r = r0 [Ozisik, HC, p.220]. g Ws⋅ In the axisymmetric case (does not depend on θ ) of cylindrical S( r,t) =δδ( r −− r) ( t t ) g 2rπ 00 m coordinates and no dependence on the z coordinate (the same source strength for any plane perpendicular to the z axis), this is a cylindrical surface heat source which can be represented with the help of the Dirac delta function g W = δ − δ − S(r,t) (r r0 ) (t t0 ) 3 2πr m W ⋅ s where g is the source strength per unit length m g W ⋅ s 2 is the source strength per unit area 2πr m r [m] the variable r appears in the denominator as a scale factor of the transformation to polar coordinates 1 δ (r − r0 ) defines the radius of the source r = r0 m 1 δ (t − t0 ) defines the moment of time t = t0 s Ws⋅ Integrated over the entire time-space domain (in the plane g m perpendicular to the z coordinate), it should yield the strength of g the heat source (conservation of energy): S( r,t) = δδ( r) ( t− t ) gW[ ⋅ s] 2rπ 0 ∞ 2π ∞ ∞ 2π ∞ g g S(r,t)rdrdθdt = δ (r − r )δ (t − t ) rdrdθdt =δδ −− δ ∫ ∫ ∫ ∫ ∫ ∫ 0 0 S( r,t) ( r) ( z z00) ( t t ) π 2rπ 0 0 0 0 0 0 2 r g ∞ 2π ∞ = δ − δ − θ ∫ ∫ ∫ (r r0 ) (t t0 )drd dt 2π 0 0 0 g ∞ 2π = δ − θ ∫ ∫ (t t0 )d dt 2π 0 0 g 2π = ∫ dθ 2π 0 g 2π = [θ ] 2π 0 = g gW[ ⋅ s] In the 3-dimensional case, the heat source is a circular ring in g S( r,t) =δδ( r −−− r0) ( z z 00) δ( t t ) = 2rπ the plane z z0 defined by g S(r,t) = δ (r − r )δ (z − z )δ (t − t ) 2πr 0 0 0 An instantaneous source is used for finding the Green function for the non-homogeneous PDE.