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# Bessel functions The Bessel function of the rst kind of order is dened by z For this is a solution of the Bessel dierential equation zy For PDF document - DocSlides

myesha-ticknor | 2014-12-11 | General

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### Presentations text content in Bessel functions The Bessel function of the rst kind of order is dened by z For this is a solution of the Bessel dierential equation zy For

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Bessel functions The Bessel function ) of the ﬁrst kind of order is deﬁned by ) = z/ 2) Γ( + 1) + 1 =0 1) Γ( + 1) (1) For 0 this is a solution of the Bessel diﬀerential equation 00 ) + zy ) + ) = 0 , (2) For ν / ∈{ ,... we have that ) is a second solution of the diﬀerential equation (2) and the two solutions ) and ) are clearly linearly independent. For ∈{ ,... we have ) = =0 1) Γ( + 1) 1) Γ( + 1) since Γ( + 1) = 0 for = 0 ,...,n This implies that ) = 1) Γ( + 1) =0 1) Γ( + 1) ( )! 2( = ( 1) =0 1) Γ( + 1) = ( 1) This implies that ) and ) are linearly dependent for ∈{ ,... A second linearly independent solution can be found as follows. Since ( 1) = cos n , we see that ) cos ) is a solution of the diﬀerential equation (2) which vanishes when ∈{ ,... . Now we deﬁne ) := ) cos sin (3) where the case that ∈{ ,... should be regarded as a limit case. By l’Hopital’s rule we have ) = lim ) = ∂J 1) ∂J This implies that ) = ( 1) ) for ∈{ ,... The function ) is called the Bessel function of the second kind of order Using the deﬁnition (1) we ﬁnd that ∂J ) ln =0 1) + 1) )!

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where ) = dz ln Γ( ) = Γ( For ν / ∈{ ,... we have ∂J ) ln =0 1) + 1) Γ( + 1) Now we use lim )Γ( ) = 1) Γ( 1) for and ) = Γ( 1) 1) for n. This implies that lim Γ( = ( 1) +1 ! for = 0 ,.... Hence lim =0 1) + 1) Γ( + 1) = ( 1) =0 1)! 1) + 1) Γ( + 1) = ( 1) =0 1)! + ( 1) =0 1) + 1) Γ( + 1) ( )! This implies that ∂J ) ln + ( 1) =0 1)! + ( 1) =0 1) + 1) )! Finally we use the fact that ) = ( 1) ) to conclude that ) = ) ln =0 1)! =0 1) )! + 1) + + 1)] for ∈{ ,... . Compare with the theory of Frobenius for linear second diﬀerential equations.

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In the theory of second order linear diﬀerential equations of the form 00 = 0 two solutions and are linearly independent if and only if ,y )( ) := = 0 This determinant is called the Wronskian of the solutions and It can (easily) be shown that this determinant of Wronski satisﬁes the diﬀerential equation ) + ) = 0 This result is called Abel’s theorem or the theorem of Abel-Liouville. In the case of the Bessel diﬀerential equation we have ) = 1 /z , which implies that ) + ) = 0 = ,y )( ) = for some constant . Now we have Theorem1. ,J )( ) = 2 sin πz and ,Y )( ) = πz (4) For 0 this implies that ) and ) are linearly independent if ν / ∈{ ,... and that ) and ) are linearly independent for all 0. Proof. Note that ,Y )( ) = ) cos sin ) cos sin sin ,J )( sin Now we use the deﬁnition (1) to obtain ) = =0 1) Γ( + 1) +2 +2 ) = =0 1) + 2 Γ( + 1) +2 +2 and ) = =0 1) Γ( + 1) +2 +2 ) = =0 1) + 2 Γ( + 1) +2 +2

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Hence we have zW ,J )( ) = =0 =0 1) + 2 Γ( + 1)Γ( + 1) +2 +2 =0 =0 1) + 2 Γ( + 1)Γ( + 1) +2 +2 =0 =0 1) (2 + 2 Γ( + 1)Γ( + 1) +2 +2 This implies that lim zW ,J )( ) = Γ( + 1)Γ( + 1) Γ( )Γ(1 2 sin Using the deﬁnition (1) we obtain dz )] = dz =0 1) Γ( + 1) +2 +2 =0 1) (2 + 2 Γ( + 1) +2 +2 =0 1) Γ( +2 +2 Hence we have dz )] = zJ ) + νJ ) = zJ (5) Similarly we have dz dz =0 1) Γ( + 1) +2 =0 1) Γ( + 1) +2 =0 1) +1 Γ( + 2) +1 +2 +1 +1 Hence we have dz +1 zJ νJ ) = zJ +1 (6) Elimination of ) from (5) and (6) gives ) + +1 ) = and elimination of ) from (5) and (6) gives +1 ) = 2

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Special cases For = 1 2 we have from the deﬁnition (1) by using Legendre’s duplication formula for the gamma function ) = =0 1) Γ( + 3 2) =0 1) (2 + 1)! πx sin x, x> and for 2 we have ) = =0 1) Γ( + 1 2) πx =0 1) (2 )! πx cos x, x> Note that the deﬁnition (3) implies that ) = ) = πx cos and ) = ) = πx sin x, x> Integral representations First we will prove Theorem2. ) = Γ( + 1 2) izt (1 dt, Re ν > (7) Proof. We start with izt (1 dt =0 iz (1 dt. Note that the latter integral vanishes when is odd. For = 2 we obtain using (1 dt = 2 (1 du (1 du + 1 , + 1 2) = Γ( + 1 2)Γ( + 1 2) Γ( + 1) Now we use Legendre’s duplication formula to ﬁnd that Γ( + 1 2) = Γ(2 Γ( Γ(2 + 1) Γ( + 1) (2 )! Hence we have izt (1 dt =0 iz (2 )! Γ( + 1 2)Γ( + 1 2) Γ( + 1) = Γ( + 1 2) =0 1) Γ( + 1) This proves the theorem.

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We also have Poisson’s integral representations Theorem3. ) = Γ( + 1 2) iz cos (sin d Γ( + 1 2) cos( cos ) (sin dθ, Re ν > Proof. Use the substitution = cos to obtain izt (1 dt iz cos (1 cos sin d iz cos (sin dθ. Further we have iz cos = cos( cos ) + sin( cos and sin( cos )(sin d = 0 This shows that Poisson’s integral representations follow from the integral representation (7). Remarks: 1. The Fourier transform is deﬁned by ) := izt dt with inversion formula ) = izt dz. This implies that the Fourier transform of the function ) = (1 | is ) = Γ( + 1 2) 2. Instead of the substitution = cos in (7) one can also use the substitution = sin which leads to slightly diﬀerent forms of Poisson’s integral representations. In fact we have ) = Γ( + 1 2) π/ π/ iz sin (cos d Γ( + 1 2) π/ π/ cos( sin ) (cos d Γ( + 1 2) π/ cos( sin ) (cos dθ, Re ν >

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Integrals of Bessel functions The Hankel transform of a function is deﬁned by ) = tf st dt for functions for which the integral converges. The inversion formula is given by ) = sF st ds. This pair of integrals is called a Hankel pair of order An example of such an integral is st dt +1 Γ( + 1) + 1 Re( It can be shown that the integral converges for Re( 0. Now we use the deﬁnition (1) to obtain st dt =0 1) Γ( + 1) +2 +2 +2 dt. Using the substitution we ﬁnd that +2 dt +1 +2 1) du +2 2) du + 2 Hence we have st dt =0 1) Γ( + 1) +2 +2 Γ( + 1) =0 1) (( 2) + 1) +1 Γ( + 1) + 1 The special case + 2 is of special interest: in that case we have ( 2 = + 1. This implies that the reduces to a which is an exponential function. The result is +1 st dt (2 +1 Re ν >

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Hankel functions The functions (1) ) and (2) ) are deﬁned by (1) ) := ) + iY ) and (2) ) := iY These functions are called Hankel functions or Bessel functions of the third kind. Note that these deﬁnitions imply that ) = (1) ) + (2) and ) = (1) (2) Further we have (1) ) = ) + iY ) = πx (cos sin ) = πx ix , x> and (2) ) = iY ) = πx (cos sin ) = πx ix , x> Similarly we have (1) ) = ) + iY ) = πx (sin cos ) = πx ix , x> and (2) ) = iY ) = πx (sin cos ) = πx ix , x> Modiﬁed Bessel functions The modiﬁed Bessel function ) of the ﬁrst kind of order is deﬁned by ) := z/ 2) Γ( + 1) + 1 =0 Γ( + 1) For 0 this is a solution of the modiﬁed Bessel diﬀerential equation 00 ) + zy ) = 0 , For ν / ∈{ ,... we have that ) is a second solution of this diﬀerential equation and the two solutions ) and ) are linearly independent. For ∈{ ,... we have ) = ). The modiﬁed Bessel function ) of the second kind of order is deﬁned by ) := 2 sin )] for ν / ∈{ ,... and ) = lim ) for ∈{ ,... Now we have for x> ) = πx sinh x, I ) = πx cosh and ) = ) =

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A generating function The Bessel function ) of the ﬁrst kind of integer order can also be deﬁned by means of the generating function exp (8) In fact, the series on the right-hand side is a so-called Laurent series at = 0 for the function at the left-hand side. Using the Taylor series for the exponential function we obtain exp = exp zt exp =0 zt =0 1) For ∈{ ,... we have =0 )! 1) =0 1) )! and =0 1) )! = ( 1) ) = This proves (8). If i , then we have i i sin exp ix sin = cos( sin ) + sin( sin Hence we have cos( sin ) + sin( sin ) = ix sin in ) [cos( n ) + sin( n )] Since ) = ( 1) ) this implies that cos( sin ) = ) cos( n ) = ) + 2 =1 ) cos(2 k and sin( sin ) = ) sin( n ) = 2 =0 +1 ) sin(2 + 1) θ. For π/ 2 this implies that cos ) + 2 =1 1) ) and sin = 2 =0 1) +1 For = 0 we also have 1 = ) + 2 =1

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The generating function (8) can be used to prove that ) = The proof is as follows: = exp = exp exp We can also ﬁnd an integral representation for the Bessel function ) of the ﬁrst kind of integral order starting from the generating function ix sin in We use the orthogonality property of the exponential function, id est ik d , k , k = 0 π, k = 0 Hence we have ix sin in d ik in d = 2 This implies that ) = sin n d cos ( sin n dθ. The special case = 0 reads ) = cos ( sin d cos( xt dt. 10

we have that is a second solution of the di64256erential equation 2 and the two solutions and are clearly linearly independent For 8712 we have 0 1 915 1 1 915 1 since 915 1 0 for 0 n This implies that 1 915 1 0 1 915 1 2 1 0 1 91 ID: 22499

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Page 1

Bessel functions The Bessel function ) of the ﬁrst kind of order is deﬁned by ) = z/ 2) Γ( + 1) + 1 =0 1) Γ( + 1) (1) For 0 this is a solution of the Bessel diﬀerential equation 00 ) + zy ) + ) = 0 , (2) For ν / ∈{ ,... we have that ) is a second solution of the diﬀerential equation (2) and the two solutions ) and ) are clearly linearly independent. For ∈{ ,... we have ) = =0 1) Γ( + 1) 1) Γ( + 1) since Γ( + 1) = 0 for = 0 ,...,n This implies that ) = 1) Γ( + 1) =0 1) Γ( + 1) ( )! 2( = ( 1) =0 1) Γ( + 1) = ( 1) This implies that ) and ) are linearly dependent for ∈{ ,... A second linearly independent solution can be found as follows. Since ( 1) = cos n , we see that ) cos ) is a solution of the diﬀerential equation (2) which vanishes when ∈{ ,... . Now we deﬁne ) := ) cos sin (3) where the case that ∈{ ,... should be regarded as a limit case. By l’Hopital’s rule we have ) = lim ) = ∂J 1) ∂J This implies that ) = ( 1) ) for ∈{ ,... The function ) is called the Bessel function of the second kind of order Using the deﬁnition (1) we ﬁnd that ∂J ) ln =0 1) + 1) )!

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where ) = dz ln Γ( ) = Γ( For ν / ∈{ ,... we have ∂J ) ln =0 1) + 1) Γ( + 1) Now we use lim )Γ( ) = 1) Γ( 1) for and ) = Γ( 1) 1) for n. This implies that lim Γ( = ( 1) +1 ! for = 0 ,.... Hence lim =0 1) + 1) Γ( + 1) = ( 1) =0 1)! 1) + 1) Γ( + 1) = ( 1) =0 1)! + ( 1) =0 1) + 1) Γ( + 1) ( )! This implies that ∂J ) ln + ( 1) =0 1)! + ( 1) =0 1) + 1) )! Finally we use the fact that ) = ( 1) ) to conclude that ) = ) ln =0 1)! =0 1) )! + 1) + + 1)] for ∈{ ,... . Compare with the theory of Frobenius for linear second diﬀerential equations.

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In the theory of second order linear diﬀerential equations of the form 00 = 0 two solutions and are linearly independent if and only if ,y )( ) := = 0 This determinant is called the Wronskian of the solutions and It can (easily) be shown that this determinant of Wronski satisﬁes the diﬀerential equation ) + ) = 0 This result is called Abel’s theorem or the theorem of Abel-Liouville. In the case of the Bessel diﬀerential equation we have ) = 1 /z , which implies that ) + ) = 0 = ,y )( ) = for some constant . Now we have Theorem1. ,J )( ) = 2 sin πz and ,Y )( ) = πz (4) For 0 this implies that ) and ) are linearly independent if ν / ∈{ ,... and that ) and ) are linearly independent for all 0. Proof. Note that ,Y )( ) = ) cos sin ) cos sin sin ,J )( sin Now we use the deﬁnition (1) to obtain ) = =0 1) Γ( + 1) +2 +2 ) = =0 1) + 2 Γ( + 1) +2 +2 and ) = =0 1) Γ( + 1) +2 +2 ) = =0 1) + 2 Γ( + 1) +2 +2

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Hence we have zW ,J )( ) = =0 =0 1) + 2 Γ( + 1)Γ( + 1) +2 +2 =0 =0 1) + 2 Γ( + 1)Γ( + 1) +2 +2 =0 =0 1) (2 + 2 Γ( + 1)Γ( + 1) +2 +2 This implies that lim zW ,J )( ) = Γ( + 1)Γ( + 1) Γ( )Γ(1 2 sin Using the deﬁnition (1) we obtain dz )] = dz =0 1) Γ( + 1) +2 +2 =0 1) (2 + 2 Γ( + 1) +2 +2 =0 1) Γ( +2 +2 Hence we have dz )] = zJ ) + νJ ) = zJ (5) Similarly we have dz dz =0 1) Γ( + 1) +2 =0 1) Γ( + 1) +2 =0 1) +1 Γ( + 2) +1 +2 +1 +1 Hence we have dz +1 zJ νJ ) = zJ +1 (6) Elimination of ) from (5) and (6) gives ) + +1 ) = and elimination of ) from (5) and (6) gives +1 ) = 2

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Special cases For = 1 2 we have from the deﬁnition (1) by using Legendre’s duplication formula for the gamma function ) = =0 1) Γ( + 3 2) =0 1) (2 + 1)! πx sin x, x> and for 2 we have ) = =0 1) Γ( + 1 2) πx =0 1) (2 )! πx cos x, x> Note that the deﬁnition (3) implies that ) = ) = πx cos and ) = ) = πx sin x, x> Integral representations First we will prove Theorem2. ) = Γ( + 1 2) izt (1 dt, Re ν > (7) Proof. We start with izt (1 dt =0 iz (1 dt. Note that the latter integral vanishes when is odd. For = 2 we obtain using (1 dt = 2 (1 du (1 du + 1 , + 1 2) = Γ( + 1 2)Γ( + 1 2) Γ( + 1) Now we use Legendre’s duplication formula to ﬁnd that Γ( + 1 2) = Γ(2 Γ( Γ(2 + 1) Γ( + 1) (2 )! Hence we have izt (1 dt =0 iz (2 )! Γ( + 1 2)Γ( + 1 2) Γ( + 1) = Γ( + 1 2) =0 1) Γ( + 1) This proves the theorem.

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We also have Poisson’s integral representations Theorem3. ) = Γ( + 1 2) iz cos (sin d Γ( + 1 2) cos( cos ) (sin dθ, Re ν > Proof. Use the substitution = cos to obtain izt (1 dt iz cos (1 cos sin d iz cos (sin dθ. Further we have iz cos = cos( cos ) + sin( cos and sin( cos )(sin d = 0 This shows that Poisson’s integral representations follow from the integral representation (7). Remarks: 1. The Fourier transform is deﬁned by ) := izt dt with inversion formula ) = izt dz. This implies that the Fourier transform of the function ) = (1 | is ) = Γ( + 1 2) 2. Instead of the substitution = cos in (7) one can also use the substitution = sin which leads to slightly diﬀerent forms of Poisson’s integral representations. In fact we have ) = Γ( + 1 2) π/ π/ iz sin (cos d Γ( + 1 2) π/ π/ cos( sin ) (cos d Γ( + 1 2) π/ cos( sin ) (cos dθ, Re ν >

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Integrals of Bessel functions The Hankel transform of a function is deﬁned by ) = tf st dt for functions for which the integral converges. The inversion formula is given by ) = sF st ds. This pair of integrals is called a Hankel pair of order An example of such an integral is st dt +1 Γ( + 1) + 1 Re( It can be shown that the integral converges for Re( 0. Now we use the deﬁnition (1) to obtain st dt =0 1) Γ( + 1) +2 +2 +2 dt. Using the substitution we ﬁnd that +2 dt +1 +2 1) du +2 2) du + 2 Hence we have st dt =0 1) Γ( + 1) +2 +2 Γ( + 1) =0 1) (( 2) + 1) +1 Γ( + 1) + 1 The special case + 2 is of special interest: in that case we have ( 2 = + 1. This implies that the reduces to a which is an exponential function. The result is +1 st dt (2 +1 Re ν >

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Hankel functions The functions (1) ) and (2) ) are deﬁned by (1) ) := ) + iY ) and (2) ) := iY These functions are called Hankel functions or Bessel functions of the third kind. Note that these deﬁnitions imply that ) = (1) ) + (2) and ) = (1) (2) Further we have (1) ) = ) + iY ) = πx (cos sin ) = πx ix , x> and (2) ) = iY ) = πx (cos sin ) = πx ix , x> Similarly we have (1) ) = ) + iY ) = πx (sin cos ) = πx ix , x> and (2) ) = iY ) = πx (sin cos ) = πx ix , x> Modiﬁed Bessel functions The modiﬁed Bessel function ) of the ﬁrst kind of order is deﬁned by ) := z/ 2) Γ( + 1) + 1 =0 Γ( + 1) For 0 this is a solution of the modiﬁed Bessel diﬀerential equation 00 ) + zy ) = 0 , For ν / ∈{ ,... we have that ) is a second solution of this diﬀerential equation and the two solutions ) and ) are linearly independent. For ∈{ ,... we have ) = ). The modiﬁed Bessel function ) of the second kind of order is deﬁned by ) := 2 sin )] for ν / ∈{ ,... and ) = lim ) for ∈{ ,... Now we have for x> ) = πx sinh x, I ) = πx cosh and ) = ) =

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A generating function The Bessel function ) of the ﬁrst kind of integer order can also be deﬁned by means of the generating function exp (8) In fact, the series on the right-hand side is a so-called Laurent series at = 0 for the function at the left-hand side. Using the Taylor series for the exponential function we obtain exp = exp zt exp =0 zt =0 1) For ∈{ ,... we have =0 )! 1) =0 1) )! and =0 1) )! = ( 1) ) = This proves (8). If i , then we have i i sin exp ix sin = cos( sin ) + sin( sin Hence we have cos( sin ) + sin( sin ) = ix sin in ) [cos( n ) + sin( n )] Since ) = ( 1) ) this implies that cos( sin ) = ) cos( n ) = ) + 2 =1 ) cos(2 k and sin( sin ) = ) sin( n ) = 2 =0 +1 ) sin(2 + 1) θ. For π/ 2 this implies that cos ) + 2 =1 1) ) and sin = 2 =0 1) +1 For = 0 we also have 1 = ) + 2 =1

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The generating function (8) can be used to prove that ) = The proof is as follows: = exp = exp exp We can also ﬁnd an integral representation for the Bessel function ) of the ﬁrst kind of integral order starting from the generating function ix sin in We use the orthogonality property of the exponential function, id est ik d , k , k = 0 π, k = 0 Hence we have ix sin in d ik in d = 2 This implies that ) = sin n d cos ( sin n dθ. The special case = 0 reads ) = cos ( sin d cos( xt dt. 10

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