They are im portant in math as well as in physical sciences physics and engineering They are especially important in solving boundary values problems in cylindrical coordi nates First we de57356ne another important function the Gamma function which ID: 22500 Download Pdf

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They are im portant in math as well as in physical sciences physics and engineering They are especially important in solving boundary values problems in cylindrical coordi nates First we de57356ne another important function the Gamma function which

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BESSEL EQUATIONS AND BESSEL FUNCTIONS Bessel functions form a class of the so called special functions . They are im- portant in math as well as in physical sciences (physics and engineering). They are especially important in solving boundary values problems in cylindrical coordi- nates. First we dene another important function: the Gamma function which is used in the series expansion of the Bessel functions, then we construct the Bessel functions and 1. The Gamma Function The Gamma function (also called Euler’s integral ) is the function dened for x > 0 by

( ) = ds : The improper integral dening is convergent for x > 0. To see why, note that for every x > 0, lim !1 = 0 Thus there exists M > 0 such that e for s > M . This implies that ds ds Also for (0 ; M ), e and ds ds =0 We have then ds ds ds This shows that ( ) is well dened for x > 0. The most important property of the Gamma function is given in the following lemma. Lemma 1. The function satises the following ( + 1) = ( x > Proof. The proof is simply an integration by parts ds ds ds By taking the limit as !1 , we get (

+ 1) = ( Date :March31,2014.

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BESSEL EQUATIONS AND BESSEL FUNCTIONS It can be shown that has derivatives of all orders for x > 0 and that has a unique extremum (global minimum) on the interval (0 ). The minimum is reached at a number (1 2) and ( 1. Furthermore, satises lim ( ) = and lim !1 ( ) = Graph of (x) over (0, 1 2 1 We can use the fundamental property to extend as a smooth functions to nf g (the whole real line except 0 and the negative integers). First we extend

to the interval ( 0) by dening ( ) = ( + 1) for 0) (note the above denition makes sense since +1 (0 1) and ( +1) is dened by the integral). Once, is dened on ( 0), we extend it to the interval ( 1) by using the same property. More precisely, if is dened on the interval ( j; 1)) with , then we extend it to the interval ( + 1) ) by using the fundamental property. We have in particular that lim ( for Now we compute some values of the Gamma function. (1) = ds = 1 By using the fundamental property of

, we get easily its values at the positive integers. (2) = (1 + 1) = 1(1) = 1 (3) = (2 + 1) = 2(2) = 2 = 2! (4) = (3 + 1) = 3(3) = 3 2! = 3! ( + 1) = The Gamma function appears as an interpolation of the factorial function.

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BESSEL EQUATIONS AND BESSEL FUNCTIONS Graph of (x) −1 −2 −3 To compute (1 2) we use the value of the Gaussian integral dt = (you have probably encountered this integral in Multivariable Calculus (MAC2313) or in Prob./Statistics class).

In the following calculation, we have made the substi- tution ds = 2 dt : The Gamma function satises many other identities such: Reection formula : ( )(1 ) = sin x = 0 Duplication formula : (2 ) = ( ) (2 = 0 2. Bessel’s Equation Bessel’s equation of order (with 0) is the second order dierential equation (1) 00 xy + ( = 0 In order to nd all solutions we need two independent solutions. We are going to construct the independent solutions for x > 0. 2.1. Construction of a rst solution. Note that = 0 is a

singular point of the equation. More precisely, it is a regular singular point (see your notes from the rst dierential equations class, MAP2302). For such dierential equations, we can use the method of Frobenius to construct series solutions. We seek a (formal) series solution (2) =0 =0

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BESSEL EQUATIONS AND BESSEL FUNCTIONS of equation (1), with , and = 0. The substitution of this series and its (formal) derivatives into equation (1) gives =0 )( 1) =0 + ( =0 = 0 We rewrite this as =0 )( 1) =0 =0 +2 =0 = 0 then as =0 )( 1) =0 =2 =0 = 0 After

grouping the like terms and simplifying, we obtain + (( + 1) +1 =2 (( = 0 In order for this series to be identically zero, each coecient must be zero. We have then = 0 + 1) = 0 = 0 ; j = 2 Since = 0, then the rst equation implies that must satisfy = 0 This is the indicial equation of the Bessel equation. The indicial roots are and : Consider the case . The second equation becomes (2 + 1) = 0 = 0 (since > 0) For 2 the recurrence relation becomes = 0 (2 Since = 0, the above relation gives 3(2 + 3) = 0 ; c 5(2 + 5) = 0 ; c = 0 That is, all coecients with

odd indices are 0 ( odd = 0). For the coecients with even indices, we have 2(2 + 2) 4(1 + 4(2 + 4) 1) (2!)(1 + )(2 + 6(2 + 6) 1) (3!)(1 + )(2 + )(3 + A proof by induction gives 1) !2 (1 + )(2 + ; j = 1

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BESSEL EQUATIONS AND BESSEL FUNCTIONS A formal solution is therefore =0 =0 1) !2 (1 + )(2 + We are going to select and use the Gamma function to rewrite the series solution in a more compact form. It follows from the fundamental property of the Gamma function that ( + 1 + ) = ( )( = ( )( 1 + )( 1 + = ( )( 1 + (1 + )(1 + Equivalently, (1 +

)(2 + ) = ( + 1 + (1 + We select as (1 + With this choice of , the particular series solution becomes ) = =0 1) !( + 1 + This solution is known as the Bessel function of the rst kind of order Now we determine the domain where the series converges. Note that ) = =0 1) !( + 1 + The last series is a power series in ( x= 2) . To nd its radius of convergence, we can use the ratio test: lim !1 1) +1 ((1 + )!( + 2 + )) 1) !( + 1 + )) = lim !1 + 1)( + 1 + = 0 The radius of convergence is innite (the power series converges

to an analytic function on ). The function ) is dened for 0. 2.2. Construction of a second solution. Recall that the indicial roots of the Bessel equation are . We have used to construct the solution ). We can redo the above construction with . However, this can be done only if 62 . In this case a second independent solution of Bessel’s equation is ) = =0 1) !( + 1 Note that is not dened at = 0. We have lim The general solution of equation in (0 ) is ) = AJ ) + BJ with and constants.

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BESSEL EQUATIONS AND BESSEL FUNCTIONS When , the situation is a little

more involved. The rst solution is ) = =0 1) !( )! If we try to dene by using the recurrence relations for the coecients, then starting with = 0, we can get 2(2 4(1 4(4 1) (2!)(1 )(2 2( 1) 1) 1)!2 2( 1) (1 )(2 At the order 2 however we get 2( 1) = 0 This is a contradiction since 2( 1) = 0. Thus, the recurrence relations will not lead to a series solution. Another attempt to dene is to dene it as ) = lim In this case, we get back either and and are dependent solutions of the equations. More precisely, we have the following lemma. Lemma. We have ) = ( 1)

Proof. For 62 (and close to ), we have ) = =0 1) !2 ( + 1 Recall that lim ! ( for = 0 or . When , ( + 1 tends to 0 or a negative integer for = 0 1). For such values of the coecients of in the series above approaches 0: lim 1) !2 ( + 1 = 0 We get then, ) = lim ) = 1) !2 ( + 1 and after using the fundamental property of the Gamma function we obtain ) = 1) !2 )! =0 1) )!2 = ( 1) Now we indicate how to construct a second independent solution of equation (1) when . Consider with 0 < < 1 (hence such 62 ). The

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BESSEL EQUATIONS AND

BESSEL FUNCTIONS corresponding Bessel equation has two independent solutions and The function dened by ) = 1) Since function is a linear combination of and , then is also a solution of the corresponding Bessel’s equation of order . We dene as: ) = lim ) = lim 1) It can be proved that the function is a solution of the Bessel equation of order and that and are independent (see for example R.Courant and D. Hilbert, Method of Mathematical Physics , vol. 2, or H. Sagan, Boundary and Eigenvalue Problems of Mathematical Physics ). This solution is called the Bessel function of the

second kind of order . It can also be proved that lim ) = 1 Another method to obtain a second solution of the Bessel equation in the excep- tional case is to seek it in the form ) = ) ln The coecients are then found by a recurrence relation. The explicit expression of the ) is given below. Its derivation can be found in advanced texts about special function. For , we have ) = + ln =0 1) !)( )! x =0 1)! !) where, for = 0 , the constants are given by = 0 ; c = 1 ; c = 1 + ;c = 1 + and where is the Euler constant given by = lim !1 1 + ln ; 57721 :::: For = 0, we have ) = +

ln =0 1) !) 2.3. General solution of the Bessel equation. We summarize the above dis- cussions in the following theorem. Theorem. Given the Bessel equation of order 00 xy + ( = 0 then we have the followings:

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BESSEL EQUATIONS AND BESSEL FUNCTIONS If 62 [f , the equation has two independent solutions and (Bessel functions of the rst kind) and the general solution is ) = AJ ) + BJ where and are constants. If with = 0 or , the equation has only one Bessel function of the rst kind , another independent solution is the Bessel function of the second kind . The general

solution of the equation is ) = AJ ) + BY 3. Remarks on Bessel functions The expansions of the functions and are ) = =0 1) !) = 1 (2!) (3!) ) = =0 1) !( + 1)! +1 (2!) (2!)(3!) (3!)(4!) The graphs of and of resemble those of cosine and sine with a decreasing 10 12 14 16 18 20 −1 −0.5 0.5 1.5 Graphs of J and J (x) (x) amplitude. Notice how the zeros of and behave. Between two consecutive zeros of there is exactly one zero of . The following table lists the approximate values of the rst 9 positive zeros of and 1 2 3 2.405 5.520 8.654 11.792 14.931 18.071 21.212 24.353 27.494

3.832 7.016 10.174 13.324 16.471 19.616 22.760 25.904 29.047

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BESSEL EQUATIONS AND BESSEL FUNCTIONS For large, the -th zero of is approximately n = 4) and the -th zero of is approximately n + ( = 4). It is shown that for large we have x cos and x cos In fact the -th Bessel function has the following behavior x cos (2 + 1) for large This approximation shows that has innitely many positive zeros that tends to innity. More precisely, we have following proposition about the zeros of Bessel functions. Proposition 1. For every ,

the positive zeros of form an increasing unbounded sequence. That is, the solution set of the equation ) = 0 ; x > forms a sequence < x < x < x < x with lim !1 The proof of this proposition is beyond the aim of this course. For + 1 , the Bessel functions are elementary functions. This means that ) can be expressed algebraically in terms of sin cos and . The following proposition gives the expressions of some Bessel functions with such indices. Proposition 2. We have the following relations ) = x sin x; ) = x cos x; ) = x sin cos ) = x cos + sin Proof. We prove

the rst relation and leave the others as an exercise. Recall that the Taylor expansion of sin is sin =0 1) (2 + 1)! +1 We need the value of ( + (3 2)). We have (1 2) = (see section about the Gamma function). By using the fundamental property of the Gamma function, we

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10 BESSEL EQUATIONS AND BESSEL FUNCTIONS get = + 1 = + 1 (3 1) = + 1 (5 1) We prove by induction that for (2 + 1)(2 1) +1 : We can simplify the product of the odd integers above as (2 + 1)(2 1) 1 = (2 + 1)! (2 )(2 2) (2 + 1)! !) Hence, (2 + 1)! +1 !) : Now we use these to

show the rst relation of the proposition. ) = =0 1) !( + (3 2)) +(1 2) =0 1) +1 !)( + (3 2)) +1 =0 1) +1 !) +1 !)(2 + 1)! +1 x =0 1) (2 + 1)! +1 x sin Analogous results about the behaviors of the Bessel functions of the second kind can be obtained.

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BESSEL EQUATIONS AND BESSEL FUNCTIONS 11 10 12 14 16 18 20 −1.6 −1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2 0.2 0.4 Bessel functions of the second kind: Y , Y , Y (x) (x) (x) 4. Some properties of the Bessel functions The bessel functions satisfy a large

number of properties. We limit ourself here to list the following. Properties of (1) (0) = 1 and (0) = 0 if > 0. (2) ) is an even function if is even and ) is an odd function in is odd. (3) ) = ( 1) ) for (4) dx +1 (5) dx )) = (6) dx )) = +1 )) (7) +1 ) + ) = ). (8) +1 dx ) + (9) dx ) + The rst two properties are easy to obtain from the series representation of and the third has already been veried.

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12 BESSEL EQUATIONS AND BESSEL FUNCTIONS Proof of 4. Multiply the series representation of by and dierentiate dx =1 1) !( + 1) =1 1)

1)!( + 1) 1+ =1 1) 1)!( + 1) 1+ =0 1) +1 !( + 1 + ( + 1)) +( +1) +1 Proof of 5. Left as an exercise Proof of 6. We have (take into account properties 4 and 5) dx )) = dx x dx x ) + +1 x +1 Similarly dx )) = dx )) x )) + dx )) x ) + )) x ) + By adding the two expressions we get dx )) = +1 Proof of 7. It follows from the proof of 6. that dx )) + x ) = dx )) x ) = +1 We get, by subtraction, x ) = ) + +1 Proof of 8. It follows from property 4 that +1 dx dx dx ) + Proof of 9. Left as an exercise. Example

1. We have proved in Proposition 2 that ) = x sin . In one of the exercises you will be asked to prove that ) = x cos We can use

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BESSEL EQUATIONS AND BESSEL FUNCTIONS 13 property 7 with = 1 2 to deduce that ) + ) = Thus, ) = x sin cos Similar arguments can be used to prove that +(1 2) is an elementary function. Example 2. We can use property 5 with = 1 to get xJ )) xJ xJ ) + ) = xJ or ) = The following table lists the values ) and of ) for some values of between 0 and 10. 0.5 1.0 1.5 2.0 2.5 3.0 1.0000 0.9385 0.7652 0.5118 0.2239 -0.0484 -0.2601 0.2423

0.4401 0.5579 0.5767 0.4971 0.3391 3.5 4.5 5.5 6.5 -0.3801 -0.3971 -0.3205 -0.1776 -0.0068 0.1506 0.2601 0.1374 -0.0660 -0.2311 -0.3276 -0.3414 -0.2767 -0.1538 7.5 8.5 9.5 10 0.3001 0.2663 0.1717 0.0419 -0.0903 -0.1939 -0.2459 -0.0047 0.1352 0.2346 0.2731 0.2453 0.1613 0.0435 By repeated use of property 7, we can get ) for any integer once and ) are known. Example 3. Let use the table to nd (3 5). We have (3 5) = 3801 and (3 5) = 0 1374. By using property 7 with = 1, then = 2 and 4, we get (3 5) + (3 5) = (3 5) (3 5) = 0 4586 (3 5) + (3 5) = (3 5) (3 5) = 0 3868 (3 5) + (3 5) = (3 5)

(3 5) = 0 2044 Example 4. We use the integral property 8 and integration by parts to nd the following integral dx )) dx )) + (2 dx ) + 2 dx ) + 2 ) +

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14 BESSEL EQUATIONS AND BESSEL FUNCTIONS 5. An integral representation of There is an interesting representation of the Bessel functions of the rst kind with integer order in terms of a denite integral. We have the following proposition. Proposition 3. For , we have ) = cos ( n sin d : Proof. Recall the Taylor expansion of the exponential function =0 (the series converges uniformly and absolutely

for j for every R > 0). We have then xt= =0 and e x= =0 1) The product is xt= x= =0 =0 1) =0 =0 1) !)( !) We rewrite this relation as a power series in (so the coecients will depend on ). x= 2)( (1 =t )) =1 ) + =1 We need to show that ) = ). The coecient ) is obtained from the double series by grouping all the coecients of . Thus all term with ) = m; j;k 1) !)( !) or equivalently (by setting ), ) = =0 1) !)( )! = ( 1) =0 1) !)( )! The last series is precisely ). We have then ) = ( 1) ) = The expansion of e x= 2)( (1 =t )) is therefore x= 2)( (1 =t )) ) + =1 1) Now we

evaluate the left side and the right side of the above expression for i = cos sin . For , we have 1) = e in + ( 1) in 2 cos( n ) if = 2 is even sin( n ) if = 2 + 1 is odd

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BESSEL EQUATIONS AND BESSEL FUNCTIONS 15 and x= 2)( (1 =t )) = e ix sin = cos( sin ) + sin( sin It follows that cos( sin )+ sin( sin ) = )+2 =1 ) cos(2 p )+2 =0 +1 ) sin(2 +1) By equating the real and imaginary parts, we get cos( sin ) = ) + 2 =1 ) cos(2 p sin( sin ) = 2 =0 +1 ) sin(2 + 1) Recall the orthogonality of the trigonometric system cos( j ) cos( k ) = cos( j ) cos( k ) = 1 if k ; 0 if k : By using

these orthogonality relations and the above series, we get cos( sin ) cos( n d ) + =1 ) cos(2 p cos( n d ) if is even if is odd Similarly, sin( sin ) sin( n d =0 +1 ) sin(2 + 1) sin( n d ) if is odd if is even By adding these relations we get for that [cos( sin ) cos( n ) + sin( sin ) sin( n )] d which proves the proposition. A immediate consequence of the integral representation is the following Corollary. For every , we have j and lim !1 ) = 0

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16 BESSEL EQUATIONS AND BESSEL FUNCTIONS 6. Exercises Exercise 1. The table bellow lists approximate values of the Gamma function

for values of in the interval [0 1]. Use the table together with the fundamental property of the Gamma function to nd the following values (9 45) (23 10) (6 05) (4 85) (8 85) ( 75) ( 65) ( 01) ( 85) ( 75) 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 ( 19.470 9.513 6.220 4.591 3.626 2.992 2.546 2.218 1.968 1.773 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00 ( 1.616 1.489 1.385 1.298 1.225 1.164 1.113 1.069 1.032 1.00 Exercise 2. The aim of this exercise is to establish the formulas

( )( ( = 2 = cos sin d x > ; y > 0 ( 1. Show that ( )( ) = ds dt dudv Hint : consider the substitutions and 2. Use polar coordinates cos sin to establish formula ( ). 3. Use formula ( ) to establish the following formula ( j;k = cos cos d 1)! ( 1)! 2 ( 1)! 4. Use formula ( ) together with ( + (1 2)) = (2 1)! 1)! . to establish = cos cos d (2 1)!( 1)!( 1)! 1)!(2 + 2 1)! Hint : Use + (1 2) and in formula ( ).) 5. Use the table of values of the Gamma function given in exercise 1 to nd

an approximation of the integral = cos sin d Exercise 3. The Psi function is dened as the logarithmic derivative of : ( ) = ( Use the fundamental property of to show that satises ( + 1) = ( ) + Exercise 4. Write the rst ve terms of the series representation of

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BESSEL EQUATIONS AND BESSEL FUNCTIONS 17 Exercise 5. Use the series expansion of to establish ) = x cos x : You can also establish this formula by using property (5) with = 1 2 and ) = x sin Exercise

6. Repeat the steps of example 1 to show that ) = x cos + sin Exercise 7. Find the expressions of and of Exercise 8. Use the table of values of and to nd the following values 5) ; J (5) ; J (8 5) Exercise 9. Prove that sJ ds xJ ). Exercise 10. Find the integrals dx; dx; dx Exercise 11. Find the integrals +1 dx; dx; dx Exercise 12. Find the integrals )] dx; ds Exercise 13. Show that x dx R Exercise 14. Show that 00 xJ +1 ) = 0 Hint : Use Bessel’s equation and property 4) Exercise 15. Show that ds = 1 Hint : Start with ) = )) and use integration by parts)

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18 BESSEL EQUATIONS AND BESSEL FUNCTIONS Exercise 16. Use the expansion of cos( sin ) involved in the proof of Proposition 3 to show that cos ) + 2 =1 1) sin = 2 =0 1) +1 1 = ) + 2 =1 Exercise 17. Use the integral representation of ) to show that ) = sin( n sin( )) sin d

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