Laplace s Equation in Cylindrical Coordinates and Bessel s Equation (II)

Transcription

1 Laplace s Equation in Cylinrical Coorinates an Bessel s Equation (II Qualitative properties of Bessel functions of first an secon kin In the last lecture we foun the expression for the general solution of Bessel s equation. More specifically, we have learnt that this solution is a linear combination of a first kin an secon kin Bessel functions. We have also seen that the orer of these functions coul be any real number. In this lecture we are mainly intereste to stuy the properties of Bessel functions whose orer is an integer, in view of later applications. Let us first look at the plots for Bessel functions of the first kin (see Figure. The following properties can be associate to these plots: all J m s are oscillating functions, crossing the x-axis an infinite number of times. The zeroes Figure : Plot of Bessel functions of the first kin, for orers, an 2.

2 Figure 2: Plot of Bessel functions of the secon kin, for orers, an 2. of each Bessel function of the first kin o not follow a perioic pattern, like for the sine an cosine functions; the amplitue of each J m ecreases at high values of x, i.e. lim J m(x =, m =,,2,3,... ( x + all J m s ten to zero when x approaches zero. Only J tens to a ifferent value, which is : lim J m(x =, m =,2,3,..., lim J (x = (2 x x specific values for Bessel functions of the first kin are tabulate in various textbooks, or can be compute using computer algebra systems, like Maple. Plots for the Bessel functions of the secon kin are illustrate in Figure 2. For these function properties similar to Bessel functions of the first kin hol. More specifically: all Y m s are oscillating functions, crossing the x-axis an infinite number of times. The zeroes of each Bessel function of the secon kin o not follow a perioic pattern; the amplitue of each Y m ecreases at high values of x, i.e. lim Y m(x =, m =,,2,3,... (3 x + all Y m s ten to when x approaches zero. In such cases we can approximate these functions with a logarithm an powers of /x: lim Y m(x x + { ln(x m = /x m m =,2,3,... (4 2

3 x J J Y Y Table : Tabulate values for some Bessel functions specific values for Bessel functions of the secon kin are tabulate in various textbooks, or can be compute using computer algebra systems. Some tabulate values for J, J, Y an Y are given in Table. 2 Four useful properties of Bessel functions There are some interesting properties relate to J m an Y m which save us the effort of looking up tabulate values too many times. These properties link in a recursive way functions of consecutive orers an their first erivatives. We can, therefore, erive values for functions of a certain orer starting from values of functions of a ifferent orer. FIRST PROPERTY. Consier the following expression: x m J m (x = n= ( n x 2n+2m 2 2n+m n!γ(m + n + Let us calculate its first erivative: x [xm J m (x] = n= ( n (2n + 2mx 2n+2m 2 2n+m n!γ(m + n + = x m x m n= ( n 2(n + mx 2n 2 2n+m n!γ(m + n + Now, using a Gamma function property, accoring to which Γ(m + n + = (n + mγ(n + m, an carrying out few simplifications, { } x [xm J m (x] = x m ( n x 2n 2 2n+m n!γ(n + m n= The quantity in brackets is J m (x. We have, thus, prove that: Property x [xm J m (x] = x m J m (x (5 SECOND PROPERTY. We can prove, in an analogous fashion, that: 3

4 Property 2 In fact, we start from the expression for x m J m (x: x [x m J m (x] = x m J m+ (x (6 x m J m (x = We then calculate the first erivative: x [x m J m (x] = n= n= ( n x 2n 2 2n+m n!γ(n + m + ( n 2nx 2n 2 2n+m n(n!γ(n + m + = ( n x 2n 2 2n+m (n!γ(n + m + We know that, thanks to the properties of Gamma function, a negative factorial is an infinite quantity. Thus the first term of the summation is zero, an it will start from n = : x [x m J m (x] = n= n= ( n x 2n 2 2n+m (n!γ(n + m + In orer to go back to a summation starting from, we can replace n by n + throughout the whole expression. Thus, { ( n+ x 2n+ } x [x m J m (x] = 2 2n+m+ n!γ(n + m + 2 = x m x m+ ( n x 2n 2 2n+m+ n!γ(n + m + 2 n= The expression in brackets is J m+ (x. We have, thus, erive property (6. THIRD AND FOURTH PROPERTY. Let us now expan the left-han sie of equation (5, an the left-han sie of equation (6 n= mx m J m + x m J m = x m J m, (7 mx m J m + x m J m = x m J m+ Let us, also, multiply both members of this last expression by x 2m, to obtain mx m J m + x m J m = x m J m+ (8 The thir property is simply foun by subtracting (8 from (7: Property 3 J m (x + J m+ (x = 2m x J m(x (9 For the fourth property we have to a ((7 to (8 an ivie the result by x m : Property 4 J m (x J m+ (x = 2J m(x ( Properties analogous to those just escribe hol for Bessel functions of the secon kin, Y m (x. Let us look at how these properties can be use, in the following three examples. 4

5 EXAMPLE. Fin the expression for J 3/2 an J 3/2, starting from the following known functions: J /2 (x = 2 2 πx sin(x, J /2(x = cos(x πx ( In this particular case it is convenient to use equation (9. Replacing in it m = /2 we get: J /2 + J 3/2 = 2 /2 J x /2 J 3/2 = x J /2 J /2 Finally, substituting the analytic expression for functions J /2 an J /2 : J 3/2 (x = [ ] 2 sin(x cos(x πx x We fin J 3/2 (x using again property 3, this time with m = /2. The final result is: EXAMPLE 2. Compute the following integral: If we re-write the above integral as [ 2 cos(x J 3/2 (x = πx x 2 2 J 4 (x x 3 x x 3 J 4 (xx, ] + sin(x we immeiately realise that property 2 coul be use avantageously. In fact: an, thus x [x 3 J 3 (x] = x 3 J 4 (x 2 2 x 3 J 4 (xx = x [x 3 J 3 (x]x = [x 3 J 3 (x] 2 = J 3 ( J 3(2 8 Suppose, now, we only have at our isposal Table. We nee, then, to fin J 3 as a function of J an J. This is reaily one by applying formula (9 twice, recursively: J 3 (x = 4 x J 2(x J (x, J 2 (x = 2 ( 8 x J (x J (x J 3 (x = x 2 J (x 4 x J (x We have, then, using values in Table, an the integral gives: J 3 ( = 7J ( 4J ( 7(.445 4(.7652 =.955 J 3 (2 = J (2 2J ( ( = x 3 J 4 (xx = J 3 ( J 3(

6 EXAMPLE 3. Fin the solution of the following Bessel s equation subject to conitions x 2 R + xr + x 2 R = R( =, R ( = We know that the general solution of the given equation is: R(x = c J (x + c 2 Y (x It is left to fin C an c 2. From the given conitions we get: { c J ( + c 2 Y ( = c J ( + c 2Y ( = To compute J an Y let us use equation (6, with m = : { J (x = J (x Y (x = Y (x (2 The given conitions are then escribe by the following system: { c J ( + c 2 Y ( = c J ( c 2 Y ( = which, after having replace values from Table, gives c an c The solution we were looking for is, thus R(x =.36576J (x.574y (x 3 Orthogonality of Bessel functions of the same orer We are alreay use to expressing a function as a series expansion of other functions. Taylor an Maclaurin series are common ways of expaning a function as an infinite summation of powers of x. Another typical application is the expansion in terms of sines an cosines, the Fourier series. If f(x is a perioic function of perio 2L, the associate Fourier series is efine as follows: f(x = a 2 + n= [ ( nπ ( nπ ] a n cos L x + b n sin L x (3 where the coefficients a, a n an b n are calculate using the following formulas: a n = L 2L ( nπ f(xcos x L x, b n = L 2L ( nπ f(xsin L x x, n =,,2,... (4 The sines an cosines in (3 form an infinite set of functions with an interesting property: 2L ( nπ ( mπ cos L x cos L x x = 2L 2L ( nπ ( mπ sin L x sin L x x = ( nπ ( mπ cos L x sin L x x = 6 { if n m L if n = m

7 i.e. the integral of the prouct of any two functions belonging to the infinite set is always zero, unless the two functions coincie. This property is calle orthogonality of functions in virtue of an analogy to the scalar prouct of two orthogonal vectors in geometry. In general, the set of functions {ϕ (x,ϕ (x,ϕ 2 (x,...}, efine in a given interval I, is sai to be orthogonal if: { if n m ϕ n (xϕ m (xp(xx = (5 K if n = m where integration is performe in the interval I, p(x is a so-calle weight function, associate to the specific set {ϕ (x,ϕ (x,ϕ 2 (x,...}, whose task is mainly to allow the convergence of integral (5, an K is a constant. Orthogonal functions are the favourite tool in mathematical physics when a series expansion is neee. This is ue to the relative easiness of fining the coefficients associate to the expansion. In fact, let us expan an arbitrary function f(x using the general set previously introuce: f(x = a n ϕ n (x (6 n= Let us now multiply both members by ϕ m (xp(x an integrate: f(xϕ m (xp(xx = a n n= ϕ n (xϕ m (xp(xx Orthogonality means that the integral on the right-han sie is zero unless n = m. We have, thus f(xϕ m (xp(xx = a m This expression gives a formula to compute all coefficients of the expansion (6: a n = f(xϕ n (xp(xx (7 You can verify that this is true, for instance, for the Fourier series coefficients. The reason why we have introuce the orthogonality of functions at this stage is because sets of orthogonal functions can be forme starting from Bessel functions of the first kin. We are going to prove now that, if α an β are two zeroes of any Bessel function J m (x, then the two functions J m (αx an J m (βx are orthogonal. More specifically: { if α β xj m (αxj m (βxx = J 2 m (α/2 if α = β (8 We can immeiately notice that (8 is an integral of type (5, introuce to efine orthogonality, with a weight function equal to x. Let us now procee to prove result (8. We know that a solution of the parametric Bessel s equation x 2 f + xf + (λ 2 x 2 m 2 f = is function J m (λx. Let us, next, choose two arbitrary zeroes, α an β, of J m (x, an introuce two new functions, u(x = J m (αx an v(x = J m (βx, both solutions of the following parametric Bessel s equations: x 2 u + xu + (α 2 x 2 m 2 u = x 2 v + xv + (β 2 x 2 m 2 v = 7

8 Let us multiply the first equation by v, the secon by u, an subtract the secon from the first. We get: x 2 (u v uv + x(u v uv + (α 2 β 2 x 2 uv = x 2 x (u v uv + x(u v uv = (β 2 α 2 x 2 uv or, iviing by x an re-arranging x [x(u v uv ] = (β 2 α 2 xuv Finally, integrating between an, an using u an v efinitions: (β 2 α 2 xu(xv(xx = [x(u v uv ] xj m (αxj m (βxx = αj m(αj m (β βj m (αj m(β β 2 α 2 (9 We sai that α an β are zeroes of J m, thus J m (α = J m (β =. If α β, then, the above integral is zero. Consequently, the first part of equation (8 has been prove. If α = β the fraction at (9 gives the ineterminate form /. We can bypass this ifficulty by using L Hopital rule an compute the limit of the erivatives of both numerator an enominator: Given that J m (α =, the above limit gives: αj m lim (αj m (β J m(αj m (β βj m(αj m (β β α 2β αj 2 m (α = 2α 2 J 2 m (α Thus, also the secon part of equation (8 has been prove. The result can be easily generalise to an interval [, a], with a a real, positive number. It will suffice to repeat all passages changing the integration extremes, or, more simply, to replace x with ax in integral (8. Thus, the orthogonality relation for the interval [,a], is: a ( xj m α x ( J m β x x = a a { if α β a 2 J 2 m (α/2 if α = β To summarise, we have foun infinite sets of functions which, starting from Bessel functions of the first kin, can be use to expan any well-behave function between an a. More specifically, if f(x is the function to be expane, an m is the orer of the Bessel functions use in the expansion process, ( x f(x = a n J m λ n (2 a n= where λ n, n =,2,..., are the zeroes of J m. Coefficients a n can be calculate using orthogonality relations (2. In fact, multiplying both members of (2 by xj m (λ l x/a an integrating between an a, we obtain: (2 8

9 a ( x xf(xj m λ l x = a a ( x ( a n xj m λ n J m a n= a ( x xf(xj m λ l x = a l a 2 a2 J 2 m (λ l So we have, in general, 2 a ( x a n = a 2 xf(xj J 2 m λ n x (22 m (λ n a Of course, we know that the only values to be tabulate are those for Bessel functions an not for their erivatives. But, using for instance properties (9 an (, we have J m (x = m x J m(x J m+ (x Now, J m (λ n =, thus the above relations give simply J m (λ n = J m+ (λ n. Eventually: 2 a ( x a n = a 2 Jm+ 2 (λ xf(xj m λ n x (23 n a EXAMPLE 4. Expan, in the interval [,], function x 2, using Bessel functions of the first kin of orer zero. We are looking for the coefficients of the following expression: x 2 = a n J (λ n x n= where λ,λ 2,λ 3,... are zeroes of J (x. using formula (23 we can compute the expression for the generic coefficient a n : a n = 2 J 2 (λ n x( x 2 J (λ n xx Let us integrate by parts, where xj (λ n x is the term to be integrate an x 2 the term to be erive. We have, first, with a change of variable, an using (5: xj (λ n xx = λ 2 tj (tt = n λ 2 tj (t = xj (λ n x n λ n Integration by parts gives, then: x( x 2 J (λ n xx = λ n [x( x 2 J (λ n x] + 2 λ n λ l x a x x 2 J (λ xx = 2 λ n Again, with a change of variable, an using (5 for a secon time, we obtain a n = 4 λ n J 2 (λ n x 2 4 J (λ xx = λ n J 2(λ n 4 = λ n J 2(λ n λ 3 n λ 3 n [(λ n x 2 J 2 (λ n x] = 4J 2(λ n λ 2 n J2 (λ n So, the final expression for x 2 expansion is: x 2 J 2 (λ n = 4 λ 2 n J2 (λ n J (λ n x n= (λ n x 2 J (λ n x(λ n x x 2 J (λ xx In Figure 3 function x 2 is plotte against the first four partial sums of the expansion. The approximation is goo enough even if we consier just the first three terms. 9

10 Figure 3: In these four plots, the thick curve represents function x 2, while the thin curves represent the first four partial sums of its expansion in series of Bessel functions of the first kin, of orer zero. The approximation is alreay goo just by consiering the first three terms of the expansion. 4 A bounary-value problem for Laplace s equation. Steaystate temperature in a cyliner Laplace s equation in cylinrical coorinates has given us the opportunity of introucing an solving Bessel s equation. After having escribe its solutions an properties, it is now time to solve Laplace s equation in a specific case, e.g. heat conuction through a semi-infinite, soli cyliner of raius a. Let us assume that the cyliner is positione as shown at Figure 4. The lateral sie of the cyliner is constantly maintaine at zero egrees centigraes, while its bottom sie (the cyliner is semi-infinite is at the uniform an constant temperature of ξ egrees centigrae. If we allow for a certain amount of time to elapse since the material at ξ egrees was first put in contact with the bottom sie, then heat propagation will be escribe by a steay-state istribution of temperature, u(r,θ,z, obeying Laplace s equation, 2 u =. The bounary-value problem is given by the following relations: 2 u = u(a,θ,z = u(r,θ, = ξ This problem, as seen in Lecture 6, is solve through separation of variables. We postulate, (24 u(r,θ,z = R(rT(θZ(z (25 an fin three ifferent equations for r, θ an z. For the variable z we have: Z(z = a e lz + a 2 e lz with l a positive constant. Only values of z in the positive irection nee to be taken care of, as the cyliner extens only in that irection. Therefore a = an the negative exponent will

11 Figure 4: The temperature on the lateral surface of this semi-infinite cyliner is kept at the constant value of zero egrees centigrae. Its bottom sie is kept at the constant temperature of ξ egrees centigrae. The propagation of heat, an therefore the temperature istribution u(r,θ,z obeys Laplace s equation insie the cyliner. make sure that the temperature ecreases as we get farer an farer from the heat source at z =, a behaviour expecte to occur in physical cases like this. For the θ variable, the general solution is: T(θ = k sin(mθ + k 2 cos(mθ where m is a positive number. The temperature at θ an θ+2π has to be the same. Therefore m nees to be an integer. Furthermore, given that the bounary conitions o not epen on θ, we shoul not notice any θ-epenence in the temperature u either, an this can be mathematically true only if this integer is zero. In such a case, in fact, the T function will become, T(θ = k 2 The thir equation, concerning variable r, is known to possess the general solution, R(r = c J (lr + c 2 Y (lr, given that the bounary conitions have selecte m =. The temperature has to be a finite quantity at r =. Thus, Y cannot be inclue in the final solution, as it is negatively infinite at r =. To summarize, the temperature function u(r,θ,z is given, using equation (25, by the following prouct, u(r,θ,z = c J (lrk 2 a 2 e lz bj (lre lz (26 where it is clear we have merge the three constants into a single constant, b. Let us now use the first bounary conition: u(a,θ,z = bj (lae lz = J (la = A Bessel function of the first kin has an infinite number of zeroes. Therefore the above equation is solve for la = λ n, n =,2,3,..., where λ n are the zeroes of J. This means that not all real values are allowe for l, only a icrete infinite number, given by, l = l n λ n a, n =,2,3,... (27

12 Using equations (26 an (27 we can, thus, affirm that for this particular problem Laplace s equation has an infinite number of solutions, u n (r,θ,z, given by: ( λn u n (r,θ,z = b n J a r e λnz/a, n =,2,3,... The general solution u(r,θ,z is, consequently, given by a linear combinations of the u n s: u(r,θ,z = n= ( λn b n J a r e λnz/a (28 The expansion coefficients can be compute using the last bounary conition, u(r,θ, = ξ n= ( λn b n J a r = ξ This is similar to equation (2. Thus, the expansion coefficients can be erive through the following integral: 2ξ a ( r b n = a 2 J 2(λ rj λ n r n a Through the substitution x = λ n r/a, the above integral is transforme into, b n = 2ξ λn λ 2 nj 2(λ xj (xx n Using the secon of the four properties of Bessel functions, we can write xj (x as [xj (x]/x. Therefore, Finally, λn b n = xj (xx = [xj (x] λn = λ n J (λ n 2ξ λ 2 nj 2 (λ n λ nj (λ n = The complete solution of bounary-value problem (24 is: u(r,θ,z = 2ξ n= 2ξ λ n J (λ n ( λ n J (λ n J λn a r e λnz/a (29 To evelop a feeling of how this solution rightly escribes a heat-propagation problem, let us limit expansion (29 to the first three terms, an fix ξ = 2 C an the raius a = cm. The approximate solution will be written as, [ u(r, θ, z 4 λ J (λ J λ 3 J (λ 3 J ( λ3 2r e λ z/ 2 + ] e λ 3z/ 2 ( λ 2r ( λ 2 J (λ 2 J λ2 2r e λ 2z/ 2 + To compute the above expression numerically at any point (r,θ,z, we nee to know the value of the first three zeroes for Bessel function J, an J at these three values. Using tabulate values from any book (or using Maple functions BesselJ an BesselJZeros, we fin: λ λ λ J (λ.5947 J (λ J (λ

13 Figure 5: Temperature istribution along half a section of the cyliner. Lighter regions correspon to temperatures closer to 2 C, arker regions to temperatures closer to C. The ouble bump of high temperature at the bottom of the cyliner is an artifact ue to series truncation. By consiering more terms in series (29, the ouble bump gives place to a continuous smooth high-temperature region. So, our truncate expansion gives: u(r,θ,z 4 [.899J ( re z.5324J (55.278re z J ( re z] From the form of this approximation, it is quite clear that each term will be ampe by the exponential as the z variable increases. Therefore the temperature will be lower an lower when we procee from the bottom to the top of the cyliner. Also, given that J is at r =, an goes to zero at the cyliner bounary, the temperature will be higher along the axis of the cyliner, an lower near its surface, because this is physically maintaine at zero egrees centigraes. A section of the cyliner, with a fille contour-plot representing temperature istribution, is shown at Figure 5. What we have just sai about the temperature istribution in the cyliner can be immeiately recognise visually in the figure. The more terms will be inclue in series (29, the closer the temperature istribution approximate its real state. 3

14 One last comment concerns cases where the bounary conitions epen on θ. In this case the solution u(r, θ, z will be a ouble summation over iscrete values of sine an cosine arguments an over iscrete values of l. We will, in such a case, eal with an infinite set of Bessel functions an, for each function, with an infinite set of zeroes. The calculations become massive, but the conceptual level of ifficulty is similar to the one of the calculation carrie out here. 4