This ODE arises in many physical systems that we shall investigate. + ( + 1)u = 0. (λ + s)x λ + s + ( + 1) a λ. (s + 1)(s + 2) a 0

Transcription

1 Legendre equation This ODE arises in many physical systems that we shall investigate We choose We then have Substitution gives ( x 2 ) d 2 u du 2x 2 dx dx + ( + )u u x s a λ x λ a du dx λ a λ (λ + s)x λ + s λ d 2 u dx a 2 λ (λ + s)(λ + s )x λ + s 2 λ a λ (λ + s)(λ + s )x λ + s 2 λ a λ (λ + s)(λ + s )x λ + s λ 2 a λ (λ + s)x λ + s + ( + ) a λ x λ + s λ [ a s(s ) ]x s 2 + [ a s(s + ) ]x s + a 2 (s + )(s + 2) a (s(s ) + 2s ( + )) λ [ ]x s [ a n+2 (s + n + 2)(s + n + ) a n ((s + n)(s + n ) + 2(s + n) ( + )) ]x s+n +...

2 or a s(s ) s, a s(s + ) s, a and s, a a n+2 (s + n + 2)(s + n + ) a n ((s + n)(s + n ) + 2(s + n) ( + )) or (s + n)(s + n ) + 2(s + n) ( + ) a n+2 (s + n + 2)(s + n + ) For s we have For s we have a n+2 a n+2 n(n + ) ( + ) (n + 2)(n + ) a n (n + 2)(n + ) ( + ) (n + 3)(n + 2) a n a n The even and odd solutions are then ( + ) ( 2) ( + )( + 3) u x 2 + x ! 4! ( )( + 2) ( 3)( )( + 2)( + 4) u 2 x x 3 + 3! 5! x

3 As before the other case does not contribute anything new. If is an integer, then the series terminate at a finite number of terms and the solutions are polynomials. We have λ u, u 2 infinite series λ u infinite series, u 2 x λ 2 u 3x 2, u 2 infinite series λ 3 u infinite series, u 2 x 5 3 x3 λ 4 u x x4, u 2 infinite series λ 5 u infinite series, u 2 x 4 3 x x5 If we multiply a solution by a constant, then it is still a solution. 3

4 If we multiply each polynomial above by a constant so that they have the value at x, then they are called the Legendre polynomials P, where In general we write P where P P x P 2 ( 2 3x2 ) P 3 ( 2 5x3 3x) P 4 ( 8 35x4 3x 2 + 3) P 5 ( 8 63x5 7x 3 + 5x) N n ( ) n (2 2n)! 2 n!( n)!( 2n)! x 2n N / 2 even ( ) / 2 odd The infinite series and the polynomials generate Legendre's functions of the second kind Q where Q u ()u 2 u 2 ()u even odd 4

5 and the most general solution of Legendre's equations is u c P + c 2 Q All Legendre's functions of the second kind are singular at x; this will be important in physical problems. If we make the substitution x cosθ Legendre's equation becomes d du sinθ sinθ dθ dθ + ( + )u which is the form that will arise most often when we solve problems in spherical-polar coordinates. Hermite Equation Now we turn our attention to Hermite' equation d 2 u du 2x + 2αu, α nonnegative integer 2 dx dx As we shall see, the solutions are polynomials called Hermite polynomials. Legendre and Hermite polynomials are the first two examples of a large class of special functions that play a major role in the solutions of partial differential equations representing physical systems. We choose 5

6 We then have u x s a λ x λ a du dx λ a λ (λ + s)x λ + s λ d 2 u dx a 2 λ (λ + s)(λ + s )x λ + s 2 λ Substitution gives a λ (λ + s)(λ + s )x λ + s 2 2 a λ λ λ (λ + s)x λ + s + 2α a λ x λ + s [ a s(s ) ]x s 2 + [ a s(s + ) ]x s + [ a 2 (s + )(s + 2) a (s 2α) ]x s [ a n+2 (s + n + 2)(s + n + ) a n ((s + n) 2α) ]x s+n +... or a s(s ) s, a s(s + ) s, a and s, a a n+2 (s + n + 2)(s + n + ) a n ((s + n) 2α) or a n+2 (s + n) 2α (s + n + 2)(s + n + ) a n λ 6

7 For s we have For s we have a n+2 a n+2 n 2α (n + 2)(n + ) a n (+ n) 2α (n + 3)(n + 2) a n The even and odd solutions are then u 2α 2! x α(2 α) x ! 2( α) u 2 x + x ( α)(3 α) x ! 5! As before the other case does not contribute anything new. Clearly the infinite series terminate if α is an integer. The Hermite polynomials are H H 2x H 2 4x 2 2 H 3 8x 3 2x or in general 7

8 where H n n! N k ( ) k (2x) n 2k k!(n 2k)! N n / 2 n even (n ) / 2 n odd The Hermite polynomials appear in the harmonic oscillator solution of the Schrodinger equation as shown below. The Schrodinger equation for the -dimensional harmonic oscillator with a potential function V 2 kx2 is given by 2 2m d 2 ψ dx kx2 ψ Eψ The standard ODE solution of this equations goes as follows. Let ξ αx α 4 mk 2 λ 2E ω k /2 ω m This gives the new equation d 2 ψ dξ 2 + λ ξ 2 ( )ψ 8

9 ξ ± If we consider the behavior of the equation as we can neglect and the equation becomes d 2 ψ dξ ξ 2 ψ 2 which says that the large ξ behavior of the solution is ψ e ξ2 /2 This suggests that we try a solution of the form which should remove the large ψ H (ξ)e ξ2 /2 ξ behavior from the equation. We get the equation H '' 2ξH '+ (λ )H This is Hermite's equation and provided we choose λ 2n +. If we do not terminate the series the solutions are not square integrable and cannot represent wave functions. The series solutions we found earlier have the recursion relation a k +2 (2k + λ) (k + 2)(k + ) a k If the series does not terminate, then its dominant asymptotic behavior can be inferred from the coefficients of its high terms λ 9

10 lim k a k +2 a k 2 k ξ n e ξ2 This ratio is the same as that of the series for with any finite value of n. This means that the solution is terminated. ψ H (ξ)e ξ2 /2 will not be square-integrable unless the series So we terminate the series and the solutions are the Hermite polynomials. For this choice of λ with associated wave functions we get energy eigenvalues E n ω n + 2 ψ n H n (αx)e α 2 x 2 /2 That is why we are studying all these special equations - we will then recognize them when we solve the equations for real physical systems and can just quote our earlier results. Gamma or Factorial Function Consider the integrals below: e α x dx α, xe α x dx, x 2 e α x dx 2!, x 3 e α x dx 3! α 2 α 3 α 4 H n

11 or If we let α, then x n e α x dx n! α n+ n! x n e x dx where we have defined n! n(n )(n 2)...(2)() and! Generalizing this result we have the Gamma Function Γ(p) and for n integer Γ(n) Now Γ(p + ) x p e x dx for p > x n e x dx (n )! Γ(n + ) n! x p e x dx for p > x p e x This is a recursion relation for the + p x p e x dx Γ function. p x p e x dx pγ(p)

12 What about negative numbers? We can use the recursion relation: Γ(p) p Γ(p + ) which gives Now Γ(.5).5 Γ(.5 + ) 2Γ(.5) Γ(.5).5 Γ(.5 + ) 2 3 Γ(.5) 4 3 Γ(.5) Γ(.5) t /2 e t dt t y 2 dt 2ydy Γ(.5) y e y2 2ydy 2 e y2 dy [ Γ(.5) ] 2 4 e x2 dx e y2 dy 4 e (x2 + y 2 ) dxdy 4 e r2 rdrdφ π /2 2π e r2 rdr 2π d 2 e r2 π Γ(.5) π In addition, since Γ() we have 2

13 lim p Γ(p + ) p Γ( p) for all negative integers Laguerre Equation - The Laguerre equation takes the form We choose We then have Substitution gives x d 2 u du + ( x) 2 dx dx + Nu u x s a λ x λ a du dx λ a λ (λ + s)x λ + s λ d 2 u dx a 2 λ (λ + s)(λ + s )x λ + s 2 λ a λ (λ + s)(λ + s )x λ + s λ a λ (λ + s)x λ + s λ + a λ (λ + s)x λ + s + N a λ x λ + s [ a s(s ) + a s]x s + [ a s(s + ) a s + a (s + ) + Na ]x s +... λ + [ a n+ (s + n + )(s + n) a n (s + n) + a n+ (s + n + ) + Na n ]x s+n +... λ 3

14 or a s(s ) + a s s 2 s a n+ (s + n + )(s + n) a n (s + n) + a n+ (s + n + ) + Na n or a n+ For s we have or a ( ) (s + n) N (s + n + )(s + n) + (s + n + ) a n a n+ n N (n + ) 2 a n (s + n) N (s + n + ) 2 a n N, a (!) 2 2 N N(N ) a 2 2 ( ) 2, a (2!) 2 a k ( ) k N(N )...(N k + ) (k!) 2 ( ) k k! N! k!(n k)! ( )k N k k! It is clear that if k > N, then u L N + a k. Thus we get N N k N ( ) k x k ( ) k k! k k N k k! x k polynomial 4

15 These are the Laguerre polynomials: L L x L 2 2x + x2 They will appear in the hydrogen atom solution of the Schrodinger equation. Before proceeding to the final example, namely, Bessel s equation, let us digress to study generating functions. Generating Functions 2! Legendre Polynomials Redux - Let us consider the function φ(x,h) ( 2xh + h 2 ), h < /2 h f (power series) Differentiation gives ( x 2 ) 2 φ φ 2x 2 x x + h 2 (hφ) h 2 If we substitute a power series we get 5

16 ( x 2 ) h f '' 2xh f ' + h ( + )h f ( x 2 ) f '' 2xf ' + ( + ) f This says that the coefficient of each power of h must separately equal zero. We get ( x 2 ) f '' 2xf ' + ( + ) f This is Legendre s equation. So the solutions are f P Legendre polynomials and φ(x,h) ( 2xh + h 2 ) P h /2 Generating function for the Legendre polynomials Let us generate the first couple of polynomials. Let φ(x,h) ( y) /2 + 2 y y xh x2 2 h P + hp + h 2 P y 2xh h 2. Then 6

17 or P, P x, P x2 2 We can also write Recursion Relations P! x φ(x,h) h Legendre polynomials of different P xp via recursion or recurrence relations. Now consider We have φ(x,t) φ(x,t) t Rearranging we get (degree) are related to each other, i.e. ( 2xt + t 2 ) t n P /2 n n x t ( 2xt + t 2 ) nt n P 3/2 n n n ( x t)φ(x,t) ( 2xt + t 2 ) nt n P n x t n ( ) t n P n (2n + )tt n P n P + t n (np n + (n + )P n+ ) n Equating coefficients of t n n we have the recursion relation 7

18 Therefore, given that (2n + )xp n np n + (n + )P n+ P and P x 3P P + 2P 2 P 2 3x 2 In a similar manner we can find Orthogonality Relations We have np n xp' n P' n φ 2 (x,t)dx ( + t 2 2tx) dx where we have used n m t n+ m Equating coefficients we get t P n P m dx n n( + h) ( ) n+ we get ( n( + t) n( t) ) 2 2k + t 2k h n n k P n P m dx 2 2n + δ nm 8

19 Thus, it is clear that generating functions are very powerful tools. Hermite Polynomials As we have seen, families of functions can be defined by ODEs. We have also seen that they can be defined by means of a generating function. Finally, they can be defined by a differential formula of the form such as the Rodriguez formula for Hermite polynomials. H n ( ) n e x2 which gives H d n 2 dx n e x H 2x H 2 4x 2 2, n,,2,3,... We can get the generating function from this formula by introducing a dummy variable t as follows. H n e x2 Earlier we saw that Therefore, we have n 2 t n e (x t ) t P! x φ(x,h) n 2 t n e2 x t h t 9

20 n t n e 2 x t2 H n n! φ(x,t) generating function In the same way as before we can derive recursion and orthogonality relations. H n+ 2xH n 2nH n H ' n 2nH n H n+ 2xH n + H ' n Differentiating this last relation we get H ' n+ 2H n 2xH ' n + H '' n H '' n 2xH ' n 2nH n which is the Hermite ODE. The solutions are the Hermite polynomials. We also have the orthogonality relations Laguerre Polynomials A similar procedure gives e x2 H n H m dx 2 n n! πδ nm L n n! ex d n dx n φ(x,t) t e xt t ( e x x n ) n L n t n 2

21 (n + )L n+ (2n + )L n xl n nl n xl ' n nl n nl n Differentiation then gives the Laguerre ODE xl '' n + ( x)l ' n + nl n and finally, we have the orthogonality relation Bessel's Equation e x L n L m dx δ nm We now consider the one of the most important equation in physics. ( ) x 2 d 2 dx 2 + x d dx + x2 µ 2 We choose u x s a λ x λ a We then have du dx u L L( x) exists even and odd solutions λ a λ (λ + s)x λ + s λ d 2 u dx a 2 λ (λ + s)(λ + s )x λ + s 2 λ 2

22 Substitution gives or or a λ (λ + s)(λ + s )x λ + s + λ a s(s ) + a s µ 2 a x s + a λ (λ + s)x λ + s λ + a λ x λ + s+2 µ 2 a λ x λ + s λ a s(s + ) + a (s + ) µ 2 a x s a n (s + n)(s + n ) + a n (s + n) + a n 2 µ 2 a n x s+n +... a s(s ) + a s µ 2 a s 2 µ 2 s ±µ a s(s + ) + a (s + ) µ 2 a (s 2 + 2s + µ 2 )a a a n (s + n)(s + n ) + a n (s + n) + a n 2 µ 2 a nn or a n (s + n )(s + n) + (s + n) µ 2 a n 2 λ (s + n) 2 µ 2 a n 2 a Since this last relation implies that all odd coefficients. 22

23 For s µ, we have a n (n + µ) 2 µ 2 a n 2 or after some algebra a 2n and the solution is J µ ( ) n (2) 2n+ µ n!γ(µ + n + ), a ( ) n x 2n+ µ (2) 2n+ µ n!γ(n + µ + ) n 2 µ Γ(µ + ) µ which is the Bessel function of the first kind of order. For s -µ, we have a n (n µ) 2 µ 2 a n 2 or after some algebra and the solution is J µ ( ) n a 2n (2) 2n µ n!γ(n µ + ) ( ) n x 2n µ (2) 2n µ n!γ(n µ + ) n 23

24 which is the Bessel function of the second kind of order which will be important in many physical problems. The most general solution is then u aj µ + bj µ µ. It is singular at x, Bessel functions appear as solutions in an amazing number of physical systems. Bessel Function Properties Orthogonal polynomials such as Legendre, Hermite and Laguerre polynomials are well-suited(can be used as basis functions) for expansions of functions or the description of physical system that are localized near the origin of coordinates. Like Taylor expansions, the further we go away from the origin, the more terms we need. Bessel functions help us get around this difficulty. The Bessel function J n of integral order n satisfies the ODE x 2 J '' n + xj ' n + (x 2 n 2 )J n The generating function for Bessel functions is G(x,t) exp 2 x t t n t n J n 24

25 To explicitly obtain the J n we expand G(x,t) in powers of t. exp 2 x t t exp 2 xt exp x 2 r,s ( ) s x 2 r + s t r s r!s! t r x r t r 2 r! x 2 s r t s r! We change the summation indices to n r-s or r n+s and r+s n+2s which gives or n+2s x G(x,t) ( ) s 2 J n n s n+2s t n ( ) s x (n + s)!s! 2 for n s J n x n 2 n! x2 4 For n < a similar analysis gives J n (n + s)!s! t n J n n! (n + )! + x2 4 2s n 2! (n + 2)!... 2 ( ) s x (s n )!s! 2 for n < s n since Γ(n) for the negative integers. 25

26 We also have J n ( ) n J n for integer n. The generating function gives the recursion relation ( 2 J n + J n+ ) n x J n ( ) J ' n 2 J J n n+ Adding and subtracting we get the ladder operators Since R n J n L n J n n x d dx n x + d dx J n J n+ J n J n L n+ R n J n L n+ J n+ J n we get (L n+ R n )J n which is Bessel s equation. 26

27 One can easily verify that the expansion for J n can have substituted for n where v is any real number and then direct differentiation shows that J ν satisfies Bessel s equation with v in place of n. The factorials in the expansion do not present a problem since they are defined for non-integer values by z! Γ(z + ) e t t z dt In many cases, instead of the J ν functions, one uses the Neuman functions defined by N ν cosνπ J J ν ν sinνπ N ν is linearly independent of J ν. This means it is also a solution of Bessel s equation. It is the second linearly independent solution, i.e., This also holds for W Wronskian J ν N ' v J ' ν N ν ( J ν J ' ν + J ' ν J ν ) / sinνπ ν n 2 independent of ν π x which says the two solutions are linearly independent. v We also have the relations d dx x p J p ( ) x p J p x J p(ax)j p (bx)dx a b 2 J 2 p+(a) 2 J 2 p (a) 2 J ' 2 p(a) a b 27

28 Relationship of the Cylindrical and Spherical Bessel Functions The spherical Bessel functions are defined by: j (z) η (z) π 2z π 2z /2 /2 J +/2 (z) N +/2 (z) Some simple ones are: In general: j (z) sin z z η (z) cos z j (z) j ' (z) sin z z 2 j 2 (z) j z j ' (z) j + j z j cos z z 3 z 3 z sin z 3 z cos z 2 Very important in quantum mechanics. j ( + ) j + (2 + ) j ' 28

29 Some Odd Topics Beta Function [] B(p,q) B(q, p) [2] x y / a B(p,q) y p (a y) q dy [3] x sin 2 θ B(p,q) 2 (sinθ) p (cosθ) q dθ a π /2 [4] x y + y B(p,q) y p [5] B(p,q) Γ(p)Γ(q) Γ(p + q) dy q (+ y) Error Function x erf 2 π e t 2 dt Stirling's Formula (Factorial function for large arguments) From the definition Γ(p + ) p! x p e x dx e p nx x dx Now let x p + y p : x y p, dx pdy 29

30 We then have Now so that or more exactly p p! e p n( p+ y p ) p y p n(p + y p) np + y p pdy np + n + y For p very large n(p + y p) np + n + y p np + p y p y2 2 p +... p p! e p np p p e y2 /2 dy p p e p p e y2 /2 dy e y2 /2 dy p p p e p p e y2 /2 dy p p e p 2π p for large p Γ(p + ) p! p p e p 2π p + 2 p p And now for something completely different... Expansions in Orthonormal Functions Gram-Schmidt in place of ODEs (partial repeat of earlier material from week ) 3

31 Definitions:. Expansion interval [a,b] : the variable x is confined to the interval a x 2. Given two functions f and g defined on the interval [a,b], the inner product of f and g is denoted by (f,g) and is given by b a ( f, g) f * gdx Over the space of complex functions. Note that order is important since ( f, g) ( g, f ) *. 3. If (f,g) the functions f and g are orthogonal. 4. The norm of a function is (f,f). If the norm is finite, the function is square-integrable. If the function has norm, then it is normalized. Any square-integrable function can be normalized by multiplication by a constant. ϕ,ϕ 2, A set of functions defined on the interval [a,b] is said to be orthonormal if ( ϕ i,ϕ ) j δ ij If we have an orthonormal set of functions on the interval [a,b] then we can write any other square-integrable function f in terms of that set by f a j ϕ j where ( ϕ k, f ) a j (ϕ k,ϕ j ) a j δ jk j If this is true for any function f, then set of functions and is called a basis set. j j a k ϕ,ϕ 2,... is complete 3

32 Gram-Schmidt Orthogonalization Method. It turns out that to be a basis set, a set of functions only has to be linearly independent. Linear independence implies that there exists no relation n c λ f λ λ c λ valid for all x, except the trivial one of all. They do not have to be orthonormal...that is only a convenience. Suppose we have a set of n linearly independent, square integrable functions f λ λ,2, 3, 4,...,n ϕ,ϕ 2,...,ϕ n on the interval [a,b]. How do we construct an orthonormal set? There is a systematic process for doing this. It goes as follows:. Choose ϕ N f where N ( f, f ) 2. Choose ϕ 2 as a linear combination of f and f 2 that is orthogonal to ϕ and normalized to. ϕ 2 [ f 2 (ϕ, f 2 )ϕ ] where N 2 ( f 2, f 2 ) (ϕ, f 2 ) 2 N 2 We have just subtracted off the non-orthogonal part! 32

33 3. This generalizes as k ϕ k [ f k (ϕ j, f k )ϕ j ] where N k ( f k, f k ) (ϕ j, f k ) 2 N k Legendre Polynomials Redux Redux j k j Consider the set of linearly independent functions which are the non-negative integral powers of x on the interval [-,] f λ x λ λ,,2, 3, 4,...,n Using the orthogonalization procedure above we get the new set of functions ϕ f where N ( f, f ) dx 2 ϕ N N 2 ϕ [ f (ϕ, f )ϕ ] x xdx N N 2 2 x N where N ( f, f ) (ϕ, f ) 2 and in the same manner ϕ ϕ ϕ x 2 dx () 2 2 / 3 ϕ 2 x x3 3 2 x 35 8 x4 5 4 x x 33

34 Apart from normalization these are the Legendre polynomials, i.e., ϕ j 2 j + P j 2 Thus, we can construct the orthogonal polynomials using linear algebra instead of ODEs. Finally, let us look at some weird numbers that have a habit of popping up in real world systems. Bernoulli Numbers Consider the series x e x n B n n! xn We can use the relation B n d n dx n x e x x B n to determine the Bernoulli numbers. This is a difficult direct process, however. An easier method is to use the uniqueness property of the power series expansions. We can write e x x x + 2! x2 + 3! x x + 2! x + 3! x2 + 4! x

35 We then get + 2! x + 3! x2 + 4! x B + B x + 2! B2x which gives(by uniqueness or equating coefficients of like powers of x) and so on... B In general we have, 2! B + B B 2 3! B + 2! B + 2! B 2 B 2 6 n n k B 2k k We obtain B 2n+, n,2,3,... B 2n ( )n 2(2n)! p 2n, n,2,3,... (2π ) 2n p These numbers are very divergent, i.e., B , B The Bernoulli numbers appear all over the place, i.e., tan x x + 6 x x ( )n 2 2n (2 2n ) B 2n x 2n +... (2n)! 35

36 Bernoulli Functions The functions are defined by It is clear that xe xs e x n B n (s) xn n! B n Bernoulli number B n () Other useful properties are: B' n (s) nb n (s), n,2,3,... B n () ( ) n B n (), n,,2,3,... 36