1 Boas, problem p.564,

Transcription

1 Physics 6C Solutions to Homewor Set # Fall 0 Boas, problem p.564,.- Solve the following differential equations by series and by another elementary method and chec that the results agree: xy = xy +y () by series: substituting the power series y(x) = a nx n in () we have ( ( xy xy y = 0 x na n x ) x n a n x ) n a n x n = 0 () calling n = m+ and substituting, n= (m+)a m+ x m+ a n x n+ a n x n = 0 (3) [(m+)a m+ a m x m+ a 0 a m+ x m+ = 0 (4) The only term with a 0-th power of x is a 0, which tells us a 0 = 0. Asing for the coefficient of the (m+)-th power of x in (4) to be zero we get So the solution for () is a m+ = m a m = mm a m =... = m! a (5) y(x) = a m+ x m+ = a m! xm+ (6) Factoring out one power of x, one recognizes the power series of the exponential e x and writes the solution as y(x) = a xe x (7) by separation of variables: which is the same as in (7). dy y = (+ x )dx = lny = x+lnx+lnc = y(x) = cxex (8) Boas, problem p.564,.-0 Solve the following differential equations by series and by another elementary method and chec that the results agree: y 4xy +(4x )y = 0 (9)

2 by series: substituting the power series y(x) = a nx n in (9) we have n(n )a n x n 4x na n x n +(4x ) a n x n = 0 (0) n= a +6a 3 x+ (n+)(n+)a n+ x n 4a x 4 na n x n +4 a n x n+ a 0 a x a n x n = 0 a +6a 3 x 4a x a 0 a x+ [(n+)(n+)a n+ (+n)a n +4a n x n = 0 the first terms give while the recursion formula is a = a 0, a 3 = a () (n+)(n+)a n+ (+n)a n +4a n = 0, for n. () Consider first the case of n = p even. Using (), we can use the recursion formula to obtain a 4. By repeated use of the recursion formula, we can obtain a 6, a 8,... After computing a few values, it appears that the general form is a p = a 0 p!. (3) Note that (3) is also valid for p = 0 and p =. To test the validity of (3), we insert this equation into (): (p+)(p+) (+4p) 4? + = 0. (4) (p+)! p! (p )! Simple algebra verifies the validity of the equation above. Next, consider the case of n = p+ odd. Using (), we can use the recursion formula to obtain a 5. By repeated use of the recursion formula, we can obtain a 7, a 9,... After computing a few values, it appears that the general form is a p+ = a p!. (5) Note that (3) is also valid for p = 0 and p =. To test the validity of (5), we insert this equation into (): (p+)(p+3) (3+4p) 4? + = 0. (6) (p+)! p! (p )! Simple algebra verifies the validity of the equation above. Hence, we conclude that y(x) = a n x n = a p x p + a p+ x p+ = a 0 x p p! +a x p+ p! [x p [x p = a 0 +a x p! p! = (a 0 +a x)e x. (7)

3 reduction of the order: one checs that the equation is solved by y 0 (x) = e x ; then we loo for another solution of the form y(x) = u(x)y 0 (x): y = u y 0 +uy 0, y = u y 0 +u y 0 +uy 0 (8) y 4xy +(4x )y = u[y 0 4xy 0 +(4x )y 0 +u y 0 +u y 0 4xu y 0 = 0 (9) e x [u 4xu +4xu = 0 = u = 0 = u = A+Bx (0) 3 Boas, problem p.586,.-5 Solve the following differential equations by the method of Frobenius: y = (A+Bx)e x () xy +y +y = 0 () Substituting the generalized power series y(x) = a nx n+s in () we have x (n+s)(n+s )a n x n+s + (n+s)a n x n+s + a n x n+s = 0 (3) the n = 0 term gives [s(s )a 0 +sa 0 x n+s = 0 = s s = 0 = s = 0, (4) while the other terms are [(n+s)(n+s )a n +(n+s)a n +a n x n+s = 0 (5) n= (6) For s = 0 we have a n = n= [n(n )a n +a n x n = 0 (7) n(n ) a n =... = ( ) n n!(n )!! a 0 (8) Where the double factorial is m!! = m(m )(m 4)... Also, we note that (n)! = n(n )(n )(n 3)... = n (n )!!n!. Inserting these coefficients bac in the series gives y(x) = For s = n= a n x n = we have instead ( x) n n!(n )!! a 0 = [n(n+)a n +a n x n = 0 = a n = ( 4x) n ( ) n ( x) n a 0 = a 0 (n)! (n)! n(n+) a n =... = = y(x) = a 0 ( ) n ( x) n+ (n+)! The general solution is then given by the linear combination of (9), (3): = a 0 cos(x / ). (9) ( ) n n!(n+)!! a 0 (30) = a 0 sin(x / ) (3) y = Acos(x / )+Bsin(x / ) (3) 3

4 4 Boas, problem p.587,.-4 Solve y = y by the Frobenius method. We tae the generalized power series y(x) = a nx n+s so that y +y = [(n+s)(n+s )a n x n+s +a n x n+s Taing the n = 0 term gives s(s )a 0 = 0, that is, s = 0,. For s = 0 we have = 0 (33) n(n )a n +a n = 0 = a n = a n n(n ) = { an+ = ( )n (n+)! a a n = ( )n (n)! a 0 (34) These two series are those defining the sine and the cosine, so that we have found the well nown result y = a 0 cosx+a sinx. If we now tae s = we have (n+)nb n +b n = 0 = b n = b n n(n+) (35) The solution for b 0 0 is then y(x) = b n x n+ = b sinx+b 0 n= ( ) n (n)! xn (36) In this expression we are missing the x 0 term that would give the expansion of the cosine, and one easily chec that this is not a solution of the differential equation. That happens because in (33) for s = the first term coefficient reads (n+)(n+0)b n ; then, b 0 coefficient has a 0 in front, which cancels it out from the rest of the problem, so we are calculating the solution modulo the constant b 0. 5 Boas, problem p.567,.- Show that P l ( ) = ( ) l. Using eq. (.6) on p. 565 of Boas, the general solution to the Legendre differential equation is: [ l(l +) y(x) = a 0 x l(l+)(l )(l +3) + x 4...! 4! [ (l )(l +) +a x x 3 (l )(l +)(l 3)(l +4) + x (37) 3! 5! If l is even, then the Legendre polynomial is defined to be the polynomial proportional to a 0 (up to an overall normalization determined by convention). If l is odd, then the Legendre polynomial is defined to be the polynomial proportional to a (up to an overall normalization determined by convention). It immediately follows that if l is even, then P l (x) is an even function of x, whereas if l is odd, then P l (x) is an odd function of x. This means that P l ( x) = ( ) l P l (x). (38) The normalization convention for the Legendre polynomials defines P l () =. Hence, inserting x = into (38) yields P l ( ) = ( ) l (39) 4

5 Note that eq. (38) is also an immediate consequence of the Rodrigues formula, and provides another way of deriving (39). 6 Boas, problem p.567,.-4 We will solve Legendre equation d l P l (x) = l l! dx l(x ) l, (40) ( x )y xy +l(l+)y = 0 (4) using the method of reduction of order: given the nown solution P l (x), we loo for an independent solution of the form y(x) = P l (x)v(x) and then solve for v(x) in (4): ( x )(v P l +v P l +vp l ) x(v P l +vp l )+l(l +)P l = 0 (4) ( x )(v P l (x)+v P l (x)) xp l(x)v = 0 = ( x )P l (x)v +(( x )P l (x) xp l(x))v = 0 = v v = xp l(x) ( x )P l x ( x = )P l (x) x P l = P l x +x P l (43) P l which is solved by lnv = ln( x) ln(+x) lnp l = ln dx v(x) = ( x)(+x)pl The second solution of the Legendre equation is then We evaluate this expression for the two cases l = 0,: ( x)(+x)pl, that is, (44) (45) Q l (x) = P l (x)v(x) (46) l = 0: P 0 (x) =, so the other solution is Q 0 (x) = dx ( x)(+x) = ( dx x + ) = +x +x ln x (47) l = : P (x) = x, so the other solution is Q (x) = x dx ( x)(+x)x = x = x ( dx x + +x + ) x = (48) +x ln (49) x 5

6 7 Boas, problem p.568,.3- We will use the hint in problem.3-6: if we write where D u, D v are operators that act only separately on u,v, we have n ( ) n n ( D n ud v n uv = d n dx nuv = (D u +D v ) n uv = =0 d dx uv = D(uv) = (D u +D v )uv (50) =0 ) D uud n v v = n =0 ( n ) d dn dxudx n v where we have used the expansion ( of the ) n-th power of a binomial formed by the two operators D u,d v n (which commute with each other). = n(n )...(n +)! is the binomial coefficient. 8 Boas, problem p.569,.4- By Rodrigues formula P l (x) = l l! dx l(x ) l (5) we have, after applying Leibniz rule (5) P l (x) = l l! l =0 ( l d l ) d (5) dl dx(x+)ldx l (x )l (53) Now, every time we differentiate (x ) l we lower the exponent by one; in particular, when we differentiate l times, we are left with a constant; when we calculate P l () any factor of (x ) will become zero, so that the only non zero contribution comes from the 0-th term in the sum: this gives P l () = l l! l l! = (54) wheretheterm l comes from (x+) l for x = and dl (x ) l = l dl (x ) l = l(l ) dl (x ) l = dx l dx l dx l... = l!. 9 Boas, problem p.569,.4-4 We want to prove that Substituting Rodrigues formula (5) we have x m d l x m P l (x)dx = 0, for m < l (55) x m P l (x)dx = l l! dx l(x ) l dx (56) x m dl dx l (x ) l m dl mx dx l (x ) l dx = (57) m dl = 0 mx dx l (x ) l m dl + m(m )x dx l (x ) l dx =... (58) 6

7 in the first passage we have neglected the constant and integrated by parts; in the second passage, we l l! see that the first term is null because after we have differentiated (l ) times we will still have at least one factor of (x ) and one of (x+) (as you can quicly chec by using Leibniz rule), which are zero when evaluated at ±. The same happens for all the other terms evaluated at ±, so that, after we have integrated by parts m times (assuming m < l), we are left with d l m m! dx l m(x ) l dx = m! d l m dx l m (x ) l = 0 (59) for the same argument we used above. Note that this does not hold for m > l, because in that case between (58) and (59) we reach a step in which l = 0 and we have x m (x ) l 0 0 Boas, problem p.574,.5-0 Express the following polynomial as a linear combination of the Legendre polynomials: The first five Legendre polynomials are: f(x) = x 4 (60) P 0 =, P = x, P = (3x ), P 3 = (5x3 3x), P 4 = 8 (35x4 30x +3) (6) We aregoing toexpandx 4 as a linear combination of thelegendrepolynomials, withunnowncoefficients; these will be found imposing that the factors for the different powers of x coincide. Because we have x 4, f(x) = 4 0 c np n must contain P 4 ; in particular, c 4 = 8 35, so that the coefficient of x4 is. Then we must put to zero the coefficient of x 3 : x 3 only appears in P 3 so we can put c 3 = 0. Right now, our function is written as f(x) = c n P n P 4(x) (6) Now we fix to zero the coefficient of x : it appears in P 4 and P and it is 0 3 c = 0 = c = 4 7 A term linear in x appears only in P, so we can set c = 0. Finally, the constant term is given by c 0 c c 4 = 0 = c 0 = 5 (63) Then we have found x 4 = 5 P 0(x)+ 4 7 P (x) P 4(x) (64) 7

8 Boas, problem p.577,.6-6 We want to show that P l and P l are orthogonal on [-, in two ways: we can use the fact that the Legendre polynomials are either even or odd functions of x (depending on whether l is even or odd, respectively), as shown in problem 5. Then, if P l is odd, its derivative P l is even, and vice versa. In general, if f(x) is an even function of x and g(x) is an odd function of x, then a a f(x)g(x) = 0. (65) This is easily proven by changing the integration variable to y = x, in which case a a f(x)g(x) = a +a f( y)g( y)dy = a a f( y)g( y)dy = a a f(y)g(y)dy, (66) As the integral is equal to minus itself, it must be equal to zero. Hence, we conclude that P l (x)p l (x) = 0. (67) we can also use the result of problem 9: remember that P l is a polynomial of order l and P l is a polynomial of order (l ). Then P l (x)p l (x) (68) is given by a sum of terms which have the form c n xm P l (x)dx, where m = 0,,...,l, that is, m < l, so they are all zero and the two functions are orthogonal. Boas, problem p.65,.3- The generating functional of the Legendre polynomials is Φ(x,h) = = h l P xh+h l (x); (69) l=0 for x = 0 this gives Φ(0,h) = h l P l (0) = But the function Φ(0,h) loos exactly lie the power of a binomial: l=0 Φ(0,h) = (+h ) / = Here we can read the Legendre polynomials in zero as +h = c l h l, (70) ( ) / h n (7) n P n+ (0) = 0 ; (7) ( ) / P n (0) = = ( )...( n+) = ( n )(n )!! n (n)! n. (73) n! 8