UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 116C. Problem Set 2. Benjamin Stahl. October 22, d dx J 0(x) = J 1 (x) (1.

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1 UNIVERSITY OF CALIFORNIA - SANTA CRUZ DEPARTMENT OF PHYSICS PHYS 6C Problem Set Benjamin Stahl October, 4 BOAS, P. 59, PROBLEM.-4 Using Equation.9 from Chapter of Boas, it will be shown that: Starting from Equation.9 an then ifferentiating both sies gives: x J (x) J (x) (.) x J p(x) ( ) n ) n+p x n Γ(n + )Γ(n + + p) ( ) n ( ) (n + p) n+p x n+p n Γ(n + )Γ(n + + p) (.) ( ) n ( ) (n) n+p x n+p ( ) n ( ) (p) n+p + x n+p Γ(n + )Γ(n + + p) Γ(n + )Γ(n + + p) n n Now multiplying both sies of the equation by x gives: x x J ( ) n ( ) (n) n+p p(x) x x n+p ( ) n ) n+p + p n Γ(n + )Γ(n + + p) n Γ(n + )Γ(n + + p) ( ) n ( ) (n) n+p x x n+p + p J p (x) Γ(n + )Γ(n + + p) n (.) Using properties of the Gamma function, this result is further simplifie: x x J ( ) n ( ) (n) n+p p(x) x x n+p + p J p (x) n nγ(n)γ(n + + p) ( ) n ) n+p x Γ(n)Γ(n + + p) n (.4)

2 Now re-inexing this result using n n + gives: x x J ( ) n ) (n+)+p p(x) x n Γ(n + )Γ(n + + p + ) ( ) n+ ) n+p+ x n Γ(n + )Γ(n + + p + ) ( ) n ) n+p+ x Γ(n + )Γ(n + + p + ) n x J p+ (x) + p J p (x) (.5) Plugging in p an simplifying proves the esire result: x x J (x) x J (x) + J (x) x J (x) J (x) (.6) BOAS, P. 59, PROBLEM.-6 Using Equation. from Chapter of Boas, it will shown that: N n+ (x) ( ) n+ J n+ (x) (.) Equation. states: Make the substitution p n+ an simplifying yiels : N p (x) cos(πp)j p(x) J p (x) sin(πp) (.) N n+ (x) cos π ( n+ ) J n+ sin π ( n+ (x) J n+ (x) ) (.) Since the argument of the trigonometric functions is always a half integer multiple of π, the cosine terms goes to an the sine term becomes ±. Using these facts yiels the esire result: N n+ (x) J n+ (x) J n+ (x) ( ) n N n+ (x) ( ) n+ J n+ (x) (.4) BOAS, P. 59, PROBLEM.5-8 Using Equation 5.4 from Chapter of Boas, it will be shown that: J (x)x J (x)x J n+ (x)x & J (x)x J (x)x J n (x)x (.) Equation 5.4 states: J p (x) J p+ (x) J p (x) (.)

3 Starting with p an then integrating both sies from to yiels: J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) lim J (x) J () J (x) J (x) J (x) (.) Now starting with p 4 an then integrating both sies from to yiels: J (x) J (x) J (x) J (x) J (x) J 5 (x) J 4 (x) J 5 (x) J 4 (x) J 5 (x) J 4 (x) J 5 (x) lim J 4(x) J 4 () J 5 (x) J (x) J 5 (x) (.4) Thus, this process will continue inefinitely for o values of p. Therefore it has been shown that: J (x)x J (x)x Starting with p an then integrating both sies from to yiels: J n+ (x)x (.5) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) J (x) lim J (x) J () J (x) J (x) J (x) (.6)

4 Now starting with p an then integrating both sies from to yiels: J (x) J (x) J (x) J (x) J (x) J 4 (x) J (x) J 4 (x) J (x) J 4 (x) J (x) J 4 (x) lim J (x) J () J 4 (x) J (x) Thus, this process will continue inefinitely for even values of p. Therefore it has been shown that: J 4 (x) (.7) J (x)x J (x)x It was presumably shown in Problem 7 of this section in Boas that: J (x)x Thus by the proven relations presente in Equations.5 &.8: & J n (x)x (.8) J (x)x (.9) J n (x)x for all integer values of n (.) 4 BOAS, P. 594, PROBLEM.6-9 The solutions to the following ifferential equation will be foun in terms of Bessel functions: x y + y + y (4.) Rewriting this equation so that it matches most closely to Equation 6. of Boas Chapter : y + x y + 4 x y (4.) Matching this equation term by term to Equation 6. yiels the following relations which are then solve to fin the variables that will be use in the expression of the solution: a a a (bcx c ) 4 x b c x c 4x c c c b c 4 b ( ) 4 b 6 b 4 a p c ( ) ( ) p p 4 9 p Thus using Equation 6. from Boas Chapter, the solution can be written as: ( ) y x a Z p (bx c ) x Z 4x Therefore, in terms of Bessel functions the solution can be expresse as: ( ) ( ) y x AJ 4x + BN 4x (4.) (4.4) (4.5) 4

5 5 BOAS, P. 594, PROBLEM.6-: SECOND PART ONLY As instructe, it will be shown that Equation 6. (y x a Z p (bx c )) from Chapter of Boas is inee a solution to Equation 6. (y" + a x y + (bcx c ) + a p c y ). First, the erivatives of y are taken as follows: x y x a Z p (bx c ) y ax a Z p (bx c ) + bcx a+c Z p (bxc ) y" a(a )x a Z p (bx c ) + abcx a+c Z p (bxc ) + (a + c )bcx a+c Z p (bxc ) + b c x a+c Z p "(bx c ) (5.) Now substituting y an its erivatives into the ifferential equation yiels: a(a )x a Z p (bx c ) + abcx a+c Z p (bxc ) + (a + c )bcx a+c Z p (bxc )+ b c x a+c Z p "(bx c ) + ( a)ax a Z p (bx c ) + ( a)bcx a+c Z p (bxc ) + (bcx c ) x a Z p (bx c )+ (a p c )x a Z p (bx c ). (5.) Simplifying this result gives: bc x a+c Z p (bxc ) + (bcx c ) x a Z p "(bx c ) + (bcx c ) x a p c x a Z p (bx c ) (5.) Multiplying this result by by x a yiels: Now multiplying by x yiels: c bc x c Z p (bxc ) + (bcx c ) Z p "(bx c ) + (bcx c ) p c Z p (bx c ) (5.4) bx c Z p (bxc ) + (bx c ) Z p "(bx c ) + (bx c ) p Z p (bx c ) (5.5) Now, if the substitution (bx c ) w is mae, the above simplifies into Bessel s ifferential equation, an thus 6. is in a fact a solution. w Z p (w) w + w Z p(w) w + w p Z p (w) (5.6) x 6 BOAS, P. 65, PROBLEM.- Using the table presente on pg 64 of Boas, the following limit will be evaluate: J 4 (x) lim x J (x) (6.) Substituting the relationships from the table an simplifying gives the esire result: J 4 (x) lim x J (x) lim x lim x ( x Γ(4+) +O(x 4+ ) Γ(+) ( x (!) ( (!) ( ) lim +O(x + x ) +! (!)! + Ax 6 ) Bx 4 + B x lim 8! 4 6 x + Ax 6 ) + Bx 4 x 4 ( ) 4 + Ax x 4 ( (!) +! ) B + B x 4 (6.) 5

6 7 BOAS, P. 66, PROBLEM.- For a Bessel function of the first kin, J p (x), the small x approximation is given in Boas to be: ) p ( +O x p+ ) (7.) Γ(p + ) The the same Bessel function, asymptotic approximation is given in Boas to be: ( πx cos x p + ) ( ) π +O x 4 (7.) Using these formulae, the function J (x) is plotte along with its small x approximation an asymptotic expansion in Figure 7.:.6.4. J (x) & Approximations J (x) Exact Small x Approximation Asymptotic Approximation Figure 7.: J (x) an its approximations over a large interval in x As can be seen in Figure 7., the asymptotic approximation agrees very well with the actual function for suitably larger values of x. It woul also seem that the small x approximation hols well for sufficiently low values of x; to see this better the plot is regenerate over a smaller interval in x:.6.4. J (x) & Approximations (Zoome) J (x) Exact Small x Approximation Asymptotic Approximation Figure 7.: J (x) an its approximations over a small interval in x From Figure 7., it is clear that the small x approximation hols well for low values of x. 6

7 8 BESSEL PROBLEM Originally, this problem aske that the following equation be ifferentiate an substitute into Bessel s ifferential equation to prove that it was a solution: y π cos(x cosh(t)) t (8.) However, it was announce in class that instea it woul be sufficient to show that oing this resulte in the following: x sin(x cosh(t))sinh(t)t (8.) t First, the above result is re-expresse by evaluating the erivative within the integral: x sin(x cosh(t))sinh(t)t t x cos(x cosh(t))x sinh(t)sinh(t) + x sin(x cosh(t))cosh(t)t x sinh (t)cos(x cosh(t)) + x cosh(t)sin(x cosh(t)) t (8.) Next, the first equation will be ifferentiate to fin y an y : y y x π y y x π sin(x cosh(t))cosh(t)t π cosh(t)cos(x cosh(t))cosh(t)t π cosh(t) sin(x cosh(t)) t cosh (t)cos(x cosh(t))t (8.4) Now, y an its erivatives are substitute into Bessel s ifferential equation. It is assume that p because the problem hints that the ultimate result is an integral representation of N. x π x y + x y + (x p )y x y + x y + x y cosh (t)cos(x cosh(t))t + x cosh(t)sin(x cosh(t))t + x cos(x cosh(t))t π π x cosh (t)cos(x cosh(t))t + x cosh(t)sin(x cosh(t))t x cos(x cosh(t))t x cosh (t)cos(x cosh(t))t cos(x cosh(t)) t + x cosh(t)sin(x cosh(t))t x cosh (t) cos(x cosh(t))t + x cosh(t)sin(x cosh(t))t x sinh (t)cos(x cosh(t))t + x cosh(t)sin(x cosh(t))t x sinh (t)cos(x cosh(t)) + x cosh(t)sin(x cosh(t)) t (8.5) Thus as require it has been shown that the original function oes result in the given integral when it is substitute into Bessel s ifferential equation. 7