Math 300 Winter 2011 Advanced Boundary Value Problems I. Bessel s Equation and Bessel Functions

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1 Math 3 Winter 2 Avance Bounary Value Problems I Bessel s Equation an Bessel Functions Department of Mathematical an Statistical Sciences University of Alberta Bessel s Equation an Bessel Functions We use the following bounary value-initial value problem satisfie by a vibrating circular membrane in the plane to introuce Bessel s equation an its solutions. 2 ( u 2 t 2 = u c2 r 2 + u r r + 2 ) u r 2 θ 2, u(r, π, t) = u(r, π, t), u u (r, π, t) = (r, π, t), θ θ u(a, θ, t) =, u(a, θ, t) < as r + u(r, θ, ) = f(r, θ) for < r < a, π < θ < π, an t >. u (r, θ, ) = g(r, θ), t We look for a separate solution of the form u(r, θ, t) = R(r) φ(θ) G(t), an substituting this into the wave equation, we have RφG = c 2 ( R + r R ) φg + c2 r 2 Rφ G, iviing by c 2 RφG, we have G c 2 G = rr + R + φ rr r 2 φ = λ (constant) since the left-han sie epens only on t an the right-han sie epens only on r an θ. Thus, we have the two ifferential equations G + λc 2 G = an r 2 R φ + rr φ + λr 2 Rφ = Rφ, an separating variables again in the secon equation, we have so that r 2 R + rr + λr 2 R R = φ φ = µ (constant) r 2 R + rr + (λr 2 µ)r = an φ + µφ =.

2 We can meet the perioicity conitions an the bounary an bouneness conitions by requiring that φ( π) = φ(π) an φ ( π) = φ (π) an R(a) = an R(r) < as r + Thus, we have the following orinary ifferential equations: (a) Angular problem: φ + µφ =, π < θ < π φ( π) = φ(π) φ ( π) = φ (π) with eigenvalues an corresponing eigenfunctions µ n = n 2 an φ n (θ) = a n cos nθ + b n sin nθ for n. (b) Raial problem: r 2 R + rr + (λr 2 n 2 )R = R(a) = R(r) < as r + (c) Temporal problem: G + λc 2 G = We can put the raial equation into Sturm-Liouville form: ( r R ) + r r with (λr n2 r ) R =, p(r) = r, q(r) = n2, σ(r) = r, r an this eigenvalue problem is a singular Sturm-Liouville problem since p() = σ() =, an q(r) as r +, as well, the bounary conitions are not of Sturm-Liouville form. However, we can still fin the eigenvalues an eigenfunctions. If (λ, R) is an eigenpair of the raial equation, the Rayleigh quotient is a a ] rr(r)r (r) + [rr (r) 2 + n2 r R(r)2 r λ = a. R(r) 2 r r From the bounary conition an the bouneness conition, we have a rr(r)r (r) = ar(a)r (a) + lim r rr(r)r (r) =, + an therefore all the eigenvalues are nonnegative.

3 For λ =, the raial problem is r 2 R + rr n 2 R =, < r < a R(a) =, R(r) < as r +, an the ifferential equation is a Cauchy-Euler equation or an equiimensional equation with solution R (r) = A log r + B, for n =, an R n (r) = A n r n + B n, for n. rn For λ >, we can transform the raial equation into an equation that is inepenent of λ by letting x = λ r, then R r = λ R 2 R an x r 2 = R λ2 x 2, an the equation becomes that is, r 2 2 R r 2 Bessel s equation of orer n (nonparametric). + r R r + (λr2 n 2 )R = x 2 2 R x 2 + xr x + (x2 n 2 )R =, x 2 2 R x 2 + xr x + (x2 n 2 )R =, We will use a power series metho calle the Metho of Frobenius, to fin two linearly inepenent solutions to Bessel s equation. Bessel Functions We look for a solution to Bessel s equation where n is an integer, of the form x 2 2 y x 2 + x y x + (x2 n 2 )y = y = a x m + a x m+ + + a k x m+k +, that is, where a. y = a k x m+k k= Differentiating the series, we have y x = (m + k)a k x m+k an 2 y x 2 = (m + k)(m + k )a k x m+k 2, k= k=

4 substituting this into the ifferential equation we have [ (m + k)(m + k ) + (m + k) n 2 ] a k x m+k + a k x m+k+2 =, k= k= an reinexing the last sum, we have [ (m + k)(m + k ) + (m + k) n 2 ] a k x m+k + a k 2 x m+k =. k= k=2 Therefore, that is, [ (m + k) 2 n 2] a k x m+k + a k 2 x m+k =, k= (m 2 n 2 )a x m + [ (m + ) 2 n 2] a x m+ + This must be an ientity in x, hence, all coefficients must vanish, so that for k 2. k=2 {[ (m + k) 2 n 2] } a k + a k 2 x m+k =. k=2 (m 2 n 2 )a = [ (m + ) 2 n 2] a = [ (m + k) 2 n 2] a k + a k 2 = Since a, we get the inicial equation from the first term m 2 n 2 =, so that m = ±n, an for the moment we choose m = n. Also, with this choice of m, (m + ) 2 n 2, we must have a =. Finally, we get the recurrence relation for k 2. a k = k(2n + k) a k 2 Using the fact that a =, from the recurrence relation we fin that is, a 2k+ = for k. a 3 =, a 5 =, a 7 =, a 9 =,..., Also, from the recurrence relation a 2 = (n + ) 2 2 a a 4 = 2 (n + 2) 2 2 a 2 a 6 = 3 (n + 3) 2 2 a 4. k a 2k = (n + k) 2 2 a 2k 2.

5 If we multiply the terms on the left an multiply the terms on the right, an cancel the common factor a 2 a 4 a 6 a 2k 2, we get for k. a 2k = ( ) k k!(n + )(n + 2) (n + k) 2 2k a = ( )k n! k!(n + k)! 2 2k a Therefore, the solution is (recall that m = n) y = a x n + a 2k x n+2k, k= an it is an easy exercise to show that this series converges absolutely an uniformly for all real numbers x (use the ratio test). Note that the constant a can have any nonzero value, an if we chose then the solution becomes y = an we enote this function by J n (x), so that that is, J n (x) = n! a = n! 2 n, xn n! 2 n + ( ) k n! ( x ) n+2k, k!(n + k)! 2 J n (x) = k= ( x 2 ) n + k= k= ( ) k ( x ) n+2k, k!(n + k)! 2 ( ) k ( x ) n+2k k!(n + k)! 2 for n. The function J n (x) is calle the Bessel function of the st kin of orer n. Note that so that J n ( x) = ( ) n J n (x), J n (x) is an even function if n =, 2, 4,... J n (x) is an o function if n =, 3, 5,... Also, J () =, while J n () = for n. The Bessel function of orer is given by J (x) = k= an since (k!) 2 = (2k) 2, we can rewrite this as J (x) = + k= ( ) k ( x ) 2k, (k!) 2 2 x 2 k (2k) 2 = x x x ,

6 which looks similar to the Maclaurin series for cos x. There are also similarities between the erivatives of J (x) an cos x; for example, J ( ) k (2k) ( x ) 2k ( ) k ( x ) 2k ( ) l+ ( x ) 2l+ (x) = = = (k!) k!(k )! 2 l!(l + )! 2 k= l= that is, k= J (x) = k= for all x. Graphs of J (x) an J (x) are shown below: ( ) k ( x ) 2k+ = J (x) k!(k + )! 2 There are two linearly inepenent solutions to Bessel s equation of orer m x 2 2 y x 2 + x y x + (x2 m 2 )y =. One of them we foun using series solutions: J m (x) = k= ( ) k ( x ) m+2k k!(m + k)! 2 for m =,, 2,..., calle a Bessel function of the st kin of orer m. The Bessel functions J (x), J (x), J 2 (x), an J 3 (x) are plotte below. A secon linearly inepenent solution is Y m (x) = 2 [ ( J m (x) γ + log x ) m (m k )! ( x ) 2k m π 2 2 m! 2 k= ( x ) m+2k ( ) k [φ(k) + φ(k + m)] ] 2, 2 k!(m + k)! k=

7 - & & & & & where φ(k) = k, an γ is the Euler-Mascheroni constant γ = lim ( k ) log k. k The function Y m (x) is calle a Bessel function of the 2n kin of orer m, or a Neumann function, or a Weber function. The functions Y (x), Y (x), an Y 2 (x) are plotte below. # ' # ' # ' # # ' # ' # ' & # '! & "$' # "$' "$'! "# "$ "% ( ) * +,-,$(,%).,. (.. ). / Note: The functions J ν (x) an Y ν (x) can also be efine for values of ν which are not integers, an satisfy Bessel s equation of orer ν : x 2 y + xy + (x 2 ν 2 )y = where ν is a constant (not necessarily an integer). One solution is the Bessel function of the st kin of orer ν ( ) n ( x ) 2n+ν J ν (x) =, Γ(n + )Γ(n + ν + ) 2 n= where Γ is the gamma function efine by for ν >. Γ(ν) = x ν e x x A secon linearly inepenent solution is the Bessel function of the 2n kin of orer ν. Y ν (x) = cos πν J ν(x) J ν (x), sin πν Properties of Bessel Functions Recall: For a nonnegative integer m, for < x <. J m (x) = k= ( ) k ( x ) m+2k ( ) k!(m + k)! 2

8 The following relations hol among Bessel functions an their erivatives, an are true for J m (x) as well as Y m (x), whether or not m is an integer. Theorem. (i) (ii) x [xm J m (x)] = x m J m (x) for m, an x [x m J m (x)] = x m J m+ (x) for m. Proof. We will prove part (i) an leave part (ii) as an exercise. Differentiating ( ), we have [ x [xm J m (x)] = ] ( ) k ( x ) 2m+2k x k!(m + k)! 2 = k= k= = x m ( ) k 2(m + k) k!(m + k)! k= = x m J m (x). ( x 2 ) 2m+2k 2 m ( ) k ( x ) 2k+m k!(m + k)! 2 We also have the following recurrence relations for J m (x) an J m (x) : Theorem 2. (i) J m (x) + m x J m(x) = J m (x) (ii) J m (x) m x J m(x) = J m+ (x) (iii) 2J m (x) = J m (x) J m+ (x) (iv) 2m x J m(x) = J m (x) + J m+ (x) Proof. From the previous theorem we have an x [xm J m (x)] = x m J m (x) + mx m J m (x) = x m J m (x), [ x m J m (x) ] = x m J m (x) mx m J m (x) = x m J m+ (x). x Diviing the first equation by x m an multiplying the secon equation by x m, we get (i) an (ii). Aing an subtracting (i) an (ii), we get (iii) an (iv).

9 Note: From the above, it is clear that every Bessel function J n (x) with n an integer can be expresse in terms of J (x) an J (x), for example, taking m = in (iv), we have an taking m = 2 in (iv), we have J 2 (x) = 2 x J (x) J (x), J 3 (x) = 4 x J (x) ( 8x ) 2 J (x). Also, we can write the ifferentiation formulas an as integral formulas x [xm J m (x)] = x m J m (x) for m, [ x m J m (x) ] = x m J m+ (x) for, x x m J m (x) x = x m J m (x) + C, an where C is constant. x m J m+ (x) x = x m J m (x) + C For example, when m =, the first equation yiels x J (x) x = x J (x) + C. Fourier - Bessel Series Given a fixe nonnegative integer m, the function J m (x) has an infinite number of positive zeros: z nm, n =, 2, 3,..., so that J m (z nm ) = for n =, 2, 3,.... Suppose that we want to expan a given function f(x) in terms of a fixe Bessel function, that is, f(x) = a n J m (z nm x) = a J m (z m x) + a 2 J m (z 2m x) + a n J m (z nm x) +, n= where f(x) is efine for x an the z nm s are the positive zeros of J m (x). In orer to etermine the coefficients a n, we nee an orthogonality relation just as we i for Fourier series. In this case, however, we are expaning the function f(x) in terms of a fixe Bessel function J m (x) an the series is summe over the positive zeros of J m (x). Recall that the Bessel function J m (x) was a solution to a singular Sturm-Liouville problem, an we will show that the eigenfunctions J m (z nm x), for n =, 2, 3,..., are orthogonal on the interval [, ] with respect to the weight function σ(x) = x.

10 Theorem 3. For a fixe integer m, for n k, an where z nm is the n th positive zero of J m (x). xj m (z nm x)j m (z km x) x = xj m (z nm x) 2 x = 2 J m+(z nm ) 2 Thus, the functions x J m (z nm x) are orthogonal on the interval x. Proof. Note that y = J m (x) is a solution to the ifferential equation y + ) x y + ( m2 y =, an if a an b are istinct positive constants, then the functions x 2 u(x) = J m (ax) an v(x) = J m (bx) satisfy the ifferential equations an u + ) x u + (a 2 m2 u = x 2 v + ) x v + (b 2 m2 v =, x 2 respectively. Multiplying the first of these equations by v an the secon by u, an subtracting, we have an multiplying by x, we have Integrating from to, an since x [u v v u] + x (u v v u) + (a 2 b 2 )uv =, (b 2 a 2 ) x [x (u v v u)] = (b 2 a 2 )xuv. xu(x)v(x) x = [x(u v v u)], u() = J m (a) an v(v) = J m (b), if a an b are istinct positive zeros of J m (x), say z nm an z km, then an since z nm z km, then ( z 2 nm zkm) 2 xj m (z nm x)j m (z km x) x =, J m (z nm x)j m (z km x), x x =, that is, the functions x J m (z nm x) are orthogonal on the interval x. To fin the normalization constant, we note again that for a >, the function J m (ax) satisfies the ifferential equation u + ) x u + (a 2 m2 u =, x 2

11 an multiplying this equation by 2x 2 u, we get 2x 2 u u + 2xu u + 2a 2 x 2 uu 2m 2 uu =, that is, [ ] x 2 u 2 + [ a 2 x 2 u 2] 2a 2 xu 2 [ m 2 u 2] =. x x x Integrating this equation from to, we have an since u() = J m () for m, then at x =, for all m. 2a 2 xu 2 x = [x 2 u 2 + (a 2 x 2 m 2 )u 2], x 2 u 2 + (a 2 x 2 m 2 )u 2 = Also, at x =, u () = x [J m(ax)] = aj m (a), x= so that J m (ax) 2 x x = 2 J m (a) 2 + ) ( m2 2 a 2 J m (a) 2 for m. Now put a = z nm, the m th positive zero of J m (x), then for m. J m (z nm x) 2 x x = 2 J m (z nm ) 2 = 2 J m+(z nm )) 2 Now suppose that we have a function f efine on the interval x, an that it has a Fourier-Bessel series expansion given by f(x) = a k J m (z km x), k= we can fin the coefficients in this expansion by multiplying the equation by xj m (z nm x) an integrating, to get Therefore, if then for n. a n = f(x)j m (z nm x)x x = a n 2 J m+(z nm ) 2. f(x) = a n J m (z nm x), n= 2 J m+ (z nm ) 2 f(x)j m (z nm x) x x We have a convergence theorem for Fourier-Bessel series similar to Dirichlet s theorem:

12 Theorem 4. (Fourier-Bessel Expansion Theorem) If f an f are piecewise continuous on the interval x, then for < x <, the series a n J m (z nm x ) converges to [ f(x + 2 ) + f(x )]. n= At x =, the series converges to, since every J m (z nm ) =. At x =, the series converges to if m, an to f( + ) if m =. Vibrating Circular Membrane We now return to the vibrating circular membrane problem we starte with: 2 ( u 2 t 2 = u c2 r 2 + u r r + 2 ) u r 2 θ 2, u(r, π, t) = u(r, π, t), u u (r, π, t) = (r, π, t), θ θ u(a, θ, t) =, u(a, θ, t) < as r +, u(r, θ, ) = f(r, θ), u (r, θ, ) = g(r, θ), t for < r < a, π < θ < π, an t >. Separating variables, we assume that u(r, θ, t) = R(r)φ(θ)G(t), an obtaine three orinary ifferential equations Temporal equation: G + λc 2 G = t > ; Angular equation: φ + µφ =, π < θ < π φ( π) = φ(π), φ ( π) = φ (π); Raial equation: r 2 R + rr + (λr 2 µ)r =, R(a) =, R(r) < as r + ; where λ an µ are separation constants. < r < a For the angular equation, the eigenvalues an eigenfunctions are for m. µ m = m 2 an φ m (θ) = a m cos mθ + b m sin mθ

13 We saw earlier using the Rayleigh quotient, that all the eigenvalues of the raial problem are nonnegative. For λ =, the raial equation is r 2 R + rr m 2 R =, R(a) =, R(r) < as r +, < r < a an the ifferential equation is a Cauchy-Euler equation or an equiimensional equation with solution R (r) = A log r + B, for m =, an R m (r) = A m r m + B m, for m. rm Since the solutions must be boune at r =, then we have A = an B m = for all m, so that R (r) = B an R m (r) = A m r m, an applying the bounary conition R(a) =, we see that B = an A m = for all m. Thus, λ = is not an eigenvalue. For λ >, the raial equation is with general solution r 2 R + rr + (λr 2 m 2 )R = R(a) = R(r) < as r +, R(r) = AJ m ( λ r) + BY m ( λ r), an since Y m is not boune as r +, then we must have B =, an the solution is The bounary conition R(a) = gives us R(r) = AJ m ( λ r). AJ m ( λ a) =, an in orer to get a nontrivial solution we must have J m ( λ a) =. The eigenvalues are those values of λ for which J m ( λ a) =, that is, for which λ a = z nm, where z nm is the n th positive zero of J m : λ nm = for n an m. ( znm ) 2 a The corresponing eigenfunctions are for n an m. R nm (r) = J m ( znm a r ) Note: For a fixe m, these eigenfunctions form an orthogonal basis for the linear space P W S[, a] of piecewise smooth functions on [, a] with weight function σ(r) = r, so that a ( znm ) ( J m a r zkm ) J m a r r r = for n k.

14 The generalize Fourier series is now calle a Fourier-Bessel series, an for w P W S[, a] we have w(r) = n= a n J m ( znm a r ), where for n. a n = a a ( znm ) f(x)j m a r r r J m ( znm a r ) 2 r r We only nee to solve the time equation for those values of λ for which we have a nontrivial solution to the raial equation, that is, λ = λ nm, an the time equation is G + c 2 λ nm G =, with general solution for n an m. G nm (t) = A cos(c λ nm t) + B sin(c λ nm t) For each m, n, the prouct solution u nm (r, θ, t) = R nm (r) φ m (θ) G nm (t) satisfies the PDE, the bounary conition, an the bouneness conition. In orer to satisfy the initial conitions we use the superposition principle to write u(r, θ, t) = m= n= for < r < a, π < θ < π, t >. { J m λnm r )[ a nm cos mθ cos ( c λ nm t ) + b nm cos mθ sin ( c λ nm t ) + c nm sin mθ cos ( c λ nm t ) + nm sin mθ sin ( c λ nm t )]} Now we apply the initial conitions to etermine the constants a nm, b nm, c nm, an nm. For simplicity, we assume that u (r, θ, ) = g(r, θ) =, t an this implies, after ifferentiating term-by-term an setting t =, that for m an n, so that the solution becomes u(r, θ, t) = Setting t =, we have m= n= f(r, θ) = u(r, θ, ) = b nm = an nm = J m λnm r ) cos ( c λ nm t )[ a nm cos mθ + c nm sin mθ ]. m= n= J m λnm r )[ a nm cos mθ + c nm sin mθ ], which is a Fourier series for f(r, θ) on the interval [ π, π] holing r fixe.

15 Therefore, n= a n J λn r ) = π f(r, θ) θ, for m = 2π π a nm J m λnm r ) = π n= c nm J m λnm r ) = π n= π π π π f(r, θ) cos mθ θ, for m f(r, θ) sin mθ θ, for m. Note: These Fourier series coefficients are actually Fourier-Bessel series, so that a n = a nm = c nm = 2π π a π π π π π π π a a a f(r, θ)j λn r ) r r θ J λn r ) 2 r r, for m =, n f(r, θ)j m λnm r ) cos mθ r r θ a J m λnm r ) 2 r r, for m, n f(r, θ)j m λnm r ) sin mθ r r θ a J m λnm r ) 2 r r, for m, n