Bessel Function (1)

8/11/2019 Bessel Function (1)

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Math 115 (2006-2007) Yum-Tong Siu 1

Bessel Functions and Vibrating Circular Membrane

Method of Separation of Variables. For a linear partial differential equationLu = 0, we can use the method of separation of variables when the linearpartial differential operatorL can be written as the sum of two linear partialdifferential operatorsPandQso thatPdepends only on variablesx1, , xkand Q depends only on the complementary set of variables xk+1, , xn.Another condition which needs to be satisfied before the method of separationof variables can be applied is that the domain D must be the product of adomain G in the space of variables x1, , xk and a domain H in the spaceof variables xk+1, , xn.

The idea of the method of separation of variables is to consider firstunknown functions u of the formu= vw so that v depends on the variablesx1, , xk andw depends on the variablesxk+1, , xn. The equationLu =0 now becomes wP v+vQw = 0 which can be rewritten as

P v

v =Qw

w

so that the left-hand side depends only on the variables x1, , xk and theright-hand side depends only on the variables xk+1, , xn and as a conse-quence both sides must be equal to some constant . The logic is that if thepartial differential equation Lu = 0 is satisfied on the domain D = G

H

for some function u of the form

u (x1, , xn) =v (x1, , xk) w (xk+1, , xn) ,

then there exists a constant such that

P v

w =Qw

w =,

which means that

P v v= 0 on G,Qw+w = 0 on H.

We then find all such special product solutions u = vjwj (forjJ) and useyet-to-be-determined coefficientscj to find a solution u =

jJcjvjwj which

satisfies the prescribed boundary conditions.


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In practice, we already use some of the prescribed boundary conditions

when we seek special product solutions u = vjwj so as to limit the indexset J to a countable set. Then we use yet-to-be-determined coefficients cjto find a solution u =

jJcjvjwj which satisfies the remaining prescribed

boundary conditions.

Wave Equation for Vibrating Circular Membrane. To present the details ofthe method of separation of variables, we choose to work out the example ofthe wave equation for a vibrating circular membrane. The circular membraneis given by the disk{ 0rc } of radius c >0 in polar coordinates (r, ).The displacement of the membrane at timetin the direction perpendicular tothe disk is given by u (t,r,). The wave equation for the vibrating membrane

is given byutt= a

2u

with boundary conditions

u= 0 at r= c and for allt0,ut= 0 at t= 0,

u= f(r, ) at t= 0,

wherea >0 is a constant determined by the surface tension of the membraneand f(r, ) is a given function on the disk{ 0rc }. The Laplacian uofuin polar coordinates is given by

u= urr +1

rur+

1

r2u.

The wave equation utt= a2ucan be written as Lu= 0 with

L= 1a2

2

t2+

2

r2+

1

r

r+

1

r22

2.

The partial differential operator Lcan be written asP+Qso that the partialdifferential operator

P =1

a22

t2

depends only on the variable t and the partial differential operator

Q= 2

r2+

1

r

r+

1

r22

2


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depends only on the variablesr, . The domainDin the space of the variables

t,r,can be written as the product of the domain { t0 } in the space of thevariable t and the domain{0rc }of the space of the variables r, . Weapply the method of the separation of variables and consider special productfunctions u = T(t)v (r, ). For the special product functionu = T(t)v (r, )we impose the boundary conditions

u= 0 at r= c and for allt0,ut= 0 at t= 0,

and save the remaining boundary condition

u= f(r, ) at t= 0

to be used later to determine the coefficients cj. The method of separationof variables gives us two equations

d2

dt2T(t) +a2T(t) = 0,

2

r2v (r, ) +

1

r

rv (r, ) +

1

r22

2v (r, ) +v (r, ) = 0,

where is a constant and must be nonnegative, because is a nonnegativedifferential operator in the sense that the integral of the product ofgandg is nonnegative for g with compact support as one can see by transformingthe integral to the integral of the gradient square off by using integrationby parts.

The solution of the differential equation

d2

dt2T(t) +a2T(t) = 0

with the boundary conditionTt = 0 for t = 0 gives (up to a nonzero constantmultiple) the solution

T(t) = cosa

tfor each.

Though the partial differential operator

Q= 2

r2+

1

r

r+

1

r22

2


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is not the sum of two partial differential operators with each one depending

only on one of the two variables r, , yet we can change2

r2v (r, ) +

1

r

rv (r, ) +

1

r22

2v (r, ) +v (r, ) = 0

to the new partial differential equation

r2 2

r2v (r, ) +r

rv (r, ) +

2

2v (r, ) +r2v (r, ) = 0

so that the partial differential operator

r22

r2 +r

r + 2

2 +r2

is the sum of the two partial differential operator

r22

r2+r

r+r2

and2

2

with each one depending on only one of the two variables r, . We can

now apply again the method of separation of variables and consider specialproduct functions v (r, ) =R(r) (). We get two differential equations

r2d2R

dr2 +r

dR

dr +r2R R= 0,

d2

dl2+ = 0,

whereis a constant and must be nonnegative, because the operator

d2

d2

is nonnegative as one can see by integration by parts. Recall that we still havethe boundary condition R(c) = 0. There are one other boundary conditionfor R(r) and two boundary conditions for (), which are not as obvious.The other boundary condition forR(r) is that R(r) is finite at r = 0. As we


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will see later that this is indeed an important boundary condition for R(r),

because the differential equation

r2d2R

dr2 +r

dR

dr +r2R R= 0

is not regular at r = 0 in the sense that the coefficient r2 for the term d2Rdr2

ofthe highest-order differentiation is not nonzero at r= 0. The two boundaryconditions for () come from the periodicity of () with period 2. Thesetwo boundary conditions are

(0) = (2), (0) = (2).

For the solution () of a second-order differential equation these two con-ditions are equivalent to the the periodicity of () with period 2.

Second-Order Ordinary Differential Equations with Boundary Conditions In-

volving Both End-Points. Suppose we have a second-order ordinary differen-tial equation

a(x)y+b(x)y+c(x)y= 0

on a finite closed interval [, ] with a(x), b(x), c(x) smooth on [, ] anda(x) nowhere zero on [, ]. For the boundary condition y () = 0 andy () = 0 involving only one single end-point , there is only one solution

y(x) of the differential equation which is identically zero.However, when we have two homogeneous boundary conditions involv-

ing both end-points and , for example, the two homogeneous boundaryconditions

y() = 0 and y() = 0

or the two homogeneous boundary conditions

y() =y() and y() =y (),

there may exist another solutiony(x) other than the identically zero solution.

The existence of non identically zero solutions for certain homogeneouslinear second-order ordinary differential equations subject to two homoge-neous boundary conditions involving both end-points makes it possible toapply the method of separation to use complete system of eigenfunctions fordifferential operators.


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In our analysis of the wave equation for a vibrating circular membrane,

for a nonnegative number we have the second order differential equationd2

d2 + = 0

for the unknown function () on [0, 2] with the two homogeneous boundaryconditions

(0) = (2), (0) = (2).

involving both end-points 0 and 2 of the interval [0, 2]. A general solutionof the second-order differential equation is of the form

sin ( ) +cos( ) with , R.The two homogeneous boundary conditions are equivalent to the conditionthat the solution is periodic of period 2. In order to have a non identicallyzero solution, it is necessary and sufficient that

is an integer. Thus the

constant must be of the form n2 for some nonnegative integer n. When = n2, the two linearly independent solutions are sin n and cos n. Eachof the two functions sin n and cos n is an eigenfunction for the operator

d2

d2

corresponding to the eigenvalue n2, because

d2

d2sin n= n2 sin n,

d2

d2cos n= n2 cos n,

We know from the theory of Fourier series that the set of all eigenfunctions

1, cos nx, sin nx for n N

form a complete system of functions on [0, 2] in the sense that any L2

function (i.e., square-integrable function) on [0, 2] is an infinite linear com-bination of them in the sense ofL2.


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We now use = n2 consider the differential equation

r2d2R

dr2 +r

dR

dr +r2R n2R= 0

for the unknown functionR(r) on [0, c] with the two boundary conditions

R(x) is finite at x= 0 and R(c) = 0.

We get rid of with the rescaling which replaces r byr as the variable. In

other words, we definey = y(x) byR(r) =y

r

so thaty(r) =R

1

r

.

Theny= y(x) satisfies the differential equation

() x2y+xy+ x2 n2 y= 0,because

xdy

dx(x) =

1

x

R

1

x

,

x2d2y

dx2(x) =

1

x

2R

1

x

and

x2y+xy+

x2 n2 y=

1

x

2R

1

x

+

1

x

R

1

x

+

1

x

2 n2

R

1

x

.

The boundary conditions become y(x) finite at x = 0 and y(c) = 0. Thedifferential equation () is known as Bessels differential equation for theBessel function of order n. We are going to solve this equation by using a

generating function which is a Laurent series in a new indeterminate t whosen-th coefficient is the Bessel function of order n. We are going to write downthe generating function and verify that its n-th coefficient satisfies Besselsdifferential equation for the Bessel function of order n.


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Generating Function for Bessel Functions. We now introduce the generating

function for the Bessel functions. It is e

x2

(t

1

t ) and we denotes the n-thcoefficient by Jn(t) for n Zso that

() ex2 (t 1t ) =

n=Jn(x)t

n.

First we are going to verify that the function Jn(x) satisfies Bessels differ-ential equation (). We will do the verification by differentiating () withrespect to x twice and differentiating () with respect to t twice. Differenti-ating () with respect to xonce, we get

() 12

t 1

t

ex2 (t 1t ) =

n=

Jn(x)tn.

Differentiating () with respect to x one more time, we get

()

1

2

t 1

t

2e

x2 (t

1

t ) =

n=Jn(x)t

n.

Differentiating () with respect to t once, we getx

21 + 1

t2 ex2 (t 1t ) =

n=

nJn(x)tn1.

and multiplying it by t, we get

() x2

t+

1

t

e

x2(t

1

t ) =

n=nJn(x)t

n.

Differentiating () with respect to t one more time, we get

x

2 1 1

t2+

x

2 t+

1

tx

2 1 +

1

t2e

x2 (t

1

t ) =

n=n2Jn(x)t

n1.

and multiplying it by t, we get

()

x

2

t 1

t

+

x

2

t+

1

t

2e

x2(t

1

t ) =

n=n2Jn(x)t

n.


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We now form the equation

x2()+x()+x2() ()

whose left-hand side is ex2 (t

1

t ) timesx

2

t 1

t

2+

x

2

t 1

t

x

2

t 1

t

+

x

2

t+

1

t

2+x2

=

x

2

t 1

t

2

x

2

t+

1

t

2+x2

= x22

t2 2 + 1

t2

x

22

t2 + 2 + 1t2 +x2 = 0

and whose right-hand side is

n=

x2Jn(x) +xJ

n(x) +

x2 n2 Jn(x) tn.

This is finishes the verification of the differential equation () for the BesselfunctionsJn(x).

Relation Between Bessel Functions for an Integer and the Negative of the

Integer. The generating functione

x

2 (t

1

t ) is unchanged whent is replaced by 1twhich from () means that

Jn(x) = (1)nJn(x) for n Z.

Since Bessels differential equation () is invariant under n n, the rela-tion Jn(x) = (1)nJn(x) for nZprecludes the easy way of using Jn(x)to get another solution which is not a scalar multiple ofJn(x). For n Z itwill take a more involved procedure to get another solution which is not ascalar multiple ofJn(x). We will indicate how this additional solution is ob-tained but for the problem of the vibrating circular membrane the additional

solution is not needed.

Power Series Expansion for Bessel Functions. We now compute the powerseries expansion for Bessel functions directly from the power series expansion


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Math 115 (2006-2007) Yum-Tong Siu 10

of the generating function ex2(t

1

t ). We write the generating function as the

product two functions so that

ex2(t

1

t ) =ext2 e

x2t .

The power series expansion of the first factor is

ext2 =

j=0

xj

j!2jtj

and the power series expansion of the second factor is

e x2t = k=0

(1)k

xk

k!2k tk.

We are interested in the coefficient of tn in the product of these two powerseries expansion. In order to get tn as a product we have to consider it asthe product oftj and tk with j k = n. These the coefficient oftn in theproduct of the two power series expansion is the sum of the coefficient oftj inthe first series times the coefficient oftk in the second series withj k= n.Thus

Jn(x) = jk=n

xj

j!2jtj

(1)kxkk!2k

tk

which can be rewritten as

(&) Jn(x) =k=0

(1)kk!(n+k)!

x2

n+2k

when we remove the index j by using j = n+k. We would like to remarkthat at x = 0 the value ofJ0 is 1 but the value ofJn is 0 for any positiveinteger n.

Bessel Functions of Non-Integral Order and the Other Solution of the Bessel

Differential Equation in the Case of Integral Order. When we substitutethe power series (&) into the Bessel differential equation, we can readily seethat Jn(x) given by (&) satisfies the Bessel differential equation. The samecomputation shows that if we allown to be a non-integer and replace (n+k)!by the Gamma function (n+k+ 1), then Bessels differential equation is


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Math 115 (2006-2007) Yum-Tong Siu 11

satisfied also by Jn(x) given by (&) even when n is not an integer. Recall

that the Gamma function is defined by

(x) =

t=0

ettx1dt.

For any real number we define the Bessel function by the power series

J(x) =k=0

(1)kk! (+k+ 1)

x2

+2k.

Then the Bessel function J(x) satisfies the Bessel differential equation

x2y+xy+

x2 2 y= 0.When is not an integer, the two solutions J(x) and J(x) are linearlyindependent. We would like to mention without going into any further detailsthat for an integer n we can use the following Hankels function

Yn(x) = lim0

1

(Jn+(x) (1)nJn(x))

as the second solution of Bessels differential equation for n. This is thestandard technique of taking the limit, as 0, of a normalized linearcombination of two solutions for Bessels differential equation for n+ . Itturns out that another less obvious linear combination gives a more elegantsimpler expression.

1

2Yn(x)+Jn(x)(log2 ) =J0(x)log x+2

J2(x) 1

2J4(x) +

1

3J6(x) +

,

where

= limk

1 +

1

2+ +1

k log k

is Eulers constant. This formula shows that the singularity order ofYn(x)

is log x as x 0, which is to be expected, because Yn(x) is obtained bydifferentiating

J(x) =k=0

(1)kk! (+k+ 1)

x2

+2k.


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Math 115 (2006-2007) Yum-Tong Siu 12

with respect to andd

dttx =tx log x.

We conclude that the only solution of Bessels differential equation () foran integer n which has no singularity at x = 0 is Jn(x) up to a constantmultiple. Thus, to specify that the solution of () without singularity hasthe same effect specifying an initial condition to single out certain solutionsof differential equations. The reason why the initial condition takes the formof specifying no singularity at x= 0 is because the coefficient of the higher-order term in Bessels differential equation vanishes at x = 0 and is thus adifferential equation which is singular at x= 0.

Derivatives of Bessel Functions and Recurrent Relation of Bessel Functions.

We now discuss the derivatives of Bessel functions and recurrent relation ofBessel functions. In two ways this discussion is needed for the problem of thevibrating circular membrane. The first is that we need some properties ofzeroes ofJn(x) for x0. The second is that we need the L2 norm ofJn(x)with respect to the weight function x over [0, c].

Bessels differential equation expresses the second-order derivative of theBessel function in terms of the function and its first-order derivative, butdoes not give immediately an explicit expression for the first-order derivativeof the Bessel function. On the other hand, the generating function, whenits first-order derivative with respect to x is compared with its first-order

derivative with respect to t, can give formulas for first-order derivatives ofthe Bessel function as follows. Differentiating the generating function in ()with respect to xgives () and multiplying both sides by x gives

() x

2

t 1

t

e

x2(t

1

t ) =

n=xJn(x)t

n.

Differentiating the generating function in () with respect to t and multi-plying both sides by x gives (). In order to compare () with (), weadd

x

t

ex2 (t

1

t ) =

n=

x

t

Jn(x)tn =

n=

xJn+1(x)tn

to () so that we have

() x

2

t+

1

t

e

x2(t

1

t ) =

n=(xJn(x) +xJn+1(x)) t

n.


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Math 115 (2006-2007) Yum-Tong Siu 13

We now can equate the right-hand side of () with the right-hand side of() and get nJn(x) =xJ

n(x) +xJn+1(x)

or equivalently

(%) xJn(x) =nJn(x) xJn+1(x).

Replacingn byn, we get

xJn(x) =nJn(x) xJn+1(x).

using Jn= (1)nJn(x), we get

($) xJn(x) =nJn(x) +xJn1(x).

A more compact way of writing these two formulas for the first-order deriva-tives of Bessel functions is

Jn

xn

= Jn+1

xn ,

(xnJn) = xnJn1.

EliminatingJn(x) from (%) and ($), we get the following algebraic recurrent

formula for Bessel functions.

xJn+1(x) = 2nJn(x) xJn1(x).

Zeros of Bessel Function of Order Zero. Recall that we have the boundary

conditionR(c) = 0. WithR(c) =y

c

the boundary condition becomes

y

c

= 0. Now y(x) = Jn(x). So we have to formulate the boundary

condition in terms ofJnand it becomesJn

c

= 0. In order to determine

the constant, we have to locate the zeroes ofJnfor each nonnegative integern. Let us first consider the simplest case of J0(x). The Bessel differential

equation for it isx2y+xy+

x2

y= 0

which is the same asxy+y+xy = 0.


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Math 115 (2006-2007) Yum-Tong Siu 14

We can get rid of the term involving the first-order derivative by introducing

a new dependent variable u(x) =x y. Then u= 1

2xy + x y andu= 1

4x

xy +

1x

y +

x y

so that

u= 14x

x

y + 1

x (y+xy) =

x+

1

4x

x

y

which is the same as

u(x) =1 + 14x2

u(x).

Whenx is large, this can be compared with

v(x) =v(x)

which admits v(x) = sin x as a solution. The key technique is to use thederivative of the Wronskian

u vu v

.

In general the Wronskian ofn functions f1, , fn is

f1 f2 fnf1 f

2 fn

f(n1)1 f

(n1)2 f(n1)n

and is used to investigate the linear independence off1, , fn. The deriva-tive of the Wronskian u vu vis

(uv vu) = uv vu,


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Math 115 (2006-2007) Yum-Tong Siu 15

which is equal to

uv v1 + 14x2 u=uv4x2 .Integrating over [a, b] yields

ba

uv

4x2dx = vu uv

x=b

x=a

.

We choose a= 2k and b = (2k+ 1) for k Z. Since

v(a) =v(b) = 0, v(a) = 1, v(b) =1

for our choice ofv(x) = sin x, it follows that (2k+1)2k

u(x)sin x

x2 dx= 4 (u(2k) +u ((2k+ 1))) .

Sincesin x

x2 >0 for 2k < x


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Math 115 (2006-2007) Yum-Tong Siu 16

and the two boundary conditions that R(r) is finite at r = 0 and R(c) = 0.

After we rewrite the differential equation as

d2R

dr2 r dR

dr =0,R

and conclude that R (r) = J0(0,r) is an eigenfunction of the differentialoperator

d2

dr2 1

r2d

dr

for the eigenvalue0, = 0,2 with the two boundary conditions that R(r) is

finite at r= 0 and R(c) = 0.

By the Sturm-Liouville theorem which we will introduce later, the familyof functionsJ0(0,x) forNis a complete orthogonal family of functionson [0, c] with respect to the weight function x. This completeness propertywill be used to express our sought-after solution of the vibrating circularmembrane as an infinite sum of special product solutions obtained by themethod of separation of variables.

Zeroes of Bessels Function for Higher Integral Order. We now use Rollestheorem and induction on n to investigate the zeroes of Jn(x). The key isthe formula for the first-order derivative ofJn(x), which is

()

Jn

xn

= Jn+1

xn .

By Rolles theorem, there is at least one zero ofJn between two consecutivezeroes of Jn. The above formula () tells us that there is at least zero ofJn+1 between two consecutive zeroes ofJn. Note that from the power seriesexpansion (&) ofJn(x) centered at x= 0 we know that the vanishing orderofJn(x) at x = 0 is precisely n. Recall that for the problem of the vibratingcircular member only the nonnegative zeroes ofJn(x) are of interest.

SinceJn is the solution of a linear homogeneous second-order differential

equation, its derivative cannot vanish at any one of its zeros. This meansthat Jn must alternate its sign between two consecutive zeroes of Jn forthe following reason. For example, if after the zero ofJn(x) at x = andbefore the next zero at = > , the sign ofJn(x) is positive, then from theconsideration of the difference quotient the derivative Jn(x) at x = must


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Math 115 (2006-2007) Yum-Tong Siu 18

Orthogonality of the Family Jn(n,x) for N. While the completenessof the family Jn(n,x) for Nover [0, c] with weight function x dependson the theorem of Sturm-Liouville with some sophisticated arguments, theorthogonality property of the family comes from the orthogonality of twoeigenfunctions of a self-adjoint operator for two distinct eigenvalues. Wenow verify the orthogonality property of the family.

Fix a nonnegative integer n. Take < m. Let y(x) = Jn(n,x) andz(x) = Jn(n,mx) on [0, c]. These two functions satisfy the following twodifferential equations

x2d2y

x2+x

dy

dx+ n,x

2y n2y= 0,

x2d2z

dx2+x

dz

dx+ n,mx

2z n2z= 0,

which we can put in divergence form

(xy) n

2

xy =n,x y.

(xz) n

2

x z=n,mx y.

Multiplying the first equation by zand the second one by y and taking their

difference, we get

(xy(x))

z(x) (xz(x)) y(x) = (n, n,m) xy(x)z(x).

Integrating over 0xc and using integration by parts, we get

xy(x)z(x) xz(x)y(x)x=c

x=0

= (n, n,m) x=cx=0

xy(x)z(x) dx.

From the vanishing ofy(x) andz(x) at x = c and their finiteness at x = 0 itfollows that the left-hand side is zero and

x=cx=0

xy(x)z(x) dx= 0,

which is the orthogonality ofJn(n,x) andJn(n,mx) over [0, c] with respectto the weight function x for 1 < m.


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Math 115 (2006-2007) Yum-Tong Siu 19

Norm of Eigenfunctions of Bessels Differential Equation for Integral Order.

In order to determine the coefficients in the infinite sum which expressesa given function in terms of the family Jn(n,x) for N over [0, c] withweight functionx, we need to have theL2 norm of each member of the familyover [0, c] with weight function x. We now compute this L2 norm.

Fix a nonnegative integer n and fix a positive integer . Let y(x) =Jn(n,x). The functiony(x) satisfies the differential equation

x2y+xy+

(n,)2

x2 n2 y = 0.After multiplying the equation by 2y, we can rewrite it as

ddx

(xy)2 +

(n,)2 x2 n2 ddx

y2 = 0.

Integrating from x= 0 to x= c yields

(xy)2

+

(n,)2

x2 n2 y2x=c

x=0

= 2 (n,)2

cx=0

xy2 dx.

Since y(c) = 0 and since both y(x) and y(x) is finite at x= 0 and y(0) = 0whenn >0, it follows that

(n,)

2

x

2

n2

y

2x=c

x=0 = 0

and

(xy)2

x=c

x=0

= (cy(c))2

.

Thus we have

(cy(c))2

= 2 (n,)2

cx=0

xy2 dx

and the formula for the L2 is given by

cx=0

x (Jn(n,x))2

dx=c2

2 (Jn(n,c))2

.

We can get an alternative form by using the formula (%)

xJn(x) =nJn(x) xJn+1(x)


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Math 115 (2006-2007) Yum-Tong Siu 20

at x= n,c so that

n,c Jn(n,c) =nJn(n,c) (n,c) Jn+1(n,c)and

Jn(n,c) = Jn+1(n,c) ,because Jn(n,c) = 0. Thus c

x=0

x (Jn(n,x))2

dx=c2

2 (Jn+1(n,c))

2.

Final Answer of Problem of Vibrating Circular Membrane. We now have thespecial product solutions

T(t)R(r)() = cos (an,t) Jn(n,r)

1 if n= 0cos n if n Nsin n if nN

for N. To get to this point, we have already used up the following fiveboundary conditions

T(0) = 0,

(0) = (2),

(0) = (2),

R(r) finite at r= 0,

R(c) = 0.

There is the following boundary condition u(t,r,) =f(r, ) at t= 0, whichhas not been used. Observe that cos (an,t) = 1 att = 0. Now we form theR-linear combination of all the special product functions given above withyet-to-be-determined coefficientsAn, (for integersn0 and1) andBn,(for integers n1 and 1) so that

f(r, ) ==1

A0,

2 J0(0,r)+

n=1

=1

Jn(n,r) (An,cos (n) +Bn,sin (n)) .

Note that here the case of n = 0 is singled out, because the Fourier seriesexpansion for f() is of the form

A0

2 +

n=1

(Ancos (n) +Bnsin (n)) ,


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Math 115 (2006-2007) Yum-Tong Siu 21

where

An= 1

2

=0

f()cos(n) d for n0,

Bn= 1

2=0

f()sin(n) d for n1,

and the case for n = 0 has to be treated separately. Here for the vibratingcircular membrane, we determine the coefficients An, (for integersn0 and1) and Bn, (for integers n1 and 1) by first using the formula forthe Fourier series coefficients to get

gn(r) =

1

2=0 f(r, )cos(n) d for n0,

hn(r) = 1

2=0

f(r, )sin(n) d for n1,

and then for any fixedn using the expansion in terms of the complete familyJn(n,x) for N to get An, from gn(r) and to get Bn, from hn(r). Weuse the formula c

x=0

x (Jn(n,x))2

dx=c2

2 (Jn+1(n,c))

2

for the L2 norm ofJn(n,x) to get

An, = 1

c2

2 (Jn+1(n,c))2

cr=0

r gn(r)Jn(n,r) dr,

Bn, = 1

c2

2 (Jn+1(n,c))

2

cr=0

r hn(r)Jn(n,r) dr

to get the following final answer u(t,r,) for the problem of the vibratingcircular membrane.

u(t,r,) =

=1

A0,

2 cos(a0,t) J0(0,r)

+n=1

=1

cos(an,t) Jn(n,r) (An,cos (n) +Bn,sin (n)) .

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Bessel Function (1)