337 Week 0910weeklying

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Weeks 09 10: Sturm-Liouville Theory and Special Functions

We recall the basic steps of the method of separation of variables.

1. Search for basic solutions that are the products of one-variable functions using the equation and anappropriate subset of the initial/boundary conditions. Ideally, only countably many (that is, canbe numbered by natural or integer numbers) such solutions exist.

2. Use the remaining initial/boundary conditions to determine an innite sum involving the basicsolutions.

3. Verify that this innite sum indeed corresponds to a function and this function indeed satises theequation together with the initial/boundary conditions.

The rst two steps are emphasized in our lectures, and the third step involves much mathematical theoryand interested readers should take higher level PDE courses or consult more advanced books.

1. General boundary value problem.We have seen that the method of separation of variables, with the help of the theory of Fourier series,

can be applied to a wide variety of PDEs. However a closer inspection reveals that all of our previousexamples have the following boundary conditions:

u(0, t ) = u(l , t ) = 0 . (1)

How about problems with other boundary conditions? Lets check some examples.

Example 1. Consider the heat equation

u t = u xx 0 < x < l , t > 0 (2)u(x, 0) = f (x) 0 x l, (3)ux (0, t ) = 0 t 0, (4)u x (l, t ) = 0 t 0. (5)

This system models the change of temperature along a rod of length l whose both ends are insulated.Solution. We apply the method of separation of variables. As we have seen, the method consists of thefollowing steps:

1. First we try to nd non-zero basic solutions whose variables are separated:

u = X (x ) T (t). (6)

Substituting this into the equation we obtain

T (t ) X (x) = T (t ) X (x) T (t)T (t )

= X (x)X (x)

. (7)

Thus we need to solve the ODEX (x) X (x) = 0 (8)

with boundary conditionsX (0) = X (t) = 0 . (9)

The solutions are

cosn

l x , n = 0 , 1, 2,

(10)Note that the sum starts from 0.

2. Represent the solution by an innite sum.As n =

n l

2, we see thatT n (t ) = e

nl t (11)

and therefore formally

u(x, t ) =n =0

a n e

nl t cos n

lx . (12)


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The coefficients a n is determined by requiring

f (x) =n =0

a n cos

n l

x . (13)

Noticing that this is simply a cosine series, we see that the coefficients can be computed through

a n = 2l 0l

f (x) cos n lx n > 0 (14)

and

a 0 =1l 0

lf (x ) dx. (15)

We will see now that for other, more complicated, boundary conditions, the theory of Fourier series is notenough anymore.

Example 2. We still consider the heat equation modeling a rod. This time the temperature at 0 is kept0 while the other end ( x = l) is in contact with a medium of tempature 0.

u t = u xx 0 < x < l , t > 0 (16)

u(x, 0) = f (x) 0 x l, (17)u (0, t ) = 0 t 0, (18)

u x (l, t ) = h u (l, t ) . t 0. (19)Here h > 0.Solution. Applying the method of separation of variables, we reach

X X = 0 , X (0) = 0 , h X (l) + X (l) = 0 . (20)We discuss the cases:

i. > 0. The general solution isA e x + B e x . (21)

NowX (0) = 0 A + B = 0 (22)

h X (l) + X (l) = 0 h + e l A + h e l B = 0 . (23)

The two equations can be written

1 1h + e l h e

lAB =

00

. (24)

For the solution to be non-zero, we have to have

0 = det1 1

h + e l h e l = h e l h + e l . (25)

As h > 0 and > 0, this is not possible.ii. = 0 . The general solution is

A + B x (26)

The boundary conditions lead to

A = 0 , h A + (h l + 1 ) B = 0 A = B = 0 . (27)


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iii. < 0. The general solution is

A cos x + B sin x . (28)Now

X (0) = 0 A = 0 , (29)

h X (l) + X (l) = 0

h sin

l +

cos

l = 0 . (30)Therefore the solution is of the form X = sin( p x) with p satisfying

tan ( p l) = p/ h. (31)It is easy to see that the solutions form an innite series

0 < p 1 < p 2 < < (32)

Therefore our solution to the PDE can be written as

1

bn ep n

2 t sin( pn x) (33)

where bn is determined by

f (x) =1

bn sin( pn x). (34)

Now how should we determine bn ? And furthermore how can we know whether the innite sum gives thesolution or equivalently whether similar properties as those hold for the Fourier series hold for our serieswith sin ( pn x)? Keep in mind that it is not possible to obtain a formula for the pn s.

Mimicking what we have done before, we compute, for n

m ,

0l

sin( pn x) sin( pm x) dx =12 0

l

[cos( pn pm ) x cos( pn + pm ) x]dx= 1

2sin( pn pm ) l

pn pm sin( pn + pm ) l

pn + pm

= 12

sin( pn l) cos( pm l) sin( pm l) cos( pn l) p

n p

m

sin( pn l) cos( pm l) + sin( pm l) cos( pn l)

pn + pm. (35)

Now using the fact that

h sin( pn l) + pn cos( pn l) = 0 sin( pn l) = pnh

cos( pn l) (36)we have

0l

sin( pn x) sin( pm x) dx = 0 . (37)

Therefore we can determine bn by

bn

=

0

lf (x ) sin( pn x) dx

0l

sin2( pn x ) dx. (38)

But a convergence theory similar to that of the Fourier series is clearly beyond us here.

Example 3. Consider the heat equation in a 2D disc x2 + y2 1:

u t = (u xx + u yy ) (39)u(x ,y, 0) = f (x, y ) (40)u(x, y, t ) = 0 x2 + y2 = 1 . (41)


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Solution. Due to the special geometry of the domain, it is natural to consider the problem using polarcoordinates (r, ) satisfying

x = r cos, y = r sin. (42)

Now we change the variables from x, y to r, . Differentiating the above relation we have

(cos) r x r (sin) x = 1 (43)(cos) r y r (sin) y = 0 (44)(sin) r x + r (cos) x = 0 (45)(sin) r y + r (cos) y = 1 (46)

consequently

r x =xr

, r y =yr

, r xx =1r

x2

r 3, r yy =

1r

y2

r 3; (47)

x = sin

r=

yr 2

, y =cos

r= x

r 2, xx =

2 x yr 4

, yy = 2 x y

r 4Therefore

uxx = u rrx2

r 2 u r2 x y

r 3+ u

y2

r 4+ u r

1r

x2

r 3+ u

2 x yr 4

, (48)

u yy = u rr y2

r 2 + u r 2 x yr 3 + u x2

y4 + u r 1r y2

r 3 + u 2 x yr 4 . (49)The equation and the initial-boundary conditions in polar coordinate form are

u t = u rr +1r

u r +1r 2

u (50)

u (r, , 0) = f (r, ) (51)u(1, , t ) = 0 . (52)

We apply separation of variables to solve this equation.First we try to nd non-trivial basic solutions of the form

u(r, , t ) = R (r ) () T (t ). (53)

Substituting this into the equation we reach

R (r ) () T (t ) = R (r ) () + 1rR (r ) () + 1r 2

R (r ) () T (t ). (54)

Dividing both sides by R (r ) () T (t ) we reach

T (t)T (t )

= R (r )R (r )

+ 1r

R (r )R (r )

+ 1r 2

()()

. (55)

As the LHS only involves t and the RHS only r, there is a constant such that

T (t )T (t )

= (56)and

R (r )R (r ) +

1r

R (r )R (r ) +

1r 2

()() = . (57)

Multiply both sides by r 2 we have

r 2 R (r )R (r )

+ r R (r )R (r )

+ r 2 = ()()

. (58)

The LHS only involves r and the RHS only . Thus there is a constant such that

()()

= ,r 2 R (r )

R (r )+

r R (r )R (r )

+ r 2 = . (59)


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As () is obviously 2 periodic, we have

= n 2, n = 1 , 2, 3, (60)and

() = A cos(n ) + B sin(n ). (61)

On the other hand, the equation forR

now becomesr 2 R + r R + r 2 n 2 R = 0 , (62)

with the boundary condition

R (1) = 0 . (63)

Now it is clear that the success of our method depends on the following:

1. For each n , we have n,k , such that the above equation has a solution R n,k ;

2. The initial data f (r, ) have the following expansion

f (r, ) =n,k

an,k R n,k (r ) cos(n ) + bn,k R n,k sin(n ). (64)

As we can always expand f (r, ) into Fourier series

f (r, ) =n

An (r ) cos(n ) + B n (r ) sin(n ), (65)

The requirement becomes expanding

An (r ) =k

a n,k R n,k (r ), B n (r ) =k

bn,k R n,k (r ). (66)

3. The resulting innite double summation

n,k

[a n,k R n,k (r ) cos(n ) + bn,k R n,k sin(n )] e n,k t (67)

indeed gives the solution.Clearly we see that a Fourier-type theory of the functions R n,k is crucial to the success of our method.We will see soon that these R n,k s are the so-called Bessel functions, which often arise in PDEs on discsand cylinders.

From the above example we see that as soon as the system becomes more and more complicated, thetheory of Fourier series helps less and less. For simpler ones, we can still develop ad hoc theories followingthe idea of Fourier series, but for more complicated ones, it seems very hard to do things on the y. Inparticular, a complete understanding of solutions to equations like

r 2 R + r R + r 2 n 2 R = 0 (68)is needed. Such understanding is obtained from the following Sturm-Liouville theory.

2. Sturm-Liouville theory.The standard Sturm-Liouville (SL) problem is of the form

( p(x) y ) + q (x) y + r (x ) y = 0 , a < x < b (69) 0 y(a ) + 1 y (a ) = 0 , (70) 0 y(b) + 1 y (b) = 0 . (71)

where all the functions and numbers are real. For simplicity we assume the coefficients are as smooth aswe need.


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The problem is called

regular when p, q , r are bounded on [a, b ] (that is the interval a x b), p, r > 0 for all a x b,and 0, 1 real, not both 0, and 0, 1 real, not both 0. singular when any one or more of the following happens

The interval (a, b ) is innite, that is either a =

or b= +

or both occurs.

p(x ) = 0 for some x [a, b] or r (x ) = 0 for some x [a, b]. One or several coefficient function becomes at a or b, or both.

Example 4. We check the systems we have dealt with

y + y = 0 , y(0) = y(l) = 0 (72)

We havea = 0 , b= l; p(x) = 1 , q (x) = 0 , r (x) = 1; 0 = 1 , 1 = 0 , 0 = 1 , 1 = 0 . (73)

The system is a regular SL problem.

y + y = 0 , y (0) = y (l) = 0 (74)We have

a = 0 , b= l; p(x) = 1 , q (x) = 0 , r (x) = 1; 0 = 0 , 1 = 1 , 0 = 0 , 1 = 1 . (75)

This is also a regular SL problem.

y + y = 0 , y(0) = 0 , y (l) = h y (l) . (76)

We havea = 0 , b= l; p(x ) = 1 , q (x ) = 0 , r (x) = 1; 0 = 1 , 1 = 0 , 0 = h, 1 = 1 . (77)

x 2 y + x y + x 2 n 2 y = 0 , y(0) bounded, y(1) = 0 . (78)At rst sight this problem is not an SL problem. However we can transform it through the fol-lowing operations:

We search for a multiplier h(x) such that

h(x) x 2 y + x y + x 2 n 2 y = ( p y ) + q y + ry. (79)Comparing the two sides, we have

h (x) x2 = p(x), h(x) x = p(x) (80)which leads to

p(x) = 1

xp(x) p(x) = x h (x ) = 1

x. (81)

Thus we see that the equation is equivalent to

(x y ) n 2

xy + x y = 0 (82)

which corresponds to

a = 0 , b= 1; p(x) = x, q (x ) = n 2

x, r (x) = x ; 0 = 1 , 1 = 0 . (83)

This is a singular SL problem.


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Any that the problem has non-trivial solutions is called an eigenvalue, the corresponding solutions arecalled eigenfunctions.

2.1. Properties of regular Sturm-Liouville problems.We see from the following theorem that the solutions to a SL problem enjoy similar properties as the

functions sin n l x and cosn

l x in the Fourier series.

Theorem 5. A regular SL problem has the following properties.1. It has nonzero solutions for a countably innite set of values of . These eigenvalues are all real.

The set of eigenvalues does not have any limit points. These eigenvalues are bounded from below if 0 1 0 and 0 1 0. These eigenvalues are bounded from below by 0 if furthermore q 0.

2. For each xed eigenvalue , the solution space is one-dimensional. That is, there is y such that all other solutions for the same is a multiple of y .

3. If we enumerate the eigenvalues as 1, 2, , then for each n we can pick one eigenfunction n .These eigenfunctions satisfy

a ) ab

n (x) m (x ) r (x ) dx = 0 for any n

m .

b) For any f having two continuous derivatives on [a, b ] and satisfying the boundary conditions,the innite sum

n =1

cn n (84)

where

cn = ab

f (x ) n (x) r (x) dx

ab

n (x)2 r (x ) dx(85)

converges absolutely uniformly to f (x ). By absolutely uniformly we mean

1

|cn

| |n

| 0.

Finally,

y z z y = 0 yy

= zz

ln y ln z = constant y/ z = constant . (113)3. We enumerate the eigenvalues by 1, 2, and denote the corresponding eigenfunctions by 1,

2, . .

a) ab

n (x) m (x ) r (x ) dx = 0 for any n

m .

It suffices to show that if , are two distinct eigenvalues, and y, z the correspondingeigenfunctions, then a

b y z r dx = 0 .


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Using the equations we have

ab

( p y ) + q y + r y z ( p z ) + q z + r z y dx = 0 . (114)After using the boundary conditions, we can show that

LHS = ( ) y z r dx. (115)Therefore( ) a

by(x) z(x) r (x) dx = 0 . (116)

As

, we have

ab

y(x) z(x) r (x) dx = 0 . (117)

b) Omitted.

c) Omitted.

d) Omitted.

2.2. Properties of singular Sturm-Liouville problems.Recall that the problem is singular when any one (or more) of the following is true:

The interval (a, b ) is innite, that is either a = or b= + or both occurs. p(x ) = 0 for some x [a, b ] or r (x ) = 0 for some x [a, b ]. One or several coefficient function becomes at a or b, or both.

For singular SL problems, appropriate boundary conditions should be specied. In particular, if p van-ishes at a or b, we should require y and y to be bounded at a or b respectively.

Similar to the regular SL problems, the eigenfunctions for singular SL problems are also orthogonalwith respect to weight r (x).

Example 6. Consider the Legendres equation

1 x2 y + y = 0 , 1 < x < 1. (118)As p(x) = 1 x2 vanishes at both ends, the boundary conditions should be taken as

y, y remain bounded as x 1. (119)The eigenvalues are n = n (n + 1 ). Here r (x) = 1 , so the corresponding eigenfunctions satisfy

11

P m (x) P n (x) dx = 0 , n

m. (120)

Example 7. Consider the Bessels equation

(x y ) + x 2

xy = 0 , 0 < x < a (121)

p(x) = x vanishes at x = 0 . Therefore we can assign the usual boundary condition

0 y + 1 y = 0 (122)

at x = a but need to require

y, y remain bounded as x 0 + . (123)


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The eigenvalues are n 2. As r (x) = x , the corresponding eigenfunctions satisfy

0a

yn (x ) ym (x) x dx = 0 , n

m. (124)

Example 8. Consider the Hermites equation

u 2 x u + u = 0 , < x < (125)To write this problem into a SL problem, we multiply the equation by ex

2

to obtain

ex2

u + e x2

u = 0 , < x < . (126)Now that we have p(x) = ex

2

which tends to 0 as x , the boundary conditions should beu, u remain bounded as x . (127)

The eigenvalues are n = 2 n for nonnegative integers n . Since r (x) = ex2

, the orthogonality propertyreads

H n (x) H m (x ) ex2

dx = 0 , n

m. (128)

3. Special functions.To apply this theory to solve PDEs, we need to obtain more information of the eigenfunctions. This

leads to the theory of special functions.

3.1. Bessel function.Recall our example problem: Consider the heat equation in a 2D disc x2 + y2 1:

u t = (uxx + u yy ) (129)u (x , y, 0) = f (x, y ) (130)u(x , y, t ) = 0 x 2 + y2 = 1 . (131)

which leads to basic solutions of the form

R (r ) () T (t ) (132)where R solves

r 2 R + r R + r 2 n 2 R = 0 , (133)with the boundary condition

R (0) bounded, R (1) = 0 . (134)

We have seen that the above equation can be written into the Sturm-Liouville system

(r R ) n 2

rR + r R = 0 (135)

which means that after detemining the eigenvalues m and the corresponding eigenfunctions R n,m , we canexpand any function of r into

f (r ) =m

Am R n,m (r ) (136)with

Am = 01

f (r ) R n,m (r ) r dr

01

R n,m (r )2 r dr. (137)

One problem to this approach is that neither m nor R n,m has a formula. However, we can qualitativelysolve the equation as follows.


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First we determine the general solutions. Notice that, if R (r ) solves the following equation

r 2 R + r R + r 2 n 2 R = 0 , (138)then we have (replacing each r by r )

r 2 R r + r R r + r 2

n 2 R r = 0 (139)

which means R (r ) R r solvesr 2 R + r R + r 2 n 2 R = 0 . (140)

Therefore we rst consider the singular SL equation

x2 y + x y + x 2 2 y = 0 (141)where is a non-negative real number. We have seen that this is closely related to solving equationsinvolving the 2D Laplacian xx + yy in a disc.

As we have mentioned, it is not possible to obtain a formula for the solutions. We have to rely on theso-called Frobenius method: We search for solutions of the form

y(x) =n =0

a n x s + n (142)with s to be determined.

Substituting this formula into the equation, we obtain

0 = x2 y + x y + x 2 2 y= x2

n =0

an (s + n ) ( s + n 1) x s + n 2

+ xn =0

a n (s + n ) x s + n 1 + x 2 2

n =0

a n x s + n

= n =0

a

n (s

+n

) (s

+n

1)x s + n

+ n =0

a

n (s

+n

)x s + n

+n =2

an 2 x

s + n n =0

2 a n x s + n

=n =0

a n (s + n )2 2 x s + n +

n =2

a n 2 x

s + n

= s 2 2 a 0 x s + (s + 1 )2 2 a 1 x s +1 +

n =2

an (s + n )2 2 + a n 2 x s + n . (143)

Therefore the solution should satisfy

s2 2 a 0 = 0 (144)(s + 1 )2

2 a 1 = 0 (145)

a n (s + n )2 2 + a n 2 = 0 n = 2 , 3, (146)As we are discussing the general case here, we assume a 0

0. As a consequence,

s 2 = 2 s = . (147)We see that close to x = 0 , the solution behaves as either x or x . The former satises y(0) = 0 whilethe latter is unbounded. Thus we discuss the two cases separately.

s = .


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In this case, (s + n )2

2 for all n > 0. Consequently we have

a 1 = 0 (148)a 2 =

a 02 (2 + 2 )

(149)

a 3 = a 1

3 (2 + 3 )= 0 (150)

a 4 = a 24 (2 + 4 ) (151)

a 2n = a 2n 2

2 n (2 + 2 n )=

a 2n 24 n ( + n )

, (152)

a 2n +1 = 0 (153)

From the above we have the formula

a 2n =( 1)

n a 04n n ! ( + n ) ( + 1 )

. (154)

The formula for y(x) is then

y(x ) = a 0n =0 ( 1)

n

4n n ! ( + n ) ( + 1 )x2n + . (155)

This is called the Bessel function of the rst kind of order , denoted by J (x ).

s = .In this case

(s + n )2 2 = ( + n )2 2 = n 2 2 n = 0 (156)

when n = 2 . Thus we discuss two cases.

2 is not an integer. In this case none of (s + n )2 2 is 0 unless n = 0 . Thus we have theinterative relation

a2 n = a2n 22 n (2 n 2 ) = a2n 24 n (n ) =

= ( 1)n

a 04n n !(n ) (1 ) , a2n 1 = 0 . (157)

Thus

J (x) y(x) = a 0n =0

( 1)n a 0

4n n ! (n ) (1 )x 2n . (158)

2 is an integer. is an integer. For convenience we denote by m 0. Then from

a n (s + n )2 2 + a n 2 = 0 (159)we have

a 2m 0

2 = 0 a2m 0

4 = 0 a 0 = 0 . (160)

On the other hand, the same iteration process gives a 2n 1 = 0 for all n > 0. As a con-sequence, the rst term in the series is actually x 2m 0 = x . In other words, when is an integer, the solution for s = is regular. In fact one can show that the solu-tion J n (x ) is simply ( 1)

n J n (x).

is not an integer but 2 is. Thus necessarily 2 is odd, denote it by 2 n 0 1. Inthis case, using the iteration relationa n (s + n )2 2 + a n 2 = 0 (161)


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we can show that a2n 1 = 0 for all n < n 0. Now if we dene

J (x) y(x) = a 0n =0

( 1)n a 0

4n n ! (n ) (1 )x2n , (162)

it can be shown that the solution y(x ) is in fact a linear combination of J and J .

Summarizing, we haveJ (x) y(x) = a0

n =0

( 1)n a 0

4n n !(n ) (1 )x 2n (163)

when is not an integer, andJ (x) = ( 1)

n J (x ) (164)when is an integer.

Recall that our purpose is to nd out the general behavior of the solutions to the Bessel equation

x 2 y + x y + x2 2 y = 0 . (165)When is not an integer, we have two linear independent solutions J therefore the general solution canbe written as

y(x) = A J (x ) + B J (x). (166)

However when is an integer, J are linearly dependent of each other and we have to nd another solu-tion which is linearly independent of J .

To remedy this, we dene

Y (x) =(cos ) J (x) J (x)

sin , (167)

which should be explained as the limit n when is an negative integer. That isY n = lim n

Y . (168)

It turns out that J and Y are always linearly independent. That is

y(x) = A J (x) + B Y (x). (169)The coefficients A, B are determined by the boundary conditions, taking advantage of the fact thatJ (0) = 0 , Y (0) = .Recall our example which leads to the discussion of Sturm-Liouville problems: Consider the heat equa-tion in a 2D disc x 2 + y2 1:

u t = (uxx + u yy ) (170)u (x , y, 0) = f (x, y ) (171)u(x , y, t ) = 0 x 2 + y2 = 1 . (172)

Separating the variables in polar coordinates

u (x , y, t ) = R (r ) () T (t ) (173)

we reach the following problem for R :

r 2 R + r R + r 2 n 2 R = 0 , (174)with the boundary condition

R (0) bounded, R (1) = 0 . (175)

We know that the general solution is of the form

R (r ) = A J n r + B Y n . (176)


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The boundary condition can also be written as

(0) = (2 ), (0) = (2 ) (192)

as this, together with the equation, guarantees (k )(0) = (k )(2 ) for all k which in turn guarantees peri-odicity.

We discuss the three cases.

i. < 0. The general solution is = A e + B e (193)

which cannot be periodic unless A = B = 0 .

ii. = 0 . The general solution is = A + B (194)

which again cannot be periodic unless A = B = 0 .

iii. > 0. The general solution is = A cos + B sin (195)

which is 2 -periodic if and only if

= n 2 (196)for some integer n . As the cosine function is even and the sine function is odd, it suffices to con-sider the case n 0.

Thus the possible (, ) pairs are

n = n 2, n = cos n and sin n . (197)

Now that we have , we turn to the equation for R . Replacing by n 2 we have

r 2 R (r )R (r )

+r R (r )R (r )

+ r 2 = n 2, R (1) = 0 (198)which becomes

r 2 R (r ) + r R (r ) + r 2

n 2 R (r ) = 0 , R (1) = 0 . (199)

The theory of Bessel functions tells us that

R (r ) = A J n r + B Y n r (200)where J n and Y n are Bessel functions of the rst and the second kinds or order n .

Integrating the boundary condition R (0) bounded, we conclude B = 0 . On the other hand, requiringR (1) = 0 leads to = m where m is the m -th root of J n . Therefore the possible (, R ) pairs are

m , R m = J n m r . (201)Finally we solve the equation for T . For each m , solving

T (t )

T (t )=

k m . (202)

would give us a function T m and nally the solution can be written as a sum

u(r, , t ) =m =1

n =0

[T m,n, 1(t) cos(n ) + T m,n, 2(t ) sin(n )] J n m r . (203)

Setting t = 0 we have

f (r, ) =m =1

n =0

[T m,n, 1(0) cos(n ) + T m,n, 2(0) sin(n )] J n m r . (204)


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To obtain these initial values, notice that J n m r depens on both indices while sin (n ) and cos (n )only depends on n , we rst expand

f (r, ) =n =0

[f n, 1(r ) cos(n ) + f n, 2(r ) sin(n )] (205)

with

f n, 1(r ) =

12 0

2

f (r, ) d n = 01 0

2f (r, ) cos(n ) d n > 0

, f n, 2(r ) =1 0

2f (r, ) sin(n ) d. (206)

Now writing

f (r, ) =m =1

n =0

[T m,n, 1(0) cos(n ) + T m,n, 2(0) sin(n )] J n m r (207)

=n =0

m =1

T m,n, 1(0) J n m r cos(n ) +

m =1

T m,n, 2(0) J n m r sin(n ). (208)

Therefore

f n, 1(r ) = m =1

T m,n, 1(0) J n m r (209)and

f n, 2(r ) =m =1

T m,n, 2(0) J n m r . (210)

Recalling the orthogonality property

01

J n m r J n k r r dr = 0 , (211)we conclude

T m,n, 1(0) = 01

f n, 1(r ) J n m r r dr

0

1J n m r 2 r dr

, m = 1 , 2, (212)

and

T m,n, 2(0) = 01

f n, 2(r ) J n m r r dr

01

J n m r 2 r dr, m = 1 , 2, (213)

Now inserting the formulas for f n, 1(r ) and f n, 2(r ), we reach

T m,n, 1(0) = a m,n, 1 =

02

01

f (r, ) J 0 m r r dr2

0

1J 0 m r 2 r dr

n = 0

02

01

f (r, ) J n m r r cos(n ) dr d 0

1J n m r 2 r dr

n 1

, (214)

T m,n, 2(0) = a m,n, 2 = 02

01

f (r, ) J n m r r sin(n ) dr d 0

1J n m r 2 r dr

. (215)


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SolvingT (t )T (t )

= k m,n, 1 and k m,n, 2 (216)with initial values a m,n, 1 and a m,n, 2 we obtain

T m,n, 1(t ) = a m,n, 1 ek m t , T m,n, 2(t ) = a m,n, 2 ek m t . (217)

Finally, putting everything together, we reach

u(r, , t ) =m =1

n =0

[a m,n, 1 cos(n ) + am,n, 2 sin(n )] J n m r ek m t (218)

with

a m,n, 1 =

02

01

f (r, ) J 0 m r r dr2 0

1J 0 m r 2 r dr

n = 0

02

01

f (r, ) J n m r r cos(n ) dr d

0

1J n m r 2 r dr

n 1

, (219)

a m,n, 2 = 02

01

f (r, ) J n m r r sin(n ) dr d 0

1J n m r 2 r dr

. (220)

The formula above is the same as the one in the book given on page 733. The equivalence follows from(8.6.30).

3.2. Legendre function.Legendre functions arise from PDEs on domains with sperical symmetry. 1

Example 10. We consider the Laplace equation in a sphere:

u xx + u yy + u zz = 0 x 2 + y2 + z2 < a 2, (223)u = f x 2 + y2 + z2 = a 2. (224)

Consider the case where f = f () is independent of the longitudinal coordinate in the spherical coordi-nates

r = x2 + y2 + z2 , 0 < < , 0 < < 2 . (225)In this case u is also a function of r, only.The problem now becomes

r 2 u r r +1

sin [(sin ) u ] = 0 , (226)

u(a, ) = f (). (227)

1. Spherical coordinates. Consider a vector in R 3 (x -y-z space). Let r be its length, and let be the angle betweenthis vector and the z -axis. Then the projection of this vector onto the x -y plane is r sin . Now we introduce a second angle

which is the angle in the polar coordinates of the x -y plane. Thus we have

x = r cos sin , y = r sin sin , z = r cos (221)

with r > 0 , 0 < 2 , 0 < .One can also try to make measuring the (signed) angle between the vector and the x -y plane, in that case the change-

of-variable relation changes tox = r cos cos, y = r sin cos, z = r sin . (222)

This means the spherical coordinate form of u xx + u yy + u zz also changes.


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We look for a basic solution with separated variables

u (r, ) = R (r ) (). (228)

Substituting into the equation, we have

1R

r 2 R = 1

sin [(sin ) ] . (229)

Thus there is a constant such that

r 2 R R = 0 , 0 < r < a (230)[(sin ) ] + (sin ) = 0 , 0 < < . (231)

The R equation can be solve through the following. Write the equation as

r (r R ) + r R R = 0 , (232)and set

x = ln r. (233)Then we have

dRdr = dRdx dxdr = r 1 dRdx r R = dRdx (234)

therefore the equation becomesd2Rdx 2 +

dRdx R = 0 (235)

whose general solution isR (r ) = A r + B r (1+ ) (236)

with

= 1 + 1 + 4 2

solves 2 + = 0 . (237)For the equation, a change of variable and unknown

x = cos , y(x) = () (238)leads to

dd

= dydx

dxd

= y ( sin ) (239)therefore

dd [(

sin) ]= dd sin2 y =d

dxx2 1 y

dxd

= x2 1 y ( sin ). (240)Thus the equation becomes

1x 2 y + y = 0 (241)which can be expanded to

1x 2 y 2 x y + y = 0 , 1 < x < 1. (242)or equivalently

1x2 y 2 x y + ( + 1 ) y = 0 , 1 < x < 1 (243)as

[(1 + )] = . (244)The solutions are called Legendre functions.

Remark 11. When the problem is not independent of the variable , the resulting equation wouldlead to the so-called associated Legendre functions.


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The Legendre equation reads

1x2 y 2 x y + ( + 1 ) y = 0 . (245)We search for solutions of the form

y(x) =m =0

a m x m . (246)

Substituting into the equation and collecting the same powers, we have

m =0

[(m + 1 ) (m + 2 ) a m +2 + ( m ) ( + m + 1 ) a m ]xm = 0 . (247)

As a consequence, we must have

a m +2 = ( m ) ( + m + 1 )

(m + 1 ) (m + 2 )a m , m 0. (248)

Iterating, we have the formulas for the coefficients

a 2 k =( 1)

k ( 2) ( 2 k + 2 ) ( + 1 ) ( + 3 ) ( + 2 k 1)(2 k)!

a 0, (249)

a 2k +1 =(

1)k (

1) (

3) (

2 k + 1 ) ( + 2 ) ( + 4 ) ( + 2 k)

(2 k + 1 )! a 0. (250)It can be shown that when n is an integer, the solution is a sum of one polynomial of order n , and aninnite series solution. The polynomial solution is denoted P n (x ), and called the Legendre polynomial of degree n or the Legendre function of the rst kind of order n . The innite series is denoted Q n (x) andcalled the Legendre function of the second kind .

Therefore the general solutions to the Legendre equation are

y(x) = A P (x) + B Q (x). (251)

In practice one usually requires y(x) to be bounded at x = 1, this is true if and only if = n is aninteger and B = 0 , that isy(x) = P n (x) (252)

A simple formula for the Legendre polynomials isP n (x) =

12n n !

dn

dxnx2 1

n . (253)

From this the rst few Legendre polynomials can be easily computed as

P 0(x) = 1 (254)P 1(x) = x (255)

P 2(x) =12

3 x2 1 (256)P 3(x) =

12 5

x2 3 x . (257)From the Sturn-Liouville theory, we know that P n (x) enjoys the orthogonality relation

11

P n (x) P m (x) dx = 0 n

m. (258)Or equivalently

0

P n (cos) P m (cos) sin d = 0 n

m. (259)

Now we state one fact that is very useful when solving equations. Observe that any polynomial of degreek can be represented as a linear combination of the rst k Legendre polynomials. Therefore the orthogo-nality condition leads to

11

P n (x ) p(x ) dx = 0 (260)


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for any polynomial p(x ) with degree k < n .Finally it can be shown that

11

P n2(x ) dx =2

2 n + 1(261)

therefore the coefficients in

f (x) =0

An P n (x) (262)

is determined by

An = 11

f (x) P n (x) dx

11

P n (x )2 dx= 2 n + 1

2 11

f (x) P n (x ) dx. (263)

Example 12. (10.13 5) Find the solution of the Dirichlet problem for a sphere

2u = 0 , r < a, 0 < < , 0 < < 2 (264)

u (a , , ) = cos2. (265)

Solution. As the boundary value is cos2

which is independent of , and furthermore the Laplacian oper-ator is invariant with respect to translations in , the solution is also independent of , that is u = u(r,).

In this case the equation becomes

r 2 u r r +1

sin [(sin ) u ] = 0 , (266)

u(a, ) = cos2(). (267)

Setting u = R (r ) (), we reach

1R

r 2 R = 1

sin [(sin ) ] . (268)

Thus there is a constant such that

r 2 R R = 0 , 0 < r < a (269)[(sin ) ] + (sin ) = 0 , 0 < < . (270)

Settingx = cos , y(x) = () (271)

we have

1x 2 y 2 x y + y = 0 , 1 < x < 1. (272)Therefore

n = n (n + 1 ), yn = P n (x), n = 0 , 1, 2, (273)Returning to , we have

n = n (n + 1 ), n () = P n (cos ), n = 0 , 1, 2,

(274)Now back to the R equation we have

R (r ) = A r n + B r (n +1 ) (275)

where B is forced to 0 as we require R (0) to be bounded.Finally the solution can be written as

n =0

An r n P n (cos ). (276)


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To determine An we use the boundary value. Setting r = a we have

cos2 =n =0

An a n P n (cos) (277)

or equivalently

x 2 =n =1

An a n P n (x ). (278)

The coefficients are computed by

An a n =2 n + 1

2 11

x 2 P n (x) dx. (279)

As x2 is a polynomial of degree 2, we immediately know that An = 0 for all n 3. Now we compute

A0 =12 1

1x2 dx = 1

3; (280)

A1 a =32 1

1x3 dx = 0; (281)

A2 a2

=52 1

1 12 3 x

2

1 x2

dx =154 1

1

x4

dx 54 1

1

x2

dx =32

56 =

23 . (282)

Finally we have

u (r, ) = 13P 0(cos) +

23

r 2

a 2P 2(cos) (283)

We can further simplify it to

u(r, ) = 13+ 1

3r 2

a 23 cos2 1 =

13

1r 2

a 2+ r

2

a 2cos2. (284)

This formula is simple so we can actually check that it solves the equation and also satises the boundarycondition.

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