Oscillations In higher dimensions

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9Oscillations in Higher Dimensions

Equations of such complexity as are the equations of the gravitational eld can be found only through the discovery of a logically simple mathematical condition that determines the equations completely or at least almost completely.

What I have to say about this book can be found inside this book. Albert Einstein (1879 -1955 )

In this chapter we will explore several generic examples of thesolution of initial-boundary value problems involving higher spatialdimensions. These are described by higher dimensional partial dif-ferential equations, such as the ones presented in Table 8.1 in the lastchapter. We will solve these problems for different geometries, usingrectangular, polar, cylindrical, or spherical coordinates.

We will solve these problems using the method of separation of variables, though there are other methods which we will not considerin this text. Using separation of variables will result in a system of ordinary differential equations for each problem. Adding the bound-ary conditions, we will need to solve a variety of eigenvalue problems.

The product solutions that result will involve trigonometric or some of the special functions that we had encountered in Chapter 5.As you go through the examples in this chapter, you will see some

common features. For example, the two key equations that we havestudied are the heat equation and the wave equation. For higher di-mensional problems these take the form

u t = k 2u, (9.1)u tt = c2 2u. (9.2)

One can rst separate out the time dependence. Let u(r, t) = (r)T ( t).Inserting u into the heat and wave equations, we have

T = kT 2, (9.3)T = c2T 2. (9.4)

Separating out the time and space dependence, we nd1k

T T

= 2

= , (9.5)
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oscillations in higher dimensions 419

9.1 Vibrations of Rectangular Membranes

Our first example will be the study of the vibrations of a rectan-gular membrane. You can think of this as a drum with a rectangular

cross section as shown in Figure 9

.1. We stretch the membrane overthe drumhead and fasten the material to the boundary of the rectan-

gle. The height of the vibrating membrane is described by its heightfrom equilibrium, u(x, y, t). This problem is a much simpler exampleof higher dimensional vibrations than that possessed by the oscillatingelectric and magnetic elds in the last chapter.

Figure 9.1: The rectangular membrane of length L and width H . There are xed boundary conditions along the edges.

The problem is given by the two dimensional wave equation inCartesian coordinates,

utt = c2(uxx + u yy), t > 0, 0< x < L, 0 < y < H , (9.12)

a set of boundary conditions,

u(0, y, t) = 0, u(L, y, t) = 0, t > 0, 0< y < H ,u(x, 0,t) = 0, u(x, H , t) = 0, t > 0, 0< x < L, (9.13)

and a pair of initial conditions (since the equation is second order intime),

u(x, y, 0) = f (x, y), ut(x, y, 0) = g(x, y). (9.14)

The rst step is to separate the variables: u(x, y, t) = X (x)Y ( y)T ( t).Inserting the guess, u(x, y, t) into the wave equation, we have

X (x)Y ( y)T ( t) = c2

X (x)Y ( y)T (t) + X (x)Y ( y)T ( t) .

Dividing by both u(x, y, t) and c2, we obtain

1c2

T T

Function of t= X

X + Y

Y

Function of x and y= . (9.15)

We see that we have a function of t equals a function of x and y.Thus, both expressions are constant. We expect oscillations in time,so we chose the constant to be positive, > 0. (Note: As usual,the primes mean differentiation with respect to the specic dependentvariable. So, there should be no ambiguity.)

These lead to two equations:

T + c2 T = 0, (9.16)

andX X

+ Y Y

= . (9.17)


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interval is [0, H ]. Thus, we can quickly write down the solutions of theeigenvalue problem as

Y ( y) = sin m x H

, =m H

2, m = 1,2,3 , . . . .

We have successfully carried out the separation of variables for thewave equation for the vibrating rectangular membrane. The productsolutions can be written as

unm = ( a cosnm t + bsinnm t) sin n x

L sin m y

H . (9.24)

Recall that is given in terms of . We have that The harmonics for the vibrating rectan-gular membrane are given by

nm = c2 nL 2 + m H 2,

for n,m = 1 ,2, . . . .

mn n =m H

2

andn =

n L

2.

Therefore, nm =

n L

2+ m

H 2

. (9.25)

So,

nm = c n L 2 + m H 2. (9.26)The most general solution can now be written as a linear combi-

nation of the product solutions and we can solve for the expansioncoefcients that will lead to a solution satisfying he initial conditions.However, we will rst concentrate on the two dimensional harmonicsof this membrane.

For the vibrating string the nth harmonic corresponded to the func-tion sin n xL . The various harmonics corresponded to the pure tonessupported by the string. These then lead to the corresponding fre-quencies that one would hear. The actual shapes of the harmonicscould be sketched by locating the nodes, or places on the string thatdid not move.

In the same way, we can explore the shapes of the harmonics of thevibrating membrane. These are given by the spatial functions

nm(x, y) = sin n x

L sin m y

H . (9.27)

Instead of nodes, we will look for the nodal curves, or nodal lines. These A discussion of the nodal lines.are the points (x, y) at which nm(x, y) = 0. Of course, these dependon the indices, n and m.

For example, when n = 1 and m = 1, we have

sin xL

sin y H

= 0.


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422 mathematical physics

Figure 9.2: The rst few modes of thevibrating rectangular membrane. Thedashed lines show the nodal lines indi-cating the points that do not move forthe particular mode. Compare these thenodal lines to the 3D view in Figure 9.3

These are zero when either

sin xL = 0, or sin y H = 0.

Of course, this can only happen for x = 0,L and y = 0, H . Thus, thereare no interior nodal lines.

When n = 2 and m = 1, we have y = 0, H and

sin 2 xL

= 0,

or, x = 0, L2, L. Thus, there is one interior nodal line at x = L2 . Thesepoints stay xed during the oscillation and all other points oscillateon either side of this line. A similar solution shape results for the(1,2)-mode; i.e., n = 1 and m = 2.

In Figure 9.2 we show the nodal lines for several modes for n,m =1,2, 3 The blocked regions appear to vibrate independently. A betterview is the three dimensional view depicted in Figure 9.3 . The fre-quencies of vibration are easily computed using the formula for nm .

For completeness, we now see how one satises the initial condi-tions. The general solution is given by a linear superposition of theproduct solutions. There are two indices to sum over. Thus, the gen-eral solution is The general solution for the vibrating

rectangular membrane.

u(x, y, t) =

n= 1

m= 1

(anm cosnm t + bnm sinnm t) sin n x

L sin m y

H ,

(9.28)where

nm = c n L 2 + m H 2. (9.29)The rst initial condition is u(x, y, 0) = f (x, y). Setting t = 0 in the


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series,anm =

4LH H 0 L0 f (x, y) sin n xL sin m y H dxdy. (9.34)

The Fourier coefcients for the doubleFourier sine series.We can carry out the same process for satisfying the second initial

condition, ut (x, y, 0) = g(x, y) for the initial velocity of each point.Inserting this into the general solution, we have

g(x, y) =

n= 1

m= 1

bnmnm sin n x

L sin m y

H . (9.35)

Again, we have a double Fourier sine series. But, now we can writedown Fourier coefcients quickly using the above expression for anm:

bnm = 4

nm LH H 0 L0 g(x, y) sin n xL sin m y H dxdy. (9.36)This completes the full solution of the vibrating rectangular mem-

brane problem. Namely, we have obtained the solution The full solution of the vibrating rectan-gular membrane.

u(x, y, t) =

n= 1

m= 1

(anm cosnm t + bnm sinnm t) sin n x

L sin m y

H ,

(9.37)where

anm = 4LH H 0 L0 f (x, y) sin n xL sin m y H dxdy, (9.38)

bnm = 4

nm LH H 0 L0 g(x, y) sin n xL sin m y H dxdy, (9.39)and the angular frequencies are given by

nm = c n L 2 + m H 2. (9.40)9.2 Vibrations of a Kettle Drum

x

y

a

r

Figure 9.4: The circular membrane of ra-dius a. A general point on the membrane is given by the distance from thecenter, r, and the angle, . There are xed boundary conditions along the edge atr = a.

In this section we consider the vibrations of a circular membrane of radius a as shown in Figure 9.4. Again we are looking for the harmon-ics of the vibrating membrane, but with the membrane xed aroundthe circular boundary given by x2 + y2 = a2. However, expressing the boundary condition in Cartesian coordinates is awkward. Namely, wecan only write u(x, y, t) = 0 for x2 + y2 = a2. It is more natural touse polar coordinates as indicated in Figure 9.4. Let the height of themembrane be given by u = u(r,, t) at time t and position (r,). Nowthe boundary condition is given as u(a,, t) = 0 for all t > 0 and [0, 2 ].


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The 2D Laplacian can now be computed as

2 f x2

+ 2 f

y2 = cos

r f x

sinr

f x

+ sin r

f y

+ cos

r

f y

= cos r

cos gr

sinr

g

sinr

cos gr

sinr

g

+ sin r

sin gr

+ cosr

g

+ cosr

sin gr

+ cosr

g

= cos cos2 g

r2 +

sinr2

g

sinr

2 gr

sin

rcos

2 gr

sinr

2 g2 sin

gr

cosr

g

+ sin sin 2 g

r2 +

cosr

2 gr

cosr2

g

+cos

rsin

2 gr

+ cos

r2 g2

+ cos gr

sinr

g

= 2 g

r2 + 1r gr + 1r2

2 g

2

= 1

rr

r gr

+ 1r2

2 g2

. (9.43)

The last form often occurs in texts because it is in the form of a Sturm-Liouville operator. Also, it agrees with the result from using the Lapla-cian written in cylindrical coordinates as given in Problem 28.

Now that we have written the Laplacian in polar coordinates wecan pose the problem of a vibrating circular membrane. It is given bya partial differential equation,1 1 Here we state the problem of a vibrat-

ing circular membrane. We have chosen

0, < < , (9.45)and the initial conditions,

u(r,, 0) = f (r,), ut(r,, 0) = g(r,). (9.46)

Now we are ready to solve this problem using separation of vari-ables. As before, we can separate out the time dependence. Letu(r,, t) = T ( t)(r,). As usual, T ( t) can be written in terms of sinesand cosines. This leads to the Helmholtz equation,

2 + = 0.
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We now separate the Helmholtz equation by letting (r,) = R(r) ().This gives

1r

r

rR

r+ 1

r22R2

+ R = 0. (9.47)

Dividing by u = R , as usual, leads to

1rR

ddr

r dRdr

+ 1r2

d2d2

+ = 0. (9.48)

The last term is a constant. The rst term is a function of r. How-ever, the middle term involves both r and . This can be remedied bymultiplying the equation by r2. Rearranging the resulting equation,we can separate out the -dependence from the radial dependence.Letting be the separation constant, we have

rR

ddr

rdRdr

+ r2 = 1

d2d2

= . (9.49)

This gives us two ordinary differential equations:

d2d2

+ = 0,

r ddr

rdRdr

+ ( r2 )R = 0. (9.50)Lets consider the rst of these equations. It should look familiar by

now. For > 0, the general solution is

() = a cos + bsin .The next step typically is to apply the boundary conditions in . How-ever, when we look at the given boundary conditions in the problem,we do not see anything involving . This is a case for which the bound-ary conditions that are needed are implied and not stated outright.

We can determine the hidden boundary conditions by making someobservations. Lets consider the solution corresponding to the end-points = , noting that at these values for any r < a we are atthe same physical point. So, we would expect the solution to have thesame value at = as it has at = . Namely, the solution iscontinuous at these physical points. Similarly, we expect the slope of the solution to be the same at these points. This tells us that The boundary conditions in are peri-

odic boundary conditions. ( ) = ( ) ( ) = ( ).

Such boundary conditions are called periodic boundary conditions.Lets apply these conditions to the general solution for (). First,

we set ( ) = ( ) and use the symmetries of the sine and cosinefunctions:

a cos + bsin = a cos bsin .


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Now we are left with

R(r) = J m( r).We have set c1 = 1 for simplicity. We can apply the vanishing conditionat r = a. This gives

J m( a) = 0.Looking again at the plots of J m(x), we see that there are an innitenumber of zeros, but they are not as easy as ! In Table 9.1 we list thenth zeros of J m, which were rst seen in Table 5.3.

n m = 0 m = 1 m = 2 m = 3 m = 4 m = 51 2.405 3 .832 5 .136 6 .380 7 .588 8 .7712 5.520 7 .016 8 .417 9 .761 11 .065 12 .3393 8.654 10 .173 11 .620 13 .015 14 .373 15 .7004 11 .792 13 .324 14 .796 16 .223 17 .616 18 .980

5 14.931 16 .471 17 .960 19 .409 20 .827 22 .2186 18.071 19 .616 21 .117 22 .583 24 .019 25 .4307 21.212 22 .760 24 .270 25 .748 27 .199 28 .6278 24.352 25 .904 27 .421 28 .908 30 .371 31 .8129 27.493 29 .047 30 .569 32 .065 33 .537 34 .989

Table 9.1: The zeros of Bessel Functions, J m( jmn ) = 0.

Lets denote the nth zero of J m(x) by jmn . Then the boundary con-dition tells us that

a = jmn .This gives us the eigenvalue as

mn = jmna2 .

Thus, the radial function satisfying the boundary conditions is

R(r) = J m( jmn

a r).

We are nally ready to write out the product solutions for the vibrating circular membrane. They are given by Product solutions for the vibrating circu-

lar membrane.

u(r,, t) = cosmn tsinmn t cosmsinm J m

( jmna

r). (9.53)

Here we have indicated choices with the braces, leading to four differ-ent types of product solutions. Also, m = 0, 1,2,. . . , and

mn = jmn

a c.
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Figure 9.5: The rst few modes of the vibrating circular membrane. The dashedlines show the nodal lines indicating thepoints that do not move for the partic-ular mode. Compare these nodal lineswith the three dimensional images inFigure 9.6.

As with the rectangular membrane, we are interested in the shapesof the harmonics. So, we consider the spatial solution (t = 0)

(r,) = ( cosm) J m jmn

a r .

Including the solutions involving sinm will only rotate these modes.The nodal curves are given by (r,) = 0. This can be satised if cosm = 0, or J m( jmna r) = 0. The various nodal curves which result areshown in Figure 9.5.

For the angular part, we easily see that the nodal curves are radiallines, = const. For m = 0, there are no solutions, since cosm = 1and sinm = 1 for m = 0. in Figure 9.5 this is seen by the absence of radial lines in the rst column.

For m = 1, we have cos = 0. This implies that = 2 . Thesevalues give the vertical line as shown in the second column in Figure9.5. For m = 2, cos2 = 0 implies that = 4 , 3 4 . This results in thetwo lines shown in the last column of Figure 9.5.

We can also consider the nodal curves dened by the Bessel func-tions. We seek values of r for which jmna r is a zero of the Bessel functionand lies in the interval [0,a]. Thus, we have

jmna

r = jmj,

orr = jmj

jmna.

These will give circles of this radius with jmj jmn , or j n. Thezeros can be found in Table 9.1. For m = 0 and n = 1, there is only onezero and r = a. In fact, for all n = 1 modes, there is only one zero andr = a. Thus, the rst row in Figure 9.5 shows no interior nodal circles.


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Figure 9.6: A three dimensional view of the vibrating circular membrane for thelowest modes. Compare these imageswith the nodal line plots in Figure 9.5.

For n = 2 modes, we have two circles, r = a and r = jm1 jm2 as shownin the second row of Figure 9.5. For m = 0,

r = 2.4055.520a 0.436a

for the inner circle. For m = 1,

r = 3.8327.016a 0.546a,

and for m = 2,r =

5.1358.147a 0.630a.

For n = 3 we obtain circles of radii r = a,

r = jm1 jm3

, and r = jm2 jm3

.

For m = 0,r = a, 5.5208.654a 0.638a,

2.4058.654a 0.278a.

Similarly, for m = 1,

r = a, 3.83210.1730.377a a, 7.01610.173a 0.0.690a


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and for m = 2,

r = a, 5.13511.620a 0.442a, 8.41711.620a 0.724a.

For a three dimensional view, one can look at Figure 9.6. Imag-ine that the various regions are oscillating independently and that the

points on the nodal curves are not moving.Example 9.1. Vibrating Annulus

x

y

a

b

Figure 9.7: An annular membrane withradii a and b > a. There are xed bound-ary conditions along the edges at r = aand r = b.

More complicated vibrations can be dreamt up for this geometry. We couldconsider an annulus in which the drum is formed from two concentric circularcylinders and the membrane is stretch between the two with an annular crosssection as shown in Figure 9.7. The separation would follow as before exceptnow the boundary conditions are that the membrane is xed around the twocircular boundaries. In this case we cannot toss out the Neumann functionsbecause the origin is not part of the drum head.

With this in mind, we have that the product solutions take the form

u(r,, t) = cosmn tsinmn t cosmsinm Rm(r), (9.54)

whereRm(r) = c1 J m( r) + c2N m( r)

and = c .For this problem the radial boundary conditions are that the membrane is

xed at r = a and r = b. Taking b < a, we then have to satisfy the conditions

R(a) = c1 J m( a) + c2N m( a) = 0,R(b) = c1 J m( b) + c2N m( b) = 0. (9.55)

This leads to two homogeneous equations for c1 and c2. The coefcient de-terminant of this system has to vanish if there are to be nontrivial solutions.This gives the eigenvalue equation :

J m( a)N m( b) J m( b)N m( a) = 0.This eigenvalue equation needs to be solved numerically. Choosing a = 2 andb = 4, we have for the rst few modes

mn 1.562, 3.137, 4.709, m = 0 1.598, 3.156, 4.722, m = 1 1.703, 3.214, 4.761, m = 2. (9.56)

Note, since mn = mnc , these numbers essentially give us the frequenciesof oscillation.

For these particular roots, we then solve for c1 and c2 by setting c2 = 1and determining

c1 = N m( mn b) J m( mn b)


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(This selection is not unique. We could replace the bs in c1 with as and thatwould work as well.) This leads to the basic modes of vibration,

Rmn (r) m() = cosmN m( mn b) J m( mn b) J m( mn r) N m( mn r) .

In Figure 9.8 we show various modes for the particular choice of membranedimensions, a = 2 and b = 4.

Figure 9.8: A three dimensional view of the vibrating annular membrane for thelowest modes.

9.3 Laplaces Equation in 2D

Another of the generic partial differential equations isLaplaces equation, 2u = 0. This equation rst appeared in the chap-ter on complex variables when we discussed harmonic functions. An-other example is the electric potential for electrostatics. As we de-scribed in the last chapter, for static electromagnetic elds,

E =

/ 0. Also, E = . In regions devoid of charge, we have 2 = 0.Another example comes from studying temperature distributions.Consider a thin rectangular plate with the boundaries set at xed tem-peratures. One can solve the heat equation. The solution is time de-pendent. However, if one wait a long time, the plate reaches ther-mal equilibrium. If the boundary temperature is zero, then the plate


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Finally, we have the solution to this problem,

u(x, y) =

n= 1

an sin n x

L sinh n ( H y)

L , (9.64)

where

an = 2

Lsinh n H L L

0 f (x) sin n x

L dx. (9.65)Example 9.3. Equilibrium Temperature Distribution for a Rectangular Plate for General Boundary Conditions

Now we consider Laplaces equation in Cartesian coordinates,

uxx + u yy = 0, 0< x < L, 0 < y < H

with the non-zero boundary conditions on more than one side,

u(0, y) = g1( y),u(L, y) = g2( y),u(x, 0) = f 1(x),u(x, H ) = f 2(x).

The boundary conditions are shown in Figure 9.10x

y

u(x, 0) = f 1(x)

u(x, H ) = f 2(x)

u(0, y) = g1( y) u(L, y) = g2( y

Figure 9.10: In this gure we show thedomain and general boundary condi-tions for the example of determining theequilibrium temperature distribution fora rectangular plate.

The problem with this example is that none of the boundary conditions arehomogeneous, so we cannot specify the boundary conditions for the eigenvalue problems. However, we can express this problem as in terms of four problemswith nonhomogeneous boundary conditions on only one side of the rectangle.In Figure 9.11 we show how the problem can be broken up into four separate problems. Since the boundary conditions and Laplaces equation are linear,the solution to the general problem is simply the sum of the solutions to these four problems.

x

y

u = f 1(x)

u = 0

u = 0 u = 0

x

y

u) = 0

u = f 2(x)

u = 0 u = 0

x

y

u = 0

u = 0

u = g1( y) u = 0

x

y

u = 0

u = 0

u = 0 u = g2( y)

Figure 9.11: Breaking up the general boundary value problem for a rectangu-lar plate.We can solve each of the problems quickly, based on the solution obtained

in the last example. The solution for boundary conditions

u(0, y) = 0,u(L, y) = 0,u(x, 0) = f 1(x),u(x, H ) = 0.


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is the easiest to write down:

u(x, y) =

n= 1

an sin n x

L sinh n ( H y)

L . (9.66)

where

an = 2

Lsinh n H L L

0 f 1(x) sin n x

L dx. (9.67)For the boundary conditions

u(0, y) = 0,u(L, y) = 0,u(x, 0) = 0,u(x, H ) = f 2(x),

the boundary conditions for X (x) are X (0) = 0 and X (L) = 0. So, we getthe same form for the eigenvalues and eigenfunctions as before:

X n (x) = sin n x

L , n =

n L

2,n = 1,2, .

However, the remaining homogeneous boundary condition is now Y (0) = 0.Recalling the equation satised by Y ( y) is

Y Y = 0,we can write the general solution as

Y ( y) = c1 cosh y + c2 sinh y.Requiring Y (0) = 0, we have c1 = 0, or

Y ( y) = c2 sinh y.Then the general solution is

u(x, y) =

n= 1bn sin n x

L sinh n y

L . (9.68)

We now force the nonhomogenous boundary condition, u(x, H ) = f 2(x),

f 2(x) =

n= 1

bn sin n x

L sinh n H

L . (9.69)

Once again we have a Fourier sine series. The Fourier coefcients are givenby

bn = 2

Lsinh n H L L

0 f 2(x) sin

n xL

dx. (9.70)

Now we turn to the problem with the boundary conditions

u(0, y) = g1( y),u(L, y) = 0,u(x, 0) = 0,u(x, H ) = 0.

In this case the pair of homogeneous boundary conditions u(x, 0) = 0,u(x, H ) =0. lead to solutions

Y n ( y) = sin n y

H , n =

n H

2,n = 1, 2 .


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The condition u(L, 0) = 0 gives X (x) = sinh n (Lx) H . The general solutionis

u(x, y) =

n= 1

cn sin n y

H sinh n (Lx)

H . (9.71)

We now force the nonhomogenous boundary condition, u(0, y) = g1( y),

g1( y) =

n= 1

cn sin n y

H sinh n L

H . (9.72)

The Fourier coefcients are given by

cn = 2

H sinh n L H H

0 g1( y) sin

n y H

dy. (9.73)

Finally, we can nd the solution for

u(0, y) = 0,u(L, y) = g2( y),u(x, 0) = 0,u(x, H ) = 0.

Following the above analysis, we nd the general solution

u(x, y) =

n= 1

dn sin n y

H sinh n x

H . (9.74)

We now force the nonhomogenous boundary condition, u(L, y) = g2( y),

g2( y) =

n= 1

dn sin n y

H sinh n L

H . (9.75)

The Fourier coefcients are given by

dn =

2 H sinh n L H

H

0 g1( y

)sin

n y

H dy. (9.76

)The solution to the general problem is given by the sum of these four solu-

tions.

u(x, y) =

n= 1

an sinh n ( H y)

L + bn sinh

n yL

sin n xL

+ cn sinh n (Lx)

H + dn sinh

n x H

sin n y H

,

(9.77)

where the coefcients are given by the above Fourier integrals.

Example 9.4. Laplaces Equation on a Disk We now turn to solving Laplaces equation on a disk of radius a as shown

in Figure 9.12. Laplaces equation in polar coordinates is given by

1r

r

rur

+ 1r2

2u2

= 0, 0< r < a, < < . (9.78)


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The boundary conditions are given as

u(a,) = f (), < < , (9.79) plus periodic boundary conditions in .

x

y

a

u(a,) = f ()Figure 9.12: The circular plate of radius awith boundary condition along the edgeat r = a.

Separation of variable proceeds as usual. Let u(r,) = R(r) (). Then

1r

r

r (R )r

+ 1r2

2(R )2

= 0, (9.80)

or 1

r(rR ) +

1r2

R = 0. (9.81)

Diving by u(r,) = R(r) () and rearranging, we haverR

(rR ) =

= . (9.82)

Since this equation gives a function of r equal to a function of , we setthe equation equal to a constant. Thus, we have obtained two differentialequations, which can be written as

r(rR ) R = 0, (9.83) + = 0. (9.84)

We can solve the second equation, using periodic boundary conditions. Thereader should be able to conrm that

() = an cosn + bn sinn, = n2,n = 0,1,2, is the solution. Note that the n = 0 case just leads to a constant solution.

Inserting = n2 into the radial equation, we nd

r2R + rR

n2R = 0.

This is a Cauchy-Euler type of ordinary differential equation. Recall that wesolve such equations by guessing a solution of the form R(r) = rm. This leadsto the characteristic equation m2 n2 = 0. Therefore, m = n. So,

R(r) = c1rn + c2rn.Since we expect nite solutions at the origin, r = 0, we can set c2 = 0. Thus,the general solution is

u(r,) = a02 +

n= 1

(an cosn + bn sinn) rn. (9.85)

Note that we have taken the constant term out of the sum and put it into a familiar form.Now were are ready to impose the remaining boundary condition, u(a,) =

f (). This gives

f () = a02 +

n= 1

(an cosn + bn sinn) an . (9.86)


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This is a Fourier trigonometric series. The Fourier coefcients can be deter-mined using the results from Chapter 4:

an = 1 an f () cosn d, n = 0,1, , (9.87)

bn = 1

an

f () sinn d n = 1, 2

. (9.88)

We can put the solution from the last example in a more compactform by inserting these coefcients into the general solution. Doingthis, we have

u(r,) = a02 +

n= 1

(an cosn + bn sinn) rn

= 12 f () d+

1

n= 1

[cosn cosn + sinn sinn] ra

n f () d

= 1 12 +

n= 1

cosn( ) ra n f () d. (9.89)The term in the brackets can be summed. We note that cosn(

) = Re(ein() ). Then

cosn( )ra

n= Re ei() r

an

= Rera

ei()n

. (9.90)

Therefore,

n= 1

cosn(

)

r

a

n= Re

n= 1

r

aei()

n

The right hand side of this equation is a geometric series with commonratio ra ei() . Since ra ei() = ra < 1, the series converges. Summingthe series, we obtain

n= 1

ra

ei()n

=ra e

i()1 ra ei()

= rei()

a rei()= re

i()a

rei()

a rei()a

rei()

= arei() r2a2 + r2 2ar cos( )

. (9.91)

We have rewritten this sum so that we can easily take the real part,

Re

n= 1

ra

ei()n

= arcos( ) r2

a2 + r2 2ar cos( )


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Therefore, the factor in the brackets under the integral in Equation(9.89) is

12 +

n= 1

cosn( )ra

n=

12 +

ar cos( ) r2a2 + r2 2ar cos( )

= a2

r2

2(a2 + r2 2ar cos( )) .(9.92)Thus, we have shown that the solution of Laplaces equation on a

disk of radius a with boundary condition u(a,) = f () can be writtenin the closed form Poisson Integral Formula

u(r,) = 12 a2 r2

a2 + r2 2ar cos( ) f () d. (9.93)

This result is called the Poisson Integral Formula and

K (,) = a2 r2

a2 + r2

2ar cos(

)

is called the Poisson kernel.

9.4 Three Dimensional Cake Baking

In the rest of the chapter we will extend our studies to threedimensions. In this section we will solve the heat equation as we lookat examples of baking cakes. We consider cake batter, which is at This discussion of cake baking is

adapted from R. Wilkinsons thesiswork. That in turn was inspired by workdone by Dr. Olszewski.

room temperature of T i = 80F. It is placed into an oven, also at axed temperature, T b = 350F. For simplicity, we will assume that thethermal conductivity and cake density are constant. Of course, thisis not quite true. However, it is an approximation which simpliesthe model. We will consider two cases, one in which the cake is arectangular solid (0 x W , 0 y L, 0 z H ), such as baking itin a 13 9 2 baking pan. The other case will lead to a cylindricalcake, such as you would obtain from a round cake pan.

Assuming that the heat constant k is indeed constant and the tem-perature is given by T (r, t), we begin with the heat equation in threedimensions,

T t

= k 2T . (9.94)

We will need to specify initial and boundary conditions. Let T i be theinitial batter temperature, and write the initial condition as

T (x, y, z, 0) = T i.

We choose the boundary conditions to be xed at the oven temperatureT b. However, these boundary conditions are not homogeneous and


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would lead to problems when carrying out separation of variables.This is easily remedied by subtracting the oven temperature from alltemperatures involved and dening u(x, y, z, t) = T (x, y, z, t) T b. Theheat equation then becomes

u

t =

k 2

u (9.95

)with initial condition

u(r, 0) = T i T b.The boundary conditions are now that u = 0 on the boundary. Wecannot be any more specic than this until we specify the geometry.

Example 9.5. Temperature of a Rectangular CakeFor this problem, we seek solutions of the heat equation plus the conditions

u(x, y, z, 0) = T i T b,u(0, y, z, t) = u(W , y, z, t) = 0,u(x, 0, z, t) = u(x, L, z, t) = 0,

u(x, y, 0,t) = u(x, y, H , t) = 0.

Using the method of separation of variables, we seek solutions of the form

u(x, y, z, t) = X (x)Y ( y)Z( z)G(t). (9.96)

Substituting this form into the heat equation, we get

1k

GG

= X X

+ Y Y

+ ZZ

. (9.97)

Setting these expressions equal to , we get1k

GG

= and X

X + Y

Y + Z

Z = . (9.98)

Therefore, the equation for G(t) is given by

G + k G = 0.

We further have to separate out the functions of x, y, and z. We anticipatethat the homogeneous boundary conditions will lead to oscillatory solutionsin these variables. Therefore, we expect separation of variable will lead to theeigenvalue problems

X + 2X = 0, X (0) = X (W ) = 0,Y + 2Y = 0, Y (0) = Y (L) = 0,

Z + 2Z = 0, Z(0) = Z( H ) = 0. (9.99)

Noting thatX X

= 2, Y

Y = 2,

ZZ

= 2,


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we have the relation 2 = 2 + 2 + 2.From the boundary conditions, we get product solutions for u(x, y, z, t) in

the form

umn (x, y, z, t) = sinmx sinn ysin ze mn kt,

for

mnl = 2m + 2n +

2 = m W

2+ n

L2

+ H

2, m,n, = 1,2, . . . .

The general solution is then

u(x, y, z, t) =

m= 1

n= 1

= 1

Amnl sinmx sinn ysin ze mn kt, (9.100)

where the Amn s are arbitrary constants.We can use the initial condition u(x, y, z, 0) = T i T b to determine the

Amn s. We nd

T i T b =

m= 1

n= 1

= 1

Amnl sinmx sinn ysin z. (9.101)

This is a triple Fourier sine series. We can determine these coefcients in amanner similar to how we handled a double Fourier sine series earlier. Den-ing

bm( y, z) =

n= 1

= 1

Amnl sinn ysin z,

we obtain a simple Fourier sine series:

T i

T b =

m= 1

bm( y, z) sinmx. (9.102)

The Fourier coefcients can then be found as

bm( y, z) = 2W W 0 (T i T b) sinmx dx.

Using the same technique for the remaining sine series and noting thatT i T b is constant, we can compute the general coefcient Amnl by carryingout the needed integrations:

Amnl = 8WLH H 0 L0 W 0 (T i T b) sinmx sinn ysin z dxdydz

= (T i T b) 8 3 cos( m x

W )

m

W

0 cos( n y

L )

n

L

0cos

( z H

) H

0

= (T i T b) 8 3

cosm 1m

cosn 1n

cos 1

= (T i T b) 8 3

0, for at least one m,n, even,2m 2n 2 , for m,n, all odd.


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Figure 9.13: Temperature evolution fora 13 9 2 cake shown as verticalslices at the indicated length in feet.Choosing as the separation constant, we get

G k G = 0, (9.105)and

1rR

ddr

rR = H H

+ . (9.106)

Since negative eigenvalues yield the oscillatory solutions we expect, we con-tinue as before by setting both sides of this equation equal to2. After somerearrangement, we obtain the needed differential equations:

ddr

rR + r2R = 0 (9.107)

and H + 2 H = 0. (9.108)

Here = 2 + 2 .We can easily write down the solutions

G(t) = Ae kt


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and H n ( z) = sin

n zZ

,n = 1,2 ,3, . . . ,

where = n Z . Recalling from the rectangular case that only odd terms arisein the Fourier sine series coefcients for the constant initial condition, we proceed by rewriting H ( z) as

H n( z) = sin(2n 1) z

Z,n = 1, 2, 3, . . . (9.109)

with = (2n1) Z .The radial equation can be written in the form

r2R + rR + r22R = 0.

This is a Bessel equation of the rst kind of order zero and the general solutionis a linear combination of Bessel functions of the rst and second kind,

R(r) = c1 J 0(r) + c2N 0(r). (9.110 )Since we wish to have u(r, z, t) bounded at r = 0 and N 0(r) is not wellbehaved at r = 0, we set c2 = 0. Up to a constant factor, the solutionbecomes

R(r) = J 0(r). (9.111 )

The boundary condition R(a) = 0 gives J 0(a) = 0 and thus m = j0ma , for m = 1,2,3, . . .. Here the j0ms are the mth roots of the zeroth-order Bessel function, J 0( j0m) = 0, which are given in Table 9.1. This suggests that

Rm(r) = J 0ra j0m ,m = 1, 2, 3, . . . . (9.112 )

Thus, we have found that the general solution is given as

u(r, z, t) =

n= 1

m= 1

Anm sin(2n 1) z

Z J 0

ra j0m e nm kt (9.113 )

with

nm =(2n 1)

Z

2+ j0m

a

2,

for n,m = 1,2,3 , . . . .Using the constant initial condition to nd the Anm s, we have

T i T b =

n= 1

m= 1

Anm sin (2n 1) zZ J 0 ra j0m .

If we let bn(r) = m= 1 Anm J 0ra j0m , we have

T i T b =

n= 1

bn(r) sin(2n 1) z

Z.


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As seen previously, this is a Fourier sine series and the Fourier coefcients are given by

bn (r) = 2

Z Z0 (T i T b) sin (2n 1) zZ dz=

2(T i

T b)

Z Z

(2n 1) cos(2n

1) z

Z

Z

0

= 4(T i T b)

(2n 1) .

Then, we have

bn(r) = 4(T i T b)(2n 1)

=

m= 1

Anm J 0ra j0m .

This is a Fourier-Bessel series. Given bn(r) = 4(T iT b)(2n1) , we seek to nd theFourier coefcients Anm . Recall from Chapter 5 that the Fourier-Bessel series

is given by

f (x) =

n= 1cn J p( j pnxa ), (9.114 )

where the Fourier-Bessel coefcients are found as

cn = 2

a2 J p+ 1( j pn) 2 a

0x f (x) J p( j pn

xa

) dx. (9.115 )

For this problem, we have

Anm = 2

a2 J 21( j0m)4(T i T b)(2n 1)

a

0 J 0(mr)r dr. (9.116 )

In order to evaluate

a0 J 0(k r)r dr, we let y = k r and get

a

0 J 0(k r)rdr = k a

0 J 0( y) yk

dyk

= 12k

k a

0 J 0( y) y dy

= 1

2k k a

0d

dy ( yJ 1( y)) dy

= 1

2k (k a) J 1(k a) =

a2

j0k J 1( j0k ). (9.117 )

Here we have made use of the identity ddx (xJ 1(x)) = J 0(x).Substituting the result of this integral computation into the expression for

Anm , we nd Anm = 8(T i T b)(2n 1)

1 j0m J 1( j0m)

.

Substituting A nm into the original expression for u(r, z, t), gives

u(r, z, t) = 8(T i T b)

n= 1

m= 1

sin (2n1) zZ(2n 1)

J 0( ra j0m)e nm Dt

j0m J 1( j0m) .


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Therefore, T (r, z, t) can be found as

T (r, z, t) = T b + 8(T i T b)

n= 1

m= 1

sin (2n1) zZ(2n 1)

J 0( ra j0m)e nm kt

j0m J 1( j0m) .

We have therefore found the general solution for the three-dimensional heatequation in cylindrical coordinates with constant diffusivity. Similar to thesolutions shown in Figure 9.13 of the previous section, we show in Figure9.14 the temperature temperature evolution throughout a standard 9 roundcake pan.

Figure 9.14: Temperature evolution for astandard 9 cake shown as vertical slicesthrough the center.

9.5 Laplaces Equation and Spherical Symmetry

We have seen that Laplace s equation ,

2u = 0, arises in elec-

trostatics as an equation for electric potential outside a charge distribu-tion and it occurs as the equation governing equilibrium temperaturedistributions. As we had seen in the last chapter, Laplaces equationgenerally occurs in the study of potential theory, which also includesthe study of gravitational and uid potentials. The equation is namedafter Pierre-Simon Laplace (1749 -1827 ) who had studied the properties


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of this equation. solutions of Laplaces equation are called harmonicfunctions.

Laplaces equation in spherical coordinates is given by2 2 The Laplacian in spherical coordinatesis given in Problem 29 in Chapter 8.

1 2

2u

+ 1

2 sin

sin u

+ 1

2 sin2 2u2

= 0, (9.118 )

where u = u( ,,).

Figure 9.15: A sphere of radius r withthe boundary condition u(r,, ) = g(,).

We seek solutions of this equation inside a sphere of radius r subjectto the boundary condition u(r,,) = g(,) as shown in Figure 9.15.

As before, we perform a separation of variables by seeking productsolutions of the form u( ,,) = R( ) () (). Inserting this forminto the Laplace equation, we obtain

x

y

z

Figure 9.16: Denition of spherical coor-dinates ( ,,). Note that there are dif-ferent conventions for labeling sphericalcoordinates. This labeling is used often

in physics.

2d

d 2

dRd

+ R

2 sindd

sin d

d+ R

2 sin2 d2d2

= 0. (9.119 )

Multiplying this equation by 2

and dividing by R , yields1R

dd

2dRd

+ 1sin

dd

sind

d+

1sin2

d2d2

= 0. (9.120)

Note that the rst term is the only term depending upon . Thus, wecan separate out the radial part. However, there is still more work todo on the other two terms, which give the angular dependence. Thus,we have

1R

dd

2dRd

= 1sin

dd

sin d

d+

1sin2

d2d2

= ,(9.121)

where we have introduced the rst separation constant. This leads totwo equations:

dd

2dRd R = 0 (9.122)

and Equation (9.123) is a key equation whichoccurs when studying problems possess-ing spherical symmetry. It is an eigen-value problem for Y (, ) = () (),LY = Y , where

L = 1sin

sin

+ 1sin2

2

2 .

The eigenfunctions of this operator are

referred to as spherical harmonics.

1sin

dd

sind

d+ 1

sin2 d2d2

= . (9.123)The nal separation can be performed by multiplying the last equa-

tion by sin2 , rearranging the terms, and introducing a second sepa-ration constant:

sin

dd sin

dd + sin

2 =

1

d2d2 = . (9.124)

From this expression we can determine the differential equations sat-ised by () and ():

sin dd

sin d

d+ ( sin2 ) = 0, (9.125)
http://0.0.0.0/http://0.0.0.0/


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andd2d2

+ = 0. (9.126)

We now have three ordinary differential equations to solve. Theseare the radial equation (9.122) and the two angular equations (9.125)-(9.126). We note that all three are in Sturm-Liouville form. We willsolve each eigenvalue problem subject to appropriate boundary condi-tions.

The simplest of these differential equations is the one for (),Equation (9.126). We have seen equations of this form many timesand the general solution is a linear combination of sines and cosines.As argues in such problems, we have to impose periodic boundaryconditions. For example, we expect that

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

Since these conidtions hold for all and , we must require that

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

As we have seen before, the eigenfunctions and eigenvalues are thenfound as

() = {cosm,sinm}, = m2, m = 0, 1, . . . . (9.127)Next we turn to solving equation, (9.126). We rst transform this

equation in order to identify the solutions. Let x = cos. Then thederivatives with respect to transform as

d

d = dx

d

d

dx = sin d

dx.

Letting y(x) = () and noting that sin2 = 1x2, Equation (9.126) becomes

ddx

(1x2)dydx

+ m2

1x2 y = 0. (9.128)

We further note that x [1, 1], as can be easily conrmed by thereader.

This is a Sturm-Liouville eigenvalue problem. The solutions consistof a set of orthogonal eigenfunctions. For the special case that m = 0Equation (9.128) becomes

ddx

(1x2) dydx + y = 0. (9.129)In a course in differential equations one learns to seek solutions of thisequation in the form

y(x) =

n= 0

an xn.


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This leads to the recursion relation

an+ 2 = n(n + 1) (n + 2)( n + 1) an.

Setting n = 0 and seeking a series solution, one nds that the resultingseries does not converge for x =

1. This is remedied by choosing

= ( + 1) for = 0, 1, . . . , leading to the differential equationd

dx(1x2)

dydx

+ ( + 1) y = 0. (9.130)

We saw this equation in Chapter 5. The solutions of this Legendredifferential equation are the Legendre polynomials, denoted by P (x).

For the more general case, m = 0, the differential equation (9.128)with = ( + 1) becomes Associated Legendre Functions

d

dx

(1

x2)

dy

dx

+ ( + 1)

m2

1x2 y = 0. (9.131)

The solutions of this equation are called the associated Legendre func-tions. The two linearly independent solutions are denoted by Pm (x)and Qm (x). The latter functions are not well behaved at x = 1, cor-responding to the north and south poles of the original problem. So,we can throw out these solutions, leaving

() = Pm (cos)

as the needed solutions. In Table 9.2 we list a few of these.

Pmn (x) Pmn (cos)P01 (x) x cosP11 (x) (1x2)

12 sin

P02 (x) 1

2(3x2 1) 12(cos2 1)P12 (x) 3x(1x2)

12 3cos sin

P22 (x) 3(1x2) 3sin2 P03 (x)

12(5x3 32x) 12(5cos3 3cos)

P13 (x) 3

2(5x2 1)( 1x2)12 32(5cos2 1) sin

P23 (x) 15x(1x2) 15cos sin2 P33 (x) 15(1x2)

32 15sin3

Table 9.2: Associated Legendre Func-

tions, Pmn (x).

The associated Legendre functions are related to the Legendre poly-nomials by3 3 Some denitions do not include the

(1)m factor.Pm (x) = ( 1)m(1x2)m/2

dm

dxm P (x), (9.132)

for = 0,1,2,, . . . and m = 0,1, . . . ,. We further note that P0(x) =P (x), as one can see in the table. Since P (x) is a polynomial of degree


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37/48


for = 0,1,2,.. . and m = 0,1 , , . . . , . These solutions can becombined to give a complex representation of the product solutions as

u( ,,) = Pm (cos)eim.

The general solution is then given as a linear combination of theseproduct solutions. As there are two indices, we have a double sum:4 4 While this appears to be a complex-valued solution, it can be rewritten as

a sum over real functions. The innersum contains terms for both m = k andm = k . Adding these contributions, wehave thata k Pk (cos)eik + a (k ) P

k (cos)eik

can be rewritten as

( A k cosk + B k sink ) Pk (cos).

u( ,,) =

= 0

m=

a m Pm (cos)eim. (9.136)

The solutions of the angular parts of the problem are often com- bined into one function of two variables, as problems with sphericalsymmetry arise often, leaving the main differences between such prob-lems conned to the radial equation. These functions are referred toas spherical harmonics, Y m(,), which are dened with a special Y m(,), are the spherical harmonics.

Spherical harmonics are important inapplications from atomic electron con-

gurations to gravitational elds, plane-tary magnetic elds, and the cosmic mi-crowave background radiation.

normalization as

Y m(,) = 2 + 14 ( m)!( + m)!Pm eim. (9.137)These satisfy the simple orthogonality relation

0 2 0 Y m(,)Y m (,) sin d d = mm .As noted in an earlier side note, the spherical harmonics are eigen-

functions of the eigenvalue problem LY = Y , where

L = 1sin

sin

+ 1sin2

2

2.

This operator appears in many problems in which there is sphericalsymmetry, such as obtaining the solution of Schrdingers equationfor the hydrogen atom as we will see later. Therefore, it is customaryto plot spherical harmonics. Because the Y ms are complex functions,one typically plots either the real part or the modulus squared. Onerendition of |Y m(,)|2 is shown in Figure 9.17.

We could also look for the nodal curves of the spherical harmonicslike we had for vibrating membranes. Such surface plots on a sphereare shown in Figure 9.18. The colors provide for the amplitude of the |Y m(,)|2. We can match these with the shapes in Figure 9.17 by coloring the plots with some of the same colors. This is shownin Figure 9.19. However, by plotting just the sign of the sphericalharmonics, as in Figure 9.20, we can pick out the nodal curves mucheasier.


38/48


Figure 9.17: The rst few spherical har-monics, |Y m(,)|2

Figure 9.18: Spherical harmonic con-tours for |Y m(,)|2.


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Figure 9.19: The rst few spherical har-monics, |Y m(,)|2

Figure 9.20: In these gures we show thenodal curves of |Y m(,)|2


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Figure 9.21: Zonal harmonics,

= 1,m = 0.

Figure 9.22: Zonal harmonics, = 2,m = 0.

Figure 9.23: Sectoral harmonics, = 2,m = 2.

Spherical, or surface, harmonics can be further grouped into zonal,sectoral, and tesseral harmonics. Zonal harmonics correspond to them = 0 modes. In this case, one seeks nodal curves for which P (cos) =0. These lead to constant values such that cos is a zero of the Legen-

dre polynomial, P (x). These correspond to the rst column in Figure9.20. Since P (x) is a polynomial of degree , the zonal harmonicsconsist of latitudinal circles.

Sectoral, or meridional, harmonics result for the case that m = .For this case, we note that P (x) (1 x2)m/2 . This vanishes forx = 1, or = 0, . Therefore, the spherical harmonics can onlyproduce nodal curves for eim = 0. Thus, one obtains the meridianscorresponding to solutions of A cosm + Bsinm = 0. Such solutionsare constant values of . These can be seen in Figure 9.20 in the topdiagonal and can be described as m circles passing through the poles,or longitudinal circles.

Tesseral harmonics are all of the rest, which typically look like achecker board glued to the surface of a sphere. Examples can be seenin the pictures of nodal curves, such as Figure 9.20. Looking in Figure9.20 along the diagonals going downward from left to right, one cansee the same number of latitudinal circles. In fact, there are mlatitudinal nodal curves in these gures

Figure 9.24: Tesseral harmonics, = 3,m = 1.

Figure 9.25: Sectoral harmonics, = 3,m = 3.

Figure 9.26: Tesseral harmonics, = 4,m = 3.

In summary, the spherical harmonics have several representations,as show in Figures 9.17-9.20. Note that there are nodal lines, mmeridional curves, and m horizontal curves in these gures. Theplots in Figures 9.17 and 9.19 are the typical plots shown in physics fordiscussion of the wavefunctions of the hydrogen atom. Those in 9.18

are useful for describing gravitational or electric potential functions,temperature distributions, or wave modes on a spherical surface. Therelationships between these pictures and the nodal curves can be bet-ter understood by comparing respective plots. Several modes wereseparated out in Figures 9.21-9.26 to make this comparison easier.Example 9.7. Laplaces Equation with Azimuthal Symmetry

As a simple example we consider the solution of Laplaces equation in whichthere is azimuthal symmetry. Let

u(r,,) = g() = 1cos2.This function is zero at the poles and has a maximum at the equator. So, this

could be a crude model of the temperature distribution of the Earth with zerotemperature at the poles and a maximum near the equator.

In problems in which there is no -dependence, only the m = 0 term of the general solution survives. Thus, we have that

u( ,,) =

= 0

a P (cos). (9.138)


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Here we have used the fact that P0(x) = P (x). We just need to determinethe unknown expansion coefcients, a. Imposing the boundary condition at = r, we are lead to

g() =

= 0

a r P (cos). (9.139)

This is a Fourier-Legendre series representation of g(). Since the Legen-dre polynomials are an orthogonal set of eigenfunctions, we can extract thecoefcients. In Chapter 5 we had proven that

0 Pn(cos)Pm(cos) sin d = 11 Pn(x)Pm(x) dx = 22n + 1nm .So, multiplying the expression for g() by Pm(cos) sin and integrating,we obtain the expansion coefcients:

a = 2 + 1

2r 0 g()P (cos) sin d. (9.140)Sometimes it is easier to rewrite g() as a polynomial in cos and avoid

the integration. For this example we see that

g() = 1cos2= 2sin2 = 22cos2 . (9.141)

Thus, setting x = cos, we have g() = 22x2. We seek the form g() = c0P0(x) + c1P1(x) + c2P2(x),

where P0(x) = 1, P1(x) = x, and P2(x) = 12(3x2 1). Since g() =22x

2does not have any x terms, we know that c1 = 0. So,

22x2 = c0(1) + c212(3x

2 1) = c0 12c2 +

32c2x

2.

By observation we have c2 = 43 and thus, c0 = 2+ 12c2 = 43. This givesthe sought expansion for g():

g() = 43P0(cos)

43P2(cos). (9.142)

Therefore, the nonzero coefcients in the general solution become

a0 = 43, a2 =

43

1r2

,

and the rest of the coefcients are zero. Inserting these into the general solu-tion, we have

u( ,,) = 43P0(cos) 43

r

2P2(cos)

= 4

3 23

r

2(3cos2 1). (9.143)


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9.6 Schrdinger Equation in Spherical Coordinates

Another important eigenvalue problem in physics is the Schrdingerequation. The time-dependent Schrdinger equation is given by

ih

t = h2

2m2

+ V

. (9.144)Here (r, t) is the wave function, which is determines the quantumstate of a particle of mass m subject to a (time independent) potential,V (r). h = h2 , where h is Plancks constant. The probability of ndingthe particle in an innitesimal volume, dV , is given by | (r, t)|2 dV ,assuming the wave function is normalized,

all space| (r, t)|2 dV = 1.One can separate out the time dependence by assuming a special

form, (r, t) = (r)eiEt / h, where E is the energy of the particularstationary state solution, or product solution. Inserting this form intothe time-dependent equation, one nds that (r) satises

h22m

2 + V = E. (9.145)

Assuming that the potential depends only on distance from the ori-gin, V = V ( ), we can further separate out the radial part of this solu-tion using spherical coordinates. Recall that the Laplacian in sphericalcoordinates is given by

2 = 1

2

2

+ 1

2

sin

sin

+ 1

2

sin2

2

2. (9.146)

Then, the time-independent Schrdinger equation can be written as

h22m

1 2

2

+ 1

2 sin

sin

+ 1

2 sin2 22

= [ E V ( )]. (9.147)Lets continue with the separation of variables. Assuming that the

wave function takes the form ( ,,) = R( )Y (,), we obtain

h22m

Y 2

dd

2dRd

+ R 2 sin

dd

sindY d

+ R 2sin2

d2Y d2

= RY [E V ( )]. (9.148)Now dividing by = RY , multiplying by

2m 2h2

, and rearranging,we have1R

dd

2dRd

2m 2

h2 [V ( ) E] =

1Y

1sin

dd

sin dY d

+ 1sin2

d2Y d2

.


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We have a function of equal to a function of the angular variables.So, we set each side equal to a constant. We will judiciously set theconstant equal to ( + 1). The resulting equations are

dd

2dRd

2m 2

h2 [V ( ) E] R = ( + 1)R, (9.149)

1sin

sinY

+ 1sin2

2Y 2

= ( + 1)Y . (9.150)The second of these equations should look familiar from the last

section. This is the equation for spherical harmonics,

Y m(,) = 2 + 12 ( m)!( + m)!Pm eim. (9.151)So, any further analysis of the problem depends upon the choice of

potential, V ( ), and the solution of the radial equation. For this, weturn to the determination of the wave function for an electron in orbitabout a proton.

Example 9.8. The Hydrogen Atom - = 0 States Historically, the rst test of the Schrdinger equation was the determina-

tion of the energy levels in a hydrogen atom. This is modeled by an electronorbiting a proton. The potential energy is provided by the Coulomb potential,

V ( ) = e2

4 0 .

Thus, the radial equation becomes

dd

2dRd

+ 2m 2

h2 e2

4 0 + E R = ( + 1)R. (9.152)

Before looking for solutions, we need to simplify the equation by absorbingsome of the constants. One way to do this is to make an appropriate changeof variables. Let = ar. Then by the Chain Rule we have

dd

= drd

ddr

= 1a

ddr

.

Under this transformation, the radial equation becomes

ddr

r2dudr

+ 2ma2r2h2

e24 0ar

+ E u = ( + 1)u, (9.153)

where u(r) = R( ). Expanding the second term,

2ma2r2

h2 e2

4 0ar + E u =

mae2

2 0h2r +

2mEa2

h2 r2 u,


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we see that we can dene

a = 2 0h2

me2 , (9.154)

= 2mEa2

h2

= 2(2 0)2h2

me4 E. (9.155)

Using these constants, the radial equation becomes

ddr

r2dudr

+ ru ( + 1)u = r2u. (9.156)

Expanding the derivative and dividing by r2,

u + 2r

u + 1r

u ( + 1)

r2 u = u. (9.157)

The rst two terms in this differential equation came from the Laplacian. Thethird term came from the Coulomb potential. The next term can be thoughtto contribute to the potential and is attributed to angular momentum. Thus,

is called the angular momentum quantum number. This is an eigenvalue problem for the radial eigenfunctions u(r) and energy eigenvalues .

The solutions of this equation are determined in a quantum mechanicscourse. In order to get a feeling for the solutions, we will consider the zeroangular momentum case, = 0 :

u + 2r

u + 1r

u = u. (9.158)

Even this equation is one we have not encountered in this book. Lets see if

we can nd some of the solutions.First, we consider the behavior of the solutions for large r. For large r the

second and third terms on the left hand side of the equation are negligible. So,we have the approximate equation

u u = 0. (9.159)The solutions thus behave like u(r) = e r . For bounded solutions, wechoose the decaying solution.

This suggests that solutions take the form u(r) = v(r)e r for someunknown function, v(r). Inserting this guess into Equation (9.158 ), gives anequation for v(r) :

rv + 2 1 r v + ( 12 )v = 0. (9.160)Next we seek a series solution to this equation. Let

v(r) =

k = 0

ck rk .


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Inserting this series into Equation (9.160 ), we have

k = 1

[k (k 1) + 2k ]ck rk 1 +

k = 1

[12 (k + 1)]ck rk = 0.

We can re-index the dummy variable in each sum. Let k = m in the rst sum

and k = m 1 in the second sum. We then nd that k = 1

[m(m + 1)cm+ [12m ]cm1]rm1 = 0.

Since this has to hold for all m 1,cm =

2m 1m(m + 1) cm1.

Further analysis indicates that the resulting series leads to unbounded so-lutions unless the series terminates. This is only possible if the numerator,2m 1, vanishes for m = n, n = 1, 2 . . . . Thus,

= 14n2 .

Since is related to the energy eigenvalue, E, we have

En = me4

2(4 0)2h2n2.

Inserting the values for the constants, this gives

En = 13.6 eV

n2 .

This is the well known set of energy levels for the hydrogen atom.The corresponding eigenfunctions are polynomials, since the innite series

was forced to terminate. We could obtain these polynomials by iterating therecursion equation for the cms. However, we will instead rewrite the radialequation (9.160 ).

Let x = 2 r and dene y(x) = v(r). Thenddr

= 2 ddx

.

This gives2 xy + ( 2x)2 y + ( 12 ) y = 0.

Rearranging, we have

xy + ( 2x) y + 12 (12 ) y = 0.

Noting that 2 = 1n , this becomesxy + ( 2x) y + ( n 1) y = 0. (9.161)


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The resulting equation is well known. It takes the form

xy + ( + 1x) y + ny = 0. (9.162)Solutions of this equation are the associated Laguerre polynomials. The The associated Laguerre polynomials are

named after the French mathematician

Edmond Laguerre (1834

-1886

).solutions are denoted by Ln (x). They can be dened in terms of the Laguerre polynomials,

Ln(x) = ex ddx

n(exxn).

The associated Laguerre polynomials are dened as

Lmnm(x) = ( 1)m ddx

mLn(x).

Note: The Laguerre polynomials were rst encountered in Problem 2 in Chap-ter 5 as an example of a classical orthogonal polynomial dened on [0, ) withweight w(x) = ex. Some of these polynomials are listed in Table 9.3.

Comparing Equation (9.161 ) with Equation (9.162 ), we nd that y(x) =L1n1(x).

Lmn (x)L00(x) 1L01(x) 1xL02(x) x

2 4x + 2L10(x) 1L11(x) 42xL12(x) 3x2 18x + 18L20(x) 2

L21(x) 6x + 18L22(x) 12x2 96x + 144

L30(x) 6L31(x) 24x + 96L32(x) 60x2 600x + 1200

Table 9.3: Associated Laguerre Func-tions, Lmn (x).

In summary, we have made the following transformations: In most derivation in quantum mechan-ics a = a02 . where a0 =

4 0h2me2 is the Bohr

radius and a0 = 5.29171011m.1. R( ) = u(r), = ar.2. u(r) = v(r)e r .3. v(r) = y(x) = L1n1(x), x = 2

r.Therefore,

R( ) = e / aL1n1(2 / a).

However, we also found that 2 = 1/ n. So,R( ) = e /2 na L1n1( / na ).
http://0.0.0.0/http://0.0.0.0/


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For the general case, for all 0, we need to solve the differentialequation

u + 2r

u + 1r

u ( + 1)

r2 u = u. (9.163)

Instead of letting u(r) = v(r)e r , we let

u(r) = v(r)r e r .

This lead to the differential equation

rv + 2( + 1 r)v + ( 12( + 1) )v = 0. (9.164)as before, we let x = 2 r to obtain

xy + 2 + 1 x2 v +

12 ( + 1) v = 0.

Noting that 2 = 1/ n, we havexy + 2[2( + 1) x]v + ( n ( + 1)) v = 0.

We see that this is once again in the form of the associate Laguerreequation and the solutions are

y(x) = L2 + 1n 1(x).

So, the solution to the radial equation for the hydrogen atom is given by

R( ) = r e r L2 + 1n 1(2 r)

= 2na e

/2 naL

2 + 1n 1

na . (9.165)

Interpretations of these solutions will be left for your quantum me-chanics course.

Problems

1. Consider Laplaces equation on the unit square, uxx + u yy = 0,0 x, y 1. Let u(0, y) = 0,u(1, y) = 0 for 0< y < 1 and u y(x, 0) = 0for 0< y < 1. Carry out the needed separation of variables and writedown the product solutions satisfying these boundary conditions.

2. Consider a cylinder of height H and radius a.

a. Write down Laplaces Equation for this cylinder in cylindri-cal coordinates.

b. Carry out the separation of variables and obtain the threeordinary differential equations that result from this problem.


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