STEADY-STATE TEMPERATURE IN A SPHERE

7. STEADY-STATE TEMPERATURE IN A SPHERE

Find the steady-state temperature inside a sphere of radius a when the surface of

the upper half is held at 100 ◦ and the surface of the lower half at 0 . Inside the sphere, the temperature u satisfies Laplace’s equation. In spherical

coordinates this is (see Chapter 10, Section 9)

u= 2 r 2 ∂u

∂u

+ 2 2 2 = 0. r ∂r

+ 2 sin θ

∂r

r sin θ ∂θ

r sin θ ∂φ

648 Partial Differential Equations Chapter 13

We separate this equation following our standard procedure. Substitute (7.2)

u = R(r)Θ(θ)Φ(φ) into (7.1) and multiply by r 2 /RΘΦ to get

If we multiply (7.3) by sin 2 θ, the last term becomes a function of φ only and the other terms do not contain φ. Thus we obtain the φ equation and its solutions:

The separation constant must be negative and m an integer to make Φ a periodic function of φ [see the discussion after (5.6)].

Equation (7.3) can now be written as

sin 2 θ The first term is a function of r and the last two terms are functions of θ, so we

have two equations

r 2 (7.6) dR = k,

If you compare (7.7) with the equation of Problem 10.2 in Chapter 12, you will see that (7.7) is the equation for the associated Legendre functions if k = l(l + 1). Recall that l must be an integer in order for the solution of Legendre’s equation to

be finite at x = cos θ = ±1, that is, at θ = 0 or π; the same statement is true for the equation for the associated Legendre functions. The corresponding result for (7.7) is that k must be a product of two successive integers; it is then convenient to replace k by l(l + 1), where l is an integer. The solutions of (7.7) are then the associated Legendre functions (see Problem 10.2, Chapter 12)

Θ=P m l (cos θ).

In (7.6), we put k = l(l + 1); you can then easily verify (Problem 5.11) that the solutions of (7.6) are

R=

r −l−1 .

Since we are interested in the interior of the sphere, we discard the solutions r −l−1 because they become infinite at the origin. If we were discussing a problem (say about water flow or electrostatic potential) outside the sphere, we would use the r −l−1 solutions and discard the solutions r l because they become infinite at infinity.

Section 7 Steady-state Temperature in a Sphere 649

The basis functions for our problem are then

sin mφ,

u=r l P m l (cos θ)

cos mφ.

[The functions P m l (cos θ) sin mφ and P m l (cos θ) cos mφ are called spherical harmon- ics and are often denoted by Y m l (θ, φ); also see Problem 16.] If the surface tem- perature at r = a were given as a function of θ and φ, we would have a double series (summed on l and m). For the given surface temperatures in our problem

(100 ◦ on the top hemisphere and 0 on the lower hemisphere), the temperature is independent of φ; thus in (7.10) we must have m = 0, cos mφ = 1. The solutions

(7.10) then reduce to r l P l (cos θ). We write the solution of the problem as a series of these basis functions:

u=

c l r l P l (cos θ).

l=0

We determine the coefficients c l by using the given temperatures when r = a; that is, we must have

(7.12) u r=a =

c l a l P l (cos θ)

l=0

that is,

0 < cos θ < 1,

that is, −1 < cos θ < 0, or, with x = cos θ,

c l a P l (x) = 100f (x)

(Note that here x just stands for cos θ and is not the coordinate x.) In Section 9 of Chapter 12, we expanded this f (x) in a series of Legendre polynomials and obtained:

The coefficients c l in (7.13) are just these coefficients times 100/a l . Substituting the c’s into (7.11), we get the final solution:

(7.15) u = 100 1 P (cos θ) + 3 r

2 1 (cos θ) − 16 a P 3 (cos θ) + 11 r 32 5 a P 5 (cos θ) + . . . .

650 Partial Differential Equations Chapter 13

We can do variations of this problem. Notice that we have not even mentioned so far what temperature scale we are using (Celsius, Fahrenheit, absolute, etc.). This is a very easy adjustment to make once we have a solution in any one scale.

To see why, observe that if u is a solution of Laplace’s equation ∇ 2 u = 0 or of the heat flow equation ∇ 2 u = (1/α 2 )(∂u/∂t), then u + C and Cu are also solutions for ◦

any constant C. If we add, say, 50 to the solution (7.15), we have the temperature distribution inside a sphere with the top half of the surface at 150 ◦ and the lower half

at 50 ◦ . If we multiply the solution (7.15) by 2, we find the temperature distribution

with given surface temperatures of 200 ◦ and 0 , and so on. The temperature of the equatorial plane θ = π/2 or cos θ = 0 as given by equa-

tions (7.11) to (7.15) is halfway between the top and bottom surface temperatures, because Legendre series, like Fourier series, converge to the midpoint of a jump in the function which was expanded to get the series. To solve the problem of the temperature in a hemisphere given the temperatures of the curved surface and of the equatorial plane, we need only imagine the lower hemisphere in place and at the proper temperature to give the desired average on the equatorial plane. When ◦

the temperature of the equatorial plane is 0 , this amounts to defining the function

f (x) in (7.13) on (−1, 0) to make it an odd function.

PROBLEMS, SECTION 7

Find the steady-state temperature distribution inside a sphere of radius 1 when the surface temperatures are as given in Problems 1 to 10.

1. 35 cos 4 θ

2. cos θ − cos 3 θ

2 3 3. 2 cos θ − 3 sin θ 4. 5 cos θ − 3 sin θ 5. |cos θ|

6. π/2−θ. See Chapter 12, Problem 9.4.  cos θ,

that is, upper hemisphere,

that is, lower hemisphere.

8. 0 ◦ , otherwise. Hint: See Problem 9.8 of Chapter 12.

9. 3 sin θ cos θ sin φ. Hint: See equation (7.10) and Chapter 12, equation (10.6). 10. sin 2 θ cos θ cos 2φ − cos θ. (See Problem 9.)

11. Find the steady-state temperature distribution inside a hemisphere if the spherical surface is held at 100 ◦ and the equatorial plane at 0 ◦ . Hint: See the last paragraph of this section above.

12. Do Problem 11 if the curved surface is held at cos 2 θ and the equatorial plane at zero. Careful: The answer does not involve P 2 ; read the last sentence of this section.

13. Find the electrostatic potential outside a conducting sphere of radius a placed in an originally uniform electric field, and maintained at zero potential. Hint: Let

the original field E be in the negative z direction so that E = −E 0 k. Then since E = −∇Φ, where Φ is the potential, we have Φ = E 0 z=E 0 r cos θ (Verify this!) for the original potential. You then want a solution of Laplace’s equation ∇ 2 u=0 which is zero at r = a and becomes u ∼ Φ for large r (that is, far away from the

Section 7 Steady-state Temperature in a Sphere 651

sphere). Select the solutions of Laplace’s equation in spherical coordinates which have the right θ and φ dependence (there are just two such solutions) and find the combination which reduces to zero for r = a.

14. Find the steady-state temperature distribution in a spherical shell of inner radius 1 and outer radius 2 if the inner surface is held at 0 ◦ and the outer surface has its upper half at 100 ◦ and its lower half at 0 ◦ . Hint: r = 0 is not in the region of interest, so the solutions r −l−1 in (7.9) should be included. Replace c l r l in (7.11) by (c l r l +b l r −l−1 ).

15. A sphere initially at 0 ◦ has its surface kept at 100 ◦ from t = 0 on (for example, a frozen potato in boiling water!). Find the time-dependent temperature distribution. Hint: Subtract 100 ◦ from all temperatures and solve the problem; then add the 100 ◦ to the answer. Can you justify this procedure? Show that the Legendre function

required for this problem is P 0 and the r solution is (1/ √ r)J 1/2 or j 0 [see (17.4) in Chapter 12]. Since spherical Bessel functions can be expressed in terms of elementary functions, the series in this problem can be thought of as either a Bessel series or a Fourier series. Show that the results are identical.

16. Separate the wave equation in spherical coordinates, and show that the θ, φ solutions are the spherical harmonics Y l m (θ, φ) =P l m (cos θ)e ±imφ and the r solutions are the spherical Bessel functions j l (kr) and y l (kr) [Chapter 12, equations (17.4)].

17. Do Problem 6.6 in 3 dimensional rectangular coordinates. That is, solve the “particle in a box” problem for a cube.

18. Separate the time-independent Schr¨ odinger equation (3.22) in spherical coordinates assuming that V = V (r) is independent of θ and φ. (If V depends only on r, then we are dealing with central forces, for example, electrostatic or gravitational forces.) Hints: You may find it helpful to replace the mass m in the Schr¨ odinger equation by M when you are working in spherical coordinates to avoid confusion with the letter m in the spherical harmonics (7.10). Follow the separation of (7.1) but with the extra term [V (r) − E]Ψ. Show that the θ, φ solutions are spherical harmonics as in (7.10) and Problem 16. Show that the r equation with k = l(l + 1) is [compare (7.6)]

19. Find the eigenfunctions and energy eigenvalues for a “particle in a spherical box” r < a. Hints: See Problem 6.6. Write the R equation from Problem 18 with V = 0,

and compare Chapter 12, Problem 17.6, with y = R, x = βr where β = p2ME/¯h 2 , and n = l.

20. Write the Schr¨ odinger equation (3.22) if ψ is a function of x, and V = 1 2 mω 2 x 2 (this is a one-dimensional harmonic oscillator). Find the solutions ψ n (x) and the energy eigenvalues E n . Hints: In Chapter 12, equation (22.1) and the first equation in (22.11), replace x by αx where α = pmω/¯h. (Don’t forget appropriate factors

of α for the x’s in the denominators of D = d/dx and ψ ′′ =d 2 ψ/dx 2 .) Compare your results for equation (22.1) with the Schr¨ odinger equation you wrote above to see that they are identical if E n = (n + 1 2 )¯ hω. Write the solutions ψ n (x) of the Schr¨ odinger equation using Chapter 12, equations (22.11) and (22.12).

21. Separate the Schr¨ odinger equation (3.22) in rectangular coordinates in 3 dimensions assuming that V = 1 mω 2 2 (x 2 +y 2 +z 2 ). (This is a 3-dimensional harmonic oscillator). Observe that each of the separated equations is of the form of the one-dimensional oscillator equation in Problem 20. Thus write the solutions ψ n (x, y, z) for the 3- dimensional problem, where n = n x +n y +n z . Find the energy eigenvalues E n and their degree of degeneracy (see Problem 6.7 and Chapter 15, Problem 4.21).

652 Partial Differential Equations Chapter 13

22. Find the energy eigenvalues and eigenfunctions for the hydrogen atom. The potential energy is V (r) = −e 2 /r in Gaussian units, where e is the charge of the electron and r is in spherical coordinates. Since V is a function of r only, you know from Problem 18 that the eigenfunctions are R(r) times the spherical harmonics Y l m (θ, φ), so you only have to find R(r). Substitute V (r) into the R equation in Problem 18 and make the following simplifications: Let x = 2r/α, y = rR; show that then

„ 2 « d 2 d d r = αx/2, R(r) =

r 2 dR

2 = xy ′′ .

y(x),

dr α Let α 2 = −2ME/¯h 2 (note that for a bound state, E is negative, so α 2 is positive)

and λ = M e 2 α/¯ h 2 , to get the first equation in Problem 22.26 of Chapter 12. Do this problem to find y(x), and the result that λ is an integer, say n. [Caution: not

the same n as in equation (22.26)]. Hence find the possible values of α (these are the radii of the Bohr orbits), and the energy eigenvalues. You should have found α proportional to n; let α = na, where a is the value of α when n = 1, that is, the radius of the first Bohr orbit. Write the solutions R(r) by substituting back y = rR, and x = 2r/(na), and find E n from α.