Phy 515 class notes for Spring, 1999

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Phy 515, Week 12

12 Apr Complete the solution to the example potential problem. The radial equation has a series solution that I worked out in detail. (It has a second solution that I didn't find, though that one is singular at the origin and doesn't show up in this problem.) The general theory says that the separation constant a is real in this case, where the radial solutions are to be orthogonal. In fact it has to be positive, so that I can get oscillating radial solutions. Let a = k2.
Match the boundary conditions on the top and side surfaces, leading to solutions of the form

sinh[ k ( c - z ) ] J0( k r )
Here the radial equation has been identified as the Bessel function, R( r ) = J0( k r ). The boundary condition on the sides restricts k so that k r0 is a root of the Bessel function. Call these roots jn, where j1 = 2.404, and I don't remember the others. They are however more or less equally spaced, because the differential equation they satisfy looks like the harmonic oscillator when you get to large r. The form of the solution for the potential is then
S Bn sinh[ kn ( c - z ) ] J0( kn r ), where kn r0 = jn, n = 1, 2, 3, . . .
The infinite number of unknowns, Bn have yet to be determined. That's where all the effort in developing Sturm-Liouville theory pays off.

Translate the general form of the orthogonality condition to this notation, and you have

ó
õ 		r0

0 
 dr r J0( kn r ) J0( km r ) = 0,        if n ¹ m
Evaluate the potential at the boundary z = 0, where the potential is the known constant V0. Multiply the infinite series by r J0( km r ) and integrate from 0 to r0. All the terms but one will vanish, and you have the relation
Bm sinh( km c ) ó
õ 		r0

0 
 dr r [ J0( km r ) ]2 = V0 ó
õ 		r0

0 
 dr r J0( km r ).
This determines all the constants. (There are ways to evaluate these integrals.)
All that's left is to analyze the results. Does it have the correct dimensions? What is its behavior for various values of z?

14 Apr For a typical differential equation of the type that you get by separating Laplace's equation, when you try to solve it by a Frobenius series, you will get a recursion relation for the coefficients that is very hard to handle. It is typically a 3-term recursion relation. This means that given the coefficient a2 and a4, you get a6. Then given a4 and a6, you get a8. You need twonumbers to get the third. There's nothing wrong with this, except that it's hard to analyze. Two-term recursion relations are much easier.

For the example that I just spent a lot of time on, this difficulty didn't arise, so I'll take a case where it does.

- d2 f / dx2 + x2 f = l f.
This equation shows up in quantum mechanics when you examine the harmonic oscillator. Try a Frobenius series solution:
f( x ) = San xn + s.
Plug this into the differential equation, and you get
- San (n + s )( n + s - 1 ) xn + s - 2 + x2 San xn + s = l San xn + s.
All these sums are from zero to infinity. The lowest (most negative) order term is the n = 0 term of the first sum. It gives
a0 s ( s - 1 ) = 0.
I am at liberty to assume that a0 is non-zero (otherwise redefine s). This says that s = 0 or s = 1 are the two possibilities. For the rest do the standard trick of redefining the index in the three sums. Let n - 2 = m in the first; let n + 2 = m in the second; let n = m in the third.
- Sam + 2 (m + 2 + s )( m + s + 1 ) xm + s + Sam + 2 xm + s = l Sam xm + s.
The sums are -2 to infinity, +2 to infinity, and 0 to infinity respectively. When you demand that this differential equation be satisfied, you have to equate like terms in the series, so all the coefficients of xm + s must match.
This gives a2 in terms of a0. Then it gives a4 in terms of a2 and a0. Then it gives a6 in terms of a4 and a2. Etc.
Such three-term recursion relations are hard to handle. Though it's not at all obvious why, the problem lies in failure to analyze the behavior of the differential equation near its singular point. The singular point this time is at infinity, because of the x2. The technique to handle it is to "factor out the behavior" at the singular point.

For large x, the differential equation looks roughly like - f '' + x2f = 0. The other term is small. The solution of this vaguely resembles an inverted oscillator equation, but with a variable coefficient, so see if the behavior is anything like an exponential. A little playing around with the possibilities leads to trying something like ecx2.

d / dx ec x2 = 2 c x ecx2, then d2 / dx2 ec x2 = 4 c2 x2 ecx2 + . . .
(Neglect smaller terms.) Try this in the differential equation and you see that it works (approximately) only if c2 = 1 / 4.

Now, factor out this behavior. Substitute exactly this time

f( x ) = ec x2 g( x ).
When you plug this into the original differential equation, you get a new equation for g. It is
g '' - 2 x g + ( l - 1 ) g = 0.
Look back to the original equation and see whyit led to a two-term recursion relation for the coefficients. When you start with a power xn, the first term (f '') drops the exponent by 2. The second term raisesit by 2. The final term leaves it alone. This links three different powers. The same analysis on the new equation for g shows that it links only two different powers, so you know that you will get only a two-term recursion relation.

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