Phy 515 class notes for Spring, 1999

Jan 20	Feb 1	Mar 1	Apr 5	Phy 515 page
Jan 25	Feb 8	Mar 8	Apr 12
	Feb 15	Mar 22	Apr 19
	Feb 22	Mar 29	Apr 26

Phy 515, Week 11

5 Apr Show how to apply the general structure to a particular problem. Take the potential problem of a cylindrical can, grounded on the top and side, and at potential V₀ on the bottom. How does this particularize the problem?
There is no q dependence, so the separation constant b = 0, allowing a constant for the q dependence.
What about a? Here there is some flexibility. If it is real and positive, you get exponential solutions in z; if it is negative, you get oscillatory solutions in z. Both are acceptable, and both lead to complete solutions to the problem, but the only way that I know to see this is to try it both ways and to find that both work. I'll arbitrarily pick a > 0.
Now to examine the radial solution. This has b = 0, so s = 0 also. The differential equation is

and the recursion relation for the series solution reduces to a_m = - ( a / m² ) a_{m - 2}. This gives a series R( r ) = a₀ [ 1 - ( a / 2² ) r² + ( a² / 2²·4² ) r⁴ - ( a³ / 2²·4²·6² ) r⁶ + . . . ] This bears a vague resemblance to the series for the cosine, so it's not too surprising that this oscillates. The differential equation (without the middle term) isthe harmonic oscillator equation, and the middle term causes some kind of damping.

To see how to use any of this requires some kind of orthogonality properties, and that's where the excursion into Sturm-Liouville theory comes in.
In standard Fourier series, you use the orthogonality of the sines or cosines or complex exponentials. These are all solutions of the same type of differential equation, and everything about these functions can, in principle, be derived from the differential equation, so how do the orthogonality properties arise? When you have the answer to this, you see that there is nothing very special about the harmonic oscillator equation, and that identical results obtain from a large class of equations. This is where Sturm-Liouville theory comes from:

( p( x ) u' )' + q( x ) u = l w( x ) u. The three functions, p, q, w, are arbitrary (at this point) and this includes many common differential equations: SHO, Legendre, Hermite, and others.
The aim is to show that some integral is zero. This is what you need for Fourier series, and this is what comes out of the manipulation of this equation. Write the equation twice, once with the parameter l₁, and once with l₂. ( p( x ) u₁' )' + q( x ) u₁ = l₁ w( x ) u₁, ( p( x ) u₂' )' + q( x ) u₂ = l₂ w( x ) u₂. Multiply the first of these by u₂^*, and multiply the complex conjugate of the second one by u₁. u₂^*( p( x ) u₁' )' + q( x ) u₂^* u₁ = l₁ w( x ) u₂^* u₁, u₁ ( p( x ) u₂^*' )' + q( x ) u₁ u₂^* = l₂^* w( x ) u₁ u₂^*. I have made some assumptions already: p, q, and w are real. Now subtract these, noting that the q term cancels, and factor what you can. (This is why I had to make the reality assumption. Otherwise what comes next won't work.) u₂^*( p( x ) u₁' )' - u₁ ( p( x ) u₂^*' )' = ( l₁ - l₂^* ) w( x ) u₂^* u₁, Now integrate this equation between two limits, call them a and b. The resulting equation can be manipulated further by partial integration. Do two such partial integration on the first term on the left, moving the derivative over from the u₁ to the u₂, and the integrated terms then cancel,leaving terms depending on only the boundary conditions.

( l2* - l1 ) ó õ b a  dx w u2* u1 = [ p( x ) [ u2*( x ) u1 ( x ) - u2* ¢( x ) u1( x ) ]]ab

7 Apr Continue the Sturm-Liouville analysis. Show what hypotheses are needed to guarantee that the parameter l₁ is real. The same requirement will give you the orthogonality of the solutions.
To see what's going on, specialize to the case of the harmonic oscillator equation, where p = 1, q = 0, and w = 1: u'' = l₁ u. The constraint on the boundaries is

[ u₂^* u₁' - u₂^*' u₁ ] ( b ) minus the same thing at ( a ) equals 0. Examples of this: a = 0, b = L, u( 0 ) = 0, u( L ) = 0. These are satisfied by sines: sin( n p x / L ). You could also make the derivatives vanish at these two points, and get cosine solutions. You can make the function vanish at one end and the derivative vanish at the other. You can make the function periodic and get complex exponentials.

9 Apr Other examples of Sturm-Liouville differential equations.

[ ( 1 - x² ) u' ]' = l u, or ( 1 - x² ) u'' - 2 x u' = l u This is the Legendre equation. The orthogonality of its solutions requires that the combination ( 1 - x² ) u₂^* u₁' - ( 1 - x² ) u₂^*' u₁ evaluated at the two limits and subtracted vanish. If you take the two limits to be +1 and -1, it looks like this will vanish automatically, without any further restriction on u, but not so fast! The differential equation is singular there. (Is it a regular singular point?) This means that the solution is likely to blow up there. If you require that the function is finite at ±1, thenyou can conclude that these boundary terms vanish. That dictates a restriction on the parameter l, just as it did for the harmonic oscillator.

Another equation of this form is

[ e^-x2 u' ]' = l e^-x2 u, or u'' - 2 x u' = l u This is the Hermite equation. The boundary conditions have to be set up so that e^-x2 [ u₂^* u₁' - u₂^*' u₁ ] evaluated at the two limits and subtracted vanishes. The common case is where the limits are at ±•, and again, there are constraints on the parameter l so that the solution, u, doesn't grow too fast.

Finally, back to the problem I was trying to solve when all of this started, the example of the potential inside a can. The equation was

R'' + ( 1 / r ) R' + a R = 0, and this can be put into Sturm-Liouville form as r R'' + R' + a r R = 0 = [ r R' ]' + a r R. This is p = r, q = 0, w = r, l = - a. The constraint on R is r [ R₂^* R₁' - R₂^*' R₁ ] evaluated at the two limits and subtracted vanishes. Here the two limits are dictated by the physical problem. The domain for R is 0 £ r £ r₀. At the lower limit, the coefficient r (from p = r) gives zero, provided that the function R itself is well behaved there. At the other limit, you can require either that the function vanishes or that its derivative does. That choice is determined by the physical problem. In the present case, the potential is zero at the boundary, so that says to take the value R( r₀ ) = 0.

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