10—Partial Differential Equations 304Having done all the work to get to the answer, what can I learn from it?
What does it look like?
Are there any properties of the solution that are unexpected?
Should I have anticipated the form of the result?
Is there an easier way to get to the result?
To see what it looks like, examine some values ofz, the distance above the surface. Ifz=L, the coefficient
for successive terms is
`= 0 :
2
πe−π= 0. 028 `= 1 :2
3 πe−^3 π= 1. 7 × 10 −^5 (41)The constant term is the average potential, and the`= 0term adds only a modest ripple, about5%of the
constant average value. If you move up toz= 2Lthe first factor is 0.0012 and that’s a little more than 0.2%
ripple. The sharp jumps from+V 0 to zero and back disappear rapidly. That the oscillations vanish so quickly
with distance is perhaps not what you would guess until you have analyzed such a problem.
The graph shows the potential function at the surface,z= 0, as it oscillates betweenV 0 and zero. It then
shows successive graphs of Eq. ( 40 )atz=L/ 2 , thenatz=L, thenatz= 1. 5 L. The ripple is barely visible
at the third distance. The radiation through the screen of a microwave oven is filtered in much the same way
because the wavelength of the radiation is large compared to the size of the holes in the screen.
When you write the form of the series for the potential, Eq. ( 37 ), you can see this coming if you look for
it. The oscillating terms inxare accompanied by exponential terms inz, and the rapid damping with distance
is already apparent:e−nπz/L. You don’t have to solve for a single coefficient to see that the oscillations vanish
very rapidly with distance.