16—Calculus of Variations 40216.10 Second Order
Except for a couple of problems in optics in chapter two,2.35and2.39, I’ve mostly ignored the question
about minimum versus maximum.
- Does it matter in classical mechanics whether the integral,
∫
Ldtis minimum or not in determining
the equations of motion? No.
- In geometric optics, does it matter whether Fermat’s principle of least time for the path of the light
ray isreally minimum? Yes, in this case it does, because it provides information about the focus. - In the calculation of minimum energy electric potentials in a capacitor does it matter? No, but only
because it’salwaysa minimum. - In problems in quantum mechanics similar to the electric potential problem, the fact that you’re
dealing sometimes with a minimum and sometimes not leads to some serious technical difficulties.
How do you address this question? The same way you do in ordinary calculus: See what happens
to the second order terms in your expansions. Take the same general form as before and keep terms
through second order. Assume that the endpoints are fixed.
I[y]=
∫badxF
(
x,y(x),y′(x)
)
∆I=I[y+δy]−I[y]
=
∫badxF
(
x,y(x) +δy(x),y′(x) +δy′(x)
)
−
∫badxF
(
x,y(x),y′(x)
)
=
∫badx
[
∂F
∂y
δy+
∂F
∂y′
δy′+
∂^2 F
∂y^2
(
δy
) 2
+ 2
∂^2 F
∂y∂y′
δyδy′+
∂^2 F
∂y′^2
(
δy′
) 2 ]
(16.48)
If the first two terms combine to zero, this says the first derivative is zero. Now for the next terms.
Recall the similar question that arose in section8.11. How can you tell if a function of two
variables has a minimum, a maximum, or neither? The answer required looking at the matrix of all the
second derivatives of the function — the Hessian. Now, instead of a 2 × 2 matrix as in Eq. (8.31) you
have an integral.
〈d~r,H d~r
〉
= (dx dy)
(
fxx fxy
fyx fyy
)(
dx
dy
)
−→
∫badx(δy δy′)
(
Fyy Fyy′
Fy′y Fy′y′
)(
δy
δy′
)
In the two dimensional case∇f= 0defines a minimum if the product
〈
d~r,H d~r
〉
is positive for all