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Calculus of Variations
It is a well-known fact, first enunciated by Archimedes, that the shortest
distance between two points in a plane is a straight-line. However, suppose that
we wish to demonstrate this result from first principles. Let us consider the
length,
, of various curves,
, which run between two fixed
points,
and
, in a plane, as illustrated in Figure 35. Now,
takes the form
![\begin{displaymath}
l = \int_A^B [dx^2 + dy^2]^{1/2} = \int_a^b [1 + y'^{\,2}(x)]^{1/2}\,dx,
\end{displaymath}](img1717.png) |
(675) |
where
. Note that
is a function of the function
.
In mathematics, a function of a function is termed a functional.
Figure 35:
Different paths between points
and
.
![\begin{figure}
\epsfysize =2.5in
\centerline{\epsffile{Chapter10/fig10.01.eps}}
\end{figure}](img1719.png) |
Now, in order to find the shortest path between points
and
, we need to minimize the functional
with respect to small variations
in the function
, subject to the constraint that the end points,
and
, remain fixed. In other words, we need to solve
![\begin{displaymath}
\delta l = 0.
\end{displaymath}](img1720.png) |
(676) |
The meaning of the above equation is that if
, where
is small, then the first-order variation in
,
denoted
,
vanishes. In other words,
. The particular function
for which
obviously yields an extremum of
(i.e., either a maximum or a minimum). Hopefully,
in the case under consideration,
it yields a minimum of
.
Consider a general functional of the form
![\begin{displaymath}
I = \int_a^b F(y, y',x)\,dx,
\end{displaymath}](img1726.png) |
(677) |
where the end points of the integration are fixed.
Suppose that
. The first-order variation in
is written
![\begin{displaymath}
\delta I = \int_a^b\left(\frac{\partial F}{\partial y}\,\delta y+ \frac{\partial F}{\partial y'}\,\delta y'\right)dx,
\end{displaymath}](img1727.png) |
(678) |
where
. Setting
to zero, we
obtain
![\begin{displaymath}
\int_a^b\left(\frac{\partial F}{\partial y}\,\delta y+ \frac{\partial F}{\partial y'}\,\delta y'\right)\,dx = 0.
\end{displaymath}](img1730.png) |
(679) |
This equation must be satisfied for all possible small perturbations
.
Integrating the second term in the integrand of the above equation by
parts, we get
![\begin{displaymath}
\int_a^b\left[\frac{\partial F}{\partial y}- \frac{d}{dx}\!\...
... +\left[\frac{\partial F}{\partial y'}\,\delta y\right]_a^b=0.
\end{displaymath}](img1731.png) |
(680) |
Now, if the end points are fixed then
at
and
. Hence, the last term on the left-hand side of the
above equation is zero. Thus, we obtain
![\begin{displaymath}
\int_a^b\left[\frac{\partial F}{\partial y}- \frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)\right]\delta y\,dx =0.
\end{displaymath}](img1734.png) |
(681) |
The above equation must be satisfied for all small perturbations
. The only way in which this is possible is for the
expression enclosed in square brackets in the integral to be zero. Hence, the functional
attains an extremum value whenever
![\begin{displaymath}
\frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)-\frac{\partial F}{\partial y} = 0.
\end{displaymath}](img1735.png) |
(682) |
This condition is known as the Euler-Lagrange equation.
Let us consider some special cases. Suppose that
does not explicitly
depend on
. It follows that
. Hence,
the Euler-Lagrange equation (682) simplifies to
![\begin{displaymath}
\frac{\partial F}{\partial y'} = {\rm const}.
\end{displaymath}](img1738.png) |
(683) |
Next, suppose that
does not depend explicitly on
. Multiplying
Equation (682) by
, we obtain
![\begin{displaymath}
y'\,\frac{d}{dx}\!\left(\frac{\partial F}{\partial y'}\right)-y'\,\frac{\partial F}{\partial y} = 0.
\end{displaymath}](img1739.png) |
(684) |
However,
![\begin{displaymath}
\frac{d}{dx}\!\left(y'\,\frac{\partial F}{\partial y'}\right...
...l F}{\partial y'}\right)+ y''\,\frac{\partial F}{\partial y'}.
\end{displaymath}](img1740.png) |
(685) |
Thus, we get
![\begin{displaymath}
\frac{d}{dx}\!\left(y'\,\frac{\partial F}{\partial y'}\right...
...\partial F}{\partial y} + y''\,\frac{\partial F}{\partial y'}.
\end{displaymath}](img1741.png) |
(686) |
Now, if
is not an explicit function of
then the right-hand side of
the above equation is the total derivative of
, namely
.
Hence, we obtain
![\begin{displaymath}
\frac{d}{dx}\!\left(y'\,\frac{\partial F}{\partial y'}\right) = \frac{dF}{dx},
\end{displaymath}](img1743.png) |
(687) |
which yields
![\begin{displaymath}
y'\,\frac{\partial F}{\partial y'} - F = {\rm const}.
\end{displaymath}](img1744.png) |
(688) |
Returning to the case under consideration, we have
, according to Equation (675) and (677). Hence,
is not
an explicit function of
, so Equation (683) yields
![\begin{displaymath}
\frac{\partial F}{\partial y'} = \frac{y'}{\sqrt{1+y'^{\,2}}} = c,
\end{displaymath}](img1746.png) |
(689) |
where
is a constant. So,
![\begin{displaymath}
y' = \frac{c}{\sqrt{1-c^2}} = {\rm const}.
\end{displaymath}](img1747.png) |
(690) |
Of course,
is the equation of a straight-line. Thus, the shortest distance between two fixed points in a plane is indeed a
straight-line.
Next: Conditional Variation
Up: Hamiltonian Dynamics
Previous: Introduction
Richard Fitzpatrick
2011-03-31