1
$\begingroup$

If I want to find the minimizing function $f(t)$ over a single parameter, like time, then I take the integrand of

\int_{t}L(t,f(t),f'(t))\:\:\:\:dt

and substitute it into the Euler-Lagrange equation, and solve for $f(t)$.

But what if I need to find the minimizing area, which occurs over two parameters?

L_1=L(t_1,f(t_1),f'(t_1))

L_2=L(t_2,f(t_2),f'(t_2))

$\int_{t_2}\int_{t_1}L_1L_2\:\:\:\:dt_1dt_2$

For the 2-parameter case, I have a particular form in mind for $L_i$:

$L_i=\frac{df}{dt_i}=\sum_j\frac{df}{dx_j}\frac{dx_j}{dt_i}\:\:\:\:\:\:\:\:;\:\:i=1,2$

($j$ is positive integer, not important how high it goes)

It is assumed that the $x_j$'s are all orthogonal to each other (a.k.a. independent, inner product=0).

Thus

$L_1L_2=\sum_j \left( \frac{df}{dx_j}\right)^2 \frac{dx_j}{dt_1}\frac{dx_j}{dt_2}$

1 Answers 1

2

Since each factor only contains one of the parameters, your integral factorizes:

$\int_{t_2}\int_{t_1}L_1L_2\,\mathrm dt_1\mathrm dt_2=\int_{t_2}L_2\mathrm dt_2\int_{t_1}L_1\,\mathrm dt_1=\left(\int_{t_1}L_1\mathrm dt_1\right)^2\;.$

The square can be minimal either when the integral is extremal or when it is zero. You can find the extremal values using normal variation; the zero case may require a separate treatment.