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


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?




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


($j$ 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).


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

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up vote 2 down vote accepted

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.

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