Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.