Say you have some functional of the form $\int_0^{t_f} L(x,\dot{x},y,\dot{y},z,\dot{z}) dt$ that you're trying to minimize. Normally one can solve this using the Euler-Lagrange equations, and when you have a constraint you can add that to the Lagrangian using Lagrange multipliers. But how do you handle it when the constraint is that the path has to have a specific length, i.e. we require $\int_0^{t_f} \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2} dt = a$ for some fixed $a$? Would it suffice to add a Lagrange multiplier of $\lambda(\int_0^{t_f} \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2} dt - a)$ to $L$ and move the derivatives inside the integral when applying the Euler-Lagrange equations, or is some other method required?


Would it suffice to add a Lagrange multiplier of $\lambda(\int_0^{t_f} \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2} dt - a)$ to $L$?

Basically, yes, except you don't bother with the constant $a$ (it would make no difference to the Euler-Lagrange equations anyway), so you end up with the auxiliary functional

$$\int_0^{t_f} L(x,\dot{x},y,\dot{y},z,\dot{z}) - \lambda\left( \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}\right)\,dt$$

from which you have three Euler-Lagrange equations. You should be given three pairs of boundary conditions to get rid of the constants of integration, and you can eliminate $\lambda$ by substituting your solutions $x,y$ and $z$ into your constraint $\int_0^{t_f} \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2} dt = a$.


Constrained calculus of variations problems can be hard. Here is an example of optimizing a functional with respect to a differential equation constraint.

I can get the answer to you tomorrow; there's a book with the answer in it that I left elsewhere, but I'll get it tomorrow and look it up for you.

For what it's worth, those are called isoperimetric problems. Maybe you can find the answer before I do tomorrow :)

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    $\begingroup$ I haven't found a way to propose corrections on MSE without sounding rude, but I think this answer may have a typo: I believe such problems are called "isoperimetric" (which I believe comes from Dido's problem of maximizing the area enclosed by a closed planar curve of fixed perimeter) rather than "isoparametric." $\endgroup$ – yoknapatawpha Feb 7 '14 at 2:36
  • $\begingroup$ You're right, fixed spelling. $\endgroup$ – Logan Tatham Feb 7 '14 at 19:27

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