I am working on solving the following optimization problem. I think it is well-poised but, if not, please give me some pointers that could make the question make more sense.

We have a parametric curve $x(t),y(t)$ for $t\leq 0\leq 1$ and want to minimize $$\int_0^1\Big(x'(t)y''(t)-x''(t)y'(t)\Big)^2dt$$ subject to the constraint $$x'(t)^2+y'(t)^2=1$$

We can minimize $\int_0^1(x'(t)y''(t)-x''(t)y'(t))^2$ fairly simply using calculus of variations, but I'm not sure how to do minimize it subject to some constraint.

An analagy I'm considering is, in multivariate calculus, we may wish to find the extrema of $f(x,y)$ subject to $g(x,y)=c$ and we simply solve $$\nabla f(x,y)=\lambda\nabla g(x,y)$$ but I'm not sure what the analogue of this would be. My thoughts are that if we want to find the extrema of the functional $\int_a^b F(x,y,x',y',...)dt$ subject to $g(x,y,x',y')=c$ we would maybe solve $$\frac{\partial F}{\partial x}-\frac d{dt}\frac{\partial F}{\partial x'}=\lambda \bigg(\frac{\partial g}{\partial x}-\frac d{dt}\frac{\partial g}{\partial x'}\bigg)$$ and $$\frac{\partial F}{\partial y}-\frac d{dt}\frac{\partial F}{\partial y'}=\lambda \bigg(\frac{\partial g}{\partial y}-\frac d{dt}\frac{\partial g}{\partial y'}\bigg)$$ Does anyone know how to solve this optimization problem?

  • $\begingroup$ Your derivations are correct $\endgroup$
    – AnilB
    Apr 10, 2013 at 15:19
  • 1
    $\begingroup$ You can do this, as you already mentioned, by a Lagrange approach. For the inner product you could choose for instance the $L^{2}$ inner product...but then this will become a question of regularity of the functions $y$ and $x$, which are involved in the minimizing process. An other approach would be to minimize the reduced cost functional (using that the equality holds). Then you obtain a cost functional which is only dependent on $x$ or $y$ but not both...Nevertheless the question of initial conditions and the function spaces involved in the minimizing process has to be answered... $\endgroup$
    – Alex
    Apr 10, 2013 at 21:21

1 Answer 1


A necessary condition for optimizing a functional $\int_0^1F(t,x,y,x',y')dt$ subject to $g(x',y')=0$ is solving $$ \frac{\partial F}{\partial x}-\frac{d}{dt}\frac{\partial F}{\partial x'}=\lambda(t)\left(\frac{\partial g}{\partial x}-\frac{d}{dt}\frac{\partial g}{\partial x'}\right) $$ and $$ \frac{\partial F}{\partial y}-\frac{d}{dt}\frac{\partial F}{\partial y'}=\lambda(t)\left(\frac{\partial g}{\partial y}-\frac{d}{dt}\frac{\partial g}{\partial y'}\right) $$ Notice that $\lambda$ is a function, $\lambda(t)$.


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