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Let's say I have the functional $F(x(\cdot))=\int_a^b \sqrt{f(\dot{x}(s),x(s))}\,ds$ where $x(t):(a,b)\rightarrow\mathbb{R}^n$ and let's consider $G(x(\cdot))=\int_a^b f(\dot{x}(s),x(s))\,ds.$

Then, under what conditions are the minimun of $F$ and $G$ the same?. I asked my calculus of variations professor and he said not always. However, if this were true, solving the asociated E-L equation of $G$ is way easier than that of $F.$

Any help is appreciated! :)

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So, the EL equations are definitely going to look different for these two scenarios, and as your professor mentioned, $G$ is much easier. Let's take a look. For $G:$ \begin{align*} \frac{\partial L}{\partial x}-\frac{d}{dt}\,\frac{\partial L}{\partial \dot{x}}&=0 \\ \frac{\partial f}{\partial x}-\frac{d}{dt}\,\frac{\partial f}{\partial \dot{x}}&=0 \\ \frac{\partial f}{\partial x}-\left[\dot{x}\,\frac{\partial^2 f}{\partial x\,\partial\dot{x}}+\ddot{x}\,\frac{\partial^2 f}{\partial\dot{x}^2}\right]&=0, \end{align*} and that's pretty much it! For $F,$ things are much more complicated: \begin{align*} \frac{\partial L}{\partial x}-\frac{d}{dt}\,\frac{\partial L}{\partial \dot{x}}&=0 \\ \frac{\partial \sqrt{f}}{\partial x}-\frac{d}{dt}\,\frac{\partial \sqrt{f}}{\partial \dot{x}}&=0 \\ \frac12 \frac{1}{\sqrt{f}}\frac{\partial f}{\partial x}-\frac{d}{dt}\left[\frac12\,\frac{1}{\sqrt{f}}\frac{\partial f}{\partial \dot{x}}\right]&=0 \\ \frac{1}{\sqrt{f}}\frac{\partial f}{\partial x}-\left[\frac{\partial f}{\partial\dot{x}}\,\frac{d}{dt}\,\frac{1}{\sqrt{f}}+\frac{1}{\sqrt{f}}\,\frac{d}{dt}\,\frac{\partial f}{\partial\dot{x}}\right]&=0. \\ \end{align*} Now, whenever we have a function $h=h(\dot{x}(t),x(t)),$ we must compute $$\frac{d}{dt}\,h(\dot{x}(t),x(t))=\dot{x}\,\frac{\partial h}{\partial x}+\ddot{x}\,\frac{\partial h}{\partial\dot{x}}. $$ So, we have \begin{align*} \frac{d}{dt}\,(f)^{-1/2}&=\dot{x}\,(-1/2)f^{-3/2}\frac{\partial f}{\partial x}+\ddot{x}\,(-1/2)f^{-3/2}\frac{\partial f}{\partial\dot{x}}\\ &=-\frac{1}{2\,f^{3/2}}\left[\dot{x}\,\frac{\partial f}{\partial x}+\ddot{x}\,\frac{\partial f}{\partial\dot{x}}\right],\quad\text{and}\\ \frac{d}{dt}\,\frac{\partial f}{\partial \dot{x}}&=\dot{x}\,\frac{\partial^2 f}{\partial x\,\partial\dot{x}}+\ddot{x}\,\frac{\partial^2 f}{\partial\dot{x}^2}. \end{align*} Plugging this into the EL equation, we have \begin{align*} \frac{1}{\sqrt{f}}\frac{\partial f}{\partial x}-\left[\frac{\partial f}{\partial\dot{x}}\,\left(-\frac{1}{2\,f^{3/2}}\left[\dot{x}\,\frac{\partial f}{\partial x}+\ddot{x}\,\frac{\partial f}{\partial\dot{x}}\right]\right)+\frac{1}{\sqrt{f}}\,\left(\dot{x}\,\frac{\partial^2 f}{\partial x\,\partial\dot{x}}+\ddot{x}\,\frac{\partial^2 f}{\partial\dot{x}^2}\right)\right]&=0 \\ \frac{1}{\sqrt{f}}\frac{\partial f}{\partial x}+\frac{\partial f}{\partial\dot{x}}\,\frac{1}{2\,f^{3/2}}\left[\dot{x}\,\frac{\partial f}{\partial x}+\ddot{x}\,\frac{\partial f}{\partial\dot{x}}\right]-\frac{1}{\sqrt{f}}\,\left(\dot{x}\,\frac{\partial^2 f}{\partial x\,\partial\dot{x}}+\ddot{x}\,\frac{\partial^2 f}{\partial\dot{x}^2}\right)&=0 \\ \frac{\partial f}{\partial x}+\frac{1}{2f}\,\frac{\partial f}{\partial\dot{x}}\left[\dot{x}\,\frac{\partial f}{\partial x}+\ddot{x}\,\frac{\partial f}{\partial\dot{x}}\right]-\left(\dot{x}\,\frac{\partial^2 f}{\partial x\,\partial\dot{x}}+\ddot{x}\,\frac{\partial^2 f}{\partial\dot{x}^2}\right)&=0. \end{align*} So, if you compare this to the same expression for $G,$ you can see that you'd need the middle term to vanish: \begin{align*} \frac{1}{2f}\,\frac{\partial f}{\partial\dot{x}}\left[\dot{x}\,\frac{\partial f}{\partial x}+\ddot{x}\,\frac{\partial f}{\partial\dot{x}}\right]&=0 \\ \frac{\partial f}{\partial\dot{x}}\left[\dot{x}\,\frac{\partial f}{\partial x}+\ddot{x}\,\frac{\partial f}{\partial\dot{x}}\right]&=0. \end{align*} So you could either have \begin{align*} \frac{\partial f}{\partial\dot{x}}&=0,\quad\text{or} \\ \dot{x}\,\frac{\partial f}{\partial x}+\ddot{x}\,\frac{\partial f}{\partial\dot{x}}&=0. \end{align*} The second expression is equivalent to $$\frac{df}{dt}=0. $$ So, if either $$\frac{\partial f}{\partial\dot{x}}=0\quad\text{or}\quad \frac{df}{dt}=0, $$ then you will have the same minimizer. As these are fairly restrictive, you can see that, in general, you won't have that happen.

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  • $\begingroup$ Thank you very much, you computations were very instructive :) $\endgroup$ Jul 9 '19 at 18:28
  • $\begingroup$ You're very welcome! $\endgroup$ Jul 9 '19 at 18:31

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