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I am trying to minimize the functional

$$I[\textbf{x}] = \int ||\dot{\textbf{x}}||^2 dt$$

subject to the constraint $\textbf{x}(t) \in \{\textbf{s} \in \mathbb{R}^3 : ||\textbf{s}|| = 1 \}$. The question states:

Use the Lagrange multiplier function formalism to obtain the following Euler-Lagrange equation: $$\ddot{\textbf{x}} + ||\dot{\textbf{x}}||^2 \textbf{x} = \textbf{0}$$

I'm not totally sure what it means by the "Lagrange multiplier function formalism", but anyway, introducing a Lagrange multiplier, $\lambda(t)$, we get:

$$\int L(\textbf{x}, \dot{\textbf{x}}, \lambda) dt = \int ||\dot{ \textbf{x}}||^2 + \lambda(||\textbf{x}|| - 1) dt$$

and hence the Euler-Lagrange equations are:

$$\frac{d}{dt}\Big( \frac{\partial L}{\partial \dot{\textbf{x}}}\Big) - \frac{\partial L}{\partial \textbf{x}} = 2 \ddot{\textbf{x}} - \frac{\lambda}{||\textbf{x}||} \textbf{x} = \textbf{0}$$

and also $||\textbf{x}|| - 1 = 0$. I need to find what $\lambda(t)$ is but I'm not sure how to progress.

Thanks in advance for any help.

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  • $\begingroup$ You might find it easier (I'm guessing -- haven't worked it out) if you wrote $\lambda( \| x \|^2 - 1)$. $\endgroup$ – John Hughes Jul 14 '14 at 16:54
  • $\begingroup$ Tip: Use \| instead of ||, it looks better: $\|\mathbf x\|$ vs. $||\mathbf x||$. $\endgroup$ – Rahul Jul 14 '14 at 17:26
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Note that $\mathbf{x}^2 = ||\mathbf{x}||^2 = 1$ can be differentiated twice to obtain $\dot{\mathbf{x}}^2+\mathbf{x}\cdot\ddot{\mathbf{x}}=0.$ But one can also obtain $\mathbf{x}\cdot \ddot{\mathbf{x}}$ from the Euler-Lagrange equation by an appropriate dot product. These together fix $\lambda$.

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  • $\begingroup$ Thanks for the reply but I'm not entirely sure as to why being able to find $\textbf{x} \cdot \ddot{\textbf{x}}$ in two different ways is necessary, surely once we have $\lambda$ in terms of $\textbf{x}$ etc we're done? At any rate, it doesn't lead to the printed result. Are you sure my E-L equation is correct? $\endgroup$ – JosephML Jul 14 '14 at 17:48
  • $\begingroup$ Your EL equation looks right to me. The point I was making was that, if you find two different equations involving both that dot product and $\lambda$, then you can eliminate the dot product and obtain a solution for $\lambda$. Does that clarify? $\endgroup$ – Semiclassical Jul 14 '14 at 18:02
  • $\begingroup$ @JosephML: Actually, I'm wrong: I hadn't seen that you'd edited the problem from $\ddot{\mathbf{x}}$ to $\dot{\mathbf{x}}$. In that case, yes, you need a bit more info. Will revise. $\endgroup$ – Semiclassical Jul 14 '14 at 18:08
  • $\begingroup$ Ah yes, I see what you mean. This gives $\lambda^2 = 4 \| \ddot{\textbf{x}} \|^2$ and when put back in gives $$\ddot {\textbf{x}} \mp \frac{2 \| \ddot{\textbf{x}}\|}{\| \textbf{x} \|} \lambda \textbf{x} = 0$$ but I can't see how one could get a $\| \dot{\textbf{x}}\|^2$ from that. $\endgroup$ – JosephML Jul 14 '14 at 18:11
  • $\begingroup$ @JosephML: Right. You need the $x^2=1$ constraint for that, and that's what I've added to the solution. $\endgroup$ – Semiclassical Jul 14 '14 at 18:20

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