Wikipedia:Reference desk/Archives/Mathematics/2008 September 7

= September 7 =

solenoidal and irrotational field
Is it possible for a vector field to be simultaneously solenoidal and irrotational, that is, divergence and curl free. I don't think it us, but I can't think of a good reason why. This seems to be the essence of the Helmholtz Theorem, but I don't quite see how I could prove that. Thanks very much for any help you can provide! Nathan12343 (talk) 00:39, 7 September 2008 (UTC)
 * Well, there's the zero field, and for that matter all constant vector fields. Algebraist 00:49, 7 September 2008 (UTC)
 * See Cauchy-Riemann equations. siℓℓy rabbit  (  talk  ) 00:59, 7 September 2008 (UTC)

Calculus of variations spring problem
Hi, I am trying to learn to use calculus of variations to solve a simple hypothetical problem I made up for fun; but I am getting stuck and I don't understand why.

So I have a spring. One end of the spring is moved around so that it is at position $$g(t)$$ at time $$t$$. The other end of the spring I can control its position $$f(t)$$ at time $$t$$. The goal is to find a $$f(t)$$ so that I get the maximum work I can get out of my end. I hope that this problem makes sense. It seems that at worst, $$f(t) = 0$$ should give a solution of zero work; and it seems to me that you should be able to do better than that. Perhaps my intuition is wrong here.

So I start with the definition of work:
 * $$\frac{dW}{dt} = F(t) \cdot f'(t)$$

And then apply Hooke's law, where $$k$$ is the spring constant:
 * $$\frac{dW}{dt} = -k (f(t) - g(t)) \cdot f'(t)$$
 * $$\frac{dW}{dt} = k (g(t) - f(t)) f'(t)$$

The total work is then
 * $$W = \int_{t_1}^{t_2} \frac{dW}{dt}\,dt$$

This is in the same form as needed for the Euler–Lagrange equation:
 * $$W[f] = \int_{t_1}^{t_2} L[t,f,f']\,dt$$

where
 * $$L[t,f,f'] = k (g(t) - f) f'$$

So now I plug this into the Euler–Lagrange equation:
 * $$ -\frac{d}{dt} \frac{\partial L}{\partial f'} + \frac{\partial L}{\partial f} = 0$$
 * $$ -\frac{d}{dt}\left[ k (g(t) - f) \right] - k f' = 0$$
 * $$ - k (g'(t) - f') - k f' = 0$$

The $$f'$$ now cancel and I am left with:
 * $$ - k g'(t) = 0$$

Now this equation is simply a condition on my given $$g(t)$$ function! And the function I am trying to optimize ($$f(t)$$) completely vanished! How is this possible? Did I do something wrong? Is there some assumption I am violating? I hope this problem makes sense. If not this way then is there another way I can use to solve this problem? Thanks, --71.141.132.142 (talk) 01:27, 7 September 2008 (UTC)


 * Here are two points. First, regarding your physical model, the distance in Hooke's is distance from equilibrium position, not from the other end:  in 1D this is simply a difference of a constant, and so the mathematics works out the same, but if you are working in more than 1D then there is more than one equilibrium, and you have a bit of mess.  Second, regarding your mathematics, offhand I can't find any mistakes.  Assuming that you have correctly applied the Euler-Lagrange equation, the conclusion you reach is not necessarily impossible:  it would simply indicate that if both $$k \neq 0$$ and there exists t for which $$g'(t) \neq 0$$ (i.e., g is non-constant), then there does not exist any such extremal f.  Unfortuantely I do not understand the mathematics well enough to be more helpful.  Eric.  213.158.252.98 (talk) 13:18, 7 September 2008 (UTC)


 * Indeed, I can create an example where arbitrarily high work can be extracted. Let t1 = 0, t2 = 1, g(t) = t, and k = 1.  Here g represents the equilibrium position.  We subject f to the constraint that the spring starts and stops at equilibrium, i.e., f(0) = 0 and f(1) = 1.


 * Consider the following f. First, just after t = 0, very quickly bring f to the point -A, where A is a constant.  Leave your end of the spring at that position until just before t=1, when you very quickly restore f to equilibrium at 1.  The work gained is approximately equal to Ak times one unit of distance.  Since A is a arbitrary, we can get both arbitrarily high and arbitrarily low amounts of work:  thus in this situation we have no global extrema.


 * A similar construction can be used to make arbitrarily high or low amounts of work in any situation provided that k is nonzero and g is nonconstant. Eric.  213.158.252.98 (talk) 13:38, 7 September 2008 (UTC)


 * Oh I see. So I can force the other end to put in arbitrarily large amounts of work if they move at all; because if the other end went up, I could just hold the spring arbitrarily low to get him to put in work. It is only when $$g'(t) = 0$$ that the other end doesn't put in work at all; and then it doesn't matter what I do. --71.141.132.142 (talk) 21:06, 7 September 2008 (UTC)


 * Precisely. You could put in realistic constraints on f (disallowing going too far from equilibrium) to force a maximum to exist, but in that case I doubt that the Euler-Lagrange equation will help you find this maximum.  Alternatively, you could use a force equation other than Hooke's Law (keep in mind that Hooke's Law is an approximation best for near equilibrium) to close this loophole, and then proceed again with the Euler-Lagrange equation.  I bet you can find something interesting with the latter (athough I don't know if it'd be physically realistic -- go find a physicist and tell us what (s)he says!).  Eric.  84.215.155.88 (talk) 20:31, 8 September 2008 (UTC)

Help
How do you get from the left hand side to the right: $$\frac{x^3}{1 + x^2} = x - \frac{x}{1 + x^2}$$ --RMFan1 (talk) 20:51, 7 September 2008 (UTC)
 * Use polynomial division (or whatever) to get $$x^3=x(1+x^2)-x$$. Then just cancel a $$1+x^2$$ and you're there. Algebraist 20:56, 7 September 2008 (UTC)
 * Multiply both sides by $${(1 + x^2)}$$ .Cuddlyable3 (talk) 23:41, 8 September 2008 (UTC)
 * That proves equality, but is no use if you have the left hand side and are trying to get to something like the right. Algebraist 23:43, 8 September 2008 (UTC)
 * Is one supposed to know what expression one is trying to get to before one gets there?Cuddlyable3 (talk) 09:08, 9 September 2008 (UTC)
 * No. That was in fact my point. I interpreted the OP's question as asking how one was supposed to have got the RHS if one had the LHS, and I outlined the usual method (polynomial division with remainder). Algebraist 11:10, 9 September 2008 (UTC)