Wikipedia:Reference desk/Archives/Mathematics/2008 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)

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:

:$\backslash frac\{dW\}\{dt\}\; =\; F(t)\; \backslash cdot\; f\text{'}(t)$

And then apply Hooke's law, where $k$ is the spring constant:

:$\backslash frac\{dW\}\{dt\}\; =\; -k\; (f(t)\; -\; g(t))\; \backslash cdot\; f\text{'}(t)$

:$\backslash frac\{dW\}\{dt\}\; =\; k\; (g(t)\; -\; f(t))\; f\text{'}(t)$

The total work is then

:$W\; =\; \backslash int\_\{t\_1\}^\{t\_2\}\; \backslash frac\{dW\}\{dt\}\backslash ,dt$

This is in the same form as needed for the Euler–Lagrange equation:

:$W[f]\; =\; \backslash int\_\{t\_1\}^\{t\_2\}\; L[t,f,f\text{'}]\backslash ,dt$

where

:$L[t,f,f\text{'}]\; =\; k\; (g(t)\; -\; f)\; f\text{'}$

So now I plug this into the Euler–Lagrange equation:

:$-\backslash frac\{d\}\{dt\}\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; f\text{'}\}\; +\; \backslash frac\{\backslash partial\; L\}\{\backslash partial\; f\}\; =\; 0$

:$-\backslash frac\{d\}\{dt\}\backslash left[\; k\; (g(t)\; -\; f)\; \backslash right]\; -\; k\; f\text{'}\; =\; 0$

:$-\; k\; (g\text{'}(t)\; -\; f\text{'})\; -\; k\; f\text{'}\; =\; 0$

The $f\text{'}$ now cancel and I am left with:

:$-\; k\; g\text{'}(t)\; =\; 0$

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 $k\; \backslash neq\; 0$ and there exists t for which $g\text{'}(t)\; \backslash neq\; 0$ (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 $g\text{'}(t)\; =\; 0$ 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)

How do you get from the left hand side to the right:

$\backslash frac\{x^3\}\{1\; +\; x^2\}\; =\; x\; -\; \backslash frac\{x\}\{1\; +\; x^2\}$ --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)