April 6

Hi,

Given that partial dw/dt = laplacian(w) and w(t,x) = f(x) Does anyone have any suggestions as to how I might show that (partial d/dt)(volume integral of (grad w)²) is ≤ 0? Thanks! 131.111.222.12 (talk) 12:44, 6 April 2011 (UTC)[reply]

When you say "w(t,x) = f(x)", did you mean that that holds when t = 0? Michael Hardy (talk) 12:48, 6 April 2011 (UTC)[reply]

....and is f some particular function? Michael Hardy (talk) 12:48, 6 April 2011 (UTC)[reply]

Hi Michael, yes it holds at t=0 (in fact for all t) and no, f is not specified... —Preceding unsigned comment added by 131.111.222.12 (talk) 13:03, 6 April 2011 (UTC)[reply]

I think maybe you need to assume some decay conditions at infinity? I'd suggest first proving it for the total heat energy, the volume integral of ${\displaystyle w^{2))$. This you can do by interchanging the order of differentiation and integration, applying the equation, and then integrating by parts. For the integral of the gradient squared, use the fact that the partial derivatives each satisfy the heat equation as well. Sławomir Biały (talk) 14:13, 6 April 2011 (UTC)[reply]

Thanks, Sławomir Biały, I shall try that —Preceding unsigned comment added by 131.111.222.12 (talk) 15:30, 6 April 2011 (UTC)[reply]

This doesn't make sense: if w(t,x) = f(x) for ALL values of t, not just for t = 0, then w(t,x) does not change as t changes. The would mean dw/dt = 0 for all values of t. Michael Hardy (talk) 15:20, 6 April 2011 (UTC)[reply]

Hi, Michael, you are very right. I have just realized that I failed to specify that w(t,x)=f(x) for all t is a boundary condition not generally true on the entire volume. —Preceding unsigned comment added by 131.111.222.12 (talk) 15:29, 6 April 2011 (UTC)131.111.222.12 (talk) 15:31, 6 April 2011 (UTC)[reply]

I'm guessing that by (grad w)² you mean the square of the norm of the gradient.

The inequality certainly looks plausible. Michael Hardy (talk) 19:30, 6 April 2011 (UTC)[reply]

Let's try something in the one-dimensional case: We're trying to show that

${\displaystyle {\frac {\partial }{\partial t))\int \left({\frac {\partial w}{\partial x))\right)^{2}\,dx\leq 0.}$

So integrate by parts with respect to x and the integral becomes:

${\displaystyle {\frac {\partial }{\partial t))\int -{\frac {\partial ^{2}w}{\partial x^{2))}\cdot w\,dx}$

(provided there is no contribution from the boundaries of the x-space). If sufficient continuity/differentiability/whatever conditions hold to justify pushing that first differential operator under the integral sign, then we get

${\displaystyle \int -{\frac {\partial ^{2}w}{\partial x^{2))}\cdot {\frac {\partial w}{\partial t))\,dx.}$

Now apply the differential equation you started with to get

${\displaystyle \int -{\frac {\partial ^{2}w}{\partial x^{2))}\cdot {\frac {\partial ^{2}w}{\partial x^{2))}\,dx.}$

That is clearly non-positive. Michael Hardy (talk) 19:39, 6 April 2011 (UTC)[reply]

Michael, thanks a lot! :) 131.111.222.12 (talk) 20:52, 6 April 2011 (UTC)—Preceding unsigned comment added by 131.111.222.12 (talk) 20:50, 6 April 2011 (UTC)[reply]

If f(x) is continuous on [0,1] and f(0)=f(1) and I have to show that for any n there exists a point a(n) in [0, 1-(1/n)] s.t. f(a+(1/n))=f(a) I have defined a new function, say g(x)= f(a+(1/n))-f(a) and am thinking of using the IVT to prove that there exists a point where g(x)=0 but am not quite sure how. Thanks in advance for any help! —Preceding unsigned comment added by 131.111.222.12 (talk) 14:56, 6 April 2011 (UTC)[reply]

Extend f periodically and then see that the integral of g from 0 to 1 is 0. Apply the mean value theorem to conclude that g vanishes as some point. Sławomir Biały (talk) 15:24, 6 April 2011 (UTC)[reply]

Thanks, Sławomir Biały. Could you possibly explain what is meant by expand f periodically? Thanks again —Preceding unsigned comment added by 131.111.222.12 (talk) 19:12, 6 April 2011 (UTC)[reply]

Extend f to a function on the real line such that ${\displaystyle f(x+1)=f(x)}$ for all x. Sławomir Biały (talk) 19:27, 6 April 2011 (UTC)[reply]

Thanks, Sławomir Biały. There was a typo in my question. The closed interval should have read [0,1-1/n] instead of [1,1-1/n]. I have changed it in the question. Sorry about that.

This looks like homework, but anyway, ... there was another typo, so I assume you meant g(x)= f(x+(1/n))-f(x). I'm fairly sure the theorem only works when n is an integer, so your proof has to reflect this somehow. I suggest dividing the interval into n segments, then looking at g(m/n) with n fixed and m varying over the integers 0 to n. Then you will see that g(x) must change sign for one of these, unless it is already zero somewhere, when the problem is solved anyway. I'll leave the rest to you, that is, if you should chance back to have another look. It's been emotional (talk) 18:37, 9 April 2011 (UTC)[reply]

Actually, it's true for any (possibly irrational) period 1/n, provided you take the periodic extension of f as I suggested. In the approach I suggested, one actually needs to use induction on n to prove it for non-periodic f. Then if the point a is in the interval [0,1−1/n], we're done. Otherwise, if it's in the interval (1−1/n,1], we apply the induction hypothesis to f restricted to the subinterval [a+1/n−1, a] to obtain the required point there.

The proof you suggest is actually similar in spirit, since it ultimately involves observing the telescoping sum ${\displaystyle g(0)+g(1/n)+\cdots +g((n-1)/n)=0}$, and therefore at least two values ${\displaystyle g(m/n)}$ and ${\displaystyle g(k/n)}$ must have opposite sign, and so the intermediate value theorem can be invoked. Sławomir Biały (talk) 19:09, 9 April 2011 (UTC)[reply]

(ec) Good point. As a concrete counterexample for non-integer n, consider ${\displaystyle f(x)=\cos(5\pi x)+2x}$, which certainly is continuous with f(0)=f(1)=1. However, ${\displaystyle f(a+2/5)-f(a)=4/5\neq 0}$ for all ${\displaystyle a}$. –Henning Makholm (talk) 19:08, 9 April 2011 (UTC)[reply]

Yes, and this is not periodic of period 1, so there is no reason my proof should work with this function. Sławomir Biały (talk) 19:14, 9 April 2011 (UTC)[reply]

How about ${\displaystyle f(x)=\cos(5\pi (x-\lfloor x\rfloor ))+2(x-\lfloor x\rfloor )}$, then? Then g does have zeroes, but they are outside the allowed interval for a. –Henning Makholm (talk) 19:21, 9 April 2011 (UTC)[reply]

In honesty, I had ignored the restriction on a. It doesn't really make sense from the periodic perspective (it's just to ensure that ${\displaystyle f(a+1/n)}$ is well-defined in the non-periodic case). So I don't really consider this to be part of the problem. Sławomir Biały (talk) 19:30, 9 April 2011 (UTC)[reply]

Thanks to both of you for the interest. Accepting the problem as stated, the counterexample for all non-integer n > 1 is to draw an imaginary line y = x, and mark on this line the points x = 0, 1/n, 2/n, ... m/n as long as m/n < 1, m an integer. Do the same going backwards from the point (1,0), ie. draw an imaginary line y = x-1, and mark the points x = 1, 1-1/n, 1-2/n, .... etc until you run out. Then just join the dots, and that's your function f. I'm fairly sure g is a constant, equal to 1. Please correct me if I'm wrong, and brownie points (a cookie perhaps?) for anyone who can actually give a precise definition of the function I described. It's been emotional (talk) 19:49, 9 April 2011 (UTC)[reply]

Cool. (No cookie for me. :-) Sławomir Biały (talk) 20:03, 9 April 2011 (UTC)[reply]

That's how I constructed my example, too. I smoothed the triangles into a cosine wave for notational convenience when writing it down. :-) The triangle wave article suggests some expressions, neither of them pretty. For a formal proof, all we really need is some nontrivial continuous function h(x) with period 1/n such that h(0)≠h(1). –Henning Makholm (talk) 20:03, 9 April 2011 (UTC)[reply]