Wikipedia:Reference desk/Archives/Mathematics/2009 February 23

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February 23

Simple (i think) vector integral inequality (electrostatic capacity of a cube)

Hi there - I'm trying to finish off a question and seem to be missing something, I don't think it's particularly hard, I think I'm just being stupid and missing something obvious (hopefully);

having shown that for u(x) the unique solution to Laplace's equation in the volume V with u=f(x)on the surface S enclosing V, and v any function with continuous 1st partial deriv.s in V which =0 on S, we have $\int _{V}\nabla v\cdot \nabla udV=0$, I want to show for any w with continuous 1st partial deriv.s in V with w=f on S,

$\int _{V}|\nabla w|^{2}dV\geq \int _{V}|\nabla u|^{2}dV$. I've tried subbing in v=w-u into the original equation to get $\int _{V}\nabla w\cdot \nabla udV=\int _{V}|\nabla u|^{2}dV$, which is sortof halfway there, but I'm not sure where to go next. If anyone could point out where I was being stupid... thanks! Otherlobby17 (talk) 12:53, 23 February 2009 (UTC)Otherlobby17

I can't see a way to massage your last equation into the desired result but with the only justification that it leads to the answer, I'd start from

$w({\vec {x}})=u({\vec {x}})+v({\vec {x}})$

Then by taking gradients,

$\nabla w=\nabla u+\nabla v$

Squaring,

$|\nabla w|^{2}=|\nabla u|^{2}+|\nabla v|^{2}+2\nabla u\cdot \nabla v$

And finally integrating over the volume, we get

$\int _{V}|\nabla w|^{2}\mathrm {d} V=\int _{V}|\nabla u|^{2}\mathrm {d} V+\int _{V}|\nabla v|^{2}\mathrm {d} V+2\int _{V}\nabla u\cdot \nabla v\mathrm {d} V$

Now, by the equation you have already proved the last term vanishes and as $|\nabla v|^{2}$ is strictly positive your inequality follows. Vespertine1215 (talk) 17:29, 23 February 2009 (UTC)

Brilliant, thanks! Otherlobby17 (talk) 17:41, 23 February 2009 (UTC)Otherlobby17

Talking about inner product spaces, is it clear the relation between "minimizing the distance from a given subspace" and "orthogonality"? It's just this. --pma (talk) 21:09, 23 February 2009 (UTC)

I suspected the two may have been linked ;) A further part of the question asks me to relate the 'capacity' of an object $-\int _{S}{\frac {\partial \phi }{\partial n}}dA$, where the potential φ(x) satisfies Laplace’s equation in the volume outside the object, $\phi =1$ on S and $\phi \to 0$ at $\infty$, before showing the capacity of a cube is bounded by $2\pi a<C<2{\sqrt {3}}\pi a$ for cube of sides a (using the in/circumcircle). Are these 2 related in a similar way too then - capacity and minimizing this distance? Thanks for your time! Otherlobby17 (talk) 7:31, 24 February 2009 (UTC)Otherlobby17

Well, as you probably know, you also have, for your potential $\phi$ (by Green identities)

$-\int _{S}{\frac {\partial \phi }{\partial n}}dA=\int _{\mathbb {R} ^{3}\setminus V}|\nabla \phi |^{2}\mathrm {d} x$,

and in fact you can define the capacity of the body V as the infimum of the energy $\scriptstyle \int _{\mathbb {R} ^{3}\setminus V}|\nabla u|^{2}\mathrm {d} x$ over all smooth function $u$ taking the value 1 on the bundary of $V$ and 0 at infinity. So an upper bound for the capacity is just the energy of any of these functions, or more generally, the infimum of the energy taken in a suitable smaller class of functions (is it clear? Infimum over a smaller class gives a larger value). Now, at least in the case of a convex body $Q$, a reasonable class is: functions of the distance from $Q$, that is, functions of the form $\scriptstyle u(x)=f(\mathrm {dist} (x,Q))$, with a smooth $\scriptstyle f:[0,\infty ]\to \mathbb {R}$. For a cube $\scriptstyle V=[0,a]^{3}$ this makes particularly simple the expression of the energy of $u$. The domain of integration $\scriptstyle R^{3}\setminus V$ can be partitioned into three subdomains, corresponding respectively to all points projecting on faces, edges, vertices: you should find:

$\int _{\mathbb {R} ^{3}\setminus V}|\nabla u|^{2}\mathrm {d} x=6a^{2}\int _{0}^{\infty }f'(t)^{2}dt+6\pi a\int _{0}^{\infty }tf'(t)^{2}dt+4\pi \int _{0}^{\infty }t^{2}f'(t)^{2}dt=$

$\int _{0}^{\infty }(6a^{2}+6\pi at+4\pi t^{2})f'(t)^{2}dt$.

Now you should check this computation, and then minimize over all $f$ with $f(0)=1,f(x)=o(1)$ for $\scriptstyle x\to \infty$ (try it, it is worth; in case ask for details). As to the lower bound, there is a non trivial fact: an Euclidean ball minimizes the capacity among all bodies with the same volume. Thus a lower bound for the capacity of a cube is the capacity of a ball with same volume.--pma (talk) 17:03, 24 February 2009 (UTC)

Note: concerning the minimum problem

$\textstyle \min \int _{0}^{\infty }(6a^{2}+6\pi at+4\pi t^{2})f'(t)^{2}dt$

over all smooth functions with $\scriptstyle f(0)=1$ and $\scriptstyle f(\infty )=0$, the Euler-Lagrange equations for the minimizer $f$ give:

$\textstyle (6a^{2}+6\pi at+4\pi t^{2})f'(t)=c$ ,

where the constant $c$ is to be determined by the constraint $\scriptstyle \int _{0}^{\infty }f'(t)dt=f(\infty )-f(0)=-1$.

You don't need to compute $\textstyle f$, because the energy only depends on $\textstyle f'$. You don't either need to know that the minimum problem really admits a minimizer, and that the minimizer satisfies the Euler-Lagrange equation (although these are both necessary issues in order to conclude that $\textstyle f$ is a minimizer!). You can follow a perfectly rigorous heuristic argument: just plug the $\textstyle f$ found by the Euler-Lagrange equation into

$\textstyle \int _{0}^{\infty }(6a^{2}+6\pi at+4\pi t^{2})f'(t)^{2}dt$,

for in any case this is an upper bound for C. You'll get:

$\textstyle a\left(\int _{0}^{\infty }{\frac {dt}{6+6\pi t+4\pi t^{2}}}\right)^{-1}$,

that is $\textstyle (8.929..)a$ ; actually, a little better than $\textstyle 2{\sqrt {3}}\pi a=(10.88..)a$ (still I wonder how your bounds have been obtained) --pma (talk) 22:10, 24 February 2009 (UTC)

As a last remark, notice that with minor changes you get a bound on the capacity C(Q) of any convex polyhedron Q --pma (talk) 11:39, 25 February 2009 (UTC)

I suspect that this problem does not rely on something as complicated as the argument that the capacity is minimized for a sphere; it is a matter of recognizing that the cube of side length a contains a sphere of radius a/2, and is contained inside a sphere of radius ${\sqrt {3}}a/2$. Recalling that the capacity of a set is always smaller than the capacity of a containing set, it remains only to find the capacity of a sphere of radius r.

The capacity of a sphere can be found by considering the function $V(x)={\frac {r}{|x|}}$ for $|x|\geq r$, which is the capacitary potential for such a sphere centered at the origin. A bit of calculation should give you that this capacity is $4\pi r$. Ray (talk) 14:25, 27 February 2009 (UTC)

Right, most likely this is how the problem was intended... but proving the monotonicity of the capacity (as defined in the OP) requires also a non trivial variational argument, so I liked more a direct and more precise computation, at least for the upper bound --the computation of the upper bound that I provided is all elementary, in particular does not uses the minimality of the sphere.. --pma (talk) 17:54, 27 February 2009 (UTC)

What I like is that the same computation with u(x)=f(dist(x,Q)) works for any bounded convex Q in $\scriptstyle \mathbb {R} ^{3}$. Minimizing over all f with f(0)=1 and $\scriptstyle f(\infty )=0$ we get as before an upper bound for the capacity of $Q$:

$\textstyle 2K\left(\int _{0}^{\infty }{\frac {dt}{\pi t^{2}+t+A/K^{2}}}\right)^{-1}$,

where A and K are respectively the area and the total mean curvature of $\scriptstyle \partial Q$. For the cube, $A=6a^{2}$ and $K=3\pi a$. For a polyhedron, $\scriptstyle K={\frac {1}{2}}\sum _{e}l_{e}\alpha _{e}$, sum extended over all edges e, where $l_{e}$ is the length of the edge $e$ and $\alpha _{e}$ is the angle between the normals to the adjacent faces to e. This upper bound still has homogeneity 1, since the total mean curvature has the dimension of a lenght. And $\scriptstyle {\frac {A}{K^{2}}}$ is a shape coefficient (adimensional). --pma (talk) 23:42, 27 February 2009 (UTC)

Terminology

If x is a rational number, let me define $\|x\|=\max\{|p|,|q|\}$, where p and q are coprime integers such that x = p/q. Does this thing have a standard name? — Emil J. 15:18, 23 February 2009 (UTC)

Not that I know of. Does it have any uses? Things only get named if they are used. --Tango (talk) 15:45, 23 February 2009 (UTC)

It is a convenient measure of the complexity of x. Such an x requires at most 2 lg(∥x∥) bits, and the cost of arithmetic with such x is conveniently bounded in terms of ∥x∥. I don't recall a name being given to it. JackSchmidt (talk) 16:33, 23 February 2009 (UTC)

What are you using it for? —Ben Kovitz (talk) 15:55, 23 February 2009 (UTC)

Well, I need it in a proof I am working on now. In fact, I use a generalization of the concept to rational vectors: $\|(x_{1},\dots ,x_{k})\|=\max\{q,|qx_{1}|,\dots ,|qx_{k}|\}$, where q is the smallest positive integer such that $qx_{1},\dots ,qx_{k}\in \mathbb {Z}$. Basically, the idea is that on the one hand, $\log \|{\vec {x}}\|$ is polynomially related to the number of bits required to write the vector ${\vec {x}}$ down (in particular, there are only finitely many vectors with $\|{\vec {x}}\|\leq n$ for a given n and k), and on the other hand, it's easier to work with in the given context than the actual number of bits. The vector definition is admittedly a bit weird, however I seem to recall that I've seen the basic definition just for rational numbers used somewhere, but I can't remember how it was called, that's why I am asking. So far I've called it a "norm" of ${\vec {x}}$ (whence the $\|\|$ notation), which is short and sweet, but misleading, as it does not really satisfy the usual axioms of a norm. — Emil J. 16:21, 23 February 2009 (UTC)

Look at Height of a polynomial. In your case the polynomial would be qx − p. Michael Hardy (talk) 22:19, 23 February 2009 (UTC)

Height has a nice ring to it. Black Carrot (talk) 23:22, 23 February 2009 (UTC)

It's usually called the multiplicative height of x, denoted H(x). There's a related quantity called the absolute logarithmic height h(x) which is just the logarithm of H(x). Chenxlee (talk) 23:31, 23 February 2009 (UTC)

Thanks! This is what I was looking for. — Emil J. 11:13, 24 February 2009 (UTC)

Using a converse of the FTC part II

Hey guys, I've been working on a proof, and was wondering if you guys could check it to ensure the reasoning is valid:

File:Inversefunction.GIF

Start with a continuous increasing one-to-one inverse function f^-1, as shown in my crudely drawn image above.

As can be seen, the rectangle with sides b and f^-1(b) can be partitioned into three areas.

Meaning,

$bf^{-1}(b)=A_{I}+A_{II}+A_{III}\,$

Area one is a rectangle of sides a and f^-1(a), while Area two is cleary the definite integral from a to b of f^-1. By virtue of an inverse function being a reflection of the original function about the line y=x, Area three is the definite integral from f^-1(a) to f^-1(b) of f(y).

Substituting:

$bf^{-1}(b)=af^{-1}(a)+\int _{a}^{b}f^{-1}(x)\,dx+\int _{f^{-1}(a)}^{f^{-1}(b)}f(y)\,dy\,$

Rearranging and evaluation yields

$\int _{a}^{b}f^{-1}(x)\,dx=bf^{-1}(b)-af^{-1}(a)-F(f^{-1}(b))+F(f^{-1}(a))\,$

where F(t) is an antiderivative of f(t).

This can be rewritten as

$\int _{a}^{b}f^{-1}(x)\,dx=g(b)-g(a){\text{ , }}g(t)=tf^{-1}(t)-F(f^{-1}(t))\,$

Now here's the step I'm not quite sure about:

By what I believe would be the converse of the Fundamental Theorem of Calculus part two,

$\int f^{-1}(x)\,dx=g(x)=xf^{-1}(x)-F(f^{-1}(x))+C\,$

Is this step allowed? Did I make any other mistakes? Any input is appreciated. Tuvwxyz (T) (C) 23:27, 23 February 2009 (UTC)

I think you can get what you got by integration by parts and some substitutions:

Take $u=f^{-1}(x),dv=dx$; then $v=x$; we will not compute $du$ for simplicity; but note that $x=f(u)$:

$\int f^{-1}(x)\,dx=xf^{-1}(x)-\int xdu=xf^{-1}(x)-\int f(u)du=xf^{-1}(x)-F(u)+C=xf^{-1}(x)-F(f^{-1}(x))+C$ --Spoon! (talk) 01:37, 24 February 2009 (UTC)

What is the difference between double bar and a single bar

I know |x| means absolute values, but I also seen ||x|| notation being used, what's the difference between the two? 23:27, 23 February 2009 (UTC)

||x|| means the norm of x. For a scalar, this is usually just the absolute value, but for vectors it can be different things. The standard norms (I forget the names) are the square root of the sum of the squares of the absolute values of the components, the sum of the absolute values of the components, and the maximum of the absolute values of the components. -mattbuck (Talk) 23:51, 23 February 2009 (UTC)

Those norms are called L² (aka Euclidean, aka standard), L¹ (aka taxicab, aka Manhattan), and L^∞ (aka uniform, aka sup) respectively. See L^p space for the family these are part of. Algebraist 10:53, 24 February 2009 (UTC)