7
$\begingroup$

Given a bounded, compact, closed surface $\Gamma\subset\mathbb{R}^n$, I'm searching $$ \max_{y\in\mathbb{R}^n, \|y\|=1} \int_\Gamma \langle y, x\rangle^2. $$ Without the square, and with the enclosed volume $\Omega$, $$ \max_{y\in\mathbb{R}^n, \|y\|=1} \int_\Omega \langle y, x\rangle, $$ I'm guessing $y_\text{max}$ points towards the centroid of $\Omega$. Not sure how to prove that though.

Any hints?

$\endgroup$
3

1 Answer 1

3
$\begingroup$

Let's do the second one first. Let \begin{align*} f(y_1, \dots, y_n) &= \int_\Omega (y_1x_1 + \dots y_n x_n)\,dV \\ g(y_1, \dots, y_n) &= y_1^2 + y_2^2 + \dots + y_n^2 \end{align*} We want the critical values of $f$ subject to the constraint that $g=1$. By the method of Lagrange multipliers, there is a constant $\lambda$ such that, for each $i$ from $1$ to $n$, $\frac{\partial f}{\partial y_i} = \lambda \frac{\partial g}{\partial y_i}$. That is, $$ \int_\Omega x_i \,dV = 2\lambda y_i $$ for each $i$. By squaring and summing all $n$ of these equations, we get $$ 4\lambda^2 = \sum_{i=1}^n \left(\int_\Omega x_i \,dV\right)^2 $$

The only way the right-hand side is zero is if $\int_\Omega x_i \,dV = 0$ for each $i$. In that case $f$ is identically zero and any $y$ will do. Otherwise, $\lambda \neq 0$, so $$ y_i = \frac{1}{2\lambda} \int_\Omega x_i \,dV $$ Since the $i$th coordinate of the centroid of $\Omega$ is $\bar x_i =\frac{1}{\operatorname{Vol}(\Omega)}\int_\Omega x_i \,dV$, I agree that $y$ is a multiple of $\bar x$.


The first one is trickier, at least, to me. Let $$ h(y_1,\dots,y_n) = \int_\Gamma (x_1y_1+\dots+x_ny_n)^2\,ds $$ Again, we want to maximize $h$ subject to $g=1$. We have $$ 2\int_\Gamma (x_1y_1+\dots+x_ny_n)x_i\,ds = 2\lambda y_i \tag{$*$} $$ for each $i$. If we set $$ M_{ij} = \int_\Gamma x_i x_j \,ds $$ then $(*)$ is equivalent to $$ \sum_{j=1}^n M_{ij} y_j = \lambda y_i $$ In other words, $y$ is an eigenvector of the matrix $M$, corresponding to the eigenvalue $\lambda$.

The matrix $M$ is symmetric, and I think, positive definite. I am guessing the latter can be shown by squaring each equation and adding them up like before. So there is an orthonormal basis of eigenvectors. Therefore the maximum value of $f$ is the maximum of the eigenvalues of $M$.

I'm not sure if we can say more than that. This is a very interesting problem but I've already spent way too much time that I should be doing something else! \smiley

$\endgroup$
1
  • 1
    $\begingroup$ The first one is clear without Lagrange: $\int (x^\text{T} y)^2 =\int y^\text{T} x \cdot x^\text{T} y = y^\text{T} \left(\int x x^\text{T}\right) y$. $\endgroup$ Oct 1, 2016 at 11:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.