6
$\begingroup$

If I can find a solution to the following set of equations then, with a bit of luck, I should be able to derive all sorts of nifty new results in non-equilibrium statistical mechanics. However, I'm not sure if there is a general solution.

The equations are as follows. There is one equation for every $k \in \mathbb{Z}^\star$ (the positive ingtegers). $$ e^{-\eta_k}\sum_{j=0}^\infty e^{\eta_j-\lambda f_{kj}} = e^{\eta_k}\sum_{i=0}^\infty e^{-\eta_i-\lambda f_{ik}}. $$ What I want is an expression for $\eta_k$ in terms of $k, \lambda$ and $\{f_{ij}\}$, so this is an infinite number of equations in an infinite number of unknowns, with an infinite number of parameters. (It could also be seen as a singe functional equation.)

The numbers $f_{ij}$ are arbitrary real parameters. We can assume they are bounded below (i.e. there exists $g$ such that for every $i$ and every $j$, $f_{ij}>g$) but not necessarily bounded above. $\lambda$ is also a real parameter; if necessary we can assume that it is sufficiently large that the infinite sums in the equations converge.

Equivalently, if we let $x_k = e^{-\eta_k}$ and $A_{ij}=e^{-\lambda f_{ij}}>0$ then we're trying to find positive solutions (for $\{x_k\}$) of $$ \sum_{j=0}^\infty A_{kj} \frac{x_k}{x_j} = \sum_{i=0}^\infty A_{ik} \frac{x_i}{x_k}. $$ I've tried everything I can think of to solve this, but I'm not much of a mathematician and haven't really got very far. Can anyone else see how to do it?

$\endgroup$
  • $\begingroup$ I really wasn't sure how to tag this - help with that would be appreciated. $\endgroup$ – Nathaniel Jan 7 '14 at 7:14
  • $\begingroup$ Do you need an analytical solution or a numerical one will do? $\endgroup$ – dexter04 Jan 13 '14 at 13:00
  • $\begingroup$ @dexter04 I was hoping for an analytical solution - when I posted I thought one would exist, but now I'm doubting that, so I'll settle for a way to sce it numerically (with a finite but large number of variables) $\endgroup$ – Nathaniel Jan 14 '14 at 1:37
3
$\begingroup$

Here is an approach that might work. First, it is obvious that if $\{x_i\}$ is a solution to the above system, then so is $\{kx_i\}$ for all $k>0$.

So, define $$\alpha_{ij} = \frac{x_i}{x_j} \quad \forall i\ne j$$ Now youcan reformulate your problem into a (infinite) system of linear equations in $\alpha_{ij}$. According to this query at Math.SO this system will have a unique solution provided the system is well behaved. Now, you can check whether the unique solution satisfies $\alpha_{ij}\alpha_{ji} = 1$. If it does, you have a solution. Else there won't be a solution as existence of another solution will violate the uniqueness requirement.

I'm at a loss about the method to find the unique solution as I don't have access to those books but there are enough pointers to relevant literature at the thread to find out the exact as well as truncated solutions.

Hope it helps!

$\endgroup$
2
+50
$\begingroup$

I wasn't going to post this, but noone seems to have any ideas. Note that your equation is equivalent to:

$$ x_k^2= \frac{\sum_{i=0}^\infty A_{ik} x_i}{\sum_{j=0}^\infty A_{kj}\frac{1}{x_j}}. $$

Try defining the function that takes in a sequence $\{y_k\}$ and puts out a sequence $\{x_k\}$ defined by $$ x_k= \sqrt{\frac{\sum_{i=0}^\infty A_{ik} y_i}{\sum_{j=0}^\infty A_{kj}\frac{1}{y_j}}}. $$

The solution you seek is a fixed point of this mapping. If you can find a suitable function space, you may be able to show its contracting map, but it seems hard. If you are interested in numerical solutions, you could try iterating this map on a large finite subsequence.

$\endgroup$
  • $\begingroup$ Thanks, this is a good idea. You'll get the bounty if no-one has a better idea. I guess there probably isn't an analytical solution in the general case, so this kind of iterative numerical solution will be the next thing to try. $\endgroup$ – Nathaniel Jan 14 '14 at 2:02
2
$\begingroup$

I often find myself answering my own questions; this time it's a negative answer.

The question as I wrote it is equivalent to a known problem called "matrix scaling", for which there are efficient numerical algorithms (because it's used as an important preprocessing step before numerically solving eigenproblems) but no analytical solution, at least not in the general case.

The matrix scaling problem is as follows: given a matrix $A$, find a diagonal matrix $D$ such that the row sums of $D^{-1}A D$ are equal to the corresponding column sums. If the diagonal elements of $D$ are $\{x_i\}$ then expanding this out gives the second equation in my post.

The simplest algorithm for solving this numerically is simply to scale one row/column pair at a time, iterating over the whole matrix, and then repeat until it converges. This will probably suffice for my purposes, though if it doesn't I believe I can find more sophisticated algorithms implemented as part of LAPACK.

$\endgroup$

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.