This intergral equation can be reduced to the Fredholm integral equation with a separable (degenerate) kernel. 

First, you can use the identity $\cosh(\beta(b+\tilde{b})) = \cosh(\beta b)\cosh(\beta\tilde{b}) + \sinh(\beta b)\sinh(\beta\tilde{b})$ and then use the binomial theorem for $\cosh^n(x)$.

This allows you to represent the kernel $K(b, \tilde{b}) = N \exp(-(n\beta b^2)/2)\cosh^n(\beta(b+\tilde{b}))$ in the form $K(b, \tilde{b}) = \sum_{i} u_i(b) v_i(\tilde{b})$. See the next steps, for example, here (from p.6): http://www.maths.manchester.ac.uk/~wparnell/MT34032/34032_IntEquns.pdf

This is analytical approach. But if n grows, the size of the matrix A from the paper above grows too. Thus, this approach cannot be used for large n.

This intergral equation can be reduced to the Fredholm integral equation with a separable (degenerate) kernel. 

First, you can use the identity $\cosh(\beta(b+\tilde{b})) = \cosh(\beta b)\cosh(\beta\tilde{b}) + \sinh(\beta b)\sinh(\beta\tilde{b})$ and then use the binomial theorem for $\cosh^n(x)$.

This allows you to represent the kernel $K(b, \tilde{b}) = N \exp(-(n\beta b^2)/2)\cosh^n(\beta(b+\tilde{b}))$ in the form $K(b, \tilde{b}) = \sum_{i} u_i(b) v_i(\tilde{b})$. See the next steps, for example, here (from p.6): http://www.maths.manchester.ac.uk/~wparnell/MT34032/34032_IntEquns.pdf

This is analytical approach. But if n grows, the size of the matrix A from the paper above grows too. Thus, this approach cannot be used for large n.

Hi Maxim,

Thank you for your suggestions!

For the case of integer n I was using binomial theorem, as you suggested, but to solve this equation numerically. One can even  perhaps try to find analytic solution for this equation (we know solution for n=1,for Gaussian with mean 0 and variance 1 boundary condition ,  so perhaps it is possible to find solution for n=2 and then guess solution for any integer n )  but it seems to me that this is not Fredholm integral because  the normalisation constant N is a functional of P[b]. The case of large n can be treated by the Laplace method. 

What I am really interested is how to solve for non-integer n.

Best,

Alexander

Hi Alex

I thought about this problem and I haven't solved it, but here are a few comments/questions/ideas:

1. Your equation is a LINEAR integral equations of the second kind. I am not sure this classification helps, but technically it is linear because it is linear in the function P[b].

2. I am not sure in what sense does N stand for a normalization constant. If P[b] is a solution to the equation, then so does C*P[b], irrespective of N, and therefore you can normalize P with any N in the equation.

3. Furthermore, probably this equation does not have a solution for any N, because N serves as an eigenvalue of the integral operator. therefore there is at most a discrete set that makes this equation solvable.

4. I am not sure how you can actually solve it for n->infinity using the Laplace method, but if you can, and if you can solve it for n=1, maybe write it up here and we'll try playing with it. Could be useful to see some results for n=2,3... - or at least the structure of it (i.e. what kind of functions appear in the series).

best

Eytan

Sir you can use MATHEMATICA s/w for solving this equation.

This is a point fixed-type problem. So one can use fixed point

theorems (may be  iterations for sufficiently small N).

Hi Eytan,

Thank you for your interest! See below for my reply to your comment.

“2. I am not sure in what sense does N stand for a normalization constant. If P[b] is a solution to the equation, then so does C*P[b], irrespective of N, and therefore you can normalize P with any N in the equation.”

N=1/\int_{-\infty}^{\infty} d \tilde{b} P[\tilde{b}] \int_{-\infty}^{\infty} d b \exp(-(n\beta b^2)/2)\cosh^n(\beta(b+\tilde{b}))

“4. I am not sure how you can actually solve it for n->infinity using the Laplace method, but if you can, and if you can solve it for n=1, maybe write it up here and we'll try playing with it. Could be useful to see some results for n=2,3... - or at least the structure of it (i.e. what kind of functions appear in the series).”

For n=1 solving the equation for P[b] iteratively with the initial condition P_0[b]= \exp(-( b^2)/2) gives us P[b]=exp(\beta/2)\exp(-(\beta b^2)/2)\cosh(\beta b) which is a “Gaussian mixture”. The equation converges to this solution after after 1st iteration. Also using the Binomial theorem to write the term 2^n\cosh^n(\beta(b+\tilde{b})) as a sum one can show that a solution of this equation is a Gaussian mixture. Regarding the Laplace method: Let P_0[b]=\delta(b – B) on the RHS of and compute the integral

N\int_{-\infty}^{\infty} d \tilde{b} P[\tilde{b}] \int_{-\infty}^{\infty} d b \exp(-(n\beta b^2)/2)\cosh^n(\beta(b+\tilde{b})) g(b)

for some test function g(b). For n\rightarrow\infty this integral is equals to g(b^*), where b^* is a solution of the equation

b=\tanh\beta(b + B), that is (LHS) P[b]=\delta(b - b^*). From this follows that if B is a solution then it must satisfy the equation

B=\tanh(2\beta B) i.e. mean-field equation of 1d Ising.

Hi Alex

Thanks for the detailed answer. I will think further about it. I still have the following comments:

1. regarding N being the normalization constant - my claim is still that it is not quite the case - any solution you may find to the equation - with any N! - could be normalized later, and still be a solution to the equation with that given N.

The only limit on the values of N, which you state, is that it may obey

N=1/\int_{-\infty}^{\infty} d \tilde{b} P[\tilde{b}] \int_{-\infty}^{\infty} d b \exp(-(n\beta b^2)/2)\cosh^n(\beta(b+\tilde{b}))

for any P[b] which solves the equation.

I therefore go back to my original argument that N is like an eigenvalue of the integral operator - there could be several N's such that the equation could be solvable, and it is not obvious which one you are after. It is a-priori not unique!

My question is therefore - in the derivation of this equation - how did this N show up?

2. your large N argument is interesting and means that the distribution goes to a delta function, i.e. becomes narrower. I agree with the integral argument, not sure how rigorous it is, but assuming this is true, it more or less gives you the solution you need numerically. For any finite n (yet large) start your iterations with this delta function and you will probably converge quickly to the value you want.

A similar thing can be done for finite n's - if n is not large and not integer. You can solve analytically the equation for the integer part of n - n'=floor(n), as well as for n''=floor(n)+1, and interpolate between the two. my guess is that a simple linear superposition of the two would do excellent job. If you want more a more accurate answer plug this into the equation and iterate numerically - you should converge quickly.

What I think is that the only real issue you have is determining N

Hi Eytan, 

Regarding you question( “My question is therefore - in the derivation of this equation - how did this N show up?”) the equation was derived as follows:

1) We start with the distribution P(b_1,..,b_M)&=& \exp [ -\frac{1}{2} n \beta \sum_\mu b^2_\mu + n\sum_\mu \log \cosh \beta ( b_\mu + b_{\mu+1} ) ]/Z of M variables, b_\mu\in(-\infty,\infty), interacting on chain. The distribution comes from the analysis of Langevin dynamics of two sets of variables subject to different noises, controlled by \beta and \tilde\beta=n\beta, and evolving at different time scales. One of these sets acts as a “fast noise” to the other and is integrated out leading to the above distribution (an example of such analysis would be in http://journals.aps.org/prb/abstract/10.1103/PhysRevB.48.16116 ).

2) We “cut” the chain in two (--o--o-- → --o o--) and write the recursive equation for the “partition function” Z_\mu[b_\mu] for one of these chains. Then the equation for P[b] is just a normalized version of the “stationary” solution (if we iterate, i.e. take the thermodynamic limit, without normalization then Z_\mu will be growing exponentially) of this recursive equation. This equation is exact for the open chain and one can show that it is also exact (in the thermodynamic limit) for the closed chain.

Hi

Thanks!

I more or less see the picture now.

However, I would expect that the normalization constant would be given more explicitly?

Eytan 

To solve this equation, you can use an eigenfunction system like (17) in

https://www.researchgate.net/publication/273058773_ANALYTICAL_SOLUTION_OF_AN_INTEGRODIFFERENTIAL_EQUATION_ARISING_FROM_A_COLLISION_OPERATOR?ev=prf_pub

Frank

Article ANALYTICAL SOLUTION OF AN INTEGRODIFFERENTIAL EQUATION ARISI...

Hi Eytan,

Regarding your question: "However, I would expect that the normalization constant would be given more explicitly?"

The equation for P[b] is an equation for infinitely large system.  The system is 1D so my guess is that irrespectively of a boundary condition it will have a unique solution (for finite n and \beta).  

Best,

A

Dear all,

Why should a mathematical expression be written like this

P[b]=N\int_{-\infty}^{\infty} d \tilde{b} P[\tilde{b}] \exp(-(n\beta b^2)/2)\cosh^n(\beta(b+\tilde{b})),

I can't understand it. What programming language is that?. Why can't the author provide a doc or pdf file with the integral equation written properly?

Dear Amaechi

This expression is written in LaTeX, a very commonly used language. Have a look at

https://en.wikipedia.org/wiki/LaTeX

To the point - the equation is attached as a figure

 Dear Brother Eytan Katzav 

   Shallom. Thank you for your explanation. unfortunately LaTex is not common in Nigeria though I have started learning it. Some journals insist your submissions must be in Latex. 

Indeed many journals like it for obvious reasons!

I think it is highly recommended to be fluent with it, although I must admit that it has disadvantages too.

I think you can apply the semi analytical methods like homotoly analysis method. This method is so strong for solving different kinds of integral equations in linear and non-linear forms. I have some papers based on this topic.

Dear Samad, Thank you for your interest in this problem. Perhaps you can suggest what would be the best starting point to learn this method.

Best regards,

Alexander