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I would like so find the function $n(z)$ that solves the following equation:

$ n(z) = \frac{.2 + .24 z \int_1^{\infty} \frac{n(z)}{z^2}dz}{.5 + .24 z \int_1^{\infty} \frac{n(z)}{z^2}dz} $

That is, $n(z)$ is defined in terms of it's own integral. Even if there's no nice analytic solution to this, Numerical approximations of some sort are good enough. In fact this is just a parametrized instance of a more general equation I am interested in. What methods should I be using to find such a solution? Does Mathematica or Matlab has nice built in routines for this kind of thing?

For your further information I am interested only in $n(z)$ on $[1,\infty]$, and it is required to be between 0 and 1, so those integrals lie between 0 and 1, so $n(z)$ should be monotonically increasing. It seems like it should be well behaved, but on the face of it Mathematica doesn't like it.


Background: it is the steady state of a dynamic system in a network; think epidemic diffusion. $n(z)$ is the infection rate among agents with z links, and in the steady state it is

$ n(z) = (g + (1-g)t v z )/(g + r + (1-g) t v z ), $ where $g$ is the new infection rate, $r$ is the cure rate,and $v$ is the new infections from your peers in a networks, and $f(z)$ is the distribution of that networks; the proportion of peers withg $z$ links. One such distribution often of interest is power-law, which is

$ f(z) = 2 z^{-3} $ and t is the infection rate among a random peer, given by $ t = \int_1^{\infty} z n(z) f(z)dz, $ which, along with a paramtetrization of the above rates, gave the problem above.

Other distributions of interest are geometric, where $f(z) = Log(4) 2^{-z}$. So we are solving for the steady state level of infection among guys with z peers.

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Well you know how $n(z)$ looks like. Note that $\displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz$ is just a number. Call it $a$.

So now we have

$n(z) = \frac{0.2+0.24 a z}{0.5+0.24 a z} = 1 - \frac{0.3}{0.5+0.24 a z}$

where $a = \displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz$.

Use the above equation to get an equation solely in $a$ which you can solve for numerically.

I have done the solving part below.

$a = \displaystyle \int_{1}^{\infty} \frac{n(z)}{z^2} dz = \int_{1}^{\infty} \frac{1}{z^2} dz - \int_{1}^{\infty} \frac{0.3}{0.5 z^2 + 0.24 a z^3} dz = 1 - \int_{1}^{\infty} \frac{15}{25 z^2 + 12 a z^3} dz$

$a = \frac{2}{5} + \frac{36}{125}a \log(\frac{25}{12a}+1)$

You can now try to solve this numerically or plug it in mathematica as I did, which gives me $a \approx 0.673338$.

As you expected, $a \in (0,1)$.

So the function now is $n(z) \approx 1 - \frac{15}{25+8.080056z} = \frac{10+8.080056z}{25+8.080056z}$

and the function $n(z) \in (0,1)$, $\forall z \in \mathbb{R}^+$. In fact when $z \in [1,\infty)$, $n(z) \in (\frac{6}{11},1)$.

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    @Dennis: Thanks for throwing light on the general problem in the question.2011-02-10