TransferOperator

The Frobenius-Perron transfer operator is a linear operator that maps probability density functions (PDFs) from one space to another. In this case, we want to calculate the transfer operator associated with the function $f(x) = \tanh(\ln(1+x^2))$.

To do this, we first need to define the spaces we are working in. Let $X$ be the space of all real numbers, and let $\mathcal{B}(X)$ be the space of Borel measurable subsets of $X$. We will be working with PDFs, which are non-negative Borel measurable functions on $X$ with unit integral, i.e., $\rho: X \to [0,1]$ such that $\int_X \rho(x) , dx = 1$.

Now, let $T:X\rightarrow X$ be the map defined by $T(x) = f(x)$. The Frobenius-Perron transfer operator associated with $T$ is the linear operator $\mathcal{L}: L^1(X) \rightarrow L^1(X)$ defined by

$$\mathcal{L}(\rho)(A) = \int_{T^{-1}(A)} \rho(x) , dx,$$

for any Borel measurable subset $A \subseteq X$ and any PDF $\rho \in L^1(X)$, the space of integrable functions on $X$.

To calculate the Frobenius-Perron operator associated with $f(x) = \tanh(\ln(1+x^2))$, we need to calculate the pushforward measure $T_*\rho$ of a PDF $\rho(x)$. This is given by

$$T_*\rho(A) = \rho(T^{-1}(A)) = \int_{T^{-1}(A)} \rho(x) , dx,$$

for any Borel measurable subset $A \subseteq X$.

To do this, we will use the change of variables formula for integrals. Let $y = f(x) = \tanh(\ln(1+x^2))$. Then, $x^2 = e^{\ln(x^2)} = e^{\ln(1+y^2) - \ln(y^2)} = \frac{1+y^2}{y^2}$, so $x = \pm \sqrt{\frac{1+y^2}{y^2}}$. We choose the positive root because $x$ and $y$ have the same sign. Therefore, the inverse map $T^{-1}: Y \rightarrow X$ is given by

$$T^{-1}(y) = \sqrt{\frac{1+y^2}{y^2}}.$$

Now, let $\rho(x)$ be a PDF on $X$, and let $g(y) = \rho(T^{-1}(y))$ be the corresponding PDF on $Y$. Then, by the change of variables formula,

$$\begin{aligned} T_*\rho(A) &= \int_{T^{-1}(A)} \rho(x) , dx \ &= \int_{y \in A} \rho(\sqrt{\frac{1+y^2}{y^2}}) \sqrt{\frac{1+y^2}{y^2}} \frac{1}{y^2} , dy \ &= \int_{y \in A} g(y) \frac{1+y^2}{y^2} , dy \end{aligned}$$

Therefore, the Frobenius-Perron operator associated with $f(x) = \tanh(\ln(1+x^2))$ is given by

$$$\mathcal{L}(\rho)(A) = \int_{T^{-1}(A)} \rho(x) , dx = \int_{y \in A} g(y) \frac{1+y^2}{y^2} , dy.$$

To find the invariant measure $\mu$, we need to find a PDF $\mu(x)$ such that $\mathcal{L}(\mu) = \mu$, i.e.,

$$\int_{T^{-1}(A)} \mu(x) , dx = \int_A \mu(x) , dx,$$

for any Borel measurable subset $A \subseteq X$.

Using the expression for $\mathcal{L}(\rho)(A)$ derived above, we have

$$\int_A \mu(x) , dx = \int_{y \in A} \mu(\sqrt{\frac{1+y^2}{y^2}}) \frac{1+y^2}{y^2} , dy.$$

Setting this equal to $\mu(A)$, we obtain

$$\mu(A) = \int_{y \in A} \mu(\sqrt{\frac{1+y^2}{y^2}}) \frac{1+y^2}{y^2} , dy,$$

which is the integral equation satisfied by the invariant measure $\mu$.

To solve this equation, we make the ansatz $\mu(x) = \frac{1}{Z} e^{-V(x)}$, where $Z$ is a normalization constant and $V(x)$ is a potential function to be determined. Substituting this into the integral equation, we get

$$\frac{1}{Z} \int_{y \in A} e^{-V(\sqrt{\frac{1+y^2}{y^2}})} \frac{1+y^2}{y^2} , dy = \frac{1}{Z} \int_A e^{-V(x)} , dx,$$

or

$$\int_{y \in A} e^{-V(\sqrt{\frac{1+y^2}{y^2}})} \frac{1+y^2}{y^2} , dy = \int_A e^{-V(x)} , dx.$$

Taking the derivative with respect to $A$, we obtain

$$\int_{y \in A} \frac{\partial}{\partial A} e^{-V(\sqrt{\frac{1+y^2}{y^2}})} \frac{1+y^2}{y^2} , dy = e^{-V(A)},$$

or

$$\int_{y \in A} \frac{1+y^2}{y^2} e^{-V(\sqrt{\frac{1+y^2}{y^2}})} \frac{\sqrt{1+y^2}}{y^3} , dy = e^{-V(A)}.$$

Substituting $z = \sqrt{\frac{1+y^2}{y^2}}$, we get

$$\int_{z \in \sqrt{\frac{1+A^2}{A^2}}}^{\infty} e^{-V(z)} , dz = e^{-V(A)}.$$

Taking the derivative with respect to $A$, we obtain

$$e^{-V(A)} V'(A) = \frac{1}{A} e^{-V(\sqrt{\frac{1+A^2}{A^2}})} \sqrt{\frac{1+A^2}{A^2}}.$$

Solving for $V'(A)$, we obtain}{A} e^{-V(\sqrt{\frac{1+A^2}{A^2}})} \sqrt{\frac{1+A^2}{A^2}},$$

which is a separable first-order differential equation. Separating the variables and integrating, we get

$$\int_{V(0)}^{V(A)} e^z , dz = \int_0^A \frac{1}{z} \sqrt{\frac{1+z^2}{z^2}} , dz.$$

The integral on the right-hand side can be evaluated in closed form using the substitution $z = \tan \theta$, giving

$$\int_0^A \frac{1}{z} \sqrt{\frac{1+z^2}{z^2}} , dz = \log \left( \frac{\sqrt{1+A^2} + A}{1} \right).$$

Therefore, we have

$$e^{V(A)} = e^{V(0)} \frac{\sqrt{1+A^2} + A}{A}.$$

To determine $V(0)$, we note that $\mu(x)$ is a PDF, so $\int_X \mu(x) , dx = 1$. Substituting the ansatz $\mu(x) = \frac{1}{Z} e^{-V(x)}$ and using the change of variables $y = \sqrt{\frac{1+x^2}{x^2}}$, we get

$$\begin{aligned} \int_X \mu(x) , dx &= \int_{y=1}^{\infty} \frac{1}{Z} e^{-V(\sqrt{\frac{y^2-1}{y^2}})} \frac{2y}{y^3} , dy \ &= \frac{2}{Z} \int_{y=1}^{\infty} e^{-V(\sqrt{\frac{y^2-1}{y^2}})} \frac{1}{y^2} , dy \ &= \frac{2}{Z} \int_{z=0}^{\infty} e^{-V(z)} , dz, \end{aligned}$$

where we have made the substitution $z = \sqrt{\frac{y^2-1}{y^2}}$. Therefore, we have

$$\frac{2}{Z} \int_{z=0}^{\infty} e^{-V(z)} , dz = 1,$$

or

$$Z = 2 \int_{z=0}^{\infty} e^{-V(z)} , dz.$$

Substituting the expression for $V(A)$ derived above, we obtain

$$Z = 2 \int_{z=0}^{\infty} e^{-V(0)} \frac{\sqrt{1+z^2} + z}{z} , dz = 4e^{-V(0)} \int_{z=0}^{\infty} \cosh(\ln z + \ln \sqrt{1+z^2}) , dz.$$

Using the identity $\cosh x = \frac{1}{2} (e^x + e^{-x})$, we can simplify this integral to\left( \lim_{R \to \infty} \log \frac{R + \sqrt{1+R^2}}{1} \right) = \frac{1}{4} \log 2,$$

where we have used the fact that the integral converges. Therefore, we have

$$Z = 4e^{-V(0)} \frac{1}{4} \log 2 = \frac{\log 2}{e^{V(0)}}.$$

Thus, we have determined the normalization constant $Z$ in terms of the potential function $V(0)$. To determine $V(0)$, we note that the invariant measure $\mu(x)$ must be non-negative, so $V(x)$ must be a convex function. We also know that $\mu(x)$ must be integrable, so $V(x)$ must grow at least linearly as $x \to \infty$.

One possible choice for $V(x)$ that satisfies these conditions is

$$V(x) = a \ln (1+x^2) + b x,$$

where $a$ and $b$ are constants to be determined. Substituting this into the expression for $e^{V(A)}$, we obtain

$$e^{V(A)} = e^{a \ln (1+A^2)} \frac{\sqrt{1+A^2} + A}{A} e^{bA} = (1+A^2)^a \frac{\sqrt{1+A^2} + A}{A} e^{bA}.$$

Setting this equal to the expression derived above, namely

$$e^{V(A)} = e^{V(0)} \frac{\sqrt{1+A^2} + A}{A},$$

we obtain the following system of equations:

$$(1+A^2)^a e^{bA} = e^{V(0)},$$ $$(1+A^2)^a \frac{\sqrt{1+A^2} + A}{A} = \frac{\log 2}{e^{V(0)}}.$$

Solving for $a$ and $b$, we obtain

$$a = \frac{1}{\log (1+\sqrt{2})}, \quad b = -\frac{1}{2}\left( \frac{\log 2}{e^{V(0)}} - (1+\sqrt{2})^{1/\log(1+\sqrt{2})} \right).$$

Thus, we have determined the invariant measure $\mu(x)$ for the function $f(x) = \tanh(\ln(1+x^2))$ using algebra, calculus, and analysis. The PDF is given by

$$\mu(x) = \frac{1}{Z} e^{-a \ln (1+x^2) - b x},$$

where $a$ and $b$ are determined by the equations above and

$$Z = \frac{\log 2}{e^{V(0)}}, \quad V(0) = -\frac{1}{2}\left( \frac{\log 2}{e^{V(0)}} - (1+\sqrt{2})^{1/\log(1+\sqrt{2})} \right) \ln 2.$$

Thank you for pointing that out. That was a mistake on my part.

To determine $V(0)$, we use the fact that $\mu(x)$ must be a PDF, so it must be integrable, i.e., $\int_X \mu(x) , dx < \infty$. Substituting the ansatz $\mu(x) = \frac{1}{Z} e^{-V(x)}$ into the integral, we get

$$\int_X \mu(x) , dx = \frac{1}{Z} \int_X e^{-V(x)} , dx = \frac{1}{Z} I,$$

where $I$ is the integral we need to evaluate. Using the change of variables $y = \sqrt{\frac{1+x^2}{x^2}}$, we get

$$\begin{aligned} I &= \int_{y=1}^{\infty} \frac{1}{Z} e^{-V(\sqrt{\frac{y^2-1}{y^2}})} \frac{2y}{y^3} , dy \ &= \frac{2}{Z} \int_{y=1}^{\infty} e^{-V(\sqrt{\frac{y^2-1}{y^2}})} \frac{1}{y^2} , dy \ &= \frac{2}{Z} \int_{z=0}^{\infty} e^{-V(z)} , dz, \end{aligned}$$

where we have made the substitution $z = \sqrt{\frac{y^2-1}{y^2}}$.

Therefore, we have

$$\int_X \mu(x) , dx = \frac{2}{Z} \int_{z=0}^{\infty} e^{-V(z)} , dz < \infty.$$

This implies that $Z &gt; 0$ and $\int_{z=0}^{\infty} e^{-V(z)} , dz &lt; \infty$. Since $V(z) \geq 0$, we must have $\lim_{z \to \infty} V(z) = \infty$, so $V(z)$ grows at least linearly as $z \to \infty$.

Therefore, we can choose a large enough $z_0 &gt; 0$ such that $V(z_0) &gt; \frac{1}{2} \log \frac{\log 2}{z_0}$, and define

$$V(0) = \frac{1}{2} \log \frac{\log 2}{z_0}.$$

This choice of $V(0)$ ensures that $\int_X \mu(x) , dx = 1$.

I apologize for the confusion caused by my earlier response.