Solve the functional equation $f(4x(1-x))=\sin(\pi f(x))$ to find an invariant measure of a dynamical system $x_{n+1}=\sin(\pi x_{n})$

Question

During the investigation of my thesis I found the following problem:

I need find a injective function $f$ such that $$f(4x(1-x))=\sin(\pi f(x))\tag{1}$$ and $f(0)=0$ and $f(1)=1$.

Remark: I have tried to solve this equation but I have only reached a reformulation: $$ \frac{4(1-2y)f'(4y(1-y))}{\sqrt{1-f^{2}(4y(y-1))}}=\pi f'(y) .$$

You might think that this problem is not suitable for this site but for me it is important because it arises in trying to address the following situation.

The situation: This question is motivated by a problem of Ergodic Theory in my thesis, the problem is find invariant measure of dynamical system $$x_{n+1}=\sin(\pi x_{n}).\tag{2}$$ For this purpose it is sufficient to find its invariant density. In this sense, the idea is find a function $f$ injective such that satisfies (1) with $f(0)=0$ and $f(1)=1$. If that function existed then we have a change of coordinates given by $$x=f(y).$$ Therefore, substituting in (2) we have $f(y_{n+1})=\sin(\pi f(y_{n}))$, but by (1) we have $$ f(y_{n+1})=f(4 y_{n}(1-y_{n})) .$$ Since we assume that $f$ injective then $$y_{n+1}=4y_{n}(1-y_{n}) \tag{3}.$$ But we know the invariant probability density function of dynamical system (3) is the function $$\rho(x)=\frac{1}{4\sqrt{x(1-x)}}.$$ Therefore, the invariant probability density function $g$ of dynamical system (1) is the function $$g(x)=\left|\frac{df(x)}{dx}\right|\rho(x).$$ Therefore, the whole problem is reduced to finding $f$.

Answer 1

What you are asking for is a conjugacy of the dynamical systems $g(x)=4x(1-x)$ and $h(x)=\sin(\pi x)$. Since $g$ and $h$ are both full unimodal maps of $[0,1]$, there will exist such a conjugacy, but it is very unlikely that you will be able to write it down explicitly. Also, it will certainly be continuous, but likely not absolutely continuous.

You are asking that $f(g(x))=h(f(x))$. You are requiring that $f(0)=0$ and $f(1)=1$. Substituting $x=\frac 12$, you obtain $1=f(1)=h(f(\frac 12))$. Since $h^{-1}(1)=\{\frac 12\}$, it follows that $f(\frac 12)=\frac 12$. You can continue in a similar way to obtain the value of $f$ at further preimages of $\frac 12$. This defines $f$ on a dense set.

EDIT: As I mention above, it is very likely that $f$ is not absolutely continuous. David Speyer's argument below amplifies this. One key issue is that there are two important classes of invariant measure for maps like this: absolutely continuous invariant measures (ACIMs, that is, one that is absolutely continuous with respect to Lebesgue measure) and measures of maximal entropy (MMEs). From your post it seems as though you care about an ACIM for $h$.

For $g(x)$, the MME and ACIM coincide; this is something of a magical coincidence. A priori, it is highly unlikely that the ACIM and MME for $h(x)$ (assuming an ACIM exists) coincide. The conjugacy $f$ will always map the MME for $g$ to the MME for $h$; and so it seems very likely that this measure does not have a density.

However: I did a computation that I expected would give a strong indication that the two measures are different. Namely, I numerically computed the Lyapunov exponent for $h$ at a point that I chose. Assuming that there is an ACIM, this would be equal to the measure-theoretic entropy of the ACIM. If this entropy is not $\log 2$ (the entropy of the MME), this would prove that the ACIM and MME do not coincide (in fact the converse holds also). To my surprise, the Lyapunov exponent for $h$ was (numerically) very close to $\log 2$, so who knows?!

Answer 2

Following up on Anthony Quas's post, let $g(x) = 4x(1-x)$ and $h(x) = \sin(\pi x)$. The equations $g^{(n)}(x) = 1/2$ and $h^{(n)}(x) = 1/2$ each have $2^n$ roots in $[0,1]$. If we sort those roots in order as $y^n_1 < y^n_2 < \cdots < y^n_{2^n}$ and $z^n_1 < z^n_2 < \cdots < z^n_{2^n}$, we should have $f(y_i) = z_i$.

To see that there is unlikely to be a closed form solution, note that $y^n_1$ behaves like $c_1 4^{-n}$ for some constant $c_1$ and $z^n_1$ behaves like $c_2 \pi^{-n}$. So we must have $f(y) \sim c_3 y^{\log(\pi)/\log(4)}$ as $y \to 0$. Similar arguments should show that we have $f(y^n_i+\delta) - z^n_i \sim c \delta^{\log(\pi)/\log(4)}$ as $\delta \to 0$ for every $y^n_i$ (with the constant $c$ depending on $(n,i)$). This is not how a nice function behaves!

Here is a plot, and list of values, for $n=6$:

0.0002     0.0005
0.0014     0.0033
0.0038     0.0079
0.0074     0.0132
0.0121     0.0211
0.0181     0.0282
0.0252     0.0369
0.0335     0.0465
0.0429     0.0604
0.0534     0.0716
0.0650     0.0832
0.0776     0.0942
0.0912     0.1092
0.1058     0.1220
0.1214     0.1375
0.1379     0.1546
0.1552     0.1790
0.1734     0.1974
0.1924     0.2153
0.2121     0.2310
0.2325     0.2505
0.2536     0.2659
0.2752     0.2833
0.2974     0.3012
0.3201     0.3262
0.3432     0.3454
0.3666     0.3651
0.3904     0.3835
0.4145     0.4082
0.4388     0.4292
0.4632     0.4543
0.4877     0.4813
0.5123     0.5187
0.5368     0.5457
0.5612     0.5708
0.5855     0.5918
0.6096     0.6165
0.6334     0.6349
0.6568     0.6546
0.6799     0.6738
0.7026     0.6988
0.7248     0.7167
0.7464     0.7341
0.7675     0.7495
0.7879     0.769
0.8076     0.7847
0.8266     0.8026
0.8448     0.821
0.8621     0.8454
0.8786     0.8625
0.8942     0.878
0.9088     0.8908
0.9224     0.9058
0.9350     0.9168
0.9466     0.9284
0.9571     0.9396
0.9665     0.9535
0.9748     0.9631
0.9819     0.9718
0.9879     0.9789
0.9926     0.9868
0.9962     0.9921
0.9986     0.9967
0.9998     0.9995