Find a pair of analytic functions?

Question

For $ 0 \leq x $ , Find Pairs of analytic functions $f,g$ such that 
 $$f(x+1) = 2 f(x)^2 + 3 g(x)^2 + 4 f(x) + 5 g(x) + 6$$

$$g(x+1) = f(x)^2 + 7 g(x)^2 + 8 f(x) + 9 g(x) + 10$$

hold simultanously.

I know some stuff about complex dynamics and I know people who know alot more than me about it.

But for these types of equations I am stuck and so are they.

How to handle this ?

Answer 1

Suppose you have one of the constant solutions. In this case, they are complex numbers, and there are four constant solutions. Then pick one of the solutions and move the fixed points to zero. Then temporarily ignore the $()2$ terms assuming they don't matter when f and g are small enough. Then near the fixed points of zero we have:

$$f(x+1)\approx a_1 f(x)+a_2 g(x);\;\;\;g(x+1)≈b_1 f(x)+b_2 g(x);$$

Next, assume for small x that $g(x) \approx k f(x)$. This holds exactly while we temporarily ignore the $()^2$ terms.

$$f(x+1)\approx a_1 f(x)+a_2 k f(x);\;\;\;g(x+1)≈b_1 f(x)+b_2 k f(x);$$ edit I assume both f and g have the same periodicity, and have the same $\lambda$ scaling factor. This seems to be the key to get a Koenig's style exponential solution. Otherwise, the approximation $g(x) \approx k f(x)$ does not hold even when the absolute value of f and g are small. If there were no $()^2$ terms, then the solution for $S_f$ and $S_g$ below would not have any $x^2$ or higher frequency terms. $$\lambda = a_1 +a_2 k = b_1 +b_2 k ;\;\;\; k = \frac{a_1-b_1}{b_2-a_2};\;\;\;\lambda=\frac{b_2 a_1 - b_1 a_2}{b_2-a_2}$$

Then we formally solve for the two Schroeder equations using $\lambda$ and $k$ from above, where $\lambda$ is the multiplier at the pair of fixed points. This works so long as $|\lambda| \ne 1$ although well behaved irrational multipliers with $|\lambda|=1$ work too. $$S_f(x) = x + c_2 x^2 + c_3 x^3 + ... + c_n x^n;\;\;\;S_g(x) = k x + d_2 x^2 + d_3 x^3 + ... + d_n x^n;\;\;\;$$ $$S_f(\lambda x) = a_1 S_f(x) + a_2 S_g(x) + 2 S_f(x)^2 + 3 S_g(x)^2 $$ $$S_g(\lambda x) = b_1 S_f(x) + b_2 S_g(x) + S_f(x)^2 + 7 S_g(x)^2 $$

I haven't worked through the details to show the joint formal solution exists, but I think it does. Then the desired solution to the OP's question is

$$f(x) = S_f(\lambda ^ x)\;\;\; g(x) = S_g(\lambda ^ x)$$

Answer 2

There is also a Böttcher function solution, where we solve for the super-attracting fixed point at infinity.

The Op wrote, "For $0\le x$ , Find Pairs of analytic functions f,g..." Since the equations for f and g both have real coefficients, one can guess that the Op was interested in a pair of solutions for f(x) and g(x) where f and g are real valued for all $x \ge 0$ and with a branch singularity at x=0.

$$f(x+1) = 2 f(x)^2 + 3 g(x)^2 + 4 f(x) + 5 g(x) + 6$$ $$g(x+1) = f(x)^2 + 7 g(x)^2 + 8 f(x) + 9 g(x) + 10$$

f and g grow rapidly due to the $()^2$ terms in the iteration definition; if f and g are large enough, we can ignore the other terms.

$$f(x+1) \approx 2 f(x)^2 + 3 g(x)^2;\;\;\; g(x+1) \approx f(x)^2 + 7 g(x)^2 $$

We assume that for large enough values of x, $\frac{f(x)}{g(x)} \approx k$, and we solve for k which results in a cubic equation. We solve for the real root of the cubic.

$$\frac{2 x^2 k^2 + 3 x^2}{x^2 k^2 + 7x^2} \approx k;\;\;\;k^3 - 2k^2+7k-3 =0;\;\;\;k \approx 0.4783035792338415$$

Using the value of "k", one can see that for large enough x, one expects $$k=\lim_{x \to \infty}\frac{f(x)}{g(x)}\;\;\;\;k_f=\lim_{x \to \infty}\frac{f(x+1)}{f(x)^2}\;\;\;\;k_g=\lim_{x \to \infty}\frac{g(x+1)}{g(x)^2} $$ $$f(x+1) \approx (2+\frac{3}{k^2}) \cdot f(x)^2 \approx k_f f(x)^2; \;\;\; k_f = 2+\frac{3}{k^2} \approx 15.1133602752610 $$ $$g(x+1) \approx (k^2+7) \cdot g(x)^2 \approx k_g g(x)^2; \;\;\; k_g = k^2+7 \approx 7.228774313907904 $$

At this point, I wrote a pari-gp program to solve for the inverse Böttcher for both f and g simultaneously. There were a number of steps involved, including scaling the inverse Böttcher solution for g(x) by $\frac{1}{k_g}$ and f by $\frac{1}{k_f}$ so that iterating them was in correspondence with iterating $x \mapsto x^2$, and then solving a simultaneous equation for each term in the formal solutions for the two inverse Böttcher equations to meet the defining equation for f and g. This involved using a Laurent series form for the equations with not too complicated algebra. In the iteration sequence to solve for the formal Laurent series terms, we use $$\text{inv_Böttcher}(x) = x + \sum_{n=0}^{\infty}\frac{a_n}{x^n}$$ $$f(n)=\frac{1}{k_f} \cdot \text{iBot}_f(x) \;\;\;\;f(n+1)=\frac{1}{k_f} \cdot \text{iBot}_f(x^2)\;\;\;\; \lim_{n \to \infty}\frac{f(n+1)}{f(n)^2}=k_f $$ $$g(n)=\frac{1}{k_g} \cdot \text{iBot}_g(x)\;\;\;\;g(n+1)=\frac{1}{k_g} \cdot \text{iBot}_g(x^2)\;\;\;\;\lim_{n \to \infty}\frac{g(n+1)}{g(n)^2}=k_g $$

From the resulting formal inverse Böttcher equations, the final equation I got was as follows, using the Laurent form for the inverse Böttcher equations. $$ f(x) = \frac{1}{k_f} \cdot \text{iBot}_f ( 2^{2^x}) \;\;\;\; g(x) = \frac{1}{k_g} \cdot \text{iBot}_g ( 2^{2^x}) $$

{iBotf=  x +
        -14.23800755632954
+x^- 1* -158.0113892928823
+x^- 2* -1805.626428647533
+x^- 3* -37392.08169596394
+x^- 4* -760773.1216212630
+x^- 5* -17189496.23553171
+x^- 6* -402959276.8509896
+x^- 7* -9812253313.462323
+x^- 8* -245031661062.1403
+x^- 9* -6246976651418.704
+x^-10* -161871077950928.2
+x^-11* -4.250944833814836 E15
+x^-12* -1.128854573478302 E17
+x^-13* -3.026148106231979 E18
+x^-14* -8.178201525114337 E19
+x^-15* -2.225742730331238 E21
+x^-16* -6.094840298845475 E22
+x^-17* -1.678059938616247 E24
+x^-18* -4.642477162585317 E25
+x^-19* -1.289940424214221 E27
+x^-20* -3.598158651314261 E28  }
{iBotg=  x +
        -6.157484073390565
+x^- 1* -0.8654395111982224
+x^- 2*  296.3671434733492
+x^- 3*  3988.114589952671
+x^- 4*  78296.87377421175
+x^- 5*  1542445.253631584
+x^- 6*  34213050.67503477
+x^- 7*  794665103.1235704
+x^- 8*  19264187536.48924
+x^- 9*  479776586670.8254
+x^-10*  12210168660711.76
+x^-11*  315974412934511.7
+x^-12*  8.289473466684748 E15
+x^-13*  2.199501582092067 E17
+x^-14*  5.892299500249988 E18
+x^-15*  1.591507164556888 E20
+x^-16*  4.329313024645326 E21
+x^-17*  1.185028580363595 E23
+x^-18*  3.261518894366849 E24
+x^-19*  9.020430101630897 E25
+x^-20*  2.505693371213574 E27  }

Both f and g are real valued, and working backwards, one finds an approximately square root singularity. Here is a graph of "g" near its approximately square root singularity, between x=0 and x=0.1. Presumably, this is what the Op had in mind, when the Op said, "For 0 $\le$ x." Here, $x_0 \approx 2.2955185$ in the inverse Böttcher equations above. I also had to use Newton's method to get inverse solutions for smaller values of the functions near the square root sinularity at zero.