Solve the functional equation $q \, \frac{f(x+1)}{f(x)}=\frac{h(x+1)}{h(x)}.$

Let $f(x),h(x)$ be two differentiate on $\mathbb{R}$ functions, $f(0)=h(0)=1$. Solve the functional equation $$q \, \frac{f(x+1)}{f(x)}=\frac{h(x+1)}{h(x)},$$ here $q$ is a constant.

For $q>0$ it is easy to find a solution: $f(x)=e^{ax}, h(x)=e^{bx}$ for some suitable $a,b.$

Questions. Are there another solutions for $q>0$? What about the case $q < 0?$

• Hint: You have also $q f(1)=h(1)$ so $q=\frac{h(1)}{f(1)}$ if $f(1)$ is non zero. – Khadija Mbarki Jul 20 '15 at 16:24
• For any function $g(x)$ you can find infinitely many functions such that $f(x+1)=g(x)f(x)$ – Michael Galuza Jul 20 '15 at 16:25
• Yes, @MichaelGaluza, I gave a full derivation of the expression. Probably not exactly what the OP was looking for, but I think it's necessary to mention, for the sake of completeness – frogeyedpeas Jul 20 '15 at 17:06
• @frogeyedpeas, My idea was: you can define $f(x)$ in any way on $[0, 1)$, for example, and continue $f(x)$ to $[1, 2)$ using $f(x+1)=f(x)g(x)$ – Michael Galuza Jul 20 '15 at 17:10
• That's correct and it's basically all I am doing as well, but for an exact symbolic closed form expression of that procedure one can use the techniques below (there might be a more elegant formulation out there but not that I personally am aware of). – frogeyedpeas Jul 20 '15 at 17:12

This can be transformed into a standard problem in linear finite differences. First we multiply both sides by $$f(x)$$ and divide by $$q$$ to arrive at

$$f(x+1) = \frac{1}{q} \frac{h(x+1)}{h(x)} f(x)$$

Then subtract $$f(x)$$ to yield

$$f(x+1) - f(x) = \left( \frac{h(x+1) - qh(x)}{qh(x)} \right)f(x)$$

Which can be written as

$$D_{1,x} f+ \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x) = 0$$

Whereas $$D_{h,x}f = \frac{f(x+h) - f(x)}{h}$$ the use of $$h=1$$ above naturally gives us $$f(x+1) - f(x)$$ as desired.

So given any $$h(x)$$, I mean any! we can find a find an f that satisfies above. The full proof requires the use of the theory of finite differences. First observe

$$D_{1,x}[2^x] = 2^{x+1} - 2^x = 2^x$$

Furthermore we can generalize that result to,

$$D_{1,x}[2^{g(x)}] = 2^{g(x+1)} - 2^{g(x)} = 2^{g(x)}(2^{g(x+1)-g(x)} - 1) = 2^{g(x)} (2^{D_{1,x}[g(x)]} - 1)$$

Furthermore, we create a product rule of sorts,

$$D_{1,x}[f(x)g(x)] = D_{1,x}[f(x)] g(x) + f(x)D_{1,x}[g(x)] + D_{1,x}[f(x)]D_{1,x}[g(x)]$$

To verify that last product rule, just expand each $$D_{1,x}$$ term into the defintion from above, and just algebraically simplify. One final tool to remark on is the idea of

$$D_{h,x}^{-1}[f]$$

Which is simply the function $$g(x)$$ such that $D_{h,x}[g] = f$$If you want a more rigorous treatment of how to define it, please mention in comments, I can give more intuition into it. For the remainder of this we will assume that it is a well defined operator, that can be taken of its argument. From here we wish to solve for $$f(x)$$ $$D_{1,x}[f(x)]+ \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x) = 0$$ We will utilize a pair of integration factors. Let us attempt to find two functions $$u_1(x), u_2(x)$$ such that $$D_{1,x}[u_1(x)f(x)] = u_2(x) \left( D_{1,x}[f(x)]+ \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x) \right)$$ After distributing we arrive at $$D_{1,x}[u_1(x)f(x)] = D_{1,x}[u_1(x)]f(x) + u_1(x)D_{1,x}[f(x)] + D_{1,x}[u_1(x)]D_{1,x}[f(x)]$$ $$u_2(x) \left( D_{1,x}[f(x)]+ \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x) \right) =$$ $$u_2(x) D_{1,x}[f(x)] + u_2(x) \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x)$$ So we can equate terms $$u_1(x)D_{1,x}[f(x)] + D_{1,x}[u_1(x)]D_{1,x}[f(x)] = u_2(x) D_{1,x}[f(x)]$$ $$D_{1,x}[u_1(x)]f(x) = u_2(x) \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x)$$ Then divide through by common terms to find: $$u_1(x) + D_{1,x}[u_1(x)] = u_2(x)$$ $$D_{1,x}[u_1(x)] = u_2(x) \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)$$ Subtract the second equation from the first to find $$u_1(x) = \left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right) u_2(x)$$ Now look at the second equation in the system of 2. We assume that $$u_2(x) = 2^{E(x)}$$ then it follows that $$D_{1,x}[u_1(x)] = D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]u_2(x) + \left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)D_{1,x}[u_2(x)] = u_2(x) \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)$$ Which means $$D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right] + \left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)2^{D_{1,x}[E(x)]} = \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)$$ Thus it follows, $$2^{D_{1,x}[E(x)]} = \frac{\left( \frac{qh(x) - h(x+1)}{qh(x)} \right) - D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]}{\left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)}$$ Giving us $$E(x) = D_{1,x}^{-1} \left[ \log_2 \left(\frac{\left( \frac{qh(x) - h(x+1)}{qh(x)} \right) - D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]}{\left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)} \right) \right]$$ Meaning $$u_2(x) = 2^{D_{1,x}^{-1} \left[ \log_2 \left(\frac{\left( \frac{qh(x) - h(x+1)}{qh(x)} \right) - D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]}{\left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)} \right) \right]}$$ $$u_1(x) = \left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)2^{D_{1,x}^{-1} \left[ \log_2 \left(\frac{\left( \frac{qh(x) - h(x+1)}{qh(x)} \right) - D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]}{\left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)} \right) \right]}$$ And since $$D_{1,x}[u_1(x)f(x)] = u_2(x) \left( D_{1,x}[f(x)]+ \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x) \right) = 0$$ It follows that $$u_1(x)f(x) = C$$ (C is the finite difference integration constant) and from here $$f(x) = \frac{C}{u_1(x)}$$ Giving us $$f(x) = \frac{C}{1 - \frac{qh(x) - h(x+1)}{qh(x)})}2^{-D_{1,x}^{-1} \left[ \log_2 \left(\frac{\left( \frac{qh(x) - h(x+1)}{qh(x)} \right) - D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]}{\left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)} \right) \right]}$$ So in short, name your h(x), ANY h(x), and that freakish clusterf--- of an expression will give you the f(x) that satisfies your problem. Let g(x)=e^{ax}\frac{f(x)}{h(x)}. Then the functional equation Yields:$$\frac{q}{e^{a}}g(x+1)=g(x)$$Pick a so that e^a=|q|. Then$$g(x+1)= sgn(q) g(x)$$Now, lets work backwards. If q>0 Pick any function g which is nonvanishing, differentiable and periodic with period one. There are many such functions. Let f(x) be any differentiable non-vanishing function. Then$$h(x)=e^{x \ln q} \frac{f(x)}{g(x)} =q^x \frac{f(x)}{g(x)}$$satisfies the given relation, and we seen above that any solution is of this form. If q<0 Pick any function g which is nonvanishing, differentiable and g(x+1)=-g(x). There are many such functions. Let f(x) be any differentiable non-vanishing function. Then$$h(x)=e^{x \ln q} \frac{f(x)}{g(x)} =q^x \frac{f(x)}{g(x)}$$satisfies the given relation, and we seen above that any solution is of this form. • what is the general solution of the equation$g(x+1)=g(x)?$– Leox Jul 20 '15 at 16:34 • @Leox The differentiable solutions to this equation are exactly the same functions as the set of differentiable functions on the circle$\mathbb R/\mathbb Z\$. This is a simple characterization, but also pretty vague, I doubt there is any better description. – N. S. Jul 20 '15 at 16:38