# Simple functional equation

I have a simple functional equation: $$\alpha(x) = {1 \over 2}\left[\alpha\left(x - 1\right) + \alpha\left(x + 1\right)\right]\,, \qquad \alpha\left(0\right) = 1\,,\quad\alpha\left(m\right) = 0$$

I know it has a linear solution $\alpha\left(x\right) = ax + b$ but I don't have any idea how to prove that this is the only solution. Is there any way I can derive this solution from the initial equation ?.

Hint 1: $1=\frac{1}{2}+\frac{1}{2}$ Hint 2: $a=1 \cdot a = (\frac{1}{2}+\frac{1}{2}) \cdot a$

• i has derived that the function has constant difference on unit interval: $\alpha(x) - \alpha(x-1) = const$. Is linear function the only one that satisfies this property? – artemka Dec 21 '13 at 6:46
• Can i use mean value theorem? So that $\frac{\alpha(x)−\alpha(x−1)}{1} = \alpha'(c) = const$ leads to an easy conclusion that $\alpha(x) = ax +b$? – artemka Dec 21 '13 at 6:57
• Is $\alpha(x)$ a continuous function? – Alex Dec 21 '13 at 6:58
• I think i'd prefer continuous solution, but there's nothing wrong with uncontinuous one. – artemka Dec 21 '13 at 7:09
• I don't think this approach works for continuous functions. In fact I think recurrences are for discrete functions. – Alex Dec 21 '13 at 14:18

assuming $m$ to be an integer. from your observation $$\alpha (x)-\alpha (x-1)= C$$ where $C$ is a constant. differentiating both sides $$\alpha '(x)-\alpha '(x-1)= 0$$ from this we get the condition that either $\alpha '(x)$ is a periodic function with $1$ as its period or a constant. so $\alpha (x)$ is either a periodic or an equation representing a line.

but if you consider the initial conditions. if $\alpha (0)=1$ then $\alpha (m)=0$ is contradicting for a periodic function. therefore the resultant should represent an equation of line.

• I think my observation was wrong. Only thing I get is $\alpha(x)-\alpha(x-1) = \alpha(x+1) - \alpha(x)$ – artemka Dec 21 '13 at 9:13
• @SurajM.S it doesn't say the function is necessarily continuous, so much as differentiable – Tim Ratigan Dec 21 '13 at 9:25

In your last comment to date, you rightly observe that for $b(x)=a(x)-a(x-1)$ you get the equation

$$b(x+1)=b(x)$$

Its solution are all functions with period 1.

Now

$$a(x+n)=a(x+n-1)+b(x)=a(x+n-2)+2\,b(x)=...=a(x)+nb(x)$$

using the periodicity of $b$. This leads to the idea to consider

$$c(x)=a(x)-x\,b(x).$$

It satisfies the discrete dynamic

\begin{align} c(x+1)&=a(x+1)-(x+1)\,b(x+1)=a(x)+b(x+1)-(x+1)\,b(x+1)\\ &=a(x)-x\,b(x)=c(x), \end{align}

so it is again a periodic function with period $1$. The general solution has thus the form

$$a(x)=c(x)+x\,b(x)$$

with $b$ and $c$ any 1-periodic functions. This specializes to the linear solution in the case that $b$ and $c$ are constant functions.


• You should probably note that this is only true for discrete values of $x$ (i.e. $x\in \Bbb Z$) – Tim Ratigan Dec 21 '13 at 10:18

Another way could be re-writing the original problem

$$\cosh(\frac{d}{d\alpha})\alpha(x) = \cosh(\frac{d}{dx})\alpha(x)$$

Since $\cosh(x) = \frac{1}{2}(e^{x} + e^{-x})$ and $e^{\frac{d}{dx}}\alpha(x) = \alpha(x+1)$, $$\frac{1}{2}(e^{\partial_\alpha} + e^{-\partial_\alpha})\alpha(x) = \frac{1}{2}\left(\alpha(x) + 1 + \alpha(x) - 1\right) = \alpha(x) = \frac{1}{2}(e^{\partial_x} + e^{-\partial_x})\alpha(x) = \frac{1}{2}\left(\alpha(x+1) + \alpha(x-1)\right)$$ You can then cancel all like terms on the right in the first expression to arrive at

$\frac{d}{d\alpha}\alpha = \frac{d}{dx}\alpha = 1$, so $\alpha(x) = x + C$.

Another thought is that the original equation implies that $$\alpha(x) = \frac{1}{2}(\alpha(x+\lambda) + \alpha(x-\lambda))$$ is true for all integers $\lambda$. Also $\cosh(\mu\frac{d}{d\alpha})\alpha(x) = \alpha(x)$ for any $\mu\in\mathbb{R}$ This means that $$\cosh(\mu\frac{d}{d\alpha})\alpha(x) = \cosh(\lambda\frac{d}{dx})\alpha(x)$$ works just as well to describe the problem and so the general solution to $$\mu = \lambda\frac{d\alpha}{dx}$$ is $\alpha(x) = ax + b$ for some constants $b$ and $a = \mu/\lambda$