Simple functional equation: $\frac 1 2 [\alpha(x - 1) + \alpha(x + 1)]$, $\alpha(0) = 1$, $\alpha(m) = 0$

Question

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

I know it has a linear solution $\alpha(x) = ax + b$ 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?

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Let's consider $\ds{\alpha\pars{x + 1} - 2\alpha\pars{x} + \pars{1 - \epsilon^{2}}\alpha\pars{x - 1} = 0}$ where $\epsilon > 0$. Later on, we'll recover the original equation in the limit $\ds{\epsilon \to 0^{+}}$. Solutions are $\ds{\propto \beta^{x}}$ such that $\ds{\beta^{2} - 2\beta + \pars{1 - \epsilon^{2}} = 0}$ wich leads to $$ \beta_{\pm} = {-\pars{-2} \pm \root{\pars{-2}^{2} - 4\pars{1 - \epsilon^{2}}} \over 2} =1 \pm \epsilon $$ The general solution is $\ds{\alpha\pars{x} = A\pars{1 + \epsilon}^{x} + B\pars{1 - \epsilon}^{x}}$. Let's impose the boundary conditions: $$\left\{% \begin{array}{rcrcl} A & + & B & = & 1 \\ \pars{1 + \epsilon}^{m}A & + & \pars{1 - \epsilon}^{m}B & = & 0 \end{array}\right. $$ Then $$ A = {\pars{1 - \epsilon}^{m} \over \pars{1 - \epsilon}^{m} - \pars{1 + \epsilon}^{m}} \,,\qquad B = -\,{\pars{1 + \epsilon}^{m} \over \pars{1 - \epsilon}^{m} - \pars{1 + \epsilon}^{m}} $$ and $$ \alpha\pars{x} = {\pars{1 - \epsilon}^{m}\pars{1 + \epsilon}^{x} - \pars{1 + \epsilon}^{m}\pars{1 - \epsilon}^{x} \over \pars{1 - \epsilon}^{m} - \pars{1 + \epsilon}^{m}} $$ With the limit $\epsilon \to 0^{+}$: \begin{align} &\alpha\pars{x} =\\[3mm]&\lim_{\epsilon \to 0^{+}} {-m\pars{1 - \epsilon}^{m - 1}\pars{1 + \epsilon}^{x} + x\pars{1 - \epsilon}^{m}\pars{1 + \epsilon}^{x - 1} - m\pars{1 + \epsilon}^{m - 1}\pars{1 - \epsilon}^{x} + x\pars{1 + \epsilon}^{m}\pars{1 - \epsilon}^{x - 1} \over -m\pars{1 - \epsilon}^{m - 1} - m\pars{1 + \epsilon}^{m - 1}} \\[3mm]&= {-2m + 2x \over -2m} \end{align} $$\color{#0000ff}{\large\alpha\pars{x} = 1 - {x \over m}}$$

Answer 5

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 $a = \mu/\lambda$