3
$\begingroup$

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 ?.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – artemka Dec 21 '13 at 6:46
  • $\begingroup$ Can i use mean value theorem? So that $\frac{\alpha(x)−\alpha(x−1)}{1} = \alpha'(c) = const$ leads to an easy conclusion that $\alpha(x) = ax +b$? $\endgroup$ – artemka Dec 21 '13 at 6:57
  • $\begingroup$ Is $\alpha(x)$ a continuous function? $\endgroup$ – Alex Dec 21 '13 at 6:58
  • $\begingroup$ I think i'd prefer continuous solution, but there's nothing wrong with uncontinuous one. $\endgroup$ – artemka Dec 21 '13 at 7:09
  • $\begingroup$ I don't think this approach works for continuous functions. In fact I think recurrences are for discrete functions. $\endgroup$ – Alex Dec 21 '13 at 14:18
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I think my observation was wrong. Only thing I get is $\alpha(x)-\alpha(x-1) = \alpha(x+1) - \alpha(x)$ $\endgroup$ – artemka Dec 21 '13 at 9:13
  • $\begingroup$ @SurajM.S it doesn't say the function is necessarily continuous, so much as differentiable $\endgroup$ – Tim Ratigan Dec 21 '13 at 9:25
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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} $$\color{#0000ff}{\large\alpha\pars{x} = 1 - {x \over m}}$$

$\endgroup$
  • $\begingroup$ You should probably note that this is only true for discrete values of $x$ (i.e. $x\in \Bbb Z$) $\endgroup$ – Tim Ratigan Dec 21 '13 at 10:18
0
$\begingroup$

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 $a = \mu/\lambda$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.