I'm a beginner to difference equations. I have a difference equation which I want to solve:

$V(x)=x^{\alpha}+\beta(\pi e ^ {-\Delta} V(x e^ {\Delta})+\pi e ^ {\Delta} V(x e^ {-\Delta})+(1-2\pi) V(x))$

I made the substitution $y=log\;x$ and then got the following characteristic equation for the homogeneous part:

$2 \lambda = \beta (e^ {\Delta}+e^ {-\Delta}\lambda^2)$.

Suppose the roots are $\lambda_1,\lambda_2 $. But after that I do not know how to use the $\lambda$'s in the solution, because the equation is not in terms of variables at times $t,t+1,t-1$ etc but in terms of $y, y+ \Delta, y- \Delta$.

The solution has been as (I don't know how we get this):

$V(x)=\frac{x^{\alpha}}{1-\beta(\pi e ^ {\Delta(1-\alpha)} + \pi e ^ {-\Delta(1-\alpha)}+(1-2 \pi))} + C_1 x^{\frac{log \lambda_1}{\Delta}}+C_2 x^{\frac{log \lambda_2}{\Delta}}$,

where "$\lambda_1$ and $\lambda_2$ are the roots of the characteristic equation and $C_1$ and $C_2$ are constants of integration."

Any help is deeply appreciated.


Your equation: $$V(x)=x^{\alpha}+\beta(\pi e ^ {-\Delta} V(x e^ {\Delta})+\pi e ^ {\Delta} V(x e^ {-\Delta})+(1-2\pi) V(x))$$

It's worth pointing out that you can see a particular solution straight away. Look at the form of the two "off centre" terms, $$\pi e ^ {-\Delta} V(x e^ {\Delta})+\pi e ^ {\Delta} V(x e^ {-\Delta})$$

Clearly, $V(x)=Ax^\alpha$ is a good solution, since it gives you some $x^{\alpha}$ terms. So plug this in and determine the constant $A$. As it turns out, this will give you $$A=\frac1{1-\beta(\pi e ^ {\Delta(1-\alpha)} + \pi e ^ {-\Delta(1-\alpha)}+(1-2 \pi))}$$ which is as given in the answer.

What about the homogeneous part of the solution?

$$\begin{align}V(e^y)&=\beta(\pi e ^ {-\Delta} V(e^{y+\Delta})+\pi e ^ {\Delta} V(e^{y-\Delta})+(1-2\pi) V(e^y))\\ \text{Denote }\tilde V(y)=V(e^y)\implies\tilde V(y)&=\beta(\pi e ^ {-\Delta} \tilde V({y+\Delta})+\pi e ^ {\Delta} \tilde V({y-\Delta})+(1-2\pi) \tilde V(y)) \end{align}$$ Now when you usually see a difference equation of this form, you'd look for a solution of the form $\tilde V(y)=\mu^y$. Then you get $$1=\beta\pi e^{-\Delta}\mu^{\Delta}+\beta\pi e^{\Delta}\mu^{-\Delta}+\beta(1-2\pi)$$ As you noticed, this is going up in increments of $\Delta$ rather than $1$. To fix this, let $\lambda=\mu^{\Delta}$. Then $$\frac1\beta\lambda=\pi e^{-\Delta}\lambda^2+\pi e^{\Delta}+\beta(1-2\pi)\lambda$$ This has roots $\lambda_1,\lambda_2$. Taking some steps back, this means the solution to the homogeneous equation was

$$C_1{\mu_1}^y+C_2{\mu_2}^y=C_1{\lambda_1}^{\log x/\Delta}+C_2{\lambda_2}^{\log x/\Delta}=C_1 x^{\frac{\log \lambda_1}{\Delta}}+C_2 x^{\frac{\log \lambda_2}{\Delta}}$$

as required.

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  • $\begingroup$ Thank you for the solution! $\endgroup$ – Canine360 May 25 '19 at 23:32

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