# Solution to non-homogeneous second order difference equation

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.

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