Finite differences to ODE in polar coordinates

Question

I have an equation of finite differences as follows:

$$\frac{X_1(r+\epsilon)-X_1(r)}{ \frac{\epsilon~\beta}{2~r+\epsilon} } + \frac{X_1(r-\epsilon)-X_1(r)}{ \frac{\epsilon~\beta}{2~r-\epsilon} } = \frac{2X_1(r)-X(r)}{ \frac{1}{r~\epsilon} }$$

For $\epsilon\rightarrow0$ this equation is supposed to lead to the ODE:

$$ \frac{d^2X_1(r)}{dr^2} +\frac{1}{r}\frac{dX_1(r)}{dr} - \beta X_1(r) = -\frac{\beta}{2} X(r) $$

But I absolutely don't see how. Can you tell me which rule to follow to obtain this step? I have solved a similar rectangular problem in cartesian coordinates before without any problems, but I haven't found anything to achieve this step for a cylindrical problem in polar coordinates. Any advice?

The author were I got these equations from didn't made any further explanation on this step, so I assume it must be something simple, I just cannot see.

Thank you!

Answer 1

Taylor expansion:

\begin{align} \text{LHS} &= \frac{X_1(r+\epsilon)-X_1(r)}{ \frac{\epsilon~\beta}{2~r+\epsilon} } + \frac{X_1(r-\epsilon)-X_1(r)}{ \frac{\epsilon~\beta}{2~r-\epsilon} } \\ &= \frac{X_1(r) +\epsilon X_1'(r) +\frac 12 \epsilon^2 X_1''(r) +O(\epsilon^3) -X_1(r)}{ \frac{\epsilon~\beta}{2~r+\epsilon} } + \frac{X_1(r) -\epsilon X_1'(r) +\frac 12\epsilon^2 X_1''(r) +O(\epsilon^3)-X_1(r)}{ \frac{\epsilon~\beta}{2~r-\epsilon} } \\ &= \frac{\epsilon X_1'(r) +\frac 12\epsilon^2 X_1''(r) +O(\epsilon^3)}{ \frac{\epsilon~\beta}{2~r+\epsilon} } + \frac{-\epsilon X_1'(r) +\frac 12\epsilon^2 X_1''(r) +O(\epsilon^3)}{ \frac{\epsilon~\beta}{2~r-\epsilon} } \\ &= \frac 1 \beta \left( \frac{ X_1'(r) +\frac 12\epsilon X_1''(r) +O(\epsilon^2)}{ \frac{1}{2~r+\epsilon} } + \frac{- X_1'(r) +\frac 12\epsilon X_1''(r) +O(\epsilon^2)}{ \frac{1}{2~r-\epsilon} } \right)\\ &= \frac 1 \beta \left( (2r+\epsilon)( X_1'(r) +\frac 12\epsilon X_1''(r) +O(\epsilon^2) ) + (2r-\epsilon)(- X_1'(r) +\frac 12\epsilon X_1''(r) +O(\epsilon^2)) \right)\\ &= \frac 1 \beta \left( 2r \epsilon X_1''(r) + 2\epsilon X_1'(r) +O(\epsilon^2) \right)\\ &= \frac {2\epsilon} \beta \left( r X_1''(r) + X_1'(r) +O(\epsilon) \right)\\ \end{align}

Plugging back in,

\begin{align} \frac {2\epsilon} \beta \left( r X_1''(r) + X_1'(r) +O(\epsilon) \right) = r\epsilon (2X_1(r)-X(r)) \end{align}

Dividing through by $\frac {2r\epsilon}\beta$ and taking $\epsilon\to0$,

\begin{align} X_1''(r) + \frac 1r X_1'(r) - \beta X_1(r) = -\frac \beta 2 X(r) \end{align}

This method works.