A nonequidistant mesh

Let us suppose, that the spatial steps fulfill the following rule:

$$\triangle x_i=\alpha\triangle x_{i-1}.$$

If $\alpha=1$ the mesh is said to be equidistant. Let us now calculate the first derivative of the function $u(x)$ of the second-order accurance.

$$u(x+\alpha\triangle x)=u(x)+\alpha\triangle x\frac{\partial u}{\p... ... x^2}+\frac{(\alpha\triangle x)^3}{3!} \frac{\partial^3 u}{\partial x^3}+\ldots$$ (1.13)

Adding the last equation with Eq. (1.4) multiplied by $\alpha$ one obtains the expression for the second derivative

$$\frac{\partial^2 u}{\partial x^2}=\frac{u(x+\alpha\triangle x)-(1... ...triangle x)}{\frac{1}{2}\alpha(\alpha+1)\triangle x^2}+\mathcal{O}(\triangle x)$$ (1.14)

After substitution of the last equation to Eq. (1.4) one obtains

$$\boxed{\frac{\partial u}{\partial x}=\frac{u(x+\alpha\triangle x)... ...ha^2 u(x-\triangle x)}{\alpha(\alpha+1)\triangle x}+\mathcal{O}(\triangle x^2)}$$ (1.15)