Derivative of Lattice Laplacian

Question

The lattice Laplacian is defined as,

$$ \nabla_L^2x_j \equiv \frac{x_{j+1} - 2x_j + x_{j-1}}{a^2} $$ where the lattice spacing, $a$, is a constant.

The derivative, with respect to $x_i$, then gives, $$ \frac{\partial}{\partial x_i} \nabla_L^2x_j = \frac{\delta_{i(j+1)} - 2\delta_{ij} + \delta_{i(j-1)}}{a^2} = \begin{cases} a^{-2}\ &\text{for}\ i=j+1\\ -2a^{-2}\ &\text{for}\ i=j\\ a^{-2}\ &\text{for}\ i=j-1 \end{cases} $$

I am not sure intuitively why this is the case. I expected the derivative of the laplacian to be a function rather than a piecewise constant?

Answer 1

The discrete Laplacian is a vector valued function of several variables: its argument is $(x_1,\dots,x_n)$, and the output is another vector.

The function is a linear one (Laplacian is a linear operator), so, its partial derivative with respect to any $x_i$ is a constant vector, specifically the vector in which three entries are $1/a^2$ or $-2/a^2$, and the rest are zeros. This is no more surprising than the derivative of $g(x)=2x$ being constant.

Informally, this means that changing the value at the $i$th lattice point affects the Laplacian only at that point and its neighbors, and describes how it affects them.

Answer 2

$\def\m#1{\left[\begin{array}{r}#1\end{array}\right]}\def\p#1#2{\frac{\partial #1}{\partial #2}}$You can approach this problem in a purely matrix form using the Up Shift Matrix.
For example, for $\,n=3$ $$U = \m{0&1&0\\0&0&1\\0&0&0}$$ Then the lattice vector could be written as $$\eqalign{ y &= \left(\frac{U-2I+U^T}{a^2}\right)x \;\doteq\; Lx \\ }$$ This is a linear equation and the gradient with respect to $x$ is simply $$\eqalign{ \p{y}{x} &= L \\\\ }$$ NB: If your lattice is circulant, then you may want to use a cyclic permutation instead of a shift, i.e.
$$U = \m{0&1&0\\0&0&1\\1&0&0}$$