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In here the Laplacian $u_{xx}+u_{yy}$ is said to:

The Laplacian measures the degree by which the value at a point differs from the average of its neighbors.

whereas here the intuition departs from a more stringent set up: there is a scalar-valued function $f(x,y),$ from which the Laplacian is defined as the divergence of the gradient, $\nabla\cdot\nabla f.$ The intuition is:

A measure of how much of a minimum a point is in a scalar valued multivariate function.

The question is where the average appears in the first interpretation - is it truly the average as understood in statistics? Because the second derivative in different directions seems to directly speak about the curvature of the graph - there is no averaging.

And is the gradient also implied in the first interpretation? Or there is no real need for a gradient to understand the Laplacian?

Antoni Parellada
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  • Use the finite difference formulation to see the averaging. Using $h$ to denote the fixed step size in both directions \begin{align} 0 &= \Delta u \ &\to \frac{u^{i+1}{j} + u^{i-1}{j} + u^{i}{j+1} + u^{i}{j-1} - 4u^{i}{j}}{h^{2}} \ \implies u^{i}{j} &= \frac{u^{i+1}{j} + u^{i-1}{j} + u^{i}{j+1} + u^{i}{j-1}}{4} \end{align} – Matthew Cassell Oct 24 '19 at 13:23

1 Answers1

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Consider the square centered at $(0,0)$ with coordinates $(\pm h,\pm h)$. Average the value of $u$ at these four corners and measure the difference from $u(0,0)$: $$ \frac{u(h,h) + u(-h,h) + u(h,-h) + u(-h,-h)}{4} - u(0,0) = \frac{u(h,h) + u(-h,h) + u(h,-h) + u(-h,-h) - 4u(0,0)}{4} $$ by Taylor's theorem, \begin{align*} u(h,h) &= u(0,0) + h u_x(0,0) + h u_y(0,0) + \frac{1}{2}h^2(u_{xx}(0,0) + 2u_{xy}(0,0) + u_{yy}(0,0)) +R(h,h) \\\\ u(h,-h) &= u(0,0) + h u_x(0,0) - h u_y(0,0) + \frac{1}{2}h^2(u_{xx}(0,0) - 2u_{xy}(0,0) + u_{yy}(0,0)) +R(h,-h) \\\\ u(-h,h) &= u(0,0) - h u_x(0,0) + h u_y(0,0) + \frac{1}{2}h^2(u_{xx}(0,0) - 2u_{xy}(0,0) + u_{yy}(0,0)) +R(-h,h) \\\\ u(-h,-h) &= u(0,0) + h u_x(0,0) - h u_y(0,0) + \frac{1}{2}h^2(u_{xx}(0,0) + 2u_{xy}(0,0) + u_{yy}(0,0)) +R(-h,-h) \\\\ \end{align*} You see that if we add these four together, $$ \frac{u(h,h) + u(-h,h) + u(h,-h) + u(-h,-h) - 4u(0,0)}{4} = h^2 \left(u_{xx}(0,0)+u_yy(0,0)\right) + \text{remainders} $$ So $$ \lim_{h\to 0} \frac{u(h,h) + u(-h,h) + u(h,-h) + u(-h,-h) - 4u(0,0)}{4h^2} = \nabla^2 u(0,0) $$

Matthew Leingang
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