How do I intuitively process what's physically going on with the Planar Laplace/Poisson equations?

Question

By definition the PDEs are used to model equilibrium phenomena, involving only spatial variables. Say for example I was given the two-dimensional Laplace equation,$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0$ whose solution is defined on the bounded, open domain $\Omega$ with either of the three types of boundary conditions(Dirichlet,Neumann or Robin) on $\partial \Omega$. How should I be physically interpreting the equation, boundary conditions, and solution? Given an equation with the time variable, such as the wave equation $\frac{\partial^2 u}{\partial t^2}-\frac{\partial^2 u}{\partial y^2}=0$, I can physically infer that we might be modeling a problem relating acceleration and curvature and whose solution provides us with the position along the x-axis at any given time.

Answer 1

You can think of the solution to the Poisson equation as the steady state solution of the heat equation, notice that they are similar, except in the Poisson/Laplace equations, $\frac{\partial u}{\partial t}$ equals zero; that is, solution no longer changes with time. It is what happens to the heat equation if you let it run for infinite time.

Answer 2

I will take a discrete analogy.

It is well known that the discrete analog to the second derivative $f''(a)$ when dealing with sampled (discretized) functions on a regular 1D grid is:

$$f''(a)\approx\dfrac{\dfrac{f(a+h)-f(a)}{h}-\dfrac{f(a)-f(a-h)}{h}}{h}$$

i.e.

$$\tag{1}f''(a)\approx\dfrac{f(a+h)-2f(a)+f(a-h)}{h^2}$$

where $h$ is the discretization step.

In particular, expressing that $f''(a)=0$ ($a$ is an inflection point) is equivalent, using (1), to

$$f(a)\approx\dfrac{f(a+h)+f(a-h)}{2}$$

which reads: $f''(a)$ is $0$ is equivalent to the fact that the central value is approximately the average of its neighboring values.

Let us now turn to 2D. Consider a 2 variables function $f(x,y)$, discretized as shown on the graphics below, i.e., with discretization steps $h$ (resp $k$) along the $x$ (resp. $y$) direction.

Using (1), a discrete analog to $\Delta f$ is, by taking $h=k$ for simpler computations:

$$\Delta f(x_0,y_0)\approx\dfrac{f(x_0+h,y_0)-2f(x_0,y_0)+f(x_0-h,y_0)}{h^2}+\dfrac{f(x_0,y_0+h)-2f(x_0,y_0)+f(x_0,y_0-h)}{h^2}$$

$$\tag{2}\Delta f(x_0,y_0)\approx\dfrac{f(x_0,y_0+h)+f(x_0,y_0-h)+f(x_0+h,y_0)+f(x_0-h,y_0)-4f(x_0,y_0)}{h^2}$$

Now, let us express that the Laplacian operator is zero:

$$\Delta f(x_0,y_0)=0\Longleftrightarrow$$ $$\Longleftrightarrow f(x_0,y_0)\approx\dfrac{1}{4}\left(f(x_0,y_0+h)+f(x_0,y_0-h)+f(x_0+h,y_0)+f(x_0-h,y_0)\right)$$

i.e., at such a point, the central value is the average of its neighboring values, as in the 1D case.

This representation connects the fact that the Laplacian is zero whenever we are at an equilibrium point.