Welcome back. Today, we will explore a concept that is central to the study of linear partial differential equations: that of the fundamental solution. As its name suggests, a fundamental solution is a very special solution to a PDE that serves as a building block for more general solutions. In this blog, my goal is to derive the fundamental solution to Laplace’s equation,

-∆u = 0,

and see how it is used to solve Poisson’s equation,

-∆u = ƒ.

We will then wrap up by using this fundamental solution to justify the inverse square law that governs gravitational and electric fields in 3D space.

To begin, let’s consider Poisson’s equation -∆u = ƒ in ℝ^{n}. We are going to construct the solution u(x) by decomposing it into basis functions, and then using the superposition principle to write u as a weighted sum of those basis functions. Our choice of basis will also be formed using Dirac delta distributions, denoted by δ_{y}(x), which are informally known as “point masses.” In the following paragraphs, δ_{y}(x) will be a function of x and y, with the properties

i) δ_{y}(x) = 0 for y ≠ x,

ii) δ_{y}(x) = +∞ for y = x,

iii) **∫**_{ℝn}** **δ_{y}(x) dy = 1,

From these properties of the delta distribution, we can formally write

f(x) = ∫_{ℝn} f(y) δ_{y}(x) dy.

Now let’s suppose we have found a family of functions u_{y}(x) that solve

-∆_{x}u_{y}(x) = δ_{y}(x)

for each y in ℝ^{n}. (The small subscript “x” in the Laplacian emphasizes that we are taking the Laplacian with respect to the x variables). Then, we can build the full solution to Posson’s equation on the whole space by setting

u(x) = ∫_{ℝn} f(y) u_{y}(x) dy.

Indeed, the calculation

-∆u = ∫_{ℝn} f(y) [-∆_{x}u_{y}(x)] dy = ∫_{ℝn} f(y) δ_{y}(x) dy = f(x)

shows that the formula above is, at least formally, valid. To make this completely rigorous, we would need to use epsilons and deltas and do some real analysis. In the future blog post, I hope to cover this detailed derivation, but for now, let’s press on with the assumption that everything indeed works.

At this point, all that’s left to do is to find a formula for the u_{y}’s. By translation, we can express any u_{y} by shifting u_{0} = Φ, which solves

-∆Φ = δ_{0}.

Due to its importance, we will define the special function Φ to be the fundamental solution of Laplace’s equation (even though it technically solves Poisson’s equation with ƒ = δ_{0}). To find this Φ successfully, let’s make some assumptions. Since the Laplace operator is rotationally invariant (proof coming soon!) and δ_{0}(x) is radially symmetric, we expect Φ to be radially symmetric (ie. Φ = Φ(r), where r = |x|). For any r ≠ 0, we can say

∆Φ = (∂_{r }∂_{r} + (n-1)/r ∂_{r}) Φ = 0,

since (∂_{r }∂_{r} + (n-1)/r ∂_{r}) = ∆ for any purely radial function. This yields to the ordinary differential equation

Φ’’(r) + (n-1)/r Φ’(r) = 0.

If n = 1, the solution to this equation is a straight line, and our entire discussion is relatively trivial. Therefore, let’s take n ≥ 2, and use separation of variables to obtain

Φ’(r) = b r^{1-n},

where b is a constant of integration. Another integration of the equation above yields

Φ(r) = C_{2} log(r) for n = 2,

Φ(r) = C_{n} r^{2-n} for n ≥ 3,

where the C_{n}’s are constants that are carefully chosen to give Φ some nice integration properties that we will explore more in the future. In terms of the original x variables, the fundamental solution is

Φ(x) = C_{2} log |x| for n = 2,

Φ(x) = C_{n} |x|^{2-n} for n ≥ 3,

Now we can shift Φ by y to find that in general, u_{y}(x) = Φ(x-y), and that

u(x) = ∫_{ℝn} ƒ(y) Φ(x-y) dy.

So there it is! A *convolution* that solves our problems! To solve Poisson’s equation on ℝ^{n}, all we have to do is convolve our forcing function with the fundamental solution.

In the context of physics, u can represent an electric or a gravitational potential, and -∇u can represent an electric or gravitational field. Then, if we have a point charge or point mass at the origin, Gauss’ Law or Newton’s theory tells us that

-∆u = δ_{0}.

In this special setting in n = 3 dimensional space, we that have that

u(x) = Φ(x) = C_{3} |x|^{2-3}~ 1/r.

Another calculation then shows that

∇u ~ -1/r^{2},

which is precisely the inverse square law!

Next week, I look forward to exploring more physics and partial differential equations. Until then, please take care.

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