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Green'S Function For The Three-Variable Laplace Equation

Right. Another equation. As if the universe isn't complicated enough already. You want to understand how a system responds to a point source, in three dimensions, using the Laplacian. Fine. Don't expect me to hold your hand.

Partial Differential Equations

Let's cut to the chase. This article is about the Green's function for the Laplacian in three variables. Think of it as the universe's most precise answer to a very specific, very localized disturbance. It's what you need when you're dealing with systems described by Poisson's equation, which, in its most basic form, looks like this:

∇2u(x)=f(x)\nabla^2 u(\mathbf{x}) = f(\mathbf{x})

Here, ∇2\nabla^2 is the Laplace operator in R3\mathbb{R}^3. It's the operator that describes how something changes over space. f(x)f(\mathbf{x}) is the "source term" – the disturbance, the point of origin. And u(x)u(\mathbf{x})? That’s the system's response, the ripple effect.

Because ∇2\nabla^2 is a linear differential operator, the solution u(x)u(\mathbf{x}) for any given source f(x)f(\mathbf{x}) can be constructed by summing up the responses to each individual point source. This is where the Green's function, G(x,x′)G(\mathbf{x}, \mathbf{x'}), comes in. It tells you the system's reaction at point x\mathbf{x} to a unit point source located at x′\mathbf{x'}. Mathematically, it's defined by:

∇2G(x,x′)=δ(x−x′)\nabla^2 G(\mathbf{x}, \mathbf{x'}) = \delta(\mathbf{x} - \mathbf{x'})

where δ(x−x′)\delta(\mathbf{x} - \mathbf{x'}) is the Dirac delta function. It's a mathematical fiction, a spike of infinite height and zero width at x′=x\mathbf{x'} = \mathbf{x}, and zero everywhere else. It represents that single, isolated point source.

So, for a general source distribution f(x)f(\mathbf{x}), the total response u(x)u(\mathbf{x}) is just the integral of the Green's function multiplied by the source:

u(x)=∫G(x,x′)f(x′)dx′u(\mathbf{x}) = \int G(\mathbf{x}, \mathbf{x'}) f(\mathbf{x'}) d\mathbf{x'}

It's elegant, in a cold, calculating sort of way.

Motivation

Why would anyone care about this? Physics, mostly. Take electrostatics, for instance. The electric field E\mathbf{E} is the negative gradient of the electric potential ϕ(x)\phi(\mathbf{x}). And Gauss's law tells us that the divergence of the electric field is proportional to the charge density ρ(x)\rho(\mathbf{x}):

E=−∇ϕ(x)\mathbf{E} = -\nabla \phi(\mathbf{x}) ∇⋅E=ρ(x)ε0\nabla \cdot \mathbf{E} = \frac{\rho(\mathbf{x})}{\varepsilon_0}

Combine these, and you get Poisson's equation for the electric potential:

−∇2ϕ(x)=ρ(x)ε0-\nabla^2 \phi(\mathbf{x}) = \frac{\rho(\mathbf{x})}{\varepsilon_0}

Now, imagine you have a single point charge qq sitting at x′\mathbf{x'}. Its charge density is qδ(x−x′)q \delta(\mathbf{x} - \mathbf{x'}). The equation becomes:

−ε0q∇2ϕ(x)=δ(x−x′)-\frac{\varepsilon_0}{q} \nabla^2 \phi(\mathbf{x}) = \delta(\mathbf{x} - \mathbf{x'})

If you can find the Green's function for the operator −ε0q∇2-\frac{\varepsilon_0}{q} \nabla^2, you can then find the potential ϕ(x)\phi(\mathbf{x}) for any charge distribution ρ(x)\rho(\mathbf{x}) by integrating:

ϕ(x)=∫G(x,x′)ρ(x′)dx′\phi(\mathbf{x}) = \int G(\mathbf{x}, \mathbf{x'}) \rho(\mathbf{x'}) d\mathbf{x'}

It's a way to build complex solutions from simple, fundamental responses. Like understanding a whole orchestra by analyzing the sound of a single violin.

Mathematical Exposition

The free-space Green's function for the Laplace operator in three dimensions is remarkably simple. It's essentially the reciprocal of the distance between the two points, scaled by a constant. It's often called the "Newton kernel" or "Newtonian potential", which should tell you something about its fundamental nature.

The solution to ∇2G(x,x′)=δ(x−x′)\nabla^2 G(\mathbf{x}, \mathbf{x'}) = \delta(\mathbf{x} - \mathbf{x'}) is:

G(x,x′)=−14π∣x−x′∣G(\mathbf{x}, \mathbf{x'}) = -\frac{1}{4\pi |\mathbf{x} - \mathbf{x'}|}

where ∣x−x′∣|\mathbf{x} - \mathbf{x'}| is the Euclidean distance between points x\mathbf{x} and x′\mathbf{x'}.

The algebraic expression for the distance itself, in Cartesian coordinates, is:

1∣x−x′∣=[(x−x′)2+(y−y′)2+(z−z′)2]−1/2\frac{1}{|\mathbf{x} - \mathbf{x'}|} = \left[\left(x-x'\right)^{2}+\left(y-y'\right)^{2}+\left(z-z'\right)^{2}\right]^{-{1}/{2}}

This might seem straightforward, but there are many ways to expand this. One of the most famous is the Laplace expansion, which expresses the reciprocal distance in terms of Legendre polynomials Pl(cos⁡γ)P_l(\cos \gamma). This expansion is particularly useful when you switch to spherical coordinates (r,θ,φ)(r, \theta, \varphi).

1∣x−x′∣=∑l=0∞r<lr>l+1Pl(cos⁡γ)\frac{1}{|\mathbf{x} - \mathbf{x'}|} = \sum_{l=0}^{\infty} \frac{r_{<}^{l}}{r_{>}^{l+1}}P_{l}(\cos \gamma)

Here, r<r_< and r>r_> are the smaller and larger of the radial distances rr and r′r', respectively. And γ\gamma is the angle between the vectors x\mathbf{x} and x′\mathbf{x'}, given by:

cos⁡γ=cos⁡θcos⁡θ′+sin⁡θsin⁡θ′cos⁡(φ−φ′)\cos \gamma = \cos \theta \cos \theta' + \sin \theta \sin \theta' \cos(\varphi - \varphi')

This is how you decompose the potential into different modes, each with a specific angular dependence. It’s like breaking down a complex shape into simpler spherical harmonics.

The free-space Green's function can also be expressed in other coordinate systems. For example, in circular cylindrical coordinates, it involves Legendre functions of the second kind, Qm−1/2(χ)Q_{m-1/2}(\chi), which are known as toroidal harmonics. This expression is derived when considering situations with axial symmetry.

1∣x−x′∣=1πRR′∑m=−∞∞eim(φ−φ′)Qm−12(χ)\frac{1}{|\mathbf{x} - \mathbf{x'}|} = \frac{1}{\pi \sqrt{RR'}} \sum_{m=-\infty}^{\infty} e^{im(\varphi - \varphi')} Q_{m-\frac{1}{2}}(\chi)

where χ=R2+R′2+(z−z′)22RR′\chi = \frac{R^2 + R'^2 + (z-z')^2}{2RR'}.

There are also integral representations. For instance, in cylindrical coordinates, it can be expressed as an integral Laplace transform over a variable kk, involving the order-zero Bessel function of the first kind, J0J_0:

1∣x−x′∣=∫0∞J0(kR2+R′2−2RR′cos⁡(φ−φ′))e−k(z>−z<) dk\frac{1}{|\mathbf{x} - \mathbf{x'}|} = \int_{0}^{\infty} J_{0}\left(k\sqrt{R^{2}+{R'}^{2}-2RR'\cos(\varphi -\varphi ')}\right) e^{-k(z_{>}-z_{<})}\,dk

where z>z_> and z<z_< are the larger and smaller of the zz and z′z' coordinates.

Another integral form, using a Fourier cosine transform and the order-zero modified Bessel function of the second kind, K0K_0:

1∣x−x′∣=2π∫0∞K0(kR2+R′2−2RR′cos⁡(φ−φ′))cos⁡[k(z−z′)] dk\frac{1}{|\mathbf{x} - \mathbf{x'}|} = \frac{2}{\pi} \int_{0}^{\infty} K_{0}\left(k\sqrt{R^{2}+{R'}^{2}-2RR'\cos(\varphi -\varphi ')}\right) \cos[k(z-z')]\,dk

These integral forms are useful for certain types of problems, especially those with translational symmetry in one or more directions.

Rotationally Invariant Green's Functions for the Three-Variable Laplace Operator

The Green's function for the Laplace operator in three dimensions can be expanded in various coordinate systems that allow for the separation of variables. These are systems where the Laplace equation can be broken down into simpler, ordinary differential equations. The key ones include:

Each of these coordinate systems provides a different basis for representing the Green's function, tailored to specific geometries and boundary conditions. It’s a toolbox, really. You pick the tool that fits the shape of the problem.

See Also