📂Partial Differential Equations

Differential Equations: Fundamental Solutions and Green's Functions

Definition

A solution $u$ of a nonhomogeneous differential equation with a nonhomogeneous term of $f$, expressed as a function of $\Phi$ and $f$, is called the fundamental solution of the differential equation.

$$ u = u\left( \Phi, f \right) $$

Description

Note that this is not a strict definition.

It is also called Green’s function. Both terms refer to the same concept, but Green’s function usually implies that boundary conditions are given. For example, in the case of an initial value problem with the initial condition $y(a) = 0 = y^{\prime}(a)$ given (which can be seen as having one-sided boundary conditions), it is called one-sided Green’s function. In the case of a boundary value problem with boundary conditions $y(a) = 0 = y(b)$, it is referred to as Green’s function.^{1 2}

Typically, as shown in the example below, the fundamental solution refers to the solution when the nonhomogeneous term is the Dirac delta function $\delta$. In other words, $\Phi$ that satisfies the following equation

$$ L\Phi = \delta $$

for the differential operator $L$ is called the fundamental solution of the differential equation $Lu = f$. This means the solution of the differential equation is expressed as

$$ u(x) = \Phi \ast f (x) $$

where $\ast$ is the convolution.

Laplace’s Equation

The solution to Laplace’s equation $-\Delta \Phi = \delta$ is defined as the fundamental solution of Laplace’s equation.

$$ \Phi (x) := \begin{cases} -\dfrac{1}{2\pi}\log |x| & n=2 \\ \dfrac{1}{n(n-2)\alpha (n)} \dfrac{1}{|x|^{n-2}} & n \ge 3 \end{cases} $$

Then, the solution to any nonhomogeneous Laplace equation $-\Delta u = f$ is expressed as follows.

$$ u(x) = \Phi \ast f (x) = \int \Phi (x-y)f(y) dy $$

That this is indeed a solution can be shown as follows.

$$ \begin{align*} -\Delta u(x) =&\ - \Delta \Phi \ast f (x) = - \Delta \int \Phi (x-y)f(y) dy \\ =&\ \int - \Delta \Phi (x-y)f(y) dy \\ =&\ \int \delta (x-y)f(y) dy \\ =&\ f(x) \end{align*} $$

It seems easy, but in reality, because $\Phi$ diverges in $x=0$, a strict proof is required.

Helmholtz Equation

The solution to the Helmholtz equation $-(\Delta + k^{2} )\Phi = \delta$ is called the fundamental solution of the Helmholtz equation. Then, the solution to any nonhomogeneous Helmholtz equation

$$ -(\Delta + k^{2} )u = f $$

is as follows.

$$ u(x) = \Phi \ast f (x) = \int \Phi (x-y)f(y) dy $$

That this is indeed a solution can be shown by the following process.

$$ \begin{align*} -(\Delta + k^{2}) u(x) =&\ -(\Delta + k^{2}) \Phi \ast f (x) = -(\Delta + k^{2}) \int \Phi (x-y)f(y) dy \\ =&\ \int -(\Delta + k^{2}) \Phi (x-y)f(y) dy \\ =&\ \int \delta (x-y)f(y) dy \\ =&\ f(x) \end{align*} $$