Introduction to Green Functions

1Introduction

This lecture explains how to construct solutions of linear PDEs by superposing responses to point sources. Green functions also apply to time-evolution problems, where time and causality require additional treatment. Here the scope is restricted to elliptic boundary-value problems.

2Terminology and Definition

A Green functionGreen function is the point-source response determined by a linear differential operator $L$L, a domain, and boundary conditions. First suppose that $L$L with homogeneous boundary conditions is invertible, so every admissible input determines a unique solution. If the source point is $\xi$\xi and the observation point is $x$x, then

and $G$G satisfies the prescribed homogeneous boundary conditions in the variable $x$x. The symbol $\delta (x-\xi )$\delta(x-\xi) denotes a point source at $\xi$\xi. It is a distribution rather than an ordinary function and acts under an integral by

A distribution is an object defined by its action on smooth test functions with compact support, rather than by ordinary pointwise values.

3Strategy

Linearity permits a general input to be represented as a continuous superposition of point sources. Once a Green function is known, the solution can be expressed by an integral or, in translation-invariant settings, by a convolution.

4Dependence on the Boundary-Value Problem

A Green function depends not only on the equation but also on the domain and boundary conditions. Different boundary conditions for the same differential operator generally produce different Green functions. Moreover, if the homogeneous problem $Lv=0$Lv=0 has a nonzero solution, then $L$L is not directly invertible.

For example, let $\mathrm{\Omega }$\Omega be a bounded connected domain with $C^1$C^1 boundary. For the homogeneous Neumann problem for $-\mathrm{\Delta }u=f$-\Delta u=f on $\mathrm{\Omega }$\Omega, constant functions lie in the kernel. The divergence theorem gives the necessary solvability condition

and the solution is determined only up to an additive constant. The unmodified equation $-{\mathrm{\Delta }}_{x}G=\delta (x-\xi )$-\Delta_xG=\delta(x-\xi) is incompatible after integration over the domain. One instead uses

The term $-1/\left|\mathrm{\Omega }\right|$-1/|\Omega| makes the right-hand side have zero mean, and the final condition fixes the freedom to add a constant to $G$G. Thus a nontrivial kernel requires both a compatibility condition on the input and a normalization of the solution.

5A One-Dimensional Poisson Problem

On $0<x<1$0<x<1, fix an interior source point $\xi \in (0,1)$\xi\in(0,1) and consider

where $f$f is continuous. The Green function is a piecewise-linear function that vanishes at the boundary and whose derivative jumps at $x=\xi$x=\xi:

This function vanishes at $x=0,1$x=0,1, is continuous at $x=\xi$x=\xi, and its derivative jump represents $-G_{xx}=\delta (x-\xi )$-G_{xx}=\delta(x-\xi). Interpreting $f$f as a superposition of point sources gives

Indeed, applying the operator under the integral in the distributional sense gives

Since $G(0,\xi )=G(1,\xi )=0$G(0,\xi)=G(1,\xi)=0, the integral also satisfies the boundary conditions. For continuous $f$f, differentiating the explicit piecewise integral verifies the equation classically as well.

6Derivation of the Green Function

For $x\ne \xi$x\ne\xi, there is no point source and $-G_{xx}=0$-G_{xx}=0. Hence $G$G is linear on each side of $\xi$\xi. Write

Continuity at $x=\xi$x=\xi requires

Integrating $-G_{xx}=\delta (x-\xi )$-G_{xx}=\delta(x-\xi) from $\xi -\epsilon$\xi-\varepsilon to $\xi +\epsilon$\xi+\varepsilon gives the jump condition

Because $G_x(\xi -,\xi )=A$G_x(\xi-,\xi)=A and $G_x(\xi +,\xi )=-C$G_x(\xi+,\xi)=-C, this condition is $A+C=1$A+C=1. Solving it together with continuity gives

which yields the stated formula for $G$G.

7Symmetry

The explicit formula satisfies $G(x,\xi )=G(\xi ,x)$G(x,\xi)=G(\xi,x). This symmetry corresponds to the symmetry of $-d^2/d{x}^2$-d^2/dx^2 with Dirichlet boundary conditions under the integral pairing. For functions $v,w$v,w satisfying the same homogeneous boundary conditions, integration by parts gives

Green functions for general operators need not be symmetric.

8Example: A Constant Load

For $f\left(\xi \right)=1$f(\xi)=1,

This solution satisfies $-u^{\prime }\text{'}=1$-u''=1 and $u\left(0\right)=u\left(1\right)=0$u(0)=u(1)=0. Direct integration gives the same result, but the Green function provides one kernel that can be reused for any right-hand side $f$f.

9Convolution and Translation Invariance

On the whole space, translation invariance can make a Green function depend only on $x-\xi$x-\xi. The solution then takes the convolution form $G*f$G*f. In a domain with a boundary, one generally needs a two-variable function $G(x,\xi )$G(x,\xi). This distinction separates the practical roles of a fundamental solution and a Green function.

10Difference from a Fundamental Solution

A fundamental solutionFundamental solution is a point-source response on the whole space satisfying $L\mathrm{\Phi }=\delta$L\Phi=\delta. A Green function incorporates a domain and boundary conditions. In some cases it can be constructed by adding a homogeneous solution to a fundamental solution so that the required boundary conditions are satisfied. Green functions are therefore adapted to boundary-value problems.

11Related Lectures

This completes the discussion of qualitative properties and solution constructions for second-order linear PDEs. The next lecture returns to first-order transport equations and derives conservation laws from local flow.