Green's Function — Mathematics Knowledge System

Overview

A Green's function G(x, x') for a linear differential operator L with given boundary conditions is a function satisfying L G(x,x') = δ(x−x'), where δ is the Dirac delta. It represents the response of the system to a unit point source at x'. Once G is known, the solution to L u = f is given by the integral u(x) = ∫ G(x,x') f(x') dx', which reduces any forcing function to a superposition of point-source responses. Green's functions connect ODEs to integral equations and are fundamental in physics (electrostatics, quantum mechanics, wave propagation) and engineering.

Intuition

Think of G(x, x') as the deformation of a string at position x when a unit force is applied at position x'. The full deformation under a distributed force f(x) is then the superposition (integral) of these point-force responses. The Green's function encodes all the information about the operator and its boundary conditions — once you have it, solving L u = f for any forcing f reduces to a single integration. The symmetry G(x, x') = G(x', x) reflects the self-adjointness of L.

Formal Definition

Definition

For a self-adjoint operator L on [a,b] with homogeneous boundary conditions, the Green's function satisfies:

L_x G(x, x') = \delta(x - x')

Defining equation

G \text{ satisfies the homogeneous BCs in }x

Boundary conditions

u(x) = \int_a^b G(x, x')\,f(x')\,dx'

Solution via Green's function

G(x, x') = \begin{cases} c\,y_1(x)y_2(x') & x < x' \\ c\,y_1(x')y_2(x) & x > x' \end{cases}

Construction (y₁, y₂ = homogeneous solutions satisfying each BC)

c = \frac{-1}{p(x')W(y_1, y_2)(x')}

Constant from jump condition, W = Wronskian

Worked Examples

For x ≠ x′, −G″ = 0 so G is linear in x on each interval.
$G(x,x') = \begin{cases} Ax & x < x' \\ B(1-x) & x > x' \end{cases}$
Continuity at x=x′: Ax′ = B(1−x′).
$Ax' = B(1-x')$
Jump condition on G′: the jump in G′ at x=x′ must equal −1 (from −G″=δ). G′ changes from A to −B, so −B − A = −1, i.e., A + B = 1.
$-B - A = -1 \Rightarrow A + B = 1$
Solving: A = 1−x′, B = x′.
$A = 1-x',\quad B = x'$
Green's function: G(x,x′) = x(1−x′) for x<x′ and x′(1−x) for x>x′.
$G(x,x') = \begin{cases} x(1-x') & x \leq x' \\ x'(1-x) & x \geq x' \end{cases}$

Answer: G(x,x′) = x<(1 − x>) where x< = min(x,x′) and x> = max(x,x′).

Practice Problems

Difficulty 7/10

Verify the answer y(x) = (x−x³)/6 to −y″=x, y(0)=y(1)=0 by direct substitution.

Difficulty 8/10

Find the Green's function for the Laplacian −∇²u = f on ℝ³ (free-space Green's function).

Difficulty 9/10

Express the Green's function of a self-adjoint SL operator L in terms of its eigenfunctions yₙ and eigenvalues λₙ.

Common Mistakes

Common Mistake

The Green's function exists for any linear ODE.

G exists only when the homogeneous problem has only the trivial solution (the operator is invertible), i.e., λ=0 is not an eigenvalue.

Common Mistake

Green's functions are always smooth.

G(x,x′) has a jump discontinuity in its first derivative at x=x′ (from the delta function source) and is only piecewise smooth.

Quiz

What equation does a Green's function G(x,x′) satisfy?

L G = 0L G = δ(x−x′)G = ∫ f dxL G = f(x)

For a self-adjoint operator L, the Green's function satisfies:

G(x,x′) = −G(x′,x)G(x,x′) = G(x′,x)G(x,x′) = 0 for x≠x′G depends only on x−x′

Summary

G(x,x′) satisfies L G = δ(x−x′) with homogeneous BCs.
The solution to L u = f is u(x) = ∫ G(x,x′) f(x′) dx′.
G is symmetric: G(x,x′) = G(x′,x) when L is self-adjoint.
G is constructed from two homogeneous solutions satisfying each BC, joined at x=x′.
The spectral expansion G = ∑ yₙ(x)yₙ(x′)/(λₙ ‖yₙ‖²) connects G to SL theory.

References

WebsiteWikipedia — Green's function