---
title: "Green's Functions for Boundary Value Problems"
description: "For an invertible linear boundary value problem with homogeneous boundary conditions, a Green's function gives the response to a unit point source. Integrating it against the forcing gives the solutio"
canonical: https://lightmysky.com/learn/mathematics/greens-functions-for-boundary-value-problems-mt_2ctFrfKIgW
source: https://lightmysky.com/learn/mathematics/greens-functions-for-boundary-value-problems-mt_2ctFrfKIgW.md
retrieved: 2026-11-12
---

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# Green's Functions for Boundary Value Problems

For an invertible linear boundary value problem with homogeneous boundary conditions, a Green's function gives the response to a unit point source. Integrating it against the forcing gives the solution. For separated second-order problems, its two pieces come from homogeneous solutions satisfying their respective endpoint conditions.

Subject: Mathematics · Area: Differential Equations · Ages 21 to 22
Page: https://lightmysky.com/learn/mathematics/greens-functions-for-boundary-value-problems-mt_2ctFrfKIgW

## Ready when they can

- Constructs the Green's function for -c u'' on [0,1], with constant c>0 and stated homogeneous Dirichlet or mixed endpoint conditions
- Recovers the solution for a given forcing term as an integral against the Green's function
- Explains the jump condition on the derivative and where it comes from; checks the zero-total-forcing compatibility condition for -u'' on [0,1] with u'(0)=u'(1)=0

## Lesson: Build a Green's function for fixed boundaries

Sturm-Liouville problems used homogeneous equations with conditions at both endpoints. Now solve -u''=f on [0,1], with u(0)=u(1)=0. G(x,s) is the response at x to a unit point source at s, where 0<s<1. Away from s, solve -G_xx=0. Choose pieces meeting their respective endpoint conditions. Match their values at s and impose the source's slope jump.

The homogeneous pieces are straight lines: Ax for x<s and B(1-x) for x>s. Continuity gives As=B(1-s). Their x slopes are A and -B. A unit source for -u'' requires right slope minus left slope to equal -1, so A+B=1. Solve the two equations: A=1-s and B=s. Continuity alone would leave their scale undetermined.

*(drawing: Match values, impose the jump, then solve for A and B.)*

Thus G(x,s)=x(1-s) for 0<=x<=s, and G(x,s)=s(1-x) for s<=x<=1. Both expressions give s(1-s) at x=s. The endpoint values are G(0,s)=G(1,s)=0. To evaluate G, compare x with s before substituting. To differentiate in x, hold s fixed: the left slope is 1-s and the right slope is -s.

For continuous or piecewise continuous f, use u(x)=∫ from 0 to 1 of G(x,s)f(s) ds. Fix x and split at s=x: u(x)=(1-x)∫ from 0 to x of sf(s) ds + x∫ from x to 1 of (1-s)f(s) ds. This solves -u''=f where f is continuous. Use f(s), not f(x). At either endpoint every kernel value is zero, so the solution integral is zero.

**Example.** For f(s)=1, the first integral contributes (1-x)x^2/2. The second contributes x(1-x)^2/2. Adding gives u(x)=x(1-x)/2. At x=1/3, the contributions are 1/27 and 2/27, giving u(1/3)=1/9. Check: u''=-1 and both endpoint values are zero. Multiplying the forcing by a constant multiplies the solution by that constant.

*(drawing: Choose the kernel piece separately on each source interval.)*

A unit point source δ(x-s) has integral 1 across s. Integrating -G_xx=δ(x-s) gives -[G_x(s+,s)-G_x(s-,s)]=1. Here s+ and s- mean right and left limits. G stays continuous, while its x derivative jumps by -1. For -c u''=f with constant c>0 and the same homogeneous conditions, divide the kernel by c. Its derivative jump is -1/c.

Replace the right condition by u'(1)=0, keeping u(0)=0. The right homogeneous piece must now be constant. Write G=Ax for x<s and G=B for x>s. Continuity gives As=B. The jump gives 0-A=-1, so A=1 and B=s. Thus G=x for x<=s and G=s for x>=s. Changing this endpoint condition changes the kernel.

The two-zero-endpoint kernel is symmetric: G(x,s)=G(s,x). For x<=s, both sides equal x(1-s); exchange x and s for the other case. For piecewise forcing, integrate only where it is nonzero. If f(s)=1 on [0,a] and zero elsewhere, then for x>a every contributing source satisfies s<x. Hence u(x)=(1-x)∫ from 0 to a of s ds=(1-x)a^2/2.

*(drawing: Zero slopes at both ends cannot support nonzero constant forcing.)*

An ordinary inverse Green's function requires a unique solution. With u'(0)=u'(1)=0, integrating -u''=f requires ∫ from 0 to 1 of f(x) dx=0. Even when this holds, adding a constant gives another solution. The problem has no ordinary inverse. The kernels above use homogeneous boundary data. For nonzero data, first subtract a function meeting those data and adjust the forcing.

**Recap.** Choose pieces for the endpoint conditions, match their values, set the operator's slope jump, then integrate against the forcing.

*This lesson, question bank and flashcards were written for this dataset.*

## Practice

17 questions on this page, each with its working shown.

## Needs first

- [Sturm-Liouville Problems and Eigenfunction Expansions](https://lightmysky.com/learn/mathematics/sturm-liouville-problems-and-eigenfunction-expansions-mt_oqadAPaSsW)

## Optional connections

- [First-Order Linear Equations and the Integrating Factor](https://lightmysky.com/learn/mathematics/first-order-linear-equations-and-the-integrating-factor-mt_BXmt2pbWp8)