Green's Functions for Boundary Value Problems

Green's functions are the computational backbone of quantum mechanics and quantum field theory, where the propagator (Green's function of the Schrödinger or Klein–Gordon operator) encodes how a particle created at one spacetime point evolves to another; Feynman diagrams are literally built from products and convolutions of Green's functions. In electromagnetism and acoustics, the Green's function for the Helmholtz operator gives the radiation pattern of a point source (an antenna element or a point acoustic driver), and antenna arrays or reflector antennas are analyzed by integrating this kernel against the source distribution. In structural engineering, the beam-deflection Green's function (the influence function for a point load) is a standard tool for computing deflections under arbitrary distributed loads, directly generalizing Example 1. In probability and stochastic processes, the Green's function of a diffusion generator is essentially the expected occupation density of a stochastic process before it exits a domain, linking directly to Brownian motion and potential theory. In numerical analysis, Green's functions underlie boundary element methods, which discretize boundary integrals rather than the whole domain, using exactly the fundamental-solution-plus-correction idea of Example 2. Finally, the existence theory here — non-resonance $\Rightarrow$ unique bounded solution operator — is the ODE/PDE prototype for the general study of Well-Posedness of PDE Problems, where existence, uniqueness, and continuous dependence on data are examined systematically.

:::quiz[If a student solves $u''+u=f$ on $[0,\pi]$ with $u(0)=u(\pi)=0$ for arbitrary continuous $f$ using an "ordinary" Green's function formula, the error is...]

• No error; the formula always works for linear ODEs
• Ignoring that the homogeneous problem has a nontrivial solution $\sin x$, so no such $G$ exists and solvability requires $f\perp \sin x$ ✓
• The formula only fails if $f$ is negative somewhere
• The formula only fails if $p$ is not smooth

Ask the tutor: How does the Green's function construction change for singular Sturm–Liouville problems, such as Bessel's equation on $(0,\infty)$?

Think of the string again: pin it at $x=a$ and $x=b$, and pluck it with a unit point force at $x=s$. On each side of $s$ the string is unloaded, so it must be a straight line (or, for a general Sturm–Liouville operator, a solution of the homogeneous equation) satisfying the boundary condition on that side. At $x=s$ the string cannot tear, so the two pieces must meet continuously; but the point force produces a kink — a jump in slope proportional to the force divided by the tension $p(s)$. These four requirements (homogeneous away from $s$, correct boundary behavior, continuity, prescribed jump) pin down $G(x,s)$ completely, provided the unforced, unloaded string cannot itself sag into a nonzero shape satisfying both boundary conditions (i.e., the homogeneous problem is non-resonant). Once $G(x,s)$ is known, superposition does the rest. A distributed load $f(s)\,ds$ over $[s, s+ds]$ produces, by linearity, a deflection $G(x,s) f(s)\, ds$ at the observation point $x$. Summing over all point loads gives the total deflection as an integral, $u(x) = \int_a^b G(x,s) f(s)\, ds$. This is exactly analogous to how a discrete linear system $A\mathbf{u} = \mathbf{f}$ is solved by $\mathbf{u} = A^{-1}\mathbf{f}$: the Green's function is the integral kernel of the inverse operator $L^{-1}$.

:::svg[The Green's function as a "tent" solving $-u'' = \delta(x-s)$ on $[0,1]$ with $u(0)=u(1)=0$]

Example 1 — asymmetric boundary conditions and non-constant loading. Solve

Here $p\equiv 1$, $q\equiv 0$, $B_1[u]=u(0)$, $B_2[u] = u'(1)+u(1)$. Homogeneous problem check. $u''=0 \Rightarrow u = c_1 + c_2 x$. $B_1[u]=c_1=0$. Then $u=c_2 x$, and $B_2[u] = c_2 + c_2 = 2c_2 = 0 \Rightarrow c_2=0$. So only the trivial solution — the hypothesis of the Theorem holds, and $G$ exists uniquely. Building $y_1,y_2$. Take $y_1(x) = x$ (satisfies $y_1(0)=0$). Take $y_2(x) = 1 - x$; check $B_2[y_2] = y_2'(1) + y_2(1) = (-1) + 0 = -1 \ne 0$ — wrong choice, try again: we need $y_2'(1)+y_2(1)=0$. Try $y_2(x) = 2 - x$: then $y_2'(1)+y_2(1) = -1 + 1 = 0$. Good, so $y_2(x) = 2-x$. Wronskian. $W(x) = y_1 y_2' - y_1' y_2 = x(-1) - 1(2-x) = -x - 2 + x = -2$. Since $p\equiv1$, $c_0 = p\,W = -2$. Green's function.

Check continuity at $x=s$: both give $-s(2-s)/2$. ✓. Check jump: $\partial_xG(s^+,s) = -\frac{-s}{2}=\frac{s}{2}$... let's compute carefully. For $x>s$, $G=-\frac{s(2-x)}{2} = \frac{s x}{2} - s$; so $\partial_x G = \frac{s}{2}$. For $x<s$, $G = -\frac{x(2-s)}{2} = \frac{x(s-2)}{2}$; so $\partial_xG = \frac{s-2}{2}$. Jump $=\frac{s}{2} - \frac{s-2}{2} = \frac{2}{2}=1 = 1/p(s)$. ✓ (since $p\equiv1$). Solution.

First integral:

Second integral:

So

Combine: $-\frac{2x^3}{6} + \frac{x^4}{6} - \frac{x^4}{6} - \frac{x^3}{3} + \frac{x^3}{2} = -\frac{x^3}{3} - \frac{x^3}{3} + \frac{x^3}{2} = x^3\left(-\frac23 + \frac12\right) = x^3\left(-\frac16\right) = -\frac{x^3}{6}$. So $u(x) = -x^3/6 +$ nothing else? Let's recheck: the $x^4$ terms canceled entirely, leaving $u(x) = -\dfrac{x^3}{6}$. Direct check. If $u=-x^3/6$, then $u''=-x$. But we need $u''=x$! There's a sign slip — let's recheck by direct integration instead, as a cross-check on the whole computation. Direct solving: $u'' = x \Rightarrow u' = x^2/2 + A \Rightarrow u = x^3/6 + Ax + B$. $u(0)=0 \Rightarrow B=0$. $u'(1)+u(1) = (1/2+A) + (1/6+A) = 2/3 + 2A = 0 \Rightarrow A = -1/3$. So

This is the correct answer via direct integration; comparing to the Green's function computation, the sign error traces to the definition $c_0=p(s)W(s)$ used with the correct-sign jump — the jump condition requires $\partial_xG(s^+,s)-\partial_xG(s^-,s) = 1/p(s) = 1$ which we verified holds, and the construction is provably correct by the Theorem, so the arithmetic error is in the final integral evaluation. Redoing the second integral:

First integral again: $-\frac{x^3(2-x)}{6} = -\frac{2x^3}{6}+\frac{x^4}{6} = -\frac{x^3}{3}+\frac{x^4}{6}$. Sum: $\left(-\frac{x^3}{3}+\frac{x^4}{6}\right) + \left(-\frac{x}{3}+\frac{x^3}{2}-\frac{x^4}{6}\right) = -\frac{x}{3} + x^3\left(-\frac13+\frac12\right) + x^4(0) = -\frac{x}{3} + \frac{x^3}{6}.$ So correctly, $u(x) = \dfrac{x^3}{6} - \dfrac{x}{3}$, matching the direct-integration check exactly (the earlier combination step mis-copied a term). This agreement confirms the Green's function construction and the representation formula.

Example 2 — Poisson's equation on a disk via the method of images (sketch of the PDE analogue). Consider $-\Delta u = f$ in the unit disk $D=\{(x,y): x^2+y^2<1\}\subset\mathbb R^2$, with $u=0$ on $\partial D$. Write points as $\mathbf{x}, \mathbf{s} \in D$. The free-space fundamental solution of $-\Delta$ in two dimensions is

satisfying $-\Delta_{\mathbf x}\Phi(\mathbf x,\mathbf s) = \delta(\mathbf x - \mathbf s)$ but not vanishing on $\partial D$. To correct the boundary behavior, reflect $\mathbf s$ to its inverse point $\mathbf s^* = \mathbf s/|\mathbf s|^2$ (for $\mathbf s\ne 0$) and set

One verifies directly: (i) for $\mathbf x\ne \mathbf s$ in $D$, the correction term is harmonic in $\mathbf x$ (since $\mathbf s^*\notin D$), so $-\Delta_{\mathbf x} G = \delta(\mathbf x-\mathbf s)$ inside $D$; (ii) on $|\mathbf x|=1$, using $|\mathbf x - \mathbf s^*| = |\mathbf x||\mathbf s^{*}-\mathbf x/|\mathbf x|^2|\cdots$, a direct computation with $|\mathbf x|=1$ gives $|\mathbf x-\mathbf s^*| = |\mathbf s|^{-1}|\mathbf x - \mathbf s|$, so

Thus $G$ vanishes on $\partial D$ as required, and it is the Green's function for the Dirichlet problem on the disk, playing exactly the same role as the ODE Green's function: the solution of $-\Delta u = f$, $u|_{\partial D}=0$ is

This is the method of images construction underlying Laplace's Equation and Harmonic Functions, and it is the two-dimensional analogue of the matched-branch construction of Example 1: a fundamental (point-source) solution plus a homogeneous correction chosen to enforce the boundary condition.