Green's Function

A Green's function is the standard tool for solving a linear differential equation with a source term, once the boundary or initial conditions are fixed. In plain language, $G(x,\xi)$ tells you how the system responds to a unit point source placed at $\xi$. For a linear operator $L$ it is defined by the point-source problem

in $x$, together with the same conditions the final solution must satisfy. Here $\delta(x-\xi)$ is the Dirac delta, a unit source concentrated at $\xi$. The payoff: solve that one point-source problem, then handle any source $f$ by superposition,

When this method applies

Use a Green's function when the problem is linear and driven by a source term, with boundary or initial conditions pinned down. The delta source is what makes the definition useful: instead of attacking the full equation at once, you first ask for the response to the most concentrated input possible. You can then read $G(x,\xi)$ as "the effect at $x$ of a unit impulse applied at $\xi$," and treat a complicated source $f$ as a continuous superposition of such point sources. That superposition step — reading $f$ as a continuous pile of point sources — only works because the problem is linear. The same idea appears across ODEs and PDEs: electrostatics, diffusion, wave problems, quantum mechanics, and elasticity, and in linear time-invariant settings it coincides with the impulse response. The language changes across subjects, but the strategy is constant: understand the response to one concentrated input, then build the response to a general input from it.

The steps

1. Fix the problem setup. State the linear operator $L$ and the boundary or initial conditions before hunting for $G$. The Green's function belongs to the full problem, not the operator alone — change the conditions and $G$ changes too.
2. Solve the point-source problem. Find $G(x,\xi)$ so that $L\,G = \delta(x-\xi)$ with those same conditions.
3. Build the full solution as the integral of $G(x,\xi)$ against the source $f(\xi)$.
4. Check the conditions — both the differential equation and the boundary or initial conditions — since the formula is valid only for that exact setup.

The reason step 1 cannot be skipped is that a Green's function is not determined by the differential operator alone. For example, the operator $-u''$ on $0 < x < 1$ with $u(0) = u(1) = 0$ has a different Green's function from the same operator with Neumann boundary conditions: the equation looks identical, but the allowed solutions change, so the kernel changes too. There is no single Green's function for "the equation" unless the conditions are already part of the statement.

Full example: the Dirichlet problem for $-u'' = f$

Take

For this specific problem the Green's function is

It is continuous in $x$, vanishes at $x=0$ and $x=1$, and its first derivative has the right jump at $x=\xi$ — that jump produces the delta source. With the kernel known, every source for this same problem reuses it:

Pick $f(x) = 1$:

Check it: $u'(x) = \tfrac{1-2x}{2}$, $u''(x) = -1$, so $-u'' = 1 = f$, and $u(0) = u(1) = 0$. You did not rebuild the method for the new $f$; you kept the kernel and changed only the integral.

Where you get stuck, and how to verify

• Assuming linearity is optional. The superposition step fails for nonlinear problems.
• Forgetting the conditions are part of the definition. $-u''$ with Dirichlet data has a different $G$ from the same operator with Neumann data.
• Assuming $G$ must be symmetric. Symmetry needs extra structure, such as a self-adjoint setup.
• Confusing a Green's function with a fundamental solution. A fundamental solution is tied to the operator; a Green's function also builds in the chosen conditions.
• Using the integral formula without checking that a Green's function exists for that particular problem.

If you forget the formal definition, keep: Green's function = response to one point source, and the full solution sums those responses across the domain.

Practice with a new source

Keep the same boundary conditions but replace $f(x) = 1$ with $f(x) = x$. The kernel stays fixed; only the last integral changes:

This cleanly separates the two moving parts — the kernel belongs to the setup, the source varies case to case. Check your result against the boundary conditions.