Green's Function

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, it tells you how the system responds to a unit point source placed at $\xi$.

For a linear operator $L$, a Green's function $G(x,\xi)$ is defined by the point-source problem

$$L\,G(x,\xi) = \delta(x-\xi)$$

in the variable $x$, together with the same boundary or initial conditions that the final solution must satisfy. Here $\delta(x-\xi)$ is the Dirac delta, a distribution that represents a unit source concentrated at $\xi$.

If such a Green's function exists and the problem is linear, then the solution of

$$L\,u(x) = f(x)$$

can often be written as

$$u(x) = \int G(x,\xi) f(\xi)\, d\xi.$$

That is the main idea: solve the point-source problem once, then combine those point responses to handle a general source $f$.

What a Green's function means

The delta source is what makes the definition useful. Instead of solving the full equation immediately, you first ask for the response to the most concentrated input possible.

So you can read $G(x,\xi)$ as "the effect at $x$ of a unit impulse applied at $\xi$." A more complicated source $f$ is then treated as a continuous superposition of point sources. This step only works because the problem is linear.

Why boundary conditions change the answer

A Green's function is not determined by the differential operator alone. It belongs to the full problem, including the conditions.

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 may look the same, but the allowed solutions change, so the kernel changes too.

This is one of the most common mistakes. There is no single Green's function for "the equation" unless the conditions are already part of the statement.

Worked example: the Dirichlet problem for $-u''(x)=f(x)$

Consider the boundary-value problem

$$-u''(x) = f(x), \qquad 0 < x < 1,$$

with

$$u(0)=0, \qquad u(1)=0.$$

For this specific problem, the Green's function is

$$G(x,\xi) = \begin{cases} x(1-\xi), & x \le \xi, \\ \xi(1-x), & x \ge \xi. \end{cases}$$

It is continuous in $x$, vanishes at $x=0$ and $x=1$, and its first derivative has the correct jump at $x=\xi$. That jump is what produces the delta source in the equation $-G''(x,\xi)=\delta(x-\xi)$.

Once the kernel is known, every source term for this same boundary-value problem uses the same formula:

$$u(x)=\int_0^1 G(x,\xi) f(\xi)\, d\xi.$$

Choose one simple source term:

$$f(x)=1.$$

Then

$$u(x)=\int_0^1 G(x,\xi)\, d\xi = \int_0^x \xi(1-x)\, d\xi + \int_x^1 x(1-\xi)\, d\xi.$$

Compute each part:

$$\int_0^x \xi(1-x)\, d\xi = (1-x)\frac{x^2}{2},$$ $$\int_x^1 x(1-\xi)\, d\xi = x\frac{(1-x)^2}{2}.$$

Adding them gives

$$u(x)=\frac{x(1-x)}{2}.$$

A quick check confirms it:

$$u'(x)=\frac{1-2x}{2}, \qquad u''(x)=-1,$$

so

$$-u''(x)=1=f(x),$$

and the boundary values are

$$u(0)=0, \qquad u(1)=0.$$

This example shows the payoff clearly. You do not rebuild the method for each new $f$; you keep the same kernel and only change the integral.

Common Green's function mistakes

1. Treating Green's functions as if they worked for nonlinear problems in the same way. The superposition step depends on linearity.
2. Forgetting that the boundary or initial conditions are part of the definition.
3. Assuming the kernel must always be symmetric. Symmetry usually needs extra structure, such as a suitable self-adjoint setup.
4. Mixing up a Green's function with a fundamental solution. They are closely related, but they are not always the same object.
5. Using the integral formula without checking that a Green's function exists for that particular problem.

Where Green's functions are used

Green's functions appear in ordinary and partial differential equations whenever a linear problem is driven by a source term. They are common in electrostatics, diffusion, wave problems, quantum mechanics, and elasticity.

In linear time-invariant settings, they are also closely related to impulse responses. The language changes across subjects, but the idea is similar: understand the response to one concentrated input, then build the response to a general input from that.

A quick way to remember it

If you forget the formal definition, keep this version:

$$\text{Green's function} = \text{response to one point source}.$$

Then the full solution comes from summing those point-source responses across the domain, which is what the integral does.

Try a similar source term

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

$$u(x)=\int_0^1 G(x,\xi)\,\xi\, d\xi.$$

That is a good next step because it separates the two moving parts cleanly: the Green's function belongs to the setup, while the source term can change from case to case. If you want to go further, try your own version with a different source and check the resulting integral against the boundary conditions.