---
title: "Fundamental solution"
sort_title: "Fundamental solution"
date: 2021-11-02
categories:
- Mathematics
- Physics
layout: "concept"
---

Given a linear operator $$\hat{L}$$ acting on $$x \in [a, b]$$,
its **fundamental solution** $$G(x, x')$$ is defined as the response
of $$\hat{L}$$ to a [Dirac delta function](/know/concept/dirac-delta-function/)
$$\delta(x - x')$$ for $$x \in ]a, b[$$:

$$\begin{aligned}
    \boxed{
        \hat{L}\{ G(x, x') \}
        = A \delta(x - x')
    }
\end{aligned}$$

Where $$A$$ is a constant, usually $$1$$.
Fundamental solutions are often called **Green's functions**,
but are distinct from the (somewhat related)
[Green's functions](/know/concept/greens-functions/)
in many-body quantum theory.

Note that the definition of $$G(x, x')$$ generalizes that of
the [impulse response](/know/concept/impulse-response/).
And likewise, due to the superposition principle,
once $$G$$ is known, $$\hat{L}$$'s response $$u(x)$$ to
*any* forcing function $$f(x)$$ can easily be found as follows:

$$\begin{aligned}
    \hat{L} \{ u(x) \}
    = f(x)
        \quad \implies \quad
    \boxed{
        u(x)
        = \frac{1}{A} \int_a^b f(x') \: G(x, x') \dd{x'}
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-solution"/>
<label for="proof-solution">Proof</label>
<div class="hidden" markdown="1">
<label for="proof-solution">Proof.</label>
$$\hat{L}$$ only acts on $$x$$, so $$x' \in ]a, b[$$ is simply a parameter,
meaning we are free to multiply the definition of $$G$$
by the constant $$f(x')$$ on both sides,
and exploit $$\hat{L}$$'s linearity:

$$\begin{aligned}
    A f(x') \: \delta(x - x')
    = f(x') \hat{L}\{ G(x, x') \}
    = \hat{L}\{ f(x') \: G(x, x') \}
\end{aligned}$$

We then integrate both sides over $$x'$$ in the interval $$[a, b]$$,
allowing us to consume $$\delta(x \!-\! x')$$.
Note that $$\int \dd{x'}$$ commutes with $$\hat{L}$$ acting on $$x$$:

$$\begin{aligned}
    A \int_a^b f(x') \: \delta(x - x') \dd{x'}
    &= \int_a^b \hat{L}\{ f(x') \: G(x, x') \} \dd{x'}
    \\
    A f(x)
    &= \hat{L} \int_a^b f(x') \: G(x, x') \dd{x'}
\end{aligned}$$

By definition, $$\hat{L}$$'s response $$u(x)$$ to $$f(x)$$
satisfies $$\hat{L}\{ u(x) \} = f(x)$$, recognizable here.
</div>
</div>

While the impulse response is typically used for initial value problems,
the fundamental solution $$G$$ is used for boundary value problems.
Suppose those boundary conditions are homogeneous,
i.e. $$u(x)$$ or one of its derivatives is zero at the boundaries.
Then:

$$\begin{aligned}
    0
    &= u(a)
    = \frac{1}{A} \int_a^b f(x') \: G(a, x') \dd{x'}
    \qquad \implies \quad
    G(a, x') = 0
    \\
    0
    &= u_x(a)
    = \frac{1}{A} \int_a^b f(x') \: G_x(a, x') \dd{x'}
    \quad \implies \quad
    G_x(a, x') = 0
\end{aligned}$$

This holds for all $$x'$$, and analogously for the other boundary $$x = b$$.
In other words, the boundary conditions are built into $$G$$.

What if the boundary conditions are inhomogeneous?
No problem: thanks to the linearity of $$\hat{L}$$,
those conditions can be given to the homogeneous solution $$u_h(x)$$,
where $$\hat{L}\{ u_h(x) \} = 0$$,
such that the inhomogeneous solution $$u_i(x) = u(x) - u_h(x)$$
has homogeneous boundaries again,
so we can use $$G$$ as usual to find $$u_i(x)$$, and then just add $$u_h(x)$$.

If $$\hat{L}$$ is self-adjoint
(see e.g. [Sturm-Liouville theory](/know/concept/sturm-liouville-theory/)),
then the fundamental solution $$G(x, x')$$
has the following **reciprocity** boundary condition:

$$\begin{aligned}
    \boxed{
        G(x, x') = G^*(x', x)
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-reciprocity"/>
<label for="proof-reciprocity">Proof</label>
<div class="hidden" markdown="1">
<label for="proof-reciprocity">Proof.</label>
Consider two parameters $$x_1'$$ and $$x_2'$$.
The self-adjointness of $$\hat{L}$$ means that:

$$\begin{aligned}
    \int_a^b G^*(x, x_1') \Big( \hat{L} \{ G(x, x_2') \} \Big) \dd{x}
    &= \int_a^b \Big( \hat{L} \{ G(x, x_1') \} \Big)^* G(x, x_2') \dd{x}
    \\
    \int_a^b G^*(x, x_1') \: \delta(x - x_2') \dd{x}
    &= \int_a^b \delta^*(x - x_1') \: G(x, x_2') \dd{x}
    \\
    G^*(x_2', x_1')
    &= G(x_1', x_2')
\end{aligned}$$

</div>
</div>



## References
1.  O. Bang,
    *Applied mathematics for physicists: lecture notes*, 2019,
    unpublished.