---
title: "Fundamental solution"
firstLetter: "F"
publishDate: 2021-11-02
categories:
- Mathematics
- Physics

date: 2021-11-01T14:57:46+01:00
draft: false
markup: pandoc
---

# Fundamental solution

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">
<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">
<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.