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+---
+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.