💼 Case: lets try and find the electric potential due to charges
1️⃣ Lets start with a single charge
The potential at $\vec r$ due to a single charge $q_0$ at $\vec r_0$ is
$$ \Phi(\vec r)=\frac{q_0}{4\pi \epsilon_0|\vec r- \vec r_0|} $$
2️⃣ Now for many charges
Using the principle of superposition, for a series of charges $q_i$ or a region $V$ of density $\rho(\vec r)$
$$ \Phi(\vec r)=\frac{1}{4\pi \epsilon_0} \sum_i \frac{q_i}{|\vec r-\vec r_i|} \qquad \Phi(\vec r)=\frac{1}{4\pi \epsilon_0} \int \frac{\rho(\vec r')}{|\vec r- \vec r'|}\,\text d^3 \vec r $$
The potential is a solution to Poisson’s equation $-\nabla^2 \Phi(\vec r)=\rho(\vec r) /\epsilon_0$
❇️ Greens formalism
The Green function here can be extracted as
$$ G(\vec r,\vec r')=\frac{1}{4\pi |\vec r-\vec r'|} \qquad \Phi (\vec r)=\int G(\vec r,r')\frac{\rho(\vec r')}{\epsilon_0} \,\text d^3 \vec r' $$
It is the potential due to a unit point charge at $\vec r'$, it satisfies Laplace equation everywhere except $\vec r=\vec r'$ and satisfies the boundary that $\Phi$ vanishes at $|\vec r|\to \infin$
🗒️ Note: here $\vec r'$ is fixed but mathematically it is a parameter of the Green function
Looking directly at the Laplace equation we have
$$ \begin{aligned} \frac{\rho(\vec r)}{\epsilon_0}=-\nabla ^2_{\vec r} \Phi(\vec r)&=-\int \nabla ^2_{\vec r} G(\vec r,\vec r')\frac{\rho(\vec r')}{\epsilon_0} \,\text d^3 \vec r' \\ \Rightarrow \qquad -\nabla^2_{\vec r} G(\vec r,\vec r')&=\delta (\vec r-\vec r') \end{aligned} $$
This is another way to represent the Green function
💼 Case: Consider the 1 dimensional case, with a linear differential operator $\mathcal L_x$ subject to boundary conditions. We define our Green function $G(x,z)$ as follows
$$ \mathcal L_xG(x,z)=\delta (x-z) $$
For some function $f(x)$ on an interval $[a,b]$ we can write
$$ \begin{aligned} f(x)&=\int ^b_a \delta (x-z) f(z) \,\text dz \\ &= \int ^b_a \mathcal L_x G(x,z)f(z)\,\text dz = \mathcal L_x \int ^b_aG(x,z)f(z)\,\text dz \\
\end{aligned} $$
Hence we get a way to rewrite out differential equation
$$ \mathcal L_x y(x)=f(x) \quad \Rightarrow \quad y(x)=\int ^b_a G(x,z) f(z)\,\text dz $$
💎 Conclusion: differential equation turned to integral which is easier to solve