Below is a university question and the corresponding solution for which I do not understand small parts of:
Question:
Show that the Green’s function for the range $x \ge 0$, satisfying $$\frac{\partial^2 G(x,z)}{\partial x^2}+G(x,z)=\delta(x-z)$$ with boundary conditions:$$G(x,z)=\frac{\partial G(x,z)}{\partial x}=0\quad\text{at}\quad x=0$$ is $$G(x,z)=\begin{cases}\cos z\sin x-\sin z\cos x, &\text{for} & x\gt z \\ 0 &\text{for} & x\lt z\end{cases}$$
Solution:
For $x \ne z$
$$\frac{\partial^2 G(x,z)}{\partial x^2}+G(x,z)=0$$ with solution
$$G(x,z)=\begin{cases}A(z)\sin x+B(z)\cos x, &\text{for} & x\lt z \\ C(z)\sin x+D(z)\cos x &\text{for} & x\gt z\end{cases}$$
$$G(x=0,z)=0\implies B(z)=0$$ The derivative is (for $x\le z$)
$$\frac{\partial G(x,z)}{\partial x}=A(z)\cos x$$
Since this is zero at $x=0$, we conclude that $A(z)=0$. Hence, $$G(x,z)=0\qquad (x\lt z)$$
$\color{red}{\mathrm{For}}$ $\color{red}{x\gt z,}$ $\color{red}{\text{we use continuity of}}$ $\color{red}{G(x,z)}$ $\color{red}{\mathrm{at}}$ $\color{red}{x=z,}$ $\color{red}{\mathrm{ie.}}$
$$\color{red}{C(z)\sin z+D(z)\cos z=0\tag{1}}$$
$\color{red}{\text{The first derivative changes by}}$ $\color{red}{1,}$ $\color{red}{\text{so since}}$ $\color{red}{G(x,z)=0}$ $\color{red}{\text{for}}$ $\color{red}{x\lt z,}$
$$\color{red}{C(z)\cos z -D(z)\sin z=1\tag{2}}$$
We can eliminate $D$ (or $C$) from equations $(1)$ and $(2)$:
$$C(z)\cos z + C(z)\frac{\sin^2z}{\cos z}=1\tag{3}$$
Multiplying equation $(3)$ by $\cos z$ gives $$C(z)\cos^2 z+C(z)\sin^2 z=\cos z$$ and since $\cos^2 z+\sin^2 z=1$,
$$C(z)=\cos z$$ and we then find $$D(z)=-\sin z$$
Thus,
$$G(x,z)=\begin{cases}\cos z\sin x-\sin z\cos x, &\text{for} & x\gt z \\ 0 &\text{for} & x\lt z\end{cases}$$
$\fbox{}$
I understand everything apart from the part marked red;
Why is equation $\color{red}{(1)}$ equal to zero? What does this have to do with continuity?
Also equation $\color{red}{(1)}$ has products: $C(z)\sin z$ and $D(z)\cos z$ since $C(z)$ and $D(z)$ are both functions of $z$. So why isn't the product rule being used to yield $$C'(z)\sin z + C(z)\cos z + D'(z)\cos z -D(z)\sin z\text{?}$$
Why is it that the "first derivative changes by $1$"?
Best Answer
The Green (or Green's) function satisfies the partial differential equation
$$\frac{\partial^2 G(x,z)}{\partial x^2}+G(x,z)=\delta(x-z) \tag 1$$
If $G$ were discontinuous at $x=z$, then its second derivative would result in a Dirac Doublet. Inasmuch as the right-hand side of $(1)$ contains only the Dirac Delta, not the doublet, then $G$ is continuous.
Next, we integrate $(1)$ with respect to $x$ from $z-\epsilon$ to $z+\epsilon$, where $0<\epsilon<z$. Exploiting the continuity of $G$, we find that
$$\begin{align} \lim_{\epsilon\to0}\int_{z-\epsilon}^{z+\epsilon}\left(\frac{\partial^2 G(x,z)}{\partial x^2}+G(x,z)\right)\,dx&=\color{blue}{\lim_{\epsilon\to0}\left.\frac{\partial G(x,z)}{\partial x}\right|_{x=z+\epsilon}^{x=z+\epsilon}}\\\\ &=\lim_{\epsilon\to0}\int_{z-\epsilon}^{z+\epsilon}\delta (x-z)\,dx\\\\ &=\color{blue}{1} \end{align}$$
which results in the second boundary condition.