Verifying properties of Green's function

[psie]

TL;DR Summary

I want to verify properties of the Green's function and its derivatives, such as continuity, discontinuity and being a solution to a linear homogeneous ODE.

I'm reading about fundamental solutions to differential operators in Ordinary Differential Equations by Andersson and Böiers. There is a remark that succeeds a theorem that I struggle with verifying. First, the theorem:

Theorem 6. Let $$L(t,\lambda)=\lambda^n+a_{n-1}(t)\lambda^{n-1}+\ldots+a_1(t)\lambda+a_0(t)\quad\text{and }D=\frac{d}{dt}.\tag1$$
Denote by $E(t,\tau)$ the uniquely determined solution $u(t)$ of the initial value problem
\begin{align}
&L(t,D)u=0 \tag2\\
&u(\tau)=u'(\tau)=\ldots=u^{(n-2)}(\tau)=0,\quad u^{(n-1)}(\tau)=1. \tag3
\end{align}
Then,
$$y(t)=\int_{t_0}^t E(t,\tau)g(\tau)d\tau\tag4$$
is the solution of the problem
\begin{align}
&L(t,D)y=g(t) \tag5\\
&y(t_0)=y'(t_0)=\ldots=y^{(n-1)}(t_0)=0. \tag6
\end{align}

If the leading coefficient in $(1)$ is not $1$ but $a_n(t)$, then the last condition in $(3)$ reads $u^{(n-1)}(\tau)=1/a_n(\tau)$ and $(4)$ changes to $$y(t)=\int_{t_0}^t E(t,\tau)\frac{g(\tau)}{a_n(\tau)}d\tau.\tag7$$ Put $\overline{E}(t,\tau)=\frac{E(t,\tau)}{a_n(\tau)}$ and define $$F(t,\tau)=\begin{cases} \overline{E}(t,\tau) &\text{when } t\geq\tau \\ 0 &\text{when } t<\tau.\end{cases}\tag8$$ then $F(t,\tau)$ satisfies the following properties:

1. $\frac{d^kF}{dt^k}(t,\tau)$ is a continuous function of $(t,\tau)$ when $k=0,1,\ldots,n-2.$
2. $\frac{d^{n-1}F}{dt^{n-1}}(t,\tau)$ is continuous when $t\neq\tau$, and has a step discontinuity of height $1/a_n(\tau)$ across the line $t=\tau$.
3. $L(t,D)F(t,\tau)=0,\quad t\neq \tau$.

The authors note that this is easily verified by noting that $E(t,\tau)$ solves the IVP $(2)$ and $(3)$, yet I have hard time verifying this to myself.

First of all, I'm confused about them writing $\frac{d^k}{dt^k}$ instead of $\frac{\partial^k}{\partial t^k}$. Is this because we view the function as a function of $t$ only? If so, then 1. makes very little sense to me. How can the $k$th derivative ($0\le k\le n-2$) of $F$ with respect to $t$ be continuous?

Second, I do not see how either 2. or 3. follows from the fact $E(t,\tau)$ solves $(1)$ and $(2)$. I'd be very grateful if someone could share their understanding on the matter.

 

[pasmith]

$\tau$ is regarded as a parameter of the IVP (2,3), so $d/dt$ here means $\partial/\partial t$.

We know that both $\bar{E}$ and 0 are $n - 1$ times differentiable with respect to $t$: 0 trivially, and $\bar E$ because it is the solution of the IVP (2,3), so its first $n - 1$ derivatives with respect to $t$ exist and are continuous in $t$. Continuity of $\bar E$ and its $t$-derivatives in $(t,\tau)$ jointly follows from the fact that if you write $$\bar E(t,\tau) = A_1(\tau)u_1(t) + \dots + A_n(\tau)u_n(t)$$ for $n$ linearly independent solutions $u_k$ of (2), then the $A_k$ can be shown to be continuous in $\tau$.

$F$ is defined as either $\bar E$ or zero so $F$ and its first $n - 1$ derivatives with respect to $t$ can fail to be continuous in $(t,\tau)$ only at the boundary of the regions where those definitions are applied, ie. when $t = \tau$. By construction $$\frac{\partial^k F}{\partial t^k}(\tau,\tau) = \begin{cases} 0 & k = 0, \dots, n-2 \\ 1/a_n(\tau) & k = n - 1. \end{cases}$$ but $$\lim_{t \to \tau^{-}} \frac{\partial^k F}{\partial t^k}(t,\tau) = 0.$$