- #1
psie
- 268
- 32
- 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:
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:
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.
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:
- ##\frac{d^kF}{dt^k}(t,\tau)## is a continuous function of ##(t,\tau)## when ##k=0,1,\ldots,n-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##.
- ##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.