

\begin{frame}
\frametitle{Technical Lemma (History Integrals)}
\scriptsize
\begin{block}{\bf Lemma}
Let $\color{eqncolor}r=r_B$ and $\color{eqncolor}f \in C_0^\infty (D)$, and define
$$\color{eqncolor}
I^{(p)}[b,r,f]=\int_0^t e^{(b(t-t')/r)}f^{(p)}(t'-r)dt'
$$
Then, we have
$$\color{eqncolor}
I^{(p)}[b,r,f]=(b/r)^p I^{(0)}[b,r,f] + \sum_{j=1}^{p}(b/r)^{p-j}f^{(j-1)}(t-r)
$$
\end{block}
\end{frame}
\begin{frame}
\frametitle{Technical Lemma (History Integrals)}
\scriptsize
\begin{block}{\bf Proof by induction}
We show:
$$\color{eqncolor}
I^{(p)} [b,r,f] = (b/r)^p I^{(0)}[b,r,f] +\sum_{j=1}^p (b/r)^{p-j}\big[f^{(j-1)}(t-r)-e^{(bt/r)}f^{(j-1)}(-r)\big] \hspace{2em}(**)
$$
The second term within the square brackets vanishes since $\color{eqncolor}r=r_B \notin D$.  Note there are no boundary terms at $\color{eqncolor}t'=0$, only $\color{eqncolor}t'=t$. Integration by parts establishes the last formula for $\color{eqncolor}p=1$. Similarly,
$$\color{eqncolor}
I^{(p)} [b,r,f] = (b/r)I^{(p-1)}[b,r,f] + f^{(p-1)}(t-r)-e^{(bt/r)}f^{(p-1)}(-r).
$$
Assuming now that $\color{eqncolor}(**)$ holds with $\color{eqncolor}p$ replaced by $\color{eqncolor}p-1$, we insert the $\color{eqncolor}p-1$ result into the last equation, thereby recovering $\color{eqncolor}(**)$ and verifying the induction step.
\end{block}
\end{frame}

\begin{frame}
\frametitle{{\bf Main Result}}
\scriptsize
\begin{block}{\bf Theorem}
The roots $\color{eqncolor}{b_{\ell j} : j = 1,...,\ell}$ of the $\color{eqncolor}\ell^{th}$ degree polynomial 
$$\color{eqncolor}p_{\ell}(z)= \sum_{k=0}^{\ell}c_{\ell k}z^{\ell-k}
$$
also obey the following set of $\color{eqncolor}\ell$ algebraic equations: (Newton's identities!)
$$\color{eqncolor}
-kc_{\ell k} = \sum_{n=1}^{\ell} b_{\ell n}\sum_{q=1}^k c_{\ell,q-1}(b_{\ell n})^{k-q}, \hspace{2em} k=1,...,\ell
$$
Here we assume $\color{eqncolor}\ell \geq 1$.
\end{block}
\end{frame}

\begin{frame}
\frametitle{{\bf Main Result:} careful proof (1)}
\scriptsize
\begin{block}{\bf proof}
Taking $\color{eqncolor}r=r_B$ and $\color{eqncolor}t>0$, we get
$$\color{eqncolor}
\frac{1}{r}\int_0^t \Omega_{\ell}(t-t',r)\Psi_{\ell}(t',r)dt' = \sum_{n=1}^\ell (b_{\ell n}/r^2)\int_0^t \exp(b_{\ell n}(t-t')/r)\Psi_{\ell}(t',r)dt'
$$
$$\color{eqncolor}
= \sum_{n=1}^{\ell}(b_{\ell n}/r^2)\sum_{k=0}^\ell r^{-k} c_{\ell k} I^{(\ell-k)}[b_{\ell n},r,f].
$$
Previous lemma gives
$$\color{eqncolor}
I^{(\ell - k)} [b_{\ell  n},r,f] = (b_{\ell  n}/r)^{\ell -k} I^{(0)} [b_{\ell n},r,f] + \sum_{j=1}^{\ell -k} (b_{\ell n}/r)^{\ell - k - j} f^{(j-1)}(t-r).
$$
\end{block}
\end{frame}
\begin{frame}
\frametitle{{\bf Main Result:} careful proof continued...(2)}
\scriptsize
\begin{block}{}
Combination of the last two equations yields
$$\color{eqncolor}
\frac{1}{r}\int_0^t \Omega_{\ell}(t-t',r)\Psi_{\ell}(t',r)dt' = r^{-(\ell+2)} \sum_{n=1}^\ell b_{\ell n}I^{(0)}[b_{\ell n},r,f] \sum_{k=0}^{\ell} c_{\ell k}(b_{\ell n})^{\ell -k}
$$
$$\color{eqncolor}
 + \sum_{n=1}^{\ell}(b_{\ell n}/r^2)\sum_{k=0}^\ell r^{-k} c_{\ell k} \sum_{j=1}^{\ell-k}(b_{\ell n}/r)^{\ell - k - j} f^{(j-1)}(t-r).
$$
\end{block}
\end{frame}
\begin{frame}
\frametitle{{\bf Main Result:} careful proof continued...(3)}
\scriptsize
\begin{block}{}
Since $\color{eqncolor}\sum_{k=0}^{\ell} c_{\ell k}(b_{\ell n})^{\ell-k} = p_{\ell}(b_{\ell n})$ is the Bessel polynomial evaluated at one of its roots, the first term on the right-hand side of the last expression vanishes. Whence, up to now
$$\color{eqncolor}
\frac{1}{r}\int_0^t \Omega_{\ell}(t-t',r)\Psi_{\ell}(t',r)dt' = \sum_{n=1}^{\ell}(b_{\ell n}/r^2)\sum_{k=0}^\ell r^{-k} c_{\ell k} \sum_{j=1}^{\ell-k}(b_{\ell n}/r)^{\ell - k - j} f^{(j-1)}(t-r).
$$
\end{block}
\end{frame}

\begin{frame}
\frametitle{{\bf Main Result:} careful proof continued...(4)}
\scriptsize
\begin{block}{}
Within the sum over $\color{eqncolor}k$, the sum over $\color{eqncolor}j$ is empty when $\color{eqncolor}k=\ell$. Therefore, here we may replace $\color{eqncolor}\sum_{k=0}^{\ell}$ by $\color{eqncolor}\sum_{k=0}^{\ell-1}$. Re-indexing $\color{eqncolor}q = k+1$, yields
$$\color{eqncolor}
\frac{1}{r}\int_0^t \Omega_{\ell}(t-t',r)\Psi_{\ell}(t',r)dt' 
$$
$$\color{eqncolor}
= \sum_{n=1}^{\ell}(b_{\ell n}/r^2)\sum_{q=1}^\ell r^{-(q-1)} c_{\ell, q-1} \sum_{j=1}^{\ell-q+1}(b_{\ell n}/r)^{\ell - q - j +1} f^{(j-1)}(t-r).
$$
\end{block}
\end{frame}
\begin{frame}
\frametitle{{\bf Main Result:} careful proof continued...(5)}
\scriptsize
\begin{block}{}
The double sum $\color{eqncolor}\sum_{q=1}^\ell \sum_{j=1}^{\ell - q + 1}(terms)_{jq}$ is equivalent to $\color{eqncolor}\sum_{j=1}^\ell \sum_{q=1}^{\ell - j + 1}(terms)_{jq}$.  We may group the two inner sums and make this exchange, thereby reaching
$$\color{eqncolor}
\frac{1}{r}\int_0^t \Omega_{\ell}(t-t',r)\Psi_{\ell}(t',r)dt' =
$$
$$\color{eqncolor}
\sum_{n=1}^{\ell}b_{\ell n}\sum_{j=1}^\ell r^{j-(\ell+2)} \sum_{q=1}^{\ell-j+1}c_{\ell, q-1}(b_{\ell n})^{\ell - q - j +1} f^{(j-1)}(t-r).
$$
\end{block}
\end{frame}
\begin{frame}
\frametitle{{\bf Main Result:} careful proof continued...(6)}
\scriptsize
\begin{block}{}
Through re-indexing of sums we get to
$$\color{eqncolor}
\frac{1}{r}\int_0^t \Omega_{\ell}(t-t',r)\Psi_{\ell}(t',r)dt' = \sum_{n=1}^{\ell}b_{\ell n}\sum_{k=1}^\ell r^{-(k+1)} \sum_{q=1}^{k}c_{\ell, q-1}(b_{\ell n})^{k - q} f^{(\ell -k)}(t-r),
$$
which is the desired form. Recalling that we have set $\color{eqncolor}r=r_B$,
$$\color{eqncolor}
-\sum_{k=1}^\ell kr^{-(k+1)}c_{\ell k} f^{(\ell-k)}(t-r) = \sum_{k=1}^{\ell}r^{-(k+1)} \sum_{n=1}^\ell b_{\ell n} \sum_{q=1}^{k} c_{\ell , q-1} (b_{\ell n})^{k-q}f^{(\ell-k)}(t-r),
$$
and we may express that last equation as
$$\color{eqncolor}
0 = \sum_{k=1}^\ell r^{-(k+1)}E_k f^{(\ell-k)}(u), \hspace{2em} E_k = kc_{\ell k} + \sum_{n=1}^\ell b_{\ell n} \sum_{q=1}^k c_{\ell ,q-1} (b_{\ell n})^{k-q}.
$$
Here $\color{eqncolor}u=t-r$ is retarded time. 
\end{block}
\end{frame}
%\begin{frame}
%\frametitle{proof...continued(7)}
%\scriptsize
%\begin{block}{}
%Introducing the operator $\color{eqncolor}Q \equiv (\partial_t + \partial_r) r^2$,
%we then use induction to show that
%$$\color{eqncolor}
%Q^p \sum_{k=1}^\ell r^{-(k+1)}E_k f^{(\ell-k)}(u)
%= \sum_{k=p+1}^\ell (-1)^p \frac{(k-1)!}{(k-p-1)!}
%r^{-(k-p+1)}E_k f^{(\ell-k)}(u).
%$$
%For $\color{eqncolor}p=\ell-1$, this identity implies $\color{eqncolor}E_\ell = 0$.
%Then assuming $\color{eqncolor}E_{p+2} = \dots = E_\ell = 0$, we find
%$$\color{eqncolor}
%Q^p \sum_{k=1}^\ell r^{-(k+1)}E_k f^{(\ell-k)}(u)
%= (-1)^p \frac{p!}{r^2} E_{p+1} f^{(\ell-p-1)}(u),
%$$
%yielding $\color{eqncolor}E_{p+1} = 0$. Therefore, backwards iteration
%$\color{eqncolor}p = \ell-1,\ell-2,\dots,0$ establishes that $\color{eqncolor}E_k=0$ for $\color{eqncolor}k=1,\dots,\ell$.
%\end{block}
%\end{frame}
%\begin{frame}
%\small
%\frametitle{Sample calculation for $\ell=1$}
%\begin{itemize}
%Sample calculation for $\ell=1$
%\end{frame}
