Stochastic epidemics: Markov process satisfies PDE While studying a book about percolation theory I encounter a chapter about the contact process and stochastic epidemics, the following problem emerges.
Let $X(t)$ be the number of healthy individuals at time $t$ and suppose that $X(0) = N$. We assume that, if $X(t)=n$, the probability of a new infection during a short time-interval $(t,t+h)$ is proportional to the number of possible encounters between ill folk and healthy folk. That is,
\begin{align}
\mathbb{P}(X(t+h) = n-1 | X(t) = n) = \lambda n (N+1 -n)h + o(h) \qquad \text{ as $h \downarrow 0$}. 
\end{align}
In the simplest situation, we assume that nobody recovers. We would like to show that
\begin{align}
G(s,t) = \mathbb{E}[s^{X(t)}] = \sum_{n=0}^N s^n \mathbb{P}(X(t) = n )
\end{align}
satisfies 
\begin{align}
\frac{\partial G}{\partial t} = \lambda (1-s) \bigg(N \frac{\partial G}{\partial s} - s \frac{\partial^2 G}{\partial s^2} \bigg).
\end{align}
However, I face some difficulties to demonstrate the above statement. Some help would be appreciated. 
 A: When $n\in\{0,\dots,N-1\}$,
$
\begin{align*}
\frac{d}{dt}\mathbb{P}(X(t)=n)&=\lim_{h\downarrow0}\frac{1}{h}(\mathbb{P}(X(t+h)=n)-\mathbb{P}(X(t)=n)) \\
&=\lim_{h\downarrow0}\frac{1}{h}(\sum_{k=0}^N\mathbb{P}(X(t+h)=n\vert X(t)=k)\mathbb{P}(X(t)=k)-\mathbb{P}(X(t)=n)) \\
&=\lim_{h\downarrow0}\frac{1}{h}((1-\lambda n(N+1-n)h+o(h))\mathbb{P}(X(t)=n) \\
&\quad\quad\quad\quad+(\lambda(n+1)(N-n)h+o(h))\mathbb{P}(X(t)=n+1) \\
&\quad\quad\quad\quad\quad-\mathbb{P}(X(t)=n)) \\
&=-\lambda n(N+1-n)\mathbb{P}(X(t)=n)+\lambda(n+1)(N-n)\mathbb{P}(X(t)=n+1).
\end{align*}
$
In fact the formula above also holds true for $n=N$.
Use the above and differentiate $G$ with respect to $t$. The idea then is to convert all probabilities into the same form by changing the index of summation. After that group the terms according to whether they have a factor of $n$ or $n^2$ in front. Finally write the terms in the form of a partial derivative with respect to $s$ by pulling out factors of $s$ as needed.
$
\begin{align*}
\frac{\partial}{\partial t}G(s,t)&=\sum_{n=0}^N s^n\frac{d}{dt}\mathbb{P}(X(t)=n) \\
&=\sum_{n=0}^N s^n(-\lambda n(N+1-n)\mathbb{P}(X(t)=n)+\lambda(n+1)(N-n)\mathbb{P}(X(t)=n+1)) \\
&=\sum_{n=0}^N-\lambda n(N+1-n)s^n\mathbb{P}(X(t)=n)+\sum_{n=1}^{N+1}\lambda n(N+1-n)s^{n-1}\mathbb{P}(X(t)=n) \\
&=\sum_{n=0}^N-\lambda n(N+1-n)s^n\mathbb{P}(X(t)=n)+\sum_{n=0}^N\lambda n(N+1-n)s^{n-1}\mathbb{P}(X(t)=n) \\
&=\sum_{n=0}^N\left(-\lambda n(N+1)s^n\mathbb{P}(X(t)=n)+\lambda n(N+1)s^{n-1}\mathbb{P}(X(t)=n)\right) \\
&\quad+\sum_{n=0}^N\left(\lambda n^2s^n\mathbb{P}(X(t)=n)-\lambda n^2s^{n-1}\mathbb{P}(X(t)=n)\right) \\
&=\sum_{n=0}^N\left(-\lambda n(N+1)s^n\mathbb{P}(X(t)=n)+\lambda n(N+1)s^{n-1}\mathbb{P}(X(t)=n)\right) \\
&\quad+\sum_{n=0}^N\left((\lambda n(n-1)+\lambda n)s^n\mathbb{P}(X(t)=n)-(\lambda n(n-1)+\lambda n)s^{n-1}\mathbb{P}(X(t)=n)\right) \\
&=\sum_{n=0}^N\left(-\lambda nNs^n\mathbb{P}(X(t)=n)+\lambda nNs^{n-1}\mathbb{P}(X(t)=n)\right) \\
&\quad+\sum_{n=0}^N\left(\lambda n(n-1)s^n\mathbb{P}(X(t)=n)-\lambda n(n-1)s^{n-1}\mathbb{P}(X(t)=n)\right) \\
&=-\lambda Ns\sum_{n=0}^Nns^{n-1}\mathbb{P}(X(t)=n)+\lambda N\sum_{n=0}^N ns^{n-1}\mathbb{P}(X(t)=n) \\
&\quad+\lambda s^2\sum_{n=0}^N n(n-1)s^{n-2}\mathbb{P}(X(t)=n)-\lambda s\sum_{n=0}^N n(n-1)s^{n-2}\mathbb{P}(X(t)=n) \\
&=-\lambda Ns\frac{\partial}{\partial s}G(s,t)+\lambda N\frac{\partial}{\partial s}G(s,t)+\lambda s^2\frac{\partial^2}{\partial s^2}G(s,t)-\lambda s\frac{\partial^2}{\partial s^2}G(s,t) \\
&=\lambda(1-s)(N\frac{\partial G}{\partial s}(s,t)-s\frac{\partial^2 G}{\partial s^2}(s,t)).
\end{align*}
$
