这是一份liverpool利物浦大学MATH366的成功案例

$$
p_{n, n}(t, t+h)=e^{-\alpha h},
$$
the function $h \mapsto p_{n, n}(t, t+h)$ is continuous, which proves (ii). Finally, we have
$$
\begin{aligned}
\lim {h \rightarrow 0} \frac{1}{h}\left(1-p{n, n}(t, t+h)\right) &=\lim {h \rightarrow 0} \frac{1}{h}\left(1-e^{-\alpha h}\right) \ &=\alpha \end{aligned} $$ as well as $$ \begin{aligned} \lim {h \rightarrow 0} \frac{1}{h} p_{n, n+1}(t, t+h) &=\lim _{h \rightarrow 0} \frac{1}{h} e^{-\alpha h} \alpha h \
&=\alpha .
\end{aligned}
$$

MATH366 COURSE NOTES ：

(1) We have already noticed in the preceding part of the proof that the differential equation
$$
\frac{d}{d t} p_{k, k}(r, t)=-p_{k, k}(r, t) \lambda_{k+1}(t)
$$
with initial condition $p_{k, k}(r, r)=1$ has the unique solution
$$
p_{k, k}(r, t)=e^{-\int_{r}^{t} \lambda_{k+1}(s) d s} .
$$
(2) Assume now that $k<n$. Then the function $t \mapsto 0$ is the unique solution of the homogeneous differential equation
$$
\frac{d}{d t} p_{k, n}(r, t)=-p_{k, n}(r, t) \lambda_{n+1}(t)
$$
with initial condition $p_{k, n}(r, r)=0$. This implies that the inhomogeneous differential equation
$$
\frac{d}{d t} p_{k, n}(r, t)=p_{k, n-1}(r, t) \lambda_{n}(t)-p_{k, n}(r, t) \lambda_{n+1}(t)
$$
with initial condition $p_{k, n}(r, r)=0$ has at most one solution.