这里假设所讨论的函数$\,u\in C^{1}(I),\,c,f\in C^{0}(I),$ 其中$\,I:=[a,T),\,-\infty<a<T\leqslant+\infty.$

$\mathbf{\text{第}\,1\,\text{节}\quad\text{微分形式的}\,}\mathbf{Gronwall}\,\mathbf{\text{不等式}}$

$\mathbf{\text{定理}\,}\mathbf{1.}$ 若$\,u(t)$ 满足形式$\,u'(t)=c(t)u(t)+f(t),\,t\in I,$
则$\,u(t)$ 必具有形式
\[
u(t)=u(a)e^{\int_{a}^{t}c(\rho)d\rho}+\int_{a}^{t}f(s)e^{\int_{s}^{t}c(\rho)d\rho}ds,\,t\in I.
\]
特别地, 若$\,u'(t)=Au(t)+C,\,t\in I,$ 则有
\[
u(t)=u(a)e^{A(t-a)}+\frac{C}{A}\left(e^{A(t-a)}-1\right),\,t\in I.
\]
$\mathit{Proof}:$ 证明就是两端乘以$\,e^{-\int_{a}^{t}c(\rho)d\rho}$ 然后积分.$\quad\quad\square$

$\mathbf{\text{定理}\,}\mathbf{2.}$ 若定理$\,1$ 中的等号改成不等号, 相应的结论仍成立,
即若
\[
u'(t)\leqslant(\geqslant)\,c(t)u(t)+f(t),\,t\in I,
\]
则
\[
u(t)\leqslant(\geqslant)\,u(a)e^{\int_{a}^{t}c(\rho)d\rho}+\int_{a}^{t}f(s)e^{\int_{s}^{t}c(\rho)d\rho}ds,\,t\in I.
\]

$\mathbf{\text{从定理}\,2\,\text{就能得到许多推论}}$.

$\mathbf{\text{推论}\,}\mathbf{3.(\text{ODE比较原理})}$ 若$\,u'(t)+Au(t)\leqslant C,\,t\in I,$
则
\[
u(t)\leqslant u(a)e^{-A(t-a)}+\frac{C}{A}(1-e^{-A(t-a)}),\,t\in I.
\]
特别地, 当$\,A>0,\,u(a)>0$ 时有
\[
u(t)\leqslant\max\{u(a),\frac{C}{A}\},\,t\in I.
\]

$\mathbf{\text{推论}\,}\mathbf{4.}$ 若$\,u'(t)+c(t)u(t)\leqslant(\geqslant)\,0,\,t\in I,$
则
\[
u(t)\leqslant(\geqslant)\,u(a)e^{-\int_{a}^{t}c(\rho)d\rho},\,t\in I.
\]

$\mathbf{\text{推论}\,}\mathbf{5.1.}$ 若$\,u'(t)+c(t)u(t)\geqslant0,\,t\in I,$
且$\,u(a)>0,$ 则
\[
u(t)>0,\,t\in I.
\]

$\mathbf{\text{推论}\,}\mathbf{5.2.}$ 若$\,u'(t)+c(t)u(t)\leqslant0,\,t\in I,$
且$\,u(a)>0,\,c(t)>0,\,t\in I,$ 则
\[
u(t)\leqslant u(a),\,t\in I.
\]

$\mathbf{\text{推论}\,}\mathbf{6.}$ 若$\,u'(t)\leqslant c(t)u(t)+f(t),\,t\in I,$
且$\,c(t),f(t)\geqslant0,\,t\in I,$ 则
\[
u(t)\leqslant e^{\int_{a}^{t}c(\rho)d\rho}\left[u(a)+\int_{a}^{t}f(s)ds\right],\,t\in I.
\]

$\mathbf{\text{推论}\,}\mathbf{7.}$ 若$\,u(t)$ 非负, 且$\,u'(t)\leqslant c(t)u(t),\,t\in I,$
$u(a)=0,$ 则
\[
u(t)\equiv0,\,t\in I.
\]

$\mathbf{\text{定理}\,}\mathbf{8.}$ 若$\,u'(t)\leqslant c(t)u(t)+f(t),\,t\in I,$
且$\,u(t),c(t),f(t)$ 都非负, 并且存在$\,r>0$ 和常数$\,M_{i}\,(i=1,2,3)$ 使得
\[
\int_{t}^{t+r}c(s)ds\leqslant M_{1},\quad\int_{t}^{t+r}f(s)ds\leqslant M_{2},\,\,t\in I,
\]
\[
\int_{t}^{t+r}u(s)ds\leqslant M_{3},\,\,t\in I,
\]
其中$\,t+r<T,$ 则
\[
u(t+r)\leqslant e^{M_{1}}\left(\frac{M_{3}}{r}+M_{2}\right),\,\,t\in I.
\]
$\mathit{Proof}:$ 任意固定$\,t_{0}\in I,$ 由积分中值定理知存在$\,\xi\in[t_{0},t_{0}+r]$
使得
\[
u(\xi)=\frac{1}{r}\int_{t_{0}}^{t_{0}+r}u(s)ds\leqslant\frac{M_{3}}{r}.
\]
由$\,u'(t)\leqslant c(t)u(t)+f(t),\,t\in[\xi,t_{0}+r]$ 及推论$\,6$ 得
\begin{eqnarray*}
u(t_{0}+r) & \leqslant & e^{\int_{\xi}^{t_{0}+r}c(\rho)d\rho}\left[u(\xi)+\int_{\xi}^{t_{0}+r}f(s)ds\right]\\
& \leqslant & e^{M_{1}}(\frac{M_{3}}{r}+M_{2}),
\end{eqnarray*}
由$\,t_{0}$ 的任意性得证.$\quad\quad\square$
\[
\]

$\mathbf{\text{第}\,2\,\text{节}\quad\text{积分形式的}\,}\mathbf{Gronwall}\,\mathbf{\text{不等式}}$

$\mathbf{\text{定理}\,}\mathbf{9.}$ 若$\,u(t)\leqslant\alpha(t)+\int_{a}^{t}\beta(s)u(s)ds,\,t\in I,$
且$\,\beta(t)\geqslant0,\,t\in I,$ 则
\[
\int_{a}^{t}\beta(s)u(s)ds\leqslant\int_{a}^{t}\alpha(s)\beta(s)e^{\int_{s}^{t}\beta(\rho)d\rho}ds,\,t\in I,
\]
\[
u(t)\leqslant\alpha(t)+\int_{a}^{t}\alpha(s)\beta(s)e^{\int_{s}^{t}\beta(\rho)d\rho}ds,\,t\in I.
\]
$\mathit{Proof}:$ 令$\,\eta(t)=\int_{a}^{t}\beta(s)u(s)ds,\,t\in I,$
则$\,\eta(a)=0,\,\eta'(t)=\beta(t)u(t),$ 从而
\[
\frac{\eta'(t)}{\beta(t)}\leqslant\alpha(t)+\eta(t),\,t\in I,
\]
由于$\,\beta(t)\geqslant0,$ 所以$\,\eta'(t)\leqslant\beta(t)\eta(t)+\alpha(t)\beta(t),\,t\in I.$
使用定理$\,2$ 的结论有
\[
\eta(t)\leqslant0+\int_{a}^{t}\alpha(s)\beta(s)e^{\int_{s}^{t}\beta(\rho)d\rho}ds,\,t\in I.
\]
再用一下条件可得
\[
\,u(t)\leqslant\alpha(t)+\eta(t)\leqslant\alpha(t)+\int_{a}^{t}\alpha(s)\beta(s)e^{\int_{s}^{t}\beta(\rho)d\rho}ds,\,t\in I.
\]
证毕.$\quad\quad\square$

$\mathbf{\text{推论}\,}\mathbf{10.}$ 若$\,u(t),\beta(t)$ 非负, 且$\,u(t)\leqslant\int_{a}^{t}\beta(s)u(s)ds,\,t\in I,$
则
\[
u(t)\equiv0,\,t\in I.
\]

$\mathbf{\text{推论}\,}\mathbf{11.}$ 若$\,u(t)\leqslant\alpha+\int_{a}^{t}\beta(s)u(s)ds,\,t\in I,$
且$\,\alpha>0,\,\beta(t)\geqslant0,\,t\in I,$ 则
\[
u(t)\leqslant\alpha e^{\int_{a}^{t}\beta(s)ds},\,t\in I.
\]
$\mathit{Proof}:$ 根据定理$\,9$ 有
\[
u(t)\leqslant\alpha\left[1+\int_{a}^{t}\beta(s)e^{\int_{s}^{t}\beta(\rho)d\rho}ds\right],
\]
由于恒等式$\,\int_{a}^{t}v'(s)e^{v(s)}ds=e^{v(t)}-e^{v(a)},$ 所以
\[
-\int_{a}^{t}\beta(s)e^{\int_{s}^{t}\beta(\rho)d\rho}ds=1-e^{\int_{a}^{t}\beta(\rho)d\rho},
\]
代入上面不等式即证.$\quad\quad\square$

$\mathbf{\text{推论}\,}\mathbf{11.}$ 若$\,u(t)\leqslant\alpha+\beta\int_{a}^{t}u(s)ds,\,t\in I,$
且$\,\alpha,\beta>0,$ 则
\[
u(t)\leqslant\alpha\left[1+\beta(t-a)e^{\beta(t-a)}\right],\,t\in I.
\]
$\mathit{Proof}:$ 根据推论$\,10$ 有
\[
u(t)\leqslant\alpha e^{\beta(t-a)},\,t\in I.
\]
由于$\,e^{x}\leqslant1+xe^{x},$ 所以得证.$\quad\quad\square$