Gronwall型不等式

Problem. Suppose $x(t)\in C[0,T]$, and satisfies $$\bex t\in [0,T]\ra 1\leq x(t)\leq C_1+C_2\int_0^t x(\tau)[1+\log x(\tau)]\rd \tau. \eex$$ Prove:   

(1)    $x(t)$ is bounded on $[0,T].$

(2)    This is in stark contrast to the estimates like: $$\bex x(t)\leq C_1+C_2\int_0^t x^{1+\ve}(\tau)\rd \tau, \eex$$ which allows blowup of $x$ in finite time. Show that such blowup can happen for $\ve=2$.

 

Proof. We first show the Gronwall inequality: $$\bee\label{182.Gronwall} \left.\ba{rr} f(t)\leq C_1+C_2\int_0^t g(s)f(s)\rd s\\ g\geq 0,\ \int_0^T g(t)\rd t<\infty \ea\right\}\ra f(t)\leq C_1e^{C_2\int_0^t g(s)\rd s}<\infty. \eee$$ Indeed, $$\beex \bea &\quad\ f(t)\leq C_1+C_2\int_0^t g(s)f(s)\rd s\\ &\ra \frac{C_2g(t)f(t)}{C_1+C_2\int_0^t g(s)f(s)\rd s}\leq C_2g(t)\\ &\ra \ln \frac{C_1+C_2\int_0^t g(s)f(s)\rd s}{C_1}\leq C_2\int_0^t g(s)\rd s\quad\sex{integrating}\\ &\ra C_1+C_2\int_0^t g(s)f(s)\rd s\leq C_1e^{C_2\int_0^tg(s)\rd s}\\ &\ra f(t)\leq C_1e^{C_2\int_0^tg(s)\rd s}. \eea \eeex$$    Then we return to the problem.   (1) $$\beex \bea &\quad\ x(t)\leq C_1+C_2\int_0^t x(\tau)[1+\log x(\tau)]\rd \tau\\ &\ra x(t)\leq C_1e^{C_2\int_0^t[1+\ln x(\tau)]\rd \tau}\quad(\eqref{182.Gronwall})\\ &\ra \ln x(t)\leq \ln C_1+C_2\int_0^t[1+\ln x(\tau)]\rd \tau\\ &\ra \ln x(t)\leq \ln C_1+C_2T+\int_0^t \ln x(\tau)\rd \tau\\ &\ra \ln x(t)\leq (\ln C_1+C_2T)e^{\int_0^t\rd \tau}\quad(\eqref{182.Gronwall}\ again)\\ &\ra x(t)\leq e^{(\ln C_1+C_2T)e^T}<\infty. \eea \eeex$$

(2) Suppose now $$\bex x(t)\leq C_1+C_2\int_0^t x^{2}(\tau)\rd \tau. \eex$$ Let $$\bex f(t)=C_1+C_2\int_0^t x^{2}(\tau)\rd \tau. \eex$$ Then $$\beex \bea &\quad f'(t)=C_2x^2(t)\leq C_2f^2(t)\\ &\ra -\frac{f'(t)}{f^2(t)}\geq -C_2\\ &\ra \frac{1}{f(t)}-\frac{1}{f(0)}\geq -C_2t\\ &\ra f(t)\leq \frac{f(0)}{1-C_2f(0)t}\\ &\ra x(t)\leq f(t)\leq \frac{C_1}{1-C_2C_1t}. \eea \eeex$$ Thus $x(t)$ may blowup at $\dps{t=\frac{1}{C_2C_1}}$.   

 

来源: 家里蹲大学数学杂志第3卷第182期_Blowup_or_Bounded