I am reading the paper "Asymptotic behaviour of solutions of the hydrodynamic model of semiconductors, Proceedings of the Royal Society of Edinburgh, 132A, 359-378, 2002"You can download this paper here and I am stumped by the following statement on Page11 in the paper :
"We have known that $$ \frac{d}{dt}(\int_0^1 \eta^2 dx)+(\frac{3}{2}-O(1)\delta_0)\int_0^1 \eta^2 dx\leq O(1)\int_0^1 (\psi_t^2+\psi^2)dx. \qquad\qquad (3.30) $$ Integrating (3.30) over [0, t] gives $$ \int_0^1 \eta^2 dx \leq e^{-c_0 t}\int_0^1 \eta_0^2 dx+O(1)(1-e^{-c_0 t})\int_0^1 (\psi_t^2+\psi^2)dx. \qquad\qquad\qquad (3.31) $$ with a constant $0<c_0<\frac{3}{2}-O(1)\delta_0.$"
This is my try, but failed :
"Let $u(t)=\int_0^1 \eta^2 dx>0,\quad f(t)=\int_0^1 (\psi_t^2+\psi^2)dx>0,\quad b=\frac{3}{2}-O(1)\delta_0>0$, then for all$0<c_0<b,$ we have $$ \frac{du(t)}{dt}+c_0u(t)\leq \frac{du(t)}{dt}+bu(t)\leq O(1)f(t) $$ i.e. $$ \frac{du(t)}{dt}+c_0u(t)\leq O(1)f(t)\qquad\qquad\qquad\qquad\qquad\qquad(m1) $$ and multiply equation (m1) by $e^{c_0t}$ and integrat the resultant equation with respect to $t$ over $[0, t]$ gives $$ u(t) \leq e^{-c_0t} u(0)+O(1) e^{-c_0t} \int_0^t e^{c_0s}f(s) ds, $$ where $u(0)=\int_0^1 \eta_0^2 dx.$
I want to use the Integral Mean-Value Theorem to deal with the term $\int_0^t e^{c_0s}f(s) ds$, but failed. Since I am sure that the function $f(t)$ defined above by me is not a monotonic function according to the information in the paper which I am reading now. "
I hope someone can help me to answer this question!! Thanks! :-)