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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"

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 all0 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! :-)

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    I agree with you. Thank you for your attention! :-)2012-04-23

1 Answers 1

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You have to integrate by parts. Indeed, let us consider your solution $ u(t) \le e^{-c_0t} u(0)+O(1) e^{-c_0t} \int_0^t e^{c_0s}f(s) ds, $ where $u(0)=\int_0^1 \eta_0^2 dx.$ Now, we integrate by parts the last term on the rhs obtaining $ u(t)\le e^{-c_0t}u(0)+O(1)e^{-c_0t}\frac{1}{c_0}(e^{c_0t}f(t)-f(0))-O(1)e^{-c_0t}\frac{1}{c_0}\int_0^tdse^{c_0s}f'(s). $ Now, it is easy to see that, when $f(t)=f(0)$ you reach authors' conclusion. This means that $ \frac{d}{dt}\int_0^1dx(\psi_t^2+\psi^2)=0. $

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    To Jon: I'm glad to discuss this question with you!If we admit f>0 is monotonically increasing,then I don't need to ask this question. I wonder that the condition,$f$ is monotonically increasing, is necessary? As I mentioned in the question, we can't assure that $f$ is monotonically increasing in the paper I read. :-)2012-04-25