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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 $0<c_0<\frac{3}{2}-O(1)\delta_0.$"

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! :-)

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I've been trying a little bit, with no success. The authors should have written some more detail, I guess... –  Siminore Apr 23 '12 at 11:24
    
I agree with you. Thank you for your attention! :-) –  Darry Apr 23 '12 at 12:19

1 Answer 1

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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First of all,thank you for your answer to my question, but in fact I thought of assuming that the $\int_0^1(\psi_t^2+\psi^2)dx$ is independent on $t$. Unfortunately, the integration $\int_0^1(\psi_t^2+\psi^2)dx$ must depend on time $t$ in the paper. So we can't use the condition: $\frac{d}{dt}\int_0^1 dx(\psi_t^2+\psi^2)=0.$ –  Darry Apr 23 '12 at 13:53
    
@DarryHu: I have checked authors' paper and they are working on semiconductors assuming $j=constant$. Now, $\psi=\rho-\rho_0$, being $\rho_0$ the static solution, and it seems quite reasonable to take this as a constant for this case. –  Jon Apr 23 '12 at 13:56
    
Please allow me to think about it. Thanks :-) –  Darry Apr 23 '12 at 14:09
    
To Jon:I am unable to find clues that can help us rigorously prove the condition $\frac{d}{dt}\int_0^1dx(\psi_t^2+\psi^2)=0$ holds. –  Darry Apr 23 '12 at 14:35
    
@DarryHu: I will check this out again and eventually I will improve my answer. But of course, from a mathematical standpoint, this is correct! –  Jon Apr 23 '12 at 16:00

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