2017-2018-2偏微分方程复习题解析4

Problem: For any positive $s$, we have $$\bex \sup_{t>0}\sum_{j\in\bbZ} t^s2^{2js} \e^{-ct2^{2j}}<\infty. \eex$$

Proof: For any $t>0$, $$\beex \bea &\quad\sum_{j\in\bbZ} t^s2^{2js} \e^{-ct2^{2j}} =\sum_{j\in\bbZ} t^s 2^{2js} \e^{-\f{ct}{4}2^{2(j+1)}}\\ & \leq \sum_{j\in\bbZ} \int_j^{j+1} t^s 2^{2\tau s}\e^{-\f{ct}{4} 2^{2\tau}} \rd\tau =\int_{\bbR} t^s 2^{2\tau s}\e^{-\f{ct}{4}2^{2\tau}}\rd \tau\\ &=\int_0^\infty \sex{\f{4v}{s}}^c \e^{-v}\f{\rd v}{2v\ln 2}\qx{\f{ct}{4}2^{2\tau} =v\ra \rd \tau=\f{\rd v}{2v\ln 2}}\\ &=\sex{\f{4}{c}}^s \f{1}{2\ln 2} \int_0^\infty v^{s-1}\e^{-s}\rd s =\sex{\f{4}{c}}^s \f{\vGa(s)}{2\ln 2}<\infty. \eea \eeex$$

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转载自www.cnblogs.com/zhangzujin/p/9012169.html