2017-2018-2偏微分方程复习题解析4

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

Proof: For any $t>0$, $$eex ea &quadsum_{jinbZ} t^s2^{2js} e^{-ct2^{2j}} =sum_{jinbZ} t^s 2^{2js} e^{-f{ct}{4}2^{2(j+1)}}\ & leq sum_{jinbZ} int_j^{j+1} t^s 2^{2 au s}e^{-f{ct}{4} 2^{2 au}} d au =int_{bR} t^s 2^{2 au s}e^{-f{ct}{4}2^{2 au}} d au\ &=int_0^infty sex{f{4v}{s}}^c e^{-v}f{ d v}{2vln 2}qx{f{ct}{4}2^{2 au} =v a d au=f{ d v}{2vln 2}}\ &=sex{f{4}{c}}^s f{1}{2ln 2} int_0^infty v^{s-1}e^{-s} d s =sex{f{4}{c}}^s f{vGa(s)}{2ln 2}<infty. eea eeex$$

原文地址:https://www.cnblogs.com/zhangzujin/p/9012169.html