{"id":609,"date":"2018-05-25T08:30:59","date_gmt":"2018-05-25T08:30:59","guid":{"rendered":"https:\/\/blog.vanabel.cn\/?p=609"},"modified":"2018-05-25T08:32:37","modified_gmt":"2018-05-25T08:32:37","slug":"gammahanshuyusobolevqianrudingli","status":"publish","type":"post","link":"https:\/\/blog.vanabel.cn\/?p=609","title":{"rendered":"Gamma\u51fd\u6570\u4e0eSobolev\u5d4c\u5165\u5b9a\u7406"},"content":{"rendered":"

\u56de\u5fc6, Gamma\u51fd\u6570\u7684\u5b9a\u4e49
\n$$
\n\\Gamma(z)=\\int_0^\\infty t^{z-1}e^{-t}dt,\\quad z\\in\\mathbb{R}^2\\setminus\\left(\\mathbb{Z}_-\\bigcup\\{0\\}\\right).
\n$$
\n\u5bb9\u6613\u9a8c\u8bc1\u5982\u4e0b\u57fa\u672c\u6027\u8d28, \u5bf9$z>0$, \u6211\u4eec\u6709
\n$$
\n\\Gamma(z+1)=\\int_0^\\infty t^ze^{-t}dt
\n=-\\int_0^\\infty t^zd e^{-t}
\n=-t^ze^{-t}|_0^\\infty+z\\int_0^\\infty t^{z-1}e^{-t}dt=z\\Gamma(z).
\n$$
\n\u5bb9\u6613\u9a8c\u8bc1\u5bf9$\\Re(z)>0$\u6709Cauchy–Riemann \u65b9\u7a0b\u6210\u7acb, \u6545\u7531Looman\u2013Menchoff<\/a><\/strong>\u5b9a\u7406, \u6211\u4eec\u77e5\u9053$\\Gamma$\u5728\u53f3\u8fb9\u5e73\u9762$\\{z=(x,y):x>0\\}$\u65f6\u662f\u5168\u7eaf\u7684\u3002
\n<\/p>\n

\u4e0b\u9762, \u6211\u4eec\u901a\u8fc7\u9010\u6b65\u7684\u8fde\u7eed\u5ef6\u62d3, \u5c06$\\Gamma$\u5ef6\u62d3\u5230$\\mathbb{R}^2\\setminus\\left(\\mathbb{Z}_-\\bigcup\\{0\\}\\right)$. \u9996\u5148, \u5bf9$\\Re (z)\\in(-1,0)$, \u6211\u4eec\u77e5\u9053, $\\Re(z+1)\\in(0,1)$, \u6545\u5982\u4e0b\u5b9a\u4e49\u7684\u5ef6\u62d3
\n$$
\n\\Gamma(z):=\\frac{\\Gamma(z+1)}{z}
\n$$
\n\u662f\u89e3\u6790\u5ef6\u62d3\u3002\u7c7b\u4f3c\u5730\u5bf9$\\Re(z)\\in(-2,-1)$, \u6b64\u65f6$\\Re(z+1)\\in(-1,0)$, \u6545\u5229\u7528\u4e0a\u9762\u5ef6\u62d3\u540e\u7684$\\Gamma$, \u6211\u4eec\u53ef\u4ee5\u5f97\u5230\u5982\u4e0b\u89e3\u6790\u5ef6\u62d3
\n$$
\n\\Gamma(z):=\\frac{\\Gamma(z+1)}{z}.
\n$$
\n\u5982\u6b64\u7ee7\u7eed\u4e0b\u53bb\u5c31\u5f97\u5230\u4e86$\\mathbb{R}^2\\setminus\\left(\\mathbb{Z}_-\\bigcup\\{0\\}\\right)$\u4e0a\u5b9a\u4e49\u7684$\\Gamma$\u51fd\u6570\u3002<\/p>\n

\u6211\u4eec\u77e5\u9053, Sobolev\u7a7a\u95f4$H^1(\\Omega)$, \u8fd9\u91cc$\\Omega\\subset\\mathbb{R}^n$\u662f\u4e00\u4e2a\u5149\u6ed1\u6709\u754c\u5f00\u533a\u57df, \u5b9a\u4e49\u4e3a
\n$$
\nf\\in H^1(\\Omega)\\iff f\\in L^2(\\Omega),\\quad\\int_\\Omega |f|^2<+\\infty\\,\\&\\,\\int_\\Omega|\\nabla f|^2<+\\infty,\n$$\n\u5176\u4e2d$\\nabla f$\u662f$f$\u7684\u5206\u5e03\u5bfc\u6570(\u5f53$f$\u53ef\u5bfc\u65f6\u5c31\u662f\u666e\u901a\u5bfc\u6570)\u3002\n\n\u6211\u4eec\u60f3\u8981\u8bf4\u660e\u7684\u662f\u5982\u4e0b\u547d\u9898\n

\u547d\u9898 1<\/span>.<\/span> \u5b58\u5728\u51fd\u6570$f\\in H^1(B)$, \u8fd9\u91cc$B\\subset\\mathbb{R}^2$\u662f\u5355\u4f4d\u7403, \u4f7f\u5f97$f$\u4e0d\u662f\u8fde\u7eed\u7684\u3002\n<\/div>\n
\u8bc1\u660e<\/span>.<\/span> \u6211\u4eec\u5c06\u8981\u6784\u9020\u7684$f$\u662f\u5982\u4e0b\u7684\u51fd\u6570\n$$\nf(x)=(-\\ln|x|^2)^\\alpha,\\quad \\alpha>0,\\,x\\in B.
\n$$
\n\u4e8b\u5b9e\u4e0a, \u76f4\u63a5\u8ba1\u7b97\u53ef\u77e5
\n$$
\n\\int_B |f|^2=\\int_0^{2\\pi}\\int_0^1(-2\\ln r)^\\alpha rdrd\\theta
\n=2\\pi\\int_0^\\infty (2t)^\\alpha e^{-2t}dt
\n=\\pi\\int_0^\\infty t^\\alpha e^{-t}dt
\n=\\pi\\Gamma(\\alpha+1).
\n$$
\n\u800c
\n$$
\n|\\nabla f|^2=(\\partial_1f)^2+(\\partial_2f)^2=(\\partial_rf)^2=4\\alpha^2(-2\\ln r)^{2\\alpha-2}r^{-2}.
\n$$
\n\u56e0\u6b64, \u5bf9$\\alpha\\in(0,1\/2)$, \u6211\u4eec\u6709
\n$$
\n\\int_{B_{1\/2}}|\\nabla f|^2=2\\pi\\int_0^{1\/2}4\\alpha^2(-2\\ln r)^{2\\alpha-2}r^{-1}dr
\n=4\\pi\\alpha^2\\int_{2\\ln 2}^\\infty t^{2\\alpha-2}dt
\n=2^{2\\alpha+1}\\frac{\\pi\\alpha^2}{1-2\\alpha}.
\n$$
\n\u7531\u6b64\u53ef\u89c1\u51fd\u6570
\n$$
\nF(x)=f(x\/\\sqrt{2})=\\left(-\\ln(|x|^2\/2)\\right)^\\alpha,\\quad \\alpha\\in(0,1\/2),\\,x\\in B.
\n$$
\n\u6ee1\u8db3\u8981\u6c42\u3002
\n<\/div><\/p>\n","protected":false},"excerpt":{"rendered":"

\u56de\u5fc6, Gamma\u51fd\u6570\u7684\u5b9a\u4e49 $$ \\Gamma(z)=\\int_0^\\infty t^{z-1}e^{-t}… \u7ee7\u7eed\u9605\u8bfbGamma\u51fd\u6570\u4e0eSobolev\u5d4c\u5165\u5b9a\u7406<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[127,128,129,24],"class_list":["post-609","post","type-post","status-publish","format-standard","hentry","category-math","tag-gammahanshu","tag-sobolevkongjian","tag-bulianxu","tag-24","entry"],"_links":{"self":[{"href":"https:\/\/blog.vanabel.cn\/index.php?rest_route=\/wp\/v2\/posts\/609","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blog.vanabel.cn\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blog.vanabel.cn\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blog.vanabel.cn\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/blog.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=609"}],"version-history":[{"count":2,"href":"https:\/\/blog.vanabel.cn\/index.php?rest_route=\/wp\/v2\/posts\/609\/revisions"}],"predecessor-version":[{"id":611,"href":"https:\/\/blog.vanabel.cn\/index.php?rest_route=\/wp\/v2\/posts\/609\/revisions\/611"}],"wp:attachment":[{"href":"https:\/\/blog.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=609"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blog.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=609"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blog.vanabel.cn\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=609"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}