关于函数极值点的定义及判别

关于函数极值点的定义及判别

在分析中，我们了解了零函数极值点的定义和评价方法。这里准备就更一般的情形加以讨论。定义1:设函数f(x,y)在点(x0,y0)处有各阶连续的偏导数,如果函数f(x,y)在点(x0,y0)处的所有不大于(≤)k(k=1,2,3,…)阶的偏导数全为零,而k+1阶偏导数不全为零,则称点(x0,y0)为函数f(x,y)的k(k=1,2,3,…)阶稳定点。定理1:设函数f(x,y)在点(x0,y0)处有各阶连续的偏导数,且点(x0,y0)为函数f(x,y)的一阶稳定点,则(1)当[f″xx(x0,y0)][f″yy(x0,y0)]-[f″xy(x0,y0)]2>0时,函数f(x,y)在点(x0,y0)处取到极值。并且:①[f″xx(x0,y0)]>0时,f(x,y)在点(x0,y0)处取极小值;②当[f″xx(x0,y0)]<0时,f(x,y)在点(x0,y0)处取极大值。(2)当[f″xx(x0,y0)][f″yy(x0,y0)]-[f″xy(x0,y0)]2<0时,函数f(x,y)在点(x0,y0)处不取极值。(3)当Δ=[f″xx(x0,y0)][f″yy(x0,y0)]-[f″xy(x0,y0]2=0时,结论不定。这就是我们熟知的教科书中给出并证明的结论。定理2:设函数f(x,y)在点(x0,y0)处有各阶连续的偏导数,且点(x0,y0)为函数f(x,y)的2k(k=1,2,3,…)阶稳定点,则点(x0,y0)一定不是f(x,y)的极值点。证明:为方便叙述,引入记号:Ai=f(2k+1)x2k+1-iyi(2k+1)x2k+1−iyi(x0,y0)(i=0,1,2,…2k+1)由于函数f(x,y)在点(x0,y0)处的各阶偏导数连续,点(x0,y0)为函数f(x,y)的2k阶稳定点,所以,由2k阶稳定点的定义及2k+1阶泰勒展开式,对任意的s和t有Δf=f(x0+s,y0+t)-f(x0,y0)=1(2k+1)!2k+1∑i=0[Ci2k+1Ais2k+1-iti]+1(2k+1)!2k+1∑i=0[Ci2k+1αis2k+1-iti]Δf=f(x0+s,y0+t)−f(x0,y0)=1(2k+1)!∑2k+1i=0[Ci2k+1Ais2k+1−iti]+1(2k+1)!∑2k+1i=0[Ci2k+1αis2k+1−iti]其中当s→0,t→0时,αi→0(i=0,1,2,…2k+1),于是当|s|及|t|充分小时,Δf的符号就由ω=12k+1!2k+1∑i=0[Ci2k+1Ais2k+1-iti]ω=12k+1!∑2k+1i=0[Ci2k+1Ais2k+1−iti]的符号决定。由于Ai(i=0,1,2,…2k+1)不全为零,所以可将2k+1∑i=0[Ci2k+1Aizi]=0∑2k+1i=0[Ci2k+1Aizi]=0视为以z为变量的至多2k+1次的代数方程,根据代数基本定理知它最多有2k+1个根。于是总可以取z0使得2k+1∑i=0[Ci2k+1Aiz0i]≠0∑2k+1i=0[Ci2k+1Aiz0i]≠0取t=z0s则ω=s2k+1(2k+1)!2k+1∑i=0[Ci2k+1Aizi0]ω=s2k+1(2k+1)!∑2k+1i=0[Ci2k+1Aizi0]于是ω的符号就取决于s的符号,所以在t=z0s的条件下,Δf的符号就取决于s的符号,故在点(x0,y0)的任何邻域内总存在点(x0+s,y0+z0s)及点(x0-s,y0-z0s)使得:f(x0+s,y0+z0s)-f(x0,y0)与f(x0-s,y0-z0s)-f(x0,y0)符号相反,所以函数f(x,y)在点(x0,y0)处不取极值。例、证明:f(x,y)=x2(1-y2)在点(0,-1)及点(0,1)不取极值。证明:f′x(x,y)=2x(1-y2)f′y(x,y)=-2x2y令它们等于零,解得(x0,y0)为(0,-1)及(0,1),不难验证:f′x(x0,y0)=f′y(x0,y0)=f″xx(x0,y0)=f″xy(x0,y0)=f″yy(x0,y0)=0但fue087xxy(0,-1)=4fue087xxy(0,1)=-4所以(0,-1)及点(0,1)均为f(x,y)的二阶稳定点,故由定理2知f(x,x)在点(0,-1)及点(0,1)不取极值。为方便定理3的叙述,引入记号:Ai=f(2k)x2k-iyi(x0,y0)(i=0,1,2,…,2k)定理3:设函数f(x,y)在点(x0,y0)处有各阶连续的偏导数,且点(x0,y0)为函数f(x,y)的2k-1(k=1,2,3…)阶稳定点,如果对任意的i=1,2,…,2k-1均有Ai=0,而A0·A2k≠0则:(1)当A0·A2k>0时,函数f(x,y)在点(x0,y0)处取极值。且A0>0时取极小值,A0<0时取极大值;(2)当A0·A2k<0时,函数f(x,y)在点(x0,y0)处不取极值。证明:由于点(x0,y0)为函数f(x,y)的2k-1阶稳定点,所以,由2k-1阶稳定点的定义及2k阶泰勒展开式和定理条件可知:当|s|及|t|充分小时,Δf=f(x0+s,y0+t)-f(x0,y0)的符号就由ω=1(2k)![A0s2k+A2kt2k]的符号决定,显然,(1)当A0·A2k>0时,对任意不全为零的s,t,ω保持与A0同号。所以,函数f(x,y)在点(x0,y0)处取极值。且A0>0时取极小值;A0<0时取极大值。(2)当A0·A2k<0时,取s≠0,t=0则ω保持与A0同号;而取s=0,t≠0则ω保持与A2k同号(与A0异号),所以,函数f(x,y)在点(x0,y0)处不取极值。为方便定理4的叙述,引入记号:Ai=f(4k)x4k-iyi(x0,y0)(i=0,1,2,…,4k)。定理4:设函数f(x,y)在点(x0,y0)处有各阶连续的偏导数,且点(x0,y0)为函数f(x,y)的4k-1(k=1,2,3…)阶稳定点,如果A2k≠0而对于任意的i≠2k且0≤i≤4k,均有Ai=0,则函数f(x,y)在点(x0,y0)处取极值。且A2k>0时取极小值;A2k<0时取极大值。证明:由于点(x0,y0)为函数f(x,y)的4k-1阶稳定点,所以,由4k-1阶稳定点的定义及4k阶泰勒展开式和定理条件可知:当|s|及|t|充分小时,Δf=f(x0+s,y0+t)-f(x0,y0)的符号就由ω=1(4k)!C2k4kA2ks2kt2k的符号决定,显然,当A2k≠0时,ω保持与A2k同号。所以,函数f(x,y)在点(x0,y0)处取极值。且A2k>0时取极小值;A2k<0时取极大值。为方便下面的叙述,引入记号:A=f(4)x4(x0,y0)B=f(4)x3y(x0,y0)C=f(4)x2y2(x0,y0)D=f(4)xy3(x0,y0)E=f(4)y4(x0,y0)定理5:设函数f(x,y)在点(x0,y0)处有各阶连续的偏导数,且点(x0,y0)为函数f(x,y)的三阶稳定点,则:(1)当A≠0,AC-B2>0,3(AC-B2)(A3E-B4)-2(A2D-B3)2>0时,函数f(x,y)在点(x0,y0)处取极值。且当A>0时取极小值;当A<0时取极大值;(2)当A≠0,AC-B2<0,3(AC-B2)(A3E-B4)-2(A2D-B3)2>0时,函数f(x,y)在点(x0,y0)处不取极值。证明:因为,点(x0,y0)为函数f(x,y)的三阶稳定点,所以,由三阶稳定点的定义及四阶泰勒展开式可知,当|s|及|t|充分小时,Δf=f(x0+s,y0+t)-f(x0,y0)符号就由ω=14![As4+4Bs3t+6Cs2t2+4Dst3+Et4]的符号决定。设s=rcosαt=rsinα则r=√s2+t2,当A≠0,AC-B2≠0时,ω=r44![(Acos4α+4Bcos3αsinα+6Ccos2αsin2α+4Dcosαsin3α+Esin4α]=r44!A3[(Acosα+Bsinα)4+6(A3C-A2B2)cos2αsin2α+4(A3D-AB3)cosαsin3α+(A3E-B4)sin4α]=r424A3(Acosα+Bsinα)4+r4sin2α36A3(AC-B2)[3A(AC-B2)cosα+(A2D-B3)sinα]2+r4sin4α72A3(AC-B2)[3(AC-B2)(A3E-B4)-2(A2D-B3)2](1)由于s,t不同时为零时,r≠0;且当sinα=0时,cosα=±0,于是,当A≠0,AC-B2>0,3(AC-B2)(A3E-B4)-2(A2D-B3)2>0时,ω保持与A同号。所以,函数f(x,y)在点(x0,y0)处取极值。且当A>0时取极小值;当A<0时取极大值。(2)当A≠0,AC-B2<0,3(AC-B2)(A3E-B4)-2(A2D-B3)2>0时,若取sinα=0,则ω与A同号;若取ctgα=-BA,则ω与A异号,所以,函数f(x,y)在点(x0,y0)处不取极值。推论:设函数f(x,y)在点(x0,y0)处有各阶连续的偏导数,且点(x0,y0)为函数f(x,y)的三阶稳定点,则:(1)当E≠0,EC-D2>0,3(EC-D2)(E3A-D4)-2(E2B-D3)2>0时,函数f(x,y)在点(x0,y0)处取极值。且当E>0时取极小值;当E<0时取极大值;(2)当E≠0,EC-D2<0,3(EC-D2)(E3A-D4)-2(E2B-D3)2>0时,函数f(x,y)在点(x0,y0)处不取极值。定理6:设函数f(x,y)在点(x0,y0)处有各阶连续的偏导数,点(x0,y0)为函数f(x,y)的三阶稳定点,且A=E=0,则当B≠0(或D≠0)16BD-9C2>0时,函数f(x,y)在点(x0,y0)处不取极值。证明:由定