關于函數(shù)極值點的定義及判別

關于函數(shù)極值點的定義及判別

在分析中，我們了解了零函數(shù)極值點的定義和評價方法。這里準備就更一般的情形加以討論。定義1:設函數(shù)f(x,y)在點(x0,y0)處有各階連續(xù)的偏導數(shù),如果函數(shù)f(x,y)在點(x0,y0)處的所有不大于(≤)k(k=1,2,3,…)階的偏導數(shù)全為零,而k+1階偏導數(shù)不全為零,則稱點(x0,y0)為函數(shù)f(x,y)的k(k=1,2,3,…)階穩(wěn)定點。定理1:設函數(shù)f(x,y)在點(x0,y0)處有各階連續(xù)的偏導數(shù),且點(x0,y0)為函數(shù)f(x,y)的一階穩(wěn)定點,則(1)當[f″xx(x0,y0)][f″yy(x0,y0)]-[f″xy(x0,y0)]2>0時,函數(shù)f(x,y)在點(x0,y0)處取到極值。并且:①[f″xx(x0,y0)]>0時,f(x,y)在點(x0,y0)處取極小值;②當[f″xx(x0,y0)]<0時,f(x,y)在點(x0,y0)處取極大值。(2)當[f″xx(x0,y0)][f″yy(x0,y0)]-[f″xy(x0,y0)]2<0時,函數(shù)f(x,y)在點(x0,y0)處不取極值。(3)當Δ=[f″xx(x0,y0)][f″yy(x0,y0)]-[f″xy(x0,y0]2=0時,結論不定。這就是我們熟知的教科書中給出并證明的結論。定理2:設函數(shù)f(x,y)在點(x0,y0)處有各階連續(xù)的偏導數(shù),且點(x0,y0)為函數(shù)f(x,y)的2k(k=1,2,3,…)階穩(wěn)定點,則點(x0,y0)一定不是f(x,y)的極值點。證明:為方便敘述,引入記號:Ai=f(2k+1)x2k+1-iyi(2k+1)x2k+1?iyi(x0,y0)(i=0,1,2,…2k+1)由于函數(shù)f(x,y)在點(x0,y0)處的各階偏導數(shù)連續(xù),點(x0,y0)為函數(shù)f(x,y)的2k階穩(wěn)定點,所以,由2k階穩(wěn)定點的定義及2k+1階泰勒展開式,對任意的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]其中當s→0,t→0時,αi→0(i=0,1,2,…2k+1),于是當|s|及|t|充分小時,Δf的符號就由ω=12k+1!2k+1∑i=0[Ci2k+1Ais2k+1-iti]ω=12k+1!∑2k+1i=0[Ci2k+1Ais2k+1?iti]的符號決定。由于Ai(i=0,1,2,…2k+1)不全為零,所以可將2k+1∑i=0[Ci2k+1Aizi]=0∑2k+1i=0[Ci2k+1Aizi]=0視為以z為變量的至多2k+1次的代數(shù)方程,根據(jù)代數(shù)基本定理知它最多有2k+1個根。于是總可以取z0使得2k+1∑i=0[Ci2k+1Aiz0i]≠0∑2k+1i=0[Ci2k+1Aiz0i]≠0取t=z0s則ω=s2k+1(2k+1)!2k+1∑i=0[Ci2k+1Aizi0]ω=s2k+1(2k+1)!∑2k+1i=0[Ci2k+1Aizi0]于是ω的符號就取決于s的符號,所以在t=z0s的條件下,Δf的符號就取決于s的符號,故在點(x0,y0)的任何鄰域內總存在點(x0+s,y0+z0s)及點(x0-s,y0-z0s)使得:f(x0+s,y0+z0s)-f(x0,y0)與f(x0-s,y0-z0s)-f(x0,y0)符號相反,所以函數(shù)f(x,y)在點(x0,y0)處不取極值。例、證明:f(x,y)=x2(1-y2)在點(0,-1)及點(0,1)不取極值。證明:f′x(x,y)=2x(1-y2)f′y(x,y)=-2x2y令它們等于零,解得(x0,y0)為(0,-1)及(0,1),不難驗證: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)及點(0,1)均為f(x,y)的二階穩(wěn)定點,故由定理2知f(x,x)在點(0,-1)及點(0,1)不取極值。為方便定理3的敘述,引入記號:Ai=f(2k)x2k-iyi(x0,y0)(i=0,1,2,…,2k)定理3:設函數(shù)f(x,y)在點(x0,y0)處有各階連續(xù)的偏導數(shù),且點(x0,y0)為函數(shù)f(x,y)的2k-1(k=1,2,3…)階穩(wěn)定點,如果對任意的i=1,2,…,2k-1均有Ai=0,而A0·A2k≠0則:(1)當A0·A2k>0時,函數(shù)f(x,y)在點(x0,y0)處取極值。且A0>0時取極小值,A0<0時取極大值;(2)當A0·A2k<0時,函數(shù)f(x,y)在點(x0,y0)處不取極值。證明:由于點(x0,y0)為函數(shù)f(x,y)的2k-1階穩(wěn)定點,所以,由2k-1階穩(wěn)定點的定義及2k階泰勒展開式和定理條件可知:當|s|及|t|充分小時,Δf=f(x0+s,y0+t)-f(x0,y0)的符號就由ω=1(2k)![A0s2k+A2kt2k]的符號決定,顯然,(1)當A0·A2k>0時,對任意不全為零的s,t,ω保持與A0同號。所以,函數(shù)f(x,y)在點(x0,y0)處取極值。且A0>0時取極小值;A0<0時取極大值。(2)當A0·A2k<0時,取s≠0,t=0則ω保持與A0同號;而取s=0,t≠0則ω保持與A2k同號(與A0異號),所以,函數(shù)f(x,y)在點(x0,y0)處不取極值。為方便定理4的敘述,引入記號:Ai=f(4k)x4k-iyi(x0,y0)(i=0,1,2,…,4k)。定理4:設函數(shù)f(x,y)在點(x0,y0)處有各階連續(xù)的偏導數(shù),且點(x0,y0)為函數(shù)f(x,y)的4k-1(k=1,2,3…)階穩(wěn)定點,如果A2k≠0而對于任意的i≠2k且0≤i≤4k,均有Ai=0,則函數(shù)f(x,y)在點(x0,y0)處取極值。且A2k>0時取極小值;A2k<0時取極大值。證明:由于點(x0,y0)為函數(shù)f(x,y)的4k-1階穩(wěn)定點,所以,由4k-1階穩(wěn)定點的定義及4k階泰勒展開式和定理條件可知:當|s|及|t|充分小時,Δf=f(x0+s,y0+t)-f(x0,y0)的符號就由ω=1(4k)!C2k4kA2ks2kt2k的符號決定,顯然,當A2k≠0時,ω保持與A2k同號。所以,函數(shù)f(x,y)在點(x0,y0)處取極值。且A2k>0時取極小值;A2k<0時取極大值。為方便下面的敘述,引入記號: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:設函數(shù)f(x,y)在點(x0,y0)處有各階連續(xù)的偏導數(shù),且點(x0,y0)為函數(shù)f(x,y)的三階穩(wěn)定點,則:(1)當A≠0,AC-B2>0,3(AC-B2)(A3E-B4)-2(A2D-B3)2>0時,函數(shù)f(x,y)在點(x0,y0)處取極值。且當A>0時取極小值;當A<0時取極大值;(2)當A≠0,AC-B2<0,3(AC-B2)(A3E-B4)-2(A2D-B3)2>0時,函數(shù)f(x,y)在點(x0,y0)處不取極值。證明:因為,點(x0,y0)為函數(shù)f(x,y)的三階穩(wěn)定點,所以,由三階穩(wěn)定點的定義及四階泰勒展開式可知,當|s|及|t|充分小時,Δf=f(x0+s,y0+t)-f(x0,y0)符號就由ω=14![As4+4Bs3t+6Cs2t2+4Dst3+Et4]的符號決定。設s=rcosαt=rsinα則r=√s2+t2,當A≠0,AC-B2≠0時,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不同時為零時,r≠0;且當sinα=0時,cosα=±0,于是,當A≠0,AC-B2>0,3(AC-B2)(A3E-B4)-2(A2D-B3)2>0時,ω保持與A同號。所以,函數(shù)f(x,y)在點(x0,y0)處取極值。且當A>0時取極小值;當A<0時取極大值。(2)當A≠0,AC-B2<0,3(AC-B2)(A3E-B4)-2(A2D-B3)2>0時,若取sinα=0,則ω與A同號;若取ctgα=-BA,則ω與A異號,所以,函數(shù)f(x,y)在點(x0,y0)處不取極值。推論:設函數(shù)f(x,y)在點(x0,y0)處有各階連續(xù)的偏導數(shù),且點(x0,y0)為函數(shù)f(x,y)的三階穩(wěn)定點,則:(1)當E≠0,EC-D2>0,3(EC-D2)(E3A-D4)-2(E2B-D3)2>0時,函數(shù)f(x,y)在點(x0,y0)處取極值。且當E>0時取極小值;當E<0時取極大值;(2)當E≠0,EC-D2<0,3(EC-D2)(E3A-D4)-2(E2B-D3)2>0時,函數(shù)f(x,y)在點(x0,y0)處不取極值。定理6:設函數(shù)f(x,y)在點(x0,y0)處有各階連續(xù)的偏導數(shù),點(x0,y0)為函數(shù)f(x,y)的三階穩(wěn)定點,且A=E=0,則當B≠0(或D≠0)16BD-9C2>0時,函數(shù)f(x,y)在點(x0,y0)處不取極值。證明:由定