向量值映照微分学11逆定理

微分学——逆映照定复旦力学谢锡麟2016419知识要定理1.1(逆映照定理/微分同胚局部存在性定理).设有映f(x):Rm⊃Dx∋x7→f(x)∈满足f(x)∈C1(Dx;x0DxDf(x0Rm×m可逆,U(x0)⊂DxV(f(x0))⊂f(Dx),使得f(x)U(x0)V(f(x0))上为双射,f−1(y)满足f−1(y):V(f(x0))∋y7→f−1(y)∈f−1(y)∈C1(V(f(x0));证明直接利用有界闭集上的压缩映照定理首先Bλ0(x0DxBµ0(y0Rm以及辅助映 ψy(x):Bλ(x0)∋x7→ψy(x)=x+(Df)−1(x0)(y−f(x))∈ ∀y∈Bµ 如果对yBµ0(y0,xyBλ0(x0),满足ψy(xy)=xy等价于y=f现设定ψy(x)的不动点唯一存在,则可η(y):Bµ0(y0)∋y7→η(y)∈满藉此具有如下性质

y=f ∀y∈Bµ0U{x∈Bλ0(x0)|f(x)∈Bµ0(y0)}U为开集.考虑f(x)Bλ0(x0)上的连续性,故对:x∈Uε>0δε>0,成立f(Bδε(xBε(f(xBµ0(y0)即有Bδε(xUf(xU上为单射.利用反证法x1,x2∈U,x1̸=f(x1)=f(x2)∈yf(x1f(x2则按上述求解的唯一性,!xyBλ0(x0)y=f(xy),而x1=x2=xy,同假设f(U)=Bµ0(y0).显然f(U)⊂Bµ0(y0).y∈Bµ(y0),!xy∈Bλ0(x0).满f(xy)=yBµ0(y0).由此Bµ(y0)f(U综上f(x)Bµ0(x0)⊃Uf(U)=Bµ0(y0)之间的双射,故存在逆映照.又由yBµ(y0)y=f(η(y)),所以yBµ(y0)f−1(y)=f−1fη(y)=η(y).因此f−1(y)= ∀y∈第一步,估计ψy(x)−x0Rm=ψy(x)−ψy0(x0)Rm=|ψ(y,x)−ψ(y0,|ψ(y,x)−ψ(y0,x)|Rm+|ψ(y0,x)−ψ(y0,

|Dyψ(y0+θ2(y−y0),x)|Rm×m·|y−

|Dxψ(y0,x0+θ1(x−x0))|Rm×m·|x−x0|Rm此 ψ(y,x)ψy(x)=x+(Df)−1(x0)(y−f ∀y∈Bµ(y0),x∈Bλ Dyψ(y,x)=(DfDxψ(y,x)=IRm−(Df)−1(x0)Df(x)=(Df)−1(x0)·[Df(x0)−Df故ψy(x)− (Df

Rm×m|y−+(Df ·|Df(x0)−Df(x)|Rm×m·|x−x0|R考虑f(xC1(DxRm),故可缩λ0λ1,使再缩小µ0µ1,

(Df

·|Df(x0)−Df(x)|Rm×m<2(Df)−1(x ·|y−y 0

0Rm<2综上,

ψy(x)−x0Rm<2λ1+2|x−x0|Rm<λ ∀x∈Bλ1(x0),y∈Bµ1yBµ(y0),ψy(Bλ1(x0Bλ1(x0)Bλ1第二步,估计压缩性,ψy(x1)−ψy(x2)Rm=|ψ(y,x1)−ψ(y,1

|Dxψ(y,x1+θ1(x2−x1))|Rm×m·|x1−x2|Rm<2|x1−x2|Rm综上有:ψy(x)Bλ1(x0)上的压缩映照,yBµ1(y0).xyBλ1(x0),ψy(xy)=xy∈Bλ1

η(y):Bµ1(y0)∋y7→η(y)=xy∈y=f◦ ∀y∈Bµ1以下η(y的连续性η(yC(Bµ1(y0Rm).估|η(y+∆y)−η(y)|Rm=ψy+∆y(η(y+∆y))−ψy(η(y))=|ψ(y+∆y,η(y+∆y))−ψ(y,|ψ(y+∆y,η(y+∆y))−ψ(y,η(y+∆y))|Rm+|ψ(y,η(y+∆y))−ψ(y,

|Dyψ(y+θ2∆y,η(y+∆y))|Rm×m

|Dxψ(y,η(y)+θ1(η(y+∆y)−η(y)))|Rm×m|η(y+∆y)−1<(Df

|∆y|

|η(y+∆y)− 即所以下考

|η(y+∆y)−η(y)|Rm<2(Dfη(y)∈C(Bµ1(y0);η(y)∈C1(Bµ1(y0);f(x)∈C1(Dx;

Rm×m|∆y|Rm故f(x+∆x)−f(x)=Df(x)·∆x+o(|∆x|Rm ∀x∈现x= ∀y∈Bµ0∆x=η(y+∆y)−则∆x=f◦η(y+∆y)−f◦η(y)=y+∆y−y=∆y=Df(x)·(η(y+∆y)−η(y))+o(|η(y+∆y)−η(y)|Rm此处要求

1∃(Df)−1(x)∈ ∀x∈Bλ1 由于Df(x0Rm×m可逆,故在选定Bµ(x0)时,就可使得对x∈Bµ(x0),都有(Df 由|η(y+∆y)−η(y)|Rm<2(Df所

Rm×m|∆y|Rm|o(|η(y+∆y)−η(y)|Rm)|Rm=|o(|η(y+∆y)−η(y)|Rm)|Rm·|η(y+∆y)− |η(y+∆y)− |o(|η(y+∆y)−η(y)|Rm |η(y+∆y)−

0Rm×m→(∆y→

·2(Df (xη(yBλ1(y0上单射,∆y̸0Rm时,η(yyη(yRm满足非接触性条综上,

η(y+∆y)−η(y)=(Df)−1(x)∆y+ x=f定理1.2(逆映照定理/微分同胚局部存在性定理).设有映f(x):Rm⊃Dx∋x7→f(x)∈满足f(x)∈C1(Dx;x0DxDf(x0Rm×m可逆,U(x0)⊂DxV(f(x0))⊂f(Dx),使得f(x)U(x0)V(f(x0))上为双射,f−1(y)满足f−1(y):V(f(x0))∋y7→f−1(y)∈f−1(y)∈C1(V(f(x0));证明利用隐映照定理,F(y,x):Rm×Dx∋{y,x}7→F(y,x)y−f(x)∈f(xC1(DxRm),F(yx)C1(RmDxRm).Df(x0)Rm×m可逆,DxF(y0x0)=−Df(x0)∈Rm×m可逆y0:=f(x0)Rm.按隐映照定理,Bµ(y0)RmBλ(x0)Dx,yBµ(y0),xyBλ(x0满足F(yxy)=0由此可作η(yBµ(y0)y7→η(y)Rm,满η(y)∈Bλ(x0);F(y,η(y))=0∈η(y)∈C1(Bµ(y0);U{xBλ(x0)|f(x)Bµ(y0)Bλ(x0),具有如下性质U为开集考虑xUf(xBµ(y0)f(xx连续则ε0δε0满足f(Bδε(x))∈Bε(f(x)),自然可Bε(f(x))⊂Bµ(y0),Bδε(x)⊂Bλ(x0),故有Bδε(x)⊂U,即证得U为开集.bebef(UBµ(y0).f(U)Bµ(y0)则只需证Bµ(y0)f(U)即yBµ(y0),xyU,f(xy)=y.按隐映照定理,η(y)∈Bλ(x0),f(η(y))=y.考虑到y∈Bµ(y0),η(y)∈U.综上,f(x)U(Bλ(x0))f(U)=Bµ(y0)之间的双射.ff−1(y):Bµ(y0)∋y7→f−1(y)∈满f−1(y)∈f◦f−1(y)=y,∀y∈考虑到fη(y)=y,y∈Bµ(y