数学分析简明教程答案17隐函数存在定理

第十七章隐函数存在定理 单个方程的情形Dx0axx0ay0byy0bF(x0,y0)0存在0，使得在点(x0y0F(xy0唯一地确定了一个定义在(x0x0yf(xF(x,f(x))0y0f(x0；yf(x在(x0x0xx0，由条件(2F(x0y00F(x0,y0b)0，F(x0,y0b)0F(xyyy0byy0bF(xy0bF(xy0

F(x,y0b)0 同理，存在202axx02x02F(x,y0b)0 取min(1,2，则xx0x0时，（1）（2）两式同时成立，对任xx0x0F(xyyy0by0bF(x,y0b)0，F(x,y0b)0yy0by0bF(xy0F(xyy在y0by0by是唯一的，这样就确定了一个定义在区间(x0x0yf(xy0f(x0xx

x0x0yf(xf(xx对0，不妨让充分小使得y,yy0b,y0by的一元F(xy在y,y上严格单调F(x,y)0，所F(x,y)0,F(x,y)0x的一元函数F(xyF(xy)0xx0x010，满足(x1x1(x0x0xx1x1F(x,y)0 同理，存在20，满足(x2x2x0x0xx2x2F(x,y)0 取min{1,20xxx时，(3)(4)y

f(xyyy，即f(xf(x)，从而结论（ii）x2ysin(xy)0yf(xxgyF(x)x2ysin(xyF(0,0)0Fx2xycos(xy),Fy1xcos(xy)yf(xFx(0,00，因而据此无法判定是否在(0,0)F(xy)y2x21x20在哪些点的附近可以唯一地确定单值、连续且有yf(x.)

0因而在方程F(x,y)0的除去(0,0)，(1,0)的解点处，均可唯一地确定单值、连续、且有yf(x.ee中sinhy 2F(xy)sinysinhyxFx(xy)1，Fy(xy)cosycoshy值不等式，coshyeyey

(xy)cosycoshy0， eye 而在方F(xy)0的任一解点附近，可确定唯一可导的函yy(x，y(x)Fx(x,y)Fy(x,

.cosycoshxyzlnye个变量的函数

1在点P0(0,1,1)的某邻域内能否确定出某一个变量是另F(x,y,z)yzexz,F(x,y,z)xz,F(x,y,z)lnyxe 某邻域内，可以确定出隐函数xxyz，亦可确定出隐函数yy(xz)，但由Fz(0,1,1)0P0(0,1,1)点的zz(xy存在f是一元函数，试f2f(xy)f(xfy在点(1,1F(xy)2f(xyf(xfyF(1,1)0fFx(x,y)2yf(xy)f(x),Fy(x,y)2xf(xy)f(y)

f(xfy在点(1,1)fx01f(10时，方2f(xy)f(xfy在点(1,1的邻域内能确定出唯一的yx的函数.xyy，其中(0)0，且当ayay)k1证明：存在0，当xyy(xxy(y(0)0F(xy)xyyF(0,0)0Fx(x,y)1,Fy(x,y)1(y)0Fy(xy0F(xyy在(0,0)1知存在0，当xyy(xxyy)y(0)0.

方程组的情形x2y21z2 P0(1,1,2xf(zyg(zF(x,y,z)x2y21z22G(x,y,z)xyz(F,(x,

2xGyP01

2 1

2401 xxy设ux ,vxx

xuvuv 设uexsinyvy

xxyxcosyuvuvy解(1)由于ux ,vxsin， ucosy yu yv

y yvcosysin, sin, cos, y

xy

x

x 即函数组ux ,vx J(u,v)

y

cosyysiny siny

10(x,

siny

ycos

cos x1vcosy x1usiny J J y1vycosysiny，y1uysinycos J J x(2)uexxsinyvexxcosyxy

uexsiny,uxcosy,vexcosy,vxsiny J(u,v)(x,

exsinyexcos

xsin

(sinycosy)1)J0的任何点的邻域内，都x1v sin x1u cos J ex(sinycosy) J ex(sinycosy)，y1v cosy y1u exsin ， J xe

(sinycosy)

J

xe

(sinycosy)设u

x,v

y,w

z，其中r

x2y

z2r r r试求以uvw为自变量的反函数组 (x,y,解(1)根据已知条件uxvr

y,wzr r

r x2y2z2知r2r4(u2v2w2)，化简1r2

u2v2xur2 ,yvr2zwr2

u2v2 u2v2 u2v2222z(2)根据222z12xr r

r4(u,v,w)(x,y,

r4

12yr rr42x2r

2 r 4.fi,i2y2r222z2r1rFi(x1,x2,,xn4.fi,i2y2r222z2r1r(F1F2Fn)(x1,x2,,xn将1(x12(x2n(xn(F1,F2,,Fn

F2

f11f21

f1d2 2f2d2

f1nf2 (x1,x2,,xn

Fn

fn1

fnd2

fnnnnd1

dnf

f1 f2

f

fn

(x)(x)(x)(f1,f2,,fn) (1,2,,

f(x,y)3xg(x,y)y0g(x,y)3yf(x,y)x0F(x,y,u,v)u3xvG(x,y,u,v)v3yuF,G关于各个变元在P0(0,1,1,1)附近有连续偏

3uy

3v

在连续可微函数uf(xy和vg(x,yf(0,1)1g(0,1)1f(x,y)3xg(x,y)y0g(x,y)3yf(x,y)x0设uf(x,y,z,g(y,z,t)h(z,t)u在什么条件下uxyxygyzt)0和h(zt0gyzt),h(zt在某一点P0y0z0t0附近对各变量有一阶连续偏导数g(y0,z0,t0)h(y0,z0,t0)0(g,(z(g,(z,g(y,z,t) zz(则在y0点附近方程组h(z,t) 唯一地确定一组函数tt(y)，而且这组函数在 点附近连续可微，从而uf(xyztf(x,yzyty就是关x,y的函数，并有 f f 1f(g, 1f(g,xx，yyzdy

dyyJz(y,t)

t(z,J

(g,.(z,设函数uu(xdud

uf(x,y,g(x,y,z)h(x,y,z)2 2dx由于原方程组能确定函数uu(x，根据方程组中ug(xyz0h(xyz0yzx (g, (g,h),dzdx

(g, (g,(y, (y,duffdyfdzff f y z

(g,(x,

z(g,(g,(g,(g,(g,(y,duffdy

y

zdx

dfdx

dzdx2x

2

2

dz

2

2y

2

d2ydx2

2

2z

d2 dx d其中 , 2的表达式中dx d2 (g,h)(g, (g, (g,h) (g,h)dx

x(x,z)(y,

(x,z)x(y,z)

(y,z) d2 (g,h)(g, (g, (g,h) (g,h)dx

x

(y,x)

(y,

x

(y,z)

(y,z) zz(xydz

f(x,y,z,t)g(x,y,z,t)zz(xytt(xyz(f, (x,

(f,g),z(f, (f,g)(z, (z,t)

(f,

(f,g)

(f,

(f,g)dzxdxydy

(z,t)dx

(y,

(z,t)dy 设u

uf(xut,yut,zg(x,y,z)xy.这时t

g(xyz0xyg1g3x0,g2g3y0zg1,zg2 uf1tuftufzu x

2 x

t 3 x

g1f1f1txf2txf3txf3gu

f1g3f3

，同样

u

3f2g3f3g

，中t是自变

g31t(f1f2f3 g31t(f1f2f3设(x0y0z0u0满足方程f(x)f(y)f(z)Fg(x)g(y)g(z)h(x)h(y)h(z)Hf(x)xg(x)x2h(xx3的情形下，上述条件相当于什么？解（1）P0(x0y0z0u0，则根据已知条件可知，当条件f(x0)f(y0)f(z0)F(u0g(x0)g(y0)g(z0)G(u0h(x)h(y)h(z)H(u f f( f f(x0 f(y0 f(z0PJP0

g

g

g(x0 h(x0

g(y0h(y0

g(z0h(z0

0（2）f(x)xg(x)x2h(x)x3x0y0z0F(u0 2y

0 G(u0xx32z3H00111111(ii)J2z06(y0x0)(z0x0)(z0y0)003