Analysis on Manifolds 9. The Implicit Function Theorem

本节是另一个重要的定理：隐函数定理，隐函数表示的函数在某一点(a,b)可微的充要条件是隐函数关于因变量的导数在(a,b)点是非奇异的，并且可以显式计算出隐函数的导数。

Exercises

Exercise 1. Let $f:\mathbf{R}^3\to\mathbf{R}^2$ be of class $C^1$ ; write $f$ in the form $f(x,y_1,y_2)$ . Assume that $f(3,-1,2)=0$ and

$\displaystyle{Df(3,-1,2)=\begin{bmatrix}1&2&1\\1&-1&1\end{bmatrix}.}$

( a ) Show there is a function $g:B\to\mathbf{R}^2$ of class $C^1$ defined on an open set $B$ in $\mathbf{R}$ such that $f(x,g_1(x),g_2(x))=0$ for $x\in B$ , and $g(3)=(-1,2)$ .
( b ) Find $Dg(3)$ .
( c ) Discuss the problem of solving the equation $f(x,y_1,y_2)=0$ for an arbitrary pair of the unknowns in terms of the third, near the point $(3,-1,2)$ .

Solution:
( a ) Since

$\displaystyle{\det\frac{\partial f(x,y_1,y_2)}{\partial(y_1,y_2)}(3,-1,2)=\det\begin{bmatrix}2&1\\-1&1\end{bmatrix}=5\neq 0}$

by the implicit function theorem, the conclusion holds.
( b ) We have

$\displaystyle{\begin{aligned}Dg(3)&=-\left[\frac{\partial f(x,y_1,y_2)}{\partial(y_1,y_2)}(3,-1,2)\right]^{-1}\cdot\frac{\partial f(x,y_1,y_2)}{\partial(x)}(3,-1,2)\\&=-\begin{bmatrix}2&1\\-1&1\end{bmatrix}^{-1}\cdot\begin{bmatrix}1\\1\end{bmatrix}\\&=\frac{1}{3}\begin{bmatrix}-1&1\\-1&2\end{bmatrix}\cdot\begin{bmatrix}1\\1\end{bmatrix}=\frac{1}{3}\begin{bmatrix}0\\1\end{bmatrix}\end{aligned}}$

( c ) By some easy calculation we can know that

$\displaystyle{\det\frac{\partial f(x,y_1,y_2)}{\partial(y_1,y_2)}(3,-1,2)\neq 0,\quad\det\frac{\partial f(x,y_1,y_2)}{\partial(x,y_1)}(3,-1,2)\neq0}$

thus the equation $f(x,y_1,y_2)=0$ can be solved for $(y_1,y_2)$ in terms of $x$ , and for $(x,y_1)$ in terms of $y_2$ , however, as

$\displaystyle{\det\frac{\partial f(x,y_1,y_2)}{\partial(x,y_2)}(3,-1,2)=\det\begin{bmatrix}1&1\\1&1\end{bmatrix}=0}$

the equation $f(x,y_1,y_2)=0$ cannot be solved for $(x,y_2)$ in terms of $y_1$ .

Exercise 2. Given $f:\mathbf{R}^5\to\mathbf{R}^2$ , of class $C^1$ . Let $\mathbf{a}=(1,2,-1,3,0)$ ; suppose that $f(\mathbf{a})=\mathbf{0}$ and

$\displaystyle{Df(\mathbf{a})=\begin{bmatrix}1&3&1&-1&2\\0&0&1&2&-4\end{bmatrix}.}$

( a ) Show there is a function $g:B\to\mathbf{R}^2$ of class $C^1$ defined on an open set $B$ of $\mathbf{R}^3$ such that

$\displaystyle{f(x_1,g_1(x),g_2(x),x_2,x_3)=0}$

for $x=(x_1,x_2,x_3)\in B$ , and $g(1,3,0)=(2,-1)$ .
( b ) Find $Dg(1,3,0)$ .
( c ) Discuss the problem of solving the equation $f(\mathbf{x})=\mathbf{0}$ for an arbitrary pair of the unknowns in terms of the others, near the point $\mathbf{a}$ .
Solution:
( a ) Since

$\displaystyle{\det\frac{\partial f}{\partial(g_1,g_2)}(\mathbf{a})=\det\begin{bmatrix}3&1\\0&1\end{bmatrix}=3\neq 0}$

by the implicit function theorem, the conclusion holds.
( b ) We have

$\displaystyle{\begin{aligned}Dg(1,3,0)&=-\left[\frac{\partial f}{\partial(g_1,g_2)}(\mathbf{a})\right]^{-1}\cdot\frac{\partial f}{\partial(x_1,x_2,x_3)}(\mathbf{a})\\&=-\begin{bmatrix}3&1\\0&1\end{bmatrix}^{-1}\cdot\begin{bmatrix}1&-1&2\\0&2&-4\end{bmatrix}\\&=\frac{1}{3}\begin{bmatrix}-1&1\\0&-3\end{bmatrix}\cdot\begin{bmatrix}1&-1&2\\0&2&-4\end{bmatrix}\\&=\begin{bmatrix}-1/3&1&-2\\0&-2&4\end{bmatrix}\end{aligned}}$

( c ) The pairs that cannot solve the equation $f(\mathbf{x})=\mathbf{0}$ in terms of the others, denoting $\mathbf{x}=(x_1,x_2,x_3,x_4,x_5)$ , are: $(x_1,x_2)$ and $(x_4,x_5)$ .

Exercise 3. Let $f:\mathbf{R}^2\to\mathbf{R}$ be of class $C^1$ , with $f(2,-1)=-1$ . Set

$\displaystyle{\begin{aligned}G(x,y,u)&=f(x,y)+u^2,\\H(x,y,u)&=ux+3y^3+u^3.\end{aligned}}$

The equations $G(x,y,u)=0$ and $H(x,y,u)=0$ have the solution $(x,y,u)=(2,-1,1)$ .
( a ) What conditions on $Df$ ensure that there are $C^1$ function $x=g(y)$ and $u=h(y)$ defined on an open set in $\mathbf{R}$ that satisfy both equations, such that $g(-1)=2$ and $h(-1)=1$ ?
( b ) Under the conditions of (a), and assuming that $Df(2,-1)=[1,-3]$ , find $g'(-1)$ and $h'(-1)$ .
Solution:
( a ) We have $DG=\begin{bmatrix}\partial{f}/\partial{x}&\partial{f}/\partial{y}&2u\end{bmatrix}$ and $DH=\begin{bmatrix}2u&9y^2&x+3u^2\end{bmatrix}$ , thus

$\displaystyle{\frac{\partial{G}}{\partial{y}}(2,-1,1)=\frac{\partial f}{\partial y}(2,-1),\quad \frac{\partial{H}}{\partial{y}}(2,-1,1)=9}$

thus we need $\partial{f}/\partial{y}(2,-1)\neq 0$ .
( b ) From $G(2,-1,1)=0$ we have

$\displaystyle{\frac{\partial{f}}{\partial{x}}(2,-1)g'(-1)+\frac{\partial{f}}{\partial{y}}(2,-1)+2h(-1)h'(-1)=0}$

which gives $g'(-1)+2h'(-1)=3$ . Similarly, from $H(2,-1,1)=0$ we have

$\displaystyle{h(-1)g'(-1)+h'(-1)g(-1)+9(-1)^2+3[h(-1)]^2h'(-1)=0}$

which gives $g'(-1)+5h'(-1)=-9$ . Together we solve to get $g'(-1)=15,h'(-1)=-6$ .

Exercise 4. Let $F:\mathbf{R}^2\to\mathbf{R}$ be of class $C^2$ , with $F(0,0)=0$ and $DF(0,0)=[2,3]$ , Let $G:\mathbf{R}^3\to\mathbf{R}$ be defined by the equation

$\displaystyle{G(x,y,z)=F(x+2y+3z-1,x^3+y^2-z^2)}$

( a ) Note that $G(-2,3,-1)=F(0,0)=0$ . Show that one can solve the equation $G(x,y,z)=0$ for $z$ , say $z=g(x,y)$ , for $(x,y)$ in a neighborhood $B$ of $(-2,3)$ , such that $g(-2,3)=-1$ .
( b ) Find $Dg(-2,3)$ .
( c ) If $D_1D_1F=3$ and $D_1D_2F=-1$ and $D_2D_2F=5$ at $(0,0)$ , find $D_2D_1g(-2,3)$ .
Solution:
( a ) Let $H(x,y,z)=(x+2y+3z-1,x^3+y^2-z^2)$ , then $G(x,y,z)=(F\circ H)(x,y,z)$ , and by the chain rule

$\displaystyle{DG(-2,3,-1)=DF(0,0)\cdot DH(-2,3,-1)=[2,3]\cdot\begin{bmatrix}1&2&3\\12&6&2\end{bmatrix}=\begin{bmatrix}38&22&12\end{bmatrix}}$

We can see that $\partial G/\partial z(-2,3,1)=12$ , which satisfies the condition of the implicit function theorem.
( b ) $Dg(-2,3)=-\dfrac{1}{12}[38,22]$ .
( c ) We have $DF=[D_1F,D_2F]$ and

$\displaystyle{DH=\begin{bmatrix}1&2&3\\3x^2&2y&-2z\end{bmatrix}}$

this gives

$\displaystyle{DG(x,y,z)=DF\cdot DH=\begin{bmatrix}D_1F+3x^2D_2F&2D_1F+2yD_2F&3D_1F-2zD_2F\end{bmatrix}}$

from this and the expression of $Dg(x,y)$ we have, suppose the condition of implicit funtion theorem is satisfied, that for $z=g(x,y)$ :

$\displaystyle{Dg(x,y)=-\frac{1}{3D_1F-2zD_2F}\begin{bmatrix}D_1F+3x^2D_2F&2D_1F+2yD_2F\end{bmatrix}}$
$\displaystyle{D_1g(x,y)=-\frac{D_1F+3x^2D_2F}{3D_1F-2zD_2F}}$

thus to calculate $D_2D_1g(-2,3)$ , we notice $D_1F(0,0)=2,D_2F(0,0)=3$ , and as $F$ is of class $C^2$ , we get $D_1D_2F=D_2D_1F=-1$ at $(0,0)$ . Let $H=D_1F+3x^2D_2F$ , we have $D_2H=D_2D_1F+3x^2D_2D_2F$ and

$\displaystyle{H(-2,3,1)=2+36=38,D_2H(-2,3,1)=-1+60=59}$

Let $K=3D_1F-2zD_2F$ , we have $D_2K=3D_2D_1F-2D_2gD_2F-2gD_2D_2F$ , thus

$\displaystyle{K(-2,3,1)=6-2(-1)3=12,D_2K(-2,3,1)=3(-1)+11+2\times 5=18}$

Now

$\displaystyle{D_2D_1g(-2,3)=-\frac{(D_2H)K-H(D_2K)}{K^2}(-2,3,1)=-\frac{59\times 12-38\times 18}{144}=-\frac{24}{144}}$

Exercise 5. Let $f.g:\mathbf{R}^3\to\mathbf{R}$ be functions of class $C^1$ . “In general”, one expects that each of the equations $f(x,y,z)=0$ and $g(x,y,z)=0$ represents a smooth surface in $\mathbf{R}^3$ , and that their intersection is a smooth curve. Show that if $(x_0,y_0,z_0)$ satisfies both equations, and if $\partial (f,g)/\partial (x,y,z)$ has rank 2 at $(x_0,y_0,z_0)$ , then near $(x_0,y_0,z_0)$ , one can solve these equations for two of $x,y,z$ in terms of the third, thus representing the solution set locally as a parametrized curve.
Solution: Since $\partial (f,g)/\partial (x,y,z)$ has rank 2 at $(x_0,y_0,z_0)$ , we see that at least one of $\det\partial (f,g)/\partial (x,z)$ , $\det\partial (f,g)/\partial (x,y)$ and $\det\partial (f,g)/\partial (y,z)$ is nonzero, which means we can apply the implicit function theorem to get the result.

Exercise 6. Let $f:\mathbf{R}^{k+n}\to\mathbf{R}^n$ be of class $C^1$ ; suppose that $f(\mathbf{a})=\mathbf{0}$ and that $Df(\mathbf{a})$ has rank $n$ . Show that if $\mathbf{c}$ is a point of $\mathbf{R}^n$ sufficiently close to $\mathbf{0}$ , then the equation $f(\mathbf{x})=\mathbf{c}$ has a solution.
Solution: Since $Df(\mathbf{a})$ has rank $n$ , we can find $n$ columns of $Df(\mathbf{a})$ to be linearly independent, we assume the first $n$ columns are so, and discuss the general case at last.
Let $F:\mathbf{R}^{k+2n}\to\mathbf{R}^{k+n}$ be defined as follows:

$\displaystyle{F(\mathbf{x},\mathbf{y})=\big(f(\mathbf{a}+\mathbf{y})-\mathbf{x},\pi_{n+1}(\mathbf{y}),\dots,\pi_{n+k}(\mathbf{y})\big),\quad\mathbf{x}\in\mathbf{R}^{n},\mathbf{y}\in\mathbf{R}^{k+n}}$

then $F(\mathbf{0}_n,\mathbf{0}_{n+k})=\mathbf{0}_{n+k}$ , and

$\displaystyle{\frac{\partial F}{\partial \mathbf{y}}(\mathbf{0}_n,\mathbf{0}_{n+k})=\begin{bmatrix}D_1f(\mathbf{a})&\cdots&D_{n+1}f(\mathbf{a})&\cdots&D_{n+k}f(\mathbf{a})\\&&1&&\\&&&\ddots&\\&&&&1\end{bmatrix}}$

this is an $(n+k)\times (n+k)$ matrix which is non-singular, as we assume $D_1f(\mathbf{a}),\dots,D_nf(\mathbf{a})$ are linearly independent. Thus by the implicit function theorem ,there is a neighborhood $B$ of $\mathbf{0}_n$ in $\mathbf{R}^n$ and a unique continuous function $g:B\to\mathbf{R}^{k+n}$ such that $g(\mathbf{0}_n)=\mathbf{0}_{n+k}$ and

$\displaystyle{F(\mathbf{x},g(\mathbf{x}))=\mathbf{0},\quad\forall\mathbf{x}\in B}$

Now if $\mathbf{c}$ is close enough to $\mathbf{0}$ such that $\mathbf{c}\in B$ , we can have $F(\mathbf{c},g(\mathbf{c}))=\mathbf{0}$ , which means $f(\mathbf{a}+g(\mathbf{c}))=\mathbf{c}$ .
Finally, in the general case, for any $n$ columns of $Df(\mathbf{a})$ which are linearly independent, which means there are $k$ columns remaining, we substitute $\pi_{n+1}(\mathbf{y}),\dots,\pi_{n+k}(\mathbf{y})$ with the corresponding $k$ column coordinate functions, and the conclusion follows.

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