Tutorial note 3 (inverse and implicit function theorem)

Example 1. (*) says that the surface \[
F^{-1}(\{0\})=\{(x,y,z)\in \R^3:F(x,y,z)=0\}
\] near $p$ can be parametrized as the graph of a $C^1$ function $G$ defined near $(p_1,p_2)$.

Thus generally implicit function theorem is nothing but parametrization of "abstract surfaces" near some point, in such general cases $G$ will be vector-valued and cannot be easily visualized, unless we are god :).

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Example 2. Define \[
(u,v)=F(x,y).
\] To express $x,y$ as a differentiable function of $(u,v)$ is to find differentiable $F^{-1}$ such that $(x,y)=F^{-1}(u,v)$. For this, we use inverse function theorem.

Step 1: We need to show that $\bm F$ is $\bm{C^1}$ near $\bm{(0,1)}$

Procedure: The standard way to do this is to find an $r>0$ such that all partial derivatives of $F$ exist and are continuous on $B((0,1),r)$. In many cases it is easy to choose suitable $r$.

Now \[
u_x=1,\quad u_y=1
\] and \[
v_x=\cos x,\quad v_y=-\sin y.
\] As they exist and are continuous on $\R^2$, they exist and are continuous on $B((0,1),{\LARGE 1})$, so $F$ is $C^1$ near $(0,1)$.

Step 2: We need to show that $\bm{F'(0,1)}$ is invertible
This is a routine calculation: \[
\det F'(0,1) = \matrixx{
1&1\\
1&-\sin 1
} = -\sin 1-1\neq 0.
\]

Step 3: Make conclusion (=.=)
By inverse function theorem near $(0,1)$ $F$ has a $C^1$ inverse $F^{-1}$, thus \[(x,y)=(x(u,v),y(u,v)) =F^{-1}(u,v):\text{near $F(0,1)$}\to \text{near $(0,1)$}\] is now a differentiable function of $(u,v)$. Moreover, \begin{align*}
\matrixx{
x_u&x_v\\
y_u&y_v
}(1,\cos 1) &= (F^{-1})'(F(0,1)) \\
&= (F'(0,1))^{-1}\\
&=\matrixx{
1&1\\1&-\sin 1
}^{-1}\\
&=\frac{1}{1+\sin 1}\matrixx{
\sin 1&1\\
1&-1
}.\end{align*}

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Example 3.
(a) We define \[
F(x,y,u,v)=(f,g)=\matrixx{
x^2+2y+2+u^2+v-6\\
2x^3+4y^2+u+v^2-9
}.
\] To "solve" $(u,v)$ out from the equations for $(x,y,u,v)$ near $p$, we try to prove $F_{u,v}(p)$ is an invertible matrix plus some extra conditions.

Step 1: Show that $\bm F$ is $\bm{C^1}$ near $\bm p$
It is similar to Example 2, indeed, \[
F'(x,y,u,v)=\matrixx{
f_x&f_y&f_u&f_v\\
g_x&g_y&g_u&g_v
}=\matrixx{
2x&4y&2u&1\\
6x^2&8y&1&2v
},
\] clearly all partial derivatives exist and are continuous on $B(p,1)$, so $F$ is $C^1$ near $p$.

Step 2: Show that $\bm{F(p)=0}$
This can be done by direct calculation, make sure that you finish this step in your midterm/final exams.

Step 3: Show that $\bm{F_{u,v}(p)}$ is invertible
We have \[
F'(p)=F'(1,-1,-1,2)=\matrixx{
2&-4&-2&1\\
6&-8&1&4
}.
\] Therefore $F_{u,v}(p)=\matrixx{-2&1\\1&4}$ with $\det F_{u,v}(p)=-9$, thus $F_{u,v}(p)$ is invertible. By implicit fucntion theorem $u,v$ can be expressed as a function $G(x,y)$ defined near $(1,-1)$ (such that $F(x,y,G(x,y))=0$ for $(x,y)$ near $(1,-1)$).

(b) Method 1. We use the equation on the first page of tutorial note 3 (or the exact version in Example 4) directly to get \[
G'(1,-1)=-\matrixx{
-2&1\\
1&4}^{-1}\matrixx{2&-4\\6&-8
}=\matrixx{
2/9&-8/9\\
-14/9&12/9
}.
\] Method 2. We can try to create a system of linear equations and then solve for $u_x$ and $v_x$. Indeed, we differentiate (1) and (2) w.r.t. $x$ to obtain \[
2x+2uu_x+v_x=0\quad \text{and}\quad 6x^2+u_x+2vv_x=0.
\] Now we put $(x,y)=(1,-1)$ and use the fact that $u(1,-1)=-1$ and $v(1,-1)=2$, then \[
2u_x-v_2=2\quad \text{and}\quad u_x+4v_x=-6,
\] we solve them to get $u_x(1,-1)=2/9$ and $v_x(1,-1)=-14/9$.

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Example 4. (i), (ii) and (iii) are just the conditions in implicit function theorem, thus near $p_0$, $(u,v)$ can be expressed as a $C^1$ function $G(x,y)$ such that \[
F(x,y,G(x,y))=0,
\] for $(x,y)$ near $(x_0,y_0)$. Now we differentiate both sides (taking Jacobian matrix) to get \[
0=F'(x,y,G(x,y))\cdot\left[
\begin{array}{c}
1\quad 0\\
0\quad 1\\
G'(x,y)
\end{array}
\right] =F_{x,y}(x,y,G(x,y)) + F_{u,v}(x,y,G(x,y))G'(x,y),
\] here last equality follows from block matrix multiplication, note that $G'(x,y)$ is a $2\times 2$ matrix, now we solve from the above equation to get \[
G'(x,y) = - F_{u,v}(x,y,G(x,y))^{-1}F_{x,y}(x,y,G(x,y)).
\] When $(x,y)=(x_0,y_0)$, then \[
\boxed{G'(x_0,y_0) = - F_{u,v}(p_0)^{-1}F_{x,y}(p_0),}
\] this is the version we use in Example 3.

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Example 5. We note that the map from $\R^n$ to $\R$ defined by \[
x\mapsto \|F'(x)\|
\] is continuous since $F$ is continuously differentiable (i.e., $C^1$). Therefore if $F'(x_0)=0$, for some $x_0$, then for every $\epsilon>0$ there is a $\delta>0$ such that \[
\|x-x_0\|<\delta \implies \|F'(x)\|<\epsilon.
\] Therefore by Exercise 4 of tutorial note 2 we have for every $x,y\in B(x_0,\delta)$, \[
\|F(x)-F(y)\|\leq \epsilon\|x-y\|.
\] By hypothesis of this example we have \[
c\|x-x_0\|\leq \epsilon\|x-x_0\|,
\] therefore if we take $x\in B(x_0,\delta)\setminus \{x_0\}$, then \[
c\leq \epsilon.
\] But $\epsilon>0$ is arbitrary, we have $0<c\leq 0$ by taking $\epsilon\to 0^+$, a contradiction.