The Implicit Function Theorem For Functions from Rn to Rn Examples 1

Recall from The Implicit Function Theorem for Functions from Rn to Rn page that if $A \subseteq \mathbb{R}^{n+k}$ is open and $\mathbf{f} : A \to \mathbb{R}^n$ is a continuously differentiable function on $A$ ($\mathbf{f}$ is $C^1$ on $A$) then if there exists an $(\mathbf{x}_0; \mathbf{t}_0) \in A$ for which $\mathbf{f}(\mathbf{x}_0; \mathbf{t}_0) = 0$ and the $n \times n$ Jacobian determinant $\mathrm{det} (D_j f_i (\mathbf{x}_0, \mathbf{t}_0) \neq 0$ then there exists a $k$-dimensional open set $T_0$ with $\mathbf{t}_0 \in T$ and a unique $\mathbf{g} : T_0 \to \mathbb{R}^n$ such that:

• 1. $\mathbf{g}$ is continuously differentiable on $T_0$ ($\mathbf{g}$ in $C^1$ on $T_0$).

• 2. $\mathbf{g}(t_0) = \mathbf{x}_0$.

• 3. $\mathbf{f}(\mathbf{g}(\mathbf{t}); \mathbf{t}) = \mathbf{0}$ for all $\mathbf{t} \in T_0$.

We will now look at some examples of applying the Implicit Function theorem.

Example 1

Determine whether the equation $(x^2 + y^2 + 2z^2)^{1/2} = \cos z$ can be solved implicitly for $y$ in terms of $x$ and $z$ in a neighbourhood of the point $(x, y, z) = (0, 1, 0)$.

We have that $n = 1$ (the number of dependent variables, here we have $1$, $y$), and $k = 2$ (the number of independent variables, here we have $2$, $x$ and $z$).

Set $\mathbf{x} = (y)$ and $\mathbf{t} = (x, z)$. Then $(\mathbf{x}, \mathbf{t}) = (y; x, z)$, and in this display of coordinates we're consider neighbourhoods around the point $(\mathbf{x}_0; \mathbf{t}_0) = (y; x, z) = (1; 0, 0)$.

We have that:

(1)

\begin{align} \quad f(y; x, z) = (x^2 + y^2 + 2z^2)^{1/2} - \cos z \end{align}

Note that $f$ is continuous differentiable ($C^1$) everywhere except at the origin $(0; 0, 0)$ and that $f(1; 0, 0) = (0^2 + 1^2 + 2(0)^2)^{1/2} - \cos 0 = 1 - 1 = 0$. The $n \times n = 1 \times 1$ Jacobian determinant of $f$ in this example, $\mathrm{det} (D_j f_i)$, is simply $\displaystyle{\frac{\partial f}{\partial y}}$:

(2)

\begin{align} \quad \frac{\partial f}{\partial y} = \frac{2y}{2(x^2 + y^2 + 2z^2)^{1/2}} = \frac{y}{(x^2 + y^2 + 2z^2)^{1/2}} \end{align}

Evaluating this at the point $(\mathbf{x}_0; \mathbf{t}_0) = (y; x, z) = (1; 0, 0)$ gives us:

(3)

\begin{align} \quad \frac{\partial f}{\partial y} \biggr \lvert_{(y; x, z) = (1; 0, 0)} = \frac{1}{1} = 1 \neq 0 \end{align}

So the conditions of the Implicit Function theorem are satisfied so it can be applied, i.e., $(x^2 + y^2 + 2z^2)^{1/2} = \cos z$ can be solved implicitly for $y$ in terms of $x$ and $z$ is a neighbourhood around the point $(x, y, z) = (0, 1, 0)$.