Solutions to Systems of Equations by Brouwer's Fixed Point Theorem

# Solutions to Systems of Equations by Brouwer's Fixed Point Theorem

Recall from the Brouwer's Fixed Point Theorem page the very important Brouwer's fixed point theorem which states that if $D^2$ denotes the closed unit disk then if $f : D^2 \to D^2$ is continuous then $f$ has a fixed point, that is, there exists a point $(x, y) \in D^2$ such that $f(x, y) = (x, y)$.

The general fixed point theorem states that if $f : D^n \to D^n$ is continuous then $f$ has a fixed point. And even more generally, if $A$ is homeomorphic to $D^n$ and $f : A \to A$ is continuous then $f$ has a fixed point.

Let $A \subset \mathbb{R}^n$ be homeomorphic to $D^n$. Consider a system of $n$ equations in $n$ variables:

(1)
\begin{align} \quad f_1(x_1, x_2, ..., x_n) &= 0 \\ \quad f_2(x_1, x_2, ..., x_n) &= 0 \\ \quad & \vdots \\ \quad f_n(x_1, x_2, ..., x_n) &= 0 \quad (*) \end{align}

Suppose further that $f_1, f_2, ..., f_n : \mathbb{R}^n \to \mathbb{R}$ are all continuous so that the function $f : \mathbb{R}^n \to \mathbb{R}^n$ defined by:

(2)
\begin{align} \quad f(x_1, x_2, ..., x_n) = (f_1(x_1, x_2, ..., x_n) + x_1, f_2(x_1, x_2, ..., x_n) + x_2, ..., f_n(x_1, x_2, ..., x_n) + x_n) \end{align}

is continuous. Suppose further that:

(3)
\begin{align} \quad f(A) \subseteq A \end{align}

Then we can apply the Brouwer's fixed point theorem to the function $f |_A : A \to A$ to obtain a fixed point $(y_1, y_2, ..., y_n) \in A$ such that:

(4)
\begin{align} \quad f(y_1, y_2, ..., y_n) = (y_1, y_2, ..., y_n) \end{align}

That is, for each $1 \leq i \leq n$:

(5)
\begin{align} \quad f_i(y_1, y_2, ..., y_n) + y_i &= y_i \\ \quad f_i(y_1, y_2, ..., y_n) &= 0 \end{align}

Therefore $(y_1, y_2, ..., y_n)$ is a solution to the system $(*)$. Observe that method allows us to determine the existence of solutions to particular systems of equations but that it doesn't actually guarantee that the solution is easy to obtain. Let's look at an example.

## Example 1

Show that the system $\left\{\begin{matrix}\sin(xy + 1) - x = 0\\ \cos(x + 2y^2 + 1) - y= 0 \end{matrix}\right.$ has a solution.

Consider the closed unit disk of radius $2$, $2D^2 = \{ (x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq 4 \}$. Define a function $f : \mathbb{R}^2 \to \mathbb{R}^2$ by:

(6)
\begin{align} \quad f(x, y) = (\sin (xy + 1), \cos(x + 2y^2 + 1)) \end{align}

Then for all $(x, y) \in 2D^2$ we have that:

(7)
\begin{align} \quad |f(x, y)| &= \sqrt{[\sin (xy + 1)]^2 + [\cos(x + 2y^2 + 1)]^2} \\ & \leq |\sin(xy + 1) | + |\cos(x + 2y^2 + 1)| \\ & \leq 2 \end{align}

Therefore $f (2D^2) \subseteq 2D^2$. Furthermore, $f$ is continuous. Hence by Brouwer's fixed point theorem there exists a point $(a, b) \in 2D^2$ such that $f(a, b) = (a,b)$, that is:

(8)
\begin{align} \quad \sin (ab + 1) = a \\ \quad \cos (a + 2b^2 + 1) = b \end{align}

Therefore:

(9)
\begin{align} \quad \sin (ab + 1) - a &= 0 \\ \quad \cos (a + 2b^2 + 1) - b &= 0 \end{align}

So $(a, b)$ is a solution to the original system.