Banach fixed-point theorem : Existence of solution

In summary, the conversation discusses a system of equations and a set $G$ where the system has a solution. The Banach fixed-point theorem is used to show that the system has a solution in $G$. The Banach fixed-point theorem states that a complete metric space with a contraction mapping will have a unique fixed-point. To prove this, it is necessary to show that $\Phi(G)\subseteq G$ and that $\Phi$ is a contraction mapping.
  • #1
mathmari
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Hey! :eek:

We have the system \begin{align*}&x_1=\left (5+x_1^2+x_2^2\right )^{-1} \\ &x_2=\left (x_1+x_2\right )^{\frac{1}{4}}\end{align*} and the set $G=\{\vec{x}\in \mathbb{R}^2: \|\vec{x}-\vec{c}\|_{\infty}\leq 0.2\}$ where $\vec{c}=(0.2,1)^T$.

I want to show with the Banach fixed-point theorem that the system has a solution in $G$.

I have done the following:

Let \begin{equation*}\Phi (x_1, x_2)=\begin{pmatrix}\left (5+x_1^2+x_2^2\right )^{-1} \\ \left (x_1+x_2\right )^{\frac{1}{4}}\end{pmatrix}\end{equation*} The jacobi matrix is \begin{equation*}\nabla \Phi =\begin{pmatrix}-\frac{2x_1}{(x_1^2+x_2^2+5)^2} & -\frac{2x_2}{(x_1^2+x_2^2+5)^2} \\ \frac{1}{4(x_1+x_2)^{\frac{3}{4}}} & \frac{1}{4(x_1+x_2)^{\frac{3}{4}}} \end{pmatrix}\end{equation*}

Then we get \begin{equation*}\|\nabla \Phi \|_{\infty}=\max \left \{\frac{2x_1+2x_2}{(x_1^2+x_2^2+5)^2}, \frac{1}{2(x_1+x_2)^{\frac{3}{4}}}\right \}\end{equation*}

We want to get an upper bound for that, don't we?

How can we do that? Could you give me a hint?

(Wondering)
 
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  • #2
Hey mathmari!

Can't we find an upper bound by using that $(x_1,x_2)$ is in $G$?
That is, $0\le x_1 \le 0.4$ and $0.8\le x_2 \le 1.2$. (Thinking)
 
  • #3
Klaas van Aarsen said:
Can't we find an upper bound by using that $(x_1,x_2)$ is in $G$?
That is, $0\le x_1 \le 0.4$ and $0.8\le x_2 \le 1.2$. (Thinking)

We have that $$\|\vec{x}-\vec{c}\|_{\infty}\leq 0.2 \Rightarrow \|(x_1-0.2, x_2-1)\|_{\infty}\leq 0.2$$ That means that both are less than $0.2$, or not?
So we get $|x_1-0.2|\leq 0.2$ and $|x_2-1|\leq 0.2$, right? (Wondering)
 
  • #4
mathmari said:
We have that $$\|\vec{x}-\vec{c}\|_{\infty}\leq 0.2 \Rightarrow \|(x_1-0.2, x_2-1)\|_{\infty}\leq 0.2$$ That means that both are less than $0.2$, or not?
So we get $|x_1-0.2|\leq 0.2$ and $|x_2-1|\leq 0.2$, right?

Yep. (Nod)
 
  • #5
We have $$\frac{2x_1+2x_2}{(x_1^2+x_2^2+5)^2}\leq 2x_1+2x_2\leq 3.2 $$ or not? What about the second term? (Wondering)
 
  • #6
mathmari said:
We have $$\frac{2x_1+2x_2}{(x_1^2+x_2^2+5)^2}\leq 2x_1+2x_2\leq 3.2 $$ or not? What about the second term?

Can't we do a little better?
$$\frac{2x_1+2x_2}{(x_1^2+x_2^2+5)^2}
\le \frac{2\cdot 0.4 + 2\cdot 1.2}{(0^2+0.8^2+5)^2} < 0.1006
$$
(Thinking)
 
  • #7
Klaas van Aarsen said:
Can't we do a little better?
$$\frac{2x_1+2x_2}{(x_1^2+x_2^2+5)^2}
\le \frac{2\cdot 0.4 + 2\cdot 1.2}{(0^2+0.8^2+5)^2} < 0.1006
$$
(Thinking)

Ah yes! And for the other one we have $$\frac{1}{2(x_1+x_2)^{\frac{3}{4}}}\leq \frac{1}{2(0+0.8)^{\frac{3}{4}}}=0.591089$$

So the maximum is less than $0.591089$ which is less than $1$. Does this mean that there is a solution is $G$ ? (Wondering)
 
  • #8
mathmari said:
Ah yes! And for the other one we have $$\frac{1}{2(x_1+x_2)^{\frac{3}{4}}}\leq \frac{1}{2(0+0.8)^{\frac{3}{4}}}=0.591089$$

So the maximum is less than $0.591089$ which is less than $1$. Does this mean that there is a solution is $G$ ?

Well, wiki lists:

Banach Fixed Point Theorem. Let (X,d) be a non-empty complete metric space with a contraction mapping $T\colon X\to X$. Then T admits a unique fixed-point x* in X (i.e. T(x*) = x*). Furthermore, x* can be found as follows: start with an arbitrary element $x_0$ in X and define a sequence $\{x_n\}$ by $x_n = T(x_{n-1})$, then xn → x*.

So if we consider $G$ to be the set of the metric space and $\Phi$ the mapping, then we would need that $\Phi(G)\subseteq G$, wouldn't we? (Wondering)
And we also need to prove that $\Phi$ is a contraction mapping, don't we? (Thinking)
 
  • #9
Klaas van Aarsen said:
Well, wiki lists:

Banach Fixed Point Theorem. Let (X,d) be a non-empty complete metric space with a contraction mapping $T\colon X\to X$. Then T admits a unique fixed-point x* in X (i.e. T(x*) = x*). Furthermore, x* can be found as follows: start with an arbitrary element $x_0$ in X and define a sequence $\{x_n\}$ by $x_n = T(x_{n-1})$, then xn → x*.

So if we consider $G$ to be the set of the metric space and $\Phi$ the mapping, then we would need that $\Phi(G)=G$, wouldn't we? (Wondering)
And we also need to prove that $\Phi$ is a contraction mapping, don't we? (Thinking)

So for these two do we have to show that $\Phi (G)\subseteq G$ ? (Wondering)
 
  • #10
mathmari said:
So for these two do we have to show that $\Phi (G)\subseteq G$ ?

I believe so yes. (Thinking)
 

FAQ: Banach fixed-point theorem : Existence of solution

What is the Banach fixed-point theorem?

The Banach fixed-point theorem is a mathematical theorem that guarantees the existence of a unique solution to certain types of equations, known as fixed-point equations. It is named after the Polish mathematician Stefan Banach, who first proved it in the early 1920s.

What is a fixed-point equation?

A fixed-point equation is an equation of the form x = f(x), where x is an unknown variable and f is a function. The solution to this type of equation is a value of x that, when substituted into the equation, results in the same value of x. In other words, the function f maps the value of x back to itself.

What are the conditions for the Banach fixed-point theorem to hold?

The Banach fixed-point theorem holds if the following conditions are met:

  • The equation x = f(x) must have a solution, i.e. there must exist a value of x that satisfies the equation.
  • The function f must be a contraction mapping, meaning that it reduces the distance between two points in its domain. This is typically expressed using the concept of a Lipschitz constant.
  • The domain of the function f must be a complete metric space, which means that it is a space where all Cauchy sequences (sequences that get arbitrarily close to each other) converge to a limit that is also in the space.

What are some examples of fixed-point equations that the Banach fixed-point theorem can be applied to?

The Banach fixed-point theorem can be applied to a variety of equations in different fields of mathematics and science. Some examples include:

  • Finding the roots of a polynomial equation by rewriting it as a fixed-point equation.
  • Solving differential equations by converting them into integral equations, which can then be rewritten as fixed-point equations.
  • Proving the existence of solutions to optimization problems, such as finding the maximum or minimum value of a function.

What are some applications of the Banach fixed-point theorem?

The Banach fixed-point theorem has numerous applications in mathematics and other fields, including:

  • Proving the existence of solutions to differential equations and integral equations.
  • Proving the convergence of iterative numerical methods, such as the Newton-Raphson method for finding roots of equations.
  • Proving the existence of solutions to optimization problems.
  • Constructing fractals, which are geometric objects that exhibit self-similar patterns and can be generated using fixed-point equations.

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