MHB Banach fixed-point theorem : Existence of solution

[mathmari]

Hey!

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)

 

[I like Serena]

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)

 

[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)

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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 x_n → 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)

 

[mathmari]

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 x_n → 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)

 

[I like Serena]

So for these two do we have to show that $\Phi (G)\subseteq G$ ?

I believe so yes. (Thinking)