- #1

mathmari

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MHB

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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)