# Proving single solution to an initial value problem y' = \left|\frac{1}{1+x^2}

iyiswad9k 2022-05-01 Answered

Proving single solution to an initial value problem

or each $\left({x}_{0},{y}_{0}\right)\in \mathbb{R}×\mathbb{R}$ I need to prove that there is a single solution defined on $\mathbb{R}$

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## Answers (1)

Zemmiq34
Answered 2022-05-02 Author has 11 answers

Step 1
Let
$g\left(x,y\right)=\mathrm{sin}|{x}^{2}+\mathrm{arctan}\left({y}^{2}\right)|=\mathrm{sin}\left({x}^{2}+\mathrm{arctan}\left({y}^{2}\right)\right)$
This function is derivable, hence
${g}_{y}\left(x,y\right)=\frac{2y}{{y}^{4}+1}\mathrm{cos}\left\{\left({x}^{2}+\mathrm{arctan}\left\{\left({y}^{2}\right)\right\}\right)\right\}$
${g}_{y}$ is continuous thus for each J there is a constant ${L}_{J}$ so that
$|g\left(x,{y}_{1}\right)-g\left(x,{y}_{2}\right)|\le {L}_{J}|{y}_{1}-{y}_{2}|$
therefore
$|f\left(x,{y}_{1}\right)-f\left(x,{y}_{2}\right)|$
$=||\frac{1}{1+{x}^{2}}+g\left(x,{y}_{1}\right)|-|\frac{1}{1+{x}^{2}}+g\left(x,{y}_{2}\right)\mid \mid$
$\le |\frac{1}{1+{x}^{2}}+g\left(x,{y}_{1}\right)-\frac{1}{1+{x}^{2}}-g\left(x,{y}_{2}\right)|$
$=|g\left(x,{y}_{1}\right)-g\left(x,{y}_{2}\right)|\le {L}_{J}|{y}_{1}-{y}_{2}|$
and $f\left(x,y\right)$ is Lipschitz continuous on the 𝑦 variable, and this is true for each box $J×\left(-\mathrm{\infty },\mathrm{\infty }\right)$ and therefore there is a single solution on $\mathbb{R}$ for each $\left({x}_{0},{y}_{0}\right)\in \mathbb{R}×\mathbb{R}$

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