Let the initial value problem :
$$x''' + \frac{1}{1+t^2}x'' + \sin t\cdot x' + \frac{t}{x^2 + y^2 + 1}=0,$$
$$y'' + e^{-t}y' + \cos(x+y)=0,$$
$$x(0)=1,x'(0)=0,x''(0)=5,y(0)=0,y'(0)=5$$
Prove that it has a unique solution, which has a domain that contains a neighborhood of $t=0$.
We've only approached theorems and their proofs so I have completely no experience over exercises, I would really appreciate a thorough solution or explanation on how to handle such problems. I know that initial value problems have unique solutions if the functions are Lipschitz but it doesn't seem to be that simple right here.
Thanks in advance !
What you need is to check if the Lipschitz condition holds or not. Let $x=x_1, x_1'=x_2,x_2'=x_3,y_1=x_4,x_4'=x_5$ and then the system of ODEs become $$ X'=f(t,X) $$ where $$ X=(x_1,x_2,x_3,x_4,x_5)^T $$ and \begin{eqnarray} f(t,X)&=&\left(\begin{matrix}x_2\\x_3\\ -\frac{1}{1+t^2}x_3 -\sin t\cdot x_2 -\frac{t}{x_1^2 + x_4^2 + 1}\\x_5\\ -e^{-t}y_2 - \cos(x_1+x_4) 0\end{matrix}\right). \end{eqnarray} It is not hard to see, for $|t|\le\delta,$, for $X,\bar{X}\in\mathbb{X}$, \begin{eqnarray} \|f(t,X)-f(t,\bar{X})\|&=&\left\|\left(\begin{matrix}x_2-\bar{x}_2\\x_3-\bar{x}_3\\ -\frac{1}{1+t^2}(x_3-\bar{x}_3) -\sin t\cdot (x_2-\bar{x}_2)- t\left(\frac{1}{x_1^2 + x_4^2 + 1}-\frac{1}{\bar{x}_1^2 + \bar{x}_4^2 + 1}\right)\\x_5-\bar{x}_5\\ -e^{-t}(x_5-\bar{x}_5) - (\cos(x_1+x_4)-\cos(\bar{x}_1+\bar{x}_4))=0\end{matrix}\right)\right\|\\ &\le&|x_2-\bar{x}_2|+|x_3-\bar{x}_3|+\frac{1}{1+t^2}|x_3-\bar{x}_3| +|\sin t||x_2-\bar{x}_2|\\ &&+|t|\left|\frac{1}{x_1^2 + x_4^2 + 1}-\frac{1}{\bar{x}_1^2 + \bar{x}_4^2 + 1}\right|+|x_5-\bar{x}_5|+ e^{-t}|x_5-\bar{x}_5|\\ &&+|\cos(x_1+x_4)-\cos(\bar{x}_1+\bar{x}_4)|\\ &\le&2|x_2-\bar{x}_2|+2|x_3-\bar{x}_3|+\delta(|x_1-\bar{x}_1|+|x_4-\bar{x}_4|)+(1+e^{\delta})|x_5-\bar{x}_5|+|x_1-\bar{x}_1|+|x_4-\bar{x}_4|\\ &\le& L\|X-\bar{X}\| \end{eqnarray} for some constant $L>0$.