For which values of $\alpha$ does the following initial value problem have a unique soltion?
$$y'''+(y'')^2-|y'|^{\alpha}+ \sin y=f(t)$$
where $\alpha$ is a constant.
The above problem is written equivalently as follows:
We set $x_1=y, x_2=y', x_3=y''$. $$\frac{d x_1}{dt}=x_2 \\ \frac{d x_2}{dt}=x_3 \\ \frac{dx_3}{dt}=f(t)-x_3^2+|x_2|^{\alpha}- \sin (x_1)$$
The problem will have a unique solution if there is a $K$ such that $|\overline{\Phi}(\overline{x}, t)- \overline{\Phi}(\overline{y},t)| \leq K |\overline{x}-\overline{y}|$.
$\overline{\Phi}=(x_2,x_3, f(t)-x_3^2+|x_2|^{\alpha}- \sin (x_1))$
$$|\overline{\Phi}(\overline{x},t)-\overline{\Phi}(\overline{y},t)|=\sqrt{(x_2-y_2)^2+(x_3-y_3)^2+ (|x_2|^{\alpha}-|y_2|^{\alpha}) + (y_3^2-x_3^2)+ ( \sin (y_1)- \sin (x_1))^2}$$
We want to show that the latter is $\leq K \sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}$
From the intermediate value theorem we have that there is a $\xi \in [x_1 , y_1]$ such that $|\cos \xi|=\left| \frac{\sin (y_1)- \sin (x_1)}{y_1-x_1}\right| \Rightarrow |\sin (y_1)- \sin (x_1)| \leq |y_1-x_1|$
We have $(y_3^2-x_3^2)=(y_3-x_3) (y_3+x_3)$
$|y|^a$ is Lipschitz for $a \geq 1$. How can we justify this?
So $||x_2|^a-|y_2|^a|| \leq \overline{K} |x_2-y_2|$.
$$\sqrt{(x_2-y_2)^2+(x_3-y_3)^2+ (|x_2|^{\alpha}-|y_2|^{\alpha}) + (y_3^2-x_3^2)+ ( \sin (y_1)- \sin (x_1))^2} \leq \sqrt{(x_2-y_2)^2+(x_3-y_3)^2+ \overline{K} |x_2-y_2|+2g (|x_3-y_3|)+|y_1-x_1|}$$
Do you have an idea how we could bound it so that we have an upper bound of the form $K \sqrt{(x_1-y_1)^2+(x_2-y_2)^2+(x_3-y_3)^2}$.
It doesn't always hold $|x_2-y_2| \leq |x_2-y_2|^2$, right?
I if we assume that $\Phi : [a_1 , b_1 ]\times [a_2 , b_2 ]\times [a_3 , b_3 ]\times [a_4 , b_4]\to \mathbb{R} $, then there exists a constants $K_1 , K_2 , K_3 $ such that $|\sin x_1 -\sin y_1 |\leq K_1|x_1 -y_1|, |x_3^2 -y_3^2 |\leq K_2 |x_3 -y_3|, ||x_2|^{\alpha} -|y_2|^{\alpha} |\leq K_3 |x_2 -y_2| \mbox{ for } \alpha \geq 1 \mbox{ and } (x_1 , x_2 , x_3 )\in [a_1 , b_1 ]\times [a_2 , b_2 ]\times [a_3 , b_3 ].$ Hence $|\Phi (\overline{x} , t ) -\Phi (\overline{y} , t )|\leq \sqrt{3}\max \{K_1 , K_2 , K_3 \} \sqrt{ (x_1 -y_1 )^2 + (x_2 -y_2)^2 +(x_3 - y_3 )^2 }$ so $\Phi $ is Lipschitz map with respect to $\overline{x} .$