Uniqueness of 2nd order IVP

Question

How do I show that the IVP

$x''(t)=\arctan{(x(t))}\arctan{(t)}$ with $x(0)=e$, $x'(0)=e^2$

is or isn't uniquely solvable and defined on $\mathbb{R}$?

Answer 1

The solution is unique on $\mathbb{R}$.

The most popular way to verify whether a solution of dynamical system is unique is to show Lipschitzness condition as stated below.

• [1, Theorem 3.2] (Global Existence and Uniqueness) Suppose that $f(t, x)$ is piecewise continuous in $t$ and satisfies $\lVert f(t, x) - f(t, y) \rVert \leq L \lVert x-y \rVert$ $\forall x, y \in \mathbb{R}^n$, $\forall t \in [t_0, t_1]$. Then, the state equation $\dot{x} = f(t, x)$, with $x(t_0) = x_0$, has a unique solution over $[t_0, t_1]$.

In this case, $f(t, x) = [x_2 \ \arctan(t) \arctan(x_1)]^T$ where $x = [x_1 \ x_2]^T$. Observe that $\lvert \arctan(t) \arctan(x_1) - \arctan(t) \arctan(y_1) \rvert \leq (\pi/2)\lvert x_1 - y_1 \rvert$ (Note that $g(x) := \arctan(x) \Rightarrow \lvert g'(x) \rvert=\frac{1}{1+x^2} \leq 1 \Rightarrow g$ is Lipschitz continuous with Lipschitz constant of $1$).

Hence, $\lVert f(t, x) - f(t, y) \rVert \leq (\pi/2)\lVert [(x_2-y_2) \ (x_1-y_1)]^T \rVert = (\pi/2) \lVert x - y\rVert$ implies that the dynamical system has a unique solution on $\mathbb{R}$ (Hint: applying this to forward and backward cases).

Reference

[1] H. K. Khalil, Nonlinear Systems, 3rd ed. Upper Saddle River, NJ: Prentice Hall, 2002.