Question
Let $f \in \mathcal C(\mathbb R_+ \times \mathbb R^n, \mathbb R^n)$ and $g \in \mathcal C(\mathbb R_+ \times \mathbb R, \mathbb R_+)$. Assume that
- $g(t,u)$ is non-decreasing in $u$ for each fixed $t$ and
$\vert f(t,x)\vert \lt g(t,\vert x \vert), (t,x)\in \mathbb R_+ \times \mathbb R^n$ - Every solution $u(t) = u(t,t_0,u_0)$ of the IVP $u^\prime(t) = g(t,u),
u(t_0)=u_0$ is bounded.
Show that every solution of the IVP $x^\prime(t) = f(t,x), x(t_0) = x_0$ exists on $[t_0,\infty)$ and converges to a constant vector as $t \to \infty$.
My Attempt
I have completed the first part of the question (show every solution of the IVP $x' = f(t,x), x(t_0) = x_0$ exists on $[t_0,\infty)$) using following corollary.
Corollary: Let $f \in \mathcal C(J \times \mathbb R^n, \mathbb R^n)$ where $J=[a,b)$ and $\vert f(t,x) \vert \lt g(t,\vert x \vert)$ for $(t,x)\in J \times R^n$ and $g \in \mathcal C(\mathbb R_+ \times \mathbb R, \mathbb R_+)$. If $\gamma(t)$ is the maximal solution of $u^\prime = g(t,u), u(t_0)=u_0$ existing on $J$, then the solution $x(t) = x(t,t_0,x_0)$ of $x^\prime = f(t,x), x(t_0) = x_0$ where $\vert x_0 \vert \leq u_0$ exists on $J$ and $\vert x(t) \vert \leq \gamma(t)$ for all $t\in J$.
Simply let $\gamma(t)$ be the maximal solution of $u^\prime = g(t,u), u(t_0)=\vert x_0 \vert$. Since we know all solutions of this IVP are bounded (by (2)), we know that $\gamma(t)$ must exist on $J = [t_0,\infty)$. By the corollary, $x(t)$ also exists on $[t_0,\infty)$. Moreover, $\vert x(t) \vert<\gamma(t)\forall t\in[t_0,\infty)$.
Having said that, I do not know how to prove that $x(t)$ must converge. I have the following results:
- Since $g(t,u)$ is always positive, we know that $\gamma(t)$ is an increasing bounded continuous function of $t$ (by (2)). Thus, $\gamma(t)$ must converge. Hence, $\vert x(t) \vert$ is bounded from above by a convergent function.
- $g(t,u)$ is a non-decreasing function of $u$ for every fixed $t$, thus $\vert x^\prime(t) \vert \leq \vert f(t,x) \vert \leq \vert g(t,\vert x(t) \vert) \vert \leq \vert (g(t,\gamma(t)) $
but I'm not sure what they imply about the convergence of $x(t)$. Any help would be greatly appreciated.
Best Answer
We have for all $t \ge0$
$$\vert x^\prime(t) \vert = \vert f(t,x(t)) \vert \leq g(t,\vert x(t) \vert)\leq g(t,\gamma(t))=\gamma^\prime(t)$$
Hence for $ 0 \lt u \lt v$
$$0 \le \left\vert x(v) - x(u) \right\vert = \left\vert\int_u^v x^\prime(t) \ dt\right\vert \le \int_u^v \vert x^\prime(t) \vert \ dt \le \gamma(v) - \gamma(u)$$
As $\gamma$ has a limit at $\infty$, $\gamma(v) - \gamma(u)$ can be as small as desired for $u,v$ large enough. Which implies that it is also the case for $\vert x(v) - x(u) \vert$ and according to Cauchy criterion that $\lim\limits_{x \to \infty} x(t)$ exists.