Unbounded solution of a ODE Let $f,g:[0,\infty)\to \mathbb{R}$ be two continuous functions such that $\lim\limits_{x\to\infty}f(x)=1$ and $\int_0^\infty|g(x)|dx<\infty$. Consider the ODE $$\begin{pmatrix} y_1'\\ y'_2\end{pmatrix}=\begin{pmatrix} 0 & f(x)\\ g(x) & 0\end{pmatrix}\begin{pmatrix} y_1\\ y_2\end{pmatrix}.$$ Suppose that $\Phi(x)=\begin{pmatrix} \phi_1(x)\\ \phi_2(x)\end{pmatrix}$ is a solution of the above ODE such that $\phi_1$ is bounded. Prove that $$\lim\limits_{x\to\infty}\phi_2(x)=0.$$
Deduce that the above ODE has an unbounded solution.
I really do not know where to start. Any hint? 
 A: This is not an answer but I don't have enough reputation to comment. Note that if $\Phi_1(x)$ is a solution then $\Phi_1'(x) = f(x) \Phi_2(x)$. Taking the limit
$$\lim_{x \to \infty} \Phi_1'(x) = \lim_{x \to \infty} f(x) \Phi_2(x) = \lim_{x \to \infty} \Phi_2(x)$$
But if $\Phi_1(x)$ is bounded, and its limit exists, then the limit of its derivative must approach zero, and hence 
$$\lim_{x \to \infty} \Phi_2(x) = 0$$
However, you would need to show the limit exists. Under the assumption that the limit is indeed zero, and letting $\Phi_2(0)=0$,
$$\lim_{x \to \infty} \Phi_2(x) = \Phi_2(t_0) + \lim_{x \to \infty} \int_0^x g(z) \Phi_1(z) dz  = \int_0^\infty g(z) \Phi_1(z) dz = 0$$
which implies 
$$\lim_{x \to \infty} \Phi_1(x) = 0 \quad or \quad \lim_{x \to \infty} g(x) = 0$$
A: Since $\phi_{1}$ is bounded, say, by a constant $L>0$, you have that
$|\phi_{2}^{\prime}(x)|\leq L|g(x)|$ for all $x$ and so $\int_{0}^{\infty
}|\phi_{2}^{\prime}(x)|\,dx<\infty$. Hence, given $\varepsilon>0$, there
exists $M_{\varepsilon}>0$ such that $\int_{M_{\varepsilon}}^{\infty}|\phi
_{2}^{\prime}(x)|\,dx\leq\varepsilon$. In turn, if $t\geq s\geq M_{\varepsilon
}$, then
$$
|\phi_{2}(t)-\phi_{2}(s)|\leq\int_{s}^{t}|\phi_{2}^{\prime}(x)|\,dx\leq
\int_{M_{\varepsilon}}^{\infty}|\phi_{2}^{\prime}(x)|\,dx\leq\varepsilon,
$$
which implies that there exists $\lim_{x\rightarrow\infty}\phi_{2}(t)=\ell
\in\mathbb{R}$. To prove that $\ell=0$, we claim that
$$
\liminf_{x\rightarrow\infty}|\phi_{1}^{\prime}(x)|=0.
$$
If not, then $\liminf_{x\rightarrow\infty}|\phi_{1}^{\prime}(x)|=c>0$, which
implies that $|\phi_{1}^{\prime}(x)|\geq\frac{c}{2}$ for all $x$ large, say
$\phi_{1}^{\prime}(x)\geq\frac{c}{2}$ for all $x\geq T$. Then $\phi
_{1}(x)=\phi_{1}(T)+\int_{T}^{x}\phi_{1}^{\prime}(s)|\,ds\geq\phi_{1}
(T)+\frac{c}{2}(x-T)\rightarrow\infty$ as $x\rightarrow\infty$, which contradicts the fact that $\phi_1$ is bounded. The case
$\phi_{1}^{\prime}(x)\leq-\frac{c}{2}$ for all $x$ large is similar. 
This shows that $\liminf_{x\rightarrow\infty}|\phi_{1}^{\prime}(x)|=0$ and so
there exists a sequence $x_{n}\rightarrow\infty$ such that $\phi_{1}^{\prime
}(x_{n})\rightarrow0$. Then from $f(x_{n})\phi_{2}(x_{n})=\phi_{1}^{\prime
}(x_{n})$, letting $n\rightarrow\infty$ we get $1\ell=0$, which shows that
$\ell=0$.
