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?

  • 1
    $\begingroup$ use $y_2(x) - y_2(x_0) = \int_0^x g(t)y_1(t) \, dt$ and the boundeness of $y_1$ implies that $|y_2(x) - y_2(x_0) | \le |y_1|_\infty \int_0^\infty |g(x)|\, dx.$ this in turn shows you that $y_2$ is bounded. $\endgroup$ – abel Mar 14 '17 at 18:45
  • $\begingroup$ is it true that if $\int_0^\infty |g(t)|dt$ converges then $g(t)$ must go to $0$ as $t\to \infty?$ $\endgroup$ – Jax Mar 14 '17 at 18:57
  • $\begingroup$ It's not true. Consider a function that's graph is a sequence of triangles with height 2 and base $1/2^n$, connected by line segments along the $x$-axis. You need $g(x)$ to be uniformly continuous to guarantee convergence. $\endgroup$ – Philip Hoskins Mar 14 '17 at 19:11

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 $\ell=0$.

  • $\begingroup$ About the existence of the $\lim\limits_{x\to\infty}\phi_2(x)$, can you explain more? $\endgroup$ – Jax Mar 19 '17 at 17:00
  • $\begingroup$ I am using Cauchy criterion for the existence of finite limits. Take a look here (it is done for the case $x\to c$ but the same proof works for $x\to\infty$) math.stackexchange.com/questions/569036/… $\endgroup$ – Gio67 Mar 19 '17 at 17:04
  • $\begingroup$ $\int_a^\infty|\phi_2(x)|dx$ converges if and only if $\forall\epsilon>0,\ \exists M\geq a$ such that for all $A,B\geq M$, $\int_A^B|\phi_2(x)|dx<\epsilon$ $\endgroup$ – Jax Mar 19 '17 at 17:07
  • $\begingroup$ yes, because $\int_{a}^{\infty} |\phi_2(x)|dx=lim_{t\to\infty}\int_{a}^{t} |\phi_2(x)|dx$. Same criterion $\endgroup$ – Gio67 Mar 19 '17 at 17:09
  • $\begingroup$ From $|\phi_2(t)-\phi_2(s)|\leq \epsilon$, how can we conclude that the limit of $\phi_2$ exists? Sorry for asking too many questions. $\endgroup$ – Jax Mar 19 '17 at 17:13

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

$$\lim_{x \to \infty} \Phi_1(x) = 0 \quad or \quad \lim_{x \to \infty} g(x) = 0$$

  • $\begingroup$ Thanks for the answer. Le me try to make it work, if possible. $\endgroup$ – Jax Mar 14 '17 at 18:40

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