Consider the differential equations
\[ \frac{dx}{dt}=-x+y,\qquad \frac{dy}{dt}=\log(20+x)-y. \]
Let \(x(t)\) and \(y(t)\) be a solution defined for all \(t\ge0\) with \(x(0)>0\) and \(y(0)>0\). Prove that \(x(t)\) and \(y(t)\) are bounded.
Solution
Set \[ U(x)=\frac{x^2}{2}-(x+20)\log(x+20)+(x+20), \qquad x>-20, \] so that \[ U'(x)=x-\log(20+x), \] and set \[ E(t)=\frac12(x'(t))^2+U(x(t)). \]
1. One has \(x(t)>0\) and \(y(t)>0\) for every \(t\ge0\).
Variation of constants gives \[ x(t)=e^{-t}\left(x(0)+\int_0^t e^s y(s)\,ds\right) \] and \[ y(t)=e^{-t}\left(y(0)+\int_0^t e^s\log(20+x(s))\,ds\right). \] Suppose some variable vanishes at a positive time, and let \(t_*\) be the first such time. On \([0,t_*)\) both variables are positive, and \(x(s)>0\) gives \(\log(20+x(s))>0\). The two displayed formulas at \(t=t_*\) then give \(x(t_*)>0\) and \(y(t_*)>0\), a contradiction.
2. The function \(x\) satisfies\[ x''+2x'+U'(x)=0, \]and \(E(t)\le E(0)\) for every \(t\ge0\).
From \[ x'=-x+y \] we have \[ y=x'+x. \] Differentiating and using the second equation gives \[ x''+x'=y'=\log(20+x)-y =\log(20+x)-x'-x. \] Therefore \[ x''+2x'+x-\log(20+x)=0, \] which is the displayed equation. Multiplying it by \(x'\) yields \[ E'(t)=\frac{d}{dt}\left(\frac12(x')^2+U(x)\right)=-2(x')^2\le0. \]
3. The function \(x\) is bounded on \([0,\infty)\).
As \(x\to\infty\), \[ U(x)=\frac{x^2}{2}-(x+20)\log(x+20)+(x+20) \longrightarrow +\infty, \] because \((x+20)\log(x+20)=o(x^2)\). Thus the sublevel set \[ \{x\ge0:U(x)\le E(0)\} \] is bounded. Since \(x(t)>0\) by step 1 and \(U(x(t))\le E(t)\le E(0)\) by step 2, the function \(x\) takes values in this set.
4. The functions \(x'\) and \(y\) are bounded on \([0,\infty)\).
Let \[ m=\min_{x\ge0}U(x), \] which exists because \(U\) is continuous on \([0,\infty)\) and tends to \(+\infty\) as \(x\to\infty\). By step 2, \[ \frac12(x'(t))^2 =E(t)-U(x(t)) \le E(0)-m, \] so \(x'\) is bounded. Finally, \[ y(t)=x'(t)+x(t), \] which is bounded by step 3.
5. Q.E.D.
Steps 3 and 4 show that \(x(t)\) and \(y(t)\) are bounded for \(t\ge0\).