Let \(y(t)\) be a real-valued solution, defined for \[ 0<t<\infty, \] of \[ \frac{dy}{dt}=e^{-y}-e^{-3y}+e^{-5y}. \] Show that \[ y(t)\longrightarrow+\infty \qquad\text{as }t\to+\infty. \]
Solution
Define \[ F(y)\coloneqq e^{-y}-e^{-3y}+e^{-5y}. \]
1. One has\[ F(y)>0 \]for every \(y\in\RR\).
Set \[ u=e^{-2y}>0. \] Then \[ F(y) = e^{-y}(1-u+u^2). \] Moreover, \[ 1-u+u^2 = \left(u-\frac12\right)^2+\frac34 >0. \] Since \(e^{-y}>0\), it follows that \(F(y)>0\).
2. The solution \(y(t)\) is strictly increasing on \((0,\infty)\).
The differential equation and step 1 give \[ y'(t)=F(y(t))>0 \] for every \(t>0\).
3. The function \(y\) cannot be bounded above.
Suppose instead that \(y\) were bounded above. By step 2, \(y\) is increasing, so there would be a finite limit \[ L\coloneqq\lim_{t\to\infty}y(t)\in\RR. \] By continuity of \(F\) and step 1, \[ F(L)>0. \] Hence there exist \(c>0\) and \(T>0\) such that \[ F(y(t))\geq c \] for all \(t\geq T\). Therefore \[ y'(t)\geq c \] for \(t\geq T\), and integration gives \[ y(t)\geq y(T)+c(t-T). \] The right-hand side tends to \(+\infty\), contradicting the assumed boundedness of \(y\).
4. One has\[ \lim_{t\to\infty}y(t)=+\infty. \]
By step 2, \(y\) is increasing. By step 3, it is not bounded above. An increasing real-valued function that is unbounded above tends to \(+\infty\).
5. Q.E.D.
Step 4 is the required conclusion.