Prove that the initial value problem \[ \frac{dx}{dt}=3x+85\cos x,\qquad x(0)=77, \] has a solution \(x(t)\) defined for all \(t\in\mathbb R\).
Solution
Let \[ F(x)=3x+85\cos x. \]
1. There is a unique maximal solution\[ x:(\alpha,\beta)\longrightarrow\RR, \qquad x(0)=77, \]with \(-\infty\le\alpha<0<\beta\le\infty\).
The function \(F\) is smooth on \(\RR\), so the local existence and uniqueness theorem and continuation of solutions give the maximal solution.
2. For every \(x\in\RR\),\[ \abs{F(x)}\le3\abs{x}+85. \]
One has \(\abs{3x+85\cos x}\le3\abs{x}+85\abs{\cos x}\le3\abs{x}+85\).
3. If \(\beta<\infty\), then\[ \abs{x(t)}\le(77+85\beta)e^{3\beta} \qquad(0\le t<\beta). \]
For \(0\le t<\beta\), \[ x(t)=77+\int_0^t F(x(s))\,ds, \] so step 2 gives \[ \abs{x(t)} \le77+85\beta+3\int_0^t\abs{x(s)}\,ds. \] Gronwall’s inequality gives \[ \abs{x(t)}\le(77+85\beta)e^{3t} \le(77+85\beta)e^{3\beta}. \]
4. If \(\alpha>-\infty\), then\[ \abs{x(t)}\le(77+85\abs{\alpha})e^{3\abs{\alpha}} \qquad(\alpha<t\le0). \]
2. The argument of step 3applied to \(y\) on \([0,\abs{\alpha})\) gives the bound.
5. \(\alpha=-\infty\) and \(\beta=\infty\).
For a smooth vector field on \(\RR\), a maximal solution with a finite endpoint satisfies \(\abs{x(t)}\to\infty\) as that endpoint is approached. Steps 3 and 4 show that \(x\) stays bounded near any finite endpoint, so neither endpoint is finite.
6. Q.E.D.
By steps 1 and 5, the maximal solution is defined for every \(t\in\RR\).