Consider \[ \frac{dy}{dx}=y-\sin y. \] Show that there is \(\varepsilon>0\) such that, whenever \(|y_0|<\varepsilon\), the solution \(y=f(x)\) satisfying \(f(0)=y_0\) obeys \[ \lim_{x\to-\infty}f(x)=0. \]
Solution
Set \[ F(y)\coloneqq y-\sin y. \]
1. Every initial value \(y(0)=y_0\) has a unique solution defined forall \(x\in\RR\).
Since \[ F'(y)=1-\cos y, \] one has \[ \abs{F'(y)}\leq2 \] for every \(y\in\RR\). Thus \(F\) is globally Lipschitz. The global existence and uniqueness theorem for autonomous ODEs therefore gives a unique solution on all of \(\RR\) for every initial value.
2. For every \(y\neq0\), the numbers \(F(y)\) and \(y\) have the samesign.
For \(y>0\), the standard strict inequality \[ \sin y<y \] gives \(F(y)>0\). Since \(F\) is odd, \(F(y)<0\) for \(y<0\).
3. If \(y_0>0\), then the corresponding solution satisfies\[ \lim_{x\to-\infty}f(x)=0. \]
By uniqueness, the solution cannot meet the equilibrium solution \(y=0\) unless it is identically zero. Hence \(f(x)>0\) for all \(x\). Step 2 gives \[ f'(x)=F(f(x))>0, \] so \(f\) is strictly increasing. In particular, for \(x\leq0\), \[ 0<f(x)\leq f(0)=y_0. \] Therefore the limit \[ L\coloneqq\lim_{x\to-\infty}f(x) \] exists and satisfies \(0\leq L\leq y_0\).
If \(L>0\), then \(F(L)>0\). By continuity of \(F\), there are \(c>0\) and \(X<0\) such that \[ f'(x)=F(f(x))\geq c \] for all \(x<X\). Hence, for \(x<X\), \[ f(X)-f(x) = \int_x^X f'(t)\,dt \geq c(X-x), \] whose right-hand side tends to \(\infty\) as \(x\to-\infty\). This contradicts the boundedness of \(f\) on \((-\infty,0]\). Thus \(L=0\).
4. If \(y_0<0\), then the corresponding solution also satisfies\[ \lim_{x\to-\infty}f(x)=0. \]
The function \(F\) is odd, so if \(f\) is the solution with initial value \(y_0<0\), then \(-f\) is the solution with initial value \(-y_0>0\). Step 3 applied to \(-f\) gives \[ \lim_{x\to-\infty}(-f(x))=0, \] and hence the same limit for \(f\).
5. The conclusion holds for every initial value \(y_0\in\RR\).
Steps <1>3–<1>4 cover \(y_0\neq0\). If \(y_0=0\), uniqueness gives the constant solution \(f\equiv0\).
6. In particular, one may take\[ \boxed{\varepsilon=1}. \]
Step 5 proves the desired limit for every \(y_0\), hence certainly for all \(\abs{y_0}<1\).
7. Q.E.D.
Step 6 supplies the required positive \(\varepsilon\).