Given the initial value problem

$$

y'(x)=y(x)-\sin{y(x)}, y(0)=1

$$

I need to prove that there is a solution defined on $\mathbb{R}$ and that the solution, $u(x)$, is an increasing function where $\lim_{x\to -\infty}{u(x)=0}$

The first part is quite easy, let $f(x, y) = y – \sin{y}$ $$f_y(x, y)=1-\cos{y}\Longrightarrow \left|f_y(x,y)\right| \le 2$$

hence $f(x, y)$ is Lipschitz continuous on $y$ for every $(x, y)\in (-\infty, \infty)\times (-\infty, \infty)$ therefore the initial value problem has a single solution on $\mathbb{R}$

I know that the solution, $u(x)$ has the form $$u(x)=y_0+\int_{x_0}^x{f(t, u(t))dt}=1+\int_0^x{(u(t)-\sin{u(t))}dt}$$

and also $$u(x)=\lim_{n\to\infty}u_n(x)$$ where $$u_n(x)=y_0+\int_{x_0}^x{f(t, u_{n-1}(t))}dt=1+\int_0^x{(u_{n-1}-\sin{(u_{n-1})})}dt,\space\space\space u_0(x)=y_0=1$$

But I'm wasn't able to find a way to prove the solution is increasing and has the desired limit.

EDIT:

I got a hint to look at $y\equiv 0$

EDIT2:

Let assume that the solution, $u(x)$ is decreasing at $(x_1, u(x_1))$ because $u(x)$ is continuous, there must be $(x_2, u(x_2)$ where $u'(x_2)=0=u(x_2)-\sin{u(x_2)}\Longrightarrow u(x_2)=0$

now if I look at the initial value problem

$$y'(x)=y(x)-\sin{y(x)}, y(x_2)=0$$

I can prove that it has a single solution in $\mathbb{R}$ like I already did, and $u(x)$ is my solution, but $u_1(x)\equiv 0$ is also a solution to this problem, contradiction, hence $u'(x)>0$ for $x\in\mathbb{R}$, i.e $u(x)$ is increasing.

But I still don't know how I can show the limit at $-\infty$

## Best Answer

We use the simple fact that $v(y)=y-\sin y$ is increasing and positive on $y>0$.

The fact that $y_0(x) = 0$ is a solution implies that $y(x)>0$ for all $x$ (by uniqueness). Then $$ y'(x) = v(y(x)) >0.$$

Thus $\lim_{x\to -\infty} y(x)$ exists and is nonnegative. If it's not zero, then $y(x)\ge y_1>0$ for some fixed $y_1$ and so $$ y'(x)= v(y(x)) \ge v(y_1)>0.$$ That would imply for any $x<0$ (by the mean value theorem and that $y(0)=1$) $$ y(x) = y(x) - y(0)+ y(0) = y'(x_0) x + 1 \le v(y_1)x+1$$ this is impossible as $y(x) >0$ for all $x$.