# Proving single solution of initial value problem is increasing

initial-value-problemsordinary differential equations

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 $$-\infty$$

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 $$x$$.