# Proving single solution of initial value problem is increasing

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 $$x$$.

• I don't understand the last equation where does $y(0)$ and $y(1)$ come from? Or $y'(x_0)x +1$? Dec 18, 2021 at 23:25
• That's a typo @CforLinux I have made an edit. Dec 18, 2021 at 23:42

Let us introduce the new variable $$z$$ defined by $$z = -x$$. Then

$$y'(x) = \frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx} = -\frac{dy}{dz}$$ Therefore, we have the differential equation with the reversed variable direction $$\frac{dy}{dz} = -y + \sin y$$

This is in fact the dynamics of a simple damped pendulum. Let $$V(y) = y^2/2$$ be the Lyapunov function of the new ODE. Then, $$\frac{dV}{dz} = y\frac{dy}{dz} = -y^2 + y\sin y$$ and $$dV/dz < 0$$ for all $$y \in \mathbb{R}\setminus \{0\}$$ and $$dV/dz = 0$$ at $$y = 0$$. Hence, $$\mathbb{R}$$ is positively invariant with respect to the ODE, meaning that every solution starting at $$z = z_0$$ is defined for all $$z \ge z_0$$. Let $$v$$ be the solution of the ODE with $$y(0) = 1$$. Then, we have that $$\lim_{z \to \infty}v(z) = 0$$. Finally, we conclude that $$\lim_{x \to -\infty}u(x) = 0$$.

We can also show that $$v(z)$$ cannot change its sign and monotonically decreases whenever $$v(z)>0$$.