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 $-\infty$


2 Answers 2


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 $x$.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.