# Prove that every solution $y(t)$ of the ordinary differential equation $\frac{dy}{dt} = \cos(y)$ is bounded above and below for all t.

Let $t\in\mathbb{R}$ and let $y:\mathbb{R}\rightarrow\mathbb{R}$. Prove that every solution $y(t)$ of the ordinary differential equation

$\frac{dy}{dt} = \cos(y)$ is bounded above and below for all t.

Is the solution $y(t)$ of the ODE

$\frac{dy}{dt} = t^3\cos(y)$ bounded above and below for all t?

My attempt: Solving the ODE directly is rather messy, so I'm assuming there is a more elegant solution. Intuitively, I understand for $\frac{dy}{dt} = \cos(y)$ that the derivative of $y$ is bounded above by $1$ and below by $-1$, and there are equilibrium points to the equation at $y=\frac{-3\pi}{2},\frac{-\pi}{2},\frac{\pi}{2},\cdots$, so every point $t\in\mathbb{R}$ is between two equilibrium points. I feel like this, along with the uniqueness theorem for ODEs should be enough to give my answer, but I'm having trouble piecing it together.

Further, I'm not really sure how to go about solving the second part at all.

Any help appreciated!

• 15 minutes to accept a solution which errs on the first part and which does not address the second part... Sure you are using the site optimally? – Did Jan 23 '17 at 15:56

You equation is equivalent to the integral equation $$y=y_0+\int_{t_0}^t \cos y\,d t.$$ Hence you have $$|y(t)|\leq |y_0|+\left|\int_{t_0}^t\cos y\,dt\right|\leq |y_0|+|t-t_0|$$ and hence your solution is bounded.

As @Did mentioned in the comments, my original argument is not valid.

Here is a valid one:

Equation $$\dot y=\cos y$$ has infinitely many equilibria, i.e., the constant solutions of the form $$y(t)=\pi/2+\pi k,\quad k\in Z.$$ For any initial condition $y_0$ you can find two equalibria $\hat y_1$ and $\hat y_2$ such that $\hat y_1\leq y_0\leq \hat y_2$. Since for this particular equation the uniqueness theorem holds, then the solution starting at $y_0$ will be bounded by $\hat y_1$ and $\hat y_2$.

Added: And the second problem is solved in exactly the same way: note that any constant function $y(t)=\pi/2+2\pi k,\,k\in\mathbf Z$ is a solution and invoke the uniqueness theorem.

• Sorry but $|y(t)|\leqslant C+C'|t|$ is not what one calls a bounded solution. The result to prove is that $|y(t)|\leqslant C$ for every $t$. – Did Jan 23 '17 at 15:54
• @Did Of course, I was too quick, thank you. – Artem Jan 23 '17 at 16:47

If $y(0)=\pi/2+k\,\pi$ for some $k\in\mathbb{Z}$, then the solution is constant and hence bounded. If not, let $k$ be the unique integer such that $\pi/2+k\,\pi<y(0)<\pi/2+(k+1)\,\pi$. Then by uniqueness of solution we have $\pi/2+k\,\pi<y(x)<\pi/2+(k+1)\,\pi$ for all $x$.

• May I ask why the down vote? – Julián Aguirre Jan 23 '17 at 17:26