# Cauchy problem with specific value of starting point, find limit

I have the following Cauchy problem:

$$y' = \vert y \vert - \arctan{e^x}$$ $$y(0) = y_0$$

I want to prove that there exists a specific value of $$y_0$$ such that:

$$\lim_{x\to\infty} y(x) = \frac{\pi}{2}.$$

I am not too sure how to even start on this, beside saying that f(x,y) is Lipschizt and therefore I have local uniqueness and existence of solution which can be extended globally to $$\mathbb{R}$$. Any help?

• As long as $y \geq 0$ you can study the linear ODE $y' = y - \arctan(e^x)$. – Christoph Jan 29 at 12:10

## 1 Answer

Let $$y($$ be a solution and suppose that $$y(x)=0$$ for some $$x\ge0$$. Then $$y'(x)<0$$. This implies that once a solution takes a negative value, it remains negative, Thus, the only way to solve the problem is to look for positive solutions. This means solving the linear equation $$y'=y-\arctan(e^x)$$. Its solution is $$y(x)=e^x\Bigl(y_0-\int_0^xe^{-t}\arctan(e^t)\,dt\Bigr).$$ This will be positive if and only if $$y_0\ge \int_0^\infty e^{-t}\arctan(e^t)\,dt=a.$$ If $$y_0>a$$, then $$\lim_{x\to\infty}y(x)=\infty$$. If $$y_0=a$$, then $$\lim_{x\to\infty}y(x)=\lim_{x\to\infty}e^x\int_x^\infty e^{-t}\arctan(e^t)\,dt=\frac\pi2.$$ Note. Although it is not needed for the argument, the actual value of $$a$$ is $$\frac{\pi+\log4}{4}.$$

• If $y_0 = a$ wouldn't y(x) be always 0? – qcc101 Jan 31 at 10:04
• No. If $x<\infty$, then $y_0>\int_0^xe^{-t}\arctan(e^t)\,dt$. – Julián Aguirre Jan 31 at 12:30