1
$\begingroup$

Let $f \colon \mathbb R \to \mathbb R$ be defined by $ f(y)=\begin{cases} y & \text{ if } y \le 1 \\ 1 & \text{ if } y>1 \end{cases} $

I have to determine explicity the solution $y_a(\cdot)$ of the Cauchy problem $\tag{CP} \begin{cases} y'=f(y) \\ y(0)=a \end{cases} $ where $a>0$.

Well, the function $ y_1(t) = \begin{cases} e^t & \text{ if } t \le 0 \\ t+1 & \text{ if } t > 0 \end{cases} $

works for $a=1$ (it is continuous and differentiable and satisfies (CP) for $a=1$).

Have you got any ideas to treat the case $a\ne 1$? What can we do?

  • 0
    @RodCarvalho Thanks, it was a typo, I've edited. Now I read your answer.2012-09-10

2 Answers 2

1

If $a > 1$, then we have the following initial value problem (IVP)

$\dot{y} (t) = 1, \qquad{} y (0) = a$

whose solution is $y (t) = a + t$ for all $t \geq 0$. If $a \leq 1$, the we have the IVP

$\dot{y} (t) = y, \qquad{} y (0) = a$

whose solution is $y (t) = a \, e^t$ for all $t \in \ [0, t_*]$, where the time instant $t_*$ is when $y$ reaches $1$, i.e., $y (t_*) =1$. We obtain $t_* = \ln(1 / a)$. For $t \geq t_*$ we have another IVP

$\dot{y} (t) = 1, \qquad{} y (t_*) = 1$

whose solution is $y (t) = 1 + (t- t_*)$ for all $t \geq t_*$. Finally, we have that for $a \leq 1$ the solution is

$y (t) = \left\{\begin{array}{cl} a \, e^t & \textrm{if} \quad{} t \in [0, \ln(1/a)]\\ 1 + (t - t_*) & \textrm{if} \quad{} t \geq \ln(1/a)\end{array}\right.$