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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?

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@ Romeo: This looks like an initial value problem (IVP). Why do you care about $y (t)$ for $t < 0$? –  Rod Carvalho Sep 10 '12 at 17:17
    
@ Romeo: note that the $y_1 (t)$ you propose does not work due to the discontinuity at $t = 0$. It should be $y_1 (t) = 1 + t$ for $t > 0$. –  Rod Carvalho Sep 10 '12 at 17:35
    
@RodCarvalho Thanks, it was a typo, I've edited. Now I read your answer. –  Romeo Sep 10 '12 at 18:34
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2 Answers

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Let $G_a:(0,\infty) \to \mathbb{R}$ be such that $$ G_a(a)=0,\ G_a'(s)=\frac{1}{f(s)} \ \forall s>0 $$ i.e. \begin{eqnarray} G_a(s)&=&\begin{cases} \ln(s/a) & \text{ if } 0<s\le 1\\ s-1-\ln a & \text{ if } s>1 \end{cases} \ \text{ for } 0<a \le 1\\ G_a(s)&=&\begin{cases} \ln s -a+1 & \text{ if } 0<s\le 1\\ s-a & \text{ if } s>1 \end{cases} \ \text{ for } a>1 \end{eqnarray} Then $G_a$ is bijective with $G_a^{-1}: \mathbb{R} \to (0,\infty)$ defined by \begin{eqnarray}\tag{1} G_a^{-1}(\tau)&=&\begin{cases} ae^\tau & \text{ if } \tau \ge -\ln a\\ \tau+1+\ln a & \text{ if } \tau < -\ln a \end{cases} \ \text{ for } 0 < a \le 1\\ G_a^{-1}(\tau)&=&\begin{cases} e^{\tau+a-1} & \text{ if } \tau \ge 1-a\\ \tau+a & \text{ if } \tau < 1-a \end{cases} \ \text{ for } a > 1 \end{eqnarray} Hence $$ y_a'=f(y_a),\ y_a(0)=a \iff t=\int_a^{y_a(t)}G_a'(s)ds=G(y_a(t)), $$ i.e. $$ y_a(t)=G_a^{-1}(t), $$ where $G_a^{-1}$ is given by (1).

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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

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

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