I am new to ODEs and I'm stuck on this problem. I will write down my attempt. The IVP in question is:
\begin{cases} x'(t) = x(t)(1-x(t)), \\ x(0) = x_{0}. \end{cases}
I applied separation of variables. I defined the function $f: X \mapsto \mathbb{R}$, $X := X_{1} \times X_{2},$ \begin{align} f(t,x) := g(t)\cdot h(x) = x(t)(1-x(t)) \end{align} such that $g: X_{1} \mapsto \mathbb{R}$ and $h: X_{2}\mapsto \mathbb{R}$, with $g(t) := 1$ and $h(x) := x(1-x), X_{1} := \mathbb{R}$ and $X_{2} := \mathbb{R}$. Then I defined the solution $x : I \mapsto \mathbb{R}$, defined on an interval $I \subset \mathbb{R}$ such that $0 = t_{0} \in I$ too.
Firstly, I checked when $h(x) = 0$: \begin{align} h(x_{0}) = 0 \implies x_{0} = 0 \text{ or } x_{0} = 1. \end{align} Hence, \begin{align} \text{for } x_{0} = 0 \implies x(t) = 0; \\ \text{for } x_{0} = 1 \implies x(t) = 1. \end{align}
Afterwards, I solved the equation given that $h(\xi) \neq 0$ ($\implies x_{0} \neq 0$ and $x_{0} \neq 1$),
$$ \int_{x_{0}}^{x}\frac{1}{h(\xi)} d\xi = \int_{t_{0}}^{t} g(\eta) d\eta, $$ i.e. \begin{align*} \int_{x_{0}}^{x}\frac{1}{\xi(1-\xi)} d\xi = \int_{0}^{t} d\eta &\implies (\ln\lvert\xi\rvert - \ln\lvert 1-\xi\rvert)\big\rvert_{x_{0}}^{x} = t \\ &\implies \ln\Big\lvert\frac{\xi}{1-\xi}\Big\rvert\Bigg\rvert_{x_{0}}^{x} = t \\ &\implies \ln\Big\lvert\frac{x}{1-x}\Big\rvert - \ln\Big\lvert\frac{x_{0}}{1-x_{0}}\Big\rvert = t \\ &\implies \ln\Big\lvert\frac{x(1-x_{0})}{x_{0}(1-x)}\Big\rvert = t \\ &\implies \Big\lvert\frac{x(1-x_{0})}{x_{0}(1-x)}\Big\rvert = e^{t}. \end{align*} I distinguished two cases:
- When $h(x_{0}) > 0 \implies h(x) > 0$, so \begin{align*} &\frac{x(1-x_{0})}{x_{0}(1-x)} = e^{t} \implies &x(t) = \frac{e^{t}x_{0}}{1-x_{0}(1-e^{t})}. \end{align*} Here is where I'm having issues with finding the maximal interval. Since, $h(x_{0}) > 0$ then \begin{align*} x_{0}(1-x_{0})>0 \implies \end{align*} \begin{cases} x_{0}>0 \\ 1-x_{0}>0 \end{cases} $$ \textbf{or} $$ \begin{cases} x_{0}<0 \\ 1-x_{0}<0 \end{cases} which results in $$ 0 < x_{0} < 1 .$$ Additionally, $$x_{0} \neq \frac{1}{1-e^{t}}.$$ However, in this case, $x(t)$ exists for $t \in \mathbb{R} \setminus \{\ln(1- \frac{1}{x_{0}})\}$, since $$ 1-x_{0}(1-e^{t}) \neq 0 \implies t \neq \ln\Big(1- \frac{1}{x_{0}}\Big) $$ which is a problem because the bounds on $x_{0}$ make it so that $\ln(1- \frac{1}{x_{0}})$ can't exist. So, I'm wondering what I did wrong and it's confusing me a lot.
- For $h(x_{0})<0$ I got the same $x(t)$ but the bounds on $x_{0}$ are $$ x_{0} < 0 \cup x_{0} > 1 $$ which in this case works with $t \in \mathbb{R} \setminus \{\ln(1- \frac{1}{x_{0}})\}$. What confuses me even more, is that by differentiating $x(t)$ you get the ODE in the IVP, so it's presumably right (?). I'm not sure if I get the wrong interval because I used the wrong approach, or I messed up the signs, or something else I'm not noticing. Any help/suggestions would be appreciated.
In the first case, where $\ 0<x_0<1\ $, your conclusion that $\ \ln\left(1-\frac{1}{x_0}\right)\ $ cannot exist isn't really a problem. It simply tells you that $\ 1-x_0\left(1-e^t\right)=x_0e^t+1-x_0\ $ can never be zero for any real value of $\ t\ $. In fact, it's easy to show that if $\ 0<x_0<1\ $, then that expression is always strictly positive. In other words, your solution $$ x(t)=\frac{x_0e^t}{1-x_0\left(1-e^t\right)} $$ of the differential equation holds good for all $\ t\ $ in the infinite interval $\ (-\infty,\infty)\ $. You appear to have implicitly assumed that this interval had to be finite. If so, that's probably what has given rise to your confusion.