# Interval of the maximum solution of a nonlinear equation

Mentor note: Thread moved from the Technical Math section, so there is no template.
@yamata1, , in the future, please post homework problems or exercises in the Homework & Coursework sections, not in the Technical Math sections. I have moved your post.

Hello,

I would like some help with an exercise, specifically the last question:
Let $y_0 \in \mathbb{R}$ and $y'(t)=exp(y(t))(1-exp(y(t)))$ with the initial condition $y(0)=y_0$
$1-$Justify that this equation has a unique maximum solution $y$ on an interval $I$ that contains $0$.
$2-$Show that if $y_0=0$ then $I=\mathbb{R}$ and $u(t)=0$, $\forall t \in \mathbb{R}$.
$3-$Show that if $y_0>0$ then $y(t)>0$ with $\forall t \in I$
$4-$Let $y_0>0$
$a)$Show that $\frac{1}{exp(x)(1-exp(x))}=exp(-x)+\frac{exp(-x)}{exp(x)(exp(-x)-1)}$.
$b)$ $H(x)=-exp(-x)-ln(1-exp(-x))$ for $x>0$ show that $H(y(t))=H(y_0)+t$,$\forall t \in I$.
$c)$ Show that the function H is a bijection of $\mathbb{R}^*_+$ in $\mathbb{R}^*_+$.
$d)$Deduce that $I=]-H(y_0),+\infty[$.

$1-$Apply Picard's existence theorem or Cauchy–Lipschitz theorem
$2-$If $y_0=0$ then $y'(0)=0$ then $y(t)=0$
we consider the maximal solution Xmax of the equation , defined on an interval $I_{max}$.
We suppose there exists a function$f : \mathbb{R} \rightarrow \mathbb{R}+$, continuous,such that
$||X(t)|| \leqslant f(t) \forall t$ $\in I_{max}$ : then $I_{max}=\mathbb{R}$.
3-The solution would be monotonous from $y'(t)$>0 ,increasing from 0 but I'm not certain on the way to prove it.
4-a)simplify exp(-x) on both sides , put the R.H.S terms to a common denominator add them and multiply the R.H.S ;both the numerator and denominator; by exp(x).
b)Prove $H(y(t))'=y'(t)*H'(y(t))$=$\frac{H(y(t))-H(y_0)}{t-0}=1$
c)The monotony of H(y(t)) is enough I think.
d) I would like help for this , all I can find is $H(y(t))>0 \Longrightarrow H(y_0)+t >0\Longrightarrow t> -H(y_0)$

I welcome your suggestions or answers ,especially $4-d)$.

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Ray Vickson
Homework Helper
Dearly Missed
Mentor note: Thread moved from the Technical Math section, so there is no template.
@yamata1, , in the future, please post homework problems or exercises in the Homework & Coursework sections, not in the Technical Math sections. I have moved your post.

Hello,

I would like some help with an exercise, specifically the last question:
Let $y_0 \in \mathbb{R}$ and $y'(t)=exp(y(t))(1-exp(y(t)))$ with the initial condition $y(0)=y_0$
$1-$Justify that this equation has a unique maximum solution $y$ on an interval $I$ that contains $0$.
$2-$Show that if $y_0=0$ then $I=\mathbb{R}$ and $u(t)=0$, $\forall t \in \mathbb{R}$.
$3-$Show that if $y_0>0$ then $y(t)>0$ with $\forall t \in I$
$4-$Let $y_0>0$
$a)$Show that $\frac{1}{exp(x)(1-exp(x))}=exp(-x)+\frac{exp(-x)}{exp(x)(exp(-x)-1)}$.
$b)$ $H(x)=-exp(-x)-ln(1-exp(-x))$ for $x>0$ show that $H(y(t))=H(y_0)+t$,$\forall t \in I$.
$c)$ Show that the function H is a bijection of $\mathbb{R}^*_+$ in $\mathbb{R}^*_+$.
$d)$Deduce that $I=]-H(y_0),+\infty[$.

$1-$Apply Picard's existence theorem or Cauchy–Lipschitz theorem
$2-$If $y_0=0$ then $y'(0)=0$ then $y(t)=0$
we consider the maximal solution Xmax of the equation , defined on an interval $I_{max}$.
We suppose there exists a function$f : \mathbb{R} \rightarrow \mathbb{R}+$, continuous,such that
$||X(t)|| \leqslant f(t) \forall t$ $\in I_{max}$ : then $I_{max}=\mathbb{R}$.
3-The solution would be monotonous from $y'(t)$>0 ,increasing from 0 but I'm not certain on the way to prove it.
4-a)simplify exp(-x) on both sides , put the R.H.S terms to a common denominator add them and multiply the R.H.S ;both the numerator and denominator; by exp(x).
b)Prove $H(y(t))'=y'(t)*H'(y(t))$=$\frac{H(y(t))-H(y_0)}{t-0}=1$
c)The monotony of H(y(t)) is enough I think.
d) I would like help for this , all I can find is $H(y(t))>0 \Longrightarrow H(y_0)+t >0\Longrightarrow t> -H(y_0)$

I welcome your suggestions or answers ,especially $4-d)$.
I would suggest you look at the DE for ##z(t) = \exp(-y(t))##; it is easily solvable.