Interval of the maximum solution of a nonlinear equation

[yamata1]

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<br /> \mathbb{R}.
3-Show that if y_0&gt;0 then y(t)&gt;0 with \forall t \in I
4-Let y_0&gt;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&gt;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[.

Ideas for answers:
1-Apply Picard's existence theorem or Cauchy–Lipschitz theorem
2-If y_0=0 then y&#039;(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 functionf : \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&#039;(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))&#039;=y&#039;(t)*H&#039;(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))&gt;0 \Longrightarrow H(y_0)+t &gt;0\Longrightarrow t&gt; -H(y_0)

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

 

[Ray Vickson]

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&#039;(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<br /> \mathbb{R}.
3-Show that if y_0&gt;0 then y(t)&gt;0 with \forall t \in I
4-Let y_0&gt;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&gt;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[.

Ideas for answers:
1-Apply Picard's existence theorem or Cauchy–Lipschitz theorem
2-If y_0=0 then y&#039;(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 functionf : \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&#039;(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))&#039;=y&#039;(t)*H&#039;(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))&gt;0 \Longrightarrow H(y_0)+t &gt;0\Longrightarrow t&gt; -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.