Interval of the maximum solution of a nonlinear equation

In summary, the conversation discusses a homework exercise involving a differential equation with initial conditions. The problem asks for justification for a unique maximum solution on an interval, and solutions for different initial conditions. The conversation also provides some ideas for solutions, including applying the Picard's existence theorem or Cauchy-Lipschitz theorem. Additionally, it suggests looking at the DE for z(t) = exp(-y(t)) for easier solution.
  • #1
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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 [itex]y_0 \in \mathbb{R} [/itex] and [itex]y'(t)=exp(y(t))(1-exp(y(t)))[/itex] with the initial condition [itex]y(0)=y_0[/itex]
[itex]1-[/itex]Justify that this equation has a unique maximum solution [itex]y[/itex] on an interval [itex] I[/itex] that contains [itex]0[/itex].
[itex]2-[/itex]Show that if [itex]y_0=0[/itex] then [itex]I=\mathbb{R}[/itex] and [itex]u(t)=0[/itex], [itex]\forall t \in
\mathbb{R}[/itex].
[itex]3-[/itex]Show that if [itex]y_0>0[/itex] then [itex]y(t)>0[/itex] with [itex]\forall t \in I[/itex]
[itex]4-[/itex]Let [itex]y_0>0[/itex]
[itex]a)[/itex]Show that [itex]\frac{1}{exp(x)(1-exp(x))}=exp(-x)+\frac{exp(-x)}{exp(x)(exp(-x)-1)}[/itex].
[itex]b)[/itex] [itex]H(x)=-exp(-x)-ln(1-exp(-x))[/itex] for [itex]x>0[/itex] show that [itex]H(y(t))=H(y_0)+t[/itex],[itex]\forall t \in I[/itex].
[itex]c)[/itex] Show that the function H is a bijection of [itex]\mathbb{R}^*_+[/itex] in [itex]\mathbb{R}^*_+[/itex].
[itex]d)[/itex]Deduce that [itex]I=]-H(y_0),+\infty[[/itex].

Ideas for answers:
[itex]1-[/itex]Apply Picard's existence theorem or Cauchy–Lipschitz theorem
[itex]2-[/itex]If [itex]y_0=0[/itex] then [itex]y'(0)=0[/itex] then [itex]y(t)=0[/itex]
we consider the maximal solution Xmax of the equation , defined on an interval [itex]I_{max}[/itex].
We suppose there exists a function[itex] f : \mathbb{R} \rightarrow \mathbb{R}+[/itex], continuous,such that
[itex]||X(t)|| \leqslant f(t) \forall t [/itex] [itex]\in I_{max}[/itex] : then [itex]I_{max}=\mathbb{R}[/itex].
3-The solution would be monotonous from [itex]y'(t)[/itex]>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 [itex]H(y(t))'=y'(t)*H'(y(t))[/itex]=[itex]\frac{H(y(t))-H(y_0)}{t-0}=1[/itex]
c)The monotony of H(y(t)) is enough I think.
d) I would like help for this , all I can find is [itex] H(y(t))>0 \Longrightarrow H(y_0)+t >0\Longrightarrow t> -H(y_0)[/itex]

I welcome your suggestions or answers ,especially [itex]4-d)[/itex].
 
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  • #2
yamata1 said:
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 [itex]y_0 \in \mathbb{R} [/itex] and [itex]y'(t)=exp(y(t))(1-exp(y(t)))[/itex] with the initial condition [itex]y(0)=y_0[/itex]
[itex]1-[/itex]Justify that this equation has a unique maximum solution [itex]y[/itex] on an interval [itex] I[/itex] that contains [itex]0[/itex].
[itex]2-[/itex]Show that if [itex]y_0=0[/itex] then [itex]I=\mathbb{R}[/itex] and [itex]u(t)=0[/itex], [itex]\forall t \in
\mathbb{R}[/itex].
[itex]3-[/itex]Show that if [itex]y_0>0[/itex] then [itex]y(t)>0[/itex] with [itex]\forall t \in I[/itex]
[itex]4-[/itex]Let [itex]y_0>0[/itex]
[itex]a)[/itex]Show that [itex]\frac{1}{exp(x)(1-exp(x))}=exp(-x)+\frac{exp(-x)}{exp(x)(exp(-x)-1)}[/itex].
[itex]b)[/itex] [itex]H(x)=-exp(-x)-ln(1-exp(-x))[/itex] for [itex]x>0[/itex] show that [itex]H(y(t))=H(y_0)+t[/itex],[itex]\forall t \in I[/itex].
[itex]c)[/itex] Show that the function H is a bijection of [itex]\mathbb{R}^*_+[/itex] in [itex]\mathbb{R}^*_+[/itex].
[itex]d)[/itex]Deduce that [itex]I=]-H(y_0),+\infty[[/itex].

Ideas for answers:
[itex]1-[/itex]Apply Picard's existence theorem or Cauchy–Lipschitz theorem
[itex]2-[/itex]If [itex]y_0=0[/itex] then [itex]y'(0)=0[/itex] then [itex]y(t)=0[/itex]
we consider the maximal solution Xmax of the equation , defined on an interval [itex]I_{max}[/itex].
We suppose there exists a function[itex] f : \mathbb{R} \rightarrow \mathbb{R}+[/itex], continuous,such that
[itex]||X(t)|| \leqslant f(t) \forall t [/itex] [itex]\in I_{max}[/itex] : then [itex]I_{max}=\mathbb{R}[/itex].
3-The solution would be monotonous from [itex]y'(t)[/itex]>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 [itex]H(y(t))'=y'(t)*H'(y(t))[/itex]=[itex]\frac{H(y(t))-H(y_0)}{t-0}=1[/itex]
c)The monotony of H(y(t)) is enough I think.
d) I would like help for this , all I can find is [itex] H(y(t))>0 \Longrightarrow H(y_0)+t >0\Longrightarrow t> -H(y_0)[/itex]

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

What is the interval of the maximum solution of a nonlinear equation?

The interval of the maximum solution of a nonlinear equation refers to the range of values for the independent variable where the function reaches its highest point.

How do you find the interval of the maximum solution of a nonlinear equation?

To find the interval of the maximum solution, you can use various methods such as graphing, differentiation, or setting the derivative of the function equal to zero and solving for the independent variable.

Why is the interval of the maximum solution important in nonlinear equations?

The interval of the maximum solution is important because it allows us to determine the maximum or minimum value of a function and identify the behavior of the function in that specific interval.

Can the interval of the maximum solution be negative?

Yes, the interval of the maximum solution can be negative if the function has a negative coefficient or if the function is reflected across the x-axis.

Is the interval of the maximum solution always a single point?

No, the interval of the maximum solution can be a single point or a range of values depending on the shape and behavior of the function. In some cases, there may not be a maximum solution at all.

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