Maximal Interval of Existence for a Differential Equation with Initial Condition

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Homework Help Overview

The discussion revolves around an initial value problem (IVP) involving a differential equation. The equation is presented in a form that relates the rate of change of a variable \(x\) with respect to time \(t\) to a cosine function, with parameters defined for two specific values of \(\alpha\). Participants are tasked with finding the maximal solution given the initial condition \(x(0) = 0\).

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the interpretation of "maximal solution" and whether it refers to the maximum interval of existence for the solution. There are attempts to solve the equation and questions about the implications of different values of \(\alpha\) on the solution's behavior. Some participants express confusion about the oscillatory nature of the solution and the conditions under which it remains defined.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some guidance has been offered regarding the oscillatory behavior of the solution and the implications of the parameter \(\alpha\). There is no explicit consensus on the definition of "maximal solution," and multiple interpretations are being considered.

Contextual Notes

Participants note the potential for confusion regarding the terminology used in the problem statement, particularly the phrase "maximal solution." There are also discussions about the constraints imposed by the values of \(\alpha\) and how they affect the solution's characteristics.

mathboy20
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Homework Statement



Given the IVP problem

\rm{(1+x^2)^{-1} \frac{dx}{dt} = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)} \iff

\rm{(1+x^2)^{-1} dx = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)} dt

Find the maximal solution with the initial conditio condition x(0) = 0.

Then alpha is either \alpha_{1} = \frac{1}{4} or \alpha_{2} = \frac{1}{2}

The Attempt at a Solution



I am fairly new to differential equation theory like this, but couldn't my professor mean "Maximum interval of existence"? By that meaning "maximal solution" is equivalent to the statement that finding the maximal interval on which the solution curve is defined.

Thusly if I try to solve the above equation I get

\rm{tan(x)^{-1} = \frac{\alpha \cdot 2 \pi}{4} \cdot sin(t) + C}

x = \rm{tan(\frac{\alpha \cdot \pi}{2} \cdot sin(t) + C)}

What I don't get here is that if I insert the values of first or second alpha then I get an interval which is centered around zero.

But that can be right can it?? Here I am stuck and need help...

Best Regards
Mathboy
 
Last edited:
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mathboy20 said:
Given the IVP problem

\rm{(1+x^2)^{-1} \frac{dx}{dt} = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)} \iff

\rm{(1+x^2)^{-1} dx = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)} dt

Find the maximal solution with the initial conditio condition x(0) = 0.

Then alpha is either \alpha_{1} = \frac{1}{4} or \alpha_{2} = \frac{1}{2}
…
I am fairly new to differential equation theory like this, but couldn't my professor mean "Maximum interval of existence"? By that meaning "maximal solution" is equivalent to the statement that finding the maximal interval on which the solution curve is defined.

Thusly if I try to solve the above equation I get
…
x = \rm{tan(\frac{\alpha \cdot \pi}{2} \cdot sin(t) + C)}

Hi mathboy20! :smile:

(what's IVP? :confused:)

(Your C is obviously 0.)

It's not "the maximal interval on which the solution curve is defined" … if alpha < 1, the solution is defined for all t.

Hint: sin(t) oscillates between ±1, so x also oscillates …

what is the interval within which it oscillates? :smile:
 
tiny-tim said:
(what's IVP? :confused:):


Initial value problem = IVP

tiny-tim said:
(Your C is obviously 0.)

But I need to find C in each of the two alpha cases?


what is the interval within which it oscillates? :smile:

Don't shoot me, but since its for all t then x oscillates on the interval \pm \rm{t}??

Then as you say alpha < 1.

Best Regards
Mathboy

p.s. If there existed a case where alpha > 1 then the solution would exeed the interval on which the previous two solutions with regards to alpha where defined?
 
Last edited:
Hi Mathboy! :smile:
I'm very sorry I didn't reply earlier … I seem to have somehow deactivated my email notification for this thread :redface:​
mathboy20 said:
But I need to find C in each of the two alpha cases?

(have a pm: ± and an alpha: α and a pi: π :smile:)

Both zero … when t = 0, sint = 0, and so tan (απ/2 sint) = 0 , for any α. :smile:
Don't shoot me, but since its for all t then x oscillates on the interval \pm \rm{t}??

Bang! :biggrin:

sint oscillates between ±1

so απ/2 sint oscillates between ±απ/2

so tan(απ/2 sint) oscillates between ±tan(απ/2) so long as that doesn't "include infinity".

So if α ≥ 1, it doesn't oscillate … it jumps wildly. :smile:

™
 
tiny-tim said:
Hi Mathboy! :smile:
I'm very sorry I didn't reply earlier … I seem to have somehow deactivated my email notification for this thread :redface:​


(have a pm: ± and an alpha: α and a pi: π :smile:)

Both zero … when t = 0, sint = 0, and so tan (απ/2 sint) = 0 , for any α. :smile:


Bang! :biggrin:

sint oscillates between ±1

so απ/2 sint oscillates between ±απ/2

so tan(απ/2 sint) oscillates between ±tan(απ/2) so long as that doesn't "include infinity".

So if α ≥ 1, it doesn't oscillate … it jumps wildly. :smile:

™


So anyway if so just to stress that I have understood You correct Mister Tiny Tim.

The maximum solution for the original equation lies on the interval between ±tan(απ/2)?

Best Regards
Mathboy
 
mathboy20 said:
So anyway if so just to stress that I have understood You correct Mister Tiny Tim.

The maximum solution for the original equation lies on the interval between ±tan(απ/2)?

hmm … the original question was …
mathboy20 said:
Find the maximal solution with the initial conditio condition x(0) = 0.

Then alpha is either \alpha_{1} = \frac{1}{4} or \alpha_{2} = \frac{1}{2}

I've never heard of the phrase "maximal solution" before :frown:, but I assume what it means is the solution with the maximum oscillation … which would be tan(απ/2) = ∞, or α = 1.

I don't see where α = 1/4 or 1/2 comes from. :confused:
 
tiny-tim said:
hmm … the original question was …I've never heard of the phrase "maximal solution" before :frown:, but I assume what it means is the solution with the maximum oscillation … which would be tan(απ/2) = ∞, or α = 1.

I don't see where α = 1/4 or 1/2 comes from. :confused:

Just looked at the again.

There are three possible alphas \alpha = 1/2, \alpha = \sqrt{2} , \alpha = 1 But that doesn't change anything does it?
 

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