Maximal Interval of Existence for a Differential Equation with Initial Condition

[mathboy20]

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

 

[tiny-tim]

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!

(what's IVP? )

(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?

 

[mathboy20]

(what's IVP? ):

Initial value problem = IVP

(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?

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?

 

[tiny-tim]

Hi Mathboy!

I'm very sorry I didn't reply earlier … I seem to have somehow deactivated my email notification for this thread ​

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

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

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

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

Bang!

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.

™

 

[mathboy20]

Hi Mathboy!

I'm very sorry I didn't reply earlier … I seem to have somehow deactivated my email notification for this thread ​

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

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


Bang!

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.

™

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

 

[tiny-tim]

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 …

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 , 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.

 

[mathboy20]

hmm … the original question was …I've never heard of the phrase "maximal solution" before , 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.

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?