Limit of unspecified solution

**hatsoff** · September 18th 2010, 10:10 AM

Consider the scalar initial value problem

$x'=\frac{t}{1+e^{\mu x^2}}$ , $x(0)=1$ ,

where $\mu\in\mathbb{R}$ is a parameter. Clearly, for each $\mu\in\mathbb{R}$ , the solution $x_\mu(t)$ of the initial-value problem exists on $\mathbb{R}$ and is unique. Find

$\displaystyle \lim_{\mu\to 0} x_\mu(t)$

(with justification).

I'm pretty stuck on this one. I believe the limit is $t^2+1$ , but I cannot prove it. The underlying sequence does not appear to be uniformly convergent, which is the only way I'd be able to show the result.

Any help would be much appreciated!

**zzzoak** · September 18th 2010, 12:50 PM

May be this way is good:
$when {\:}{\;}\mu {\:}{\;}tends{\:}{\;} to{\:}{\;} zero$
$x'=\frac{t}{2+\mu x^2}$
x(0)=1

**hatsoff** · September 18th 2010, 02:38 PM

I'm not sure I follow... $x'$ is $\frac{t}{1+e^{\mu x^2}}$ , not $\frac{t}{2+\mu x^2}$ .

**zzzoak** · September 18th 2010, 03:02 PM

$e^x=1+x+x^2/2+...$
Consider two terms.

**hatsoff** · September 18th 2010, 04:14 PM

EDIT: This doesn't work after all.

Okay, let's see...

$x'=\frac{t}{1+e^{\mu x^2}}=\frac{t}{1+1+\mu x^2+\mu^2 x^4/2+\cdots}$

So $(2+\mu x^2+\mu^2 x^4/2+\cdots)dx=t\;dt$

which means $2x+\frac{\mu x^3}{3}+\frac{\mu^2 x^5}{10}+\cdots=\frac{t^2}{2}+C$

To find $C$ , we plug in $(t,x)=(0,1)$ :

$2+\frac{\mu}{3}+\frac{\mu^2}{10}+\cdots=C$ .

So $2x+\frac{\mu x^3}{3}+\frac{\mu^2 x^5}{10}+\cdots=\frac{t^2}{2}+2+\frac{\mu}{3}+\fra c{\mu^2}{10}+\cdots$ , which means

$\lim_{\mu\to 0}\left(2x+\frac{\mu x^3}{3}+\frac{\mu^2 x^5}{10}+\cdots\right)=\lim_{\mu\to 0}\left(\frac{t^2}{2}+2+\frac{\mu}{3}+\frac{\mu^2} {10}+\cdots\right)$

Or $2\lim_{\mu\to 0}x=\frac{t^2}{2}+2$ , that is, $\lim_{\mu\to 0}x=t^2/4+1$ .

Okay, that looks like it will fly! Thanks!

**zzzoak** · September 18th 2010, 04:43 PM

Sorry from your last line I see
$x=t^2/4+1$

**hatsoff** · September 18th 2010, 05:06 PM

Oops, yes, you're right. Also...

Suppose $x_\mu(t)\geq k/\sqrt{\mu}$ for some $t$ in the domain. Then $\lim_{\mu\to 0}\left(2x+\frac{\mu x^3}{3}+\frac{\mu^2 x^5}{10}+\cdots\right)\neq2\lim_{\mu\to 0}x$ .

So I guess this argument doesn't work after all.

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