# Limit of unspecified solution

• Sep 18th 2010, 10:10 AM
hatsoff
Limit of unspecified solution
Quote:

Consider the scalar initial value problem

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

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

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

(with justification).
I'm pretty stuck on this one. I believe the limit is $\displaystyle t^2+1$, 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!
• Sep 18th 2010, 12:50 PM
zzzoak
May be this way is good:
$\displaystyle when {\:}{\;}\mu {\:}{\;}tends{\:}{\;} to{\:}{\;} zero$
$\displaystyle x'=\frac{t}{2+\mu x^2}$
x(0)=1
• Sep 18th 2010, 02:38 PM
hatsoff
I'm not sure I follow... $\displaystyle x'$ is $\displaystyle \frac{t}{1+e^{\mu x^2}}$, not $\displaystyle \frac{t}{2+\mu x^2}$.
• Sep 18th 2010, 03:02 PM
zzzoak
$\displaystyle e^x=1+x+x^2/2+...$
Consider two terms.
• Sep 18th 2010, 04:14 PM
hatsoff
EDIT: This doesn't work after all.

Okay, let's see...

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

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

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

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

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

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

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

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

Okay, that looks like it will fly! Thanks!
• Sep 18th 2010, 04:43 PM
zzzoak
Sorry from your last line I see
$\displaystyle x=t^2/4+1$
• Sep 18th 2010, 05:06 PM
hatsoff
Oops, yes, you're right. Also...

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

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