# Thread: Limit of unspecified solution

1. ## Limit of unspecified solution

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!

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

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

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

5. 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!

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

7. 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.