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Problem

Let $u(x,t) = \frac{e^\frac{-x^2}{4t}}{\sqrt{4 \pi t}} $ for $ t > 0, > x \in \mathbb{R} $. If a > 0, prove that $u(x,t) \rightarrow 0$ as $t > \rightarrow 0+$, uniformly for $x \in [a, \infty ]$

I proved like this, but I didn't understand well.

$|u(x,t)| \le u(a,t)$ for all $ x \in \mathbb{R} $ and $u(a,t) = \frac{e^\frac{-a^2}{4t}}{\sqrt{4 \pi t}} $.
As t -> 0+, $e^\frac{-a^2}{4t}$ converges to zero much faster than $ \frac{1}{\sqrt{4 \pi t}} $ to infinity. Thus u(a,t) uniformly converges to zero, so u(x,t) uniformly converges to zero for all $ x \in \mathbb{R} $.

Is my proof correct?

alryosha
  • 563

1 Answers1

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Showing that $u(a,t) \to 0$ is equivalent to showing that $u^{2}(a,t) \to 0$. Make the substitution $s=\frac {a^{2}} {2t}$ and reduce to the proof to $se^{-s} \to 0$ as $ s\to \infty$ Apply L'Hopital's Rule to the ration $ \frac s {e^{s}}$.

Kavi Rama Murthy
  • 311,013