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Knowing that $b,a \in C^{0}((0,L]) \cap C^{1}((0,L))$, are positive and $b(x) = \dfrac{1}{\sqrt{xa(x)}}$. Assume that $0 < \alpha < 1$ and $$ \int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{x}}{\sqrt{a(x)}}, \ \ \text{when} \ \ x \to 0 $$ Then $$\dfrac{a(x)}{x^{2}}\int_{0}^{x}b(\tau)d\tau \sim \dfrac{2}{1-\alpha}\dfrac{\sqrt{a(x)}}{x^{3/2}}, \ \ \text{when} \ \ x \to 0.$$

I'm trying to understand the above implication which is a passage from the proof of Lemma 5.12 of Martinez & Vancostenoble - Carleman estimates for one-dimensional degenerate heat equations.

It seems to be simple, I'm trying to apply L'hopital's rule. Basically what the author of the article did was multiply $\dfrac{a(x)}{x^{2}}$ by the limit of $\displaystyle\int_{0}^{x}b(\tau)d\tau$. Why is this possible?

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  • $\begingroup$ If you're going to \dfrac your fractions, please also \displaystyle your integrals: $\displaystyle\frac{a(x)}{x^2}\int_0^x b(\tau)\mathrm d\tau$ \displaystyle\frac{a(x)}{x^2}\int_0^x b(\tau)\mathrm d\tau looks better than $\dfrac{a(x)}{x^2}\int_0^x b(\tau)d\tau$ \dfrac{a(x)}{x^2}\int_0^x b(\tau)d\tau (though it's better not to \displaystyle titles at all). I have edited accordingly. $\endgroup$
    – LSpice
    Commented Sep 19, 2022 at 16:40
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    $\begingroup$ I don't think there's anything involved here. You just multiplied both sides by $a/x^2$, which is certainly justified if $\sim$ has the standard meaning $A\sim B \iff A/B\to 1$. $\endgroup$ Commented Sep 19, 2022 at 16:56
  • $\begingroup$ @ChristianRemling Thank you! $\endgroup$
    – user253963
    Commented Sep 19, 2022 at 19:17

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