given the function $$g(x,t)=\frac{e^{\frac{Y}{2}}}{2 \epsilon [1+e^{\frac{Y}{2}}]^3}$$ with $Y=\frac{x-\frac{t}{2}}{\epsilon}$ where $x\in \mathbb{R}$ and $t>0$.

I'm looking for $C_{\epsilon}>0$ where $$||g(.,t)||_2\leq C_{\epsilon}$$ where $C_{\epsilon} \to 0 $ as $\epsilon \to 0^+.$

My idea, I used the following variable change $u=x/\epsilon$ and I ve got:

$$||g(.,t)||_2^2=\frac{e^{\frac{-t}{2 \epsilon}}}{4 \epsilon} \int_{\mathbb{R}}\frac{ e^u}{[1+e^{\frac{-t}{4 \epsilon}} e^{\frac{u}{2}}]^6}du$$

Now I only have to prove that the integral in bounded by a constant that doesn't depend on $\epsilon$

So I used the following variable change: $z=e^{-\frac{t}{4 \epsilon}}e^{\frac{u}{2}}$ but I only got:

$$||g(.,t)||_2^2=\frac{1}{2 \epsilon}\int_0^{+\infty}\frac{z}{(1+z)^6}dz$$


1 Answer 1


Integration-by-parts results in

$$\int\frac{z}{(1+z)^6}\text{d}z = - \frac{z}{5(1+z)^5} + \frac{1}{5}\int\frac{1}{(1+z)^5}\text{d}z,$$

where the second integral is straightforward,

$$\int\frac{1}{(1+z)^5}\text{d}z = -\frac{1}{4(1+z)^4}.$$

Taking limits, the improper integral is

$$\int_{0}^{+\infty}\frac{z}{(1+z)^6}\text{d}z = \frac{1}{20}.$$

  • $\begingroup$ That s not what I’m asking , I’m asking to prove that the $L^2$ norm of $g$ converges to $0$ as $\epsilon$ goes to $0$ $\endgroup$
    – BrianTag
    Jan 2, 2020 at 14:34
  • $\begingroup$ @AnasBOUALII One of the main issues is the change of variables. If you want to prove that $||g||_2 \rightarrow 0$ as $\epsilon \rightarrow 0$, the variable of transformation, named either z or u, shouldn't contain the variable $\epsilon$. I'll think about the problem and I'll edit my comment if I solve it. $\endgroup$
    – V.J.
    Jan 2, 2020 at 22:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .