I need help solving this (I suggest something hereafter but I am not sure if it's ok):

I would like to find an approximate solution of the function $\bar{f}(s)$ defined in the Laplace space. At long times in the real time domain, I suppose that $t \to \infty$ so in the Laplace space $s \to \infty$. So, I am looking for a limit/asymptote in the laplace space when $s \to \infty$ and would find the inverse Laplace transform to get a solution in the real time domain.

Before defining $\bar{f}(s)$ let's define the following functions:

$\bar{u}(s) = \cfrac{s\omega (\omega -1)+\lambda }{s(\omega -1)+\lambda } \qquad\mathrm{with} \qquad\omega \quad \mathrm{and} \quad\lambda \in \mathbb{R^{+}} \\ \sigma = \sqrt{s\bar{u}(s)}\\ \nu = 1- \cfrac{n}{2} \quad \mathrm{with} \quad n \in ]0;3]$

Now, $\bar{f}(s) = \cfrac{\sigma}{\Gamma\left(\frac{n}{2} \right)}\cfrac{K_{\nu-1}\left(\sigma \right)}{K_\nu\left(\sigma\right)}$

In which $K_\alpha\left(\mathcal{Z}\right)$ is the modified Bessel function of second kind of order $\alpha$ and argument $\mathcal{Z}$ and $\Gamma$ the $\Gamma$ function.

If $\lim\limits_{s \to 0} u \left( s \right) = 1$, then we should have $\lim\limits_{s \to 0} \sigma \left( s \right) = 0$.

For small arguments, the modified Bessel function can be approximated like this: $K_{\alpha} \left( \mathcal{Z} \right) \approx \begin{cases} \cfrac{\Gamma \left( \alpha \right)}{2} \left( \cfrac{2}{\mathcal{Z}} \right)^{\alpha} & \qquad \mathrm{if } \qquad~\alpha > 0 \in \mathbb{R} \\ - \left( \gamma +\ln \cfrac{\mathcal{Z}}{2} \right) & \qquad \mathrm{if} \qquad~\alpha = 0, \mathrm{with~ } \gamma = \mathrm{Euler's~constant} \end{cases}$

• The first case (i.e. $\alpha > 0$) is verified when $0 < \nu < 2$ and an approximative value of $\bar{f}(s)$ can be found. Replacing $\alpha$ and $\mathcal{Z}$ we get: $\bar{f}(s) \approx \cfrac{ s \omega \left( 1 - \omega \right) + \lambda }{s \left( 1 - \omega \right)+ \lambda}\cfrac{1}{2\Gamma \left( n/2 \right)} \cfrac{\Gamma \left( \nu -1\right)}{\Gamma \left( \nu \right)}$

If I numerically evaluate the inverse Laplace transform (with DeHoog's algorithm) of $\cfrac{ s \omega \left( 1 - \omega \right) + \lambda }{s \left( 1 - \omega \right)+ \lambda}$, I find a good approximation to my problem.

If I take the inverse Laplace transform $\mathcal{L}^{-1}_s \left\{ \cfrac{ s \omega \left( 1 - \omega \right) + \lambda }{s \left( 1 - \omega \right)+ \lambda} \right\}\left(t_D\right) = \omega \delta(t_D) + \lambda e^{(\lambda t_D)/(\omega-1)}$

We should have $\bar{f}(t_D) \approx \cfrac{\lambda e^{(\lambda t_D)/(\omega-1)}}{2\Gamma \left( n/2 \right)} \cfrac{\Gamma \left( \nu -1\right)}{\Gamma \left( \nu \right)} \qquad \mathrm{when} \qquad 0< n <2$.

As $t_D \to \infty$ the term $\omega \delta(t_D)$ $\to 0$ but I am not too sure about (i) this assumption and (ii) the previous inverse Laplace transform. Are these 2 points correct?

Also when $n =2$, we get the second case of approximation for the Bessel function. But what can we say anything for $n > 2$ (can we approx something at late times)??

Well, any help and corrections will be appreciated (!!)

Thank you

Viv

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