4
$\begingroup$

What is the inverse Laplace transform of L(s)=$exp⁡[-(1/2)sI]$*$I_0[(1/2)sI]$, where $I_0$ is the modified bessel function of first kind.

I am told that the answer is $\frac{1}{π}$ $\left(\frac{1}{(sI-s^2)}\right)^\frac{1}{2}$ but I am struggling to achieve this answer.

Can anyone help?

$\endgroup$
2
  • $\begingroup$ please use mathjax to format your question. Here is a link to a quick tutorial on how to use it . $\endgroup$ May 21 '18 at 20:24
  • $\begingroup$ Thank you, I will do that ASAP $\endgroup$
    – Mlo27
    May 21 '18 at 20:24
1
$\begingroup$

Well, we are looking at the inverse Laplace transform of:

$$\text{y}_{\space\text{n}}\left(t\right):=\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\text{n}\cdot\text{s}\right)\cdot\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(t\right)}\tag1$$

Where $\mathcal{I}_{\space\text{p}}\left(\text{z}\right)$ is the modified Bessel function of the first kind.

Now, using the convolution theorem, of the Laplace transform, we can write:

$$\text{y}_{\space\text{n}}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\text{n}\cdot\text{s}\right)\right]_{\left(t-\tau\right)}\space\text{d}\tau\tag2$$

Using the table of selected Laplace transforms, we can write:

$$\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\text{n}\cdot\text{s}\right)\right]_{\left(t-\tau\right)}=\delta\left(t-\tau-\text{n}\right)\tag3$$

Where $\delta\left(x\right)$ is the Dirac Delta function.

So, we can rewrite equation $\left(2\right)$ as follows:

$$\text{y}_{\space\text{n}}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(\tau\right)}\cdot\delta\left(t-\tau-\text{n}\right)\space\text{d}\tau\tag4$$

Using the definition of the modified Bessel function of the first kind, we can write:

$$\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)=\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\left(\frac{\text{n}\cdot\text{s}}{2}\right)^{2\text{k}}\tag5$$

Where $\Gamma\left(\text{s}\right)$ is the Gamma function.

So using the table of selected Laplace transforms, we can write:

$$\mathscr{L}_\text{s}^{-1}\left[\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(\tau\right)}=\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\left(\frac{\text{n}\cdot\text{s}}{2}\right)^{2\text{k}}\right]_{\left(\tau\right)}=$$ $$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\frac{1}{\Gamma\left(-2\text{k}\right)}\cdot\left(\frac{\text{n}}{2}\right)^{2\text{k}}\cdot\frac{1}{\tau^{1+2\text{k}}}\tag6$$

Now, we can rewrite equation $\left(4\right)$ as follows:

$$\text{y}_{\space\text{n}}\left(t\right)=\int_0^t\left\{\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\frac{1}{\Gamma\left(-2\text{k}\right)}\cdot\left(\frac{\text{n}}{2}\right)^{2\text{k}}\cdot\frac{1}{\tau^{1+2\text{k}}}\right\}\cdot\delta\left(t-\tau-\text{n}\right)\space\text{d}\tau=$$ $$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\frac{1}{\Gamma\left(-2\text{k}\right)}\cdot\left(\frac{\text{n}}{2}\right)^{2\text{k}}\int_0^t\frac{\delta\left(t-\tau-\text{n}\right)}{\tau^{1+2\text{k}}}\space\text{d}\tau\tag7$$

When $\text{n}>0\space\wedge\space\text{k}\ge0\space\wedge\space t\ge0$, we can write:

$$\int_0^t\frac{\delta\left(t-\tau-\text{n}\right)}{\tau^{1+2\text{k}}}\space\text{d}\tau=\frac{2\cdot\theta\left(t\right)-1}{\left(t-\text{n}\right)^{1+2\text{k}}}\cdot\theta\left(t-\text{n}-t\cdot\theta\left(-t\right)\right)\cdot\theta\left(\text{n}-t+t\cdot\theta\left(t\right)\right)\tag8$$

Where $\theta\left(\text{a}\right)$ is the Heaviside step function.

$\endgroup$
1
$\begingroup$

This is pretty straightforward if one writes

$$I_0(z) = \frac1{\pi} \int_{-1}^1 dv \, (1-v^2)^{-1/2} \, e^{z v} $$

Then, letting $a=I/2 \gt 0$,

$$\begin{align} f(t) &= \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, I_0(a s) e^{(t-a) s}\\ &= \frac1{\pi} \int_{-1}^1 dv \, (1-v^2)^{-1/2} \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{(t-a+a v)s} \\ &= \frac1{\pi} \int_{-1}^1 dv \, (1-v^2)^{-1/2} \delta(a v+t-a) \\ &= \frac1{\pi a} \left (1- \left (1-\frac{t}{a} \right )^2 \right )^{-1/2} \\ &= \frac1{\pi} (2 a t-t^2)^{-1/2} \end{align} $$

That said, note that for the argument of the delta function inside the $v$-integral to vanish, $t \in [0,2 a]$. Noting that I defined $a=I/2$, the ILT for the given Laplace transform is

$$f(t) = \frac1{\pi} (I t-t^2)^{-1/2} \theta(t) \theta(I-t) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.