# Inverse Laplace Transform involving $\cosh$.

While doing an assignment on solving a PDE I stumbled into the following inverse Laplace transform question (involving $\cosh$? I can't believe it). Mathematica gives no solution and I have no idea where to start, and don't know how to use $\mathcal L[f_1\star f_2] = \mathcal L[f_1] \mathcal L[f_2]$, if it's of any help.

$$\mathcal L_{t}^{-1} \left[ \frac{1}{s(s^2+\omega^2)} \frac{\cosh(sx/a)}{\cosh(sl/a)} \right]$$

I have to change $s$ back to $t$, which will allow me to finish off this problem. Please share your ideas.

## 1 Answer

Usually these require the Residue theorem. The inverse Laplace transform is given by

$$\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{\exp{(s t)}}{s (s^2 + \omega^2)} \frac{\cosh{(s \frac{x}{a})}}{\cosh{(s \frac{\ell}{a})}}$$

where $c$ is a real number such that all of the singularities of the integrand lie to the left.

I will refer you to other posts to view a picture of the contour, which involves a semicircle along the line of integration oriented to the left, enclosing all of the poles of the integrand.

The poles of the integrand here are at $s=0$ and $s = \pm i \omega$. The residues corresponding to these poles (i.e., $\lim_{x \rightarrow p} (x-p) f(x)$ for integrand $f(x)$ and pole $p$) are as follows: $\frac{1}{\omega^2}$ and

$$\frac{\exp{(\pm i \omega t)} \cos{(\frac{x}{a})}}{2 \omega^2 \cos{(\frac{\ell}{a})}}$$

respectively. Adding the residues up, we get

$$\frac{1}{\omega^2} \left( 1 - \frac{\cos{(\frac{x}{a})} \cos{(\omega t)}}{\cos{(\frac{\ell}{a})}} \right )$$

There are also poles resulting from the $\cosh{(s \frac{\ell}{a})}$ term in the integrand at $s=\pm i (2 k+1) (\pi/2) (a/\ell)$ for all integers $k$. The residues from these poles are

$$\sum_{k=-\infty}^{\infty} (-1)^{k+1} \frac{\exp{(i (2 k+1) \frac{\pi}{2} \frac{a}{\ell} t)} \cos{((2 k+1) \frac{\pi}{2} \frac{x}{\ell} t)}}{i (2 k+1) \frac{\pi}{2} \frac{a}{\ell} \left ( \omega^2 - (2 k+1)^2 (\frac{\pi}{2} \frac{a}{\ell})^2 \right )}$$

$$= 2 \sum_{k=0}^{\infty} (-1)^{k+1} \frac{\sin{((2 k+1) \frac{\pi}{2} \frac{a}{\ell} t)} \cos{((2 k+1) \frac{\pi}{2} \frac{x}{\ell} t)}}{(2 k+1) \frac{\pi}{2} \frac{a}{\ell} \left ( \omega^2 - (2 k+1)^2 (\frac{\pi}{2} \frac{a}{\ell})^2 \right )}$$

I do not know of a closed-form expression for this latter sum.

• You may want to check to see if there are singularites of the cosh term in the denominator; I am a little suspicious that there is a pole at $\omega = 0$. – Ron Gordon Dec 26 '12 at 14:49
• I can't comprehend becase I have no basics in complex analysis, and it will take me some time to learn it. Also, the Laplace transform of your result is off by 2h's, as in $\cosh$ to $\cos$. Have you switched $\cosh$ to $\cos$ in the middle? – Frenzy Li Dec 26 '12 at 15:03
• @rigordonma - The forward Laplace transform of the last expression in your answer gives something that is quite different from the original problem. So something is not quite right. In particular, $\cosh z = 0$ for $z = (k + 1/2) i \pi, k \in \mathbb{Z}$, so there are additional singularities. – Hans Engler Dec 26 '12 at 15:20
• @rlgordonma Appreciate your attention, but Hans Engler's statement is true. Please take your time to elaborate on your answer, because I'm going to sleep. – Frenzy Li Dec 26 '12 at 15:23
• @Hans, yes, I was aware of this as soon as I hot the send button. (See my previous comment.) I will make a correction. – Ron Gordon Dec 26 '12 at 18:22