Numerical inverse Laplace

Question

I have this function \begin{equation} F(s) = \exp\left(a s + \frac{b}{s+\gamma}\right), \end{equation} for which I need to compute the inverse Laplace transform, at least numerically. I can compute the inverse Laplace transform for the Log of the function, namely: \begin{equation} \mathcal{L}^{-1}\left\{\ln[F(s)]\right\} = a H(t) + b e^{-\gamma t}, \end{equation} Is there any way to use this result to numerically compute $f(t) = \mathcal{L}^{-1}\{F(s)\}$ over a finite interval $[0,T]$? I am using Matlab but so far the only algorithms I have found are extremely unstable and blow up at $t=0$. I should also point out that my time interval is quite small (T is of the order of $10^{-6}$ or so), so even some kind of asymptotic approximation would be useful.

Answer 1

Well, first of all we are trying to find the inverse Laplace transform of:

$$\text{y}_{\alpha,\beta,\gamma}\left(t\right):=\mathcal{L}_\text{s}^{-1}\left[\exp\left(\alpha\cdot\text{s}+\frac{\beta}{\gamma+\text{s}}\right)\right]_{\left(t\right)}\tag1$$

Now, we rewrite as follows:

$$\exp\left(\alpha\cdot\text{s}+\frac{\beta}{\gamma+\text{s}}\right)=\exp\left(\alpha\cdot\text{s}\right)\cdot\exp\left(\frac{\beta}{\gamma+\text{s}}\right)\tag2$$

Using the time shifting property of the Laplace transform, we can write:

$$\text{y}_{\alpha,\beta,\gamma}\left(t\right)=\mathcal{L}_\text{s}^{-1}\left[\exp\left(\alpha\cdot\text{s}\right)\cdot\exp\left(\frac{\beta}{\gamma+\text{s}}\right)\right]_{\left(t\right)}=\theta\left(t+\alpha\right)\cdot\mathcal{L}_\text{s}^{-1}\left[\exp\left(\frac{\beta}{\gamma+\text{s}}\right)\right]_{\left(t+\alpha\right)}\tag3$$

Where $\theta\left(x\right)$ is the Heaviside step function.

Now, using the formal definition of the exponential function:

$$\exp\left(\frac{\beta}{\gamma+\text{s}}\right)=\sum_{\text{k}=0}^\infty\frac{\left(\frac{\beta}{\gamma+\text{s}}\right)^\text{k}}{\text{k}!}=\sum_{\text{k}=0}^\infty\frac{\beta^\text{k}}{\text{k}!}\cdot\frac{1}{\left(\gamma+\text{s}\right)^\text{k}}\tag4$$

So, we can rewrite equation $(3)$:

$$\text{y}_{\alpha,\beta,\gamma}\left(t\right)=\theta\left(t+\alpha\right)\cdot\mathcal{L}_\text{s}^{-1}\left[\sum_{\text{k}=0}^\infty\frac{\beta^\text{k}}{\text{k}!}\cdot\frac{1}{\left(\gamma+\text{s}\right)^\text{k}}\right]_{\left(t+\alpha\right)}=$$ $$\theta\left(t+\alpha\right)\cdot\sum_{\text{k}=0}^\infty\frac{\beta^\text{k}}{\text{k}!}\cdot\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\left(\gamma+\text{s}\right)^\text{k}}\right]_{\left(t+\alpha\right)}\tag5$$

Now, using the table of selected Laplace transforms, we find that:

$$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\left(\gamma+\text{s}\right)^\text{k}}\right]_{\left(t+\alpha\right)}=\frac{\exp\left(-\left(\alpha+t\right)\cdot\gamma\right)}{\left(\alpha+t\right)^{1-\text{k}}\cdot\Gamma\left(\text{k}\right)}\tag6$$

So, in the end we get:

$$\text{y}_{\alpha,\beta,\gamma}\left(t\right)=\theta\left(t+\alpha\right)\cdot\sum_{\text{k}=0}^\infty\frac{\beta^\text{k}}{\text{k}!}\cdot\frac{\exp\left(-\left(\alpha+t\right)\cdot\gamma\right)}{\left(\alpha+t\right)^{1-\text{k}}\cdot\Gamma\left(\text{k}\right)}=$$ $$\theta\left(t+\alpha\right)\cdot\exp\left(-\left(\alpha+t\right)\cdot\gamma\right)\cdot\sum_{\text{k}=0}^\infty\frac{\beta^\text{k}}{\text{k}!}\cdot\frac{1}{\left(\alpha+t\right)^{1-\text{k}}\cdot\Gamma\left(\text{k}\right)}\tag7$$

Answer 2

Let $a, b \in \mathbb R$. Then $$\mathcal L^{-1}[s \mapsto e^{b/s}](t) = \sqrt {\frac b t} \, I_1(2 \sqrt {b t}) H(t) + \delta(t), \\ \mathcal L^{-1}[s \mapsto e^{a s} F(s + \gamma)](t) = e^{-\gamma (t + a)} \mathcal L^{-1}[F](t + a).$$ If $a > 0$, we need to take the two-sided transform to recover $F$.