# A tricky inverse Laplace transform

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1. Jan 13, 2017

### Hoplite

1. The problem statement, all variables and given/known data
I want to invert a function from Laplace transform space to normal space.
2. Relevant equations
In Laplace transform space, the function takes the form $$\bar f (s) = \frac{\exp\left[ x (-a +\sqrt{a^2+ b +c s} )\right]}{-a +\sqrt{a^2+ b +c s}}.$$
Here, $s$ is the Laplace transform space parameter $(t\rightarrow s)$, and $a$, $b$, $c$, and $x$ are all positive constants.

3. The attempt at a solution
Naturally, my first method for inverting a Laplace transform space function is to ask Mathematica, but it can't solve this one. My second method is to break the function into to pieces and use the inverse Laplace transform property:
$$\bar h (s) \bar g(s) \mapsto \int_0^t h(\tau ) g(t-\tau ) d\tau .$$
However, while Mathematica can find the Laplace Transform of the numerator by itself
$$\exp\left[ x (-a +\sqrt{a^2+ b +c s} )\right] \mapsto \frac{x e^{-\frac{t \left(a^2+b\right)}{c}+a x-\frac{c x^2}{4 t}}}{2 \sqrt{\pi } c \sqrt{\frac{t^3}{c^3}}} ,$$
the other part of the function, it cant do:
$$\frac{1}{-a +\sqrt{a^2+ b +c s}} \mapsto ?? .$$
(It does give an inverse Laplace transform for this function, but only for the case where $a<0$, and I need $a>0$.)

I also tried to solve this by finding the residues. I figure there's only one poll (at $s =-b/c$). And Mathematica gives the residue of
$$\frac{\exp[s t]}{-a +\sqrt{a^2+ b +c s}}$$
at $s=-b/c$ as $0$. Could there by any other poles for this function?

I also tried calculating the residue myself by taking a series expansion of
$$\frac{\exp[s t](s+\frac{b}{c})}{-a +\sqrt{a^2+ b +c s}}$$
around $s=-b/c$. But there is no constant term in that series, which I understand mean that the residue is indeed zero.

Does anyone have any suggestions for how I could inverse Laplace transform this function?

2. Jan 13, 2017

### Ray Vickson

You can write
$$\frac{1}{-a +\sqrt{a^2+ b +c s}}$$
as
$$\frac{C}{-A+\sqrt{B+s}},$$
and Maple gets its inverse transform as
$$C \left[ A e^{(A^2-B)t} \text{erfc}(-A\sqrt{t}) +\frac{1}{\sqrt{\pi t}} e^{-Bt} \right] .$$
Here, $\text{erfc}(u)$ is the complementary error function, defined as
$$\text{erfc}(u) = 1 - \text{erf}(u) \equiv 1-\frac{2}{\sqrt{\pi}} \int_0^u e^{-t^2} \, dt .$$

3. Jan 13, 2017

### Hoplite

Hi Ray. Thanks for your response. However, if I rearrange the function as you've suggested, and then ask Mathematica to find the inverse Laplace transform, it also gives the function you've written there, but only as a conditional expression for the case where $A<0$. The problem is that my $a>0$ and therefore my $A>0$.

(I checked, and that function does not work as an inverse Laplace transform for $A>$. I guess it's to be expected that their inverse transforms will be different for positive or negative $A$ since their polls will be in different places.)

Last edited: Jan 13, 2017
4. Jan 13, 2017

### Ray Vickson

Maple imposes no such conditions. In fact, when I asked Maple to find the inverse laplace transform I told it to assume A>0 and B>0.

However, we can check that when we perform the integration $\int_0^{\infty} e^{-st} f(t) \, dt$ we get back the original $F(s)$ only if we assume $B+s > A^2$ (assuming $s > 0$). Either sign for $A$ is OK if that restriction holds, so the easiest restriction is $B \geq A^2$.

5. Jan 14, 2017

### Hoplite

I tried that, and it gives a function that can only be rearranged into the function I'm looking for if $A<0$.

However, this did lead me to find a solution. First rearrange into
$$\frac{1}{-A+\sqrt{B+s}} = \sum_{n=0}^\infty \frac{A^n}{(B+s)^{(n+1)/2 }}.$$
Then take the inverse Laplace transform of every component of the sum. Summing these components gives
$$e^{-B t} \left(A e^{A^2 t} \left(\text{erf}\left(A \sqrt{t}\right)+1\right)+\frac{1}{\sqrt{\pi } \sqrt{t}}\right) ,$$
which Laplace transforms into the original function.

6. Jan 15, 2017

### Ray Vickson

I find it mysterious that you seem to need a restriction on the sign of $A$. Assuming $B+s > A^2$ I get
$$\int_0^{\infty} e^{-st} \left[ A e^{(A^2-B)t} \text{erfc}(-A\sqrt{t}) +\frac{1}{\sqrt{\pi t}} e^{-Bt} \right] \, dt \\ = \frac{\sqrt{B+s}+A}{B+s-A^2} = \frac{1}{-A + \sqrt{B+s}},$$
with no need to assume anything about the sign of $A$.

Last edited: Jan 15, 2017
7. Jan 15, 2017

### Hoplite

I think the difference relates to the process of rearranging the function. If we first make the substitution
$$A=\pm \alpha, \qquad \alpha > 0.$$
Then the left-hand-side becomes
$$\frac{\sqrt{B+s}\pm \alpha}{B+s-\alpha^2} =\frac{\sqrt{B+s}\pm \alpha}{(\sqrt{B+s}+ \alpha)(\sqrt{B+s}- \alpha)} .$$
This therefore gives
$$\frac{1}{\sqrt{B+s}\mp \alpha} .$$
So what it rearranges to depends on the sign of $A$.

8. Jan 15, 2017

### Ray Vickson

You have $A = \pm\, \alpha$, so your answer is
$$\frac{1}{\sqrt{B+s}\, \mp\, \alpha} = \frac{1}{\sqrt{B+s} - A},$$
which is exactly what you want!

Last edited: Jan 15, 2017
9. Jan 16, 2017

### Hoplite

Oh yeah, I didn't look closely enough at that function. That inverse Laplace transform I posted does work too though, which is odd. I guess it's probably just a rearranged form of the other inverse Laplace transform though.