Laplace inverse of a function

**Ehsan** · September 16th 2008, 02:17 AM

Hi,
Can anybody help me to find inverse laplace transform of one of these functions.
Thanks,

$ f(s)=\frac{1}{s+a+b\sqrt{s+c}} $

$ f(s)=\frac{1}{s(s+a+b\sqrt{s+c})} $

**TKHunny** · September 16th 2008, 07:09 AM

Currently, I am of the opinion that you will not find a convenient expression for those. It may be time for the fundamental definition and figuring out how to deal with the integral expressions.

However, for the first, you may wish to consider a transformation of variables, maybe:

$w = \sqrt{s+c}+\frac{b}{2}$

$v = \sqrt{a-c-\frac{b^{2}}{4}}$

Then consider the transform with respect to "w", rather than 's', and see if that leads to anything even close to

$\frac{\sin(v \cdot t)}{v}$

**shawsend** · September 16th 2008, 11:05 AM

Hi. Those look like a challenge. Isn't someone in here taking Honors Complex Analysis? Hummmm . . . Anyway, Mathematica can't even compute $\mathcal{L}^{-1}\left\{\frac{1}{s+\sqrt{s-1}}\right\}$ . Doesn't mean there isn't one of course. Even though I'm not certain the integral converges, I found it interesting that I could (apparently) get a reasonably accurate numerical approximation to what I think is the inverse transform by computing numerically $\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}\frac{e^{st}}{s+\sqrt{s-1}}ds$ over the interval $t\in (0,4)$ . The first plot is the data points I calculated and the blue line is a fit to the equation $a+bt+ct^2$ . Granted that's probably not exactly the inverse transform. But I'm encouraged by next taking the Laplace transform of my fit:

$\mathcal{L}\left\{a+bt+ct^2\right\}=\frac{a}{s}+\f rac{b}{s^2}+\frac{2c}{s^3}$ (in blue) against $\frac{1}{s+\sqrt{s-1}}$ (in red) which is the second plot. Granted it's not perfect, but the agreement is surprisingly close in the range (1.5,4).

I'm surprised I could get even this close with numerical methods and I'm optimistic that the numerical result is approximately the inverse transform in this range of $t$ .

**Ehsan** · September 17th 2008, 08:27 AM

Hi, Thanks for your help and interest to solve this problem, but I could solve it this way.

$ f(s)=\frac{1}{s+a+b\sqrt{s+c}}=\frac{1}{(\sqrt{s+c }+B_1)(\sqrt{s+c}+B_2)} $

$ =\frac{1}{B_1-B_2}(\frac{1}{\sqrt{s+c}+B_2}-\frac{1}{\sqrt{s+c}+B_1}) $

where

$ B_1,B_2=\frac{1}{2}[b\pm\sqrt{\Delta}] $

and

$\Delta=b^2-4(a-c)$

**shawsend** · September 17th 2008, 09:02 AM

you da' man Ehsan!

**shawsend** · September 17th 2008, 09:41 AM

I was curious how close I came with the numerical approximation. So using you expression, I used Mathematica and calculated the inverse transform of $\frac{1}{s+\sqrt{s-1}}$ :

$f(t)=\frac{1}{b_1-b2}\left[e^t\left(\frac{1}{\sqrt{\pi t}}-b_2e^{b_2^2 t}\textbf{Erfc}(b_2\sqrt{t})\right)-e^t\left(\frac{1}{\sqrt{\pi t}}-b_1e^{b_1^2 t}\textbf{Erfc}(b_1\sqrt{t})\right)\right]$

The plot below shows $f(t)$ above in red, and the numerical fit I calculated in an earlier post still in blue with the blue data points. I'm really surprised I guess the agreement is pretty good outside the end points.

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