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I have a Laplace tranform in the form given below

$$\mathcal{L}(s)=\text{exp}(-As^{2/\alpha}-Bs^{3/\beta})$$

which is an multiplication of two stretched exponential decay function

where $A,B >0$

I need to find the inverse Laplace transform of this.

$\textit{I know that this cannot be solved symbolically using elementary functions.}$

Can some one suggest and/or refer me to some solution?

Here, $\alpha$ and $\beta$ can take values like $3,4,5...$

Any approximation is also welcome

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Hint: Try to expand the exponential function as a power sum.

$$\exp(-As^{2/\alpha}-Bs^{3/\beta})=1+\frac{-As^{2/\alpha}-Bs^{3/\beta}}{1!}+\frac{(-As^{2/\alpha}-Bs^{3/\beta})^2}{2!}+\cdots$$

You will need the following formula for inverting the powers of s.

$$\mathcal{L}^{-1}\left[ s^\alpha\right]=\frac{t^{-\alpha-1}}{\Gamma(-\alpha)}$$

Where $\Gamma$ is the Gamma function.

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  • $\begingroup$ Thanks for your hint. Frankly speaking, I am not a math person. I just came across this problem and do not know how to solve this. Would you please give me some solution. $\endgroup$ Mar 30 '16 at 8:31
  • $\begingroup$ The solution is an infinite series. You just expand the series and apply the inverse Laplace transform for every individual term. $\endgroup$
    – MrYouMath
    Mar 30 '16 at 8:34
  • $\begingroup$ Should there not be something else added to your final formula...perhaps an $\mathcal{L}^{-1}$? $\endgroup$ Mar 30 '16 at 9:30
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    $\begingroup$ Yeah, you are right :D. Btw how did you do the curly L? $\endgroup$
    – MrYouMath
    Mar 30 '16 at 12:05
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    $\begingroup$ \mathcal{L} is how $\endgroup$
    – Moo
    Mar 30 '16 at 12:14

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