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I am struggling on solving a PDE group by Laplace transform method. I could not find similar inverse equations to the question I listed.

I would appreciate if you could give me some suggestions.

Han
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  • Welcome. Do you want to find the $L^{-1}\left{e^{\sqrt{\frac{s+1}{s}}}\right}$ or $L^{-1}\left{e^{\sqrt{s+\frac1s}}\right}$? – Sebastiano Jun 27 '20 at 17:48
  • It is s plus (1/s). I am sorry, the first time to post a question and not familiar with the code – Han Jun 27 '20 at 17:57
  • Don't worry :-) – Sebastiano Jun 27 '20 at 17:59
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    Thank you for your reply. – Han Jun 27 '20 at 18:02
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    Are you sure the inverse laplace exists in terms of standard mathematical functions : https://www.wolframalpha.com/input/?i=inverse+laplace+transform+e%5E%28sqrt%28%28s%5E2%2B1%29%2Fs%29%29 – The Dragonborn Jun 27 '20 at 18:04
  • I tried solving this by Wolfram Mathmatica 12.0. No solution is given. But I think this can be calculated by other methods. Because some inverse Laplace transform pairs that exist on the tables, are also can not computed by Wolfram. – Han Jun 27 '20 at 18:10
  • Have you tried solving this directly, using the contour integral? If you can find the poles of your function in the complex plane, you may be able to use residues to calculate the integral. – Alex Jones Jun 27 '20 at 22:11
  • Hi, thank you for your reply. I am thinking about contour integral. But I am not sure about pole of this expression, is infinity? If the pole is infinity, that means no pole in complex plane and the inverse Laplace transform is not exist? – Han Jun 27 '20 at 22:44
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    @Han Yes, it appears that the function blows up at infinity, which is an indication that the contour integral will not converge, so there is likely no inverse Laplace transform. – Alex Jones Jun 27 '20 at 22:52
  • Sorry, it seems I forget to type a negative symbol in the front of square. Then the pole is 0 and the inverse Laplace transform might exist, I think. If it is convenient for you to show me how to calculate contour integrate? – Han Jun 27 '20 at 22:59
  • The closest thing I got is $$\mathcal{L}^{-1} e^{-\sqrt{s}} = \frac{1}{2\sqrt{\pi}x^{3/2}}\exp\left(-\frac{1}{4x}\right) $$ but the extra $\frac{1}{s}$-term inside the square root is troublesome, $0$ and $\infty$ are not poles but ugly singularities for $\exp\left(-\sqrt{s+1/s}\right)$. – Jack D'Aurizio Jun 27 '20 at 23:57
  • Additionally, if $\mathcal{L}^{-1}\exp\left(-\sqrt{s+1/s}\right)$ exists it must have a real zero (and probably infinite real zeroes), since the Laplace transform of a non-negative function is log-convex and $-\sqrt{s+1/s}$ is not convex. – Jack D'Aurizio Jun 28 '20 at 00:00
  • Thank you. It seems no inverse transform. In fact, the original expression is e^{-{z\over 2}}\sqrt{{(-B*(s-A)^2+C)\over s}}}, all letters are constant except s. May be something wrong during calculating process. Anyway, thank you for your reply. It is helpful. – Han Jun 28 '20 at 00:22
  • I have answer only with regularized generalized hypergeometric function in Infinite sum,calculated by CAS. – Mariusz Iwaniuk Jul 15 '20 at 15:16

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