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Make InverseLaplaceTransform?

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Why can't this function with respect to s take InverseLaplaceTransform? enter image description here

POSTED BY: Zhonghui Ou
Answer
4 months ago

It's seems Mathematica can't find Inverse Laplace Transform.It is therefore best to use some mathematical programs like Maple or Matlab.

Maple only find symbolic solution in special caseses if $a>0$

 InverseLaplaceTransform[(Sqrt[s]*Sqrt[a + s])/(a + s), s, t] == a/2*(-BesselI[0,a*t/2] + BesselI[1, a*t/2])*Exp[-a*t/2]+DiracDelta[t]
 InverseLaplaceTransform[(Sqrt[s]*Sqrt[2 a + s])/(a + s), s, t] == 1/2 a E^(-a t) (BesselI[1, a t] (2 + a \[Pi] t StruveL[0, a t]) - 
 a t BesselI[0, a t] (2 + \[Pi] StruveL[1, a t]))

and for: $a<0$

  InverseLaplaceTransform[(Sqrt[s]*Sqrt[a + s])/(a + s), s, t] == a/2*(BesselI[0, a*t/2] + BesselI[1,a*t/2])*Exp[a*t/2]+DiracDelta[t]
  InverseLaplaceTransform[(Sqrt[s]*Sqrt[2 a + s])/(a + s), s, t] == 1/2 a E^(a t) (BesselI[1, a t] (2 + a \[Pi] t StruveL[0, a t]) - 
  a t BesselI[0, a t] (2 + \[Pi] StruveL[1, a t]))

If you want to solve numericall see notebook.

Attachments:
POSTED BY: Mariusz Iwaniuk
Answer
4 months ago

Dear Mariusz,
Thanks for you reply. I think the function, InverseLaplaceTransform is from Mathematica, invlaplace from Malple, and ilaplace from MATLAB. The codes you developed was made by Mathematica or by Maple?

with best regards, J

POSTED BY: Zhonghui Ou
Answer
4 months ago

I have translated the Maple code into Mathematica.

with best regards, Mariusz

POSTED BY: Mariusz Iwaniuk
Answer
4 months ago

Would you like post the Maple codes directly? Since I can't run the codes in Mathematica. Thanks.

POSTED BY: Zhonghui Ou
Answer
4 months ago

Why can't you run code in Mathematica? Your original equation was formulated in terms of Mathematica code, so you do have the software, right?

POSTED BY: Kapio Letto
Answer
4 months ago

Maple 2017.1 code:

restart:
with(inttrans):
invlaplace(sqrt(s)*sqrt(a+s)/(a+s), s, t)
invlaplace(sqrt(s)*sqrt(-a+s)/(-a+s), s, t)
`assuming`([invlaplace(sqrt(s)*sqrt(2*a+s)/(a+s), s, t)], [a > 0])
`assuming`([invlaplace(sqrt(s)*sqrt(-2*a+s)/(-a+s), s, t)], [a > 0])

enter image description here

POSTED BY: Mariusz Iwaniuk
Answer
4 months ago

Dear Mariusz, I can get the same results as yours. But I can't get the result of eq9, which is the real problementer image description here

enter image description here

POSTED BY: Zhonghui Ou
Answer
4 months ago

Maple can't find Inverse Laplace Transform symbolic solution.You may try a numeric solution or approximation with series.

 n = 3; (*for n = 50 numeric approximation is very good*)
 InverseLaplaceTransform[Series[(Sqrt[s] Sqrt[c3 + s])/(c4 + s), {s, Infinity, n}] // Normal, s, t]

 (* 1/2 (c3 - 2 c4) + 1/8 (-c3^2 - 4 c3 c4 + 8 c4^2) t + 
  1/32 (c3^3 + 2 c3^2 c4 + 8 c3 c4^2 - 16 c4^3) t^2 + 
  1/768 (-5 c3^4 - 8 c3^3 c4 - 16 c3^2 c4^2 - 64 c3 c4^3 + 
     128 c4^4) t^3 + DiracDelta[t]*)
POSTED BY: Mariusz Iwaniuk
Answer
4 months ago

The series expansion to 3rd order is locally approximate, rather than globally.

POSTED BY: Zhonghui Ou
Answer
4 months ago

Dear Mariusz, The problem we discussed is only one step´╝î and it's not the final expression. I still need this symbolic solution to get the final expression. Thanks for your help.

with best regards, Ja

POSTED BY: Zhonghui Ou
Answer
4 months ago

With Mrs Leucippus help: $$\mathcal{L}_s^{-1}\left[\frac{\sqrt{s (s+a)}}{s+b}\right](t)=\delta (t)+\left(\frac{a}{2}-b\right) \exp (-b t)-\frac{1}{2} \exp (-b t) (a-2 b) \left(1-2 \sqrt{-\frac{(a-b) b}{(a-2 b)^2}}\right)+\exp (-b t) \sum _{k=1}^{\infty } \frac{4^{-k} a^{2 k} (a-2 b)^{1-2 k} \Gamma \left(-1+2 k,\frac{1}{2} (a-2 b) t\right)}{\Gamma (k+1) \Gamma (k)}$$

 DiracDelta[t] + (a/2 - b)*Exp[-b*t] - Exp[-b*t]*1/2 (a - 2 b) (1 - 2 Sqrt[-(((a - b) b)/(a - 2 b)^2)]) + 
Exp[-b*t]*Sum[(4^-k a^(2 k) (a - 2 b)^(1 - 2 k) * Gamma[-1 + 2 k, 1/2 (a - 2 b) t])/(Gamma[k + 1] Gamma[k]), {k, 1, Infinity}]

but I could not find closed form of sum.

POSTED BY: Mariusz Iwaniuk
Answer
4 months ago

Hello, Mariusz. Mrs. Leucippus and you calculated it by hand? Would you like to provide the deduction or the codes? Thanks.

with best regards, J

POSTED BY: Zhonghui Ou
Answer
4 months ago

Hello,

See this link

with best regards, Marusz.

POSTED BY: Mariusz Iwaniuk
Answer
4 months ago

I found the introduction to Numerical Inverse Laplace Transform in http://www.mapleprimes.com/posts/42361-Numerical-Inverse-Laplace-Transform. But I couldn't understand the program.

ILT:=proc(f,s) local k,dig; if nargs>2 and type(args[-1],'posint') then dig:=args[-1] else dig:=Digits fi; proc(T) local t,d,a; d:=dig; a:=Limit(); t:=evalf(T,dig); while op(0,a)='Limit' do a:=evalf(Limit((-1)^k/k!(k/t)^(k+1)eval(diff(f,s$k),s=k/t),k=infinity),d); d:=2*d od; evalf(a,dig) end end:

POSTED BY: Updating Name
Answer
4 months ago

Well is it a Post's inversion formula and works if you find a k-th derivative of F with respect to s, and find a limit. You can apply it to simple functions.

Let's say for example, a function: $F(s)=\frac{1}{s^2+1}$

F[s_] := 1/(s^2 + 1)
diff = Assuming[{k > 1, s > 0}, D[F[s], {s, k}]][[1, 1, 1]](*finding k-th derivative*)
(*-(1/2) I ((-I - s)^(-1 - k) - (I - s)^(-1 - k)) k!*)
Limit[FullSimplify[((-1)^k/k!*(k/t)^(k + 1)*(diff /. s -> (k/t))), Assumptions -> {t > 0, k > 0, k \[Element] Integers}], 
k -> Infinity] // FullSimplify
(* Sin[t] *)
POSTED BY: Mariusz Iwaniuk
Answer
4 months ago

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