Make InverseLaplaceTransform?

GROUPS:
 Why can't this function with respect to s take InverseLaplaceTransform?
1 year ago
15 Replies
 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:
1 year 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
1 year ago
 I have translated the Maple code into Mathematica.with best regards, Mariusz
1 year ago
 Would you like post the Maple codes directly? Since I can't run the codes in Mathematica. Thanks.
1 year 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?
1 year 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]) 
1 year ago
 Dear Mariusz, I can get the same results as yours. But I can't get the result of eq9, which is the real problem
1 year 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]*) 
1 year ago
 The series expansion to 3rd order is locally approximate, rather than globally.
1 year ago
 Dear Mariusz, The problem we discussed is only one step， 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
1 year 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.
 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$k),s=k/t),k=infinity),d); d:=2*d od; evalf(a,dig) end end: Answer 1 year ago  Well is it a Post's inversion formula and works if you find a k-th derivative of F with respect to s, and find a limit. You can apply it to simple functions.Let's say for example, a function:$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] *)