Why can’t this function with respect to s take InverseLaplaceTransform?
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.
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
I have translated the Maple code into Mathematica.
with best regards, Mariusz
Would you like post the Maple codes directly? Since I can’t run the codes in Mathematica.
Thanks.
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?
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]*)
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
The series expansion to 3rd order is locally approximate, rather than globally.
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.
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
I found the introduction to Numerical Inverse Laplace Transform in Numerical Inverse Laplace Transform - MaplePrimes.
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:=2d od;
evalf(a,dig) end end:
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] *)