A simplification of a complex expression

Question

I use mathematica to confirm a result from a paper ("Unsteady unidirectional flow of Bingham fluid between parallel plates with different given volume flow rate conditions") which should be simplified to a real expression. However, I have tried many many times but all failed. The following code showed my efforts:

ClearAll["Global`*"]
Γ[s_] := Module[{m, delta, ee},
m = Sqrt[s]/Sqrt[ν];

delta = Sinh[m h] Sinh[m h0] - Cosh[m h] Cosh[m h0];

ee = Cosh[m h0] (Sinh[m h] - Sinh[m h0]) - 
Sinh[m h0] (Cosh[m h] - Cosh[m h0]);

(m h  delta)/(
m h delta + m h0 (Cosh[m h0]^2 - Sinh[m h0]^2) + ee) ];

Simplify[Im[
E^( I ω t) Γ[I ω] + 
E^(- I ω t) Γ[-I ω] // ComplexExpand],
TimeConstraint -> Infinity]

Actually, the complex expression in the above code is the first term of dp/dx as shown in the following picture, and dp/dx is a pressure gradient which should be a expression of real values. So the image part of this expression should vanish, but the above code gives no result after a hour.

Thanks.

Edit 1: -----------------------------------

As suggested by @Bob Hanlon, I tried to include all the variable constraints in Simplify, but still failed to get a real-valued expression.

expr = E^(I \[Omega] t) \[CapitalGamma][I \[Omega]] + E^(- I \[Omega] t) \[CapitalGamma][-I \[Omega]]

Simplify[ComplexExpand[expr], {h > h0 > 0,t > 0, \[Omega] > 0, \[Nu] > 0}]

Answer 1

The key idea of this solution is to partly simplify the expression by direct substitution of some known terms rather than using functions for simplification like Simplify.

First, take out the imaginary part of the expression and observe:

impart = E^(I ω t) Γ[I ω] + E^(-I ω t) Γ[-I ω] // Im // ComplexExpand

Hmm… seems that there're many Arg terms in impart. It's natural to guess these terms will slow down the simplification process of Simplify. Let's check what kind of Arg term is contained in impart further:

cons = {h > h0 > 0, t > 0, ω > 0, ν > 0};
Cases[impart, Arg[_], Infinity] // Union
(* {Arg[ν], Arg[-I ω], Arg[I ω]} *)
imrule = # -> Simplify[#, cons] &@% // Thread
(* {Arg[ν] -> 0, Arg[-I ω] -> -(π/2), Arg[I ω] -> π/2} *)

There're 3 different Arg terms, whose values are known under the constraints. Plug them back to impart:

immid = impart /. imrule;

and simplify:

Simplify[immid, cons]
(* 0 *)

OK, the minimum goal is achieved, and with the same method, the final goal i.e. find the free-of-I form of the expression can be achieved, too:

repart = E^(I ω t) Γ[I ω] + E^(-I ω t) Γ[-I ω] // Re // ComplexExpand;

Cases[repart, Arg[_], Infinity] // Union
(* {Arg[ν], Arg[-I ω], Arg[I ω]} *)
rerule = # -> Simplify[#, cons] &@% // Thread
(* {Arg[ν] -> 0, Arg[-I ω] -> -(π/2), Arg[I ω] -> π/2} *)

remid = repart /. rerule;

Simplify[remid, cons] // AbsoluteTiming
(* {16.001111, …… *)

The result is a little long so I'd like to omit it here.