Inverse Laplace Transform of $\frac{1}{\lambda\cosh(\sqrt{as}+\sqrt{bs})+(1-\lambda)\cosh(\sqrt{as}-\sqrt{bs})}$.

Question

I am trying to find the inverse Laplace transform of some function of the form:

$$ \mathrm{F}\left(s\right) = \frac{1}{\lambda\cosh\left(\,\sqrt{\, as\,}\, + \,\sqrt{\, bs\, }\right) + \left(1 - \lambda\right)\cosh\left(\,\sqrt{\, as\,}\, -\,\sqrt{\, bs\,}\right)}, $$

where $\lambda \in \left[0,1\right]$. I've tried to use the residue theorem, however, finding the poles ( roots of the denominator ) seems quite complicated, which I have not figured out, yet.

I've also tried to calculate the integral: $$ \int_{\sigma - \mathrm{i}\infty}^{\sigma + \mathrm{i}\infty} \mathrm{F}\left(s\right)\mathrm{e}^{st}\,\mathrm{d}s, $$ by using some substitution of the form $u = \mathrm{e}^{\,\sqrt{\, s\,}\,}$, which, however, yields another complicated form from which I can proceed further.

Any ideas, hints, partial solutions, are appreciated!

Answer 1

Concerning the zeros for

$$ \lambda\cosh(\sqrt{as}+\sqrt{bs})+(1-\lambda)\cosh(\sqrt{as}-\sqrt{bs}) = 0 $$

making $s = x + i y$ and taking the real and the imaginary parts

$$ R=\lambda \cos \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin \left(\frac{1}{2} \arg (x+i y)\right)\right) \cosh \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \cos \left(\frac{1}{2} \arg (x+i y)\right)\right)-(\lambda -1) \cos \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin \left(\frac{1}{2} \arg (x+i y)\right)\right) \cosh \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2} \cos \left(\frac{1}{2} \arg (x+i y)\right)\right)=0\\ I = \lambda \sin \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin \left(\frac{1}{2} \arg (x+i y)\right)\right) \sinh \left(\left(\sqrt{a}+\sqrt{b}\right) \sqrt[4]{x^2+y^2} \cos \left(\frac{1}{2} \arg (x+i y)\right)\right)-(\lambda -1) \sin \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2} \sin \left(\frac{1}{2} \arg (x+i y)\right)\right) \sinh \left(\left(\sqrt{a}-\sqrt{b}\right) \sqrt[4]{x^2+y^2} \cos \left(\frac{1}{2} \arg (x+i y)\right)\right)=0 $$

and analyzing the root locus for some values of $a,b,\lambda$ we have good information.

For instance with $a = 3, b = 2, \lambda = 0.9$ no zeroes.

with $a = 3, b = 2, \lambda = 0.1$ the zeroes are at the intersections.

with $a = 3, b = 2, \lambda = 0.5$ no zeroes.

So we can visualize the zeroes position depending on the parameters. This can help with the residues method. Follows the MATHEMATICA script to make the plots

re0 = -(-1 + lambda) Cos[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
 Sin[1/2 Arg[x + I y]]] Cosh[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
 Cos[1/2 Arg[x + I y]]] + lambda Cos[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
 Sin[1/2 Arg[x + I y]]] Cosh[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
 Cos[1/2 Arg[x + I y]]]
im0 = -(-1 + lambda) Sin[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
 Sin[1/2 Arg[x + I y]]] Sinh[(Sqrt[a] - Sqrt[b]) (x^2 + y^2)^(1/4)
 Cos[1/2 Arg[x + I y]]] + lambda Sin[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
 Sin[1/2 Arg[x + I y]]] Sinh[(Sqrt[a] + Sqrt[b]) (x^2 + y^2)^(1/4)
 Cos[1/2 Arg[x + I y]]]

a = 3;
b = 2;
lambda = 0.1;
r = 50;
gr1 = ContourPlot[re0 == 0, {x, -r, r}, {y, -r, r}, ContourStyle -> Red, PlotPoints -> 25];
gr2 = ContourPlot[im0 == 0, {x, -r, r}, {y, -r, r}, ContourStyle -> Blue, PlotPoints -> 25];
Show[gr1, gr2]