I'm an undergraduat EE student and I'm analysing a scientific paper regarding the theory of the photoacoustic effect. Although the paper is pure physics, my problem is more of a mathematical nature. The subject of investigation is a so called photoacoustic cell, in which a higly light-absorbing solid material is placed on a backing material and the rest of the cell is filled with gas (see picture below). Only the solid material is capable of absorbing light and converting it into heat. The cell is illuminated by sinosoidal chopped monochromatic light flux. The goal is to find the settled temperature distribution inside the cell. Therefore, three heat equations have been set up in picture pg65_2, two homogenous ones for the backing material and the gas and a non homogeneous heat equation for the solid, because heat is being created inside of it because of light absorption. The general solutions to these three equations are in picture pg66_1, not taking into account the boundary conditions of temperature and heat flux continuity at the material borders, although it is assumed that the temperature $$\phi(x,t)$$ at each end of the cell is zero.
The full article is accessible on: http://www.edcc.com.cn/en/upload/newsfile/202008/1596532988i77e2ebw.pdf Of relavance are pages 65 and 66.
I understand that in steady-state conditions the solution consists of a dc "static-like" solution and an ac alternating solution component, because the forcing function is a harmonic one. If I have not been mistaken, the authors solved the three heat equations in the article for the dc-steady state separately, where they tried to find such a temperature function $$\phi(x, t),$$ where $$\frac{\partial^2 \phi}{\partial x^2}$$ would be equal to zero for the backing material/gas and equal to $$-A e^{\beta x}$$ for the solid sample.
I have tried to solve the heat equation for the gas, which is a homogenous PDE $$\frac{\partial ^2\phi}{\partial x^2}=\frac{1}{\alpha_g}\frac{\partial \phi}{\partial t},$$ using variable separation. For the time dependant function I get $$F(t) = Ae^{-\alpha_g^{2}k^{2}t}$$ and for the x dependant function I get $$G(x) = B\cos{kx}+C\sin{kx},$$ where k is some negative constant I have used during the variable separation part. I have set the boundary conditions such that at $$x=l$$ the temperature $$\phi(t, l) = 0$$ and at $$x=0$$ it has some arbitrary value $$\phi(t, 0) = T,$$ depending on the temperature of the solid sample. Putting those boundary conditions into the function F(t) and G(x), I get: $$F(t)G(0) = F(t)B = T$$ and $$F(t)G(l) = F(t)(B\cos{kl}+C\sin{kl}) = 0.$$ When substituting $$T/F(t)$$ instead of B into the latter equation, I get a transcedental function, because the F(t) is an exponential function. I suppose the component with the exponential function is the component the authors have omitted in the solution, as can be read in the picture pg66_1. So that would mean, that I'll have to take into account only the cosine component and use it to calculate the constant k, so $$T\cos(kl) = 0.$$ However, then I get $$T\cos{\frac{\pi nx}{2l}},$$ which is nowhere near the solution the authors would get: $$Te^{\sigma_gx + j\omega t},$$ where $$\sigma_g = (1+j)a_g$$ and $$a_g = (\omega/(2\alpha_g))^{1/2}.$$
I got a bit confused and started wondering whether at least the two homogenous heat equations could be solved using variable enter image description hereseparation. I would be very pleased and thankful for any intuitive tips on how to deal with equations like that, or at least understand why it is being solved inenter image description here that way.
Edit1: I apologize to those who have stumbled upon this question earlier. My LaTeX format was wrong, so I had to change it a bit.
