How can I find the following one dimensional heat conduction solution?

$$\frac{\partial T}{\partial t}=\alpha\frac{\partial^2 T}{\partial^2 t}$$

with an initial condition and boundary conditions $$T(x,0)=T_0$$ $$T(L,t)=T_0$$ $$-k\left.\frac{\partial T}{\partial x}\right|_{x=0}=2A\cos^2\left(\frac{\omega t}{2}\right)=A\cos(\omega t+1)$$

The solution for this problem is

$$T(x,t)-T_0=\frac Ak\sqrt{\frac\alpha\omega}\exp\left (-\sqrt{\frac{\omega}{2\alpha}}x\right)\\ \times\cos\left(\omega t-\sqrt{\frac{\omega}{2\alpha}}x-\frac\pi4\right)-\frac Ak(x-L)$$

I'm trying to solve Fourier heat conduction equation with the boundary conditions shown above.But the solution I'm getting contains eigenvalues and Fourier series which the solution doesn't contain. More interestingly the solution contains a phase terms. I got this from a paper but the paper doesn't contain any derivation. I searched in the internet but couldn't find how they solved it. I would appreciate any help!

• At the first glance, it does not look like the solution satisfies $T(x,0)=T_0$. Also, exponent usually contains $t$, Did you copy it correctly from the paper? – Vasya Mar 8 at 22:09
• It's correct! I don't have any clue how they solved it without any exp(t) term or satisfying the initial condition. In the paper they mentioned that initial condition is reasonably satisfied. Here's the link of the paper. doi.org/10.1063/1.1145989 – Batsy Mar 8 at 22:21
• Is the equation supposed to be $T_t = \alpha T_{xx}$? – Dylan Mar 10 at 8:26