What they have done is focus exclusively on the long-time solution when the system has reached "oscillatory steady state." This solution does not feature any exponentially decaying terms in time.
So their solution for the temperature is taken to be of the form: $$T(x,t)-T_0=\alpha(x)\cos(\om...
Show us more of the details of what you did. I've solve many oscillatory steady state problems in heat transfer and mass transfer using this approach, and it was always successful.
Tried to solve by taking $$T(x,t)−T_0={\alpha(x)}cos(\omega t)+{\beta(x)}sin(\omega t)+\frac Ak(x-L)$$ as a particular solution ${\alpha(x)}$ and ${\beta(x)}$ values was found by first two conditions ${T(x,0)=T_o}$ and ${T(L,t)=T_0}$. Where ${\beta(x)}$ become zero. But the third condition remain unsused and also it doesn't provide the solution mentioned in the question. Can you tell me if you got the same solution as the question?
Like I said. Please show us the details. Your application of the boundary conditions is definitely incorrect.
The differential equations I solved were $$\frac{d^2\alpha}{dx^2}=+\frac{\omega}{\kappa}\beta$$ and $$\frac{d^2\beta}{dx^2}=-\frac{\omega}{\kappa}\alpha$$ Is that what you ended up with?
Yes, I got something similar. After applying the particular solution to the heat equation, I get $${-a(x)\omega\sin(\omega t)+b(x)\omega\cos(\omega t) = \alpha cos(\omega t)\frac{d^2 a(x)}{dx^2}+\alpha sin(\omega t)\frac{d^2 b(x)}{dx^2}}$$ Separating the coefficient finally end up $${\alpha\frac{d^2 a(x)}{dx^2}=\omega b(x)}$$ and $${\alpha\frac{d^2 b(x)}{dx^2}=-\omega a(x)}$$ But how do I proceed from here?
Next, you eliminate b from the equations to obtain $$\frac{d^4a}{dx^4}+\left(\frac{\omega}{\alpha}\right)^2a=0$$Do you know how to solve this differential equation for the complementary solution for a (the particular solution is zero)?
Thank you for that help. I got the following equation after solving the differential equation. This equation contains 4 terms but looking at the final solution, seems like we can exclude 3 terms from the following equation. How it can be done systematically? And how can we evaluate the boundary conditions for a(x)? $$a(x) = c_1 e^{(-1)^{(\frac 34)} \sqrt{\frac {\omega}{\alpha}} x} + c_2 e^{-(-1)^{\frac 14} \sqrt{\frac {\omega}{\alpha}} x}$$ $$+ c_3 e^{-(-1)^{\frac 34} \sqrt{\frac {\omega}{\alpha}} x} + c_4 e^{(-1)^{\frac 14} \sqrt{\frac {\omega}{\alpha}} x}$$
Do you know how to evaluate the real and imaginary parts of -1 to the 1/4 power? I should also mention that the authors of your article make the unstated assumption that the ratio $\omega/\alpha$ is very large so that the length of the sample L does not come into play in the oscillatory part of the solution; the oscillatory part of the solution is damped out spatially toward the end close to x << L. We know this because L does not appear in their oscillatory terms.
Your complementary solution should feature terms like $e^{-y}\cosy$ and $e^{-y}\sin{y}$ with $y=\sqrt{\frac{\omega}{2\alpha}}x$ It's impossible to show equations in this chat. Have you ever visited PhysicsForums.com?
Discussion between Betsy and Chester …
Discussion between Betsy and Chester …