From my book:

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

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

where $A=V_0^2/(8RhL)$, $V_0$ is the voltage applied to the heater, $R$ the electrical resistance of the heater, $h$ the thickness of the thin film, $\alpha$ the thermal diffusivity of the thin film, and $\omega/2$ the heating frequency. 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 derive how he came to the solution from the boundary conditions because the final solution doesn't contain any Fourier transformation and eigenvalues. There is no derivation in the paper, and I searched books and the internet thoroughly but couldn't find anything. Any help would be appreciated!

  • 1
    $\begingroup$ Please show us what you've tried so far. $\endgroup$ – Gert Mar 8 '19 at 2:24
  • $\begingroup$ @Gert I have tried conventional method of variable separation and fourier transformation. but all my answers contains eigenvalues and a fourier series, which this solution (in pic) does not contain. $\endgroup$ – Betsy Mar 8 '19 at 2:34
  • $\begingroup$ This page (disclaimer: my site) solves a similar problem. I remember working from a great handbook on PDEs that assembled a variety of methods; will try to track it down. $\endgroup$ – Chemomechanics Mar 8 '19 at 2:43
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    $\begingroup$ You should type out relevant parts of text rather than post pictures of the text. I have done it for you this time. $\endgroup$ – BioPhysicist Mar 8 '19 at 3:18
  • $\begingroup$ @Chemomechanics . Thank you! That helps. I too got something similar to the solution in your website and my solutions contain eigenvalues. But the solution in the pic I uploaded is different because it doesn’t contain eigenvalues, but contains a phase term in cos(). I'd really appreciate if you could tell me what that handbook was :) $\endgroup$ – Betsy Mar 8 '19 at 3:22

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: $$T(x,t)-T_0=\alpha(x)\cos(\omega t-\phi)+\beta(x)\sin(\omega t-\phi)+\frac{A}{k}(L-x)$$where $\phi$ is the phase shift between the heat flux time variation and the temperature time variation. They solve for $\alpha(x)$ and $\beta(x)$ subject to the boundary conditions. So, they're basically not forcing themselves to satisfy the initial condition.

Try this form of the solution and see what you get.

  • $\begingroup$ Very interesting, Chester. +1 from me. $\endgroup$ – Gert Mar 8 '19 at 3:03
  • $\begingroup$ Thank you; this is a very good suggestion, let me try this particular solution method! $\endgroup$ – Betsy Mar 8 '19 at 3:25
  • $\begingroup$ 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. $\endgroup$ – Chet Miller Mar 8 '19 at 20:58
  • $\begingroup$ 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? $\endgroup$ – Betsy Mar 8 '19 at 21:08
  • $\begingroup$ Like I said. Please show us the details. Your application of the boundary conditions is definitely incorrect. $\endgroup$ – Chet Miller Mar 8 '19 at 21:14

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