# I Solution of the 1D heat equation

#### Betsy

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

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 got this from a paper. I'm trying to derive how the author came to the solution from the boundary conditions. There is no derivation in the paper, and I searched books and the internet thoroughly but couldn't find anything. I attached the solution I'm getting. But I don't think it's correct! Any help regarding where I'm doing wrong will be greatly appreciated.

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#### Orodruin

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That solution does not satisfy the boundary condition at x=L.

#### Betsy

That solution does not satisfy the boundary condition at x=L.
But it was published in a good journal by a very renowned professor (now). They must have some logic behind it which is unfortunately not mentioned in the paper.

#### berkeman

Mentor
But it was published in a good journal by a very renowned professor (now).
Can you provide a link to the paper in the journal? Your PDF attachment does not include the beginning of the paper with that information. Thanks.

#### fluidistic

Gold Member
From the sentence below the solution, it is said that the B.C. is reasonably well satisfied because A approaches 0 for the experimental parameters chosen.
And indeed if A equals 0, you do get the B.C. satisfied.

#### Chestermiller

Mentor
Like I told you in the other forum that you and I have been interacting in, the authors of this article implicitly assume that the oscillatory part of the solution damps out before it reaches x = L. The condition for this to happen is $\sqrt{\frac{\omega}{2\alpha}}L>>0$.

In your analysis, Eqn. 17 is incorrect. It should read (based on the above assumption), $$a(x)=C_1\exp{\left(-\sqrt{\frac{\omega}{2\alpha}}x\right)}\cos{\left(\sqrt{\frac{\omega}{2\alpha}}x\right)}+C_2\exp{\left(-\sqrt{\frac{\omega}{2\alpha}}x\right)}\sin{\left(\sqrt{\frac{\omega}{2\alpha}}x\right)}$$
You are missing factors of 2 in the denominators of the square root terms in your expression.

To obtain b(x), you substitute this corrected form of Eqn. 17 into your Eqn. 14. What do you obtain?

BvU

#### Chestermiller

Mentor
Maybe it would help also if I told you that the four roots of of $x^4=-1$ are
$$x=\frac{1+i}{\sqrt{2}}$$
$$x=\frac{-1+i}{\sqrt{2}}$$
$$x=\frac{-1-i}{\sqrt{2}}$$
and $$x=\frac{1-i}{\sqrt{2}}$$
This is where the $\sqrt{2}$ comes from in the correct form of Eqn. 17.

#### Chestermiller

Mentor
In a previous post, I indicated that the solution for a(x) is $$a(x)=C_1\exp{\left(-\sqrt{\frac{\omega}{2\alpha}}x\right)}\cos{\left(\sqrt{\frac{\omega}{2\alpha}}x\right)}+C_2\exp{\left(-\sqrt{\frac{\omega}{2\alpha}}x\right)}\sin{\left(\sqrt{\frac{\omega}{2\alpha}}x\right)}$$
If I substitute this into your Eqn. 14, I obtain: $$b(x)=C_1\exp{\left(-\sqrt{\frac{\omega}{2\alpha}}x\right)}\sin{\left(\sqrt{\frac{\omega}{2\alpha}}x\right)}-C_2\exp{\left(-\sqrt{\frac{\omega}{2\alpha}}x\right)}\cos{\left(\sqrt{\frac{\omega}{2\alpha}}x\right)}$$But, applying the boundary condition on b(x) at x = 0, we have that $C_2=0$. This means that $$a(x)=C_1\exp{\left(-\sqrt{\frac{\omega}{2\alpha}}x\right)}\cos{\left(\sqrt{\frac{\omega}{2\alpha}}x\right)}$$ and $$b(x)=C_1\exp{\left(-\sqrt{\frac{\omega}{2\alpha}}x\right)}\sin{\left(\sqrt{\frac{\omega}{2\alpha}}x\right)}$$
In your Eqn. 11, you have the wrong sign on A/k. You gotta do the math correctly.

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