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I just want to see if I completed this problem right. Here is the problem:

Consider $\frac{\partial T}{\partial t} = k \frac{\partial^{2} T}{\partial x^2} -\alpha T$ where $k,\alpha >0$ are constants and $\partial_x(0,t) = \partial_x (L,t) = 0$ and $T(x,0) = f(x)$. Find the equilibrium temperature and T(x,t). Find the long time asymptotic limit of T and compare to the equilibrium temperature.

Here is my attempt at the problem:

Assume the solution is in the form $T(x,t) = F(x)G(t)$. Then, By Separation of Variables, I got

$(\lambda +\alpha) G \frac{\partial G}{\partial t} = \frac{k}{F} \frac{\partial^{2} T}{\partial x^2}=\lambda $, which lambda is the separation constant.I have examined the cases for $\lambda$, the cases for $\lambda = 0$, and $\lambda >0$, they would yield the trivial solution for F(x). For $\lambda <0$, I would have $F(x) = A \cos \sqrt{\frac{\lambda}{k}} +B \sin \sqrt{\frac{\lambda}{k}}$ which means

$F'(X) = \sqrt{\frac{\lambda}{k}} (-A \sin (\sqrt{\frac{\lambda}{k}}) +B\cos (\sqrt{\frac{\lambda}{k}}))$.

When I initialize the initial conditions, I got the general solution to be a sequence of functions, which is $F_n(x) = A_n \sin (\frac{n \pi x}{L}) $. For G(t), I got the function to be $G(t) = C e ^{-(\lambda -\alpha)t} $.

Since $\lambda = (\frac{n \pi x}{L})^2 k$. $G(t)$ is also generalized as a sequence of functions which is

$G_n(t) = C_n e^{((\frac{n \pi x}{L})^2 k - \alpha)t} $

Thus the general solution would be a linear sum of sequence of functions:

$T(x,t) = \sum_{n=1}^{\infty} a_n \sin (\frac{n \pi x}{L})e^{((\frac{n \pi x}{L})^2 k - \alpha)t}$

From the initial condition, $T(x,0) = F(x)$, this would turn out to be

$T(x,0) = \sum_{n=1}^{\infty} a_n \sin (\frac{n \pi x}{L})=F(x) $ which is a Fourier Sine Series representation.

Am I on the right track on this? Also, for the equilibrium temperature, this is where we set $ \frac{\partial T}{\partial t} = 0$ and solve. What I got for the equilibrium temperature function was

$F(x) = A e^\sqrt{\frac{\alpha}{k}x} +B e^{-\sqrt{\frac{\alpha}{k}x}}$. From the boundary conditions, I got the solution was the trivial solution. Did I do this correctly?

Thank you for all of your help.

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Your separation of variables looks wrong to me, or at the very least a wierd typo, since your derivation after that does not seem to flow from the expression you have given.

I get $$ \dfrac{\dot{G}}{G} = k\dfrac{F''}{F} - \alpha = \lambda $$

OP what do you think?

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  • $\begingroup$ When I substituted in $T(x,t) = F(x)G(t)$ in the PDE, I have $F \frac{\partial G}{\partial t} = kG \frac{\partial^{2} F}{\partial x^2} - \alpha F G$. I divided both sides by $G$ to get $\frac{F}{G} \frac{\partial G}{\partial t} = k \frac{\partial^{2} F}{\partial x^2} - \alpha F$. When I separated the variables, I should get $\frac{1}{G} \frac{\partial G}{\partial t} = \frac{k}{F} \frac{\partial^{2} F}{\partial x^2}- \alpha $ $\endgroup$ – user179766 Oct 1 '14 at 15:46
  • $\begingroup$ You divide by $FG$ that's the benefit (and requirement) of separation of variables is making use that things are linear so we can separate the equation in to the two variables. $\endgroup$ – Chinny84 Oct 1 '14 at 15:49
  • $\begingroup$ I think I see (and hopefully you now) where you went wrong :). Good luck. $\endgroup$ – Chinny84 Oct 1 '14 at 15:49
  • $\begingroup$ How would you find the equilibrium temperature though? $\endgroup$ – user179766 Oct 2 '14 at 0:09
  • $\begingroup$ You have already stayed how you would do that..:) $\endgroup$ – Chinny84 Oct 2 '14 at 7:18

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