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 $F(x) = A \cos \sqrt{\frac{\lambda}{k} }x +B \sin \sqrt{\frac{\lambda}{k} }x$ which means

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

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?

Also, for the coefficients, $a_n$, after you are done solving them, do you have to substitute it in into $T(x,t)$?

Thank you for all of your help.


2 Answers 2


Given the pde \begin{align} \partial_{t} u(x,t) = k \partial_{x}^{2} u(x,t) - \alpha u(x,t), \end{align} with the conditions $\partial_{x}u(0,t)= \partial_{x}u(L,t) = 0$ and $u(x,0) = f(x)$, then by separation of variables let $u = F(t) G(x)$ to obtain \begin{align} \frac{F'}{F} = - \lambda^{2} = k \, \frac{G''}{G} - \alpha. \end{align} This yields the two equations \begin{align} F' + \lambda^{2} F &= 0 \\ G'' + \left( \frac{\lambda^{2} - \alpha}{k} \right) G &= 0. \end{align} The first equation provides the solution \begin{align} F(t) = e^{- \lambda^{2} t}. \end{align} The equation for $G(x)$ is seen as follows. For the case that $\lambda^{2} = \alpha$ then $G''=0$ for which $G(x) = ax +b$. For the case $\lambda^{2} \neq \alpha$ then \begin{align} G(x) = A \cos\left(\sqrt{\frac{\lambda^{2}-\alpha}{k}} \, x \right) + B \sin\left(\sqrt{\frac{\lambda^{2}-\alpha}{k}} \, x \right) \end{align} and \begin{align} G'(x) = - A \sqrt{\frac{\lambda^{2}-\alpha}{k}} \, \sin\left(\sqrt{\frac{\lambda^{2}-\alpha}{k}} \, x \right) + B \sqrt{\frac{\lambda^{2}-\alpha}{k}} \, \cos\left(\sqrt{\frac{\lambda^{2}-\alpha}{k}} \, x \right). \end{align}

Now, given the boundary conditions that $G'(0) = G'(L) = 0$ and $u(x,0) = f(x)$ then for the first case of $G$ then $G'(x) = a$ and $G'(0) = a = 0$. For the second case $G'(0) = 0 \rightarrow B = 0$ and $G'(L) = 0$ leads to \begin{align} \sin\left(\sqrt{\frac{\lambda^{2}-\alpha}{k}} \, L \right) = 0 \rightarrow \sqrt{\frac{\lambda^{2}-\alpha}{k}} \, L = n \pi \end{align} for $n \geq 1$. This leads to \begin{align} \lambda^{2} = \alpha + k \left( \frac{n \pi}{L} \right)^{2}. \end{align}

Combining the solutions it is seen that \begin{align} u(x,t) = A e^{-\alpha t} + \sum_{n=1}^{\infty} B_{n} \sin\left( \frac{n\pi x}{L} \right) \, e^{- \alpha t - k \left( \frac{n \pi}{L}\right)^{2} t}. \end{align} For $u(x,0) = f(x)$ then \begin{align} f(x) = A + \sum_{n=1}^{\infty} B_{n} \sin\left( \frac{n\pi x}{L} \right). \end{align} From Fourier series the coefficients are seen as \begin{align} A &= \frac{1}{L} \, \int_{0}^{L} f(x) \, dx \\ B_{n} &= \frac{2}{L} \, \int_{0}^{L} f(x) \, \sin\left( \frac{n \pi x}{L} \right) \, dx. \end{align}


When I initialize the initial conditions,

Too early to look at initial condition. After you separate the variables, you deal with boundary conditions first. Which brings me to ...

$F_n(x) = A_n \sin (\frac{n \pi x}{L})$

This is incorrect; you need cosines in order to have zero slope at the endpoints. Sines have zero value but nonzero slope.

Also, a typo here:

$G(t) = C e ^{-(\lambda -\alpha)t}$

Sanity check: the coefficients $\alpha$ should contribute to decay, not growth. I think you want $G(t) = C e ^{-(\lambda +\alpha)t}$. Actually, you don't need $C$, since you are multiplying by indefinite coefficient $A_n$ anyway.

At the very last step, you deal with initial condition: plug $t=0$ into the series you've got, and say what $A_n$ need to be in order for this thing to be $f$. Of course, you can't say much if $f$ is not given: just write down the relation to cosine Fourier coefficients of $f$.

  • $\begingroup$ Okay, I see what I did wrong. Now what I said for the relation of $A_n$ was that $A_n = \frac{2}{L} \int_0^L f(x) cos(\frac{n \pi x}{L}) dx$ and for $A_0= \frac{1}{L} \int_0^L f(x) dx$. Is this it? $\endgroup$
    – user179766
    Oct 2, 2014 at 4:26
  • $\begingroup$ @SSivetts Yes, that looks right $\endgroup$
    – user147263
    Oct 2, 2014 at 4:38

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