I am stuck at the beginning of an excercise of PDE. The question is the following:

A thin bar with length L, so that x=0 --> x=L. The bar is totally insulated and has a temperature of $100°C$. The temperature of the environment is $0°C$. At t=0 the insulation at the point $x=L$ is taken away. At this moment, the heat exchange with the environment is happening according to the following equation:

$\frac{\delta u}{\delta x}(L,t)=-hu(L,t)$

Now I started out with the 1 dimensional heat transfer equation. And I wrote the solution as:


$T'(t)=a^2 \lambda T(t)$

$X''(x)=\lambda X(x)$

I sort of tried to find good boundary conditions to spot the Sturm-Liouville problem and start to solve this.

$u'(0,t) = 0$

$u(L,0) = 100$

$u(0,0) = 100$

Now I am a little bit stuck, and it has something to do with the law of heat transfer that is given at the beginning.

I want to solve it myself, but could someone give me some starting hints, or a beginning where I can build on?

Thank you very much



Tangens and 1/x

  • $\begingroup$ Sorry, I didn't read the part about hints for the beginning of the problem. Just read the setup and let others learn from the rest. $\endgroup$
    – Ron Gordon
    Jan 17, 2015 at 13:47
  • $\begingroup$ No problem. I am gonna try it with the beginning of your exercise. I think it is pretty straight forward now! $\endgroup$
    – Thomas
    Jan 17, 2015 at 13:54

1 Answer 1


I think you want the heat equation with the following BC's:
$$u_{xx} = a u_t$$

$$u_x(0,t) = 0$$ $$u_x(L,t)+h u(L,t) = 0$$ $$u(x,0) = u_0=100$$

Separate variables as $u = X(x)T(t)$ to get

$$X'' +\lambda^2 X = 0$$ $$a T' + \lambda^2 T = 0$$

where $-\lambda^2$ is the separation constant. The $x$ equation yields

$$X(x) = A \cos{\lambda x} + B \sin{\lambda x} $$ $$X'(x) = -A \lambda \sin{\lambda x} + B \lambda \cos{\lambda x}$$ $$X'(0) = 0 \implies B = 0$$ $$X'(L) + h X(L) = 0 \implies \lambda \tan{\lambda L} = h $$

To this last equation, there will be an infinite sequence of solutions $\lambda = \lambda_n$. Thus, there will be an infinite sequence of solutions $X_n(x) = A_n \cos{\lambda_n x} $, with $A_n$ yet to be determined.

The $t$ equation yields

$$T_n(t) = C_n e^{-\lambda_n^2 t/a}$$

with $C_n$ yet to be determined. Thus, the full solution is a linear combination of these solutions, i.e., a sum over $n$:

$$u(x,t) = \sum_{k=1}^{\infty} D_n e^{-\lambda_n^2 t/a} \cos{\lambda_n x} $$

where $D_n$ is given by

$$\frac{\displaystyle \int_0^L dx\, u(x,0) \cos{\lambda_n x}}{\displaystyle \int_0^L dx\, \cos^2{\lambda_n x}} = \frac{2 u_0 \sin{\lambda_n L}}{\lambda_n L + \sin{\lambda_n L} \cos{\lambda_n L}}$$


Here is a plot of the solution for $t \in \{0,0.01,0.1,0.2,0.5,1.0\}$ for $a=1$, $L=1$, and $h=1$. The plot was generated in Mathematica. I used $100$ terms in the sum.

enter image description here

  • $\begingroup$ Ok thnks, I am gonna try it in a minute! (those exercises are pretty long :D) $\endgroup$
    – Thomas
    Jan 17, 2015 at 13:53
  • $\begingroup$ BTW I made an initial mistake in specifying the $D_n$'s. I have edited the solution to reflect the correct values. $\endgroup$
    – Ron Gordon
    Jan 17, 2015 at 15:30
  • $\begingroup$ Can you explain me why the equation with the tangens has an infinit amount of solutions ? $\endgroup$
    – Thomas
    Jan 17, 2015 at 16:02
  • $\begingroup$ @Thomas: Draw plots of $y=\tan{x}$ and $y=1/x$, and note that there are an infinite number of intersections. $\endgroup$
    – Ron Gordon
    Jan 17, 2015 at 16:05
  • $\begingroup$ Ok, i plotted it, I see it now. (see question) $\endgroup$
    – Thomas
    Jan 17, 2015 at 16:24

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