Short version: Below, I present a simple PDE with simple boundary conditions (BCs), which has a simple solution. I then modify one of the BCs, and end up with a transcendental equation for the wave number parameter, preventing me from going further. This surprises me, since the resulting problem doesn't seem too complicated. My main question: Am I right to conclude that an explicit solution is impossible to find in this case?

Long version: Consider the equation (EQ1) (Note: I'm starting from the heat conduction equation here but omitting some constants for clarity, consider all variables and constants dimensionless) $$ \theta_{xx} = \theta_t, \qquad 0 < x < a,$$ where $\theta = \theta(x,t)$ and the subscripts denote partial derivatives. The boundary conditions are (BC1 and BC2, resp.) $$ \theta_x = 0, \qquad x = 0, $$ $$ \theta_x = 0, \qquad x = a. $$ The initial condition IC is $$ \theta = \theta_0(x), \qquad t = 0. $$

The solution is obtained by a straightforward application of separation of variables: $\theta(x,t) = X(x)T(t) \Rightarrow$ $$ \frac{X''(x)}{X(x)} = \frac{T'(t)}{T(t)} = -\lambda^2 \Rightarrow $$ $$ T(t) = C_1\exp({-\lambda^2t}), $$ $$ X(t) = C_2\sin(\lambda x) + C_3\cos(\lambda x). $$ Now, BC1 $\Rightarrow C2 = 0$ and BC2 $\Rightarrow$ $$ -C_3\lambda\sin(\lambda a) = 0 \Rightarrow \sin(\lambda a) = 0 \Rightarrow $$ $$ \lambda_n = \frac{n\pi}{a}. $$ It follows that $$ X_n(x) = C_3\cos(\frac{n\pi x}{a}) $$ and finally $$ \theta(x,t) = \sum_{n=0}^{\infty}c_n\cos(\frac{n\pi x}{a})\exp({-\frac{n^2\pi^2 t}{a^2}}). $$ Now, the coefficients $c_n$ can be obtained by applying the initial condition and using the trigonometric orthogonality property. The latter requires knowledge of the spatial period of the solution.

Now, consider replacing BC2 with $$ \theta_x = \theta, \qquad x = a. $$ Then, instead of the $\lambda_n$ given above, we would obtain $$ -C_3\lambda\sin(\lambda a) = C_3\cos(\lambda a) \Rightarrow $$ $$ -\lambda\tan(\lambda a) = 1. $$ This is a transcendental equation in $\lambda$, so no explicit expression for $\lambda$ can be obtained.

Does this mean that we can't go any further? That more or less complicated numerical methods are the only way to proceed? Or am I missing something here?

  • $\begingroup$ I am not entirely sure about the answer, but here is something I cannot see: What makes you think that your solution would still be in the form $\cos(\lambda t)$ and $\exp(-\lambda^2 x)$, rather than something like $\exp(\lambda t)$ and trignometric in $x$ component? The double 0 condition will force it to be trignometric, but now you removed it. $\endgroup$ – Lost1 May 14 '13 at 15:17
  • $\begingroup$ Thanks for the comment. I removed some constants in order to remove clutter that seemed to be beside the point. I think I can get away with it by considering all variables and constants dimensionless. But I might have confused things more than I've simplified them. Not sure :). I've added an edit. $\endgroup$ – andreasdr May 14 '13 at 15:25
  • $\begingroup$ I don't get you what you are saying. Let me repeat what I said again: seperation of variable assumed $X''/X'=-\lambda^2$, why cant it be the case $X''/X'=\lambda^2$ or $X''= 0$? You haven't justified this. $\endgroup$ – Lost1 May 14 '13 at 15:28
  • $\begingroup$ For when the boundary condition $=0$ on both side, this is easy to justify because no linear combination of $\exp(\lambda x)$ and $\exp(-\lambda x)$ can equal to 0 at 2 different places. now you changed the condition, this is not so clear to me. $\endgroup$ – Lost1 May 14 '13 at 15:30
  • $\begingroup$ The above is a standard separation of variables procedure, of which you can find better descriptions on the web. The reason that the $-\lambda^2$ appears is that $\frac{X''(x)}{X(x)} = \frac{T'(t)}{T(t)}$. Thus, the LHS is constant or a function of $x$ and the RHS is constant or a function of $t$, so they must both be constant. The reason I choose the constant to $-\lambda^2$ (which I'm free to do since it's arbitrary at that point) is for simplicity in the notation. $\endgroup$ – andreasdr May 14 '13 at 15:45

No, you do not miss anything: no explicit $\lambda$s can be found. However, the general theory tells us:

  • there are infinitely many lambda's
  • there is minimal $\lambda$ and there is no maximal one
  • the eigenfunctions corresponding to lambdas are orthogonal on the interval

Hence we can always write the solution as an infinite series. And if you want to actually calculate the function, you will need a numerical routine to find your eigenvalues (this is really simple in your case)

Added: About the point, raised by @Lost1 in comments. Note that your boundary conditions imply (physically) that the left side of the bar is isolated, and at the right side of the bar the flux of temperature is equal to the temperature itself. This means that there will be always incoming flux of the energy in your bar, which means obviously that the temperature has to approach infinity, which implies that you have to assume that you constant of separation is positive. Take not $-\lambda^2$ but $\mu^2$ and see the implications. And this also means that $\cos$ are not the only eigenfunctions in your problem.

  • $\begingroup$ Is there any way to express $c_n$? You say that the eigenfunctions are orthogonal on the interval. So is it true to say that $\int_0^a\cos(\lambda_n)\cos(\lambda_k) = 0$? $\endgroup$ – andreasdr May 14 '13 at 15:38
  • $\begingroup$ @andreasdr Almost, $\int_{[0,a]}\cos (\lambda_n x) \cos (\lambda_k x) dx=0$ $\endgroup$ – Artem May 14 '13 at 15:42
  • $\begingroup$ Of course, my bad. Thanks for the answer, I'll go ahead and test this out numerically now! :) $\endgroup$ – andreasdr May 14 '13 at 15:46
  • $\begingroup$ @andreasdr You should also make sure that you understand comments by Lost1. He has a good point there (note, that I did not claim that $\lambda>0$) $\endgroup$ – Artem May 14 '13 at 15:48
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    $\begingroup$ @andreasdr Yup, see Sturm-Liouville theory $\endgroup$ – Artem May 14 '13 at 16:21

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