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In my exercise bundle about Sturm-Liouville problems and solving partial differential equations with the separation method there is an exercise that goes as follows:

Calculate the temperature distribution in a circular disc (infinitely thin) with radius $b$. The temperature on the boundary is equal to $f(\theta) = u(b,\theta)$. The temperature in the disc is finite.

I've solved a few of this kind of exercise, but in those you could find the homogeneous boundary conditions. In this exercise I don't see them. First I thought: in the center the temperature gradient must be zero, because it is in a minimum there. But that can't be true because the temperature on the circle (the boundary) is not constant.

I used the Laplace equation to solve this (heat equation with stationary temperature distribution).

After separation:

$$ r^2\frac{R''}{R}+r\frac{R'}{R}=-\frac{T''}{T}=\lambda, $$

with lambda of course the eigenvalue of the eigenfunctions we are looking for.

Thank you

Kind regards

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  • $\begingroup$ Just to add to @abel's answer, for the second angle variable you should use the periodic conditions $T(0)=T(2\pi),\,T'(0)=T'(2\pi)$ $\endgroup$
    – Artem
    Commented Dec 30, 2014 at 15:27
  • $\begingroup$ Ok, and can you use these boundary conditions because when you add them up the result is the same (when you are writing the solution as a Fourier serie? Like $T(0)=T(4\pi)$ $\endgroup$
    – Thomas
    Commented Dec 30, 2014 at 16:06
  • $\begingroup$ I tried to solve it this way. For $\lambda>0$ I find a singular solution For $\lambda=0$ I also find the singular solution For $\lambda<0$ I find that my constants can be anything. $T(\theta) = A Cos(\sqrt{-\lambda} \theta) + B sin(\sqrt{-\lambda} \theta)$ So I assume Lambda can be anything ? $\endgroup$
    – Thomas
    Commented Dec 30, 2014 at 16:40
  • $\begingroup$ No, $\lambda\geq 0$ in this case, and each $\lambda$ (except for zero) has two eigenfunctions $\cos \sqrt{\lambda}\theta$ and $\sin \sqrt{\lambda}\theta$. $\endgroup$
    – Artem
    Commented Dec 30, 2014 at 17:12
  • $\begingroup$ Ok, I used my minus sign in the Euler equation. ok, When $\lambda = 0$ $T(\theta) = A+B\theta$ and A can be any constant other than zero, so the eigenfunction here is a constant with eigenvalue zero ? But when $\lambda>0$ What are the eigenvalues in this case? Because normally. With homogeneous boundary conditions you can calculate them easily. $\endgroup$
    – Thomas
    Commented Dec 30, 2014 at 17:38

2 Answers 2

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To continue what I started in the comments.

Here, when you use first boundary condition, you will find that $$ A=A\cos \sqrt{\lambda}2\pi+B\sin\sqrt{\lambda}2\pi. $$ From the secomd boundary condition $$ B\sqrt{\lambda}=-A\sqrt{\lambda}\sin \sqrt{\lambda}2\pi+B\sqrt{\lambda}\cos \sqrt{\lambda}2\pi. $$ This is a homogeneous system of two equations with two unknowns $A,B$. To have a nontrivial solution, the determinant of this system has to be zero. Can you finish?

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  • $\begingroup$ Ow.. I made such a stupid mistake. I wrote $cos\sqrt{\lambda}2\pi)=1$ $\endgroup$
    – Thomas
    Commented Dec 30, 2014 at 18:13
  • $\begingroup$ mm.. I found that $\lambda=n^2$ $\endgroup$
    – Thomas
    Commented Jan 3, 2015 at 16:03
  • $\begingroup$ @Thomas, you're right $\endgroup$
    – Artem
    Commented Jan 3, 2015 at 16:49
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the equation for $R$ is called an euler equation that has solutions of the form $R = r^n.$ you would want the condition $R$ to be bounded at $r=0.$ this condition will eliminate negative exponents in $r$ like solutions of the form $R = 1/r.$

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  • $\begingroup$ I knew this, but I was searching for what Artem said in the comment. thank you very much for the fast reply $\endgroup$
    – Thomas
    Commented Dec 30, 2014 at 16:07

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