I am working through some past papers in preparation to an upcoming exam and I am stuck on the last part of this question.

We have a circular plate of radius $a$. The temperature distribution, $u(\rho,\phi)$, has boundary conditions $u(a,\phi) = T_1$ when $0<\phi<\pi $ and $T_2$ when $ \pi<\phi<2\pi$. The steady state temperature distribution satisfies the Laplace equation.

I have used separation of variables to reduce the equation to two ODE's which I solved to find the general solution to be $u(\rho,\phi)=\sum C_\lambda \exp(\lambda\phi)\phi^\lambda$

The question then asks us to find the Fourier series for $u(a,\phi)$. I did this by finding the series for the two boundary conditions which resulted in: $u(a,\phi) = \frac{(T_1-T_2)}{2}+\sum\frac{((-1^m)-1)(T_2-T_1)sin(m\phi)}{\pi m}$ (Noted that I am not 100% sure this is correct)

The final part of the question, and the source of my problem, asks us to find an expression for $u(\rho,\phi)$ as an infinite series using the previous answer. I do not understand how to form a general solution using this - I cannot see how the Fourier series is of any relevance to a general solution as it doesnt appear to help us find $C_\lambda$ or $\lambda$ itself. Any help would be much appreciated!


1 Answer 1


I will split the problem into two pieces, at the end will use both pieces to generate the answer


Imagine a function of the form

$$ f(\phi) = \left\{\begin{array}{ll} T_1,& 0 < \phi < \pi \\ T_2,& \pi < \phi < 2\pi \end{array} \right. $$

This function is defined in the interval $(0, 2\pi)$, but we can always make a periodic odd extension such that $f(-\phi) = f(\phi)$. This selection is completely arbitrary, you can try to do it with an even extension, it will work out just the same. The period of this function is then $T = 4\pi$, and the Fourier Series is simply

$$ f(\phi) = \sum_{n = 1}^{+\infty}B_n \sin\left(\frac{n \phi}{2}\right) \tag{1}\label{1} $$


$$ B_n = \frac{1}{\pi}\int_0^{2\pi} {\rm d}\phi~f(\phi)\sin\left(\frac{n \phi}{2}\right) $$

I will strongly encourage you to solve this integral, for now I will just write the result

$$ B_n = \frac{2}{n\pi}\left\{\begin{array}{ll} (1 - (-1)^{n/2})(T_1 - T_2), & n \mbox{ even} \\ T_1 + T_2, & n \mbox{ odd} \end{array}\right. $$

Below is an example with $T_1 = 1$, $T_2 = 3$ and truncating the sum at $n = 50$

enter image description here


If $u(\rho,\phi) = R(\rho)\Phi(\phi)$ represents the steady state solution for the temperature profile, then $\nabla^2u = 0$, which can be rewritten as

$$ \frac{\rho}{R(\rho)}\frac{{\rm d}}{{\rm d}\rho}\left(\rho \frac{{\rm d}R(\rho)}{{\rm d}\rho} \right) = -\frac{1}{\Phi(\phi)}\frac{{\rm d}^2\Phi(\phi)}{{\rm d}\phi^2} = m^2 $$

I assumed that the constant must be positive, otherwise you'll have a hard time trying to reconcile the boundary conditions for this geometry. Indeed, it must be true that

$$ u(a, \phi) = f(\phi) = R(a)\Phi(\phi) $$

So, if we impose that $R(a) = 1$, then $\Phi(\phi) = f(\phi)$ from the previous step

$$ \Phi_n(\phi) = B_n \sin\left(\frac{n\phi}{2}\right)\tag{2}\label{2} $$

With $m = n/2$. Now we only requiere to calculare $R(\rho)$,

$$ R(\rho) = C_-\left(\frac{\rho}{a} \right)^{-n/2} + C_+\left(\frac{\rho}{a} \right)^{n/2} \tag{3}\label{3} $$

we requiere that the temperature is finite at the origin, so $C_- = 0$, and that $R(a) = 1$ which leads to $C_+ = 1$


If we put together $\ref{2}$ and $\ref{3}$ we arrive to the conclusion that

$$ u(\rho,\phi) = \sum_{n = 1}^{+\infty}B_n \sin\left(\frac{n\phi}{2}\right) \left(\frac{\rho}{a}\right)^{n/2} $$

  • $\begingroup$ Thank you very much for your very detailed answer! Its makes good sense apart your derivation of (3). Would you be kind enough to explain where this has come from? $\endgroup$
    – Tech
    Apr 30, 2017 at 19:32
  • $\begingroup$ It comes from solving the equation $$ \frac{\rho}{R(\rho)}\frac{{\rm d}}{{\rm d}\rho}\left(\rho \frac{{\rm d}R(\rho)}{{\rm d}\rho} \right) = -m^2 $$ $\endgroup$
    – caverac
    Apr 30, 2017 at 19:42

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