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Consider a sphere of radius $a$ that is heated uniformly and then immersed in cold water. The temperature in the sphere can be modelled by $\theta(r,t)$ where $\theta(r,t)$ satisfies $$\frac{\partial \theta}{\partial t}= \frac{D}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial \theta}{\partial r} \right)$$ where $0<r<a, t>0, \theta(a,t)=0,\theta(r,0)=1$.

I'm given that the solution is $$\theta(r,t)=\frac{1}{r}\sum_{n=1}^{\infty}a_ne^{\left(\frac{\pi n}{a}\right)^2Dt}\sin\left(\frac{\pi n}{a}r\right)$$ where $$a_n=\frac{-2a(-1)^n}{\pi n}$$

I'm asked to find an expression for the heat at the centre of the sphere for $t>0$. The centre of the sphere corresponds to $r=0$ where the solution is undefined, so how do I do this?

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  • $\begingroup$ The solution is defined at $r= 0$, notice $r$ does not appear alone, but as $sin(r)/r$. $\endgroup$ – Taozi Mar 29 '17 at 17:32
  • $\begingroup$ @Taozi So how do I use that to answer the question? $\endgroup$ – Si.0788 Mar 29 '17 at 17:54
  • $\begingroup$ Try to expand $\sin x$, $\sin x=x+O(x^3)$ $\endgroup$ – Rafa Budría Mar 29 '17 at 18:19
  • $\begingroup$ @RafaBudría So will this end up giving me $\theta(0,t) \approx \sum_{n=1}^{\infty} a_n e^{-\left(\pi n/a\right)^2Dt}$? $\endgroup$ – Si.0788 Mar 29 '17 at 20:32
  • $\begingroup$ Not completely sure. The "tail" of the sum has terms that doesn't admit the expansion (they had period as little as you could wish), but by the other side $a_n$ is too as little as you can whish. I think your formula is good, but I don't know how to write the considerations I've mentioned. $\endgroup$ – Rafa Budría Mar 29 '17 at 20:57
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Hi I posted a question here

Summing $\frac{1}{a}-\frac{1}{a^4}+\frac{1}{a^9}-\cdots$

The answer to your question is far less trivial than I expected, it can only be expressed as Jacobi $\vartheta$ function. Referring to Mathematica documentation, you can plot the solution in Mathematica as (define $x = \pi^2 D t /a^2$):

Plot[1 - EllipticTheta[3, \[Pi]/2, Exp[-x]], {x, 0, 10}]

The figure is shown below (horizontal axis is "scaled" time, vertical axis is temperature). You can see the temperature is pretty much constant due to the thick, outer region of the sphere, then it quickly decreases after some point. Beautiful! Solution

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Take the limit $\lim_{r\to0} \theta(r,t)$. Note that

$$\lim_{x\to0} \frac{\sin(x)}{x} =1$$

So if a factor $C$ $(=\frac{\pi n}{a})$ is present in the argument, substitute $u=Cx$ (observing $u\to0$ as $x\to0$):

$$\lim_{x\to0} \frac{\sin(Cx)}{x} \\ =\lim_{u\to0} \frac{\sin(u)}{u/C} = C$$

I'm not going to spend several minutes formatting your series in Latex, but I hope you can see the $\sin$ and $1/r$ factors will collectively reduce to $\frac{\pi n}{a}$ and the rest of the series is independent of $r$ so remains unchanged.

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