Proving Steady State Temperature Distribution in a Disk

[Dustinsfl]

Prove that the steady state temperature in a disk is never higher than the highest temperature on the boundary.
Prove that for a steady state temperature distribution in a disk, the coldest spot is on the boundary, unless the temperature is constant throughout the disk.
How is this part suppose to help? I don't see it.
$$
\int_{-\pi}^{\pi}P(r,\theta)d\theta = 1
$$
Recall from calculus that
$$
m(b - a)\leq\int_a^bf(x)dx\leq M(b - a)
$$
where $M$ and $m$ are the maximum and minimum, respectively on the interval $[a,b]$.

 

[Jhoneine]

Let's apply this to our disk. Let $M$ be the highest temperature on the boundary and $m$ be the lowest temperature on the boundary. Applying the above theorem, we get$$m(\pi - (-\pi)) \leq \int_{-\pi}^{\pi}P(r,\theta)d\theta \leq M(\pi - (-\pi))$$Since the integral of any function is equal to 1, we can conclude that $$m \leq 1 \leq M$$Therefore, the maximum temperature is $M$, which is the highest temperature on the boundary, and the steady state temperature in a disk is never higher than the highest temperature on the boundary.To prove that for a steady state temperature distribution in a disk, the coldest spot is on the boundary, unless the temperature is constant throughout the disk, we can use the same proof as above. Since the integral of any function is equal to 1, if the temperature was constant throughout the disk, then the maximum and minimum would be equal, and they would both be equal to 1. However, since the temperature on the boundary is not constant, we can assume that the maximum temperature on the boundary is greater than the minimum temperature on the boundary. Thus, the maximum temperature in the disk is the maximum temperature on the boundary, which is the coldest spot, and the temperature in the disk is never colder than the temperature on the boundary.