# Deriving heat equation in polar coordinates

Assume that the temperature is circularly symmetric: $$u=u(r,t)$$, where $$r^2= x^2+y^2$$. Consider any circular annulus $$a\le r\le b$$. It can be shown that the total heat energy is $$2\pi \int_a^b c\rho u rdr$$.

The next step is to show that the flow of heat energy per unit time out of the annulus at $$r=b$$ is $$-2\pi b K_0 \partial u / \partial r\vert_{r=b}$$. That's where I'm stuck. To show this, I do the following:

$$\text{flux}=-K_0\nabla u\cdot \hat{n}$$

But $$\hat{n}$$ is the outward normal to the surface, so $$\nabla u\cdot \hat{n}=0$$. What am I missing?

The next step is to use the $$2\pi \int_a^b c\rho u rdr$$ and the $$-2\pi b K_0 \partial u / \partial r\vert_{r=b}$$ expressions to show that the circularly symmetric heat equation without sources comes to be

$$\frac{\partial u}{\partial t}=\frac{k}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)$$

But this part is clear to me. Would appreciate clarification of the part in the yellow colour.

Since $u$ depends on $r$ and $t$, but not on $\theta$, we see that

$$\nabla u(r,t)=\hat r \frac{\partial u(r,t)}{\partial r}\tag 1$$

Using $(1)$, we see that the flow of heat energy (per unit time) out of the annulus at $r=b$ is given by

\begin{align} \oint_{r=b} \color{blue}{\hat n}\cdot \color{red}{(-K\nabla u(r,t))}\,\color{orange}{d\ell}&=\int_0^{2\pi} \underbrace{\color{blue}{\hat r}\cdot \color{red}{\left.\left(-K\hat r \frac{\partial u(r,t)}{\partial r}\right)\right|_{r=b}}}_{\text{Independent of}\,\,\theta}\,\color{orange}{b\,d\theta}\\\\ &=-2\pi b K \left.\left( \frac{\partial u(r,t)}{\partial r}\right)\right|_{r=b} \end{align}

as was to be shown!

• Can you please clarify how does $\hat{n}$ become $\hat{r}$? Mar 13, 2018 at 22:42
• Yes. $\hat n$ is the outer unit normal to the annulus at $r=b$. The outer unit normal to a circle is $\hat r =\frac{\nabla r}{|\nabla r|}$ since the circle is defined by the level surface $r=b$. Does that suffice? Mar 13, 2018 at 22:44
• $\hat{r} = \frac{\vec{r}}{|\vec{r}|}=\frac{(r\cos\theta, r\sin\theta)}{r}=(\cos\theta, \sin\theta)$. But $\nabla r = \frac{\partial r}{\partial r} = 1$. I think I don't know something. Can you please clarify? Mar 13, 2018 at 23:02
• 0$\nabla =\hat r \frac{\partial }{\partial r}+\hat \theta \frac1r \frac{\partial }{\partial \theta}$. And yes, $hat r=\hat x \cos(\Th theta)+\hat y\sin(\theta)$. So, $\hat n=\hat r$. What do you not understand? Mar 13, 2018 at 23:12
• @sequence The normal vector and the tangent vector are orthogonal to each other. It is the vector $\hat \theta$ that is tangent to the contour $r=b$ while $\hat r$ is normal (perpendicular) to the contour $r=b$. Mar 14, 2018 at 3:06