I'm stuck trying to understand a the mathematical step in a heat transfer texbook's example (found on Step 3 of .pdf Page 77 here). Re: Radial Heat Conduction in a Tube
The PDE is simplified to:
$$ \frac{1}{r}\frac{\partial }{\partial r}\left ( r \frac{\partial T}{\partial r} \right )= 0 $$ We want to integrate to come to a conclusion, so I first try combining the left side of the equation :
$$ \frac{r}{r}\frac{\partial ^{2}T}{\partial r} = 0 $$
The first integral will give:
$$ r\frac{\partial T}{\partial r} = C_{1} $$
The second integral is where I'm having trouble; naturally I'd want to say:
$$ \frac{1}{2}r^{2} = C_{1}r + C_{2} $$
However, I think the book is saying :
$$ r \frac{\partial T}{\partial r} = C_{1} \rightarrow \frac{\partial T}{\partial r}=\frac{1}{r}C_{1}\rightarrow T=C_{1}\ln r +C_{2} $$
Can someone help explain this to me?
PS - sorry - trying to format the math nicely now; it does look alright in the linked .pdf textbook.
Your mistake is that $\frac{\partial^2 T}{\partial r^2}$ is not zero. You can multiply both sides by $r$ to get
$$\frac{\partial}{\partial r} \left ( r \frac{\partial T}{\partial r} \right ) = 0$$
but after that, the first integration gives
$$r \frac{\partial T}{\partial r}=C_1$$
so that
$$\frac{\partial T}{\partial r}=\frac{C_1}{r}.$$
Then you integrate again to get the logarithm.