using Boltzman transformation $\phi=z/\sqrt{t}$ to transform a patrial Problem:
$$\frac{\partial \theta}{\partial t} =\frac{\partial }{\partial z} \left(D{\frac{\partial \theta}{\partial z}}\right)$$
Prerequisite: $\frac{\partial \phi}{\partial z}=\frac{1}{\sqrt{t}}$; and $\frac{\partial \phi}{\partial t}=-\frac{1}{2}\frac{z}{t^{3/2}}$
On the left hand side:
$$\frac{\partial \theta}{\partial t}=\frac{\partial \theta}{\partial \phi}\frac{\partial \phi}{\partial t}=-\frac{1}{2}\frac{z}{t^{3/2}}\frac{\partial \theta}{\partial \phi}$$
On the right hand side:
$$\frac{\partial }{\partial z} \left(D{\frac{\partial \theta}{\partial z}}\right)= \frac{\partial }{\partial \phi} \left(D\frac{\partial \theta}{\partial \phi} \frac{\partial \phi}{\partial z} \right) \frac{\partial \phi}{\partial z}$$ $$= \left[ \frac{\partial }{\partial \phi} \left(D\frac{\partial \theta}{\partial \phi}\right) \frac{\partial \phi}{\partial z} + D\frac{\partial \theta}{\partial \phi} \color{red}{\frac{\partial } {\partial \phi}\left( \frac{\partial \phi}{\partial z} \right) } \right] \frac{\partial \phi}{\partial z}$$
I think the above-mentioned equations do not have problem. However, There are some issues on solving the red equation $\color{red}{\frac{\partial } {\partial \phi}\left( \frac{\partial \phi}{\partial z} \right) }$:
Method 1: $$\color{red}{\frac{\partial } {\partial \phi}\left( \frac{\partial \phi}{\partial z} \right) }= \frac{\partial } {\partial z}\left( \frac{\partial \phi}{\partial \phi} \right)=0$$
I am not sure whether this is correct. For example: if we assume $y=x^{2}$, it seems that $\frac{\partial^{2} y}{\partial y \partial x}$ has two solutions: (1)$\frac{\partial^{2} y}{\partial y \partial x}=\frac{\partial}{\partial x}\left(\frac{\partial y}{\partial y }\right)=0$; (2)$\frac{\partial^{2} y}{\partial y \partial x}=\frac{\partial}{\partial y}\left(\frac{\partial y}{\partial x }\right) =\frac{1}{\partial y}\left(2x\right)=\frac{1}{\sqrt{y}}$
Method 2:
$$\color{red}{\frac{\partial } {\partial \phi}\left( \frac{\partial \phi}{\partial z} \right) } \stackrel{\frac{\partial \phi}{\partial z}=\frac{1}{\sqrt{t}}}{=} \frac{\partial } {\partial \phi}\left( \frac{1}{\sqrt{t}} \right) \stackrel{\frac{1}{\sqrt{t}}=\frac{\phi}{z}}{=}?$$
I don't know:
(1) What is the problem with the method 1, how to explain the example given?
(2) where method 2 ends up to?
The RIGHT solution to Boltzman transformation to original equation is that (the red equation equal 0):
$$-\frac{1}{2} \frac{\partial \theta}{\partial \phi}= \left[ \frac{\partial }{\partial \phi} \left(D\frac{\partial \theta}{\partial \phi}\right)\right] \left( \frac{\partial \phi}{\partial z} \right)^{2} $$ $$-\frac{\phi}{2}\frac{\partial \theta}{\partial \phi}= \frac{\partial }{\partial \phi} \left(D\frac{\partial \theta}{\partial \phi}\right) $$
Thanks very much
Don't replace the outside $ \frac{\partial}{\partial z} $ so fast:
$$ \frac{\partial }{\partial z} \left(D{\frac{\partial \theta}{\partial z}}\right)= \frac{\partial }{\partial z} \left(D\frac{\partial \theta}{\partial \phi} \frac{\partial \phi}{\partial z} \right) = \frac{\partial }{\partial z} \left(D\frac{\partial \theta}{\partial \phi} \frac{1}{\sqrt{t}} \right) $$ So now you see that, you can just pull out the $ \frac{1}{\sqrt{t}} $ through the $ \frac{\partial}{\partial z } $ and then continue $$ \frac{1}{\sqrt{t}} \frac{\partial }{\partial z} \left(D\frac{\partial \theta}{\partial \phi} \right) = \frac{1}{t} \frac{\partial}{\partial \phi} \left(D \frac{\partial \theta}{\partial \phi} \right) $$
It's a bit of a trick, I think, because you are NOT making a coordinate transformation. If it were an honest-to-goodness coordinate transformation, lets say replacing $ z,t $ for $ \phi, \psi $, then you would indeed write the $ \frac{\partial \phi}{\partial z} $ of the red line in terms of $ \phi, \psi $ and proceed with differentiating. Rather you are making the ansatz $ \theta(z,t) = f(\frac{z}{\sqrt{t}}) $. Better to understand it this way, I think.