Symmetry method for 2d heat equation Suppose we have the pde 
$$
\frac{\partial p}{\partial t} = \frac{1}{4}\left( \frac{\partial^2 p}{\partial x^2} + \frac{\partial^2 p}{\partial y^2} \right) 
$$
Assuming a solution of the form,
$$
p(x,y,t) = \frac{1}{t}\phi(\xi), \\
\xi^2=\frac{1}{t}(x^2+y^2)
$$
deduce that
$$
p(x,y,t)=\frac{1}{\pi t}exp\left( -\frac{1}{t}(x^2+y^2)\right)
$$
I can't seem to get that after going thru my workings several times, below are my steps,
$$
\frac{\partial p}{\partial t}=\frac{1}{t}\phi'(\xi)\frac{\partial \xi}{\partial t} -\frac{1}{t^2}\phi(\xi)\\
\frac{\partial p}{\partial x}=\frac{1}{t}\phi'(\xi)\frac{\partial \xi}{\partial x}\\
\frac{\partial^2 p}{\partial x^2}=\frac{1}{t}\left[ \phi'(\xi)\frac{\partial^2 \xi}{\partial x^2} + \phi''(\xi) \left( \frac{\partial \xi}{\partial x} \right)^2 \right] $$
similarly
$$
\frac{\partial^2 p}{\partial y^2}=\frac{1}{t}\left[ \phi'(\xi)\frac{\partial^2 \xi}{\partial y^2} + \phi''(\xi) \left( \frac{\partial \xi}{\partial y} \right)^2 \right]
$$
Now we need the terms, 
$$ 
\frac{\partial \xi}{\partial t} = -\frac{\xi}{2t}\\
\frac{\partial \xi}{\partial x} = \frac{x}{t\xi},\frac{\partial \xi}{\partial y} = \frac{y}{t\xi}\\
\frac{\partial^2 \xi}{\partial x^2}+\frac{\partial^2 \xi}{\partial y^2}=\frac{1}{t\xi}
$$ 
Now we substitute all these back into the initial pde to get,
$$
-\frac{1}{t^2}\left[ \frac{\xi}{2}\phi'(\xi) +\phi(\xi)\right] = \frac{1}{4t^2}\left[ \frac{1}{\xi}\phi'(\xi)+\phi''(\xi)\right]
$$
After rearranging terms I finally get, 
$$
\phi''(\xi) + \left( 2\xi+\frac{1}{\xi}\right) \phi'(\xi) + 4\phi(\xi) = 0
$$
This doesn't look like it will produce the right answer.
Can someone help see where i went wrong?
Thanks!
 A: I think it will be much easier for you to assume:
$$
p(x,y,t) = \frac{1}{t}\phi(\xi), \\
\xi=\frac{1}{t}(x^2+y^2)
$$
so that $ \xi $ is not squared. Then it is working out much nicer for me.
EDIT: 
Here is how I did it. Assuming the above, one finds:
$$
\frac{\partial p}{\partial t} = - \frac{\phi }{t^2} + \frac{\xi^2}{t} \phi' \\
\frac{\partial p}{\partial x} = \frac{2 x \phi'}{t^2} \\
\frac{\partial^2 p}{\partial x^2} = \frac{2 \; t \; \phi' + 4 \; x^2 \; \phi''}{t^3}
$$
Putting this in to the heat equation and simplifying:
$$
-t \phi + t \phi' + (x^2 + y^2) \phi' + (x^2+y^2) \phi'' = 0
$$
Now, if the terms multiplied by $ t $ and the terms multiplied by $ x^2 + y^2 $ are separately zero, then the equation will surely be true for all $ t,x,y $. The terms multiplied by $ t $ set to zero give:
$$
\phi = -\phi' \\
$$
so we immediately have:
$$ \phi(\xi) = C e^{- \xi} $$
Note that the terms multiplied by $ x^2 +y^2 $ gives:
$$
\phi' = - \phi''
$$
which is of course the same thing.
I hope this will help you. I'm sorry that I can not spot a mistake in your algebra, and that I am just too lazy to wade through all the algebra myself! It should work out like above. 
