Change of Coordinate in Differential Equation I'm sorry, it's probably a very simple question but I'm confused between change of variable and change of coordinate in a differential equation. To take a very simple example, let's start with this equation for the function $f(x,t)$:
$$
\frac{\partial f}{\partial t} = \alpha \frac{\partial^2 f}{\partial x^2}
$$
and say that I want to find the new coordinate $y$ such that $f(y,t)$ verifies:
$$
\frac{\partial f}{\partial t} = \frac{\partial^2 f}{\partial y^2}
$$
Is the answer $y = x / \alpha$ or $y = x / \sqrt{\alpha}$? It seems to be the second one, but I can't explain rigorously why.. What if $\alpha$ was negative, would we change for a complex coordinate? This seems weird to me. What confuses me is that both the variable of f and the variable wrt which we differentiate change. The fact that both happen at once makes it unclear to me what happens exactly, and makes me suspicious about considering the $\partial \cdot^2$ simply as a "squared" $\partial \cdot$, as the notation suggests. Can anyone please explain clearly what's going on there? Thanks in advance.
 A: You are getting confused by your notation.  Instead of "changing" $f$ so that it takes a different input variable, let's just define a brand new function $g$ so that
$$
g(y(x),t) = f(x,t) \quad \quad (1)
$$
Note that while $y$ is a function of $x$, $g$ itself only cares about the value of $y$.
Get the partial derivatives of $g$.  Differentiate (1) with respect to $x$ (twice) to get
$$
g_y y_x = f_x \\
(g_{yy} y_x) y_x + g_y y_{xx} = f_{xx}
$$
and differentiate (1) with respect to $t$ to get
$$
g_t = f_t
$$
$$
$$
Finally, plug these into 
$$
f_t = \alpha f_{xx}
$$
to get the PDE in terms of $g$:
$$
g_t = \alpha \left((g_{yy} y_x) y_x + g_y y_{xx}\right)
$$
$$
$$
You want the coefficient on $g_{yy}$ to be 1 and the coefficient on $y_{xx}$ to be zero:
$$
\alpha y_x^2 = 1 \\
y_{xx} = 0
$$
Any choice of $y(x)$ that satisfies this will work for you.  One choice is $y(x) = x/\sqrt{\alpha}$.  If $\alpha < 0$, you could pick $y(x) = ix/\sqrt{-\alpha}$.  Then 
$$
\alpha y_x^2 = \alpha {\left(\frac{i}{\sqrt{-\alpha}}\right)}^2 \\
\alpha y_x^2 = \alpha \frac{-1}{-\alpha}
$$
which equals 1, as required.
A: Make the change of variable $y=\lambda\,x$ where $\lambda\ne0$. Then
$$
\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}\frac{dy}{dx}=\lambda\,\frac{\partial f}{\partial y}.
$$
Differentiating with respect to $x$ is the same as differentiating with respect to the new variable $y$ and then multiplying by $\lambda$. Then
$$
\frac{\partial^2 f}{\partial x^2}=\frac{\partial}{\partial x}\Bigl(\frac{\partial f}{\partial x}\Bigr)=\lambda\frac{\partial}{\partial y}\Bigl(\lambda\frac{\partial f}{\partial y}\Bigr)=\lambda^2\frac{\partial^2f}{\partial y^2}.
$$
Now
$$
\frac{\partial}{\partial t}-\alpha\frac{\partial^2f}{\partial x^2}=\frac{\partial}{\partial t}-\alpha\,\lambda^2\frac{\partial^2f}{\partial y^2},
$$
so you want $\alpha\,\lambda^2=1$ and $\lambda=1/\sqrt\alpha$.
