How to transform diffusion equation into Burgers equation using Cole-Hopf transformation I want to show that Burgers equation $$\frac{\partial }{\partial t}u(x,t) = 2u \frac{\partial}{\partial x}u(x,t) + \frac{\partial^2}{\partial x^2}u(x,t)$$ satisfy diffusion equation $f_t(x,t) = f_{xx}(x,t)$ using Cole-Hopef transformation $u=(\log f)_x$.
I referred this post, but I still don't understand.
I followed the steps. In this case, we can rewrite $$\psi_t = 2(\psi_x)^2+\psi_{xx}$$. Here I set $u=\psi_x$. But I don't understand why they set $\psi =-2v \ln \psi$. Can someone help me to solve this?
 A: I think the best way to understand this is to start from the beginning and derive the transformation yourself. Starting with the Burgers equation
$$u_{t} - 2uu_{x} = u_{xx}$$
we want to look for a transformation $f :u \to f(u)$ that turns the Burgers PDE into something simpler. Also, imagine that, a priori, you didn't know what the final PDE should be (even though we do here, it will be the heat equation).
Under the above transformation, then we can compute derivatives
\begin{align}
u_{t} &\to f'(u) u_{t} \\
u_{x} &\to f'(u) u_{x} \\
u_{xx} &\to f''(u) u_{x}^{2} + f'(u) u_{xx}
\end{align}
and substitution into the original equation yields
$$u_{t} f' - 2 u_{x} f f' = u_{x}^{2} f'' + u_{xx} f'$$
Now, you might notice that both $u_{t}$ and $u_{xx}$ have the same 'coefficient', $f'$. Hence, if we can solve the ODE
$$-2 u_{x} ff' = u_{x}^{2} f'' \tag 1$$
for $f \ne 0$, then we know the explicit transformation that will turn the Burgers equation into the heat equation. So, lets solve the ODE $(1)$
\begin{align}
-2 u_{x} ff' &= u_{x}^{2} f'' \\
\implies -u_{x} f^{2} &= u_{x}^{2} f' + C_{1} \\
\implies -\frac{1}{u_{x}} &= \frac{f'}{f^{2}} \quad \text{(setting $C_{1} = 0$ for convenience)} \\
\implies -\frac{u}{u_{x}} &= -\frac{1}{f} + C_{2} \\
\implies f &= \frac{u_{x}}{u} \quad \text{(setting $C_{2} = 0$ for convenience)} \\
&= (\ln u)_{x}
\end{align}
which is the Hopf-Cole transformation.
