# 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?

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.