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?


1 Answer 1


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.

  • $\begingroup$ There seem to be some errors here. For example, how did you convert u to f in the substitution? $\endgroup$
    – user572780
    Commented Mar 1 at 18:32

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