## Hopf-Cole transformation and the Burgers’ equation

The Burgers equation is the following:

$\displaystyle u_t +\frac{(u^2)_x}{2}=\mu u_{xx}. \ \ \ \ \ (1)$

Here ${u}$ can be regarded as the (scalar) velocity of a fluid and ${\mu}$ can be regarded as the viscosity. This equation can be regarded as the simplified version of the Euler’s equation without the pressure term.

This equation is non-linear. Instead it is semilinear (the highest order term has coefficient that depends on ${(x, t)}$ only. In fact, it is constant). It turns out that this equation is related with the 1-D (linear!) heat equation

$\displaystyle H_t=\mu H_{xx}$

via the so-called Hopf-Cole transformation. It is well-known that the heat equation can be solved by convolving the initial function with the heat kernel. Therefore we can actually solve the Burgers equation explicitly, although the result is a bit complicated. I will record the result here.

First of all, we integrate (1) w.r.t. the first variable of ${u(\cdot, t)}$ and define ${U(x, t)=\int_0^{x} u(s, t)ds}$. From this we can get

$\displaystyle U_t + \frac{U_x^ 2}{2}=\mu U_{xx}. \ \ \ \ \ (2)$

This is still non-linear. It turns out that we can remove the non-linearity by using logarithm. With some hindsight, let

$\displaystyle U= C \log H$

where ${C}$ is to be chosen. Then (2) becomes

$\displaystyle C \frac{H_t}{H}+ \frac{C^2}{2} \frac{H_x^2}{H^2}=\mu C \frac{H_{xx}}{H}-\mu C \frac{H_x^2}{H^2}.$

So to remove the non-linearity, set ${C=-2 \mu }$, i.e.

$\displaystyle U=-2 \mu \log H.$

Then (2) becomes

$\displaystyle H_{t}= \mu H_{xx}.$

It is well-known that the heat equation

$\displaystyle \begin{cases} H_{t}= \mu H_{xx}\\ H(x, 0)= f(x) \end{cases}$

has unique solution

$\displaystyle H(x, t)=\int \Phi (x-y, t) f(y)dy$

where ${\Phi(x, t)= \frac{1}{\sqrt {4\pi \epsilon t}} e^{-\frac{x^2 }{4\mu t}}}$.

From this, and after some non-trivial work, we have

 Theorem 1 (Hopf) Under some technical condition, the function $\displaystyle u(x,t)= \frac{\int_{-\infty}^{\infty} \frac{x-y}{t} \exp\left\{ -\frac{1}{2\mu }F (x, y, t)\right\}dy}{\int_{-\infty}^\infty \exp \left\{ -\frac{1 }{2\mu } F(x, y, t)\right\}dy}$ solves (1) with the initial condition ${u(x, 0)= u_0(x)}$, where $\displaystyle F(x , y, t)=\frac{(x-y)^2}{2t}+ \int_0^y u_0(\eta)d\eta$