Burgers Equation

Nonlinear PDE's Course
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Introduction

We have already met the conservation law for the traffic equations

${\partial }_t\rho +c\left(\rho \right){\partial }_x\rho =0$

and seen how this leads to shocks. We can smooth this equation by adding dispersion to the equation to give us

${\partial }_t\rho +c\left(\rho \right){\partial }_x\rho =\nu {\partial }_x^2\rho$

where $\nu >0.$

The simplest equation of this type is to write

${\partial }_{t}u+u{\partial }_{x}u=\nu {\partial }_x^{2}u$

(changing variables to $u$ and this equation is known as Burgers equation.

Travelling Wave Solution

We can find a travelling wave solution by assuming that

$u(x,t)=u(x-ct)=u\left(\xi \right)$

This leads to the equations

$-c{u}^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0$

We begin by looking at the phase plane for this system, writing $w=u^{\prime }$ so that

$\begin{array}{ccc}\frac{\mathrm{d}u}{\mathrm{d}\xi }& =& w\\ \frac{\mathrm{d}w}{\mathrm{d}\xi }& =& \frac{1}{\nu }\left(w(u-c)\right)\end{array}$

This is a degenerate system with the entire $u$ axis being equilibria.

We can also solve this equation exactly as follows.

$-c{u}^{\prime }+u^{\prime }u-\nu u^{\prime \prime }=0$

can be integrated to give

$-cu+\frac{1}{2}{\left(u\right)}^2-\nu u^{\prime }=c_1$

which can be rearranged to give

$u^{\prime }=\frac{1}{2\nu }({\left(u\right)}^2-2cu-2c_1)$

We define the two roots of the quadratic ${\left(u\right)}^2-2\nu u-2c_1=0$ by $u_1$ and $u_2$ and we assume that $u_2<u_1$. Note that there is only a bounded solution if we have two real roots and for the bounded solution $u_2<u<u_1$. We note that the wave speed is

$c=\frac{1}{2}(u_1+u_2)$

The equation can therefore be written as

$2\nu u^{\prime }=(u-u_1)(u-u_2)$

which has solution

$u\left(\xi \right)=\frac{1}{2}(u_1+u_2)-\frac{1}{2}(u_1-u_2)\tanh \left[\left(\frac{\xi }{4\nu }\right)(u_1-u_2)\right]$

Numerical Solution of Burgers equation

We can solve the equation using our split step spectral method. The equation can be written as

${\partial }_{t}u=-\frac{1}{2}{\partial }_x\left(u^2\right)+\nu {\partial }_x^{2}u$

We solve this by solving in Fourier space to give

${\partial }_t\hat{u}=-\frac{1}{2}ik\widehat{\left(u^2\right)}-\nu k^2\hat{u}$

Then we solve each of the steps in turn for a small time interval to give

$\begin{array}{ccc}\tilde{u}(k,t+\mathrm{\Delta }t)& =& \hat{u}(k,t)-\frac{\mathrm{\Delta }t}{2}ik\mathcal{F}\left({\left[{\mathcal{F}}^{-1}\hat{u}(k,t)\right]}^2\right)\\ \hat{u}(k,t+\mathrm{\Delta }t)& =& \tilde{u}(k,t+\mathrm{\Delta }t)\exp (-\nu k^2\mathrm{\Delta }t)\end{array}$

Phase plane for a travelling wave solution	Numerical solution of Burgers equation
Phase plane for a travelling wave solution of Burgers equation	Numerical solution of Burgers equation

Exact Solution of Burgers equations

We can find an exact solution to Burgers equation. We want to solve

$\begin{array}{ccc}{\partial }_{t}u+u{\partial }_{x}u& =& \nu {\partial }_x^{2}u\\ u(x,0)& =& F\left(x\right)\end{array}$

Frist we write the equation as

${\partial }_{t}u+{\partial }_x(\frac{u^2}{2}-\nu {\partial }_{x}u)=0$

We want to find a function $\psi (x,t)$ such that

${\partial }_x\psi =u,{\partial }_t\psi =\nu {\partial }_{x}u-\frac{u^2}{2}$

Note that because ${\partial }_x{\partial }_t\psi ={\partial }_t{\partial }_x\psi$ we will satisfy Burgers equation. This gives us the following equation for $\psi$

${\partial }_t\psi =\nu {\partial }_x^2\psi -\frac{1}{2}{\left({\partial }_x\psi \right)}^2$

We introduce the Cole-Hopf transformation

$\psi =-2\nu \log \left(\varphi \right)$

From this we can obtain the three results:

$\begin{array}{rl}{\partial }_x\psi & =-2\nu \frac{{\partial }_x\varphi }{\varphi }\\ {\partial }_x^2\psi & =2\nu {\left(\frac{{\partial }_x\varphi }{\varphi }\right)}^2-\frac{2\nu }{\varphi }{\partial }_x^2\varphi \\ {\partial }_t\psi & =-2\nu \frac{{\partial }_t\varphi }{\varphi }\end{array}$

Therefore

${\partial }_t\psi =\nu {\partial }_x^2\psi -\frac{1}{2}{\left({\partial }_x\psi \right)}^2$

becomes

$-2\nu \frac{{\partial }_t\varphi }{\varphi }=2{\nu }^2{\left(\frac{{\partial }_x\varphi }{\varphi }\right)}^2-2{\nu }^2\frac{{\partial }_x^2\varphi }{\varphi }-\frac{1}{2}{\left(2\nu \frac{{\partial }_x\varphi }{\varphi }\right)}^2$

or

${\partial }_t\varphi =\nu {\partial }_x^2\varphi$

which is just the diffusion equation. Note that we also have to transform the boundary conditions. We have

$F\left(x\right)=u(x,0)=-2\nu \frac{{\partial }_x\varphi (x,0)}{\varphi (x,0)}$

We can write this as

$\frac{\mathrm{d}}{\mathrm{d}x}(\log \left(\varphi \right))=-\frac{1}{2\nu }F\left(x\right)$

which has solution

$\varphi (x,0)=\mathrm{\Phi }\left(x\right)=\exp (-\frac{1}{2\nu }{\int }_0^{x}F\left(s\right)\mathrm{d}s)$

We need to solve

$\begin{array}{ccc}{\partial }_t\varphi & =& \nu {\partial }_x^2\varphi \\ \varphi (x,0)& =& \mathrm{\Phi }\left(x\right)\end{array}$

We take the Fourier transform and obtain

$\begin{array}{ccc}{\partial }_t\hat{\varphi }& =& -k^2\nu \hat{\varphi }\\ \hat{\varphi }(k,0)& =& \hat{\mathrm{\Phi }}\left(k\right)\end{array}$

which has solution

$\hat{\varphi }(k,t)=\hat{\mathrm{\Phi }}\left(k\right)e^{-k^2\nu t}$

We can then use the convolution theorem to write

$\begin{array}{ccc}\varphi (x,t)& =& \mathrm{\Phi }\left(x\right)\ast {\mathcal{F}}^{-1}\left[e^{-k^2\nu t}\right]\\ & =& \frac{1}{2\sqrt{\pi \nu t}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Phi }\left(y\right)\exp [-\frac{{(x-y)}^2}{4\nu t}]\mathrm{d}y\end{array}$

Which can be expressed as

$\varphi (x,t)=\frac{1}{2\sqrt{\pi \nu t}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\exp [-\frac{f}{2\nu }]\mathrm{d}y$

where

$f(x,y,t)=\frac{1}{2\nu }{\int }_0^{y}F\left(s\right)\mathrm{d}s+\frac{{(x-y)}^2}{2t}$

To find $u$ we recall that

$\begin{array}{ccc}u(x,t)& =& -2\nu \frac{{\partial }_x\varphi (x,t)}{\varphi (x,t)}\\ & =& \frac{{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left(\frac{x-y}{t}\right)\exp [-\frac{f}{2\nu }]\mathrm{d}y}{{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\exp [-\frac{f}{2\nu }]\mathrm{d}y}\end{array}$