My watch list

Burgers' equation

Burgers' equation is a fundamental partial differential equation from fluid mechanics. It occurs in various areas of applied mathematics, such as modeling of gas dynamics and traffic flow. It is named for Johannes Martinus Burgers (1895-1981).

Additional recommended knowledge

Better weighing performance in 6 easy steps

Recognize and detect the effects of electrostatic charges on your balance

How to quickly check pipettes?

For a given velocity u and viscosity coefficient $ν$ , the general form of Burgers' equation is:

$\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} = \nu \frac{\partial^2 u}{\partial x^2}$ .

When $ν = 0$ , Burgers' equation becomes the inviscid Burgers' equation:

$\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} = 0$ ,

which is a prototype for equations for which the solution can develop discontinuities (shock waves).

Solution

The inviscid Burgers' equation is a first order partial differential equation. Its solution can be constructed by the method of characteristics. This method yields that if $X (t)$ is a solution of the ordinary differential equation

$\frac{dX(t)}{dt} = u[X(t),t]$

then $U (t): = u [X (t), t]$ is constant as a function of $t$ . Hence $[X (t), U (t)]$ is a solution of the system of ordinary equations

$\frac{dX}{dt}=U$

$\frac{dU}{dt}=0.$

The solutions of this system are given in terms of the initial values by

$\displaystyle X(t)=X(0)+tU(0)$

$\displaystyle U(t)=U(0).$

Substitute $X (0) = η$ , then $U (0) = u [X (0),0] = u (η,0)$ . Now the system becomes

$\displaystyle X(t)=\eta+tu(\eta,0)$

$\displaystyle U(t)=U(0).$

Conclusion:

$\displaystyle u(\eta,0)=U(0)=U(t)=u[X(t),t]=u[\eta+tu(\eta,0),t].$

This is an implicit relation that determines the solution of the inviscid Burgers' equation.

The viscous Burgers equation can be linearized by the Cole-Hopf substitution

$u=-2\nu \frac{1}{\phi}\frac{\partial\phi}{\partial x},$

which turns it into the diffusion equation

$\frac{\partial\phi}{\partial t}=\nu\frac{\partial^2\phi}{\partial x^2}.$

That allows one to solve an initial value problem:

$u(x,t)=-2\nu\frac{\partial}{\partial x}\ln\Bigl\{(4\pi\nu t)^{-1/2}\int_{-\infty}^\infty\exp\Bigl[-\frac{(x-x')^2}{4\nu t} -\frac{1}{2\nu}\int_0^{x'}u(x'',0)dx''\Bigr]dx'\Bigr\}.$