Transport Equation

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The transport equation is one of the simplest partial differential equations and has the form

\begin{align} u_t + v \cdot u_x=0\\ \end{align}

To introduce the solution method called method of characteristics in an intuitive way at the example of the transport equation, we imagine a device that consists of a pipe with constant radius $r$ and a valve in which a chemical can be injected. This pipe contains a fluid, for example water.

Pipevalve.png

In the following we assume that once we injected the chemical, the concentration varies only in the direction of the pipe. Further we assume that the diffusion is negligible and that the concentration profile in the pipe that results form the injection at $x=x_0$ has a bell-shape $e^{-x^2}$.

Transport1.png

Thus the concentration $u(x)$ varies with $x$ along the pipe so that $u(x)=e^{-x^2}$ (for $x_0=0$). Now we ask, how the fluid containing the chemical moves forward, when pressure is applied so that the fluid starts moving. Since there is no diffusion and no turbulences in the pipe the package of fluid containing the chemical is shifted along the pipe with some velocity $v$. The shape $e^{-x^2}$ thereby is preserved. After time $\Delta t$ the package was transported the distance $\Delta x =v \cdot \Delta t$ in the direction of $v$. So we can give the concentration profile $u(x)$ after the time $\Delta t$ as simply shifting the curve $e^{-x}$ by $\Delta x$ to the right. This can be done by adding $-v \cdot \Delta t$ to the independent variable $x$, so we get $e^{-(x-x_0)^2}=e^{-(x-v\cdot \Delta t)}$. Notice, $u$ is now both a function of $x$ and $\Delta t$. For some arbitrary time $t$ the displacement of the concentration profile can be written as

\begin{align} u(x,t)=e^{-(x-v\cdot t)^2} \end{align} This equation describes a concentration profile $u(x)$ travelling with speed $v$ in the positive $x$ direction.

If we now plot the concentration as a function of space and time, we observe the following

Transport-uxt.png

Along the line $x(t)=x_0+v \cdot t$ (or $x-v \cdot t = x_0= const.$), $u(x,t)$ does not vary. The curve (in our case a linear curve) along which $u(x,t)$ do not vary are called characteristic lines or characteristics.

picture shows some characteristic lines ($x=x_0+v \cdot t$) for different constants $x_0$, that have the same velocity $v$.

What would an observer see, from a different coordinate system $S$ that moves relative to the pipe? Let the position of this observer relative to the pipe be $x(t)$ and still the concentration is $u(x,t)$, then the time differentiation by the chain rule is \begin{align} \frac{d}{dt}u(x(t),t)=u_x \frac{dx}{dt} + u_t \end{align} This means, if $\frac{d}{dt}u(x,t)=0$ the concentration profile for the observer on $S$ does not change with time. In other words, the coordinate system $S$ moves at the same speed $v$ as the fluid. This is exactly the case, when $x(t)=x_0+vt$ as we derived above. Therefore, characteristic lines are curves $x(t)$ along which the concentration profile only depends on one variable $x$. Change of coordinates into the system $x(t)$ makes the concentration independent of time.

This equation is called transport equation \begin{align} u_t + v \cdot u_x=0\\ \end{align}

By use of vector calculus, we can reformulate the idea of characteristics. To this end we parameterise the characteristic line and write it as a vector \begin{align} x(t)=v \cdot t +x_0 \text{ and } v=\frac{a_y}{a_x} \rightarrow \begin{pmatrix} a_x \\ a_y \end{pmatrix}\begin{pmatrix} 1 \\ v \end{pmatrix} \end{align} For one step $a_x=1$ in the x-direction, the linear curve goes $a_y=v$ in the y-direction. So we get \begin{align} \mathbf{x}(t)=\mathbf{x}_0 + t \cdot \begin{pmatrix} 1 \\ v \end{pmatrix} \end{align} We know that the slope of $u(x,t)$ in the direction of the characteristic line is zero, so the product of the gradient of the scalar field $u(x,t)$ with $\mathbf{\dot{x}}(t)$ (direction in which $\mathbf{x}(t)$ changes) must be zero. \begin{align} \begin{pmatrix} 1 \\ v \end{pmatrix} \cdot \nabla u(x,t)=0 \end{align} We can now use the concept of characteristic curves to solve arbitrary transport equations.


Example


Solve the equation $u_t-2u_x=0$ with initial condition $u(x,0)=cos(x)$.

The vector $(1,-2)$ is normal to $\nabla u$, thus $(1,-2)$ gives the direction of the characteristic line. We get $v=a_x/a_y=v/1=-2/1$ \begin{align} x(t)=x_0+v \cdot t=x_0-2 t \end{align} along the characteristic line $x+2t=C$ the concentration is a function of $x$ only \begin{align} u(x,0)=u(x(t)=C,0)=f(x+2t)=cos(x) \end{align} The solution to the initial value problem is \begin{align} u(x,t)=cos(x+2t) \end{align} a concentration profile, that travels to the left ($v<0$) and has a cosine-shape.




Video Lectures:

  • PDE1 - Introduction [1] (time 14:50)
  • PDE2 - Three fundamental examples [2] (time 13:20)
  • PDE3 - Transport equation: derivation [3] (time 11:31)
  • PDE4 - Transport equation: general solution [4] (time 14:12)
  • PDE5 - Method of characteristics [5] (time 14:59)
  • PDE6 - Transport with decay and nonlinear transport [6] (time 13:45)



Further Reading:

  • K. F. Riley , M. P. Hobson, S. J. Bence - Mathematical Methods for Physics and Engineering: A Comprehensive Guide
  • Walter A. Strauss - Partial Differential Equations: An Introduction