The constant velocity transport equation — characteristics

We consider the equation

$$\hspace{100pt} \frac{\partial u}{\partial t} +c \frac{\partial u}{\partial x} =0. \hspace{100pt}\textrm{(CVT)}$$

If $u(x, t)$ is a continuously differentiable solution, then by the several variable chain rule

$$\frac{du(x_0+ct, t)}{dt} = \frac{\partial u}{\partial x}\frac{d(x_0+ct)}{dt} + \frac{\partial u}{\partial t}=cu_x+u_t=0$$

for any $x_o, t\in \mathbb{R}$. This means that any solution is constant along the lines $x(t)=x_0+ct \;(t\in\mathbb{R})$. These lines are called the characteristics of the equation.

The initial value problem

Suppose we are given the values $u(x, 0) = F(x)$ of the solution at time $t=0$ for all $x\in\mathbb{R}$. Then for any $(x,t)\in\mathbb{R}^2$ one has

$$u(x, t) = u(x-ct, 0) = F(x-ct).$$

In other words, if there is a solution with the prescribed initial values then it must be $u(x, t) = F(x-ct)$.

On the other hand, if $F:\mathbb{R}\to\mathbb{R}$ is $C^1$ then you can verify by substituting that $u(x, t)=F(x-ct)$ satisfies the transport equation (CVT), i.e. $u_t+cu_x=0$.

Solutions without derivatives

Much of the modern theory of partial differential equations deals with “generalized solution” of one kind or another. Here is an example that begins to show why one would want to do that.

Consider the solutions with initial values $u_n(x, 0) = \arctan(nx)$. The solutions are

$$u_n(x, t) = \arctan(n(x-ct)).$$

What happens if we let $n\to\infty$? We get

$$u_\infty(x, 0) = \lim_{n\to\infty} u_n(x, 0) = \lim_{n\to\infty} \arctan{nx} =\begin{cases} +\pi/2 &(x\gt0), \\ -\pi/2 & (x\lt0), \end{cases}$$

and

$$u_\infty(x, t) = \lim_{n\to\infty} u_n(x, t) = \lim_{n\to\infty} \arctan{n(x-ct)} =\begin{cases} +\pi/2 &(x\gt ct), \\ -\pi/2 & (x\lt ct), \end{cases}$$

The function $u_\infty(x, t)$ is the limit of solutions, but it is not continuous and therefore not differentiable because it has a jump discontinuity at $x=ct$. It can't be a solution of the PDE $u_t+cu_x=0$ because the derivatives in the equation aren't defined at all $(x, t)$.

Nevertheless, $u_\infty$ is a function of the form $F(x-ct)$ and it arises as the limit of actual solutions, so we would like to call it a solution. The discontinuous solution $u_\infty$ is itself not a solution of the PDE, but it does give a good description of all smooth solutions that are close to $u_\infty$.

This leads to the question: how do we change the definition of solution to allow for discontinuous solutions?

Method of characteristics

There is a general method, called the method of characteristics for finding the solutions to a first order equation of the form

$$\frac{\partial u}{\partial t} + a(x, t) \frac{\partial u}{\partial x} = f(x, t, u) \tag{1}$$

in which $a:\mathbb{R}^2\to\mathbb{R}$ and $f:\mathbb{R}^3\to\mathbb{R}$ are continuously differentiable functions.

Definition. A characteristic for the equation (1) is a function $x_c:[t_0,t_1]\to\R$ that is defined on some interval $[t_0,t_1]\subset\R$ and that satisfies the characteristic equation

$$\frac{dx_c(t)}{dt} = a\bigl(x_c(t), t\bigr) \qquad \text{for all }t\in [t_0, t_1]\tag{2}$$

Often the graph of $x_c$, i.e. the curve $\{(t, x_c(t))\in\R^2 \mid t_0\leqslant t\leqslant t_1\}$ is also called a characterstic of the equation (1).

Theorem. If $x_c:[t_0, t_1]\to\R$ is a characteristic of equation (1) then the function $u_c(t)=u(x(t), t)$ satisfies

$$u_c'(t) = f\bigl(x_c(t), t, u_c(t)\bigr).\tag{3}$$

In the special case where $f(x, t, u)=0$ for all $(x, t, u)$ it follows that $u_c(t)=u_c(t_0)$ for all $t\in[t_0,t_1]$, i.e. if the right hand side $f$ in the equation (1) is just $0$, then any $C^1$ solution $u$ is constant along a characteristic.

Note that $u_c(t)$ is a function of one variable and (3) is an ordinary differential equation.

Proof of the Theorem. To derive (3) let $x_c:\mathbb{R}\to\mathbb{R}$ be a characteristic of the PDE (1), so that $x_c:\mathbb{R}\to\mathbb{R}$ satisfies (2). Then the several variable chain rule implies

$$\frac{du(x_c(t), t)}{dt} = u_x\bigl(x_c(t), t\bigr)\cdot x_c'(t) + u_t\bigl(x_c(t), t\bigr).$$

Since $u$ satisfies (1), we can write this as

$$\frac{du(x_c(t), t)}{dt} = u_x \cdot x_c'(t) - a(x_c(t), t)\cdot u_x + f(x_c(t), t, u(x_c(t), t)).$$

The characteristic equation (2) implies $x_c'(t)=a(x_c(t), t)$ so the first two terms on the left cancel. This proves (3).

We next show that if $f(x, t, u)=0$ for all $x, t, u$, then solutions are constant along characteristics. If $f=0$ and if $x_c:[t_0, t_1]\to\R$ is a characteristic, then (3) implies that $u_c(t) = u(x_c(t), t)$ is a function of one variable that is differentiable for all $t\in[t_0, t_1]$, and that satisfies $u_c'(t)=0$ for all $t\in [t_0,t_1]$. This implies that $u_c(t)$ does not depend on $t$, i.e. $u_c(t)=u_c(t_0)$ for all $t\in[t_0, t_1]$.                  Q.E.D.

The recipe for solving initial value problems for 1st order PDE.

We can use the above Theorem to solve an initial value problem for the partial differential equation (1).

Special case where the right hand side $f(x, t, u)=0$.
In this case (1) is the equation $u_t+a(x, t)u_x=0$, and we can proceed as follows.
If we are given the initial values $u_0(x) = u(x, 0)$ of a solution for all $x\in\mathbb{R}$, and we want to find value of the solution $u$ at some point $(x_1, t_1)$ in space and time, then

• find the characteristic through $(x_1, t_1)$, i.e. find the solution $x_{\rm c}:\mathbb{R}\to\mathbb{R}$ of

$$x_c'(t) = a(x_c(t), t), \qquad x_c(t_1) = x_1$$

• Check that the characteristic is defined for all $t\in [0, t_1]$
• It follows from equation (3) that $\frac{du(x_c(t), t)}{dt}=0$ for all $t\in[0, t_1]$. Therefore $u(x_c(t), t)$ does not depend on $t$ and we have

$$u(x_1, t_1) = u\bigl(x_c(t_1), t_1\bigr)= u\bigl(x_c(0), 0\bigr) = u_0(x_c(0)).$$

General case with arbitrary right hand side $f(x, t, u)$.

• find the characteristic through $(x_1, t_1)$, i.e. find the solution $x_{\rm c}:\mathbb{R}\to\mathbb{R}$ of

$$x_c'(t) = a(x_c(t), t), \qquad x_c(t_1) = x_1$$

• Check that the characteristic is defined for all $t\in [0, t_1]$
• It follows from equation (3) that $\dfrac{du(x_c(t), t)}{dt}=f(x_c(t), t, u(x_c(t), t))$ for all $t\in[0, t_1]$. To find $u(x_c(t), t)$ define $u_c(t) = u(x_c(t),t)$ and solve the differential equation

$$u_c'(t) = f(x_c(t), t, u_c(t)), \qquad u_c(0) = u_0(x_c(0)).$$

• If the solution $u_c(t)$ is defined for all $t\in[0, t_1]$ then we have

$$u(x_1, t_1) = u\bigl(x_c(t_1), t_1\bigr)= u_c(t_1).$$

Equations for the derivatives of a solution

If $u$ is a solution of

$$u_t+\sin(x)u_x=u^2$$

then the derivative $v=u_x$ also satisfies a partial differential equation. We get this equation by differentiating the equation for $u$ with respect to $x$ on both sides:

$$\frac{\partial \bigl(u_t+\sin(x)u_x\bigr)}{\partial x} = \frac{\partial u^2}{\partial x} \implies\frac{\partial u_t}{\partial x}+\cos(x)\frac{\partial u}{\partial x}+ \sin(x)\frac{\partial u_x}{\partial x} =2u\frac{\partial u}{\partial x}$$

Rewrite the second order derivatives that appeared as derivatives of $v$ using

$$\frac{\partial u_x}{\partial x} = \frac{\partial v}{\partial x}\qquad \text {and }\qquad \frac{\partial u_t}{\partial x}= \frac{\partial^2 u}{\partial x\partial t}= \frac{\partial^2 u}{\partial t\partial x} =\frac{\partial u_x}{\partial t} = v_t.$$

We get

$$v_t + \cos(x)\frac{\partial u}{\partial x}+\sin(x)\frac{\partial v}{\partial x} = 2u \frac{\partial u}{\partial x}$$

Finally, replace every $\frac{\partial u}{\partial x}$ by $v$:

$$v_t + \sin(x) v_x = 2uv - \cos(x) v$$

A nonlinear equation

Consider the so-called inviscid Burger's equation

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

If $u(x, t)$ is a $C^1$ solution then any level set of $u$ on which $u_x\neq 0$ is a straight line.

Suppose a level set is a graph $x=x(t)$, i.e. suppose that for some function $x=x(t)$ one has $u(x(t), t) = c$ for all $t$. Then

$$\left. \begin{aligned} 0& = \frac{dc}{dt}=\frac{du(x(t), t)}{dt} \\ &= u_x(x(t), t)x'(t) + u_t(x(t), t) \\ &= u_x(x(t), t)x'(t) - u(x(t), t) u_x(x(t), t)\\ &= u_x(x(t), t)x'(t) - c u_x(x(t), t)\\ &= \bigl(x'(t) - c \bigr) u_x(x(t), t) \end{aligned} \right\} \quad \implies x'(t) = c.$$

Problems

Suppose $u:[0,\infty)\times[0, \infty) \to \mathbb{R}$ is a $C^1$ solution of

$$u_t +x u_x = 0.$$

1. Find the characteristics of the equation.
2. If you are given the initial condition $u(x, 0) = u_0(x)$ for all $x\in [0, \infty)$, can you compute $u(x, t)$ for all $x, t\geq 0$?
3. Show that the function $v(x, t) = u_x(x, t)$ satisfies the partial differential equation $v_t+xv_x=-v$.
4. Show that along a characteristic $x(t)$ of the equation for $v$ one has $u_x(x(t), t) = Ae^{Bt}$ and find $A,B$.

Suppose $u:[0,\infty)\times[0, \infty) \to \mathbb{R}$ is a $C^1$ solution of

$$u_t = x^2 u_x.$$

1. Find the characteristics of the equation for $u$.
2. If you are given the initial condition $u(x, 0) = u_0(x)$ for all $x\in [0, \infty)$, can you compute $u(x, t)$ for all $x, t\geq 0$?

About the inviscid Burger’ equation.

Let $u:\mathbb{R}\times[0, T)\to\mathbb{R}$ be a $C^2$ solution to the inviscid Burger's equation. Let $a\in\mathbb{R}$ be given, and define $c=u(a, 0)$.

1. What can you say about the level set $\{(x, t) \mid u(x, t)=c\}$?
2. Show that the function $v(x, t) = u_x(x, t)$ satisfies the partial differential equation

   $$v_t+uv_x+v^2=0.$$

3. Show that $S(t) = u_x(a+ct,t)$ satisfies $S'(t)=-S(t)^2$.
4. Assume that $u(x, 0)=\dfrac{2}{1+e^x}$ and compute $u_x(t,t)$ for all $t>0$; show that there is a $T>0$ such that

$$\lim_{t\nearrow T} u_x(t, t) = -\infty.$$