Introduction

Abstract

Hyperbolic conservation laws are partial differential equations of the form

$$\displaystyle\frac{\partial u}{\partial t}+\nabla\cdot f(u)=0.$$

If we write $f=\left(f_{1},\dots,f_{m}\right)$, $x=\left(x_{1},x_{2},\dots,x_{m}\right)\in\mathbb{R}^{m}$, and introduce initial data u ₀ at t = 0, the Cauchy problem for hyperbolic conservation laws reads

$$\displaystyle\frac{\partial u(x,t)}{\partial t}+\sum_{j=1}^{m}\frac{\partial}{\partial x_{j}}f_{j}\left(u(x,t)\right)=0,\quad u|_{t=0}=u_{0}.$$

(1.1)

In applications, t normally denotes the time variable, while x describes the spatial variation in m space dimensions. The unknown function u (as well as each f _j ) can be a vector, in which case we say that we have a system of equations, or u and each f _j can be a scalar. This book covers the theory of scalar conservation laws in several space dimensions as well as the theory of systems of hyperbolic conservation laws in one space dimension. In the present chapter we study the one-dimensional scalar case to highlight some of the fundamental issues in the theory of conservation laws.

I have no objection to the use of the term ‘‘Burgers equation’’ for the nonlinear heat equation(provided it is not written ‘‘Burger’s equation’’). — Letter from Burgers to Batchelor (1968)

I have no objection to the use of the term ‘‘Burgers’ equation’’ for the nonlinear heat equation(provided it is not written ‘‘Burger’s equation’’).

— Letter from Burgers to Batchelor (1968)

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

Hyperbolic conservation laws are partial differential equations of the form

$$\displaystyle\frac{\partial u}{\partial t}+\nabla\cdot f(u)=0.$$

If we write $f=\left(f_{1},\dots,f_{m}\right)$, $x=\left(x_{1},x_{2},\dots,x_{m}\right)\in\mathbb{R}^{m}$, and introduce initial data u ₀ at t = 0, the Cauchy problem for hyperbolic conservation laws reads

$$\displaystyle\frac{\partial u(x,t)}{\partial t}+\sum_{j=1}^{m}\frac{\partial}{\partial x_{j}}f_{j}\left(u(x,t)\right)=0,\quad u|_{t=0}=u_{0}.$$

(1.1)

In applications, t normally denotes the time variable, while x describes the spatial variation in m space dimensions. The unknown function u (as well as each f _j ) can be a vector, in which case we say that we have a system of equations, or u and each f _j can be a scalar. This book covers the theory of scalar conservation laws in several space dimensions as well as the theory of systems of hyperbolic conservation laws in one space dimension. In the present chapter we study the one-dimensional scalar case to highlight some of the fundamental issues in the theory of conservation laws.

We use subscripts to denote partial derivatives, i.e., $u_{t}(x,t)=\partial u(x,t)/\partial t$. Hence we may write (1.1) when m = 1 as

$$\displaystyle u_{t}+f(u)_{x}=0,\quad u|_{t=0}=u_{0}.$$

(1.2)

If we formally integrate equation (1.2 ) between two points x ₁ and x ₂, we obtain

$$\displaystyle\int_{x_{1}}^{x_{2}}u_{t}\,dx=-\int_{x_{1}}^{x_{2}}f(u)_{x}\,dx=f\left(u\left(x_{1},t\right)\right)-f\left(u\left(x_{2},t\right)\right).$$

Assuming that u is sufficiently regular to allow us to take the derivative outside the integral, we get

$$\displaystyle\frac{d}{dt}\int_{x_{1}}^{x_{2}}u(x,t)\,dx=f\left(u\left(x_{1},t\right)\right)-f\left(u\left(x_{2},t\right)\right).$$

(1.3)

This equation expresses conservation of the quantity measured by u in the sense that the rate of change in the amount of u between x ₁ and x ₂ is given by the difference in $f(u)$ evaluated at these points.¹ Therefore, it is natural to interpret $f(u)$ as the flux density of u. Often, $f(u)$ is referred to as the flux function.

Consider a fluid with density $\rho=\rho(x,t)$ and velocity v. Assume that there are no sources or sinks, so that amount of fluid is conserved. For a given and fixed bounded domain $D\subset\mathbb{R}^{m}$, conservation of fluid implies

$$\displaystyle\frac{d}{dt}\int_{D}\rho(x,t)\,dx=-\int_{\partial D}(\rho v)\cdot n\,dS_{x},$$

(1.4)

where n is the outward unit normal at the boundary $\partial D$ of D. If we interchange the time derivative and the integral on the left-hand side of the equation, and apply the divergence theorem on the right-hand side, we obtain

$$\displaystyle\int_{D}\rho(x,t)_{t}\,dx=-\int_{D}\mathop{\mathrm{div}}(\rho v)\,dx$$

(1.5)

which we rewrite as

$$\displaystyle\int_{D}\big(\rho_{t}+\mathop{\mathrm{div}}(\rho v)\big)\,dx=0.$$

(1.6)

Since the domain D was arbitrary, we obtain the hyperbolic conservation law

$$\displaystyle\rho_{t}+\mathop{\mathrm{div}}(\rho v)=0.$$

(1.7)

The above derivation is very fundamental, and only two assumptions are made. First of all, we make the physical assumption of conservation, and secondly, we assume sufficient smoothness of the functions to perform the necessary mathematical manipulations. The latter aspect will a recurring theme throughout the book.

As a simple example of a conservation law, consider a one-dimensional medium consisting of noninteracting particles, or material points, identified by their coordinates y along a line. Let $\phi(y,t)$ denote the position of material point y at time t. The velocity and the acceleration of y at time t are given by $\phi_{t}(y,t)$ and $\phi_{tt}(y,t)$, respectively. Assume that for each t, $\phi(\,\cdot\,,t)$ is strictly increasing, so that two distinct material points cannot occupy the same position at the same time. Then the function $\phi(\,\cdot\,,t)$ has an inverse $\psi(\,\cdot\,,t)$, so that $y=\psi(\phi(y,t),t)$ for all t. Hence $x=\phi(y,t)$ is equivalent to $y=\psi(x,t)$. Now let u denote the velocity of the material point occupying position x at time t, i.e., $u(x,t)=\phi_{t}(\psi(x,t),t)$, or equivalently, $u(\phi(y,t),t)=\phi_{t}(y,t)$. Then the acceleration of material point y at time t is

$$\begin{aligned}\displaystyle\phi_{tt}(y,t)&\displaystyle=u_{t}(\phi(y,t),t)+u_{x}(\phi(y,t),t)\phi_{t}(y,t)\\ \displaystyle&\displaystyle=u_{t}(x,t)+u_{x}(x,t)u(x,t).\end{aligned}$$

If the material particles are noninteracting, so that they exert no force on each other, and there is no external force acting on them, then Newton’s second law requires the acceleration to be zero, giving

$$\displaystyle u_{t}+\left(\frac{1}{2}u^{2}\right)_{x}=0.$$

(1.8)

The last equation, (1.8 ), is a conservation law; it expresses that u is conserved with a flux density given by $u^{2}/2$. This equation is often referred to as the Burgers equation without viscosity, ² and is in some sense the simplest nonlinear conservation law.

Burgers’s equation, and indeed any conservation law, is an example of a quasilinear equation, meaning that the highest derivatives occur linearly. A general inhomogeneous quasilinear equation for functions of two variables x and t can be written

$$\displaystyle a(x,t,u)u_{t}+b(x,t,u)u_{x}=c(x,t,u).$$

(1.9)

If the coefficients a and b are independent of u, i.e., $a=a(x,t)$, $b=b(x,t)$, we say that the equation is semilinear, while the equation is linear if, in addition, the same applies to c, i.e., $c=c(x,t)$.

We may consider the solution as the surface $S=\{(t,x,u(x,t))\in\mathbb{R}^{3}\mid(t,x)\in\mathbb{R}^{2}\}$ in $\mathbb{R}^{3}$. Let Γ be a given curve in $\mathbb{R}^{3}$ (which one may think of as the initial data if t is constant) parameterized by $(t(y),x(y),z(y))$ for y in some interval. We want to construct the surface $S\subset\mathbb{R}^{3}$ parameterized by $(t,x,u(x,t))$ such that $u=u(x,t)$ satisfies (1.9) and $\Gamma\subset S$. It turns out to be advantageous to consider the surface S parameterized by new variables $(s,y)$, thus $t=t(s,y)$, $x=x(s,y)$, $z=z(s,y)$, in such a way that $u(x,t)=z(s,y)$. We solve the system of ordinary differential equations

$$\displaystyle\frac{\partial t}{\partial s}=a,\quad\frac{\partial x}{\partial s}=b,\quad\frac{\partial z}{\partial s}=c,$$

(1.10)

with

$$\displaystyle t(s_{0},y)=t(y),\quad x(s_{0},y)=x(y),\quad z(s_{0},y)=z(y).$$

(1.11)

In this way we obtain the parameterized surface $S=\{(t(s,y),x(s,y),z(s,y)\mid(s,y)\in\mathbb{R}^{2}\}$. Assume that we can invert the relations $x=x(s,y)$, $t=t(s,y)$ and write $s=s(x,t)$, $y=y(x,t)$. Then

$$\displaystyle u(x,t)=z(s(x,t),y(x,t))$$

(1.12)

satisfies both (1.9) and the condition $\Gamma\subset S$. Namely, we have

$$\displaystyle c=\frac{\partial z}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial s}=u_{x}b+u_{t}a.$$

(1.13)

However, there are many pitfalls in the above construction: the solution (1.10) may only be local, and we may not be able to invert the solution of the differential equation to express $(s,y)$ as functions of $(x,t)$. These problems are intrinsic to equations of this type and will be discussed at length.

Equation (1.10 ) is called the characteristic equation, and its solutions are called characteristics. This can sometimes be used to find explicit solutions of conservation laws. In the homogeneous case, that is, when c = 0, the solution u is constant along characteristics, namely,

$$\displaystyle\frac{d}{ds}u(x(s,y),t(s,y))=u_{x}x_{s}+u_{t}t_{s}=u_{x}b+u_{t}a=0.$$

(1.14)

FormalPara ⋄ Example 1.1

Consider the (quasi)linear equation

$$\displaystyle u_{t}-xu_{x}=-2u,\quad u(x,0)=x,$$

with associated characteristic equations

$$\displaystyle\frac{\partial t}{\partial s}=1,\quad\frac{\partial x}{\partial s}=-x,\quad\frac{\partial z}{\partial s}=-2z.$$

The general solution of the characteristic equations reads

$$\displaystyle t=t_{0}+s,\quad x=x_{0}e^{-s},\quad z=z_{0}e^{-2s}.$$

Parameterizing the initial data for s = 0 by t = 0, $x=y$, and $z=y$, we obtain

$$\displaystyle t=s,\quad x=ye^{-s},\quad z=ye^{-2s},$$

which can be inverted to yield

$$\displaystyle u=u(x,t)=z(s,y)=xe^{-t}.$$

 ⋄

FormalPara ⋄ Example 1.2

Consider the (quasi)linear equation

$$\displaystyle xu_{t}-t^{2}u_{x}=0.$$

(1.15)

Its associated characteristic equation is

$$\displaystyle\frac{\partial t}{\partial s}=x,\quad\frac{\partial x}{\partial s}=-t^{2}.$$

This has solutions given implicitly by $x^{2}/2+t^{3}/3$ equals a constant, since after all, $\partial(x^{2}/2+t^{3}/3)/\partial s=0$, so the solution of (1.15) is any function $\varphi$ of $x^{2}/2+t^{3}/3$, i.e., $u(x,t)=\varphi(x^{2}/2+t^{3}/3)$. For example, if we wish to solve the initial value problem (1.15) with $u(x,0)=\sin|x|$, then $u(x,0)=\varphi(x^{2}/2)=\sin|x|$. Consequently, $\varphi(\zeta)=\sin\sqrt{2\zeta}$ with $\zeta\geq 0$, and the solution u is given by

$$\displaystyle u(x,t)=\sin\sqrt{x^{2}+2t^{3}/3},\qquad t\geq 0.$$

 ⋄

FormalPara ⋄ Example 1.3 (Burgers’s equation)

If we apply this technique to Burgers’s equation(1.8) with initial data $u(x,0)=u_{0}(x)$, we get that

$$\displaystyle\frac{\partial t}{\partial s}=1,\qquad\frac{\partial x}{\partial s}=z,\quad\text{and}\quad\frac{\partial z}{\partial s}=0$$

with initial conditions $t(0,y)=0$, $x(0,y)=y$, and $z(0,y)=u_{0}(y)$. We cannot solve these equations without knowing more about u ₀, but since u (or z) is constant along characteristics, cf. (1.14), we see that the characteristics are straight lines. In other words, the value of z is transported along characteristics, so that

$$\displaystyle t(s,y)=s,\quad x(s,y)=y+sz=y+su_{0}\left(\eta\right),\quad z(s,y)=u_{0}(y).$$

We may write this as

$$\displaystyle x=y+u_{0}(y)t.$$

(1.16)

If we solve this equation in terms of $y=y(x,t)$, we can use y to obtain $u(x,t)=z(s,y)=u_{0}(y(x,t))$, yielding the implicit relation

$$\displaystyle u(x,t)=u_{0}(x-u(x,t)t).$$

(1.17)

Given a point $(x,t)$, one can in principle determine the solution $u=u(x,t)$ from equation (1.17). By differentiating equation (1.16) we find that

$$\displaystyle\frac{\partial x}{\partial y}=1+tu_{0}^{\prime}\left(y\right).$$

(1.18)

Thus a solution certainly exists for all t > 0 if $u_{0}^{\prime}> 0$, since x is a strictly increasing function of η in that case. On the other hand, if $u_{0}^{\prime}(\tilde{x})<0$ for some $\tilde{x}$, then a solution cannot be found for $t> t^{*}=-1/u_{0}^{\prime}(\tilde{x})$. For example, if $u_{0}(x)=-\arctan(x)$, there is no smooth solution for t > 1.

What actually happens when a smooth solution cannot be defined? From (1.18) we see that for $t> t^{*}$, there are several y that satisfy (1.16) for each x, since x is no longer a strictly increasing function of y. In some sense, we can say that the solution u is multivalued at such points. To illustrate this, consider the surface in $(t,x,u)$-space parameterized by s and y,

$$\displaystyle\left(s,y+su_{0}(y),u_{0}(\eta)\right).$$

Let us assume that the initial data are given by $u_{0}(x)=-\arctan(x)$ and $t_{0}=0$. For each fixed t, the curve traced out by the surface is the graph of a (multivalued) function of x. In Fig. 1.1 we see how the multivaluedness starts at t = 1 when the surface ‘‘folds over,’’ and that for t > 1 there are some x that have three associated u values. To continue the solution beyond t = 1 we have to choose among these three u values. In any case, it is impossible to continue the solution and at the same time keep it continuous.  ⋄

Fig. 1.1

A multivalued solution

Now we have seen that no matter how smooth the initial function is, we cannot expect to be able to define classical solutions of nonlinear conservation laws for all time. In this case we have to extend the concept of solution in order to allow discontinuities.

The standard way of extending the admissible set of solutions to partial differential equations is to look for weak solutions rather than so-called classical solutions, by introducing distribution theory. Classical solutions are sufficiently differentiable functions such that the differential equation is satisfied for all values of the independent arguments. However, there is no unique definition of weak solutions. In the context of hyperbolic conservation laws we do not need the full machinery of distribution theory, and our solutions will be functions that may be nondifferentiable.

In this book we use the following standard notation: $C^{i}(U)$ is the set of i times continuously differentiable functions on a set $U\subseteq\mathbb{R}^{n}$, and $C^{i}_{0}(U)$ denotes the set of such functions that have compact support in U. Then $C^{\infty}(U)=\bigcap_{i=0}^{\infty}C^{i}(U)$, and similarly for $C^{\infty}_{0}$. Where there is no ambiguity, we sometimes omit the set U and write only C ⁰, etc.

If we have a classical solution to (1.2 ), we can multiply the equation by a function $\varphi=\varphi(x,t)\in C^{\infty}_{0}(\mathbb{R}\times[0,\infty))$, called a test function, and integrate by parts to get

$$\begin{aligned}\displaystyle 0&\displaystyle=\int_{0}^{\infty}\int_{\mathbb{R}}\big(u_{t}\varphi+f(u)_{x}\varphi\big)\,dx\,dt\\ \displaystyle&\displaystyle=-\int_{0}^{\infty}\int_{\mathbb{R}}\big(u\varphi_{t}+f(u)\varphi_{x}\big)\,dx\,dt-\int_{\mathbb{R}}u(x,0)\varphi(x,0)\,dx.\end{aligned}$$

Observe that the boundary terms at $t=\infty$ and at $x=\pm\infty$ vanish, since $\varphi$ has compact support, and that the final expression incorporates the initial data. Now we define a weak solution of (1.2) to be a measurable function $u(x,t)$ such that

$$\displaystyle\int_{0}^{\infty}\int_{\mathbb{R}}\big(u\varphi_{t}+f(u)\varphi_{x}\big)\,dxdt+\int_{\mathbb{R}}u_{0}\varphi(x,0)\,dx=0$$

(1.19)

holds for all $\varphi\in C^{\infty}_{0}(\mathbb{R}\times[0,\infty))$. We see that the weak solution u is no longer required to be differentiable, and that a classical solution is also a weak solution. We will spend considerable time in understanding the constraints that the equation (1.19) puts on u.

We employ the usual notation that for $p\in[0,\infty)$, $L^{p}(U)$ denotes the set of all measurable functions $F\colon U\to\mathbb{R}$ such that the integral

$$\displaystyle\int_{U}|F|^{p}\,dx$$

is finite. The set $L^{p}(U)$ is equipped with the norm

$$\displaystyle{\left\|F\right\|}_{p}={\left\|F\right\|}_{L^{p}}={\left\|F\right\|}_{L^{p}(U)}=\left(\int_{U}|F|^{p}dx\right)^{1/p}.$$

If $p=\infty$, $L^{\infty}(U)$ denotes the set of all measurable functions F such that

$$\displaystyle\mathop{\mathrm{ess\,sup}}_{U}|F|$$

is finite. The space $L^{\infty}(U)$ has the norm ${\left\|F\right\|}_{\infty}=\mathop{\mathrm{ess\,sup}}_{U}|F|$. As is well-known, the spaces $L^{p}(U)$ are Banach spaces for $p\in[1,\infty]$, and $L^{2}(U)$ is a Hilbert space. In addition, we will frequently use the spaces

$$\displaystyle L^{p}_{\mathrm{loc}}(U)=\{f\colon U\to\mathbb{R}\mid\text{$f\in L^{p}(K)$ for every compact set $K\subseteq U$}\}.$$

So what kind of discontinuities are compatible with (1.19)? If we assume that u is constant outside some finite interval, the remarks below (1.2) imply that

$$\displaystyle\frac{d}{dt}\int_{-\infty}^{\infty}u(x,t)\,dx=0.$$

Hence, the total amount of u is independent of time, or equivalently, the area below the graph of $u(\,\cdot\,,t)$ is constant.

Fig. 1.2

Different solutions with u conserved

FormalPara ⋄ Example 1.4 (Burgers’s equation (cont’d.))

We now wish to determine a discontinuous function such that the graph of the function lies on the surface given earlier with $u(x,0)=-\arctan x$. Furthermore, the area under the graph of the function should be equal to the area between the x-axis and the surface. In Fig. 1.2 we see a section of the surface making up the solution for t = 3. The curve is parameterized by x ₀, and explicitly given by $u=-\arctan\left(x_{0}\right)$, $x=x_{0}-3\arctan\left(x_{0}\right)$.

The function u is shown by a thick line, and the surface is shown by a dotted line. A function $u(x)$ that has the correct integral, $\int u\,dx=\int u_{0}\,dx$, is easily found by making any cut from the upper fold to the middle fold at some negative x _c with $x_{c}\geq-\sqrt{2}$, and then making a cut from the middle part to the lower part at $-x_{c}$. We see that in all cases, the area below the thick line is the same as the area bounded by the curve $\left(x\left(x_{0}\right),u\left(x_{0}\right)\right)$. Consequently, conservation of u is not sufficient to determine a unique weak solution.  ⋄

Let us examine what kind of discontinuities are compatible with (1.19) in the general case. Assume that we have an isolated discontinuity that moves along a smooth curve $\Gamma\colon x=x(t)$. The discontinuity being isolated means that the function $u(x,t)$ is differentiable in a sufficiently small neighborhood of $x(t)$ and satisfies equation (1.2) classically on each side of $x(t)$. We also assume that u is uniformly bounded in a neighborhood of the discontinuity.

Now we choose a neighborhood D around the point $(x(t),t)$ and a test function $\phi(x,t)$ whose support lies entirely inside the neighborhood. The situation is as depicted in Fig. 1.3. The neighborhood consists of two parts D ₁ and D ₂, and is chosen so small that u is differentiable everywhere inside D except on $x(t)$. Let $D_{i}^{\varepsilon}$ denote the set of points

$$\displaystyle D_{i}^{\varepsilon}=\left\{(x,t)\in D_{i}\mid\mathrm{dist}\big((x,t),(x(t),t)\big)> \varepsilon\right\}.$$

The function u is bounded, and hence

$$\displaystyle 0=\int_{D}\left(u\phi_{t}+f(u)\phi_{x}\right)\,dx\,dt=\lim_{\varepsilon\to 0}\int_{D^{\varepsilon}_{1}\cup D^{\varepsilon}_{2}}\left(u\phi_{t}+f(u)\phi_{x}\right)\,dx\,dt.$$

(1.20)

Fig. 1.3

An isolated discontinuity

Since u is a classical solution inside each $D_{i}^{\varepsilon}$, we can use Green’s theorem and obtain

$$\int_{D^{\varepsilon}_{i}}\left(u\phi_{t}+f(u)\phi_{x}\right)\,dx\,dt =\int_{D^{\varepsilon}_{i}}\left(u\phi_{t}+f(u)\phi_{x}+(u_{t}+f(u)_{x})\phi\right)\,dx\,dt$$

$$ =\int_{D^{\varepsilon}_{i}}\left((u\phi)_{t}+(f(u)\phi)_{x}\right)\,dx\,dt$$

$$ =\int_{D^{\varepsilon}_{i}}(\partial_{x},\partial_{t})\cdot\big(f(u)\phi,u\phi\big)\,dx\,dt$$

$$ =\int_{\partial D^{\varepsilon}_{i}}\phi\left(f(u),u\right)\cdot n_{i}\,ds. $$

(1.21)

Here n _i is the outward unit normal at $\partial D^{\varepsilon}_{i}$. But ϕ is zero everywhere on $\partial D_{i}^{\varepsilon}$ except in the vicinity of $x(t)$. Let $\Gamma_{i}^{\varepsilon}$ denote this part of $\partial D_{i}^{\varepsilon}$. Then

$$\begin{aligned}\displaystyle\lim_{\varepsilon\to 0}\int_{\Gamma_{i}^{\varepsilon}}\phi\left(f(u),u\right)\cdot n_{i}\,ds&\displaystyle=\int_{I}\phi\left(-u_{l}x^{\prime}(t)+f_{l}\right)\,dt\\ \displaystyle&\displaystyle=-\int_{I}\phi\left(-u_{r}x^{\prime}(t)+f_{r}\right)\,dt\end{aligned}$$

for some suitable time interval I. Here u _l denotes the limit of $u(x,t)$ as $x\to x(t)-$, and u _r the limit as x approaches $x(t)$ from the right, i.e., $u_{r}=\lim_{x\to x(t)+}u(x,t)$. Similarly, $f_{l}=f(u_{l})$ and $f_{r}=f(u_{r})$. The reason for the difference in sign is that according to Green’s theorem, we must integrate along the boundary counterclockwise. Therefore, the positive sign holds for i = 1, and the negative for i = 2. Using (1.20 ) we obtain (slightly abusing notation by writing $u(t)=u(x(t),t)$, etc.)

$$\displaystyle\int_{I}\phi\left[-\left(u_{l}(t)-u_{r}(t)\right)x^{\prime}(t)+\left(f_{l}(t)-f_{r}(t)\right)\right]\,dt=0.$$

Since this is to hold for all test functions ϕ, we must have

$$\displaystyle s\left(u_{r}-u_{l}\right)=f_{r}-f_{l},$$

(1.22)

where $s=x^{\prime}(t)$. This equality is called the Rankine–Hugoniot condition or the jump condition, and it expresses conservation of u across jump discontinuities. It is common in the theory of conservation laws to introduce a notation for the jump in a quantity. Write

$$\displaystyle[\![a]\!]=a_{r}-a_{l}$$

(1.23)

for the jump in any quantity a. With this notation the Rankine–Hugoniot relation takes the form

$$\displaystyle s[\![u]\!]=[\![f]\!].$$

(1.24)

FormalPara ⋄ Example 1.5 (Burgers’s equation (cont’d.))

For Burgers’s equation we see that the shock speed must satisfy

$$\displaystyle s=\frac{[\![u^{2}/2]\!]}{[\![u]\!]}=\frac{\left(u_{r}^{2}-u_{l}^{2}\right)}{2\left(u_{r}-u_{l}\right)}=\frac{1}{2}\left(u_{l}+u_{r}\right).$$

Consequently, the left shock in parts a and b in Fig. 1.2 above will have greater speed than the right shock, and will, eventually, collide. Therefore, solutions of type a or b cannot be isolated discontinuities moving along two trajectories starting at t = 1. Type c yields a stationary shock.  ⋄

FormalPara ⋄ Example 1.6 (Traffic flow)

I am ill at these numbers. — W. Shakespeare, Hamlet (1603)

Rather than continue to develop the theory, we shall now consider an example of a conservation law in some detail. We will try to motivate how a conservation law can model the flow of cars on a crowded highway.

Consider a road consisting of a single lane, with traffic in one direction only. The road is parameterized by a single coordinate x, and we assume that the traffic moves in the direction of increasing x.

Suppose we position ourselves at a point x on the road and observe the number of cars $N=N(x,t,h)$ in the interval $\left[x,x+h\right]$. If some car is located at the boundary of this interval, we account for that by allowing N to take any real value. If the traffic is dense, and if h is large compared with the average length of a car, but at the same time small compared with the length of our road, we can introduce the density given by

$$\displaystyle\rho(x,t)=\lim_{h\to 0}\frac{N(x,t,h)}{h}.$$

Then $N(x,t,h)=\int_{x}^{x+h}\rho(y,t)\,dy$.

Let now the position of some vehicle be given by $r(t)$, and its velocity by $v(r(t),t)$. Considering the interval $\left[a,b\right]$, we wish to determine how the number of cars changes in this interval. Since we have assumed that there are no entries or exits on our road, this number can change only as cars are entering the interval from the left endpoint, or leaving the interval at the right endpoint. The rate of cars passing a point x at some time t is given by

$$\displaystyle v(x,t)\rho(x,t).$$

Consequently,

$$\displaystyle-\left(v(b,t)\rho(b,t)-v(a,t)\rho(a,t)\right)=\frac{d}{dt}\int_{a}^{b}\rho(y,t)\,dy.$$

Comparing this with (1.3) and (1.2), we see that the density satisfies the conservation law

$$\displaystyle\rho_{t}+\left(\rho v\right)_{x}=0.$$

(1.25)

In the simplest case we assume that the velocity v is given as a function of the density ρ only. This may be a good approximation if the road is uniform and does not contain any sharp bends or similar obstacles that force the cars to slow down. It is also reasonable to assume that there is some maximal speed $v_{\text{max}}$ that any car can attain. When traffic is light, a car will drive at this maximum speed, and as the road gets more crowded, the cars will have to slow down, until they come to a complete standstill as the traffic stands bumper to bumper. Hence, we assume that the velocity v is a monotone decreasing function of ρ such that $v(0)=v_{\text{max}}$ and $v\left(\rho_{\text{max}}\right)=0$. The simplest such function is a linear function, resulting in a flux function given by

$$\displaystyle f(\rho)=v\rho=\rho v_{\text{max}}\left(1-\frac{\rho}{\rho_{\text{max}}}\right).$$

(1.26)

For convenience we normalize by introducing $u=\rho/\rho_{\text{max}}$ and $\tilde{x}=v_{\text{max}}x$. The resulting normalized conservation law reads

$$\displaystyle u_{t}+\left(u(1-u)\right)_{x}=0.$$

(1.27)

Setting $\tilde{u}=\frac{1}{2}-u$, we recover Burgers’s equation, but this time with a new interpretation of the solution.

Let us solve an initial value problem explicitly by the method of characteristics described earlier. We wish to solve (1.27), with initial function $u_{0}(x)$ given by

$$\displaystyle u_{0}(x)=u(x,0)=\begin{cases}\frac{3}{4}&\text{for $x\leq-a$,}\\ \frac{1}{2}-x/(4a)&\text{for $-a<x<a$,}\\ \frac{1}{4}&\text{for $a\leq x$.}\end{cases}$$

The characteristics satisfy $t^{\prime}(\xi)=1$ and $x^{\prime}(\xi)=1-2u(x(\xi),t(\xi))$. The solution of these equations is given by $x=x(t)$, where

$$\displaystyle x(t)=\begin{cases}x_{0}-t/2&\text{for $x_{0}<-a$,}\\ x_{0}+x_{0}t/(2a)&\text{for $-a\leq x_{0}\leq a$,}\\ x_{0}+t/2&\text{for $a<x_{0}$.}\end{cases}$$

Inserting this into the solution $u(x,t)=u_{0}\left(x_{0}(x,t)\right)$, we find that

$$\displaystyle u(x,t)=\begin{cases}\frac{3}{4}&\text{for $x\leq-a-t/2$,}\\ \frac{1}{2}-x/(4a+2t)&\text{for $-a-t/2<x<a+t/2$,}\\ \frac{1}{4}&\text{for $a+t/2\leq x$.}\end{cases}$$

This solution models a situation in which the traffic density initially is small for positive x, and high for negative x. If we let a tend to zero, the solution reads

$$\displaystyle u(x,t)=\begin{cases}\frac{3}{4}&\text{for $x\leq-t/2$,}\\ \frac{1}{2}-x/(2t)&\text{for $-t/2<x<t/2$,}\\ \frac{1}{4}&\text{for $t/2\leq x$.}\end{cases}$$

As the reader may check directly, this is also a classical solution everywhere except at $x=\pm t/2$. It takes discontinuous initial values:

$$\displaystyle u(x,0)=\begin{cases}\frac{3}{4}&\text{for $x<0$,}\\ \frac{1}{4}&\text{otherwise.}\end{cases}$$

(1.28)

This initial function may model the situation when a traffic light turns green at t = 0. The density of cars facing the traffic light is high, while on the other side of the light there is a small constant density.

Initial value problems of the kind (1.28 ), where the initial function consists of two constant values, are called Riemann problems. We will discuss Riemann problems at great length in this book.

If we simply insert $u_{l}=\frac{3}{4}$ and $u_{r}=\frac{1}{4}$ in the Rankine–Hugoniot condition (1.22), we find another weak solution to this initial value problem. These left and right values give s = 0, so the solution found here is simply $u_{2}(x,t)=u_{0}(x)$. A priori, this solution is no better or worse than the solution computed earlier. But when we examine the situation the equation is supposed to model, the second solution u ₂ is unsatisfactory, since it describes a situation in which the traffic light is green, but the density of cars facing the traffic light does not decrease!

In the first solution the density decreased. Examining the model a little more closely, we find, perhaps from experience of traffic jams, that the allowable discontinuities are those in which the density is increasing. This corresponds to the situation in which there is a traffic jam ahead, and we suddenly have to slow down when we approach it.

When we emerge from a traffic jam, we experience a gradual decrease in the density of cars around us, not a sudden jump from a bumper to bumper situation to a relatively empty road.

We have now formulated a condition, in addition to the Rankine–Hugoniot condition, that allows us to reduce the number of weak solutions to our conservation law. This condition says that every weak solution u has to increase across discontinuities. Such conditions are often called entropy conditions. This terminology comes from gas dynamics, where similar conditions state that the physical entropy has to increase across any discontinuity.

Let us consider the opposite initial value problem, namely,

$$\displaystyle u_{0}(x)=\begin{cases}\frac{1}{4}&\text{for $x<0$,}\\ \frac{3}{4}&\text{for $x\geq 0$.}\end{cases}$$

Now the characteristics starting at negative x ₀ are given by $x(t)=x_{0}+t/2$, and the characteristics starting on the positive half-line are given by $x(t)=x_{0}-t/2$. We see that these characteristics immediately will run into each other, and therefore the solution is multivalued for every positive time t. Thus there is no hope of finding a continuous solution to this initial value problem for any time interval $(0,\delta)$, no matter how small δ is. When inserting the initial values $u_{l}=\frac{1}{4}$ and $u_{r}=\frac{3}{4}$ into the Rankine–Hugoniot condition, we see that the initial function is already a weak solution. This time, the solution increases across the discontinuity, and therefore satisfies our entropy condition. Thus, an admissible solution is given by $u(x,t)=u_{0}(x)$.

Now we shall attempt to solve a more complicated problem in some detail. Assume that we have a road with a uniform density of cars initially. At t = 0 a traffic light placed at x = 0 changes from green to red. It remains red for some time interval $\Updelta t$, then turns green again and stays green thereafter. We assume that the initial uniform density is given by $u=\frac{1}{2}$, and we wish to determine the traffic density for t > 0.

When the traffic light initially turns red, the situation for the cars to the left of the traffic light is the same as when the cars stand bumper to bumper to the right of the traffic light. So in order to determine the situation for t in the interval $[0,\Updelta t)$, we must solve the Riemann problem with the initial function

$$\displaystyle u^{l}_{0}(x)=\begin{cases}\frac{1}{2}&\text{for $x<0$,}\\ 1&\text{for $x\geq 0$.}\end{cases}$$

(1.29)

For the cars to the right of the traffic light, the situation is similar to the situation in which the traffic abruptly stopped at t = 0 behind the car located at x = 0. Therefore, to determine the density for x > 0 we have to solve the Riemann problem given by

$$\displaystyle u^{r}_{0}(x)=\begin{cases}0&\text{for $x<0$,}\\ \frac{1}{2}&\text{for $x\geq 0$.}\end{cases}$$

(1.30)

Returning to (1.29), here u is increasing over the initial discontinuity, so we can try to insert this into the Rankine–Hugoniot condition. This gives

$$\displaystyle s=\frac{f_{r}-f_{l}}{u_{r}-u_{l}}=\frac{\frac{1}{4}-0}{\frac{1}{2}-1}=-\frac{1}{2}.$$

Therefore, an admissible solution for x < 0 and t in the interval $[0,\Updelta t)$ is given by

$$\displaystyle u^{l}(x,t)=\begin{cases}\frac{1}{2}&\text{for $x<-t/2$,}\\ 1&\text{for $x\geq-t/2$.}\end{cases}$$

This is indeed close to what we experience when we encounter a traffic light. We see the discontinuity approaching as the brake lights come on in front of us, and the discontinuity has passed us when we have come to a halt. Note that although each car moves only in the positive direction, the discontinuity moves to the left.

In general, we have to deal with three different speeds when we study conservation laws: the particle speed, in our case the speed of each car; the characteristic speed; and the speed of a discontinuity. These three speeds are not equal if the conservation law is nonlinear. In our case, the speed of each car is nonnegative, but both the characteristic speed and the speed of a discontinuity may take both positive and negative values. Note that the speed of an admissible discontinuity is less than the characteristic speed to the left of the discontinuity, and larger than the characteristic speed to the right. This is a general feature of admissible discontinuities.

It remains to determine the density for positive x. The initial function given by (1.30 ) also has a positive jump discontinuity, so we obtain an admissible solution if we insert it into the Rankine–Hugoniot condition. Then we obtain $s=\frac{1}{2}$, so the solution for positive x is

$$\displaystyle u^{r}(x,t)=\begin{cases}0&\text{for $x<t/2$,}\\ \frac{1}{2}&\text{for $x\geq t/2$.}\end{cases}$$

Piecing together u ^l and u ^r, we find that the density u in the time interval $[0,\Updelta t)$ reads

$$\displaystyle u(x,t)=\begin{cases}\frac{1}{2}&\text{for $x\leq-t/2$,}\\ 1&\text{for $-t/2<x\leq 0$,}\\ 0&\text{for $0<x\leq t/2$,}\\ \frac{1}{2}&\text{for $t/2<x$,}\end{cases}\qquad t\in[0,\Updelta t).$$

What happens for $t> \Updelta t$? To find out, we have to solve the Riemann problem

$$\displaystyle u(x,\Updelta t)=\begin{cases}1&\text{for $x<0$,}\\ 0&\text{for $x\geq 0$.}\end{cases}$$

Now the initial discontinuity is not acceptable according to our entropy condition, so we have to look for some other solution. We can try to mimic the example above in which we started with a nonincreasing initial function that was linear on some small interval $(-a,a)$. Therefore, let $v(x,t)$ be the solution of the initial value problem

$$\begin{aligned}\displaystyle v_{t}+\left(v(1-v)\right)_{x}&\displaystyle=0,\\ \displaystyle v(x,0)&\displaystyle=v_{0}(x)=\begin{cases}1&\text{for $x<-a$,}\\ \frac{1}{2}-x/(2a)&\text{for $-a\leq x<a$,}\\ 0&\text{for $a\leq x$.}\end{cases}\end{aligned}$$

As in the above example, we find that the characteristics are not overlapping, and they fill out the positive half-plane exactly. The solution is given by $v(x,t)=v_{0}\left(x_{0}(x,t)\right)$:

$$\displaystyle v(x,t)=\begin{cases}1&\text{for $x<-a-t$,}\\ \frac{1}{2}-x/(2a+2t)&\text{for $-a-t\leq x<a+t$,}\\ 0&\text{for $a+t\leq x$.}\end{cases}$$

Letting $a\to 0$, we obtain the solution to the Riemann problem with a left value 1 and a right value 0. For simplicity we also denote this function by $v(x,t)$.

This type of solution can be depicted as a ‘‘fan’’ of characteristics emanating from the origin, and it is called a centered rarefaction wave, or sometimes just a rarefaction wave. The origin of this terminology lies in gas dynamics.

We see that the rarefaction wave, which is centered at $(0,\Updelta t)$, does not immediately influence the solution away from the origin. The leftmost part of the wave moves with a speed −1, and the front of the wave moves with speed 1. So for some time after $\Updelta t$, the density is obtained by piecing together three solutions, $u^{l}(x,t)$, $v(x,t-\Updelta t)$, and $u^{r}(x,t)$.

The rarefaction wave will of course catch up with the discontinuities in the solutions u ^l and u ^r. Since the speeds of the discontinuities are $\mp\frac{1}{2}$, and the speeds of the rear and the front of the rarefaction wave are ∓1, and the rarefaction wave starts at $(0,\Updelta t)$, we conclude that this will happen at $(\mp\Updelta t,2\Updelta t)$.

It remains to compute the solution for $t> 2\Updelta t$. Let us start with examining what happens for positive x. Since the u values that are transported along the characteristics in the rarefaction wave are less than $\frac{1}{2}$, we can construct an admissible discontinuity using the Rankine–Hugoniot condition (1.22). Define a function that has a discontinuity moving along a path $x(t)$. The value to the right of the discontinuity is $\frac{1}{2}$, and the value to the left is determined by $v(x,t-\Updelta t)$. Inserting this into (1.22), we get

$$\displaystyle x^{\prime}(t)=s=\frac{\frac{1}{4}-\left(\frac{1}{2}+\frac{x}{2(t-\Updelta t)}\right)\left(\frac{1}{2}-\frac{x}{2(t-\Updelta t)}\right)}{\frac{1}{2}-\left(\frac{1}{2}-\frac{x}{2(t-\Updelta t)}\right)}=\frac{x}{2(t-\Updelta t)}.$$

Since $x(2\Updelta t)=\Updelta t$, this differential equation has solution

$$\displaystyle x_{+}(t)=\sqrt{\Updelta t(t-\Updelta t)}.$$

The situation is similar for negative x. Here, we use the fact that the u values in the left part of the rarefaction fan are larger than $\frac{1}{2}$. This gives a discontinuity with a left value $\frac{1}{2}$ and right values taken from the rarefaction wave. The path of this discontinuity is found to be $x_{-}(t)=-x_{+}(t)$.

Now we have indeed found a solution that is valid for all positive time. This function has the property that it is a classical solution at all points x and t where it is differentiable, and it satisfies both the Rankine–Hugoniot condition and the entropy condition at points of discontinuity. We show this weak solution in Fig. 1.4 , both in the $(x,t)$-plane, where we show characteristics and discontinuities, and u as a function of x for various times. The characteristics are shown as gray lines, and the discontinuities as thicker black lines. This concludes our example. Note that we have been able to find the solution to a complicated initial value problem by piecing together solutions from Riemann problems. This is indeed the main idea behind front tracking, and a theme to which we shall give considerable attention in this book.  ⋄

Fig. 1.4

A traffic light on a single road. To the left we show the solution in $(x,t)$, and to the right the solution $u(x,t)$ at three different times t

1.1 Linear Equations

I don’t make unconventional stories; I don’t make nonlinear stories. I like linear storytelling a lot. — Steven Spielberg

We now make a pause in the exposition of nonlinear hyperbolic conservation laws and take a brief look at linear transport equations. Many of the methods and concepts introduced later in the book are much simpler if the equations are linear.

Let $u\in\mathbb{R}$ be an unknown scalar function of $x\in\mathbb{R}$ and $t\in[0,\infty)$ satisfying the Cauchy problem

$$\displaystyle\begin{cases}u_{t}+a\leavevmode\nobreak\ u_{x}=0,&x\in\mathbb{R},\ \ t> 0,\\ u(x,0)=u_{0}(x),\end{cases}$$

(1.31)

where a is a given (positive) constant, and u ₀ is a known function. Recall the theory of characteristics. Since this case is particularly simple, we can use t as a parameter, and we will here use $(t,x_{0})$ rather than $(s,y)$ as parameters. Thus the characteristics $x=\xi(t;x_{0})$ are defined as

$$\displaystyle\frac{d}{dt}\xi(t;x_{0})=a,\ \ \ \xi(0;x_{0})=x_{0},$$

with solution

$$\displaystyle\xi(t;x_{0})=at+x_{0}.$$

We know that $\tfrac{d}{dt}u\left(\xi(t;x_{0}),t\right)=0$, and thus $u(\xi(t;x_{0}),t)=u(\xi(0;x_{0}),0)=u(x_{0},0)=u_{0}(x_{0})$. We can use the solution of ξ to write

$$\displaystyle u(at+x_{0},t)=u_{0}(x_{0}).$$

If we set $x=at+x_{0}$, i.e., $x_{0}=x-at$, we get the solution formula

$$\displaystyle u(x,t)=u_{0}(x-at).$$

Thus (1.31) expresses that the initial function u ₀ is transported with a constant velocity a.

The same reasoning works if now $a=a(x,t)$, where the map $x\mapsto a(x,t)$ is Lipschitz continuous for all t. In this case let $u=u(x,t)$ satisfy the Cauchy problem

$$\displaystyle\begin{cases}u_{t}+a(x,t)u_{x}=0,\ \ x\in\mathbb{R},\ \ t> 0,\\ u(x,0)=u_{0}(x).\end{cases}$$

(1.32)

First we observe that this equation is not conservative, and the interpretation of $a(x,t)u$ is not the flux of u across a point. Now let $\xi(t;x_{0})$ denote the unique solution of the ordinary differential equation

$$\displaystyle\frac{d}{dt}\xi(t;x_{0})=a(\xi(t;x_{0}),t),\ \ \xi(0;x_{0})=x_{0}.$$

(1.33)

By the chain rule we also now find that

$$\displaystyle\frac{d}{dt}u\left(\xi(t;x_{0}),t\right)=\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}\frac{d}{dt}\xi(t;x_{0})=u_{t}(\xi,t)+a(\xi,t)u_{x}(\xi,t)=0.$$

Therefore $u(\xi(t;x_{0}),t)=u_{0}(x_{0})$. In order to get a solution formula, we must solve $x=\xi(t;x_{0})$ in terms of x ₀, or equivalently, find a function $\zeta(\tau;x)$ that solves the backward characteristic equation,

$$\displaystyle\frac{d}{d\tau}\zeta(\tau;x)=-a(\zeta(\tau;x),t-\tau),\ \ \zeta(0;x)=x.$$

(1.34)

Then

$$\displaystyle\frac{d}{d\tau}u(\zeta(\tau;x),t-\tau)=0,$$

which means that $u(x,t)=u(\zeta(0;x),t)=u(\zeta(t;x),0)=u_{0}(\zeta(t;x))$.

⋄ Example 1.7

Let us study the simple example with $a(x,t)=x$. Thus

$$\displaystyle u_{t}+xu_{x}=0,\ \ u(x,0)=u_{0}(x).$$

Then the characteristic equation is

$$\displaystyle\frac{d}{dt}\xi=\xi,\ \ \xi(0)=x_{0},$$

with solution

$$\displaystyle\xi(t;x_{0})=x_{0}e^{t}.$$

Solving $\xi(t;x_{0})=x$ in terms of x ₀ gives $x_{0}=xe^{-t}$, and thus

$$\displaystyle u(x,t)=u_{0}\left(xe^{-t}\right).$$

 ⋄

⋄ Example 1.8

Let us look at another example:

$$\displaystyle a(x)=\begin{cases}0&x<0,\\ x&0\leq x\leq 1,\\ 1&1<x.\end{cases}$$

(1.35)

In this case the characteristics are straight lines $\xi(t;x_{0})=x_{0}$ if $x_{0}\leq 0$, and $\xi(t;x_{0})=x_{0}+t$ if $x_{0}\geq 1$. Finally, whenever $0<x_{0}<1$, the characteristics are given by

$$\displaystyle\xi(t;x_{0})=\begin{cases}x_{0}e^{t}&t\leq-\ln(x_{0}),\\ 1+t+\ln(x_{0})&t> -\ln(x_{0}).\end{cases}$$

See Fig. 1.5 for a picture of this. In this case a is increasing in x, and therefore the characteristics are no closer than they were initially. Since u is constant along characteristics, this means that

$$\displaystyle\max_{x}\left|u_{x}(x,t)\right|\leq\max_{x}\left|u_{0}^{\prime}(x)\right|.$$

If a is decreasing, such a bound cannot be found, as the next example shows.  ⋄

Fig. 1.5

Characteristics in the $(x,t)$-plane for (1.35)

⋄ Example 1.9

Let now

$$\displaystyle a(x)=\begin{cases}1&x<0,\\ 1-x&0\leq x\leq 1,\\ 0&1<x.\end{cases}$$

(1.36)

In this case the characteristics are given by

$$\displaystyle\xi(t;x_{0})=\begin{cases}\begin{cases}x_{0}+t,&t<-x_{0},\\ 1-e^{-(t+x_{0})}&t\geq-x_{0},\end{cases}&x_{0}<0,\\ 1-(1-x_{0})e^{-t}&0\leq x_{0}<1,\\ x_{0}&1\leq x_{0}.\end{cases}$$

See Fig. 1.6 for an illustration of these characteristics. Let now x ₀ be in the interval $(0,1)$, and assume that u ₀ is continuously differentiable. Since u is constant along characteristics, $u(\,\cdot\,,t)$ is also continuously differentiable for all t > 0. Thus

$$\displaystyle u^{\prime}_{0}(x_{0})=\frac{\partial}{\partial x_{0}}u(\xi(t;x_{0}),t)=u_{x}(\xi(t;x_{0}),t)\frac{\partial\xi}{\partial x_{0}},$$

which, when $x_{0}\in(0,1)$, implies that $u_{x}(x,t)=u^{\prime}_{0}(x_{0})e^{t}$ for $x=\xi(t;x_{0})$. From this we see that the only bound on the derivative that we can hope for is of the type

$$\displaystyle\max_{x}\left|u_{x}(x,t)\right|\leq e^{t}\max_{x}\left|u_{0}^{\prime}(x)\right|.$$

 ⋄

Fig. 1.6

The characteristics for (1.36)

1.1.1 Numerics (I)

If we (pretend that we) do not have the characteristics, and still want to know the solution, we can try to approximate it by some numerical method.

To this end we introduce approximations to the first spatial derivative

$$\begin{aligned}\displaystyle D_{-}u(x)&\displaystyle=\frac{u(x)-u(x-{\Updelta x})}{{\Updelta x}},\\ \displaystyle D_{+}u(x)&\displaystyle=\frac{u(x+{\Updelta x})-u(x)}{{\Updelta x}},\ \ \text{and }\\ \displaystyle D_{0}u(x)&\displaystyle=\frac{u(x+{\Updelta x})-u(x-{\Updelta x})}{2{\Updelta x}},\end{aligned}$$

where ${\Updelta x}$ is a small positive number. When we deal with numerical approximations, we shall always use the notation $u_{j}(t)$ to indicate an approximation to $u(j{\Updelta x},t)$ for some integer j. We also use the notation

$$\displaystyle x_{j}=j{\Updelta x},\ \ x_{j\pm 1/2}=\Big(j\pm\frac{1}{2}\Big){\Updelta x}=x_{j}\pm\frac{{\Updelta x}}{2}.$$

Now consider the case in which a is a positive constant. As a semidiscrete numerical scheme for (1.31) we propose to let u _j solve the (infinite) system of ordinary differential equations

$$\displaystyle u_{j}^{\prime}(t)+aD_{-}u_{j}(t)=0,\ \ u_{j}(0)=u_{0}(x_{j}).$$

(1.37)

We need to define an approximation to $u(x,t)$ for every x and t, and we do this by linear interpolation:

$$\displaystyle u_{{\Updelta x}}(x,t)=u_{j}(t)+\left(x-x_{j}\right)D_{-}u_{j+1}(t),\ \ \text{ for $x\in[x_{j},x_{j+1})$.}$$

(1.38)

We want to show that (a) $u_{\Updelta x}$ converges to some function u as ${\Updelta x}\to 0$, and (b) the limit u solves the equation.

If u ₀ is continuously differentiable, we know that a solution to (1.31) exists (and we can find it by the method of characteristics). Since the equation is linear, we can easily study the error $e_{\Updelta x}(x,t)=u(x,t)-u_{\Updelta x}(x,t)$. In the calculation that follows, we use the following properties:

$$\displaystyle\text{$D_{+}u_{j}-D_{-}u_{j}={\Updelta x}D_{+}D_{-}u_{j}$ and $D_{-}u_{j+1}=D_{+}u_{j}$.}$$

Inserting the error term $e_{\Updelta x}$ into the equation, we obtain for $x\in(x_{j},x_{j+1})$,

$$\begin{aligned}\displaystyle\frac{\partial}{\partial t}e_{\Updelta x}+a\frac{\partial}{\partial x}e_{\Updelta x}&\displaystyle=-\frac{\partial}{\partial t}u_{\Updelta x}-a\frac{\partial}{\partial x}u_{\Updelta x}\\ \displaystyle&\displaystyle=-\frac{d}{dt}\left[u_{j}(t)+\left(x-x_{j}\right)D_{-}u_{j+1}(t)\right]-aD_{-}u_{j+1}(t)\\ \displaystyle&\displaystyle=-u_{j}^{\prime}(t)-\left(x-x_{j}\right)D_{-}u_{j+1}^{\prime}(t)-aD_{+}u_{j}(t)\\ \displaystyle&\displaystyle=aD_{-}u_{j}(t)-aD_{+}u_{j}(t)+a\left(x-x_{j}\right)D_{-}D_{-}u_{j+1}(t)\\ \displaystyle&\displaystyle=-a{\Updelta x}D_{+}D_{-}u_{j}(t)+a\left(x-x_{j}\right)D_{+}D_{-}u_{j}(t)\\ \displaystyle&\displaystyle=a\left(\left(x-x_{j}\right)-{\Updelta x}\right)D_{+}D_{-}u_{j}(t).\end{aligned}$$

Next let $f_{\Updelta x}$ be defined by

$$\displaystyle f_{\Updelta x}(x,t)=a\left(\left(x-x_{j}\right)-{\Updelta x}\right)D_{+}D_{-}u_{j}(t)\ \text{ for $x\in[x_{j},x_{j+1})$,}$$

so that

$$\displaystyle\left(e_{\Updelta x}\right)_{t}+a\left(e_{\Updelta x}\right)_{x}=f_{\Updelta x}.$$

(1.39)

Using the method of characteristics on this equation gives (see Exercise 1.3)

$$\displaystyle e_{\Updelta x}(x,t)=e_{\Updelta x}(x-at,0)+\int_{0}^{t}f_{\Updelta x}(x-a(t-s),s)\,ds.$$

(1.40)

(Here we tacitly assume uniqueness of the solution.) Hence we get the bound

$$\displaystyle\left|e_{\Updelta x}(x,t)\right|\leq\sup_{x}\left|e_{\Updelta x}(x,0)\right|+t\left\|f_{\Updelta x}\right\|_{L^{\infty}(\mathbb{R}\times[0,t])}.$$

(1.41)

In trying to bound $f_{\Updelta x}$, note first that

$$\displaystyle\left|f_{\Updelta x}(x,t)\right|\leq{\Updelta x}\,a\leavevmode\nobreak\ \left|D_{-}D_{+}u_{j}(t)\right|,$$

so $f_{\Updelta x}$ tends to zero with ${\Updelta x}$ if $D_{-}D_{+}u_{j}$ is bounded. Writing $w_{j}=D_{-}D_{+}u_{j}$ and applying $D_{-}D_{+}$ to (1.37), we get

$$\displaystyle w_{j}^{\prime}(t)+aD_{-}\left(w_{j}\right)=0,\ \ w_{j}(0)=D_{-}D_{+}u_{0}(x).$$

Now it is time to use the fact that a > 0. To bound w _j , observe that if $w_{j}\leq w_{j-1}$, then $D_{-}w_{j}\leq 0$. Hence, if $w_{j}(t)\leq w_{j-1}(t)$, then

$$\displaystyle\frac{d}{dt}w_{j}(t)=-aD_{-}w_{j}(t)\geq 0.$$

Similarly, if for some t, $w_{j}(t)\geq w_{j-1}(t)$, then $w_{j}^{\prime}(t)\leq 0$. This means that

$$\displaystyle\inf_{x}u_{0}^{\prime\prime}(x)\leq\inf_{k}D_{-}D_{+}u_{k}(0)\leq w_{j}(t)\leq\sup_{k}D_{-}D_{+}u_{k}(0)\leq\sup_{x}u_{0}^{\prime\prime}(x).$$

Thus w _j is bounded if $u_{0}^{\prime}$ is Lipschitz continuous. Note that it is the choice of the difference scheme (1.37) (choosing $D_{-}$ instead of $D_{+}$ or D) that allows us to conclude that we have a bounded approximation. It remains to study $e_{\Updelta x}(x,0)$. For $x\in[x_{j},x_{j+1})$,

$$\begin{aligned}\displaystyle\left|e_{\Updelta x}(x,0)\right|&\displaystyle=\Bigl|u_{0}(x)-u_{0}(x_{j})-\frac{x-x_{j}}{{\Updelta x}}\left(u_{0}(x_{j+1})-u_{0}(x_{j})\right)\Bigr|\\ \displaystyle&\displaystyle\leq 2{\Updelta x}\max_{x\in[x_{j},x_{j+1}]}\left|u_{0}^{\prime}(x)\right|.\end{aligned}$$

Then we have proved the bound

$$\displaystyle\left|u_{{\Updelta x}}(x,t)-u(x,t)\right|\leq{\Updelta x}\left(2\left\|u_{0}^{\prime}\right\|_{L^{\infty}(\mathbb{R})}+t\,a\leavevmode\nobreak\ \left\|u^{\prime\prime}_{0}\right\|_{L^{\infty}(\mathbb{R})}\right),$$

(1.42)

for all x and t > 0.

Strictly speaking, in order for this argument to be valid, we have implicitly assumed in (1.40) that equation (1.39) has only the solution (1.40). This brings us to another topic.

1.1.2 Entropy Solutions (I)

You should call it entropy …[since] … no one knows what entropy really is, so in a debate you will always have the advantage. — John von Neumann ³

Without much extra effort, we can generalize slightly, and we want to ensure that the equation

$$\displaystyle u_{t}+a(x,t)u_{x}=f(x,t)$$

(1.43)

has only one differentiable solution. If we let the characteristic curves be defined by (1.33), a solution is given by (see Exercise 1.3)

$$\displaystyle u\left(\xi(t;x_{0}),t\right)=u_{0}(x_{0})+\int_{0}^{t}f(\xi(s;x_{0}),s)\,ds.$$

In terms of the inverse characteristic ζ defined by (1.34) this formula reads (see Exercise 1.3)

$$\displaystyle u(x,t)=u_{0}\left(\zeta(t;x)\right)+\int_{0}^{t}f\left(\zeta(\tau;x),t-\tau\right)\,d\tau.$$

If u ₀ is differentiable and f is bounded, this formula gives a differentiable function $u(x,t)$.

Now we can turn to the uniqueness question. Since (1.43) is linear, to prove uniqueness means to show that the equation with f = 0 and $u_{0}=0$ has only the zero solution. Therefore, we consider

$$\displaystyle u_{t}+a(x,t)u_{x}=0.$$

Now let $\eta(u)$ be a differentiable function, and multiply the above by $\eta^{\prime}(u)$ to get

$$\displaystyle 0=\frac{\partial}{\partial t}\eta(u)+a\frac{\partial}{\partial x}\eta(u)=\eta(u)_{t}+(a\eta(u))_{x}-a_{x}\eta(u).$$

Assume that $\eta(0)=0$ and $\eta(u)> 0$ for $u\neq 0$, and that $\left|a_{x}(x,t)\right|<C$ for all x and t. If $\eta(u(\,\cdot\,,t))$ is integrable, then we can integrate this to get

$$\displaystyle\frac{d}{dt}\int_{\mathbb{R}}\eta(u(x,t))\,dx=\int_{\mathbb{R}}a_{x}(x,t)\eta(u(x,t))\,dx\leq C\int_{\mathbb{R}}\eta(u(x,t))\,dx.$$

By Gronwall’s inequality (see Exercise 1.10),

$$\displaystyle\int_{\mathbb{R}}\eta(u(x,t))\,dx\leq e^{Ct}\int_{\mathbb{R}}\eta(u_{0}(x))\,dx.$$

If $u_{0}=0$, then $\eta(u_{0})=0$, and we must have $u(x,t)=0$ as well. We have shown that if $\eta(u_{0})$ is integrable for some differentiable function η with $\eta(0)=0$ and $\eta(u)> 0$ for $u\neq 0$, and a _x is bounded, then (1.43) has only one differentiable solution.

Frequently, the model (1.43) (with f identically zero) is obtained by the limit of a physically more realistic model,

$$\displaystyle u^{\varepsilon}_{t}+a(x,t)u^{\varepsilon}_{x}=\varepsilon u^{\varepsilon}_{xx}$$

(1.44)

as $\varepsilon$ becomes small. You can think of $u^{\varepsilon}$ as the temperature in a long rod moving with speed a. In this case $\varepsilon$ is proportional to the heat conductivity of the rod. Equation (1.44) has more regular solutions than the initial data u ₀ (see Appendix B). If we multiply this equation by $\eta^{\prime}(u^{\varepsilon})$, where $\eta\in C^{2}(\mathbb{R})$ is a convex function, we get

$$\displaystyle\eta\left(u^{\varepsilon}\right)_{t}+a\leavevmode\nobreak\ \eta\left(u^{\varepsilon}\right)_{x}=\varepsilon\left(\eta^{\prime}\left(u^{\varepsilon}\right)u^{\varepsilon}_{x}\right)_{x}-\varepsilon\eta^{\prime\prime}\left(u^{\varepsilon}\right)\left(u^{\varepsilon}_{x}\right)^{2}.$$

The function η is often called an entropy. The term with $(u^{\varepsilon}_{x})^{2}$ is problematic when $\varepsilon\to 0$, since the derivative will not be square integrable in this limit. For linear equations the integrability of this term depends on the integrability of this term initially. However, for nonlinear equations, we have seen that jumps can form independently of the smoothness of the initial data, and the limit of $u^{\varepsilon}_{x}$ will in general not be square integrable.

The key to circumventing this problem is to use the convexity of η, that is, $\eta^{\prime\prime}(u)\geq 0$, and hence $\varepsilon\eta^{\prime\prime}\left(u^{\varepsilon}\right)\left(u^{\varepsilon}_{x}\right)^{2}$ is nonnegative, to replace this term by the appropriate inequality. Thus we get that

$$\displaystyle\eta\left(u^{\varepsilon}\right)_{t}+\left(a\eta\left(u^{\varepsilon}\right)\right)_{x}-a_{x}\eta\left(u^{\varepsilon}\right)\leq\varepsilon\left(\eta^{\prime}\left(u^{\varepsilon}\right)u^{\varepsilon}_{x}\right)_{x}.$$

(1.45)

Now the right-hand side of (1.45) converges to zero weakly.⁴ We define an entropy solution to be the limit $u=\lim_{\varepsilon\to 0}u^{\varepsilon}$ of solutions to (1.44) as $\varepsilon\to 0$. Formally, an entropy solution to (1.43) should satisfy (reintroducing the function f)

$$\displaystyle\eta(u)_{t}+(a\eta(u))_{x}-a_{x}\eta(u)\leq\eta^{\prime}(u)f(x,t),$$

(1.46)

for all convex functions $\eta\in C^{2}(\mathbb{R})$. We shall see later that this is sufficient to establish uniqueness even if u is not assumed to be differentiable.

1.1.3 Numerics (II)

Let us for the moment return to the transport equation

$$\displaystyle u_{t}+a(x,t)u_{x}=0.$$

(1.47)

We want to construct a fully discrete scheme for this equation, and the simplest such scheme is the explicit Euler scheme,

$$\displaystyle D_{\!+}^{t}u^{n}_{j}+a^{n}_{j}D_{-}u^{n}_{j}=0,\ \ n\geq 0,$$

(1.48)

and $u^{0}_{j}=u_{0}(x_{j})$. Here $D_{\!+}^{t}$ denotes the discrete forward time difference

$$\displaystyle D_{\!+}^{t}u(t)=\frac{u(t+{\Updelta t})-u(t)}{{\Updelta t}},$$

and $u^{n}_{j}$ is an approximation of $u(x_{j},t_{n})$, with $t_{n}=n{\Updelta t}$, $n\geq 0$. Furthermore, $a^{n}_{j}$ denotes some approximation of $a(x_{j},t_{n})$, to be determined later. We can rewrite (1.48) as

$$\displaystyle u^{n+1}_{j}=u^{n}_{j}-a^{n}_{j}\lambda\left(u^{n}_{j}-u^{n}_{j-1}\right),$$

where $\lambda={\Updelta t}/{\Updelta x}$.⁵

Let us first return to the case that a is constant. We can then use von Neumann stability analysis. Assume that the scheme produces approximations that converge to a bounded solution for almost all x and t; in particular, assume that $u^{n}_{j}$ is bounded independently of ${\Updelta x}$ and ${\Updelta t}$. Consider the periodic case. We make the ansatz that $u^{n}_{j}=\alpha^{n}e^{ij{\Updelta x}}$ with $i=\sqrt{-1}$ (the equation is linear, so we might as well expand the solution in a Fourier series). Inserting this into the equation for $u^{n+1}_{j}$, we get

$$\begin{aligned}\displaystyle\alpha^{n+1}e^{ij{\Updelta x}}&\displaystyle=\alpha^{n}e^{ij{\Updelta x}}-\lambda a\left(\alpha^{n}e^{ij{\Updelta x}}-\alpha^{n}e^{i(j-1){\Updelta x}}\right)\\ \displaystyle&\displaystyle=\alpha^{n}e^{ij{\Updelta x}}\big(1-\lambda a(1-e^{-i{\Updelta x}})\big),\end{aligned}$$

so that

$$\displaystyle\alpha=1-\lambda a\big(1-\cos({\Updelta x})+i\sin({\Updelta x})\big).$$

If $\left|\alpha\right|\leq 1$, then the sup-norm estimate will hold also for the solution generated by the scheme. In this case the scheme is called von Neumann stable.

We calculate

$$\begin{aligned}\displaystyle\left|\alpha\right|^{2}&\displaystyle=1+2\lambda^{2}a^{2}-2\lambda a(1+(1-\lambda a)\cos({\Updelta x}))\\ \displaystyle&\displaystyle=1-2a\lambda\left(1-a\lambda\right)\left(1-\cos({\Updelta x})\right).\end{aligned}$$

This is less than or equal to 1 if and only if $a\lambda(1-a\lambda)\geq 0$. Thus we require

$$\displaystyle 0\leq\lambda a\leavevmode\nobreak\ \leq 1.$$

(1.49)

This relationship between the spatial and temporal discretization (as measured by λ) and the wave speed given by a is the simplest example of the celebrated CFL condition, named after Courant–Friedrichs–Lewy. We will return to the CFL condition repeatedly throughout the book.

Returning to the scheme for the transport equation with variable and nonnegative speed, we say that the scheme will be von Neumann stable if

$$\displaystyle\lambda\max_{(x,t)}a(x,t)\leq 1.$$

(1.50)

Consider now the scheme (1.48) with

$$\displaystyle a^{n}_{j}=\frac{1}{{\Updelta t}}\int_{t_{n}}^{t_{n+1}}a(x_{j},t)\,dt.$$

We wish to establish the convergence of $u^{n}_{j}$. To this end, set

$$\displaystyle e^{n}_{j}=u(x_{j},t_{n})-u^{n}_{j},$$

where u is the unique solution to (1.47). Inserting this into the scheme, we find that

$$\begin{aligned}\displaystyle D_{\!+}^{t}e^{n}_{j}+a^{n}_{j}D_{-}e^{n}_{j}&\displaystyle=D_{\!+}^{t}u(x_{j},t_{n})+a^{n}_{j}D_{-}u(x_{j},t_{n})\\ \displaystyle&\displaystyle=\frac{1}{{\Updelta t}}\!\int_{t_{n}}^{t_{n+1}}\!u_{t}(x_{j},t)\,dt+\frac{a^{n}_{j}}{{\Updelta x}}\!\int_{x_{j-1}}^{x_{j}}\!u_{x}(x,t_{n})\,dx\\ \displaystyle&\displaystyle=\frac{1}{{\Updelta x}{\Updelta t}}\!\int_{t_{n}}^{t_{n+1}}\!\int_{x_{j-1}}^{x_{j}}\!\big(u_{t}(x_{j},t)+a(x_{j},t)u_{x}(x,t_{n})\big)\,dx\,dt\\ \displaystyle&\displaystyle=\frac{1}{{\Updelta x}{\Updelta t}}\!\int_{x_{j-1}}^{x_{j}}\!\int_{t_{n}}^{t_{n+1}}\!a(x_{j},t)\left(u_{x}(x,t_{n})-u_{x}(x_{j},t)\right)\,dt\,dx\\ \displaystyle&\displaystyle=\frac{1}{{\Updelta x}{\Updelta t}}\!\int_{x_{j-1}}^{x_{j}}\!\int_{t_{n}}^{t_{n+1}}\!a(x_{j},t)\Bigl(\int^{x}_{x_{j}}\!u_{xx}(z,t_{n})\kern 1.0ptdz\kern-1.0pt-\kern-0.5pt\!\int_{t_{n}}^{t}\!u_{xt}(x_{j},s)\kern 1.0ptds\!\Bigr)dt\kern 1.0ptdx\\ \displaystyle&\displaystyle=:R^{n}_{j}.\end{aligned}$$

Assuming now that u _xx and u _tx are bounded, which they will be if we consider a finite time interval $[0,T]$, choose M such that $\max\{\|u_{xx}\|_{L^{\infty}},\|u_{tx}\|_{L^{\infty}},\|a\|_{L^{\infty}}\}\leq M$. Then we get the bound

$$\begin{aligned}\displaystyle\left|R^{n}_{j}\right|&\displaystyle\leq\frac{M^{2}}{{\Updelta x}{\Updelta t}}\int_{x_{j-1}}^{x_{j}}\int_{t_{n}}^{t_{n+1}}\big(\left(x_{j}-x\right)+(t-t_{n})\big)\,dt\,dx\\ \displaystyle&\displaystyle=\frac{M^{2}}{2}\left({\Updelta x}+{\Updelta t}\right).\end{aligned}$$

Therefore the error will satisfy the inequality

$$\displaystyle e^{n+1}_{j}\leq{}\underbrace{e^{n}_{j}\left(1-\lambda a^{n}_{j}\right)+\lambda a^{n}_{j}e^{n}_{j-1}}_{\Upsilon}{}+{\Updelta t}\frac{M^{2}}{2}\left({\Updelta x}+{\Updelta t}\right).$$

If $\left\|a\right\|_{L^{\infty}}\lambda<1$ (recall the CFL condition), then $\Upsilon$ is a convex combination of $e^{n}_{j}$ and $e^{n}_{j-1}$, which is less than or equal to $\max\left\{e^{n}_{j},e^{n}_{j-1}\right\}$. Taking the supremum over j, first on the right, and then on the left, we get

$$\displaystyle\sup_{j}\left\{e^{n+1}_{j}\right\}\leq\sup_{j}\left\{e^{n}_{j}\right\}+{\Updelta t}\frac{M^{2}}{2}\left({\Updelta x}+{\Updelta t}\right).$$

We also have that

$$\displaystyle e^{n+1}_{j}\geq{}\underbrace{e^{n}_{j}\left(1-\lambda a^{n}_{j}\right)+\lambda a^{n}_{j}e^{n}_{j-1}}_{\Upsilon}{}-{\Updelta t}\frac{M^{2}}{2}\left({\Updelta x}+{\Updelta t}\right),$$

which implies that

$$\displaystyle\inf_{j}\left\{e^{n+1}_{j}\right\}\geq\inf_{j}\left\{e^{n}_{j}\right\}-{\Updelta t}\frac{M^{2}}{2}\left({\Updelta x}+{\Updelta t}\right).$$

With $\bar{e}^{n}=\sup_{j}\left|e^{n}_{j}\right|$, the above means that

$$\displaystyle\bar{e}^{n+1}\leq\bar{e}^{n}+{\Updelta t}\frac{M^{2}}{2}\left({\Updelta x}+{\Updelta t}\right).$$

Inductively, we then find that

$$\displaystyle\bar{e}^{n}\leq\bar{e}^{0}+t_{n}\frac{M^{2}}{2}\left({\Updelta x}+{\Updelta t}\right)=t_{n}\frac{M^{2}}{2}\left({\Updelta x}+{\Updelta t}\right),$$

since $e^{0}_{j}=0$ by definition. Hence, the approximation defined by (1.48) converges to the unique solution if u is twice differentiable with bounded second derivatives.

We have seen that if $x\mapsto a(x,t)$ is decreasing on some interval, the best bounds for u _xx and u _xt are likely to be of the form $Ce^{Ct}$, which means that the ‘‘constant’’ M is likely to be large if we want to study the solution for large (or even moderate) times.

Similarly, if $a(x,t)<0$, the scheme

$$\displaystyle D_{\!+}^{t}u_{j}^{n}+a^{n}_{j}D_{+}u^{n}_{j}=0$$

will give a convergent sequence.

1.1.4 Entropy Solutions (II)

Consider the Cauchy problem

$$\displaystyle\begin{cases}u_{t}+a(x,t)u_{x}=0,&x\in\mathbb{R},\ \ t> 0,\\ u(x,0)=u_{0}(x),\end{cases}$$

(1.51)

where a is a continuously differentiable function (in this section not assumed to be nonnegative). Recall that an entropy solution is defined as the limit of the singularly perturbed equation (1.44). For every positive $\varepsilon$, $u^{\varepsilon}$ satisfies (1.45), implying that the limit $u=\lim_{\varepsilon\to 0}u^{\varepsilon}$ should satisfy (1.46) with f identically zero. Multiplying the inequality (1.46) by a nonnegative test function ψ, and integrating by parts, we find that

$$\displaystyle \int_{0}^{\infty}\!\!\int_{\mathbb{R}}\!\big(\eta(u)\psi_{t} + a\eta(u)\psi_{x} + a_{x}\eta(u)\psi\big)\,dx\,dt + \int_{\mathbb{R}}\!\eta(u_{0}(x))\psi(x,0)\,dx \geq 0$$

(1.52)

should hold for all nonnegative test functions $\psi\in C^{\infty}_{0}(\mathbb{R}\times[0,\infty))$, and for all convex η. If $u(x,t)$ is a function in $L^{1}_{\mathrm{loc}}(\mathbb{R}\times[0,\infty))$ that satisfies (1.52) for all convex entropies η, then u is called a weak entropy solution to (1.51). The point of this is that we no longer require u to be differentiable, or even continuous. Therefore, showing that approximations converge to an entropy solution should be much easier than showing that the limit is a classical solution.

We are going to show that there is only one entropy solution. Again, since the equation is linear, it suffices to show that $u_{0}=0$ (in $L^{1}(\mathbb{R})$) implies $u(\,\cdot\,,t)=0$ (in $L^{1}(\mathbb{R})$).

To do this, we specify a particular test function. Let ω be a $C^{\infty}$ function such that

$$\displaystyle 0\leq\omega(\sigma)\leq 1,\quad\mathop{\mathrm{supp}}\omega\subseteq[-1,1],\quad\omega(-\sigma)=\omega(\sigma),\quad\int_{-1}^{1}\omega(\sigma)\,d\sigma=1.$$

Now define

$$\displaystyle\omega_{\varepsilon}(\sigma)=\frac{1}{\varepsilon}\omega\left(\frac{\sigma}{\varepsilon}\right).$$

(1.53)

Let $x_{1}<x_{2}$, and introduce

$$\displaystyle\varphi_{\varepsilon}(x,t)=\int_{x_{1}+Lt}^{x_{2}-Lt}\omega_{\varepsilon}(x-y)\,dy,$$

where L is a constant such that $L> \left\|a\right\|_{L^{\infty}(\Omega)}$ and $\Omega=\mathbb{R}\times[0,\infty)$. We fix a T such that $T<(x_{2}-x_{1})/(2L)$, and consider $t<T$. Observe that $\varphi_{\varepsilon}(\,\cdot\,,t)$ is an approximation to the characteristic function for the interval $(x_{1}+Lt,x_{2}-Lt)$.

Next introduce

$$\displaystyle h_{\varepsilon}(t)=1-\int_{0}^{t}\omega_{\varepsilon}(s-T)\,ds.$$

This is an approximation to the characteristic function of the interval $(-\infty,T]$. Finally, we choose the test function

$$\displaystyle\psi_{\varepsilon}(x,t)=h_{\varepsilon}(t)\varphi_{\varepsilon}(x,t)\in C^{\infty}_{0}(\Omega).$$

Inserting this into the entropy inequality (1.52), we get

$$\displaystyle\begin{aligned}\displaystyle&\displaystyle\iint_{\Omega}\eta(u)\varphi_{\varepsilon}h^{\prime}_{\varepsilon}(t)\,dx\,dt\\ \displaystyle&\displaystyle+\iint_{\Omega}h_{\varepsilon}(t)\eta(u)\left(\frac{\partial}{\partial t}\varphi_{\varepsilon}(x,t)+a(x,t)\frac{\partial}{\partial x}\varphi_{\varepsilon}(x,t)\right)\,dx\,dt\\ \displaystyle&\displaystyle+\iint_{\Omega}a_{x}\eta(u)h_{\varepsilon}\varphi_{\varepsilon}\,dx\,dt+\int_{\mathbb{R}}\eta(u_{0})\varphi_{\varepsilon}(x,0)\,dx\geq 0.\end{aligned}$$

(1.54)

We treat the second integral first, and calculate

$$\begin{aligned}\displaystyle\frac{\partial}{\partial t}\varphi_{\varepsilon}(x,t)&\displaystyle=-L\left(\omega_{\varepsilon}(x-x_{2}+Lt)+\omega_{\varepsilon}(x-x_{1}-Lt)\right),\\ \displaystyle\frac{\partial}{\partial x}\varphi_{\varepsilon}(x,t)&\displaystyle=-\omega_{\varepsilon}(x-x_{2}+Lt)+\omega_{\varepsilon}(x-x_{1}-Lt).\end{aligned}$$

Therefore,

$$\begin{aligned}\displaystyle\frac{\partial}{\partial t}\varphi_{\varepsilon}+a\leavevmode\nobreak\ \frac{\partial}{\partial x}\varphi_{\varepsilon}&\displaystyle=(-L+a)\omega_{\varepsilon}(x-x_{2}+Lt)+(-L-a)\omega_{\varepsilon}(x-x_{1}-Lt)\\ \displaystyle&\displaystyle\leq\left(\left|a\right|-L\right)\big(\omega_{\varepsilon}(x-x_{2}+Lt)+\omega_{\varepsilon}(x-x_{1}-Lt)\big)\leq 0,\end{aligned}$$

since L is chosen to be larger than $\left|a\right|$. Hence, if $\eta(u)\geq 0$, then the second integral in (1.54) is nonpositive. Thus we have

$$\displaystyle\begin{aligned}\displaystyle\iint_{\Omega}\eta(u)\varphi_{\varepsilon}h^{\prime}_{\varepsilon}(t)\,dx\,dt+\iint_{\Omega}&\displaystyle a_{x}\eta(u)h_{\varepsilon}\varphi_{\varepsilon}\,dx\,dt\\ \displaystyle&\displaystyle+\int_{\mathbb{R}}\eta(u_{0})\varphi_{\varepsilon}(x,0)\,dx\geq 0.\end{aligned}$$

(1.55)

Let us for the moment proceed formally. The function $h_{\varepsilon}$ approximates the characteristic function $\chi_{(-\infty,T]}$, which has derivative $-\delta_{T}$, a negative Dirac delta function at T. Similarly, $\varphi_{\varepsilon}$ approximates the characteristic function $\chi_{(x_{1}+Lt,x_{2}-Lt)}$, with derivative $L(\delta_{x_{1}+Lt}-\delta_{x_{2}-Lt})$. From (1.54) we formally obtain by sending $\varepsilon\to 0$, that

$$\displaystyle-\int_{x_{1}+LT}^{x_{2}-LT}\!\!\eta(u(x,T))\,dx+\int_{0}^{T}\int_{x_{1}+Lt}^{x_{2}-Lt}a_{x}(x,t)\eta(u(x,t))\,dx\,dt+\int_{x_{1}}^{x_{2}}\!\!\eta(u(x,0))\,dx\geq 0,$$

(1.56)

and this is what we intend to prove next.

The first integral in (1.54) reads

$$\displaystyle-\iint_{\Omega}\eta(u)\varphi_{\varepsilon}(x,t)\omega_{\varepsilon}(t-T)\,dx\,dt=-\int_{0}^{\infty}f_{\varepsilon}(t)\omega_{\varepsilon}(t-T)\,dt,$$

where

$$\displaystyle f_{\varepsilon}(t)=\int_{\mathbb{R}}\varphi_{\varepsilon}(x,t)\eta(u(x,t))\,dx.$$

Keeping t fixed, we obtain

$$\displaystyle f_{\varepsilon}(t)\to\int_{x_{1}+Lt}^{x_{2}-Lt}\!\!\eta(u(x,t))\,dx=f_{0}(t)\quad\text{ as $\varepsilon\to 0$,}$$

the limit being uniform in t for $t\in[0,T]$. If $t\mapsto u(\,\cdot\,,t)$ is continuous as a map from $[0,\infty)$ with values in $L^{1}(\mathbb{R})$, then $f_{\varepsilon}$ and f ₀ are continuous in t. In that case,

$$\begin{aligned}\displaystyle\int_{0}^{\infty}f_{\varepsilon}(t)\omega_{\varepsilon}(t-T)\,dt&\displaystyle=\int_{0}^{\infty}\big(f_{\varepsilon}(t)-f_{0}(t)\big)\omega_{\varepsilon}(t-T)\,dt+\int_{0}^{\infty}f_{0}(t)\omega_{\varepsilon}(t-T)\,dt\\ \displaystyle&\displaystyle\to f_{0}(T)\quad\text{ as $\varepsilon\to 0$,}\end{aligned}$$

since

$$\begin{aligned}\displaystyle\left|\int_{0}^{\infty}\big(f_{\varepsilon}(t)-f_{0}(t)\big)\omega_{\varepsilon}(t-T)\,dt\right|&\displaystyle\leq\left\|f_{\varepsilon}-f_{0}\right\|_{L^{\infty}}\int_{0}^{\infty}\omega_{\varepsilon}(t-T)\,dt\\ \displaystyle&\displaystyle=\left\|f_{\varepsilon}-f_{0}\right\|_{L^{\infty}}\to 0.\end{aligned}$$

In order to ensure that $t\mapsto u(\,\cdot\,,t)$ is continuous as a map from $[0,\infty)$ to $L^{1}(\mathbb{R})$, we define an entropy solution to have this property; see Definition 1.10 below. We have that

$$\displaystyle h_{\varepsilon}(t)\varphi_{\varepsilon}(x,t)\to\chi_{\Pi_{T}}(x,t)\quad\text{in $L^{1}(\Omega_{T})$,}$$

where $\Pi_{T}=\left\{(x,t)\;|\;0\leq t\leq T,\ x_{1}+Lt\leq x\leq x_{2}-Lt\right\}$ and $\Omega_{T}=\mathbb{R}\times[0,T]$. By sending $\varepsilon\to 0$ in (1.54), we then find that (cf. (1.56))

$$\displaystyle\int_{x_{1}}^{x_{2}}\!\!\eta(u(x,0))\,dx+\int_{0}^{T}\int_{x_{1}+Lt}^{x_{2}-Lt}a_{x}(x,t)\eta(u(x,t))\,dx\,dt\geq\int_{x_{1}+LT}^{x_{2}-LT}\!\!\eta(u(x,T))\,dx,$$

(1.57)

which implies that

$$\displaystyle f_{0}(T)\leq f_{0}(0)+\left\|a_{x}\right\|_{L^{\infty}(\Omega_{T})}\int_{0}^{T}f_{0}(t)\,dt,$$

assuming that η is positive.

Gronwall’s inequality then implies

$$\displaystyle f_{0}(T)\leq f_{0}(0)e^{\left\|a_{x}\right\|_{L^{\infty}(\Omega_{T})}T},$$

or, writing it out explicitly,

$$\displaystyle\int_{x_{1}+LT}^{x_{2}-LT}\!\!\eta(u(x,T))\,dx\leq\int_{x_{1}}^{x_{2}}\!\!\eta(u_{0}(x))\,dx\,e^{\left\|a_{x}\right\|_{L^{\infty}(\Omega_{T})}T},$$

for every nonnegative convex function η. Observe that this proves the finite speed of propagation.

Choosing $\eta(u)=\left|u\right|^{p}$ for $1\leq p<\infty$, assuming $\eta(u)$ to be integrable, and sending x ₁ to $-\infty$ and x ₂ to $\infty$, we get

$$\displaystyle\left\|u(\,\cdot\,,T)\right\|_{{L^{p}}(\mathbb{R})}\leq\left\|u_{0}\right\|_{L^{p}(\mathbb{R})}e^{\left\|a_{x}\right\|_{L^{\infty}(\Omega_{T})}T/p},\ \ 1\leq p<\infty.$$

(1.58)

Next, we can let $p\to\infty$, assuming $\eta(u)$ to be integrable for all $1\leq p<\infty$, to get

$$\displaystyle\left\|u(\,\cdot\,,T)\right\|_{L^{\infty}(\mathbb{R})}\leq\left\|u_{0}\right\|_{L^{\infty}(\mathbb{R})}.$$

(1.59)

In order to formalize the preceding argument, we introduce the following definition.

Definition 1.10

A function $u=u(x,t)$ in $C([0,\infty);L^{1}(\mathbb{R}))$ is called a weak entropy solution to the problem

$$\displaystyle\begin{cases}u_{t}+a(x,t)u_{x}=0,\ \ t> 0,\ x\in\mathbb{R},\\ u(x,0)=u_{0}(x),\end{cases}$$

if for all nonnegative and convex functions $\eta(u)$ and all nonnegative test functions $\varphi\in C^{\infty}_{0}(\Omega)$, the inequality

$$\displaystyle\int_{0}^{\infty}\int_{\mathbb{R}}\big(\eta(u)\varphi_{t}+a\leavevmode\nobreak\ \eta(u)\varphi_{x}+a_{x}\eta(u)\varphi\big)\,dx\,dt+\int_{\mathbb{R}}\eta(u_{0}(x))\varphi(x,0)\,dx\geq 0$$

holds.

Theorem 1.11

Assume that $a=a(x,t)$ is such that a _x is bounded. Then the problem (1.32) has at most one entropy solution $u=u(x,t)$, and the bounds (1.58) and (1.59) hold.

Remark 1.12

From the proof of this theorem (applying (1.58) for p = 1), we see that if we define an entropy solution to satisfy the entropy condition only for $\eta(u)=\left|u\right|$, then we get uniqueness in $C([0,\infty),L^{1}(\mathbb{R}))$.

1.1.5 Numerics (III)

We now reconsider the transport equation

$$\displaystyle\begin{cases}u_{t}+a(x,t)u_{x}=0,&t> 0,\\ u(x,0)=u_{0}(x),\end{cases}$$

(1.60)

and the corresponding difference scheme

$$\displaystyle D_{\!+}^{t}u^{n}_{j}+a^{n}_{j}D_{-}u^{n}_{j}=0,$$

with

$$\displaystyle a^{n}_{j}=\frac{1}{{\Updelta t}}\int_{t_{n}}^{t_{n+1}}a(x_{j},t)\,dt,\qquad u^{0}_{j}=\frac{1}{{\Updelta x}}\int_{x_{j-1/2}}^{x_{j+1/2}}\!\!u_{0}(x)\,dx,$$

where as before, we assume that $a(x,t)\geq 0$. In order to have an approximation defined for all x and t, we define

$$\displaystyle u_{\Updelta x}(x,t)=u^{n}_{j}\ \ \text{for $(x,t)\in I^{n}_{j-1/2}:=[x_{j-1},x_{j})\times[t_{n},t_{n+1})$},$$

where $t_{n}=n{\Updelta t}$. We wish to show that $u_{\Updelta x}$ converges to an entropy solution (the only one!) of (1.60). Now we do not use the linearity, and first prove that $\left\{u_{\Updelta x}\right\}_{{\Updelta x}> 0}$ has a convergent subsequence.

First we recall that the scheme can be written

$$\displaystyle u^{n+1}_{j}=\left(1-a^{n}_{j}\lambda\right)u^{n}_{j}+a^{n}_{j}\lambda u^{n}_{j-1}.$$

We aim to use Theorem A.11 to prove compactness. First we show that the approximation is uniformly bounded. This is easy, since $u^{n+1}_{j}$ is a convex combination of $u^{n}_{j}$ and $u^{n}_{j-1}$, so new maxima or minima are not introduced. Thus

$$\displaystyle\left\|u_{\Updelta x}(\,\cdot\,,t)\right\|_{L^{\infty}(\mathbb{R})}\leq\left\|u_{0}\right\|_{L^{\infty}(\mathbb{R})}.$$

Therefore, the first condition of Theorem A.11 is satisfied.

To show that the second condition holds, recall, or consult Appendix A, that the total variation of a function $u\colon\mathbb{R}\to\mathbb{R}$ is defined as

$$\displaystyle\mathrm{T.V.}\left(u\right)=\sup_{\left\{x_{i}\right\}}\sum_{i}\left|u(x_{i})-u(x_{i-1})\right|,$$

where the supremum is taken over all finite partitions $\left\{x_{i}\right\}$ such that $x_{i}<x_{i+1}$. This is a seminorm, and we also write $\left|u\right|_{BV}:=\mathrm{T.V.}\left(u\right)$.

We have to estimate the total variation of $u_{\Updelta x}$. For $t\in[t_{n},t_{n+1})$ this is given by

$$\displaystyle\left|u_{{\Updelta x}}(\,\cdot\,,t)\right|_{BV}=\sum_{j}\left|u^{n}_{j}-u^{n}_{j-1}\right|.$$

We also have that

$$\begin{aligned}\displaystyle u^{n+1}_{j}-u^{n+1}_{j-1}&\displaystyle=(1-a^{n}_{j}\lambda)u^{n}_{j}+a^{n}_{j}\lambda u^{n}_{j-1}-(1-a^{n}_{j-1}\lambda)u^{n}_{j-1}-a^{n}_{j-1}\lambda u^{n}_{j-2}\\ \displaystyle&\displaystyle=(1-a^{n}_{j}\lambda)(u^{n}_{j}-u^{n}_{j-1})+a^{n}_{j-1}\lambda(u^{n}_{j-1}-u^{n}_{j-2}).\end{aligned}$$

By the CFL condition $0\leq a^{n}_{j}\lambda\leq 1$ for all n and j, we infer

$$\displaystyle\left|u^{n+1}_{j}-u^{n+1}_{j-1}\right|\leq(1-a^{n}_{j}\lambda)\left|u^{n}_{j}-u^{n}_{j-1}\right|+\lambda a^{n}_{j-1}\left|u^{n}_{j-1}-u^{n}_{j-2}\right|.$$

Therefore

$$\begin{aligned}\displaystyle\sum_{j}\left|u^{n+1}_{j}-u^{n+1}_{j-1}\right|&\displaystyle\leq\sum_{j}(1-a^{n}_{j}\lambda)\left|u^{n}_{j}-u^{n}_{j-1}\right|+\sum_{j}\lambda a^{n}_{j-1}\left|u^{n}_{j-1}-u^{n}_{j-2}\right|\\ \displaystyle&\displaystyle=\sum_{j}\left|u^{n}_{j}-u^{n}_{j-1}\right|-\sum_{j}\lambda a^{n}_{j} \left|u^{n}_{j}-u^{n}_{j-1}\right|+\sum_{j}\lambda a^{n}_{j} \left|u^{n}_{j}-u^{n}_{j-1}\right|\\ \displaystyle&\displaystyle=\sum_{j}\left|u^{n}_{j}-u^{n}_{j-1}\right|.\end{aligned}$$

Hence

$$\displaystyle\left|u_{\Updelta x}(\,\cdot\,,t)\right|_{BV}\leq\left|u_{{\Updelta x}}(\,\cdot\,,0)\right|_{BV}\leq\left|u_{0}\right|_{BV}.$$

This shows that the second condition of Theorem A.11 is satisfied; see Remark A.12.

To show that the third condition holds, i.e., the continuity of the L ¹-norm in time, we assume that $s\in[t_{n},t_{n+1})$, and that t is such that $t-s\leq{\Updelta t}$. Then

$$\begin{aligned}\displaystyle\int_{\mathbb{R}}\left|u_{\Updelta x}(x,t)-u_{\Updelta x}(x,s)\right|\,dx&\displaystyle\leq{\Updelta x}\sum_{j}\left|u^{n+1}_{j}-u^{n}_{j}\right|\\ \displaystyle&\displaystyle={\Updelta x}\sum_{j}a^{n}_{j}\lambda\left|u^{n}_{j}-u^{n}_{j-1}\right|\\ \displaystyle&\displaystyle\leq{\Updelta t}\left\|a\right\|_{L^{\infty}(\Omega)}\sum_{j}\left|u^{n}_{j}-u^{n}_{j-1}\right|\\ \displaystyle&\displaystyle\leq{\Updelta t}\left\|a\right\|_{L^{\infty}(\Omega)}\left|u_{0}\right|_{BV}.\end{aligned}$$

If $s\in[t_{n},t_{n+1})$ and $t\in[t_{n+k},t_{n+k+1})$, we have

$$\begin{aligned}\displaystyle\int_{\mathbb{R}}\left|u_{\Updelta x}(x,t)-u_{\Updelta x}(x,s)\right|\,dx&\displaystyle={\Updelta x}\sum_{j}\left|u_{j}^{n+k}-u^{n}_{j}\right|\\ \displaystyle&\displaystyle\leq\sum_{m=n}^{n+k-1}{\Updelta x}\sum_{j}\left|u^{m+1}_{j}-u^{m}_{j}\right|\\ \displaystyle&\displaystyle=\sum_{m=n}^{n+k-1}{\Updelta x}\sum_{j}a^{m}_{j}\lambda\left|u^{m}_{j}-u^{m}_{j-1}\right|\\ \displaystyle&\displaystyle\leq\sum_{m=n}^{n+k-1}{\Updelta t}\left\|a\right\|_{L^{\infty}(\Omega)}\sum_{j}\left|u^{m}_{j}-u^{m}_{j-1}\right|\\ \displaystyle&\displaystyle\leq k{\Updelta t}\left\|a\right\|_{L^{\infty}(\Omega)}\left|u_{0}\right|_{BV}\\ \displaystyle&\displaystyle\leq(t-s+{\Updelta t})\left\|a\right\|_{L^{\infty}(\Omega)}\left|u_{0}\right|_{BV}.\end{aligned}$$

Hence, also the third condition of Theorem A.11 is fulfilled, and we have the convergence (of a subsequence) $u_{\Updelta x}\to u$ as ${\Updelta x}\to 0$. It remains to prove that u is the entropy solution.

To do this, start by observing that

$$\begin{aligned}\displaystyle\eta\left(u^{n+1}_{j}\right)&\displaystyle=\eta\left((1-a^{n}_{j}\lambda)u^{n}_{j}+a^{n}_{j}\lambda u^{n}_{j-1}\right)\\ \displaystyle&\displaystyle\leq(1-a^{n}_{j}\lambda)\eta\left(u^{n}_{j}\right)+a^{n}_{j}\lambda\eta\left(u^{n}_{j-1}\right),\end{aligned}$$

since η is assumed to be a convex function. This can be rearranged as

$$\displaystyle D_{\!+}^{t}\eta^{n}_{j}+a^{n}_{j}D_{-}\eta^{n}_{j}\leq 0,$$

where $\eta^{n}_{j}=\eta(u^{n}_{j})$, and as

$$\displaystyle D_{\!+}^{t}\eta^{n}_{j}+D_{-}\left(a^{n}_{j}\eta^{n}_{j}\right)-\eta^{n}_{j-1}D_{-}a^{n}_{j}\leq 0.$$

(1.61)

The operators $D_{-}$, $D_{+}$, and $D_{\!+}^{t}$ satisfy the following ‘‘summation by parts’’ formulas:

$$\begin{aligned}\displaystyle\sum_{j}a_{j}D_{-}b_{j}&\displaystyle=-\sum_{j}b_{j}D_{+}a_{j},\ \ \text{if $a_{\pm\infty}=0$ or $b_{\pm\infty}=0$,}\\ \displaystyle\sum_{n=0}^{\infty}a^{n}D_{\!+}^{t}b^{n}&\displaystyle=-\frac{1}{{\Updelta t}}a^{0}b^{0}-\sum_{n=1}^{\infty}b^{n}D_{\!-}^{t}a^{n}\ \ \text{if $a^{\infty}=0$ or $b^{\infty}=0$.}\end{aligned}$$

Let $\varphi$ be a nonnegative test function in $C^{\infty}_{0}(\Omega)$ and set

$$\displaystyle\varphi^{n}_{j}=\frac{1}{|I^{n}_{j-1/2}|}\iint_{I^{n}_{j-1/2}}\varphi(x,t)\,dx\,dt.$$

We multiply (1.61) by ${\Updelta t}{\Updelta x}\,\varphi^{n}_{j}$ and sum over $n\geq 0$ and $j\in\mathbb{Z}$, using the summation by parts formulas above, to get

$$\displaystyle\begin{aligned}\displaystyle&\displaystyle{\Updelta x}{\Updelta t}\sum_{n=1}^{\infty}\sum_{j}\eta(u^{n}_{j})D_{\!-}^{t}\varphi^{n}_{j}\\ \displaystyle&\displaystyle\quad+{\Updelta x}{\Updelta t}\sum_{n=0}^{\infty}\sum_{j}\left(a^{n}_{j}\eta(u^{n}_{j})D_{+}\varphi^{n}_{j}+\eta(u^{n}_{j-1})D_{-}a^{n}_{j}\varphi^{n}_{j}\right)+{\Updelta x}\sum_{j}\eta(u^{0}_{j})\varphi^{0}_{j}\geq 0.\end{aligned}$$

Call the left-hand side of the above inequality $B_{\Updelta x}$, and set

$$\displaystyle A_{\Updelta x}=\iint_{\Omega}\big(\eta(u_{\Updelta x})\varphi_{t}+a\eta(u_{\Updelta x})\varphi_{x}+a_{x}\eta(u_{\Updelta x})\varphi\big)\,dx\,dt+\int_{\mathbb{R}}\eta(u_{0})\varphi(x,0)\,dx.$$

Then we have

$$\displaystyle A_{\Updelta x}=B_{\Updelta x}+(A_{\Updelta x}-B_{\Updelta x})\geq A_{\Updelta x}-B_{\Updelta x}.$$

We find that

$$A_{\Updelta x}-B_{\Updelta x} =\sum_{n=1}^{\infty}\sum_{j}\iint_{I^{n}_{j-1/2}}\eta^{n}_{j}\left(\varphi_{t}-D_{\!-}^{t}\varphi^{n}_{j}\right)\,dx\,dt$$

(1.62a)

$$ \quad+\sum_{j}\iint_{I^{0}_{j-1/2}}\eta^{0}_{j}\varphi_{t}\,dx\,dt$$

(1.62b)

$$ \quad+\sum_{j,n}\iint_{I^{n}_{j-1/2}}\eta^{n}_{j}a\left(\varphi_{x}-D_{+}\varphi^{n}_{j}\right)\,dxdt$$

(1.62c)

$$ \quad+\sum_{j,n}\iint_{I^{n}_{j-1/2}}\eta^{n}_{j}D_{+}\varphi^{n}_{j}\left(a-a^{n}_{j}\right)\,dx\,dt$$

(1.62d)

$$ \quad+\sum_{j,n}\iint_{I^{n}_{j-1/2}}\left(\eta^{n}_{j}-\eta^{n}_{j-1}\right)a_{x}\varphi\,dx\,dt$$

(1.62e)

$$ \quad+\sum_{j,n}\iint_{I^{n}_{j-1/2}}a_{x}\eta^{n}_{j-1}\left(\varphi-\varphi^{n}_{j}\right)\,dx\,dt$$

(1.62f)

$$ \quad+\sum_{j,n}\iint_{I^{n}_{j-1/2}}\eta^{n}_{j-1}\left(a_{x}-D_{-}a^{n}_{j}\right)\varphi^{n}_{j}\,dx\,dt$$

(1.62g)

$$ \quad+\sum_{j}\int_{I_{j-1/2}}\left(\eta(u_{0})-\eta^{0}_{j}\right)\varphi(x,0)\,dx$$

(1.62h)

$$ \quad+\sum_{j}\int_{I_{j-1/2}}\eta^{0}_{j}\left(\varphi(x,0)-\varphi^{0}_{j}\right)\,dx.$$

(1.62i)

Here $I_{j-1/2}=[x_{j-1},x_{j})$. To show that the limit u is an entropy solution, we must show that all the terms (1.62a)–(1.62i) vanish when ${\Updelta x}$ becomes small. A small but useful device is contained in the following remark.

Remark 1.13

For a continuously differentiable function ϕ we have

$$\begin{aligned}\displaystyle\left|\varphi(x,t)-\varphi(y,s)\right|&\displaystyle=\left|\int_{0}^{1}\frac{d}{d\sigma}\varphi(\sigma(x,t)+(1-\sigma)(y,s))d\sigma\right|\\ \displaystyle&\displaystyle=\left|\int_{0}^{1}\nabla\varphi(\sigma(x,t)+(1-\sigma)(y,s))\cdot(x-y,t-s)d\sigma\right|\\ \displaystyle&\displaystyle\leq\left|x-y\right|\left\|\varphi_{x}\right\|_{L^{\infty}}+\left|t-s\right|\left\|\varphi_{t}\right\|_{L^{\infty}}.\end{aligned}$$

We start with the last term (1.62i). Now

$$\begin{aligned}\displaystyle&\displaystyle\int_{I_{j-1/2}}\eta^{0}_{j}\left(\varphi(x,0)-\varphi^{0}_{j}\right)\,dx\\ \displaystyle&\displaystyle\qquad=\frac{\eta^{0}_{j}}{{\Updelta x}{\Updelta t}}\int_{I_{j-1/2}}\iint_{I^{0}_{j-1/2}}\varphi(x,0)-\varphi(y,t)\,dy\,dt\,dx\\ \displaystyle&\displaystyle\qquad=\frac{\eta^{0}_{j}}{{\Updelta x}{\Updelta t}}\int_{I_{j-1/2}}\iint_{I^{0}_{j-1/2}}\Bigl(\int_{y}^{x}\varphi_{x}(z,0)\,dz+\int_{0}^{t}\varphi_{t}(y,s)\,ds\Bigr)\,dy\,dt\,dx.\end{aligned}$$

Therefore,

$$\displaystyle\left|(1.62i)\right|\leq\left\|\eta(u_{0})\right\|_{L^{1}(\mathbb{R})}\left(\left\|\varphi_{x}\right\|_{L^{\infty}(\Omega)}{\Updelta x}+\left\|\varphi_{t}\right\|_{L^{\infty}(\Omega)}{\Updelta t}\right),$$

where we used the convexity of η. Next, we consider the term (1.62h): Since η is convex, we have

$$\displaystyle\left|\eta(b)-\eta(a)\right|\leq\max\left\{\left|\eta^{\prime}(a)\right|,\left|\eta^{\prime}(b)\right|\right\}\left|b-a\right|.$$

Furthermore, if both x and y are in $I_{j-1/2}$, then

$$\displaystyle\left|u_{0}(x)-u_{0}(y)\right|\leq\left|u_{0}\right|_{BV(I_{j-1/2})}.$$

Using this and choosing $C=\left\|\eta^{\prime}(u_{0})\right\|_{L^{\infty}}$ yields

$$\begin{aligned}\displaystyle\Bigl|\int_{I_{j-1/2}}&\displaystyle\left(\eta(u_{0})-\eta(u^{0}_{j})\right)\varphi(x,0)\,dx\Bigr|\\ \displaystyle&\displaystyle\qquad\qquad\leq C\left\|\varphi\right\|_{L^{\infty}(\Omega)}\int_{I_{j-1/2}}\left|u_{0}(x)-u^{0}_{j}\right|\,dx\\ \displaystyle&\displaystyle\qquad\qquad\leq C\left\|\varphi\right\|_{L^{\infty}(\Omega)}\int_{I_{j-1/2}}\!\!\frac{1}{{\Updelta x}}\int_{I_{j-1/2}}\!\!\!\left|u_{0}(x)-u_{0}(y)\right|\,dx\,dy\\ \displaystyle&\displaystyle\qquad\qquad\leq C\left\|\varphi\right\|_{L^{\infty}(\Omega)}{\Updelta x}\left|u_{0}\right|_{BV(I_{j-1/2})}.\end{aligned}$$

Therefore,

$$\begin{aligned}\displaystyle\left|(1.62h)\right|&\displaystyle\leq C\left\|\varphi\right\|_{L^{\infty}(\Omega)}{\Updelta x}\sum_{j}\left|u_{0}\right|_{BV(I_{j-1/2})}\\ \displaystyle&\displaystyle\leq C\left\|\varphi\right\|_{L^{\infty}(\Omega)}{\Updelta x}\left|u_{0}\right|_{BV}.\end{aligned}$$

Next, we consider (1.62g). First observe that

$$\displaystyle D_{-}a^{n}_{j}=D_{-}\Bigl(\frac{1}{{\Updelta t}}\int_{t_{n}}^{t_{n+1}}a(x_{j},t)\,dt\Bigr)=\frac{1}{{\Updelta x}{\Updelta t}}\iint_{I^{n}_{j-1/2}}a_{x}(x,t)\,dx\,dt.$$

Therefore,

$$\begin{aligned}\displaystyle\eta^{n}_{j-1}\iint_{I^{n}_{j-1/2}}\!\!a_{x}(x,t)-D_{-}a^{n}_{j}\,dx\,dt&\displaystyle=\eta^{n}_{j-1}\Bigl(\iint_{I^{n}_{j-1/2}}\!\!a_{x}(x,t)\,dx\,dt-{\Updelta x}{\Updelta t}a^{n}_{j}\Bigr)\\ \displaystyle&\displaystyle=0,\end{aligned}$$

and (1.62g)${}=0$. We continue with the term (1.62f), namely

$$\begin{aligned}\displaystyle\iint_{I^{n}_{j-1/2}}&\displaystyle\left|a_{x}\right|\eta^{n}_{j-1}\left|\varphi-\varphi^{n}_{j}\right|\,dx\,dt\\ \displaystyle&\displaystyle\leq\left\|a_{x}\right\|_{L^{\infty}(\Omega)}\eta^{n}_{j-1}\frac{1}{{\Updelta x}{\Updelta t}}\iint_{I^{n}_{j}}\iint_{I^{n}_{j}}\left|\varphi(x,t)-\varphi(y,s)\right|\,dy\,ds\,dx\,dt\\ \displaystyle&\displaystyle\leq\left\|a_{x}\right\|_{L^{\infty}(\Omega)}\eta^{n}_{j-1}{\Updelta x}{\Updelta t}\left({\Updelta x}\left\|\varphi_{x}\right\|_{L^{\infty}(\Omega)}+{\Updelta t}\left\|\varphi_{t}\right\|_{L^{\infty}(\Omega)}\right).\end{aligned}$$

Recall that the test function $\varphi$ has compact support, contained in $\left\{t<T\right\}$. Furthermore, using the scheme for $\eta^{n}_{j}$, cf. (1.61), it is straightforward to show that

$$\displaystyle{\Updelta x}\sum_{j}\eta^{n}_{j}\leq e^{Ct_{n}}{\Updelta x}\sum_{j}\eta^{0}_{j}\leq e^{Ct_{n}}\left\|\eta(u_{0})\right\|_{L^{1}(\mathbb{R})},$$

where C is a bound on $D_{-}a^{n}_{j}$. Therefore,

$$\begin{aligned}\displaystyle\Bigl|\sum_{j,n}\iint_{I^{n}_{j-1/2}}a_{x}\eta^{n}_{j-1}\left(\varphi-\varphi^{n}_{j}\right)\,dx\,dt\Bigr|&\displaystyle\leq C_{T}{\Updelta x}\sum_{j,n}\eta^{n}_{j}{\Updelta t}({\Updelta x}+{\Updelta t})\\ \displaystyle&\displaystyle\leq C_{T}{\Updelta x}\sum_{n,j}\eta^{0}_{j}{\Updelta t}({\Updelta x}+{\Updelta t})\\ \displaystyle&\displaystyle\leq C_{T}T\left\|\eta(u_{0})\right\|_{L^{1}(\mathbb{R})}({\Updelta x}+{\Updelta t}),\end{aligned}$$

since the sum in n is only over those n such that $t_{n}=n{\Updelta t}\leq T$. Regarding (1.62e), and setting $M> \left\|u_{0}\right\|_{L^{\infty}(\mathbb{R})}$, we have that

$$\begin{aligned}\displaystyle&\displaystyle\Bigl|\sum_{j,n}\iint_{I^{n}_{j-1/2}}\left(\eta^{n}_{j}-\eta^{n}_{j-1}\right)a_{x}\varphi\,dx\,dt\Bigr|\\ \displaystyle&\displaystyle\quad\leq\left\|\eta^{\prime}\right\|_{L^{\infty}((-M,M))}\left\|a_{x}\right\|_{L^{\infty}(\Omega)}\left\|\varphi\right\|_{L^{\infty}(\Omega)}{\Updelta x}{\Updelta t}\sum_{j,n}\left|u^{n}_{j}-u^{n}_{j-1}\right|\\ \displaystyle&\displaystyle\quad\leq C{\Updelta x}T\left|u_{0}\right|_{BV}.\end{aligned}$$

Next, we turn to (1.62d):

$$\begin{aligned}\displaystyle\left|(1.62d)\right|&\displaystyle\leq\left\|\varphi_{x}\right\|_{L^{\infty}(\Omega)}\sum_{j,n}\eta^{n}_{j}\iint_{I^{n}_{j-1/2}}\left|a(x,t)-a(x_{j},t)\right|\,dxdt\\ \displaystyle&\displaystyle\leq\left\|\varphi_{x}\right\|_{L^{\infty}(\Omega)}\left\|a_{x}\right\|_{L^{\infty}(\Omega)}{\Updelta x}\sum_{j,n}\eta^{n}_{j}{\Updelta x}{\Updelta t}\\ \displaystyle&\displaystyle\leq\left\|\varphi_{x}\right\|_{L^{\infty}(\Omega)}\left\|a_{x}\right\|_{L^{\infty}(\Omega)}C_{T}T{\Updelta x}\left\|\eta(u_{0})\right\|_{L^{1}(\mathbb{R})}.\end{aligned}$$

We can use the same type of argument to estimate (1.62c):

$$\begin{aligned}\displaystyle\left|(1.62c)\right|&\displaystyle\leq\left\|a\right\|_{L^{\infty}(\Omega)}\left({\Updelta x}\left\|\varphi_{xx}\right\|_{L^{\infty}(\Omega)}+{\Updelta t}\left\|\varphi_{xt}\right\|_{L^{\infty}(\Omega)}\right)\sum_{n,j}\eta^{n}_{j}{\Updelta x}{\Updelta t}\\ \displaystyle&\displaystyle\leq\left\|a\right\|_{L^{\infty}(\Omega)}\left({\Updelta x}\left\|\varphi_{xx}\right\|_{L^{\infty}(\Omega)}+{\Updelta t}\left\|\varphi_{xt}\right\|_{L^{\infty}(\Omega)}\right)C_{T}T\left\|\eta(u_{0})\right\|_{L^{1}(\mathbb{R})}.\end{aligned}$$

Similarly, we show that

$$\displaystyle\left|(1.62b)\right|\leq C_{\Updelta t}{\Updelta t}\left\|\eta(u_{0})\right\|_{L^{1}(\mathbb{R})}\left\|\varphi_{t}\right\|_{L^{\infty}(\Omega)}.$$

Now the end is in sight. We estimate the right-hand side of (1.62a). This will be less than

$$\begin{aligned}\displaystyle\sum_{j,n\geq 1}\eta^{n}_{j}&\displaystyle\iint_{I^{n}_{j-1/2}}\left|\varphi_{t}-D_{\!+}^{t}\varphi^{n}_{j}\right|\,dx\,dt\\ \displaystyle&\displaystyle\leq\left({\Updelta x}\left\|\varphi_{xt}\right\|_{L^{\infty}(\Omega)}+{\Updelta t}\left\|\varphi_{tt}\right\|_{L^{\infty}(\Omega)}\right)\sum_{j,n\geq 1}\eta^{n}_{j}{\Updelta x}{\Updelta t}\\ \displaystyle&\displaystyle\leq\left({\Updelta x}\left\|\varphi_{xt}\right\|_{L^{\infty}(\Omega)}+{\Updelta t}\left\|\varphi_{tt}\right\|_{L^{\infty}(\Omega)}\right)C_{T}T\left\|\eta(u_{0})\right\|_{L^{1}(\mathbb{R})}.\end{aligned}$$

To sum up, what we have shown is that for every test function $\varphi(x,t)$,

$$\begin{aligned}\displaystyle\iint_{\Omega}\big(\eta(u)\varphi_{t}+&\displaystyle a\eta(u)\varphi_{x}+a_{x}\eta(u)\big)\,dx\,dt+\int_{\mathbb{R}}\eta(u_{0})\varphi(x,0)\,dx\\ \displaystyle&\displaystyle=\lim_{{\Updelta x}\to 0}A_{\Updelta x}\\ \displaystyle&\displaystyle\geq\lim_{{\Updelta x}\to 0}\big(A_{\Updelta x}-B_{\Updelta x}\big)=0,\end{aligned}$$

if a _x is (locally) continuous and $u_{0}\in BV(\mathbb{R})$. Hence the scheme (1.60) produces a subsequence that converges to the unique weak solution. Since the limit is the unique entropy solution, every subsequence will produce a further subsequence that converges to the same limit, and thus the whole sequence converges!

If $u_{0}^{\prime\prime}$ is bounded, we have seen that the scheme (1.60) converges at a rate $\mathcal{O}\left({\Updelta x}\right)$ to the entropy solution. The significance of the above computations is that we have the convergence to the unique entropy solution even if u ₀ is assumed to be only in $L^{1}(\mathbb{R})\cap BV(\mathbb{R})$. However, in this case we have not shown any convergence rate.

1.1.6 Systems of Equations

I have a different way of thinking. I think synergistically. I’m not linear in thinking, I’m not very logical. — Imelda Marcos

Now we generalize, and let $u\colon\mathbb{R}\times[0,\infty)\to\mathbb{R}^{n}$ be a solution of the linear system

$$\displaystyle\begin{cases}u_{t}+Au_{x}=0,&x\in\mathbb{R},\ \ t> 0,\\ u(x,0)=u_{0}(x),\end{cases}$$

(1.63)

where A is an $n\times n$ matrix with real and distinct eigenvalues $\left\{\lambda_{i}\right\}_{i=1}^{n}$. We order these such that

$$\displaystyle\lambda_{1}<\lambda_{2}<\cdots<\lambda_{n}.$$

If this holds, then the system is said to be strictly hyperbolic. The matrix A will also have n linearly independent right eigenvectors $r_{1},\ldots,r_{n}$ such that

$$\displaystyle Ar_{i}=\lambda_{i}r_{i}.$$

Similarly, it has n independent left eigenvectors $l_{1},\ldots,l_{n}$ such that

$$\displaystyle l_{i}A=\lambda_{i}l_{i}.$$

We assume r _i to be column vectors and l _i to be row vectors. However, we will not enforce this strictly, and will write, e.g., $l_{i}\cdot r_{k}$. For $k\neq m$, l _k and r _m are orthogonal, since

$$\displaystyle\lambda_{m}r_{m}\cdot l_{k}=Ar_{m}\cdot l_{k}=r_{m}\cdot l_{k}A\leavevmode\nobreak\ =\lambda_{k}r_{m}\cdot l_{k}.$$

Let

$$\displaystyle L=\begin{pmatrix}l_{1}\\ \vdots\\ l_{n}\end{pmatrix},\quad R=\begin{pmatrix}r_{1}&\cdots&r_{n}\\ \end{pmatrix}.$$

Normalize the eigenvectors so that $l_{k}\cdot r_{i}=\delta_{ki}$, i.e., $L=R^{-1}$, or $LR=I$. Then

$$\displaystyle LAR=\begin{pmatrix}\lambda_{1}&&0\\ &\ddots&\\ 0&&\lambda_{n}\end{pmatrix}.$$

We can multiply (1.63) by L from the left to get

$$\displaystyle Lu_{t}+LAu_{x}=0,$$

and defining w by $u=Rw$, we find that

$$\displaystyle w_{t}+\begin{pmatrix}\lambda_{1}&&0\\ &\ddots&\\ 0&&\lambda_{n}\end{pmatrix}w_{x}=0.$$

(1.64)

This is n decoupled equations, one for each component of $w=(w_{1},\dots,w_{n})$,

$$\displaystyle\frac{\partial w_{i}}{\partial t}+\lambda_{i}\frac{\partial w_{i}}{\partial x}=0,\ \ \text{ for $i=1,\ldots,n$.}$$

The initial data transforms into

$$\displaystyle w_{0}=Lu_{0}=(l_{1}\cdot u_{0},\dots,l_{n}\cdot u_{0}),$$

and hence we obtain the solution

$$\displaystyle w_{i}(x,t)=l_{i}\cdot u_{0}(x-\lambda_{i}t).$$

Transforming back into the original variables, we obtain

$$\displaystyle u(x,t)=\sum_{i=1}^{n}w_{i}(x,t)r_{i}=\sum_{i=1}^{n}\left[l_{i}\cdot u_{0}(x-\lambda_{i}t)\right]r_{i}.$$

(1.65)

⋄ Example 1.14 (The linear wave equation)

Now consider the linear wave equation; $\alpha\colon\mathbb{R}\times(0,\infty)\to\mathbb{R}$ is a solution of

$$\displaystyle\begin{cases}\alpha_{tt}-c^{2}\alpha_{xx}=0,&x\in\mathbb{R},\ \ t> 0,\\ \alpha(x,0)=\alpha_{0}(x),\ \ \alpha_{t}(x,0)=\beta_{0}(x),\end{cases}$$

where c is a positive constant. Defining

$$\displaystyle u=\begin{pmatrix}u_{1}\\ u_{2}\end{pmatrix}=\begin{pmatrix}\alpha_{t}\\ \alpha_{x}\end{pmatrix}$$

implies that

$$\displaystyle\begin{aligned}\displaystyle\frac{\partial u_{1}}{\partial t}-c^{2}\frac{\partial u_{2}}{\partial x}&\displaystyle=0,\\ \displaystyle\frac{\partial u_{2}}{\partial t}-\frac{\partial u_{1}}{\partial x}&\displaystyle=0,\end{aligned}\qquad\text{or}\qquad u_{t}+\begin{pmatrix}0&-c^{2}\\ -1&0\end{pmatrix}u_{x}=0.$$

The matrix

$$\displaystyle A=\begin{pmatrix}0&-c^{2}\\ -1&0\end{pmatrix}$$

has eigenvalues and eigenvectors

$$\displaystyle\lambda_{1}=-c,\ \ r_{1}=\begin{pmatrix}c\\ 1\end{pmatrix},\ \ \ \lambda_{2}=c,\ \ r_{2}=\begin{pmatrix}-c\\ 1\end{pmatrix}.$$

Thus

$$\displaystyle R=\begin{pmatrix}c&-c\\ 1&1\end{pmatrix},\ \ \ \ L=R^{-1}=\frac{1}{2c}\begin{pmatrix}1&c\\ -1&c\end{pmatrix}.$$

Hence we find that

$$\displaystyle\begin{pmatrix}w_{1}\\ w_{2}\end{pmatrix}=\frac{1}{2c}\begin{pmatrix}u_{1}+cu_{2}\\ -u_{1}+cu_{2}\end{pmatrix}.$$

Writing the solution in terms of $\alpha_{x}$ and $\alpha_{t}$, we find that

$$\begin{aligned}\displaystyle\alpha_{t}(x,t)+c\alpha_{x}(x,t)&\displaystyle=\beta_{0}(x+ct)+c\alpha_{0}^{\prime}(x+ct),\\ \displaystyle-\alpha_{t}(x,t)+c\alpha_{x}(x,t)&\displaystyle=-\beta_{0}(x-ct)+c\alpha_{0}^{\prime}(x-ct).\end{aligned}$$

Therefore,

$$\begin{aligned}\displaystyle\alpha_{x}(x,t)&\displaystyle=\frac{1}{2}\left(\alpha_{0}^{\prime}(x+ct)+\alpha_{0}^{\prime}(x-ct)\right)+\frac{1}{2c}\left(\beta_{0}(x+ct)-\beta_{0}(x-ct)\right),\\ \displaystyle\alpha_{t}(x,t)&\displaystyle=\frac{1}{2}\left(\beta_{0}(x+ct)+\beta_{0}(x-ct)\right)+\frac{c}{2}\left(\alpha_{0}^{\prime}(x+ct)-\alpha_{0}^{\prime}(x-ct)\right).\end{aligned}$$

To find α, we can integrate the last equation in t,

$$\displaystyle\alpha(x,t)=\frac{1}{2}\left(\alpha_{0}(x+ct)+\alpha_{0}(x-ct)\right)+\frac{1}{2c}\int_{x-ct}^{x+ct}\beta_{0}(y)\,dy,$$

after a change of variables in the integral involving $\beta_{0}$. This is the famous d’Alembert formula for the solution of the linear wave equation in one dimension.  ⋄

Next, we discuss the notion of entropy solutions. The meaning of an entropy solution to an equation written in characteristic variables is that for some convex function $\hat{\eta}(u)$, the entropy solution should satisfy

$$\displaystyle\hat{\eta}(u)_{t}+\hat{q}(u)_{x}\leq 0,$$

in the weak sense. Here the entropy flux $\hat{q}$ should satisfy

$$\displaystyle\nabla_{w}\hat{q}(u)=\nabla_{u}\hat{\eta}(u)\Lambda,\ \ \ \text{i.e.,}\ \ \ \frac{\partial\hat{q}}{\partial u_{i}}=\lambda_{i}\frac{\partial\hat{\eta}}{\partial u_{i}}\ \text{for $i=1,\ldots,n$.}$$

An entropy solution to (1.63) is the limit (if such a limit exists) of the parabolic reqularization

$$\displaystyle u^{\varepsilon}_{t}+Au^{\varepsilon}_{x}=\varepsilon u^{\varepsilon}_{xx}$$

as $\varepsilon\to 0$. To check whether we have a convex entropy $\eta\colon\mathbb{R}^{n}\to\mathbb{R}$, we take the inner product of the above with $\nabla\eta(u^{\varepsilon})$ to get

$$\displaystyle\eta\left(u^{\varepsilon}\right)_{t}+\nabla\eta\left(u^{\varepsilon}\right)\cdot A\leavevmode\nobreak\ u^{\varepsilon}_{x}\leq\varepsilon\left(\nabla\eta\left(u^{\varepsilon}\right)\cdot u^{\varepsilon}_{x}\right)_{x},$$

by the convexity of η. Observe that the convexity is used to get rid of a term containing $(u_{x}^{\varepsilon})^{2}$, which may not be wellbehaved (for nonlinear equations) in the limit $\varepsilon\to 0$, and we obtain an inequality rather than an equality. We want to write the second term on the left as the x derivative of some function $q(u^{\varepsilon})$. Using

$$\displaystyle q\left(u^{\varepsilon}\right)_{x}=\nabla q\left(u^{\varepsilon}\right)\cdot u^{\varepsilon}_{x},$$

we see that if this is so, then

$$\displaystyle\frac{\partial q}{\partial u_{j}}=\sum_{i}a_{ij}\frac{\partial\eta}{\partial u_{i}}\ \ \text{for $j=1,\ldots,n$.}$$

(1.66)

This is n equations in the two unknowns η and q. Thus we cannot expect any solution if n > 2. The right-hand side of (1.66) is given, and hence we are looking for a potential q with a given gradient. This problem has a solution if

$$\displaystyle\frac{\partial^{2}q}{\partial u_{k}\partial u_{j}}=\frac{\partial^{2}q}{\partial u_{j}\partial u_{k}},$$

or

$$\displaystyle\sum_{i}a_{ik}\frac{\partial^{2}\eta}{\partial u_{i}\partial u_{j}}=\sum_{i}a_{ij}\frac{\partial^{2}\eta}{\partial u_{i}\partial u_{k}}\ \ \text{for $1\leq j,k\leq n$.}$$

If we wish to find an entropy flux for the entropy $\eta(u)=\left|u\right|^{2}/2$, note that

$$\displaystyle\frac{\partial^{2}\eta}{\partial u_{i}\partial u_{k}}=\delta_{ik}.$$

Thus we can find an entropy flux if $a_{jk}=a_{kj}$ for $1\leq j,k\leq n$; in other words, A must be symmetric. In this case the entropy flux q reads

$$\displaystyle q(u)=\sum_{i,j}a_{ij}u_{i}u_{j}-\frac{1}{2}\sum_{i}a_{ii}u_{i}^{2}.$$

Hence, an entropy (using the entropy $\eta(u)=\left|u\right|^{2}/2$) solution satisfies

$$\displaystyle\left|u\right|^{2}_{t}+q(u)_{x}\leq 0\ \ \text{weakly.}$$

This means that

$$\displaystyle\left\|u(\,\cdot\,,t)\right\|_{L^{2}(\mathbb{R})}\leq\left\|u_{0}\right\|_{L^{2}(\mathbb{R})},$$

and thus there is at most one entropy solution to (1.63) if A is symmetric.

1.1.7 The Riemann Problem for Linear Systems

Returning to the general case, recall that the solution $u=u(x,t)$ to (1.63) is given by (1.65 ), namely

$$\displaystyle u(x,t)=\sum_{i=1}^{n}\left[l_{i}\cdot u_{0}(x-\lambda_{i}t)\right]r_{i}.$$

(1.67)

Now we shall look at a type of initial value problem where u ₀ is given by two constant values, namely

$$\displaystyle u_{0}(x)=\begin{cases}u_{\text{left}}&x<0,\\ u_{\text{right}}&x\geq 0,\end{cases}$$

(1.68)

where $u_{\text{left}}$ and $u_{\text{right}}$ are two constant vectors. This type of initial value problem is called a Riemann problem, (cf. (1.28)) which will a problem of considerable interest throughout the book.

For a single equation (n = 1), the weak solution to this Riemann problem reads

$$\displaystyle u(x,t)=u_{0}(x-\lambda_{1}t)=\begin{cases}u_{\text{left}}&x<\lambda_{1}t,\\ u_{\text{right}}&x\geq\lambda_{1}t.\end{cases}$$

Note that u is not continuous. Nevertheless, it is the unique entropy solution in the sense of Definition 1.10 to (1.63) with initial data (1.68) (see Exercise 1.4).

For two equations (n = 2), we write

$$\displaystyle u_{\text{left}}=\sum_{i=1}^{2}\left[l_{i}\cdot u_{\text{left}}\right]r_{i},\qquad u_{\text{right}}=\sum_{i=1}^{2}\left[l_{i}\cdot u_{\text{right}}\right]r_{i}.$$

We can find the solution of each component separately. Namely, using (1.67) for initial data (1.68), we obtain

$$\displaystyle\left[l_{1}\cdot u(x,t)\right]=\begin{cases}l_{1}\cdot u_{\text{left}}&x<\lambda_{1}t,\\ l_{1}\cdot u_{\text{right}}&x\geq\lambda_{1}t,\end{cases}\quad\left[l_{2}\cdot u(x,t)\right]=\begin{cases}l_{2}\cdot u_{\text{left}}&x<\lambda_{2}t,\\ l_{2}\cdot u_{\text{right}}&x\geq\lambda_{2}t.\end{cases}$$

Combining these we see that

$$\begin{aligned}\displaystyle u(x,t)&\displaystyle=\left[l_{1}\cdot u(x,t)\right]r_{1}+\left[l_{2}\cdot u(x,t)\right]r_{2}\\ \displaystyle&\displaystyle=\begin{cases}\left[l_{1}\cdot u_{\text{left}}\right]r_{1}+\left[l_{2}\cdot u_{\text{left}}\right]r_{2}&x<\lambda_{1}t,\\ \left[l_{1}\cdot u_{\text{right}}\right]r_{1}+\left[l_{2}\cdot u_{\text{left}}\right]r_{2}&t\lambda_{1}<x\leq t\lambda_{2},\\ \left[l_{1}\cdot u_{\text{right}}\right]r_{1}+\left[l_{2}\cdot u_{\text{right}}\right]r_{2}&x\geq\lambda_{1}t,\end{cases}\\ \displaystyle&\displaystyle=\begin{cases}u_{\text{left}}&x<\lambda_{1}t,\\ u_{\mathrm{middle}}&t\lambda_{1}<x\leq t\lambda_{2},\\ u_{\text{right}}&x\geq\lambda_{1}t,\end{cases}\end{aligned}$$

with $u_{\mathrm{middle}}=\left[l_{1}\cdot u_{\text{right}}\right]r_{1}+\left[l_{2}\cdot u_{\text{left}}\right]r_{2}$. Observe the structure of the different states:

$$\begin{aligned}\displaystyle u_{\text{left}}&\displaystyle=\left[l_{1}\cdot u_{\text{left}}\right]r_{1}+\left[l_{2}\cdot u_{\text{left}}\right]r_{2},\\ \displaystyle u_{\mathrm{middle}}&\displaystyle=\left[l_{1}\cdot u_{\text{right}}\right]r_{1}+\left[l_{2}\cdot u_{\text{left}}\right]r_{2},\\ \displaystyle u_{\text{right}}&\displaystyle=\left[l_{1}\cdot u_{\text{right}}\right]r_{1}+\left[l_{2}\cdot u_{\text{right}}\right]r_{2}.\end{aligned}$$

We can also view the solution in phase space, that is, in the $(u_{1},u_{2})$-plane. We see that for every $u_{\text{left}}$ and $u_{\text{right}}$, we have the solution $u(x,t)=u_{\text{left}}$ for $x<\lambda_{1}t$ and $u(x,t)=u_{\text{right}}$ for $x\geq\lambda_{2}t$. In the middle, $u(x,t)=u_{\mathrm{middle}}$ for $\lambda_{1}t\leq x<\lambda_{2}t$. The middle value $u_{\mathrm{middle}}$ is on the intersection of the line through $u_{\text{left}}$ parallel to r ₁ and the line through $u_{\text{right}}$ parallel to r ₂. See Fig. 1.7. In the general, nonlinear, case, the straight lines connecting $u_{\text{left}}$, u _m , and $u_{\text{right}}$ will be replaced by arcs, not necessarily straight. However, the same structure prevails, at least locally.

Fig. 1.7

The solution of the Riemann problem. a In $(x,t)$-space. b In phase space

Now we can find the solution to the Riemann problem for any n, namely

$$\displaystyle u(x,t)=\begin{cases}u_{\text{left}}&x<\lambda_{1}t,\\ u_{i}&\lambda_{i}t\leq x<\lambda_{i+1}t,\quad i=1,\ldots,n-1,\\ u_{\text{right}}&x\geq\lambda_{n}t,\end{cases}$$

where

$$\displaystyle u_{i}=\sum_{j=1}^{i}\left[l_{j}\cdot u_{\text{right}}\right]r_{j}+\sum_{j=i+1}^{n}\left[l_{j}\cdot u_{\text{left}}\right]r_{j}.$$

Observe that this solution can also be viewed in phase space as the path from $u_{0}=u_{\text{left}}$ to $u_{n}=u_{\text{right}}$ obtained by going from $u_{i-1}$ to u _i on a line parallel to r _i for $i=1,\ldots,n$. This viewpoint will be important when we consider nonlinear equations, where the straight lines will be replaced by arcs. Locally, the structure will remain unaltered.

1.1.8 Numerics for Linear Systems with Constant Coefficients

If $\lambda_{i}> 0$, then we know that the scheme

$$\displaystyle D_{\!+}^{t}w_{i,j}^{m}+\lambda_{i}D_{-}w_{i,j}^{m}=0$$

will produce a sequence of functions $\left\{w_{i,{\Updelta x}}\right\}$ that converges to the unique entropy solution of

$$\displaystyle\frac{\partial w_{i}}{\partial t}+\lambda_{i}\frac{\partial w_{i}}{\partial x}=0.$$

Similarly, if $\lambda_{i}<0$, the scheme

$$\displaystyle D_{\!+}^{t}w_{i,j}^{m}+\lambda_{i}D_{+}w^{m}_{i,j}=0$$

will give a convergent sequence. Both of these schemes will be convergent only if ${\Updelta t}\leq{\Updelta x}\left|\lambda_{i}\right|$, which is the CFL condition. In eigenvector coordinates, with

$$\displaystyle w=\begin{pmatrix}w_{1}\\ \vdots\\ w_{n}\end{pmatrix},\quad w_{j}^{m}\approx w(j{\Updelta x},m{\Updelta t}),$$

the resulting scheme for w reads

$$\displaystyle D_{\!+}^{t}w^{m}_{j}+\Lambda_{+}D_{-}w^{m}_{j}+\Lambda_{-}D_{+}w^{n}_{j}=0,$$

(1.69)

where

$$\displaystyle\Lambda_{-}=\begin{pmatrix}{\lambda_{1}}\wedge{0}&&0\\ &\ddots&\\ 0&&{\lambda_{n}}\wedge{0}\end{pmatrix}\ \text{ and }\ \Lambda_{+}=\begin{pmatrix}{\lambda_{1}}\vee{0}&&0\\ &\ddots&\\ 0&&{\lambda_{n}}\vee{0}\end{pmatrix},$$

and we have introduced the notation

$$\displaystyle{a}\vee{b}=\max\left\{a,b\right\}\ \text{and}\ {a}\wedge{b}=\min\left\{a,b\right\}.$$

Observe that $\Lambda=\Lambda_{+}+\Lambda_{-}$. If the CFL condition

$$\displaystyle\frac{{\Updelta t}}{{\Updelta x}}\max_{i}\left|\lambda_{i}\right|=\frac{{\Updelta t}}{{\Updelta x}}\max\left\{\left|\lambda_{1}\right|,\left|\lambda_{n}\right|\right\}\leq 1$$

holds, then the scheme (1.69) will produce a convergent sequence, and the limit w will be the unique entropy solution to

$$\displaystyle w_{t}+\Lambda w_{x}=0.$$

(1.70)

By defining $u=Rw$, we obtain a solution of (1.63).

We can also make the same transformation on the discrete level. Multiplying (1.69) by L from the left and using that $u=Rw$ yields

$$\displaystyle D_{\!+}^{t}u^{m}_{j}+A_{+}D_{-}u^{m}_{j}+A_{-}D_{+}u^{m}_{j}=0,$$

(1.71)

where

$$\displaystyle A_{\pm}=R\Lambda_{\pm}L,$$

and this finite difference scheme will converge directly to u.

1.2 Notes

Never any knowledge was delivered in the same order it was invented. — Sir Francis Bacon ⁶(1561–1626)

The simplest nontrivial conservation law, the inviscid Burgers equation, has been extensively analyzed. Burgers introduced the ‘‘nonlinear diffusion equation’’

$$\displaystyle u_{t}+\frac{1}{2}(u^{2})_{x}=u_{xx},$$

(1.72)

which is currently called (the viscous) Burgers’s equation, in 1940 37; (see also 38; ) as a model of turbulence. Burgers’s equation is linearized, and thereby solved, by the Cole–Hopf transformation 46; ; 98; . Both the equation and the Cole–Hopf transformation were, however, known already in 1906; see Forsyth 66; , p. 100. See also Bateman 14; . The early history of hyperbolic conservation laws is presented in 56; , pp. XV–XXX. A source of some of the early papers is 104; .

The most common elementary example of application of scalar conservation laws is the model of traffic flow called ‘‘traffic hydrodynamics’’ that was introduced independently by Lighthill and Whitham 134; and Richards 155; . A modern treatment can be found in Haberman 81; . Example 1.6 presents some of the fundamentals, and serves as a nontechnical introduction to the lack of uniqueness for weak solutions. Extensions to traffic flow on networks exist; see 94; and 68; .

The jump condition, or the Rankine–Hugoniot condition, was derived heuristically from the conservation principle independently by Rankine in 1870 152; and Hugoniot in 1886 101; ; 102; ; 103; . Our presentation of the Rankine–Hugoniot condition is taken from Smoller 169; .

The notion of ‘‘Riemann problem’’ is fundamental in the theory of conservation laws. It was introduced by Riemann in 1859 156; ; 157; in the context of gas dynamics. He studied the situation in which one initially has two gases with different (constant) pressures and densities separated by a thin membrane in a one-dimensional cylindrical tube. See 97; and 56; , pp. XV–XXX for a historical discussion.

The final section of this chapter contains a detailed description of the one-dimensional linear case, both in the scalar case and in the case of systems. This allows us to introduce some of the methods in a simpler case. Here existence of solutions is shown using appropriate finite difference schemes, in contrast to the front-tracking method used in the text proper.

There are by now several books on various aspects of hyperbolic conservation laws, starting with the classical book by Courant and Friedrichs 51; . Nice treatments with emphasis on the mathematical theory can be found in books by Lax 126; ; 127; , Chorin and Marsden 42; , Roždestvenskiı̆ and Janenko 164; , Smoller 169; , Rhee, Aris, and Amundson 153; ; 154; , Málek et al. 141; , Hörmander 99; , Liu 137; , Serre 167; ; 168; , Benzoni-Gavage and Serre 15; , Bressan 24; ; 27; , Dafermos 56; , Lu 139; , LeFloch 129; , Perthame 150; , Zheng 192; . The books by Bouchut 19; , Godlewski and Raviart 78; ; 79; , LeVeque 130; ; 131; , Kröner 116; , Toro 180; , Thomas 179; , and Trangenstein 183; focus more on the numerical theory.

1.3 Exercises

• 1.1 Determine characteristics for the following quasilinear equations:

  $$\displaystyle\begin{aligned}\displaystyle u_{t}+\sin(x)u_{x}&\displaystyle=u,\\ \displaystyle\sin(t)u_{t}+\cos(x)u_{x}&\displaystyle=0,\\ \displaystyle u_{t}+\sin(u)u_{x}&\displaystyle=u,\\ \displaystyle\sin(u)u_{t}+\cos(u)u_{x}&\displaystyle=0.\end{aligned}$$

• 1.2 Use characteristics to solve the following initial value problems:

  • (a) $uu_{x}+xu_{y}=0$, $u(0,s)=2s\text{ for $s> 0$}.$

  • (b) $e^{y}u_{x}+uu_{y}+u^{2}=0$, $u(x,0)=1/x\text{ for $x> 0$}.$

  • (c) $xu_{y}-yu_{x}=u$, $u(x,0)=h(x)\text{ for $x> 0$}.$

  • (d) $(x+1)^{2}u_{x}+(y-1)^{2}u_{y}=(x+y)u$, $u(x,0)=-1-x.$

  • (e) $u_{x}+2xu_{y}=x+xu$, $u(1,y)=e^{y}-1.$

  • (f) $u_{x}+2xu_{y}=x+xu$, $u(0,y)=y^{2}-1.$

  • (g) $xuu_{x}+u_{y}=2y$, $u(x,0)=x.$

• 1.3

  • (a) Use characteristics to show that

    $$\displaystyle u_{t}+au_{x}=f(x,t),\quad u|_{t=0}=u_{0},$$

    with a a constant, has solution

    $$\displaystyle u(x,t)=u_{0}(x-at)+\int_{0}^{t}f(x-a(t-s),s)\,ds.$$

  • (b) Show that

    $$\displaystyle u\left(\xi(t;x_{0}),t\right)=u_{0}(x_{0})+\int_{0}^{t}f(\xi(s;x_{0}),s)\,ds$$

    holds if u is the solution of

    $$\displaystyle u_{t}+a(x,t)u_{x}=f(x,t),\quad u|_{t=0}=u_{0},$$

    (1.73)

    where ξ satisfies

    $$\displaystyle\frac{d}{dt}\xi(t;x_{0})=a(\xi(t;x_{0}),t),\ \ \xi(0;x_{0})=x_{0}.$$

  • (c) Show that

    $$\displaystyle u(x,t)=u_{0}\left(\zeta(t;x)\right)+\int_{0}^{t}f\left(\zeta(\tau;x),t-\tau\right)\,d\tau$$

    holds if u is the solution of (1.73) and

    $$\displaystyle\frac{d}{d\tau}\zeta(\tau;x)=-a(\zeta(\tau;x),t-\tau),\ \ \zeta(0;x)=x.$$

• 1.4 Show that

  $$\displaystyle u(x,t)=\begin{cases}u_{\text{left}}&x<at,\\ u_{\text{right}}&x\geq at,\end{cases}$$

  is the entropy solution in the sense of Definition 1.10 for the equation $u_{t}+au_{x}=0$ (where a is constant) and $u|_{t=0}(x)=u_{\text{left}}\chi_{x<0}+u_{\text{right}}\chi_{x\geq 0}$.

• 1.5 Find the shock condition (i.e., the Rankine–Hugoniot condition) for one-dimensional systems, i.e., the unknown u is a vector $u=\left(u_{1},\dots,u_{n}\right)$ for some n > 1, and also $f(u)=\left(f_{1}(u),\dots,f_{n}(u)\right)$.

• 1.6 Consider a scalar conservation law in two space dimensions,

  $$\displaystyle u_{t}+\frac{\partial f(u)}{\partial x}+\frac{\partial g(u)}{\partial y}=0,$$

  where the flux functions f and g are continuously differentiable. Now the unknown u is a function of x, y, and t. Determine the Rankine–Hugoniot condition across a jump discontinuity in u, assuming that u jumps across a regular surface in $(x,y,t)$. Try to generalize your answer to a conservation law in n space dimensions.

• 1.7 We shall consider a linearization of Burgers’s equation. Let

  $$\displaystyle u_{0}(x)=\begin{cases}1&\text{for $x<-1$},\\ -x&\text{for $-1\leq x\leq 1$},\\ -1&\text{for $1<x$}.\end{cases}$$

  • (a) First determine the maximum time that the solution of the initial value problem

    $$\displaystyle u_{t}+\frac{1}{2}\left(u^{2}\right)_{x}=0,\quad u(x,0)=u_{0}(x),$$

    will remain continuous. Find the solution for t less than this time.

  • (b) Then find the solution v of the linearized problem

    $$\displaystyle v_{t}+u_{0}(x)v_{x}=0,\quad v(x,0)=u_{0}(x).$$

    Determine the solution also in the case $v(x,0)=u_{0}(\alpha x)$, where α is nonnegative.

  • (c) Next, we shall determine a procedure for finding u by solving a sequence of linearized equations. Fix $n\in\mathbb{N}$. For t in the interval $(m/n,(m+1)/n]$ and $m\geq 0$, let v _n solve

    $$\displaystyle(v_{n})_{t}+v_{n}\left(x,m/n\right)\left(v_{n}\right)_{x}=0,$$

    and set $v_{n}(x,0)=u_{0}(x)$. Then show that

    $$\displaystyle v_{n}\left(x,\frac{m}{n}\right)=u_{0}\left(\alpha_{m,n}x\right)$$

    and find a recurrence relation (in m) satisfied by $\alpha_{m,n}$.

  • (d) Assume that

    $$\displaystyle\lim_{n\to\infty}\alpha_{m,n}=\bar{\alpha}(t),$$

    for some continuously differentiable $\bar{\alpha}(t)$, where $t=m/n<1$. Show that $\bar{\alpha}(t)=1/(1-t)$, and thus $v_{n}(x)\to u(x)$ for t < 1. What happens for $t\geq 1$?

• 1.8

  • (a) Solve the initial value problem for Burgers’s equation

    $$\displaystyle u_{t}+\frac{1}{2}\left(u^{2}\right)_{x}=0,\quad u(x,0)=\begin{cases}0&\text{for $x<0$},\\ 1&\text{for $x\geq 0$}.\end{cases}$$

    (1.74)

  • (b) Then find the solution where the initial data are

    $$\displaystyle u(x,0)=\begin{cases}1&\text{for $x<0$},\\ 0&\text{for $x\geq 0$}.\end{cases}$$

  • (c) If we multiply Burgers’s equation by u, we formally find that u satisfies

    $$\displaystyle\frac{1}{2}\left(u^{2}\right)_{t}+\frac{1}{3}\left(u^{3}\right)_{x}=0,\quad u(x,0)=u_{0}(x).$$

    (1.75)

    Are the solutions to (1.74) you found in parts a and b weak solutions to (1.75)? If not, then find the corresponding weak solutions to (1.75). Warning: This shows that manipulations valid for smooth solutions are not necessarily so for weak solutions.

• 1.9 (169; , p. 250) Show that

  $$\displaystyle u(x,t)=\begin{cases}1&\text{for $x\leq(1-\alpha)t/2$},\\ -\alpha&\text{for $(1-\alpha)t/2<x\leq 0$},\\ \alpha&\text{for $0<x\leq(\alpha-1)t/2$},\\ -1&\text{for $x\geq(\alpha-1)t/2$}\end{cases}$$

  is a weak solution of

  $$\displaystyle u_{t}+\left(\frac{1}{2}u^{2}\right)_{x}=0,\quad u(x,0)=\begin{cases}1&\text{for $x\leq 0$},\\ -1&\text{for $x> 0$},\end{cases}$$

  for all $\alpha\geq 1$. Warning: Thus we see that weak solutions are not necessarily unique.

• 1.10 We outline a proof of some Gronwall inequalities.

  • (a) Assume that u satisfies

    $$\displaystyle u^{\prime}(t)\leq\gamma u(t).$$

    Show that $u(t)\leq e^{\gamma t}u(0)$.

  • (b) Assume now that u satisfies

    $$\displaystyle u^{\prime}(t)\leq C(1+u(t)).$$

    Show that $u(t)\leq e^{Ct}(1+u(0))-1$.

  • (c) Assume that u satisfies

    $$\displaystyle u^{\prime}(t)\leq c(t)u(t)+d(t),$$

    for $0\leq t\leq T$, where $c(t)$ and $d(t)$ are in $L^{1}([0,T])$. Show that

    $$\displaystyle u(t)\leq u(0)+\int_{0}^{t}d(s)\exp\big(\int_{s}^{t}c(\tilde{s})\,d\tilde{s}\big)\,ds$$

    for $t\leq T$.

  • (d) Assume that u is in $L^{1}([0,T])$ and that for $t\in[0,T]$,

    $$\displaystyle u(t)\leq C_{1}\int_{0}^{t}u(s)\,ds+C_{2}.$$

    Show that

    $$\displaystyle u(t)\leq C_{2}e^{C_{1}t}.$$

• 1.11 Consider the semidiscrete difference scheme (1.37). The goal of this exercise is to prove that a unique solution exists for all t > 0.

  • (a) Let $\eta(u)$ be a smooth function. Show that

    $$\displaystyle D_{-}\eta(u_{j})=\eta^{\prime}(u_{j})D_{-}u_{j}-\frac{{\Updelta x}}{2}\eta^{\prime\prime}(u_{j-1/2})\left(D_{-}u_{j}\right)^{2},$$

    where $u_{j-1/2}$ is some value between u _j and $u_{j-1}$.

  • (b) Assume now that $\eta^{\prime\prime}\geq 0$. Show that

    $$\displaystyle\frac{d}{dt}\sum_{j}\eta(u_{j})\leq\sup_{j}\left|D_{+}a_{j}\right|\sum_{j}\eta(u_{j}).$$

    Note that in particular, this holds for $\eta(u)=u^{2}$.

  • (c) Show that for fixed ${\Updelta x}$, and $u\in l_{2}$, the function $F:l_{2}\to l_{2}$ defined by $F_{j}(u)=aD_{-}u_{j}$ is Lipschitz continuous.

    If we view $u(t)_{j}=u_{j}(t)$, then the difference scheme (1.37) reads $u^{\prime}=-F(u)$. Since we know that the solution is bounded in l ₂, we cannot have a blowup, and the solution exists for all time.

• 1.12 Consider the fully discrete scheme (1.48). Show that

  $$\displaystyle\sum_{j}\eta(u^{n+1}_{j})\leq\sum_{j}\eta(u^{n}_{j})+{\Updelta t}\sum_{j}\eta(u^{n}_{j})D_{+}a^{n}_{j}.$$

  Use this to show that

  $$\displaystyle{\Updelta x}\sum_{j}\eta(u^{n}_{j})\leq e^{Ct_{n}}\left\|\eta(u_{0})\right\|_{L^{1}(\mathbb{R})},$$

  where C is a bound on a _x .

• 1.13 The linear variational wave equation reads

  $$\displaystyle\begin{gathered}\alpha_{tt}+c(x)\left(c(x)\alpha_{x}\right)_{x}=0,\qquad t> 0,\ x\in\mathbb{R},\\\alpha(x,0)=\alpha_{0}(x),\ \ \alpha_{t}(x,0)=\beta_{0}(x),\end{gathered}$$

  (1.76)

  where c is a positive Lipschitz continuous function, and $\alpha_{0}$ and $\beta_{0}$ are suitable initial data.

  • (a) Set $u=\alpha_{t}+c\alpha_{x}$ and $v=\alpha_{t}-c\alpha_{x}$. Find the equations satisfied by u and v.

  • (b) Find the solutions of these equations in terms of the characteristics.

  • (c) Formulate a difference scheme to approximate $u(x,t)$ and $v(x,t)$, and give suitable conditions on your scheme and the initial data (here you have a large choice) that guarantee the convergence of the scheme.

  • (d) Test your scheme with $c(x)=\sqrt{1+\sin^{2}(x)}$, $\alpha_{0}(x)=\max\left\{0,1-\left|x\right|\right\}$, $\beta_{0}=0$, and periodic boundary conditions in $[-\pi,\pi]$.

• 1.14 Consider the transport equation

  $$\displaystyle\begin{aligned}\displaystyle u_{t}+a(x,t)u_{x}&\displaystyle=0,\ \ t> 0,\ x\in\mathbb{R},\\ \displaystyle u(x,0)&\displaystyle=u_{0}(x).\end{aligned}$$

  We know that the (unique) solution can be written in terms of the backward characteristics, $u(x,t)=u_{0}(\zeta(t;x))$, where ζ solves

  $$\displaystyle\frac{d}{d\tau}\zeta(\tau;x)=-a(\zeta(\tau;x),t-\tau),\ \zeta(0;x)=x.$$

  We want to use this numerically. Write a routine that given t, u ₀, and a, calculates an approximation to $u(x,t)$ using a numerical method to find $\zeta(t;x)$. Test the routine for the initial function $u_{0}(x)=\sin(x)$, and for a given by (1.35) and (1.36), as well as for the example $a(x,t)=x^{2}\sin(t)$.