[under construction]

Existence of entropy solutions for 1D conservation laws

Basic assumptions

We will construct entropy solutions of the conservation law
\begin{equation} \tag{C} u_t + F(u)_x = 0, \qquad u(x, 0)=g(x). \end{equation}
where we assume that there is a constant $\delta\gt0$ such that
  1. $F\in C^2(\R)$,
  2. $F'(u)\ge 0$ for all $u\in\R$
  3. $F''(u)\ge \delta$ for all $u\ge 0$
For simplicity we assume that all functions are periodic in $x$ with period $\ell$:
\begin{equation}\tag{Per} \forall x, t : \quad u(x+\ell, t) = u(x, t), \quad g(x+\ell) = g(x). \end{equation}
In the end we want to construct an entropy solution for any bounded measurable initial data $g\in L^\infty(\R)$, but for now we will assume that $g\in C^1(\R)$. In particular, we will assume that
  1. $0 \lt g(x) \lt M$ for all $x\in\R$
  2. $g'(x) \leq M_1$ for all $x\in\R$

The discretization and the approximate solutions $u^N(x, t)$

We discretize the equation in the spatial direction by choosing a large integer $N$ and setting $\Delta x = \ell/N$. Instead of considering the function $u(x, t)$ at all real $x$, we only consider its values at the discrete set of $x$ values $x_i = i\Delta x$ ($i\in\Z$). So we set \[ u_i(t) = u(i\Delta x, t), \quad i\in\Z, 0\le t \lt T \] for some fixed $T\gt0$. Because of the periodicity condition we have $u_{i+N}(t) = u_i(t)$ for all $t$ and all $i\in\Z$. Replacing the $x$-derivative in $F(u)_x$ in the conservation law by a discrete difference quotient, we arrive at this set of ordinary differential equations for the functions $u_i(t)$:
\begin{equation}\tag{Disc} \frac{du_i}{dt} + \frac{F(u_i(t)) - F(u_{i-1}(t))}{\Delta x} = 0, \qquad u_i(0) = g(x_i). \end{equation}
Note that we made a choice here, namely we picked a left differential quotient instead of a right differential quotient: i.e. we could also have picked \[ \frac{du_i}{dt} + \frac{F(u_{i+1}(t)) - F(u_{i}(t))}{\Delta x} = 0. \] As we will see below, there will be one point where our construction falls apart if we choose the right difference instead of the left difference (Disc).
For any $N\in\N$ we get a solution $u^N_i(t)$ of the discretized conservation law. We now define a sequence of step functions by setting
\begin{equation} \tag{4} u^N(x, t) = u^N_i(t) \text{ for } x\in(x_i, x_{i+1}), \; t\gt0. \end{equation}
Our strategy for proving the existence of an entropy solution to the conservation law is to do the following:

Existence for the discretized equation

Because of our periodicity assumption we have $u_{i+N}(t) = u_i(t)$ so that we really only have to consider the unknown functions $u_1(t)$, …, $u_N(t)$, and so that (Disc) is a system of $N$ ordinary differential equations. Since the function $F:\R\to\R$ is $C^1$, the theory of ODE provides us with solution $\{u_i(t)\}_{i\in\Z}$ that is defined on some short time interval $0\leq t\lt t_*$.
Lemma 1.   $0\leq u_i(t) \leq M_0$ for all $i\in\Z$, and $t\in [0, t_*)$.
Proof. We use a maximum principle argument: if at any time $t$ we choose $j$ so that $u_j(t) = \max_i u_i(t)$, then $u_j(t)\geq u_{j-1}(t)$, so that $F'\geq0$ implies $F(u_j(t)) \geq F(u_{j-1}(t))$, and so that the discretized conservation law (Disc) implies $u_j'(t)\leq0$. This suggests that $\max_i u_i(t)$ is a nonincreasing function of $t$, which is what the Lemma claims. The complete (but less transparent) argument goes like this:
Let $M_\varepsilon(t) = M_0 + \varepsilon + \varepsilon t$ where $\varepsilon\gt0$ is a (small) constant. We show by contradiction that $u_i(t)\lt M_\varepsilon(t)$ for all $i$ and all $t\in[0, t_*)$. If this were not true, then there would be a smallest $s\in[0, t_*)$ and a $j\in\Z$ with $u_j(s) = \max_i u_i(s) = M_\varepsilon(s)$. Since $M_\varepsilon(0) \gt M_0$ it cannot be true that $s=0$, so we have $s\in(0, t_*)$. Then $u_i(s)\leq u_j(s)$ for all $i\in\Z$, and for all $t\in[0, s]$. Using $F'(u)\geq0$ we conclude that \[ \frac{F(u_j(s)) - F(u_{j-1}(s))}{\Delta x} \geq0, \qquad u_j'(s) \geq \varepsilon. \] This contradicts the differential equations (Disc).
We have shown that $u_i(t)\leq M_\varepsilon(t)$ for all $i$, $t$, and for all $\varepsilon\gt0$. Letting $\varepsilon\to0$ we conclude that $u_i(t)\leq M_0$ for all $t$, $i$, as claimed.
One proves $u_i(t)\geq 0$ in the same way. ////
Consequence. The solution $\{u_i(t)\}_{i\in\Z}$ to the discretized conservation law (Disc) exists for all $t\geq0$.
Proof. The theory of ordinary differential equations implies that the solution will exist until it becomes unbounded, while Lemma 1 shows that the solution $u_i(t)$ remains bounded. ////

The entropy condition

Lemma 2.   The solution $u_i(t)$ to (Disc) satisfies \[ \frac{u_i(t) - u_{i-1}(t)}{\Delta x} \leq \frac{2}{\delta t}, \] for all $i\in\Z$, $t\gt0$.
Proof. Consider \[ v_i(t) = \frac{u_{i+1}(t) - u_i(t)}{\Delta x}. \] The differential equations for $u_i$ and $u_{i+1}$ imply \[ v_i'(t) + \frac{F(u_{i+1}) - 2 F(u_i) + F(u_{i-1})}{(\Delta x)^2} = 0. \] The Taylor expansion of $F$ at $u_i$ together with $F''(u) \geq \delta$ implies \[ F(u) \geq F(u_i) + F'(u_i) (u-u_i) + \frac{\delta}{2}(u-u_i)^2. \] Hence
\begin{equation} \tag{1} v_i'(t) + F'(u_i) \frac{v_i - v_{i-1}}{\Delta x} \leq -\frac{\delta}{2} \bigl( v_i^2 + v_{i-1}^2\bigr) \leq -\frac{\delta}{2}v_i^2. \end{equation}
For any constant $K\gt0$ we note that $\max _i v_i(t) \leq K/t$ holds true for small enough $t\gt0$. If the inequality fails for some positive $t$, then there is again a smallest $s\gt0$ such that $\max_i v_i(s) = K/s$, and we can choose $j\in\Z$ so that $v_j(s) = \max_i v_i(s)$. In particular $v_j(s)\geq v_{j-1}(s)$, so (1) implies that
\begin{equation}\tag{2} v_j'(s) \leq -\frac{\delta}{2}v_j(s)^2 = - \frac{K^2\delta}{2s^2}. \end{equation}
On the other hand $v_j(t)\leq K/t$ for $t\leq s$ with $v_j(s) = K/s$ implies
\begin{equation}\tag{3} v_j'(s) \geq \frac{d}{ds} \frac{K}{s} = -\frac{K}{s^2}. \end{equation}
Together, (2) and (3) imply \[ \forall s\gt 0:\; -\frac{K}{s^2} \leq -\frac{K^2\delta}{2s^2}\quad \text{i.e. } K \le \frac{2}{\delta}. \] We arrived at this conclusion by assuming that $\max _i v_i(t) \gt K/t$ for some $t\gt0$. Hence, if we choose $K\gt 2/\delta$ then we have shown that $v_i(t) \leq K/t$ for all $i\in\Z$ and $t\gt0$. Since this holds for all $K\gt 2/\delta$, we can let $K\searrow 2/\delta$, and the Lemma follows. ////

A derivative estimate

The inequality in Lemma 2 is a universal upper bound for $v_i(t)$ in the sense that the same upper bound holds for all initial conditions $g$. We can improve the upper bound by taking the initial condition $g$ into account. The biggest improvement appears as $t\searrow0$, which is good because the upper bound we have right now ($v_i(t) \leq 2/\delta t$) deteriorates as $t\searrow0$.
Lemma 3.    If $\{u_i(t)\}$ is a solution of the discretized conservation law, and if $g$ satisfies $g'(x)\leq M_1$ for all $x\in\R$, then $v_i(t)\leq M_1$ for all $i\in\Z$ and $t\gt 0$.
Proof. The mean value theorem implies that at $t=0$ we have $v_i(0) = g'(\xi_i)$ for some $\xi_i\in (x_i, x_{i+1})$. Hence we have $v_i(0)\leq M_1$.
For $t\gt 0$ we again use the equation (1) for $v_j(t)$ and apply a maximum principle argument. At any time $t_1\gt 0$ there will be some $j$ such that $v_j(t_1)=\max_i v_i(t_1)$. At this $j$ it then follows from (1) that either $v_j(t_1)=0$ or $v_j'(t_1)\lt 0$. This implies that $\max_i v_i(t)$ can never increase above its initial value and there never exceeds $M_1$. ////

Compactness

We have found upper bounds for the discrete derivative $v_i(t)$, but no lower bounds. Nevertheless, the fact that $0\leq u_i(t)\leq M_0$ combined with $v_i(t)\leq M_1$ implies an $L^1$ type bound for $|v_i(t)|$.
Lemma 4.   There is a constant $C$ such that \[ \sum_{i=1}^{N} \bigl\{|v_i(t)| + |u_i'(t)|\bigr\}\Delta x \leq C \] for all $N\in\N$, and all $t\geq0$.
Proof.  It follows from $v_i(t)\leq M_1$ that $k\mapsto u_k(t) - M_1k \Delta x$ is nonincreasing. Using the periodicity of $k\mapsto u_k(t)$ and $N\Delta x=\ell$, we then get
\[ \sum_{i=1}^N \{M_1 - v_i(t)\} \Delta x =M_1\underbrace{N\Delta x}_{=\ell} + \underbrace{u_{1}(t) - u_{N+1}(t)}_{=0} =M_1\ell. \] Note that $M_1-v_i(t)\geq0$ for all $i, t$, so \[ \sum_{i=1}^N |v_i(t)| \Delta x \leq \sum_{i=1}^N \bigl\{ M_1 + |M_1-v_i(t)|\bigr\} \Delta x \leq 2M_1 \ell. \] To get the estimate for $u_i'(t)$ we observe that \[ u_i'(t) = - \frac{F(u_i) - F(u_{i-1})}{\Delta x} = F'(\tilde u_i) v_{i-1}(t) \] where $\tilde u_i$ is a number between $u_{i-1}$ and $u_i$, provided by the Mean Value Theorem. Since $F\in C^1(\R)$ there is a constant $C_0$ such that $0\leq F'(u) \leq C_0$ for all $u\in [0, M_0]$. We have already found that our solutions $u_i$ are bounded by $0\leq u_i\leq M_0$, so $0\leq F'(\tilde u_i)\leq C_0$, and thus \[ |u_i'(t)| \leq C_0 |v_{i-1}(t)|. \] Multiplying with $\Delta x$ and summing over $i$ we then get the estimate for $\sum |u_i'(t)|\Delta x$. ////
Lemma 5.   For all $\xi\in(-1, 1)$ and $\tau\in(0,1)$ one has \[ \iint_{Q} \left| u^N(x, t) - u^N(x+\xi, t+\tau)\right|\, dx\,dt \leq C_1 \bigl(|\xi|+\tau\bigr) \] where the constant $C_1$ is independent of $N\in\N$.
This implies that the sequence $u^N(x, t)$ has a subsequence that converges in $L^1(Q)$, and therefore has a further subsequence that converges pointwise almost everywhere.

The limit is an integral solution

We will prove
\[ \lim_{N\to\infty} \iint \varphi_t u^N(x, t)\,dx\,dt + \iint \varphi_x F(u^N(x, t)) dx\,dt + \int g(x) \varphi(x, 0) \,dx =0.\tag{5} \]
For the first term we have \begin{align} \iint \varphi_t u^N(x, t) \, dx\,dt &= \sum_i \int_0^T \int_{x_i}^{x_{i+1}} \varphi_t(x, t) u_i(t) dt \\ &= -\sum_i \int_{x_i}^{x_{i+1}} \varphi(x, 0) g(x_i) dt -\sum_i \int_0^T \int_{x_i}^{x_{i+1}} \varphi(x, t) u_i'(t) dx dt \end{align} Uniform continuity of $g$ implies that \[ \sum_i \int_{x_i}^{x_{i+1}} \varphi(x, 0) g(x_i) dt = \int_0^\ell \varphi(x, 0)g(x)dx + o(1) \] as $N\to\infty$, so that
\begin{equation} \iint \varphi_t u^N(x, t) \, dx\,dt = - \int g(x) \varphi(x, 0) dx - \sum_i \int_0^T \int_{x_i}^{x_{i+1}} \varphi(x, t) u_i'(t) dt + o(1). \tag{6} \end{equation}
For the second term in (5) we get
\begin{align*} \iint \varphi_x F(u^N)\, dx\, dt &= \sum_i \int_0^T \int_{x_i}^{x_{i+1}} \varphi_x(x, t) F(u_i(t))\,dx\,dt\\ &= \sum_i \int_0^T \bigl(\varphi(x_{i+1}, t) - \varphi(x_i, t)\bigr) F(u_i(t)) dt \\ &= - \sum_i \int_0^T \varphi(x_i, t) \bigl\{F(u_i(t)) - F(u_{i-1}(t))\bigr\}\, dt\\ &=-\sum_i \int_0^T \int_{x_i}^{x_{i+1}} \varphi(x_i, t) \frac{F(u_i(t)) - F(u_{i-1}(t))}{\Delta x}\,dx\,dt \\ &=-\sum_i \int_0^T \int_{x_i}^{x_{i+1}} \varphi(x_i, t) u_i'(t) \,dx\,dt. \tag{7} \end{align*}
This last expression looks almost the same as the double integrals at the end of (6); the difference is that in (7) $\varphi$ is evaluated at $x_i$ while in (6) $\varphi$ is evaluated at $x$.
By combining (6), and (7) we find that \begin{multline} J^N \stackrel{\rm def}= \iint \Bigl\{\varphi_t u^N + \varphi_x F(u^N)\Bigr\}\, dx\,dt +\int g(x) \varphi(x,0)\, dx = \\ \sum_i \int_0^T\int_{x_i}^{x_{i+1}} \bigl(\varphi(x, t) -\varphi(x_i, t)\bigr) \frac{du_i^N}{dt} \,dx\,dt + o(1). \end{multline} The test function $\varphi$ is continuously differentiable and in the above integral we always have $x_i\lt x \lt x_{i+1}$, so that $|\varphi(x,t)-\varphi(x_i,t)|\leq C (x_{i+1} - x_i) = C \Delta x$, where $C = \sup |\varphi_x|$, and therefore \[ \left| \int_{x_i}^{x_{i+1}} \bigl( \varphi(x, t) - \varphi(x_i, t)\bigr) \frac{du^N_i}{dt} \,dx \right| \leq C (\Delta x)^2 \left|\frac{du^N_i}{dt} \right|. \] Hence, summing over $i=1, \dots, N$ and integrating over $t\in(0,T)$ we get \[ |J^N| \leq C\ell \Delta x \max_i \int_0^T \left|\frac{du^N_i}{dt} \right| \,dt + o(1) \leq C' \Delta x +o(1) = o(1). \]