Conservation of mass in hyperbolic PDE [reference request]

Question

Scalar, hyperbolic conservation laws, i.e. PDEs of the type

\begin{align} u_t + f(u)_x = 0 && u(0,x) = u_0(x)\end{align}

are designed with conservation of mass in mind. For convex $f$, together with the Oleinik entropy condition

$$ \frac{f(u^-) - f(u)}{u^- - u} \ge \frac{f(u^-) - f(u^+)}{u^- - u^+} \ge \frac{f(u^+)-f(u)}{u^+ - u}$$

it is known that they admit unique weak solutions for initial data $u_0\in L^\infty(\mathbb R)$. Now what are the minimal assumptions on $u_0\in L^\infty \cap L^1$ such that it can be proven that the entropy solution $u(t,x)$ conserves mass, i.e. for all $t>0$ (what it should do by design!)

$$ \int_{-\infty}^{+\infty} u(t,x) d x = \int_{-\infty}^{+\infty} u_0(x) d x $$

The best result I could find in the literature so far is Problem 13 from Evan's book "partial Differential equations", discussed in this thread. Here the additional assumption made is that $u(t,x)$ has compact support. However there are obviously cases where this does not apply, like $u_0(x) = e^{-x^2}$.

Answer 1

This is not a direct answer to the question about conservation of mass. But the analytic solution of the PDE could help.

GENERAL SOLUTION of $\quad u_t+f(u)_x=0\quad$

where $f(z)$ is a given (known) function, any dummy variable $z$. Thus $f'(z)=\frac{df}{dz}$ is a known function, which will be involved latter.

Solving, thanks to the method of characteristics :

$f(u)_x=\frac{\partial f(u)}{\partial x}=\frac{df}{du}\frac{\partial u}{\partial x}\quad\to\quad u_t+f'(u)u_x=0\quad \text{with symbol } f'(u)=\frac{df}{du}$

The system of ODEs for the characteristic lines is : $$\frac{dt}{1}=\frac{dx}{f'(u)}=\frac{du}{0}$$ A first family of characteristic lines comes from $\quad du=0\quad\to\quad u=c_1$

A second family of characteristic lines comes from $\quad \frac{dt}{1}=\frac{dx}{f'(c_1)}\quad\to\quad x-f'(c_1)t=c_2$

The general solution expressed on the form of implicit equation is $\quad \Phi(u,x-f'(u)t)=0\quad$ with any differentiable function $\Phi$ of two variables. Or on an equivalent form : $$u=F\left( x-f'(u)t\right) \quad\text{with any differentiable function }F.$$ PARTICULAR SOLUTION according to the initial condition $u(x,0)=u_0(x)$

$u(x,0)=u_0(x)=F\left( x-f'(u)0\right)=F(x)\quad$ determines the function $F$ : $$F(X)=u_0(X)\quad\text{any dummy variable }X.$$ Putting this function $F$ into $u=F\left( x-f'(u)t\right)$ with $\quad X=x-f'(u)t\quad$ gives : $$u=u_0\left(x-f'(u)t\right)$$ This is the analytic solution expressed on implicit form, where $u_0$ and $f'$ are given (known) functions.

Answer 2

Recall that the weak solution must satisfy

$$ \int_{0}^{+\infty} \int_{-\infty}^{+\infty} u \phi_t +f(u)\phi_x d x d t = -\int_{-\infty}^{+\infty} u_0(x)\phi(0,x) d x $$

One idea that might work is the following: for a second, ignore the requirement that the test function needs to be smooth and consider $\phi(t,x) = H(t_0 -t)$ for some $t_0>0$. When restricted to $\mathbb R \times [0,\infty)$ its support is $[0,t_0]$, i.e. compact. Then since $\phi_t = -\delta(t_0 -t) = -\delta(t-t_0)$ and $\phi_x = 0$ and $\phi(0,x) = H(t_0) = 1$ it follows that

$$ - \int_{-\infty}^{+\infty} u(t_0,x) d x = -\int_{-\infty}^{+\infty} u_0(x)\phi(0,x) d x $$

Which is exactly what we want to prove! The crux is of course that $\phi$ is not smooth. However $\phi$ can be approximated arbitrarily well by smooth functions so instead of using it directly we can simply take a sequence of smooth test functions converging against it and argue that the claim holds anyway in the limit.

Is this reasonable?