Weak solution and Rankine-Hugoniot condition I know from PDE lectures that in case of the following Cauchy problem:
$$
\left\{
\begin{array}{l}
        u_{t}+uu_{x}=0\\
        u(x,0)=g(x)\\
\end{array}
\right.
$$
if a piecewise $C^{1}$ function $u$ is a weak solution of above problem, i.e. it satisfies $\int\limits_{0}^{\infty}\int\limits_{-\infty}^{\infty}\left(u\varphi_{t}+\frac{1}{2}u^{2}\varphi_{x}\right)dxdt+\int\limits_{-\infty}^{\infty}g(x)\varphi(x,0)=0$ for any $\varphi\in C^{1}_{0}(\Omega)$ (spaces of $C^{1}$ functions with the compact support contained in $\Omega:=\mathbb{R}\times[0,\infty)$), then $u$ is classical solution of given Cauchy problem in domains where it is a $C^{1}$ function and satisfies Rankine-Hugoniot condition along its every discontinuity curve. I wonder if reverse theorem is true, i.e. if $u$ is classical solution of given Cauchy problem in domains where it is a $C^{1}$ function and satisfies Rankine-Hugoniot condition along its every discontinuity curve, then $u$ is a weak solution? If so, how to prove it? Thanks
 A: It’s true! All the passages in the proof that you already know are reversible.

Edit:
Let $ [u] \stackrel{\text{df}}{=} {u_{+}}(p) - {u_{-}}(p) $ for all $ p \in \gamma_{0} = \{ (\xi(t),t) \mid t \in I \} $.
(R.H.S.): As $ \xi'(t) = \dfrac{[f(u)]}{[u]}(\xi(t),t) $ for all $ t \in I $, we have
\begin{align}
& [f(u)](\xi(t),t) - [u](\xi(t),t) \cdot \xi'(t) = 0 \\ \Longrightarrow \quad
& 0 = \int_{I}
      \Big( [f(u)](\xi(t),t) - [u](\xi(t),t) \cdot \xi'(t) \Big) \cdot
      \phi(\xi(t),t) ~
      \mathrm{d}{t}.
\end{align}
Let $ v = (v_{1},v_{2}) = (1,- \xi'(t)) $. We can then write the previous formula as follows:
\begin{align}
    0
& = \int_{I} ([f(u)] v_{1} + [u] v_{2}) \phi ~ \mathrm{d}{s} \\
& = \int_{\gamma_{0} \cap \text{supp}(\phi)}
    \Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~
    \mathrm{d}{s} -
    \int_{\gamma_{0} \cap \text{supp}(\phi)}
    \Big( f(u_{-}) v_{1} + u_{-} v_{2} \Big) \phi ~
    \mathrm{d}{s}.
\end{align}
We have taken $ \phi \in {C_{c}^{1}}(\Omega) $, so $ \text{supp}(\phi) = \omega_{-} \cup (\gamma_{0} \cap \text{supp}(\phi)) \cup \omega_{+} $.
This is the crucial point: We are going to use the Divergence Theorem in the ‘non-standard’ way:
\begin{align}
    \int_{\gamma_{0} \cap \text{supp}(\phi)}
    \Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~
    \mathrm{d}{s}
& = \int_{\partial \omega_{+}}
    \Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~\mathrm{d}{s} \\
& = \iint_{\omega_{+}}
    \left(
    u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
    \right) ~
    \mathrm{d}{x} \mathrm{d}{t} +
    \iint_{\omega_{+}}
    \left(
    \frac{\partial u}{\partial t} + \frac{\partial f(u)}{\partial x}
    \right) \phi ~ \mathrm{d}{x} \mathrm{d}{t}.
\end{align}
Note that the last integral is $ 0 $ because our solution is classical outside the shock. We now have the following equations:
$$
  \int_{\gamma_{0} \cap \text{supp}(\phi)}
  \Big( f(u_{+}) v_{1} + u_{+} v_{2} \Big) \phi ~
  \mathrm{d}{s}
= \iint_{\omega_{+}}
  \left(
  u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
  \right) ~
  \mathrm{d}{x} \mathrm{d}{t},
$$
and, similarly,
$$
  - \int_{\gamma_{0} \cap \text{supp}(\phi)}
  \Big( f(u_{-}) v_{1} + u_{-} v_{2} \Big) \phi ~ \mathrm{d}{s}
= \iint_{\omega_{-}}
  \left(
  u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
  \right) ~
  \mathrm{d}{x} \mathrm{d}{t}.
$$
Summing up term by term:
\begin{align}
    0
& = \int_{I} \Big( [f(u)] v_{1} + [u] v_{2} \Big) \phi ~ \mathrm{d}{s} \\
& = \iint_{\omega_{+} \cup \omega_{-}}
    \left(
    u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
    \right) ~
    \mathrm{d}{x} \mathrm{d}{t} \\
& = \int_{\text{supp}(\phi)}
    \left(
    u \frac{\partial \phi}{\partial t} + f(u) \frac{\partial \phi}{\partial x}
    \right) ~
    \mathrm{d}{x} \mathrm{d}{t}.
\end{align}
This is exactly what we were supposed to prove.
Hope it helps!
