Weak solution and Rankine-Hugoniot condition

Question

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

Answer 1

It’s true! All the passages in the proof that you already know are reversible.

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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!