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
[Math] Weak solution and Rankine-Hugoniot condition
hyperbolic-equationspartial differential equations
Related Solutions
If needed, here is a sketch of the characteristic curves in $x$-$t$ plane, which may help to see better what happens:
The solution obtained by the method of characteristics up to the breaking time satisfies $u = \phi (x-ut)$ for $t<1$, i.e. $$ u(x,t) = \left\lbrace\begin{aligned} &1 & &\text{if}\quad x \leq t \\ &\tfrac{1-x}{1-t} & &\text{if}\quad t \leq x\leq 1\\ &0 & &\text{if}\quad x\geq 1 \end{aligned}\right. $$ At the breaking time, the shock speed $s$ is given by the Rankine-Hugoniot condition $s = \tfrac12 (1 + 0)$. Therefore, after the breaking time $t\geq 1$, $$ u(x,t) = \left\lbrace\begin{aligned} &1 & &\text{if}\quad x < 1 +\tfrac12 (t-1) \\ &0 & &\text{if}\quad x > 1 +\tfrac12 (t-1) \end{aligned}\right. $$
To your questions:
- The Rankine-Hugoniot condition $[\![\frac{1}{2}u^2]\!] = s [\![u]\!]$ is a condition satisfied by discontinuous weak solutions of the Burgers' equation. A weak solution which satisfies the Rankine-Hugoniot condition is not necessarily a solution to the proposed initial-value problem $u(x,0) = \phi(x)$. To be a solution, it must be in agreement with the characteristics equation until they cross, i.e. the fact that $u$ is constant along the curves $u = \phi(x - ut)$. Moreover, since weak solutions are not unique, it must satisfy an additional entropy condition. In the case of Burgers' equation, the Lax entropy condition amounts to the following statement: "if characteristics cross, then a shock-wave arises; but if they separate, a rarefaction wave occurs".
- The location of a discontinuity is deduced from the characteristics equation, whereas its speed is deduced from the Rankine-Hugoniot condition. In the present case, a shock occurs at the time $t=1$ (characteristics intersect). Its speed deduced from the Rankine-Hugoniot condition is $s=\frac{1}{2}$.
Let us make the change of variable $(\xi, \tau) = (x-at,t)$, in other words $(x,t) = (\xi + a\tau,\tau)$, so that \begin{aligned} &\phi_t = \phi_\xi \xi_t + \phi_\tau \tau_t = \phi_\tau - a\phi_\xi \\ &\phi_x = \phi_\xi \xi_x + \phi_\tau \tau_x = \phi_\xi \, . \end{aligned} Thus, using Fubini's theorem and integration by parts, \begin{aligned} \iint_{\Bbb R\times\Bbb R_+} u\, (\phi_t + a \phi_x)\, \text d x\,\text d t &= \iint_{\Bbb R\times\Bbb R_+} u \phi_\tau\, \text d \xi\,\text d \tau \\ &= \iint_{\Bbb R_+\times\Bbb R} u \phi_\tau\, \text d \tau\,\text d \xi \\ &= \int_{\Bbb R}\left[u\phi\right]_{\tau\in\Bbb R_+}\text d\xi - \iint_{\Bbb R_+\times\Bbb R} \underbrace{u_\tau}_{u_t + au_x =0} \phi\, \text d \tau\,\text d \xi \\ &= -\int_{\Bbb R}\left.(u\phi)\right|_{t = 0}\text dx \, . \end{aligned} We have shown that the definition holds for all smooth $\phi$ with compact support. Hence, $u(x,t) = g(x-at)$ is a weak solution to the Cauchy problem of the advection equation. Note that there is a sign mistake in OP.
Note that $u$ does not need to be continuous to apply integration by parts as above (see Wikipedia article, §Extension to other cases, and Wikipedia article, §A concrete example).
Best Answer
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!