Consider the initial value problem,
\begin{eqnarray}
u_t+f(u)_x=0 &\quad (x,t) \in \mathbb{R} \times \mathbb{R^+}\\
u(x,0)=u_0(x) &\quad x\in \mathbb{R}
\end{eqnarray}
For $f$ Lipschitz and $u_0 \in L^{\infty}(\mathbb{R})$ the above problem may admit multiple weak solution and unique solution is chosen so as to satisfy Kruzkhov entropy condition.
However when $f$ is a linear function given by $f(u)=au$ we don't impose such additional entropy condition. How to prove the uniqueness for the above mentioned IVP with $f(u)=au$?
Does the weak solution of transport equation satisfy the entropy condition automatically, i.e.
\begin{eqnarray}
\partial_t|u-k|+\partial_x \left(sgn(u-k)a(u-k)\right) \leq 0
\end{eqnarray}
for all $k \in \mathbb{R}$, in the sense of distribution.
If so how to prove it?
Best Answer
In the case of linear advection $f(u) = a u$, the solution obtained by the method of characteristics is unique, and it satisfies $u(x,t) = u_0(x-at)$. I don't know any example where multiple weak solutions shall be considered in the linear case. Anyway, let's check the entropy restrictions. Consider the convex entropy function $\eta = u^2$ which flux is $\psi = au^2$. For smooth solutions, we have $\eta_t + \psi_x = 0$. For discontinuous solutions, let's integrate $\eta_t + \psi_x$ over a rectangle around a discontinuity which Rankine-Hugoniot speed is $a$ (as done in Section 3.8.1 of (1)). This gives \begin{aligned} \left.\int_{x_1}^{x_2} u^2\text d x \right|_{t_1}^{t_2} + a\left.\int_{t_1}^{t_2} u^2\text d t \right|_{x_1}^{x_2} &= 0\cdot \Delta t + O(\Delta t^2) \leq 0 \, . \end{aligned} Remarkably, no entropy restriction is obtained at first order for the present contact discontinuity. A weak solution with discontinuity is automatically admissible. Or if you prefer, using non-negative test functions $\phi$ in $C_0^1(\Bbb R\times \Bbb R_+)$, the entropy inequality \begin{aligned} \iint \phi_t\eta+\phi_x\psi\,\text dx\text dt &= \iint (\phi_t +a\phi_x) \eta \,\text dx\text dt \\ &= \iint \varphi_t \eta \,\text d\xi\text dt \\ &= -\int [\varphi \eta]|_{t=0} \text d\xi \quad\leq\quad -\int [\phi \eta]|_{t=0} \text dx \end{aligned} is always satisfied: it is even an equality here. The conclusion is the same: a weak solution is always admissible.
NB: the change of variable $\varphi(\xi,t) = \phi(x,t)$ with $\xi = x-at$ has been used.
(1) R.J. LeVeque, Numerical Methods for Conservation Laws, Birkhauser, 1992.