Consider the initial value problem
\begin{eqnarray}
u_t+au_x=0\\
u(x,0)=u_0(x)
\end{eqnarray}
If $u$ and $v$ are two smooth solutions of the above problem how to show that $u=v.$\
I know that by Kruzhkov doubling of the variable, we can prove this uniqueness result for any $L^{\infty}$ data.
But is there any elementary way to prove it for smooth solution by using some energy functional kind of argument?.
What about the unquess of smooth solution for the wave equation \begin{eqnarray}
u_{tt}=u_{xx}\\
u(x,0)=f(x)\\
u_t(x,0)=g(x)
\end{eqnarray}
And in general for the hyperbolic conservation law
\begin{eqnarray}
u_t+f(u)_x=0\\
u(x,0)=u_0(x)
\end{eqnarray}
Can we prove the uniqueness for smooth solution using energy functional kind of argument.(Without using advanced techniques like doubling of variable and all)
Best Answer
The energy of the second-order wave equation $u_{tt} = u_{xx}$ is the sum of the total kinetic energy and of the total potential energy: $E = \frac12 \int_{\Bbb R} (u_t)^2 + (u_x)^2\, \text d x$. For smooth solutions, we have $$ \begin{aligned} \frac{\text d}{\text dt} E &= \int_{\Bbb R} u_t u_{tt} + u_x u_{xt}\, \text d x \\ &= \int_{\Bbb R} u_t u_{xx} + u_x u_{xt}\, \text d x \\ &= \left.[u_t u_{x}]\right|_{x \in \Bbb R} + \int_{\Bbb R} u_{tx} u_{x}\, \text d x - \int_{\Bbb R} u_x u_{xt}\, \text d x \\ &= 0 \end{aligned} $$ by integration by parts, if $u$, $u_t$ and $u_x$ have compact support.
One notes that we have the following factorization of the d'Alembert operator $$ \square u = (\partial_{tt} - \partial_{xx}) u = (\partial_{t} - \partial_{x})(\partial_{t} + \partial_{x}) u \, . $$ Hence, we may think that $\frac{\text d}{\text dt} E = 0$ remains true for the scalar advection equation $u_t \pm u_x = 0$. Indeed, the same derivation as for the second-order wave equation can be made due to the fact that $u_{tt} = \mp u_{tx} = u_{xx}$. Moreover, with $\Psi = \frac12 \int_{\Bbb R} u^2 \, \text d x$, we have $$ \begin{aligned} \frac{\text d}{\text dt} \Psi &= \int_{\Bbb R} u u_t \, \text d x \\ &= \mp \int_{\Bbb R} u u_x \, \text d x \\ &= \mp \tfrac12 \left.[u^2]\right|_{x \in \Bbb R} \\ &= 0 \, , \end{aligned} $$ for smooth $u$ with compact support. In the case of the conservation law $u_t + f(u)_x = 0$, we still have $$ \begin{aligned} \frac{\text d}{\text dt} \Psi &= \int_{\Bbb R} u u_t \, \text d x \\ &= - \int_{\Bbb R} u f(u)_x \, \text d x \\ &= -\left.[u f(u)]\right|_{x \in \Bbb R} + \int_{\Bbb R} u_x f(u) \, \text d x \\ &= -\left.[u f(u)]\right|_{x \in \Bbb R} + \left.[f(u)]\right|_{x \in \Bbb R} \\ &= 0 \end{aligned} $$ by integration by parts, if $u$ and $f\circ u$ have compact support. For instance, it should work for $u$ with compact support and for a flux $f$ which vanishes at the origin --- without loss of generality, as $g:u\mapsto f(u) - f(0)$ satisfies $u_t + g(u)_x = 0$ and $g(0) = 0$. Note that the spatial density $\psi = \frac12 u^2$ of $\Psi = \int \psi\,\text d x$ is a convex function of $u$. Hence, $\psi$ is a mathematical entropy, in the sense that $\psi(u)_t + h(u)_x = 0$, for the corresponding entropy flux $ h(u) = \int^u \psi'(v) f'(v)\,\text d v $.
All these estimates can be used for the uniqueness proof.