I have to solve the following exercise: \begin{equation} \begin{cases} \partial_t u(t,x)-7\partial_x u(t,x)=0\\ u(t=0,x)=e^{-x^6} \end{cases} \end{equation} And I found the classical solution $u(t,x)=e^{-(x+7t)^6}$. Now the request is to find a solution for the problem: \begin{equation} \partial_t u(t,x)-7\partial_x u(t,x)=0\\ u(t=0,x)=e^{-x^6}\textbf{1}_{[-R,R]}(x) \end{equation} Since the initial datum is not continuous I'm looking for weak solution, hence solution that satisfy: $\forall\phi\in C^\infty_c(\mathbb{R}^2)$: \begin{equation} \int_{\mathbb{R}_+}\int_{\mathbb{R}}(\partial_t\phi-7\partial_x\phi)u(t,x)dxdt+\int_\mathbb{R}\phi(0,x)u(0,x)dx=0 \end{equation} Now my idea is to prove that: \begin{equation} u(t,x)=\begin{cases} e^{-(x+7t)^6} \ \ \ x\in[-R,R]\\ 0 \ \ \ \text{otherwise} \end{cases} \end{equation} is a weak solution,thus I need to verify that: \begin{equation} \int_{\mathbb{R}_+}\int_{-R}^R(\partial_t\phi(t,x)-7\partial_x\phi(t,x))e^{-(x+7t)^6}dxdt+\int_{-R}^R\phi(0,x)e^{-x^6}=0 \end{equation} Now use the following change of variable: $\tau=t$ and $\xi=x+7t$, we obtain that: \begin{equation} \int_{-R}^R\int_{\mathbb{R}_+}[\phi_\tau(\tau,\xi-7\tau)+7\phi_\xi(\tau,\xi-7\tau)-7\phi_\xi(\tau,\xi-7\tau)]e^{-\xi^6}d\tau d\xi=\\ =\int_{-R}^R\int_{\mathbb{R}_+}\phi_\tau(\tau,\xi-7\tau)e^{-\xi^6}d\tau d\xi=-\int_{-R}^R\phi(0,\xi)e^{-\xi^6}d\xi \end{equation} I have used that $\phi$ is compactly supported and hence the limit as $\tau\to\infty$ is zero. Now since this hold for any test function we have concluded that the function defined before is a week solution.
Now there is one last question on the existence in some sense of the limit as $R\to\infty$ of the solution
I would like to see if the weak solution $u_R(x,t)=e^{-(x+7t)^6}\textbf{1}_{[-R,R]}$ will converge to $u(t,x)$ in L^p; thus I have to compute the limit: \begin{equation} \lim_{R\to\infty}\int_0^\infty\int_{-\infty}^\infty|e^{-(x+7t)^6}-e^{-(x+7t)^6}\textbf{1}_{[-R,R]}|^pdxdt=\\ =\lim_{R\to\infty}\int_0^\infty\bigg(\int_{-\infty}^{-R}e^{-p(x+7t)^{6}}dx+\int_{R}^{\infty}e^{-p(x+7t)^{6}}dx\bigg)dt \end{equation} At this point I have the idea of bring the limit inside the integral using the continuity of the function but I'm not sure if the procedure is correct.
Finally does the computation for the first point make sense? Can someone give me an hint on how to proceed with the last point?
For this linear advection equation, the initial datum is simply shifted at speed $-7$ along the $x$-axis as times increase, independently of the regularity. For the second Cauchy problem (with the indicator function), the method of characteristics produces $$ u(t,x) = e^{-(x+7t)^6} {\bf 1}_{[-R,R]}(x+7t) . $$ Now you might want to prove that it is a weak solution of the initial-value problem. This solution is valid for $R$ arbitrarily large too. Even though it might not be the kind of proof you are looking for, the pointwise limit of $u(t,x)$ at fixed $t$, $x$ as $R$ becomes infinite is exactly $e^{-(x+7t)^6}$.