Find a self–similar solution with finite $\int_{-\infty}^\infty u\,dx$ to the following pde $(\ref{f})$.
\begin{equation} u_{t} = (u u_{x})_{x} \qquad -\infty< x <\infty, \quad t>0 \label{f} \end{equation}
Let $v(x,t) \equiv u_{\alpha,\beta,\gamma}(x,t) = \gamma u(\alpha x,\beta t)$. $v(x,t)$ satisfies $(\ref{f})$ if $\beta=\gamma\alpha^2$. Finite $\int_{-\infty}^\infty u\,dx$ implies $ \gamma=\alpha$ as $\alpha$ is assumed to be positive. Therefore, $\beta= \alpha^3$. Choose $\alpha = t^{-\frac{1}{3}}$.
\begin{equation}
v(x,t) = t^{-\frac{1}{3}} u(t^{-\frac{1}{3}} x, 1) \equiv t^{-\frac{1}{3}} \phi (t^{-\frac{1}{3}} x) \label{v}
\end{equation}
Substitute the partial derivatives of $v(x,t)$ into $(\ref{f})$ and define $\xi \equiv t^{-\frac{1}{3}} x $.
\begin{align}
v_{t} &= (v v_{x})_ {x} \\
-\frac{1}{3} t^{-\frac{4}{3}} \bigl( \xi\phi(\xi)\bigr)' &= t^{-\frac{4}{3}} \bigl(\phi(\xi) \phi’(\xi)\bigr)' \\
\phi(\xi) \bigl( \phi’(\xi) + \frac{1}{3} \xi \bigr) &= 0
\end{align}
Hence, either $\phi(\xi)=0$ or $\phi’(\xi) + \frac{1}{3} \xi=0$. We use the condition that $\int_{-\infty}^\infty v\,dx$ is finite to obtain a solution. In particular,
\begin{equation}
v(x,t) =\left\{\begin{aligned} -\frac{x^2}{6t}& & |x| \le C t^{\frac{1}{3}} \\ 0& & |x| > C t^{\frac{1}{3}} \end{aligned}\right.
\end{equation}
for some $C>0$. Note that $\int_{-\infty}^\infty v\,dx = - \bigl( \frac{C}{3} \bigr) ^ {3}$ is a finite constant.
