TITLE: How to solve this PDE $u_t - u u_x=0$?
QUESTION [1 upvotes]: \begin{equation}u_t - u u_x=0-\infty< x< \infty,\ t>0 \end{equation}
\begin{equation} u(x,0)= \frac{1}{2}x^2 \end{equation}
By the method of characteristics, we have
\begin{equation} \frac{d x}{d t} = -\mu,\frac{d \mu}{d t}=0;\end{equation}
So,
\begin{equation*} \mu=C_1,x=-\mu t+C_2; \end{equation*}
We know $C_1$ should be a function of $C_2$. Then we can write
\begin{equation} \mu=C_1=C_1(C_2)=C_1(x+\mu t) \end{equation}
Notice the initial condition, we have 
\begin{equation} \mu(x,0)=C_1(x)=\frac{1}{2}x^2 \end{equation}
which implies that
\begin{equation} \mu=C_1(x+\mu t)=\frac{1}{2}(x+\mu t)^2 \end{equation}
Rearrange the equation, we have
\begin{equation} t^2\mu^2+2(tx-1)\mu+x^2=0 \end{equation}
But now I have two solutions, just wondering whether the above process is correct.

REPLY [3 votes]: You are on the right track. The solution on implicit form is :
$$t^2u^2+2(-1+xt)u+x^2=0$$
Solving for $u$ gives the result :
$$u(x,t)=\frac{1-xt\pm \sqrt{1-2xt}}{t^2}$$
You have not two functions, but only one multivaluated function.
http://mathworld.wolfram.com/MultivaluedFunction.html
Edit : A typo corrected and in addition :
The condition $u(x,0)$ implies $\quad u(x,t)=\frac{1-xt-\sqrt{1-2xt}}{t^2}$
In $t\to 0$ the series expansion leads to : $\sqrt{1-2xt}=1-xt-\frac{1}{2}x^2t^2+O(t^3)$
$u(x,t\to 0)=\frac{1-xt+1-xt-\frac{1}{2}x^2t^2+O(t^3)}{t^2}=\frac{1}{2}x^2+O(t)$
$$u(x,t\to 0)\to \frac{1}{2}x^2$$
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