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\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.

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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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  • $\begingroup$ But not both are satisfying the initial value. So it seems reasonable to consider $\lim_{t\rightarrow 0} u(x,t)$ and to see if it will be the given initial value... $\endgroup$ – Alex Aug 22 '16 at 19:36
  • $\begingroup$ @Alex : There was a typo in my first answer, now corrected. You are right, only one satisfies the initial condition : in addition. $\endgroup$ – JJacquelin Aug 22 '16 at 20:47
  • $\begingroup$ Yeah. Well, now I see that this was asked a long time ago...sorry, sometimes I just look at questions not considering when they were asked...:). $\endgroup$ – Alex Aug 22 '16 at 23:17

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