Strong solution of inviscid Burgers' equation with initial data $u(x,0)=x^2$

I'm studying for an exam and having trouble solving the following:

Find the strong solution to the inviscid Burgers' equation $u_t+uu_x=0$ with initial data $u(x,0)=x^2$

Using the initial data, I've found $u(x,t)=f(x-ut)=(x-ut)^2=x^2-2xut+u^2t^2$

but I'm not sure how to proceed from there. Any hints or suggestions?

Solution 1:

This was asked before (here and here) but never got an explanation beyond formal manipulations, so here goes. To find a sensible solution of the Burgers' equation (to the extent that it's at all possible with your initial data) one has to consider the characteristics. From each point $(x_0,0)$ begins a characteristic of the form $x=x_0+tx_0^2$. Solving this for $x_0$, we find $$x_0=\frac{1}{2t}(-1\pm \sqrt{1+4xt}) \tag{1}$$ Therefore:

1. Through every point $(x,t)$ with $xt>-1/4$ there pass two characteristics.
2. There are no characteristics passing through $(x,t)$ with $xt<-1/t$.

Here is an illustration so far.

In order to sort this out, we have to terminate some characteristics sooner, so that such intersections do not occur. In the equation (1), the $+$ sign is preferable when $t\to0$, as it gets us $x_0\to x$ and subsequently $u(x,t)\to x^2$. The $-$ sign corresponds to characteristics coming from way out in the left. They all are tangent to the hyperbola $xt=-1/4$ and should terminate there. Here is a cleaned up picture:

And the corresponding solution:

$$u(x,t) = \frac{1}{(2t)^2}(-1+ \sqrt{1+4xt})^2,\quad x>-\frac1{4t} $$