2

Given the following inital value problem problem

$$u_t+uu_x=0, \quad x \in \mathbb{R}, 0<t<\infty \\ u(x,0)=f(x)=x^2$$

I want to find the solution. Also I want to check if the formula holds for all $x \in \mathbb{R}, t>0$ and if there is a $C^1$ solution for each $x \in \mathbb{R}, t>0$.

Using the method of characteristics, I have shown that the solution is $$u(x,t)=f(x-ut)$$

Doesnt the fact that $f$ is smooth, imply that the formula holds for all $x \in \mathbb{R}, t>0$ and that it is a $C^1$ solution?

Or am I wrong at some point?

EditPiAf
  • 20,898
Mary Star
  • 13,956

1 Answers1

1

This PDE is the inviscid Burgers' equation. With $u=f(x-ut)$ and $f(x)=x^2$, we have $$ t^2u^2-(1+2xt)u+x^2=0, $$ which solution is $$ u(x,t) = \frac{1+2tx - \sqrt{1+4tx}}{2t^2} $$ for $tx\geq {-1}/4$. The solution is smooth in the interior of that domain, but undefined outside the domain (see e.g. this post for the characteristic curves).

Note that such a nonlinear hyperbolic equation can generate non-smooth solutions from smooth initial data, as shown in this post for instance, with a blow-up of the spatial gradient $v=u_x$. Indeed, differentiating the partial differential equation w.r.t. $x$, we have $$ v_t + u v_x = -v^2 . $$ Along the characteristics, the gradient evolves according to $v' = -v^2$, which solution blows up at the breaking time, if the initial gradient $v(x,0)=f'(x)$ is smaller than $-1$ somewhere.

EditPiAf
  • 20,898