1
$\begingroup$

I'm studying the problem $u_x+u_t=x(1-u^2), u=g\in \mathcal{C}^1(\Omega)$ at $\partial\Omega$ , $g$ bounded, where $\Omega$ is $\{(x,t)\in\mathbb{R}^2|t>0\}$. I found the solution, but I don't know how to prove the uniqueness of this solution. The usual technic of taking two different solutions and proving that the difference of them is zero, doesn't work here, because the pde is not linear :(

$\endgroup$

1 Answer 1

1
$\begingroup$

You should examine the argument that led to finding $u$, to see if anything there might have a uniqueness statement. Presumably, you used the method of characteristics, in which case the argument goes like this:

  1. Fix $x_0\in \mathbb R$ and consider the restriction of $u$ to the line $x=x_0+t$, call it $v(t)=u(x_0+t,t)$.
  2. Observe that $v'(t)=(x_0+t)(1-v^2)$. This is a separable ODE you can solve, with the initial condition $v(0)=g(x_0)$.
  3. Furthermore, the solution is unique by the Picard theorem: the right hand side of the ODE is a continuously differentiable function, therefore locally Lipschitz.
  4. It follows that $u$ is uniquely determined on the characteristic line $x=x_0+t$. Since $x_0$ was arbitrary, $u$ is unique.
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .