1
$\begingroup$

Let $u(t,X)$ be a smooth solution of the heat equation on $R^2$

$u_t=\Delta u,$

where $(t,X)\in R \times R^2$. Suppose $\lim_{t \rightarrow 0} u(t,x,y)=x^2-y^2$. Can we prove that the nodal set of $u_0(x,y)=u(0,x,y)$ divides $R^2$ into at least four regions?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Can you be more specific about what you mean by "behaves like something for large $t$"? It'd be good to know what kinds of perturbations you are willing to allow. I think if you want $u(t,X)$ to behave like $v(t,X)$, it would be good if $v$ solved the heat equation. I get the idea that the asymptotic behavior of $u$ is prescribed, but I don't think it's detailed enough. $\endgroup$
    – Joonas Ilmavirta
    Commented Mar 5, 2015 at 2:30
  • 1
    $\begingroup$ might be relevant:arxiv.org/abs/1304.4562 $\endgroup$
    – Piyush Grover
    Commented Mar 5, 2015 at 2:30

1 Answer 1

Reset to default
2
$\begingroup$

No. Let $\phi$ be a compactly supported positive smooth function with $\phi(0)>0$. Now $u_0(X)=u_\infty(X)+\phi(X)$ is a possible initial value but the set $\{X;u_0(X)=0\}$ divides the plane to only three parts. The function $\phi$ essentially builds a bridge between the first and the third quadrant.

Using a different $\phi$ you can make the initial topology almost anything you like.

Improve this answer
$\endgroup$
2
  • $\begingroup$ How do we know that the solution of the heat equation with the initial value $u_0=u_{\infty}+\phi$ will converge to $u_{\infty}$? $\endgroup$
    – A random mathematician
    Commented Mar 5, 2015 at 1:29
  • 2
    $\begingroup$ @MathStudent, you can write the solution to the heat equation with initial condition $\phi$ using the Poisson kernel. From this integral you will see that the solution converges to zero pointwise. And I was assuming your final state $u_\infty$ is harmonic; replace it with a harmonic one and then everything works as I wrote. $\endgroup$
    – Joonas Ilmavirta
    Commented Mar 5, 2015 at 1:31

You must log in to answer this question.

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