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

$u_t=\Delta u,$

where $(t,X)\in R \times R^2$. Suppose $u(t,x,y)$ behaves like $e^{-\alpha t} xy$ for large $t$, meaning that the nodal set of $u(t,x,y)$ ($\{(x,y)\in R^2: u(t,x,y)=0\}$) is a small perturbation of the nodal set of $e^{-\alpha t} xy$ for large $t$ and it divides $R^2$ into precisely four regions with the same topology. Also $\lim_{t \rightarrow \infty} u(t,x,y)=0$.

  Can we prove that the nodal set of $u_0(x,y)=u(0,x,y)$ divides $R^2$ into at least four regions?

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?

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

$u_t=\Delta u,$

where $(t,X)\in R \times R^2$. Suppose $u(t,x,y)$ behaves like $e^{-\alpha t} xy$ for large $t$, meaning that the nodal set of $u(t,x,y)$ ($\{(x,y)\in R^2: u(t,x,y)=0\}$) is a small perturbation of the nodal set of $e^{-\alpha t} xy$ for large $t$ and it divides $R^2$ into precisely four regions with the same topology. Also $\lim_{t \rightarrow} u(t,x,y)=0$.

Can we prove that the nodal set of $u_0(x,y)=u(0,x,y)$ divides $R^2$ into at least four regions?

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

$u_t=\Delta u,$

where $(t,X)\in R \times R^2$. Suppose $u(t,x,y)$ behaves like $e^{-\alpha t} xy$ for large $t$, meaning that the nodal set of $u(t,x,y)$ ($\{(x,y)\in R^2: u(t,x,y)=0\}$) is a small perturbation of the nodal set of $e^{-\alpha t} xy$ for large $t$ and it divides $R^2$ into precisely four regions with the same topology. Also $\lim_{t \rightarrow \infty} u(t,x,y)=0$.

Can we prove that the nodal set of $u_0(x,y)=u(0,x,y)$ divides $R^2$ into at least four regions?

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

$u_t=\Delta u,$

where $(t,X)\in R \times R^2$. Suppose $u(t,x,y)$ behaves like $e^{-\alpha t} xy$ for large $t$, meaning that the nodal set of $u(t,x,y)$ ($\{(x,y)\in R^2: u(t,x,y)=0\}$) is a small perturbation of the nodal set of $e^{-\alpha t} xy$ for large $t$ and it divides $R^2$ into precisely four regions.

Can we prove that the nodal line of $u_0(x,y)=u(0,x,y)$ divides $R^2$ into at least four regions?

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

$u_t=\Delta u,$

where $(t,X)\in R \times R^2$. Suppose $u(t,x,y)$ behaves like $e^{-\alpha t} xy$ for large $t$, meaning that the nodal set of $u(t,x,y)$ ($\{(x,y)\in R^2: u(t,x,y)=0\}$) is a small perturbation of the nodal set of $e^{-\alpha t} xy$ for large $t$ and it divides $R^2$ into precisely four regions with the same topology. Also $\lim_{t \rightarrow} u(t,x,y)=0$.

Can we prove that the nodal set of $u_0(x,y)=u(0,x,y)$ divides $R^2$ into at least four regions?