Let $f$ be a smooth function on $R^2$, and define $N_f$ to be the set of points $p$ such that the nodal set of $f$ ($\{x\in R^2: f(x)=0\}$) divided every neighborhood of $p$ into four regions. Indeed, the nodal set of $f$ looks like $"\times"$ around $p$ and $f$ changes sign four times on every small enough circle centered at $p$.

Let $u(t,x)\in C^{\infty}(R\times R^2)$ be a solution of the heat equation $u_t=\Delta u$ with $u(0,x)=u_0(x)$, and assume that $N_{u_0}=\emptyset$. I wonder if the heat flow can stably generate such points in time, i.e. can $ \cup_{t>0} N_{u(t,x)}$ contain an unbounded continuous curve $C$ in $R^3$ such that $C \cap N_{u(t,x)} \neq \emptyset$ for all $t>t_1$ for some $0<t_1\in R$.

  • $\begingroup$ Have you tried to understand this property at the discrete level ? Considering some finite difference scheme ? $\endgroup$ Mar 9, 2015 at 12:51
  • $\begingroup$ Sounds like a good idea. $\endgroup$ Mar 10, 2015 at 0:25

3 Answers 3


Yes, this is possible (at least if you require only that $u$ be defined for $t>0$, which is the usual context for the heat equation, not for all $t \in {\bf R}$ as implied by "$u(t,x) \in C^\infty({\bf R} \times {\bf R}^2)$"):

(source: harvard.edu)

Here $u(t,x)$ is antisymmetric about the vertical axis $x_2=0$; thus the zero-set $V_{u(t,\cdot)} = \{x: u(t,x) = 0\}$ is always symmetric about that axis and contains it. The curved contours of that Sage plot, in black, red, orange, green, blue, purple, and gray, show the other component(s) of $V_{u,(t,\cdot)}$ for $t = t_1/8$, $t_1/4$, $t_1/2$, $t_1$, $2t_1$, $4t_1$, $8t_1$. The zero set $V_{u,(t,\cdot)}$ contains a $\ast$-shaped triple point for $t=t_1$, and two $+\,$-shaped double points for all $t > t_1$, at height proportional to $\pm \sqrt{t-t_1}$.

To obtain this function, start from the usual heat kernel $g(t,x) = (4\pi t)^{-1} \exp (-|x|^2/4t)$, and set $u(t,x) = \Delta_x(g(t,x-e_1)-g(t,x+e_1))$ where $e_1$ is the unit vector $(1,0)$. (To check that $u$ is a solution of the heat equation $u_t = \Delta_x u$, note that $g$ satisfies the heat equation and that the differential operators $\partial / \partial t$ and $\Delta_x$ commute with $\Delta_x$ and with translations by $\pm e_1$.) The nodal set of each term $\Delta_x(g(t,x\mp e_1))$ of $u$ is the circle of radius $2t^{1/2}$ about $\pm e_1$. For small $t$, the nodal set $V_{u(t,\cdot)}$ of the difference consists of the vertical axis and very close approximations to those circles. As $t$ increases, the circles grow and distort, eventually meeting to form a figure-eight at $t=t_1$ and a single closed curve for all $t > t_1$. The two double points for $t>t_1$ can be located as the zeros on the vertical axis $x_2=0$ of the partial derivative of $u(t,x)$ with respect to $x_1$.


Let me try to reformulate (too long for a comment) your question by specializing it to a more particular (and more stable) case. Let $u_0:\mathbb R^2\rightarrow \mathbb R$, be a (smooth) Morse function with index 0, i.e such that $$ du_0(x)=0\Longrightarrow u_0''(x) \text{ positive or negative definite}\Longleftrightarrow \det u''_0(x)>0. $$ The solution of the heat equation is $$ u(t,x)=(\Gamma(t)\ast u_0)(x),\quad \Gamma(t)(x)=(4πt)^{-1} \exp {-\frac{\vert x\vert^2}{4t}}. $$ A simple obvious result is that $u_0''(x)$ non-negative (as a two by two matrix) everywhere implies $u(t)_{xx}$ non-negative everywhere (as a two by two matrix), simply because the Gaussian kernel is non-negative. This gives a complete answer for initial data polynomial with degree less or equal than two. The same would work with a convexity (resp. concavity) assumption on $u_0$.

Now that does not prove that a Morse function with index 0 will not develop a saddle structure by the heat flow, but this reformulated question could be more tractable in this 2D setting.


If you by "can stably generate such points in time" asks if there exists some $u_0$ such that your statement holds, then just take the stationary solution $$u(t,x)=u_0(x):=x^2-y^2.$$ Then your curve $C$ is given by $(t,0,0)$, but I think that I have misunderstood your question since interpreting it as if there for any $u_0$ such that $N_{u_0}\ne \varnothing$ exists a curve $C$ such that..., it makes much more sense, and for which I don't know the answer.

  • $\begingroup$ In your example $N_{u_0} \neq \emptyset$. $\endgroup$ Mar 8, 2015 at 19:14

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