4
$\begingroup$

Let $u_0$ be a smooth function on the unit sphere $S^1$ and assume that $u(t,x)$ is a smooth solution of the heat equation with initial data $u(0,x)=u_0(x)$. How one can apply the maximum principle to prove that the number of critical points of $u(t,x)$ at any given time is not bigger than the number of critical points of $u_0$?

$\endgroup$

2 Answers 2

7
$\begingroup$

Since $u$ is nothing but a $2\pi$-periodic solution of $u_t=u_{xx}$, looking at critical points amounts to looking at the zeroes of $v:=u_x$, which is another solution of the same equation. Then your question is solved by H. Matano : Nonincrease of the lap-number of a solution for a one-dimensional semilinear parabolic equation. J. Fac. Sci. Univ. Tokyo Sect. IA Math. 29 (1982), no. 2, 401–441.

$\endgroup$
3
$\begingroup$

In arbitrary dimensions one has the following phenomenon. Start with a compact Riemann manifold $(M,g)$. Fix an orthonormal basis of $L^2(M)$ consisting of eigenfunctions $(\psi_k)_{k\geq 0}$ of the Laplacian. Assume that the spectrum is

$$ =0=\lambda_0<\lambda_1 \leq \lambda_2\leq \cdots \leq \lambda_k\leq cdots $$

where aboveeach eigenvalues appears as often as its multiplicity A smooth function $\newcommand{\bR}{\mathbb{R}}$ $u_0:M\to \bR$ has an eigenfunction expansion

$$u_0=\sum_{k\geq 0} A_k \psi_k. $$

Suppose now that the coefficients are independent normal random variables such that the above random function is almost surely smooth. (This happens if the variance of $A_k$ goes to zero very fast.) We get another random function

$$ u_t=e^{-t\Delta} u_0. $$

As $t\to\infty$ the expected number of critical points of $u_t$ converges to the number of critical points of the eigenfunction of $\Delta$ corresponding to the first nonzero eigenvalue.

On the other hand if you choose $u_0$ of the form

$$ u_{0, R}=\sum_{\lambda_k \leq R^2} C_k \psi_k, $$

where $(C_k)$ is sequence of independent standard normal random variables, then for large $R$, the expected number of critical points of the random function $u_{0, R}$ is approximatively $Z_m R^m$, where $m=\dim M$ and $Z_m$ is a universal constant that depends only on $m$. This random number of critical points is highly concentrated around its mean. On the other hand as $t\to\infty$ the function

$$ e^{-t\Delta} u_{0,R}, $$ will have, on average, fewer and fewer critical points. The animation below illustrates this phenomenon (in the case $M=S^1$).

enter image description here

Remark 1. Consider a random trigonometric polynomial of high degree $N$

$$u_0=\frac{1}{\sqrt{\pi}}\sum_{n=1}^N (A_n\cos n\theta+B_n\sin n\theta), $$

where $A_n, B_n$ are independent normal random variables.

We set $u_t=e^{-t\Delta} u_0$, and denote by $C_t(N)$ the expcted number of critical points of $u_t$. Then Kac-Rice formula implies that

$$C_t(N)=2\sqrt{\frac{\sum_{n=1}^N n^4 e^{-2tn^2}}{\sum_{n=1}^N n^2 e^{-2tn^2}}}. $$

One can prove the following.

For $t>0$ fixed we have

$$\lim_{N\to \infty}C_t(N)= 2. $$

For $t=0$ we have

$$ C_0(N)\sim 2\sqrt{\frac{3}{5}} N\;\;\mbox{as}\;\;N\to\infty. $$

Remark 2. Given a smooth function on a Riemannian manifold $(M,g)$ with eigenfunction decomposition

$$ u_0= v_1+\sum_{\lambda_k>\lambda 1}c_k\psi_k, $$

and $v_1$ is in the $\lambda_1$-eigenspace, then

$$u_t= e^{-t\Delta}u_0= e^{-t\lambda_1}v_1+\sum_{\lambda k >\lambda_1} e^{-\lambda_k t} c_k\psi_k, $$

then we observe that $u_t$ has the same number of critical points as

$$ U_t=v_1+\underbrace{\sum_{\lambda k >\lambda_1} e^{-(\lambda_k-\lambda_1) t} c_k\psi_k}_{=:R_t}. $$

We have

$$\lim_{t\to infty}\Vert R_t\Vert_{C^2(M)}=0. $$

The last condition implies that if $v_1$ is a Morse function, then the number of critical points of $U_t$ (or $u_t$) is equal to the number of critical points of $v_1$ if $t$ is sufficiently large. The above animation gives a depiction of $U_t$ for various moments of time $t$.

We typically expect $u_0$ to have a large number of critical points because the eigenfunctions $\psi_k$ for $k$ large are highly oscillatory.

$\endgroup$
5
  • $\begingroup$ All that I claimed is this will happen for a random $u_0$ with high probability. $\endgroup$ Jan 21, 2015 at 16:15
  • $\begingroup$ Thanks Liviu. Where can I find a proof for the fact that "the function $e^{-t\Delta} u_{0,R}$ will have, on average, fewer and fewer critical points" ? $\endgroup$ Jan 21, 2015 at 16:17
  • $\begingroup$ Is it true that there exists a sequence $t_n$ such that $t_n \rightarrow \infty $ and $e^{-t_n\Delta} u_0$ has fewer critical points than $u_0$ itself? $\endgroup$ Jan 21, 2015 at 18:43
  • 1
    $\begingroup$ See the new remarks I have added. $\endgroup$ Jan 21, 2015 at 21:55
  • $\begingroup$ Thanks Liviu. I just posted a related question and I would appreciate if you take a look at it: mathoverflow.net/questions/194522/… $\endgroup$ Jan 22, 2015 at 2:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.