The Wave Maps Equation and Brownian Paths

Abstract

We discuss the $(1+1)$-dimensional wave maps equation with values in a compact Riemannian manifold . Motivated by the Gibbs measure problem, we consider Brownian paths on the manifold as initial data. Our main theorem is the probabilistic local well-posedness of the associated initial value problem. The analysis in this setting combines analytic, geometric, and probabilistic methods.

Appendix A. Deterministic Ill-Posedness

We now prove the mild ill-posedness statement in Theorem 1.2, which concerns the unboundedness of the first Picard iterate.

Proof of Theorem 1.2.(ii)

We only treat the case ${\mathbb {S}}^2$, since the argument easily generalizes to ${\mathbb {S}}^{D-1}$ with $D\geqslant 3$. We split the argument into two steps. In the first step, we present several reductions which simplify the first Picard iterate. In the second step, we construct an explicit sequence of functions for which the first Picard iterate diverges.

Step 1: Reductions. We let $\phi _0,\phi _1\in C^\infty _b({\mathbb {R}}\rightarrow {\mathbb {R}}^3)$ satisfy

$$\begin{aligned} \Vert \phi _0(x) \Vert _2^2=1 \quad \text {and}\quad \langle \phi _0(x), \phi _1(x) \rangle =0 \qquad \forall x \in {\mathbb {R}}. \end{aligned}$$

(A.1)

Here, $\Vert \cdot \Vert _2$ refers to the Euclidean norm on ${\mathbb {R}}^3$. Then, the right and left-moving linear waves are given by

$$\begin{aligned} \phi ^\pm (x) = \frac{1}{2} \Big ( \phi _0(x) {\mp } \int _0^x \textrm{d}y\, \phi _1(y) \Big ). \end{aligned}$$

We recall that the second fundamental form of the sphere ${\mathbb {S}}^2$ is given by

$$\begin{aligned} \mathrm {I\!I}^k_{ij}(\phi )= \delta _{ij} \phi ^k. \end{aligned}$$

As a result, the first Picard iterate of the wave maps equation (WM) is given by

We now fix any time $t>0$. Furthermore, we let $\chi$ be our previous nonnegative, smooth cut-off function, which satisfies ${\text {supp}}(\chi ) \subseteq (-3,3)$, and let $0<\epsilon \leqslant t/100$. For any $r\in {\mathbb {R}}$, it then holds that

In order to prove the unboundedness of the first Picard iterate on $C^r\times C^{r-1}$ for any $r\leqslant 1/2$, it therefore suffices to prove that

(A.2)

We now choose $\phi _0(x)=e_3 \in {\mathbb {R}}^3$ and choose $\phi _1(x)=\psi ^\prime (x)$ for $\psi (x) \in C^\infty _c((-1,1)\rightarrow {\mathbb {R}}^3)$. In particular, it holds that $\Vert \psi \Vert _{C^{1/2}} \lesssim \Vert \phi _1 \Vert _{C^{-1/2}}$. Due to the geometric constraint in (A.1), $\psi$ has to satisfy $\langle e_3, \psi \rangle =0$. Using our assumptions, the first Picard iterate takes the form

In order to further simplify , we assume that ${\text {supp}}(\psi ) \subseteq (-2\epsilon ,2\epsilon )$. In particular, it holds that $\psi (x+t)=\psi (x-t)=0$ for all $x\in (-2\epsilon ,2\epsilon )$. Then, a direct computation yields that

$$\begin{aligned} \int _{x-t}^{x+t} \textrm{d}v^\prime \int _{x-t}^{v^\prime } \textrm{d}u^\prime \, 2e_3 \, \langle \psi ^\prime (u^\prime ), \psi ^\prime (v^\prime ) \rangle&= 0, \\ -\int _{x-t}^{x+t} \textrm{d}v^\prime \int _{x-t}^{v^\prime } \textrm{d}u^\prime \, \psi (u^\prime ) \langle \psi ^\prime (u^\prime ), \psi ^\prime (v^\prime ) \rangle&= -\frac{1}{2} \int _{x-t}^{x+t} \textrm{d}y\, \psi ^\prime (y) \Vert \psi (y) \Vert _2^2, \\ \int _{x-t}^{x+t} \textrm{d}v^\prime \int _{x-t}^{v^\prime } \textrm{d}u^\prime \, \psi (v^\prime ) \langle \psi ^\prime (u^\prime ), \psi ^\prime (v^\prime ) \rangle&= -\frac{1}{2} \int _{x-t}^{x+t} \textrm{d}y\, \psi ^\prime (y) \Vert \psi (y) \Vert _2^2. \end{aligned}$$

As a result, we obtain that

We now write $\psi (y)=\psi ^1(y) e_1 +\psi ^2(y) e_2$, which satisfies the constraint $\langle \psi ,e_3 \rangle =0$. Then, the first coordinate of is given by

Thus, it suffices to prove that

$$\begin{aligned} \sup _{\begin{array}{c} \psi ^1,\psi ^2 :\\ {\text {supp}}(\psi ^j)\subseteq (-2\epsilon ,2\epsilon ) \\ \Vert \psi ^j \Vert _{C^{1/2}}\leqslant 1 \end{array}} \Big | \int _{-\infty }^\infty \textrm{d}x \chi (x/\epsilon ) \int _{x-t}^{x+t} \textrm{d}y\, (\psi ^1)^\prime (y) (\psi ^2)^2(y) \Big |=\infty . \end{aligned}$$

(A.3)

Step 2: Proof of (A.3). The main idea in the proof of (A.3) is to create a severe high$\times$high$\times$low-interaction in the integrand. In order to cover the endpoint $C^{1/2}$, however, we need to be careful and work at multiple scales.

We let $\kappa ,\kappa _0 \in {\mathbb {N}}$ be arbitrary and define the set of frequencies

$$\begin{aligned} F_{\kappa ,\kappa _0} = \{ 2^{10k} :\kappa _0 \leqslant k \leqslant \kappa \}. \end{aligned}$$

Then, we define

$$\begin{aligned} \psi ^1(y)&= \chi (y/\epsilon ) \sum _{n\in F_{\kappa ,\kappa _0}} n^{-1/2} \sin (ny), \end{aligned}$$

(A.4)

$$\begin{aligned} \psi ^2(y)&= \chi (y/\epsilon ) \Big ( \sin (y) + \sum _{n\in F_{\kappa ,\kappa _0}} n^{-1/2} \sin ((n-1)y) \Big ). \end{aligned}$$

(A.5)

Since the frequencies in $F_{\kappa ,\kappa _0}$ are well-separated, it holds that

$$\begin{aligned} \Vert \psi ^1 \Vert _{C^{1/2}}, \Vert \psi ^2 \Vert _{C^{1/2}} \lesssim _\epsilon 1, \end{aligned}$$

where the implicit constant is uniform in $\kappa$ and $\kappa _0$. We now claim that

$$\begin{aligned}{} & {} \Big | \int _{x-t}^{x+t} \textrm{d}y\, (\psi ^1)^\prime (y) (\psi ^2)^2(y) + \, \frac{1}{2} (\kappa -\kappa _0) \int _{x-t}^{x+t} \textrm{d}y\, \chi (y/\epsilon )^3 (1-\cos (2y)) \Big |\nonumber \\{} & {} \quad \lesssim _\epsilon 2^{-5\kappa _0} (\kappa -\kappa _0) + 1. \end{aligned}$$

(A.6)

Before proving (A.6), we first show that (A.6) implies (A.3). By integrating (A.6) against $\chi (x/\epsilon )$, using that $1-\cos (2y)\geqslant 0$, and using that $0<\epsilon \leqslant t/10$, we obtain

$$\begin{aligned}&\, \bigg | \int _{-\infty }^\infty \textrm{d}x\, \chi (x/\epsilon ) \int _{x-t}^{x+t} \textrm{d}y\, (\psi ^1)^\prime (y) (\psi ^2)^2(y) \bigg | \\&\geqslant \, \frac{1}{2} (\kappa -\kappa _0) \bigg | \int _{-\infty }^\infty \textrm{d}x\, \chi (x/\epsilon ) \int _{x-t}^{x+t} \textrm{d}y\, \chi (y/\epsilon )^3 (1-\cos (2y)) \bigg | - C_\epsilon \big ( 2^{-5\kappa _0} (\kappa -\kappa _0) + 1 \big ) \\&\geqslant \, c_\epsilon (\kappa -\kappa _0) - C_\epsilon \big ( 2^{-5\kappa _0} (\kappa -\kappa _0) + 1 \big ), \end{aligned}$$

where $C_\epsilon >0$ and $c_\epsilon >0$ are sufficiently large and small constants, respectively. We now obtain the desired conclusion (A.3) by first choosing a parameter $\kappa _0=\kappa _0(\epsilon )$ such that $C_\epsilon 2^{-5\kappa _0}$ is smaller than $c_\epsilon$ and then letting $\kappa \rightarrow \infty$. Thus, it now only remains to prove the claim (A.6). By inserting (A.4) and (A.5) into the integrand, we obtain that

(A.7)

(A.8)

(A.9)

where we have already estimated terms in which the derivative hits the cut-off $\chi (y/\epsilon )$. We start by analyzing the main term (A.7). First, we treat the contribution of the diagonal case $m=n$. Using trigonometric identities, we have that

$$\begin{aligned} \sin (y) \sin ((n-1) y) \cos (ny) = \frac{1}{4} \Big ( -1 + \cos (2y) + \cos ((2n-2) y) - \cos (2ny) \Big ). \end{aligned}$$

Using integration by parts, this yields

which is the main term in (A.6). In the non-diagonal case $m\ne n$, we use that the frequencies in $F_{\kappa ,\kappa _0}$ are well-separated. Together with integration by parts, this yields

$$\begin{aligned}&\Big | \sum _{\begin{array}{c} m,n\in F_{\kappa ,\kappa _0}\\ :m \ne n \end{array}} m^{-1/2} n^{1/2} \int _{x-t}^{x+t} \textrm{d}y\, \chi (y/\epsilon )^3 \sin (y) \sin ((m-1)y) \cos (ny) \Big | \\&\lesssim _\epsilon \sum _{m,n\in F_{\kappa ,\kappa _0}} m^{-1/2} n^{1/2} \max (m,n)^{-1} \lesssim 1. \end{aligned}$$

We now estimate the first error term (A.8). Using integration by parts, it follows that

$$\begin{aligned} \Big | \sum _{n \in F_{\kappa ,\kappa _0}} n^{1/2} \int _{x-t}^{x+t} \textrm{d}y\, \chi (y/\epsilon )^3 \sin ^2(y) \cos (ny) \Big |\lesssim _\epsilon \sum _{n \in F_{\kappa ,\kappa _0}} n^{-\frac{1}{2}} \lesssim 1. \end{aligned}$$

It remains to estimate the third error term (A.9). To this end, we distinguish two cases. In the case $\max (\ell ,m,n)>{{\,\textrm{med}\,}}(\ell ,m,n)$, we use that the frequencies in $F_{\kappa ,\kappa _0}$ are well-separated, which implies that

$$\begin{aligned}&\Big | \sum _{\begin{array}{c} \ell ,m,n\in F_{\kappa ,\kappa _0}:\\ \max (\ell ,m,n)>{{\,\textrm{med}\,}}(\ell ,m,n) \end{array}} \ell ^{-1/2}m^{-1/2} n^{1/2} \\&\int _{x-t}^{x+t} \textrm{d}y\, \chi (y/\epsilon )^3 \sin ((\ell -1)y) \sin ((m-1)y) \cos (ny)\Big | \\ \lesssim _\epsilon&\sum _{\ell ,m,n\in F_{\kappa ,\kappa _0}} \ell ^{-1/2}m^{-1/2} n^{1/2} \max (\ell ,m,n)^{-1} \lesssim 1. \end{aligned}$$

In the case $\max (\ell ,m,n)={{\,\textrm{med}\,}}(\ell ,m,n)$, we only use that $\sin (\cdot )$ and $\cos (\cdot )$ are bounded, which implies that

$$\begin{aligned}&\Big | \sum _{\begin{array}{c} \ell ,m,n\in F_{\kappa ,\kappa _0}:\\ \max (\ell ,m,n)={{\,\textrm{med}\,}}(\ell ,m,n) \end{array}} \ell ^{-1/2}m^{-1/2} n^{1/2} \\&\int _{x-t}^{x+t} \textrm{d}y\, \chi (y/\epsilon )^3 \sin ((\ell -1)y) \sin ((m-1)y) \cos (ny)\Big | \\ \lesssim _\epsilon&\sum _{\begin{array}{c} \ell ,m,n\in F_{\kappa ,\kappa _0}:\\ \max (\ell ,m,n)={{\,\textrm{med}\,}}(\ell ,m,n) \end{array}}\ell ^{-1/2}m^{-1/2} n^{1/2} \lesssim \sum _{\ell ,m\in F_{\kappa ,\kappa _0}} \ell ^{-1/2} \lesssim 2^{-5\kappa _0} (\kappa -\kappa _0). \end{aligned}$$

This completes the proof of (A.6). $\square$