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Let $z_1,\ldots,z_n$ be $n\ge 1$ distinct points of $\mathbb R^2$. Define the potential function $U: \mathbb R^2 \to\mathbb R$ by

$$U(x):=\sum_{1\le i\le n} \log(|x-z_i|),$$

where $|\cdot|$ denotes the Euclidean norm. Denote by $F$ be the negative gradient of $U$, i.e.

$$F(x):=-\nabla U(x)=-\sum_{1\le i\le n} \frac{x-z_i}{|x-z_i|^2}.$$

For any $x\in\mathbb R^2$ that is different from $z_1,\ldots,z_n$, define the integral curve $Y\equiv Y_x$ by $Y(0)=x$ and

$$\frac{dY(t)}{dt}=F(Y(t)),\quad \forall t\ge 0.$$

By standard results about flows on vector fields, the curve $Y$ can be defined over some maximal domain $[0,\tau\equiv \tau_x)$ with $0<\tau\le \infty$. Can we prove

$$\lim_{t\uparrow\tau} Y(t) \mbox{ exists and belongs to } \{z_1,\ldots,z_n\}?$$

If so, the space $\mathbb R^2$ is divided into a partition $V_1,\ldots, V_n$ defined by

$$V_i:=\big\{x\in \mathbb R^2: \lim_{t\uparrow\tau_x} Y_x(t)=z_i\big\}.$$

PS: The case $n=1$ can be easily handled. Set $z_1\equiv z$ and it holds that

$$\frac{d|Y(t)-z|^2}{dt}=2(Y(t)-z)\cdot \frac{dY(t)}{dt}=-2(Y(t)-z)\cdot\frac{Y(t)-z}{|Y(t)-z|^2}=-2.$$

This implies $|Y(t)-z|^2=|x-z|^2-2t$ for all $t<\tau$. In particular, $\tau<\infty$ and $t\mapsto |Y(t)-z|^2$ is decreasing. If $\lim_{t\uparrow\tau} |Y(t)-z|^2\neq 0$, we see immediately the solution can still be extended at $t=\tau$, which contradicts the maximally of $\tau$. Hence $\lim_{t\uparrow\tau} Y(t)=z$.

Let $z_1,\ldots,z_n$ be $n\ge 1$ distinct points of $\mathbb R^2$. Define the potential function $U: \mathbb R^2 \to\mathbb R$ by

$$U(x):=\sum_{1\le i\le n} \log(|x-z_i|),$$

where $|\cdot|$ denotes the Euclidean norm. Denote by $F$ be the negative gradient of $U$, i.e.

$$F(x):=-\nabla U(x)=-\sum_{1\le i\le n} \frac{x-z_i}{|x-z_i|^2}.$$

For any $x\in\mathbb R^2$ that is different from $z_1,\ldots,z_n$, define the integral curve $Y\equiv Y_x$ by $Y(0)=x$ and

$$\frac{dY(t)}{dt}=F(Y(t)),\quad \forall t\ge 0.$$

By standard results about flows on vector fields, the curve $Y$ can be defined over some maximal domain $[0,\tau\equiv \tau_x)$ with $0<\tau\le \infty$. Can we prove for almost every $x\in\mathbb R^2$

$$\lim_{t\uparrow\tau} Y(t) \mbox{ exists and belongs to } \{z_1,\ldots,z_n\}?$$

If so, the space $\mathbb R^2$ is divided into a partition $V_1,\ldots, V_n$ defined by

$$V_i:=\big\{x\in \mathbb R^2: \lim_{t\uparrow\tau_x} Y_x(t)=z_i\big\}.$$

PS: The case $n=1$ can be easily handled. Set $z_1\equiv z$ and it holds that

$$\frac{d|Y(t)-z|^2}{dt}=2(Y(t)-z)\cdot \frac{dY(t)}{dt}=-2(Y(t)-z)\cdot\frac{Y(t)-z}{|Y(t)-z|^2}=-2.$$

This implies $|Y(t)-z|^2=|x-z|^2-2t$ for all $t<\tau$. In particular, $\tau<\infty$ and $t\mapsto |Y(t)-z|^2$ is decreasing. If $\lim_{t\uparrow\tau} |Y(t)-z|^2\neq 0$, we see immediately the solution can still be extended at $t=\tau$, which contradicts the maximally of $\tau$. Hence $\lim_{t\uparrow\tau} Y(t)=z$.

Let $z_1,\ldots,z_n$ be $n\ge 1$ distinct points of $\mathbb R^2$. Define the potential function $U: \mathbb R^2 \to\mathbb R$ by

$$U(x):=\sum_{1\le i\le n} \log(|x-z_i|),$$

where $|\cdot|$ denotes the Euclidean norm. Denote by $F$ be the negative gradient of $U$, i.e.

$$F(x):=-\nabla U(x)=-\sum_{1\le i\le n} \frac{x-z_i}{|x-z_i|^2}.$$

For any $x\in\mathbb R^2$ that is different from $z_1,\ldots,z_n$, define the integral curve $Y\equiv Y_x$ by $Y(0)=x$ and

$$\frac{dY(t)}{dt}=F(Y(t)),\quad \forall t\ge 0.$$

By standard results about flows on vector fields, the curve $Y$ can be defined over some maximal domain $[0,\tau\equiv \tau_x)$ with $0<\tau\le \infty$. Can we prove

$$\lim_{t\uparrow\tau} Y(t) \mbox{ exists and belongs to } \{z_1,\ldots,z_n\}?$$

If so, the space $\mathbb R^2$ is divided into a partition $V_1,\ldots, V_n$ defined by

$$V_i:=\big\{x\in \mathbb R^2: \lim_{t\uparrow\tau_x} Y_x(t)=z_i\big\}.$$

PS: The case $n=1$ can be easily handled. Set $z_1\equiv z$ and it holds that

$$\frac{d|Y(t)-z|^2}{dt}=2(Y(t)-z)\cdot \frac{dY(t)}{dt}=-2(Y(t)-z)\cdot\frac{Y(t)-z}{|Y(t)-z|^2}=-2.$$

This implies $|Y(t)-z|^2=|x-z|^2-2t$ for all $t<\tau$. In particular, $\tau<\infty$ and $t\mapsto |Y(t)-z|^2$ is decreasing. If $\lim_{t\uparrow\tau_x} |Y(t)-z|^2\neq 0$, we see immediately the solution can still be extended at $t=\tau$, which contradicts the maximally of $\tau$. Hence $\lim_{t\uparrow\tau_x} Y(t)=z$.

Let $z_1,\ldots,z_n$ be $n\ge 1$ distinct points of $\mathbb R^2$. Define the potential function $U: \mathbb R^2 \to\mathbb R$ by

$$U(x):=\sum_{1\le i\le n} \log(|x-z_i|),$$

where $|\cdot|$ denotes the Euclidean norm. Denote by $F$ be the negative gradient of $U$, i.e.

$$F(x):=-\nabla U(x)=-\sum_{1\le i\le n} \frac{x-z_i}{|x-z_i|^2}.$$

For any $x\in\mathbb R^2$ that is different from $z_1,\ldots,z_n$, define the integral curve $Y\equiv Y_x$ by $Y(0)=x$ and

$$\frac{dY(t)}{dt}=F(Y(t)),\quad \forall t\ge 0.$$

By standard results about flows on vector fields, the curve $Y$ can be defined over some maximal domain $[0,\tau\equiv \tau_x)$ with $0<\tau\le \infty$. Can we prove

$$\lim_{t\uparrow\tau} Y(t) \mbox{ exists and belongs to } \{z_1,\ldots,z_n\}?$$

If so, the space $\mathbb R^2$ is divided into a partition $V_1,\ldots, V_n$ defined by

$$V_i:=\big\{x\in \mathbb R^2: \lim_{t\uparrow\tau_x} Y_x(t)=z_i\big\}.$$

PS: The case $n=1$ can be easily handled. Set $z_1\equiv z$ and it holds that

$$\frac{d|Y(t)-z|^2}{dt}=2(Y(t)-z)\cdot \frac{dY(t)}{dt}=-2(Y(t)-z)\cdot\frac{Y(t)-z}{|Y(t)-z|^2}=-2.$$

This implies $|Y(t)-z|^2=|x-z|^2-2t$ for all $t<\tau$. In particular, $\tau<\infty$ and $t\mapsto |Y(t)-z|^2$ is decreasing. If $\lim_{t\uparrow\tau} |Y(t)-z|^2\neq 0$, we see immediately the solution can still be extended at $t=\tau$, which contradicts the maximally of $\tau$. Hence $\lim_{t\uparrow\tau} Y(t)=z$.

Let $z_1,\ldots,z_n$ be $n\ge 1$ distinct points of $\mathbb R^2$. Define the potential function $U: \mathbb R^2 \to\mathbb R$ by

$$U(x):=\sum_{1\le i\le n} \log(|x-z_i|),$$

where $|\cdot|$ denotes the Euclidean norm. Denote by $F$ be the negative gradient of $U$, i.e.

$$F(x):=-\nabla U(x)=-\sum_{1\le i\le n} \frac{x-z_i}{|x-z_i|^2}.$$

For any $x\in\mathbb R^2$ that is different from $z_1,\ldots,z_n$, define the integral curve $Y\equiv Y_x$ by $Y(0)=x$ and

$$\frac{dY(t)}{dt}=F(Y(t)),\quad \forall t\ge 0.$$

By standard results about flows on vector fields, the curve $Y$ can be defined over some maximal domain $[0,\tau\equiv \tau_x)$ with $0<\tau\le \infty$. Can we prove

$$\lim_{t\uparrow\tau} Y(t) \mbox{ exists and belongs to } \{z_1,\ldots,z_n\}?$$

If so, the space $\mathbb R^2$ is divided into a partition $V_1,\ldots, V_n$ defined by

$$V_i:=\big\{x\in \mathbb R^2: \lim_{t\uparrow\tau_x} Y_x(t)=z_i\big\}.$$

Let $z_1,\ldots,z_n$ be $n\ge 1$ distinct points of $\mathbb R^2$. Define the potential function $U: \mathbb R^2 \to\mathbb R$ by

$$U(x):=\sum_{1\le i\le n} \log(|x-z_i|),$$

where $|\cdot|$ denotes the Euclidean norm. Denote by $F$ be the negative gradient of $U$, i.e.

$$F(x):=-\nabla U(x)=-\sum_{1\le i\le n} \frac{x-z_i}{|x-z_i|^2}.$$

For any $x\in\mathbb R^2$ that is different from $z_1,\ldots,z_n$, define the integral curve $Y\equiv Y_x$ by $Y(0)=x$ and

$$\frac{dY(t)}{dt}=F(Y(t)),\quad \forall t\ge 0.$$

By standard results about flows on vector fields, the curve $Y$ can be defined over some maximal domain $[0,\tau\equiv \tau_x)$ with $0<\tau\le \infty$. Can we prove

$$\lim_{t\uparrow\tau} Y(t) \mbox{ exists and belongs to } \{z_1,\ldots,z_n\}?$$

If so, the space $\mathbb R^2$ is divided into a partition $V_1,\ldots, V_n$ defined by

$$V_i:=\big\{x\in \mathbb R^2: \lim_{t\uparrow\tau_x} Y_x(t)=z_i\big\}.$$

PS: The case $n=1$ can be easily handled. Set $z_1\equiv z$ and it holds that

$$\frac{d|Y(t)-z|^2}{dt}=2(Y(t)-z)\cdot \frac{dY(t)}{dt}=-2(Y(t)-z)\cdot\frac{Y(t)-z}{|Y(t)-z|^2}=-2.$$

This implies $|Y(t)-z|^2=|x-z|^2-2t$ for all $t<\tau$. In particular, $\tau<\infty$ and $t\mapsto |Y(t)-z|^2$ is decreasing. If $\lim_{t\uparrow\tau_x} |Y(t)-z|^2\neq 0$, we see immediately the solution can still be extended at $t=\tau$, which contradicts the maximally of $\tau$. Hence $\lim_{t\uparrow\tau_x} Y(t)=z$.