Machine-translated from Chinese.
Heat kernels on cyclic groups - Anders Karlsson and Markus Neuhauser
图上热核
令$X$为图, $N(x)$为$x$的全体邻居, $X$上的组合Laplacian定义为
$$
\Delta f(x)=f(x)-\frac{1}{|N(x)|} \sum_{y\in N(x)}f(y).
$$
热方程为
$$
\left(\Delta +\frac{\partial {} }{\partial {}t}\right)f(t,x)=0.
$$
热核$K^X(t,x)$即为热方程的基本解, 满足$K^X(0,x)=\delta_0(x),$
基点$0\in X.$
若$X$为有限图, 那么Laplacian为对称阵,
有实特征值$0=\lambda_0\le \lambda_1\le \cdots\le \lambda_{m-1},$
对应正规特征向量$\varphi_0,\varphi_1,\cdots,\varphi_{m-1}.$
我们可以将$\delta_0$用基表示:
$$
\delta_0=\sum_{j=0}^{m-1}\overline{\varphi_j(0)}\varphi_j.
$$
那么热核就等于
$$
K^X(t,x)=e^{-t\Delta}\delta_0(x)=\sum_{j=0}^{m-1}e^{-\lambda_jt}\varphi_j(x)\overline{\varphi_j(0)}.
$$
改进Bessel函数
对于整数$x\ge 0,$ $z\in\mathbb{C},$ 第一类改进Bessel函数为
$$
I_x(z)=\sum_{k=0}^\infty \frac{(z/2)^{x+2k} }{k!(x+k)!}=\frac{1}{\pi}\int_0^\pi e^{z\cos \theta}\cos x\theta d\theta.
$$
对于$x<0,$ 令$I_x=I_{-x}.$ 它是微分方程
$$
z^2\frac{d {}^2w}{d {}z^2}+z\frac{d {}w}{d {}z}-(z^2+x^2)w=0
$$
的解,
与经典$J$-Bessel函数的关系是
$$
I_x(z)=e^{-\pi ix/2}J_x(z e^{\pi i/2}).
$$
$I_x$满足的一个重要的性质是
$$
I_{x+1}+I_{x-1}=2I_x',\quad \,\forall\,x\in \mathbb{Z}.
$$
在$t$充分大时, 我们有渐进估计
$$
I_v(t)\sim \frac{e^t}{\sqrt{2\pi t} }\left(1-\frac{4\nu^2-1}{8t}\right).
$$
$\mathbb{Z}$上热核
$\mathbb{Z}$上Laplacian由
$$
\Delta f(x)=f(x)-\frac{1}{2}(f(x+1)+f(x-1)),\quad x\in\mathbb{Z}
$$
给出. 我们可以通过Fourier变换(关于$x$变量)的方法给出基本解
$$
K^\mathbb{Z}(t,x)=(-1)^x\sum_{n=|x|}^\infty \frac{(-t/2)^n}{n!}\binom{2n}{n-x},
$$
事实上后面将给出
$$
K^\mathbb{Z}(t,x)=e^{-t}I_x(t).
$$
我们可以直接验证$e^{-t}I_x(t)$为热核.
$$
\begin{aligned}
\left(\Delta+\frac{\partial {} }{\partial {}t}\right)e^{-t}I_x(t)&= e^{-t}\left(I_x(t)-\frac{1}{2}(I_{x+1}(t)+I_{x-1}(t))-I_x(t)+I_x'(t)\right)=0.
\end{aligned}
$$
$e^{-t}I_x(t)$是$\mathbb{R}$上热核(Gaussian)$\frac{1}{\sqrt{4\pi t} }e^{-\frac{x^2}{4t} }$的类比,
有渐进估计
$$
\frac{1}{\sqrt{2\pi t} }\left(1-\frac{4x^2-1}{8t}\right).
$$
$\mathbb{Z}_m$上热核
$\mathbb{Z}_m$上的Laplacian和$\mathbb{Z}$上一致, 取等价类即可.
取函数族$\{\chi_j\}$为
$$
\chi_j(x):=e^{2\pi ijx/m},\quad j=0,1,\cdots,m-1.
$$
$\chi_j$满足$\chi_j(x+y)=\chi_j(x)\chi_j(y).$ 从而
$$
\Delta \chi=\left(1-\frac{1}{2}(\chi(1)+\chi(-1))\right)\chi.
$$
这意味着$\chi_j$为特征值$\lambda_j=1-\cos(2\pi j/m)$对应的特征函数. 进而
$$
e^{-t\Delta}\chi_j=e^{-t(1-\cos(2\pi j/m))}\chi_j=e^{-2\sin^2(\pi j/m)t}\chi_j.
$$
由正交性,
$$
\delta_y=\frac{1}{m}\sum_{j=0}^{m-1} \overline{\chi_j(y)}\chi_j.
$$
从而热核为
$$
K^{\mathbb{Z}_m}(t,x):=e^{-t\Delta}\delta_0(x)=\frac{1}{m}\sum_{j=0}^{m-1}e^{-2\sin^2(\pi j/m)t}e^{2\pi ijx/m}.
$$
Theta反演公式
考虑函数
$$
e^{-t}\sum_{k=-\infty}^\infty I_{x+km}(t),
$$
它是$\mathbb{Z}$上的周期函数, 可以诱导$\mathbb{Z}_m$上的函数.
这对所有$t$收敛, 因为对于$x\ge 0,$
$$
\left|\sum_{k=0}^\infty I_{x+km}(t)\right|\le \sum_{k=0}^\infty \sum_{n=0}^\infty \frac{|t/2|^{x+km+2n} }{n!(x+km+n)!}\le \sum_{k=0}^\infty \sum_{n=0}^\infty \frac{|t/2|^{k+n} }{n!k!}\le e^{|t|}.
$$
于是该函数也是$\mathbb{Z}_m$上的热核. 由热核唯一性, 我们有
$$
e^{-t}\sum_{k=-\infty}^\infty I_{x+km}(t)=\frac{1}{m}\sum_{j=0}^{m-1}e^{-2\sin^2(\pi j/m)t}e^{2\pi ijx/m}.
$$
另一方面, 我们也可以由如下两引理直接证明,
将右边视为左边以$x$为自变量的Fourier展开即可.
引理 1. 函数$\omega\mapsto e^{-2\sin^2(\omega/2)t}$的第$x$个Fourier系数有级数表示$(-1)^x\sum_{n=|x|}^\infty \frac{(-t/2)^n}{n!}\binom{2n}{n-x},$ 等于$e^{-t}I_x(t)$.
直接计算即可, 利用展开
$$
e^{-2\sin^2(\omega/2)t}=\sum_{n=0}^\infty \frac{(-2t)^n}{n!}\sin^{2n}(\omega/2),
$$
$$
\sin^{2n}(\omega/2)=\left(\frac{1}{2i}(e^{i\omega/2}-e^{-i\omega/2})\right)^{2n}=(2i)^{-2n}\sum_{y=0}^{2n}\binom{2n}{y}(-1)^ye^{i\omega(n-y)}.
$$
这样可得到第$x$个Fourier系数为
$$
(-1)^x\sum_{n=|x|}^\infty \frac{(-t/2)^n}{n!}\binom{2n}{n-x}.
$$
由于
$$
e^{-2\sin^2(\omega/2)t}=e^{-t}e^{t\cos\omega},
$$
第$x$个Fourier系数又有表示
$$
\frac{1}{2\pi}\int_0^{2\pi} e^{-t}e^{t\cos \omega}e^{-i\omega x}d\omega=\frac{e^{-t} }{\pi}\int_0^\pi e^{t\cos\omega}\cos(\omega x)d\omega=e^{-t}I_x(t).
$$
这就给出了引理.
引理 2. 函数$x\mapsto \sum_{k=-\infty}^\infty (-1)^{x+km}\sum_{n=|x+km|}^\infty \frac{(-t/2)^n}{n!}\binom{2n}{n-x-km}$的第$j$个Fourier系数为$e^{-2t\sin^2(\pi j/m)}.$
直接计算$\sum_{x=0}^{m-1}\bullet \cdot e^{-2\pi ijx/m}$即可.
这样就给出了Fourier展开
$$
e^{-t}\sum_{k=-\infty}^\infty I_{x+km}(t)=\frac{1}{m}\sum_{j=0}^{m-1}e^{-2\sin^2(\pi j/m)t}e^{2\pi ijx/m}.
$$
这类似于反演公式. 当然也有如下常用的表示
$$
\sum_{k=-\infty}^\infty I_{x+km}(t)=\frac{1}{m}\sum_{j=0}^{m-1}e^{\cos(2\pi j/m)z+2\pi ijx/m},\quad z\in \mathbb{C},\quad x,m\in\mathbb{Z},\quad m>0.
$$
通过取特殊情况求导($x=0,$ $t=0,$ $n$阶导), 我们还有
推论 3. $\sum_{j=0}^{m-1}\cos^n(2\pi j/m)=\frac{m}{2^n}\sum_{l\in L(n,m)}\binom{n}{l},$ 其中$L(n,m)$为$l\in \mathbb{Z},$ 满足$0\le l\le n$且$l=\frac{n-mq}{2},$ $q\in\mathbb{Z}.$ 特别的, 若$n$为奇数, $m$为偶数, 那么$L(n,m)=\varnothing,$ 和取零.
当$n=2,$ $m\ge 3$时, 我们有
$$
\sum_{j=0}^{m-1}\cos^2(2\pi j/m)=\frac{m}{2}.
$$
文章最后更新于 2023-09-05 15:47:41
Heat kernels on cyclic groups - Anders Karlsson and Markus Neuhauser
Heat Kernel on a Graph
Let $X$ be the graph, $N(x)$ be all the neighbors of $x$, and the combined Laplacian on $X$ is defined as
$$
\Delta f(x)=f(x)-\frac{1}{|N(x)|} \sum_{y\in N(x)}f(y).
$$
The heat equation is
$$
\left(\Delta +\frac{\partial {} }{\partial {}t}\right)f(t,x)=0.
$$
The heat kernel $K^X(t,x)$ is the basic solution of the heat equation and satisfies $K^X(0,x)=\delta_0(x),$
Base point$0\in X.$
If $X$ is a finite graph, then the Laplacian is a symmetric matrix,
Has real eigenvalues $0=\lambda_0\le \lambda_1\le \cdots\le \lambda_{m-1},$
Corresponding normal feature vector $\varphi_0,\varphi_1,\cdots,\varphi_{m-1}.$
We can express $\delta_0$ in basis:
$$
\delta_0=\sum_{j=0}^{m-1}\overline{\varphi_j(0)}\varphi_j.
$$
Then thermonuclear is equal to
$$
K^X(t,x)=e^{-t\Delta}\delta_0(x)=\sum_{j=0}^{m-1}e^{-\lambda_jt}\varphi_j(x)\overline{\varphi_j(0)}.
$$
Improved Bessel function
For integers $x\ge 0,$ $z\in\mathbb{C},$ the first kind of improved Bessel function is
$$
I_x(z)=\sum_{k=0}^\infty \frac{(z/2)^{x+2k} }{k!(x+k)!}=\frac{1}{\pi}\int_0^\pi e^{z\cos \theta}\cos x\theta d\theta.
$$
For $x<0,$ let $I_x=I_{-x}.$ it is the differential equation
$$
z^2\frac{d {}^2w}{d {}z^2}+z\frac{d {}w}{d {}z}-(z^2+x^2)w=0
$$
solution,
The relationship to the classical $J$-Bessel function is
$$
I_x(z)=e^{-\pi ix/2}J_x(z e^{\pi i/2}).
$$
An important property that $I_x$ satisfies is that
$$
I_{x+1}+I_{x-1}=2I_x',\quad \,\forall\,x\in \mathbb{Z}.
$$
When $t$ is sufficiently large, we have an asymptotic estimate
$$
I_v(t)\sim \frac{e^t}{\sqrt{2\pi t} }\left(1-\frac{4\nu^2-1}{8t}\right).
$$
$\mathbb{Z}$ Upper Thermonuclear
$\mathbb{Z}$ on Laplacian by
$$
\Delta f(x)=f(x)-\frac{1}{2}(f(x+1)+f(x-1)),\quad x\in\mathbb{Z}
$$
is given. We can give the basic solution through the method of Fourier transformation (about the $x$ variable)
$$
K^\mathbb{Z}(t,x)=(-1)^x\sum_{n=|x|}^\infty \frac{(-t/2)^n}{n!}\binom{2n}{n-x},
$$
In fact, it will be given later
$$
K^\mathbb{Z}(t,x)=e^{-t}I_x(t).
$$
We can directly verify that $e^{-t}I_x(t)$ is a hot core.
$$
\begin{aligned}
\left(\Delta+\frac{\partial {} }{\partial {}t}\right)e^{-t}I_x(t)&= e^{-t}\left(I_x(t)-\frac{1}{2}(I_{x+1}(t)+I_{x-1}(t))-I_x(t)+I_x'(t)\right)=0.
\end{aligned}
$$
$e^{-t}I_x(t)$ is the analogy of the upper thermal core (Gaussian) $\frac{1}{\sqrt{4\pi t} }e^{-\frac{x^2}{4t} }$ of $\mathbb{R}$,
There is an asymptotic estimate
$$
\frac{1}{\sqrt{2\pi t} }\left(1-\frac{4x^2-1}{8t}\right).
$$
$\mathbb{Z}_m$ Upper Thermonuclear
The Laplacian on $\mathbb{Z}_m$ is consistent with that on $\mathbb{Z}$, so just take the equivalence class.
Take the function family $\{\chi_j\}$ as
$$
\chi_j(x):=e^{2\pi ijx/m},\quad j=0,1,\cdots,m-1.
$$
$\chi_j$ satisfies $\chi_j(x+y)=\chi_j(x)\chi_j(y).$ thus
$$
\Delta \chi=\left(1-\frac{1}{2}(\chi(1)+\chi(-1))\right)\chi.
$$
This means that $\chi_j$ is the eigenfunction corresponding to the eigenvalue $\lambda_j=1-\cos(2\pi j/m)$. Then
$$
e^{-t\Delta}\chi_j=e^{-t(1-\cos(2\pi j/m))}\chi_j=e^{-2\sin^2(\pi j/m)t}\chi_j.
$$
By orthogonality,
$$
\delta_y=\frac{1}{m}\sum_{j=0}^{m-1} \overline{\chi_j(y)}\chi_j.
$$
Therefore, the thermal core is
$$
K^{\mathbb{Z}_m}(t,x):=e^{-t\Delta}\delta_0(x)=\frac{1}{m}\sum_{j=0}^{m-1}e^{-2\sin^2(\pi j/m)t}e^{2\pi ijx/m}.
$$
Theta inversion formula
consider function
$$
e^{-t}\sum_{k=-\infty}^\infty I_{x+km}(t),
$$
It is a periodic function on $\mathbb{Z}$ and can induce a function on $\mathbb{Z}_m$.
This converges for all $t$, since for $x\ge 0,$
$$
\left|\sum_{k=0}^\infty I_{x+km}(t)\right|\le \sum_{k=0}^\infty \sum_{n=0}^\infty \frac{|t/2|^{x+km+2n} }{n!(x+km+n)!}\le \sum_{k=0}^\infty \sum_{n=0}^\infty \frac{|t/2|^{k+n} }{n!k!}\le e^{|t|}.
$$
So this function is also a thermal kernel on $\mathbb{Z}_m$. From the uniqueness of the thermal kernel, we have
$$
e^{-t}\sum_{k=-\infty}^\infty I_{x+km}(t)=\frac{1}{m}\sum_{j=0}^{m-1}e^{-2\sin^2(\pi j/m)t}e^{2\pi ijx/m}.
$$
On the other hand, we can also directly prove by the following two lemmas,
Just treat the right side as the Fourier expansion of the left side with $x$ as the independent variable.
Lemma 1. The $x$th Fourier coefficient of function $\omega\mapsto e^{-2\sin^2(\omega/2)t}$ has a series that means $(-1)^x\sum_{n=|x|}^\infty \frac{(-t/2)^n}{n!}\binom{2n}{n-x},$ is equal to $e^{-t}I_x(t)$.
Just calculate it directly, use expand
$$
e^{-2\sin^2(\omega/2)t}=\sum_{n=0}^\infty \frac{(-2t)^n}{n!}\sin^{2n}(\omega/2),
$$
$$
\sin^{2n}(\omega/2)=\left(\frac{1}{2i}(e^{i\omega/2}-e^{-i\omega/2})\right)^{2n}=(2i)^{-2n}\sum_{y=0}^{2n}\binom{2n}{y}(-1)^ye^{i\omega(n-y)}.
$$
In this way, the $x$th Fourier coefficient can be obtained as
$$
(-1)^x\sum_{n=|x|}^\infty \frac{(-t/2)^n}{n!}\binom{2n}{n-x}.
$$
due to
$$
e^{-2\sin^2(\omega/2)t}=e^{-t}e^{t\cos\omega},
$$
The $x$th Fourier coefficient is expressed again
$$
\frac{1}{2\pi}\int_0^{2\pi} e^{-t}e^{t\cos \omega}e^{-i\omega x}d\omega=\frac{e^{-t} }{\pi}\int_0^\pi e^{t\cos\omega}\cos(\omega x)d\omega=e^{-t}I_x(t).
$$
This gives the lemma.
Lemma 2. The $j$th Fourier coefficient of function $x\mapsto \sum_{k=-\infty}^\infty (-1)^{x+km}\sum_{n=|x+km|}^\infty \frac{(-t/2)^n}{n!}\binom{2n}{n-x-km}$ is $e^{-2t\sin^2(\pi j/m)}.$
Just calculate $\sum_{x=0}^{m-1}\bullet \cdot e^{-2\pi ijx/m}$ directly.
This gives the Fourier expansion
$$
e^{-t}\sum_{k=-\infty}^\infty I_{x+km}(t)=\frac{1}{m}\sum_{j=0}^{m-1}e^{-2\sin^2(\pi j/m)t}e^{2\pi ijx/m}.
$$
This is similar to the inversion formula. Of course, there are also commonly used expressions as follows
$$
\sum_{k=-\infty}^\infty I_{x+km}(t)=\frac{1}{m}\sum_{j=0}^{m-1}e^{\cos(2\pi j/m)z+2\pi ijx/m},\quad z\in \mathbb{C},\quad x,m\in\mathbb{Z},\quad m>0.
$$
By taking the special case derivative ($x=0,$ $t=0,$ $n$ order derivative), we also have
Corollary 3. $\sum_{j=0}^{m-1}\cos^n(2\pi j/m)=\frac{m}{2^n}\sum_{l\in L(n,m)}\binom{n}{l},$ where $L(n,m)$ is $l\in \mathbb{Z},$ satisfying $0\le l\le n$ and $l=\frac{n-mq}{2},$ $q\in\mathbb{Z}.$ In particular, if $n$ is an odd number and $m$ is an even number, then the sum of $L(n,m)=\varnothing,$ is zero.
When $n=2,$ $m\ge 3$, we have
$$
\sum_{j=0}^{m-1}\cos^2(2\pi j/m)=\frac{m}{2}.
$$
The article was last updated on 2023-09-05 15:47:41
