Monthly Archives: December 2016

Marcinkiewicz-Zygmund’s SLLN

First, we present well-known Khintchine-Kolmogorov Theorem.

Theorem (Khintchine-Kolmogorov). Let X_{1},X_{2},\dots be independent random variables with finite expectation. If \sum_{n=1}^{\infty}\mathrm{Var}\left(X_{n}\right)<\infty,
then \sum_{n=1}^{\infty}\left[X_{n}-\mathbb{E}\left[X_{n}\right]\right] converges a.e.

Now we present the Marcinkiewicz-Zygmund theorem which is our object of this article.

Theorem (Marcinkiewicz-Zygmund Theorem). Suppose that X_{1},X_{2},\dots are i.i.d with \mathbb{E}\left|X_{1}\right|^{p}<\infty for some 0<p<2. Then

    \[ \sum_{n=1}^{\infty}\left(\frac{X_{n}}{n^{\frac{1}{p}}}-\mathbb{E}\left[Y_{n}\right]\right)\quad\text{converges a.e.}, \]

where

    \[ Y_{n}=\frac{\mathbb{E}X_{n}1_{\left\{ \left|X_{n}\right|\le n^{\frac{1}{p}}\right\} }}{n^{\frac{1}{p}}}. \]

Proof. Set

    \[ A_{j}=\left\{ \left(j-1\right)^{\frac{1}{p}}<\left|X_{n}\right|\le j^{\frac{1}{p}}\right\} . \]

Then

    \begin{align*} \sum_{n=1}^{\infty}\mathbb{E}\left|Y_{n}\right|^{2} & =\sum_{n=1}^{\infty}\sum_{j=1}^{n}n^{-\frac{2}{p}}\mathbb{E}\left|X_{n}I_{A_{j}}\right|^{2}\\ & =\sum_{n=1}^{\infty}\sum_{j=1}^{n}n^{-\frac{2}{p}}\int_{A_{j}}\left|X_{n}\right|^{2}d\mathbb{P}\\ & =\sum_{j=1}^{\infty}\sum_{n=j}^{\infty}\int_{A_{j}}n^{-\frac{2}{p}}\left|X_{1}\right|^{2}d\mathbb{P}\\ & \le\sum_{j=1}^{\infty}\left(\frac{1}{j}+\frac{p}{2-p}\right)\int_{A_{j}}\left|X_{1}\right|^{2}d\mathbb{P}. \end{align*}

The last inequality comes from the fact

    \begin{align*} \sum_{n=j}^{\infty}n^{-\frac{2}{p}} & \le\int_{j}^{\infty}x^{-\frac{2}{p}}dx+\frac{1}{j}\\ & =\frac{p}{2-p}j^{-\frac{2}{p}}+\frac{1}{j}\\ & \le\frac{p}{2-p}+\frac{1}{j}. \end{align*}

Now

    \[ \sum_{n=1}^{\infty}\mathbb{E}\left|Y_{n}\right|^{2}\le\frac{2}{2-p}\mathbb{E}\left|X_{1}\right|^{p}<\infty. \]

Hence

    \[ \sum_{j=1}^{\infty}\left(Y_{n}-\mathbb{E}\left[Y_{n}\right]\right) \]

converges a.e. Note that

    \[ \sum_{n=1}^{\infty}\mathbb{P}\left(\frac{X_{n}}{n^{\frac{1}{p}}}\neq Y_{n}\right)=\sum_{n=1}^{\infty}\mathbb{P}\left(\left|X_{1}\right|>n^{\frac{1}{p}}\right)\le\mathbb{E}\left|X_{1}\right|^{p}<\infty. \]

Hence by Borel-Cantelli’s lemma, \frac{X_{n}}{n^{\frac{1}{p}}}=Y_{n} a.e. for all but finitely many n. Hence

    \[ \sum_{j=1}^{\infty}\left(\frac{X_{n}}{n^{\frac{1}{p}}}-\mathbb{E}\left[Y_{n}\right]\right) \]

converges a.e. This completes the proof.

Now we give some application of this theorem.

Problem. Given i.i.d sequence of random variables X_{n}, assume \mathbb{E}X_{n}=0 and

    \[ \mathbb{E}\left|X_{n}\right|\log\left(1+\left|X_{n}\right|\right)<\infty. \]

Show \sum_{n}\left(\frac{X_{n}}{n}\right) converges almost surely.

Proof. Note that X_{n}\in L^{1}. Indeed,

    \begin{align*} \mathbb{E}\left|X_{n}\right| & =\int_{\left\{ \left|X_{n}\right|\le1\right\} }\left|X_{n}\right|d\mathbb{P}+\int_{\left\{ \left|X_{n}\right|>1\right\} }\left|X_{n}\right|d\mathbb{P}\\ & \le\mathbb{P}\left\{ \left|X_{n}\right|\le1\right\} +\int_{\left\{ \left|X_{n}\right|>1\right\} }\left|X_{n}\right|\log\left(1+\left|X_{n}\right|\right)d\mathbb{P}\\ & \le1+\mathbb{E}\left|X_{n}\right|\log\left(1+\left|X_{n}\right|\right)<\infty. \end{align*}

Hence it suffices to show

(1)   \begin{equation*} \sum_{n=1}^{\infty}\frac{\mathbb{E}\left|X_{1}\right|1_{\left\{ \left|X_{1}\right|\ge n\right\} }}{n}<\infty. \end{equation*}

Indeed, suppose the above is true. Then note that

    \begin{align*} \sum_{n=1}^{\infty}\left|\frac{\mathbb{E}X_{1}1_{\left|X_{1}\right|\le n}}{n}\right| & =\sum_{n=1}^{\infty}\left|\frac{\mathbb{E}X_{1}\left(1-1_{\left|X_{1}\right|>n}\right)}{n}\right|\\ & =\sum_{n=1}^{\infty}\left|\frac{\mathbb{E}X_{1}1_{\left\{ \left|X_{1}\right|\ge n\right\} }}{n}\right|\\ & \le\sum_{n=1}^{\infty}\frac{\mathbb{E}\left|X_{1}\right|1_{\left\{ \left|X_{1}\right|\ge n\right\} }}{n}<\infty \end{align*}

since \mathbb{E}X_{1}=0.

Now Marcinkiewicz-Zygmund theorem gives

    \[ \sum_{n=1}^{\infty}\left(\frac{X_{n}-\mathbb{E}X_{n}1_{\left\{ \left|X_{n}\right|\le n\right\} }}{n}\right) \]

converges almost surely.

Now we left to show that (1) holds. Let E_{k}=\left\{ k\le\left|X_{1}\right|<k+1\right\}.
Then observe that

    \begin{align*} \sum_{n=1}^{\infty}\frac{\mathbb{E}\left|X_{1}\right|1_{\left\{ \left|X_{1}\right|\ge n\right\} }}{n} & =\sum_{n=1}^{\infty}\sum_{k=n}^{\infty}\frac{\mathbb{E}\left|X_{1}\right|1_{E_{k}}}{n}\\ & =\sum_{k=1}^{\infty}\sum_{n=1}^{k}\frac{\mathbb{E}\left|X_{1}\right|1_{E_{k}}}{n}\\ & \le C\sum_{k=1}^{\infty}\log k\cdot\mathbb{E}\left|X_{1}\right|1_{E_{k}}\\ & \le C\mathbb{E}\left|X_{1}\right|\log\left(1+\left|X_{1}\right|\right)<\infty. \end{align*}

So we are done.

Derivation of Navier-Stokes equation

이 글은 Euler equation에서 얻는 momentum conservation law로부터 Newtonian fluid에서의 Navier-Stokes equation을 유도하는 것을 목표로 한다.

\rho를 유체의 밀도라고 하고, \mathbf{u}(x,t)=(u^1,u^2,u^3)를 유체의 속도, p(x,t)를 유체의 압력이라 하자.  Euler equation에서 얻는 momentum conservation law는 다음과 같다:

    \[ \frac{\partial}{\partial t} (\rho u^i) +\sum_{k=1}^{3}\frac{\partial}{\partial x_k} \Pi_{ik}\]

여기서 \Pi_{ik}=\delta_{ik} p+\rho u^i u^k이다.

여기서 이 방정식은 ideal fluid에서 유도된 것이다. ideal fluid은 열 교환이나 점성을 무시하는 유체다. 그러나 Navier-Stokes equation에서 고려하는 상황은 점성(viscosity)가 있는 상황이므로 momentum conservation에 다른 변수가 추가되어야 한다.

위 식에서 \Pi_{ik} 대신 \Pi_{ik}-\sigma^\prime_{ij}을 넣는다.  여기서 \sigma^\prime_{ij}는 점성 효과를 넣기 위하여 고려된 변수로 Newtonian fluid에서는 \sigma^\prime_{ij} \propto \frac{\partial u^i}{\partial x_j}라 가정하는 것이 합리적이며,  이를 바탕으로

    \[ \sigma_{ij}^{\prime}=\eta\left(\frac{\partial u^{i}}{\partial x_{j}}+\frac{\partial u^{j}}{\partial x_{i}}\right)+\mu\delta_{ij}\left(\nabla\cdot\boldsymbol{u}\right) \]

와 같이 정의한다.

여기서 \eta,\mu는 상수이며 특히 \eta은 양수인 점성 상수다.

그럼 이를 바탕으로 Navier-Stokse equation을 유도한다. Summation 기호를 생략하면 곱의 법칙에 의하여

    \begin{align*} & \frac{\partial}{\partial t}\left(\rho u^{i}\right)+\frac{\partial}{\partial x_{k}}\left(p\delta_{ik}+\rho u^{i}u^{k}-\eta\left(\frac{\partial u^{i}}{\partial x_{k}}+\frac{\partial u^{k}}{\partial x_{i}}\right)-\mu\delta_{ik}\left(\nabla\cdot\boldsymbol{u}\right)\right)\\ & =\frac{\partial\rho}{\partial t}u^{i}+\rho\frac{\partial u^{i}}{\partial t}+\frac{\partial p}{\partial x_{i}}+\left(\frac{\partial u^{i}}{\partial x_{k}}\right)\rho u^{k}+u^{i}\frac{\partial}{\partial x_{k}}\left(\rho u^{k}\right)\\ &\relphantom{=}-\eta\triangle u^{i}-\eta\frac{\partial}{\partial x_{i}}\left(\nabla\cdot\boldsymbol{u}\right)-\mu\frac{\partial}{\partial x_{i}}\left(\nabla\cdot\boldsymbol{u}\right)\\ & =\frac{\partial\rho}{\partial t}u^{i}+u^{i}\frac{\partial}{\partial x_{k}}\left(\rho u^{k}\right)+\rho\frac{\partial u^{i}}{\partial t}+\frac{\partial p}{\partial x_{i}}+\left(\frac{\partial u^{i}}{\partial x_{k}}\right)\rho u^{k}\\ &\relphantom{=}-\eta\triangle u^{i}-\left(\eta+\mu\right)\frac{\partial}{\partial x_{i}}\left(\nabla\cdot\boldsymbol{u}\right) \end{align*}

을 얻는다.

여기서 mass conservartion law에 의하여

    \[ u^{i}\left(\frac{\partial\rho}{\partial t}+\nabla\cdot\left(\rho\boldsymbol{u}\right)\right)=0 \]

에 의하여

    \[ \rho\frac{\partial u^{i}}{\partial t}+\rho\boldsymbol{u}\cdot\nabla u^{i}+\frac{\partial p}{\partial x_{i}}-\eta\triangle u^{i}-\left(\eta+\mu\right)\frac{\partial}{\partial x_{i}}\left(\nabla\cdot\boldsymbol{u}\right)=0 \]

을 얻으며, 이를 벡터기호로 정리하면

    \[ \rho\frac{\partial\boldsymbol{u}}{\partial t}+\rho\left(\boldsymbol{u}\cdot\nabla\right)\boldsymbol{u}=-\nabla p+\eta\triangle\boldsymbol{u}+\left(\eta+\mu\right)\nabla\left(\nabla\cdot\boldsymbol{u}\right) \]

을 얻는다. 좌변은 유체의 관성을 기술하며, 우변은 유체에 가해지는 힘이다.

이제 incompressiblity을 가정하면, 즉 \rho를 상수로 취급하고 \nabla\cdot\boldsymbol{u}=0이라
하면

    \[ \frac{\partial\boldsymbol{u}}{\partial t}+\left(\boldsymbol{u}\cdot\nabla\right)\boldsymbol{u}=-\nabla\left(\frac{p}{\rho}\right)+\nu\triangle\boldsymbol{u} \]

을 얻는다.

여기서 Navier-Stokes equation을 momentum conservation law을 바탕으로 유도했다.
Incompressible Navier-Stokes equation은 점성항 때문에 일반적으로 energy conservation law가 발생하지 않는다. 적절한 가정을 위하여 \left|\boldsymbol{x}\right|\gg1에 대하여
\boldsymbol{u}\left(\boldsymbol{x},t\right)=0라 하자.

    \[ E_{\text{kin}}=\frac{\rho}{2}\int\boldsymbol{u}\cdot\boldsymbol{u}dV \]

라 하면

    \begin{align*} \frac{\partial}{\partial t}\left(\frac{1}{2}\rho\boldsymbol{u}\cdot\boldsymbol{u}\right) & =\rho u^{i}\frac{\partial u^{i}}{\partial t}\\  & =\rho u^{i}\left(-u^{k}\frac{\partial u^{i}}{\partial x_{k}}-\frac{1}{\rho}\frac{\partial p}{\partial x_{i}}+\frac{1}{\rho}\frac{\partial\sigma_{ik}^{\prime}}{\partial x_{k}}\right)\\  & =-\rho\boldsymbol{u}\cdot\left(\boldsymbol{u}\cdot\nabla\right)\boldsymbol{u}\\  & \relphantom{=}-\boldsymbol{u}\cdot\nabla p+u^{i}\frac{\partial\sigma_{ik}}{\partial x_{k}}\\  & =-\rho\left(\boldsymbol{u}\cdot\nabla\right)\left(\frac{1}{2}\boldsymbol{u}\cdot\boldsymbol{u}+\frac{p}{\rho}\right)\\  & \relphantom{=}+\nabla\cdot\left(\boldsymbol{u}\cdot\sigma^{\prime}\right)-\sigma_{ik}^{\prime}\frac{\partial u^{i}}{\partial x_{k}} \end{align*}

를 얻는다.

Incompressibility 가정 \nabla\cdot\boldsymbol{u}=0으로부터

    \[ \frac{\partial}{\partial t}\left(\frac{1}{2}\rho\boldsymbol{u}\cdot\boldsymbol{u}\right)=-\nabla\cdot\left[\rho\boldsymbol{u}\left(\frac{1}{2}\boldsymbol{u}\cdot\boldsymbol{u}+\frac{p}{\rho}\right)-\boldsymbol{u}\cdot\sigma^{\prime}\right]-\sigma_{ik}^{\prime}\frac{\partial u^{i}}{\partial x_{k}} \]

을 얻는다. 이제 이를 적분하면 divergence theorem에 의하여

    \begin{align*} \frac{\partial}{\partial t}\int_{B}\left(\frac{1}{2}\rho\boldsymbol{u}\cdot\boldsymbol{u}\right)dV & =-\oint_{\partial B}\left[\rho\boldsymbol{u}\left(\frac{1}{2}\boldsymbol{u}\cdot\boldsymbol{u}+\frac{p}{\rho}\right)-\boldsymbol{u}\cdot\sigma^{\prime}\right]\cdot d\boldsymbol{S}\\  & \relphantom{=}-\int_{B}\sigma_{ik}^{\prime}\frac{\partial u^{i}}{\partial x_{k}}dV \end{align*}

를 얻는다. B를 원점을 중심으로 하는 구라 하고 반지름이 적당히 커서 \boldsymbol{u}=0이 되도록
한다면,

    \begin{align*} \frac{\partial}{\partial t}\int_{B}\left(\frac{1}{2}\rho\boldsymbol{u}\cdot\boldsymbol{u}\right)dV & =-\int_{B}\sigma_{ik}^{\prime}\frac{\partial u^{i}}{\partial x_{k}}dV\\  & =-\frac{1}{2}\int_{B}\sigma_{ik}^{\prime}\left(\frac{\partial u^{i}}{\partial x_{k}}+\frac{\partial u^{k}}{\partial x_{i}}\right)dV \end{align*}

를 얻는다. 마지막은 \sigma_{ik}^{\prime}이 대칭텐서이기 때문에 성립한다.

여기서

    \[ \sigma_{ik}^{\prime}=\eta\left(\frac{\partial u^{i}}{\partial x_{k}}+\frac{\partial u^{k}}{\partial x_{i}}\right) \]

라 가정했으므로

    \[ \frac{\partial}{\partial t}\int_{B}\left(\frac{1}{2}\rho\boldsymbol{u}\cdot\boldsymbol{u}\right)dV=-\frac{\eta}{2}\int_{B}\left(\frac{\partial u^{i}}{\partial x_{k}}+\frac{\partial u^{k}}{\partial x_{i}}\right)^{2}dV \]

를 얻는다. 여기서 \eta>0이므로

    \[ \frac{\partial}{\partial t}E_{\mathrm{kin}}\le0 \]

임을 얻는다. 즉 에너지가 일반적으로 분산된다.