Hasegawa(1969) 1번 식 해설 4편(완)

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$$\rho {\partial }_t^2{u}_i=\sum _j{\partial }_j(\lambda {\delta }_{ij}\sum _k{\epsilon }_{kk}+\mu ({\partial }_i{u}_j+{\partial }_j{u}_i))$$

 

식이 복잡하므로 식을 전개한 뒤 우변을 둘로 나눠 정리하도록 한다.

이때 $\lambda$, $\mu$는 상수이므로 앞으로 자유롭게 이동할 수 있다.

$$\lambda \sum _j{\partial }_j{\delta }_{ij}\sum _k{\epsilon }_{kk}+\mu \sum _j{\partial }_j({\partial }_i{u}_j+{\partial }_j{u}_i)$$

 

식이 여전히 복잡하므로 조금 더 전개한 뒤 3개로 나눠 살펴보도록 한다.

$$\lambda \sum _j{\partial }_j{\delta }_{ij}\sum _k{\epsilon }_{kk}+\mu \sum _j{\partial }_j{\partial }_i{u}_j+\mu \sum _j{\partial }_j{\partial }_j{u}_i$$

 

이 정도 하면 어느 정도 식이 눈에 들어오게 된다.

First, let's apply kronecker delta in our equation.

$${\delta }_{ij}=\{\begin{array}{l}1(i=j)\\ 0(i\ne j)\end{array}$$

 

So we can simplify the equation like below.

$$\lambda \sum _j{\partial }_j\sum _k{\epsilon }_{kk}+\mu \sum _j{\partial }_j{\partial }_i{u}_j+\mu \sum _j{\partial }_j{\partial }_j{u}_i$$

 

At this point, let's bring the concepts gradient and divergence.

 

#1 Gradient

: Scalar function to Vector function

$$\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}u=\mathrm{\nabla }f=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}+\frac{\partial f}{\partial z}\mathbf{k}$$

$$\mathrm{\nabla }=\frac{\partial }{\partial x}\mathbf{i}+\frac{\partial }{\partial y}\mathbf{j}+\frac{\partial }{\partial z}\mathbf{k}$$

$$\mathrm{\nabla }=\sum _k{\partial }_k$$

$$=\sum _k{\partial }_{k}f$$

 

#2 Divergence

: Vector function to Scalar funtion

$$\mathrm{d}\mathrm{i}\mathrm{v}\mathbf{F}=\mathrm{\nabla }\cdot \mathbf{F}=(\frac{\partial }{\partial x},\frac{\partial }{\partial y},\frac{\partial }{\partial z})\cdot (F_x,F_y,F_z)=\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}$$

$$=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=\sum _k{\partial }_k{u}_k$$

 

Before applying gradient and divergence with Einstein notation, bring the equation we used before.

$${\epsilon }_{ij}=\frac{1}{2}({\partial }_i{u}_j+{\partial }_j{u}_i)$$

 

if we assume $i=j=k$, we can write it like below.

$${\epsilon }_{kk}=\frac{1}{2}({\partial }_k{u}_k+{\partial }_k{u}_k)={\partial }_k{u}_k$$

$$\sum _k{\epsilon }_{kk}=\sum _k{\partial }_k{u}_k=\mathrm{\nabla }\cdot \mathbf{u}$$

 

$$\lambda \sum _j{\partial }_j\sum _k{\epsilon }_{kk}+\mu \sum _j{\partial }_j{\partial }_i{u}_j+\mu \sum _j{\partial }_j{\partial }_j{u}_i$$

$$\lambda \sum _j{\partial }_j(\mathrm{\nabla }\cdot \mathbf{u})+\mu \sum _j{\partial }_j{\partial }_i{u}_j+\mu \sum _j{\partial }_j{\partial }_j{u}_i$$

$$\lambda \mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})+\mu \sum _j{\partial }_j{\partial }_i{u}_j+\mu \sum _j{\partial }_j{\partial }_j{u}_i$$

 

Now let's arrange the second term,

(다만, 이게 맞는 풀이인지 여전히 의문입니다.)

$$\mu \sum _j{\partial }_j{\partial }_i{u}_j=\mu \sum _j{\partial }_i{\partial }_j{u}_j$$

$${\partial }_j{u}_j=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=\mathrm{\nabla }\cdot \mathbf{u}$$

$$=\mu \mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})$$

 

Then we can arrange the eqaution

$$\lambda \mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})+\mu \mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})+\mu \sum _j{\partial }_j{\partial }_j{u}_i$$

 

Then the third term can be...

(세 번째 항도 이렇게 전개하는 것이 옳은건지 의문입니다.)

$$\sum _j{\partial }_j{\partial }_j=\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}+\frac{{\partial }^2}{\partial z^2}$$

$$=\mu \sum _j{\partial }_j{\partial }_j{u}_i=\mu {\mathrm{\nabla }}^2\mathbf{u}$$

 

$$\lambda \mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})+\mu \mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})+\mu {\mathrm{\nabla }}^2\mathbf{u}$$

 

And let's apply vector calculus

$${\mathrm{\nabla }}^2\mathbf{F}=\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{F})-\mathrm{\nabla }\times (\mathrm{\nabla }\times \mathbf{F})$$

 

$$\lambda \mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})+\mu \mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})+\mu {\mathrm{\nabla }}^2\mathbf{u}$$

$$=\lambda \mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})+\mu \mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})+\mu [\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})-\mathrm{\nabla }\times (\mathrm{\nabla }\times \mathbf{u})]$$

$$=(\lambda +2\mu )\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})-\mu \mathrm{\nabla }\times (\mathrm{\nabla }\times \mathbf{u})$$

 

$$\rho {\partial }_t^2{u}_i=(\lambda +2\mu )\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathbf{u})-\mu \mathrm{\nabla }\times (\mathrm{\nabla }\times \mathbf{u})$$

 

We can also write this equation into this way

$$(\lambda +2\mu )\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{d}\mathrm{i}\mathrm{v}\mathbf{u}-\mu \mathrm{r}\mathrm{o}\mathrm{t}\left(\mathrm{r}\mathrm{o}\mathrm{t}\mathbf{u}\right)={\rho }^{\ast }{\partial }^2\mathbf{u}/\partial t^2$$

 

글을 작성하면서 suffix notation, kronecker delta and Levi-Civita tensor에 대한 학습이 추가로 이뤄져야함을 느낌.

 

Reference

Wikipedia, S wave

Wikipedia, Einstein notation

A Primer on Index Notation, John Crimaldi. August 28, 2006

네이버 블로그, 공부가 싫은 사람 「Gradient, Divergence and Curl in Suffix Notation」