I tried to write formulas (part 4).
<Wave Equation>
Wave equation in 3 dimensions is described as below.
\begin{eqnarray}\frac{\partial^2u}{\partial t^2} &=& v^2\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}\right)\end{eqnarray}
Here, we assume as follows: \begin{eqnarray}u(x,y,z,t)=f(x)g(y)h(z)I(t)\end{eqnarray}
Then, the equation can be deformed as below.
\begin{eqnarray}f(x)g(y)h(z)\frac{\partial^2I(t)}{\partial t^2} &=& v^2\left(g(y)h(z)I(t)\frac{\partial^2f(x)}{\partial x^2}+f(x)h(z)I(t)\frac{\partial^2g(y)}{\partial y^2}\\+f(x)g(y)I(t)\frac{\partial^2h(z)}{\partial z^2}\right)\notag\\\notag\\\frac{1}{I(t)}\frac{\partial^2I(t)}{\partial t^2} &=& v^2\left(\frac{1}{f(x)}\frac{\partial^2f(x)}{\partial x^2}+\frac{1}{g(y)}\frac{\partial^2g(y)}{\partial y^2}+\frac{1}{h(z)}\frac{\partial^2h(z)}{\partial z^2}\right) \end{eqnarray}
Here, left hand side is a function of , and the right hand side is a function of . Therefore, the both sides must be equal to constant function ( is nonzero because is nozero).Moreover, when is replaced by , the equation can be deformed as below.
\begin{eqnarray}\frac{1}{I(t)}\frac{\partial^2I(t)}{\partial t^2} &=& -\omega^2 \\v^2\left(\frac{1}{f(x)}\frac{\partial^2f(x)}{\partial x^2}+\frac{1}{g(y)}\frac{\partial^2g(y)}{\partial y^2}+\frac{1}{h(z)}\frac{\partial^2h(z)}{\partial z^2}\right) &=& -\omega^2 \end{eqnarray}
Also, wave number is expressed by and .
\begin{eqnarray}\frac{\omega}{v}&=&\frac{2\pi f}{f\lambda}=\frac{2\pi}{\lambda}{\equiv}\space k\\k^2 &=&k^2_{x}+k^2_{y}+k^2_{z}\end{eqnarray}
So, the equation can be deformed into 4 equations as below. \begin{eqnarray}\frac{1}{I(t)}\frac{\partial^2I(t)}{\partial t^2} &=& -\omega^2 \\\frac{1}{f(x)}\frac{\partial^2f(x)}{\partial x^2}&=& -k^2_x \\\frac{1}{g(y)}\frac{\partial^2g(y)}{\partial y^2}&=& -k^2_y \\\frac{1}{h(z)}\frac{\partial^2h(z)}{\partial z^2} &=& -k^2_z \end{eqnarray}
Then, each equation can be solved as follows.
\begin{eqnarray}f(x)&=&A_1e^{-ik_xx}+A_2e^{ik_xx}\\g(y)&=&A_3e^{-ik_yy}+A_4e^{ik_yy}\\h(z)&=&A_5e^{-ik_zz}+A_6e^{ik_zz}\\I(t)&=&A_7e^{-i\omega t}+A_8e^{i\omega t}\end{eqnarray}
For the sake of simplicity, the solutions are expressed as below. \begin{eqnarray}f(x)&=&A_2e^{ik_xx}\\g(y)&=&A_4e^{ik_yy}\\h(z)&=&A_6e^{ik_zz}\\I(t)&=&A_7e^{-i\omega t}\end{eqnarray}
Therefore the wave equation can be solved as below. \begin{eqnarray}u(x,y,z,t)&=&f(x)g(y)h(z)I(t)\\&=&A_2e^{ik_xx}A_4e^{ik_yy}A_6e^{ik_zz}A_7e^{-i\omega t}\\&=&A_2A_4A_6A_7e^{ik_xx}e^{ik_yy}e^{ik_zz}e^{-i\omega t}\\&=&Ae^{i(k_xx+k_yy+k_zz-\omega t)}\space(\because A=A_2A_4A_6A_7) \end{eqnarray}
The solution can be expressed as follows, when we use the wave number vector and the position vector .
\begin{eqnarray}u({\boldsymbol r}, t)=Ae^{i( {\boldsymbol k}{\cdot}{\boldsymbol r}-\omega t)}\end{eqnarray}
<波動方程式>
3次元における波動方程式は次のように記述される。
\begin{eqnarray}\frac{\partial^2u}{\partial t^2} &=& v^2\left(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}\right)\end{eqnarray}
ここで、次のように仮定する。\begin{eqnarray}u(x,y,z,t)=f(x)g(y)h(z)I(t)\end{eqnarray}
すると、方程式は次のように変形できる。
\begin{eqnarray}f(x)g(y)h(z)\frac{\partial^2I(t)}{\partial t^2} &=& v^2\left(g(y)h(z)I(t)\frac{\partial^2f(x)}{\partial x^2}+f(x)h(z)I(t)\frac{\partial^2g(y)}{\partial y^2}\\+f(x)g(y)I(t)\frac{\partial^2h(z)}{\partial z^2}\right)\notag\\\notag\\\frac{1}{I(t)}\frac{\partial^2I(t)}{\partial t^2} &=& v^2\left(\frac{1}{f(x)}\frac{\partial^2f(x)}{\partial x^2}+\frac{1}{g(y)}\frac{\partial^2g(y)}{\partial y^2}+\frac{1}{h(z)}\frac{\partial^2h(z)}{\partial z^2}\right) \end{eqnarray}
ここで、左辺は時刻の関数であり、右辺は位置の関数である。それゆえ両辺は定数関数に等しくなければならない ( は0でないので、は0でない)。さらにをで置き換えれば、方程式は次のように変形できる。
\begin{eqnarray}\frac{1}{I(t)}\frac{\partial^2I(t)}{\partial t^2} &=& -\omega^2 \\v^2\left(\frac{1}{f(x)}\frac{\partial^2f(x)}{\partial x^2}+\frac{1}{g(y)}\frac{\partial^2g(y)}{\partial y^2}+\frac{1}{h(z)}\frac{\partial^2h(z)}{\partial z^2}\right) &=& -\omega^2 \end{eqnarray}
さらに波数はとを用いて次のように表される。
\begin{eqnarray}\frac{\omega}{v}&=&\frac{2\pi f}{f\lambda}=\frac{2\pi}{\lambda}{\equiv}\space k\\k^2 &=&k^2_{x}+k^2_{y}+k^2_{z}\end{eqnarray}
ゆえに、方程式は次のように4つの方程式へ変形できる。\begin{eqnarray}\frac{1}{I(t)}\frac{\partial^2I(t)}{\partial t^2} &=& -\omega^2 \\\frac{1}{f(x)}\frac{\partial^2f(x)}{\partial x^2}&=& -k^2_x \\\frac{1}{g(y)}\frac{\partial^2g(y)}{\partial y^2}&=& -k^2_y \\\frac{1}{h(z)}\frac{\partial^2h(z)}{\partial z^2} &=& -k^2_z \end{eqnarray}
それぞれの方程式は、次のように解かれる。
\begin{eqnarray}f(x)&=&A_1e^{-ik_xx}+A_2e^{ik_xx}\\g(y)&=&A_3e^{-ik_yy}+A_4e^{ik_yy}\\h(z)&=&A_5e^{-ik_zz}+A_6e^{ik_zz}\\I(t)&=&A_7e^{-i\omega t}+A_8e^{i\omega t}\end{eqnarray}
簡単のために、解を次のように表す。\begin{eqnarray}f(x)&=&A_2e^{ik_xx}\\g(y)&=&A_4e^{ik_yy}\\h(z)&=&A_6e^{ik_zz}\\I(t)&=&A_7e^{-i\omega t}\end{eqnarray}
よって、波動方程式は次のように解くことができる。\begin{eqnarray}u(x,y,z,t)&=&f(x)g(y)h(z)I(t)\\&=&A_2e^{ik_xx}A_4e^{ik_yy}A_6e^{ik_zz}A_7e^{-i\omega t}\\&=&A_2A_4A_6A_7e^{ik_xx}e^{ik_yy}e^{ik_zz}e^{-i\omega t}\\&=&Ae^{i(k_xx+k_yy+k_zz-\omega t)}\space(\because A=A_2A_4A_6A_7) \end{eqnarray}
波動方程式の解は波数ベクトルと位置ベクトルを用いれば、次のように表現することができる。
\begin{eqnarray}u({\boldsymbol r}, t)=Ae^{i( {\boldsymbol k}{\cdot}{\boldsymbol r}-\omega t)}\end{eqnarray}