91影视

Write the expression for the transmitted wave.

${\mathit{g}}_{\mathbf{t}}\mathbf{(}\mathit{z}\mathbf{-}{\mathit{v}}_{\mathbf{2}}\mathit{t}\mathbf{)}\mathbf{=}{\mathit{g}}_{\mathbf{1}}\mathbf{(}\mathit{z}\mathbf{-}{\mathit{v}}_{\mathbf{1}}\mathit{t}\mathbf{)}\mathbf{+}{\mathit{h}}_{\mathbf{R}}\mathbf{(}\mathit{z}\mathbf{-}{\mathit{v}}_{\mathbf{1}}\mathit{t}\mathbf{)}$ 鈥�. (1)

Here, $g_1(z-v_{1}t)$is the shape of the incident wave and is the shape of the reflected wave.

Consider first equation for the boundary condition.

$f(0^{-1},t)=f(0^{+},t)$ 鈥�.. (1)

Consider the first equation for the boundary condition.

${\overline{)\frac{鈭�f}{鈭�z}}}_{0-}={\overline{)\frac{鈭�f}{鈭�z}}}_{0+}$ 鈥�.. (2)

Apply the boundary condition from equation (2).

$$\begin{gathered} g_T(0-v_{2}t)=g_1(0-v_{1}t)+h_R(0+v_{1}t) \\ {g}_T(-v_{2}t)=g_1(-v_{1}t)+h_R\left(v_{1}t\right)......\left(4\right) \end{gathered}$$$$\begin{gathered} g_T(0-v_{2}t)=g_1(0-v_{1}t)+h_R(0+v_{1}t) \\ {g}_T(-v_{2}t)=g_1(-v_{1}t)+h_R\left(v_{1}t\right)......\left(4\right) \end{gathered}$$

Apply the boundary condition from equation (4).

$$\begin{gathered} -\frac{1}{v_1}\frac{鈭�g_T(-v_{1}t)}{鈭�t}+\frac{1}{v_1}\frac{鈭�h_R\left(v_{1}t\right)}{鈭�t}=-\frac{1}{v_2}\frac{鈭�g_T(-v_{2}t)}{鈭�t} \\ -\frac{鈭�g_1(-v_{1}t)}{鈭�t}+\frac{鈭�h_R\left(v_{1}t\right)}{鈭�t}=\frac{v_1}{v_2}\frac{鈭�g_T(-v_{2}t)}{鈭�t} \\ \frac{鈭�g_1(-v_{1}t)}{鈭�t}-\frac{鈭�h_R\left(v_{1}t\right)}{鈭�t}=\frac{v_1}{v_2}\frac{鈭�g_T(-v_{2}t)}{鈭�t} \end{gathered}$$

Integrate on both sides,

$g_1(-v_{1}t)-h_R(v,t)=\frac{v_1}{v_2}{g}_T(-v_{2}t)+k$ 鈥︹�� (5)

Add equations (4) and (5).

$$\begin{gathered} g_T\left(-v_{2}t\right)+\frac{v_1}{v_2}{g}_T(-v_{2}t)+k=g_1(-v_{1}t)+h_R\left(v_{1}t\right)+g_T(-v_{1}t)-h_{R(-v_{1}t)} \\ {g}_T\left(-v_{2}t\right)\left[1+\frac{v_1}{v_2}\right]+k=g_l(-v_{1}t)+g_l(-v_{1}t) \\ 2g_l\mathbf{(}\mathbf{-}{\mathit{v}}_{\mathbf{1}}\mathit{t}\mathbf{)}\mathbf{=}{\mathit{g}}_{\mathbf{T}}\left(-v_{2}t\right)\left[\frac{v_{2+}{v}_1}{v_2}\right]\mathbf{+}\mathit{k} \\ {\mathit{g}}_{\mathbf{T}}\left(-v_{2}t\right)\mathbf{=}\left(\frac{2v_2}{v_{2+}{v}_1}\right){\mathit{g}}_{\mathbf{l}}\mathbf{(}\mathbf{-}{\mathit{v}}_{\mathbf{1}}\mathit{t}\mathbf{)}\mathbf{+}{\mathit{k}}^{\mathbf{l}} \end{gathered}$$

Here, ${\mathit{k}}^{\mathbf{1}}\mathbf{=}\mathbf{-}\mathit{k}\mathbf{\left(}\frac{{\mathbf{v}}_{\mathbf{2}}}{{\mathbf{v}}_{\mathbf{1}\mathbf{+}}{\mathbf{v}}_2}\mathbf{\right)}$.

Let$(z-v_{1}t)=(z-v_{2}t)=(z-v_{1}t)=u$

Substitute the values in equation (1).

$g_T\left(u\right)=\left(\frac{2v_2}{v_1+v_2}\right)g_l\left(\frac{v_1}{v_2}u\right)+k^1$

Multiply equation (4) with

$\frac{v_1}{v_2}$.

role="math" localid="1657700056261" $\frac{v_1}{v_2}{g}_T\left(-v_{2}t\right)=\frac{v_1}{v_2}{g}_l\left(-v_{1}t\right)=\frac{v_1}{v_2}{h}_T\left(v_{1}t\right)$

Subtract equation (5) from equation (6).

$$\begin{gathered} \frac{v_1}{v_2}{g}_T\left(-v_{2}t\right)+k-\frac{v_1}{v_2}{g}_T\left(-v_{2}t\right)=g_T\left(-v_{1}t\right)-h_R\left(v_{1}t\right)-\frac{v_1}{v_2}{g}_l\left(-v_{1}t\right)-\frac{v_1}{v_2}{h}_R\left(v_{1}t\right) \\ \left[g_l(-v_{1}T)-\frac{v_1}{v_2}{g}_l\left(-v_{1}t\right)\right]-\left[h_R(-v_{1}T)-\frac{v_1}{v_2}{h}_R\left(-v_{1}t\right)\right]=k \\ \left(1-\frac{v_1}{v_2}\right)g_l\left(-v_{1}t\right)-\left(1+\frac{v_1}{v_2}\right)h_R\left(-v_{1}t\right)=k \\ {h}_R\left(u\right)=\left(\frac{v_2-v_1}{v_1+v_2}\right)g_l\left(-v_{1}t\right)+k^l \end{gathered}$$

Therefore, the shape of the reflected wave $\left(h_R\right)$is $h_R\left(u\right)=\left(\frac{v_2-v_1}{v_1+v_2}\right)g_l\left(-v_{1}t\right)+k^l$, and the shape of the transmitted wave $\left(g_T\right)$is $g_T\left(u\right)=\left(\frac{2v_2}{v_1+v_2}\right)g_l\left(\frac{v_1}{v_2}u\right)+k^l$.