This part mainly focus on ship wave with the infinite depth water, also mainly use Fourier transform method.
\(\delta = h_0 / \lambda\); \(\varepsilon=a/h_0\)
\(s=\delta \sqrt{k^2+l^2} \quad(s>0)\)
\(\beta=\sqrt{\frac{s}{\delta^2}+\frac{W_\lambda s^3}{\delta^4}}\)
\(\frac{T}{\rho g \lambda^2}=W_\lambda\)
the typical length of boat \(L\) and Froude number \(F_r\), \(V\) is the velocity of boat. \[ F r=\frac{V}{\sqrt{g L}} \]
Method I
nondimensional variables \[ \begin{aligned} & \tilde{x} \rightarrow \lambda x \quad \tilde{y} \rightarrow \lambda y \quad \tilde{z} \rightarrow h_0 z \quad \tilde{t} \rightarrow \frac{\lambda}{\sqrt{g h_0}} t \\ & \tilde {u} \rightarrow \sqrt{g h_0} u \quad \tilde{v} \rightarrow \sqrt{g h_0} v \quad \widetilde{w} \rightarrow \frac{h_0 \sqrt{g h_0}}{\lambda} w\\ &\tilde\varphi \rightarrow \sqrt{g h_0} \lambda \varphi, \quad \tilde{\eta}=a \eta, \quad a: \text { typical amplitude } \end{aligned} \] therefore the Laplace's equation of non-dimensional velocity potential \(\varphi\) is (incompressible and irrotational flow) \[ \begin{aligned} &\varphi_{x x}+\varphi_{y y}+\frac{1}{\delta^2} \varphi_{z z}=0 \quad (\text {no } \varepsilon) \\ &\varphi_{x x}+\varphi_{y y}+\frac{1}{\delta_2} \varphi_{z z}=0 \quad (\text {add } \varepsilon) \end{aligned} \]
✏️explanation (no \(\varepsilon \rightarrow\) case 1)
the original equations are: \[ \begin{aligned} & \text { (N1) }\varphi_x=u ; \varphi_y=v \quad \varphi_z=\frac{h_0^2}{\lambda^2} w \\ & \text { (N2) } u_x+v_y+w_z=0 \\ & \text { (N3) } \varphi_{x x}+\varphi_{y y}+\frac{\lambda^2}{h_0^2} \varphi_{z z}=0 \end{aligned} \] \(\varphi_z\) is not fixed but \(\tilde{u}_x+\tilde{v}_y+\tilde{w}_z \equiv 0\), we should keep in mind, here, we only give assumptions below: \[ \tilde{x}=\lambda x,\quad \tilde{y}=\lambda y,\quad z=h_0z ,\quad\tilde{t} \rightarrow \frac{\lambda}{\sqrt{g h_0}} t \] then use them into the original equations, we start from \(\tilde{u}_{\tilde{x}}\), \(\tilde{v}_{\tilde{y}}\), \(\tilde{w}_{\tilde{z}}\): \[ \begin{aligned} \tilde{u}_{\tilde{x}} & =\frac{\partial \tilde{u}}{\partial u} \frac{\partial u}{\partial x} \frac{\partial x}{\partial \tilde{x}} \\ & =\sqrt{g h_0} u_x \frac{1}{\lambda} \\ & =\frac{\sqrt{g h_0}}{\lambda} u_x \end{aligned} \] similarly: \[ \begin{aligned} &\widetilde{v}_{\tilde{y}}=\frac{\sqrt{g h_0}}{\lambda} v_y \\ &\widetilde{w}_{\tilde{z}} =\frac{\partial \widetilde{w}}{\partial w} \frac{\partial w}{\partial z} \frac{\partial z}{\partial \tilde{z}} \\ & =\frac{h_0 \sqrt{g h_0}}{\lambda} w_z \frac{1}{h_0} \\ & =\frac{\sqrt{g h_0}}{\lambda} w_z \end{aligned} \] next we go for \(\varphi\), \(\varphi_z\) \[ \begin{aligned} & \varphi_x=\frac{\partial \varphi}{\partial \tilde{\varphi}} \frac{\partial \tilde{\varphi}}{\partial \tilde{x}} \frac{\partial \tilde{x}}{\partial x} \quad(=u) \\ &=\frac{\partial \varphi}{\partial \widetilde{\varphi}} \tilde{u} \lambda \quad(=\tilde{u} \frac{1}{\sqrt{g h_0}})\\ \Rightarrow & \frac{\partial \varphi}{\partial \tilde{\varphi}} \lambda=\frac{1}{\sqrt{g h_0}} \\ & \lambda \sqrt{g h_0} \varphi=\widetilde{\varphi} \end{aligned} \] similarly, \[ \begin{aligned} & \varphi_z=\frac{\partial \varphi}{\partial z} \\ &=\frac{\partial \varphi}{\partial \widetilde{\varphi}} \frac{\partial \widetilde{\varphi}}{\partial \widetilde{z}} \frac{\partial \widetilde{z}}{\partial z} \\ &=\frac{1}{\lambda \sqrt{g h_0}} \widetilde{w}_z h_0 \\ &=\frac{1}{\lambda \sqrt{g h_0}} \frac{h_0 \sqrt{g h_0}}{\lambda} w h_0 \\ &=\frac{h_0^2}{\lambda^2} w \end{aligned} \]
on surface the kinematic condition \[ \begin{array}{ll} \varphi_z=\frac{a h_0}{\lambda^2}\left(\eta_t+u \eta_x+v \eta_y\right) & \text { (no } \varepsilon) \\ \varphi_z=\delta^2\left[\eta_t+\varepsilon\left(u \eta_x+v \eta_y\right)\right] & (\varepsilon \text { added }) \\ \varphi_z=\delta^2 \eta_t \end{array} \]
Method II
分离变量法，求解一下，就用 modern introduce...的假设解法
\(w_t=\eta_{t t} \quad\) at \(z=1 \quad t>0\) \(w_t=-\frac{1}{\delta^2} p_z=-\frac{1}{\delta^2} \eta_z \quad\) at \(z=1 \quad t>0\)
how can we get the conclusion below
\(\hat{w}=\hat{\eta}_t e^{\delta s(z-1)}\)
3. Relating \(\mathcal H_1{p}\) and \(\mathcal H_1{u}\)
From the transformed momentum equation in the rrr-direction:
\(\frac{\partial \hat{u}}{\partial t} = s H_1{p}(s, z, t)\)
But from the continuity equation:
\(H_1{u}(s, z, t) = -\frac{1}{s} \frac{\partial \hat{w}}{\partial z}\)
To relate \(H_1{p}\) and \(H_1{u}\), we need to express H_1[p} in terms of ww. However, ppp is related to ηvia boundary conditions, which we'll explore next.