流体动力学:如何求解该粒子轨迹微分方程组?
Alright, let's work through this problem step by step to derive the particle trajectory expressions $x(t)$ and $y(t)$. First, let's restate the given system of ordinary differential equations (ODEs) clearly:
\begin{align} \frac{dx}{dt} & = k \left(\frac{gA}{\omega} \cosh (ky) + \frac{A \omega}{k} \sinh (ky) \right) \cos(kx- \omega t) \\ \frac{dy}{dt} & = k \left(\frac{gA}{\omega} \sinh (ky) + \frac{A \omega}{k} \cosh (ky) \right) \sin(kx- \omega t) \\ \end{align}
Where $g$, $k$, $A$, and $\omega$ are positive constants. This system describes fluid particle motion, so we can use tricks from complex analysis and fluid wave theory to simplify and solve it.
Step 1: Simplify with a Phase Variable
First, notice both equations share the phase term $\theta(t) = kx(t) - \omega t$. Let's compute its time derivative to link it to our ODEs:
\frac{d\theta}{dt} = k\frac{dx}{dt} - \omega
Substitute the expression for $\frac{dx}{dt}$ from the first equation, and you'll see this helps tie the two equations together. But an even more powerful approach is to use complex variables.
Step 2: Use Complex Velocity for Simplification
Define the complex position $z(t) = x(t) + iy(t)$, where $i$ is the imaginary unit. The complex velocity is then $\frac{dz}{dt} = \frac{dx}{dt} + i\frac{dy}{dt}$.
Substitute the original ODEs into this complex velocity, and use the identities $\cos\theta = \frac{e{i\theta}+e{-i\theta}}{2}$ and $\sin\theta = \frac{e{i\theta}-e{-i\theta}}{2i}$. After simplifying (the imaginary units will cancel out nicely), you'll end up with:
\frac{dz}{dt} = \frac{kA}{2}\left(\frac{g}{\omega} + \frac{\omega}{k}\right)e^{ky}e^{i(kx - \omega t)} + \frac{kA}{2}\left(\frac{g}{\omega} - \frac{\omega}{k}\right)e^{-ky}e^{-i(kx - \omega t)}
We can make this even cleaner by recognizing hyperbolic function identities: $\cosh(ky) + \sinh(ky) = e^{ky}$ and $\cosh(ky) - \sinh(ky) = e^{-ky}$.
Step 3: Deep Water Gravity Wave Case (Most Common Scenario)
For deep water gravity waves, we have a key dispersion relation: $\omega^2 = gk$ (this comes from balancing gravitational and inertial forces in the wave). Substitute $g = \frac{\omega^2}{k}$ into our complex velocity equation, and you'll find the second term vanishes entirely!
The system simplifies to:
\frac{dx}{dt} = A\omega e^{ky}\cos(kx - \omega t) \\ \frac{dy}{dt} = A\omega e^{ky}\sin(kx - \omega t)
Small Amplitude Approximation
In most practical cases, the wave amplitude $A$ is much smaller than the wavelength ($A \ll 1/k$). This means the particle's vertical displacement is tiny, so $ky(t) \approx ky_0$ (where $y_0$ is the initial vertical position). The term $e^{ky}$ becomes a constant $e^{ky_0}$.
Now, let $\phi_0 = kx_0$ (the initial phase). We can approximate the phase $\theta(t) \approx \phi_0 - \omega t$ (since the change in $x(t)$ is small). Integrate the simplified ODEs:
- For $x(t)$:
x(t) = x_0 + A e^{ky_0}\sin(kx_0 - \omega t) - For $y(t)$:
y(t) = y_0 + A e^{ky_0}\cos(kx_0 - \omega t)
This describes a circular trajectory—the particle moves in a circle with radius $A e^{ky_0}$. Notice the radius decreases as depth increases (if $y$ is measured downward), which matches the behavior of deep water waves.
Step 4: General Case (No Dispersion Relation Assumption)
If we don't assume the deep water dispersion relation, we can still solve the system under the small amplitude approximation. Let $C_1 = \frac{kA}{2}\left(\frac{g}{\omega} + \frac{\omega}{k}\right)$ and $C_2 = \frac{kA}{2}\left(\frac{g}{\omega} - \frac{\omega}{k}\right)$.
Integrate the complex velocity expression, then separate the real and imaginary parts to get:
x(t) = x_0 + \frac{C_1}{\omega}\sin(kx_0 - ky_0 - \omega t) + \frac{C_2}{\omega}\sin(kx_0 + ky_0 + \omega t) \\ y(t) = y_0 - \frac{C_1}{\omega}\cos(kx_0 - ky_0 - \omega t) + \frac{C_2}{\omega}\cos(kx_0 + ky_0 + \omega t)
This is a superposition of two harmonic motions, which describes elliptical trajectories (common in shallow water waves or waves with surface tension).
内容的提问来源于stack exchange,提问作者glowstonetrees

