含离散非连续项的微分方程命名及离散ω²下方程解的差异咨询

Hey there! Let's unpack this question to clear up your confusion—great to see you digging into the nuances of differential equations.

First off, the equation you're studying: $x'' + \omega^2 x = 0$ is the classic second-order linear homogeneous ordinary differential equation (ODE) for simple harmonic motion. Let's break down the difference between continuous and discrete $\omega^2$ cases:

Continuous $\omega^2$ scenario #

When $\omega^2$ can take any positive real value, the general solution is a family of continuous frequency harmonic motions:
$$x(t) = A\cos(\omega t) + B\sin(\omega t)$$
or equivalently, written in phase form:
$$x(t) = C\cos(\omega t + \phi)$$
Here, $A, B, C, \phi$ are constants determined by initial conditions (like $x(0)$ and $x'(0)$). The key point is: any positive $\omega$ is allowed, so you can get harmonic motion at literally any frequency.

Discrete $\omega^2$ scenario #

In your case, $\omega^2$ is restricted to discrete values: $\omega_m^2 = C m(m+2)$ where $C$ is a constant and $m = 1,2,3,\dots$. This means each integer $m$ corresponds to a fixed, distinct angular frequency $\omega_m = \sqrt{C m(m+2)}$.

The solutions here are still harmonic motions, but they're limited to a countable, discrete set of frequencies:
For each $m$, the solution is:
$$x_m(t) = A_m\cos(\omega_m t) + B_m\sin(\omega_m t)$$
Since the ODE is linear and homogeneous, any linear combination of these individual solutions is also a valid solution:
$$x(t) = \sum_{m=1}^{\infty} \left(A_m\cos(\omega_m t) + B_m\sin(\omega_m t)\right)$$
The constants $A_m, B_m$ are still set by initial conditions, but you can't get a solution with a frequency that isn't one of the $\omega_m$ values.

Key differences to highlight #

• Frequency flexibility: Continuous $\omega^2$ lets you have harmonic motion at any positive frequency; discrete $\omega^2$ only allows the specific frequencies tied to each $m$.
• Solution space: With continuous $\omega^2$, the solution space is infinite-dimensional in a "continuous" sense (you can pick any $\omega$). With discrete $\omega^2$, the solution space is still infinite-dimensional but spanned by a countable set of basis solutions (one for each $m$).
• Physical context: Discrete $\omega^2$ usually comes from boundary conditions in physical problems—think things like vibrating strings with fixed ends, quantum mechanical bound states, or resonant systems where only specific frequencies are allowed (these are often called eigenfrequencies or normal modes).

If you have more context about where this discrete $\omega^2$ comes from (like specific boundary conditions that enforce this restriction), we can dive even deeper into why this discretization happens!

备注：内容来源于stack exchange，提问作者fp007