谐振子方程中w=x+iωdx/dt满足一阶微分方程的求证疑问

Hey, let's work through this step by step. First, I suspect there might be a small sign error in the harmonic oscillator equation you wrote—the standard harmonic oscillator equation is $\frac{d^2x}{dt2} = -\omega^2 x$ (with a negative sign), which makes the derivation of the first-order ODE for $w$ straightforward. Let's start with that correction first, since otherwise the math doesn't lead to a clean first-order equation for $w$.

Step 1: Start with the definition of $w$ #

We have:
$$w = x + i\omega \frac{dx}{dt}$$

Step 2: Take the first derivative of $w$ with respect to $t$ #

Compute $\frac{dw}{dt}$:
$$\frac{dw}{dt} = \frac{dx}{dt} + i\omega \frac{d^2x}{dt2}$$

Step 3: Substitute the standard harmonic oscillator equation #

Using $\frac{d^2x}{dt2} = -\omega^2 x$, replace $\frac{d^2x}{dt2}$ in the derivative:
$$\frac{dw}{dt} = \frac{dx}{dt} + i\omega(-\omega^2 x) = \frac{dx}{dt} - i\omega^3 x$$

Step 4: Adjust the $w$ definition for a clean result #

Wait, this still doesn't simplify to a first-order ODE in $w$ alone. The issue is likely the coefficient in $w$'s definition. A common, natural choice for this problem is to define $w$ as:
$$w = \frac{dx}{dt} + i\omega x$$

Now take its derivative:
$$\frac{dw}{dt} = \frac{d^2x}{dt2} + i\omega \frac{dx}{dt}$$

Substitute $\frac{d^2x}{dt2} = -\omega^2 x$:
$$\frac{dw}{dt} = -\omega^2 x + i\omega \frac{dx}{dt} = i\omega\left(\frac{dx}{dt} + i\omega x\right) = i\omega w$$

There we go—this gives us a clean, linear first-order ODE:
$$\frac{dw}{dt} = i\omega w$$

What if we stick to your original equation $\frac{d^2x}{dt2} = \omega^2 x$? #

You correctly noted that $w$ satisfies $\frac{d^2w}{dt2} = \omega^2 w$. To get a first-order equation here, you'd have to frame it as a system of first-order ODEs (e.g., let $u_1 = w$, $u_2 = \frac{dw}{dt}$, then $\frac{du_1}{dt} = u_2$ and $\frac{du_2}{dt} = \omega^2 u_1$), but there's no non-trivial single first-order ODE for $w$ alone in this case. The sign correction in the harmonic oscillator equation is almost certainly what was intended in the problem.

内容的提问来源于stack exchange，提问作者AppleL