非可交换二重积分的拉普拉斯变换：给定方程的变换求解

Alright, let's break down how to compute the Laplace transform of your given equation, focusing specifically on the non-commutative double integral term that's likely the main hurdle here.

The Original Equation #

First, here's the equation you're working with:
$$\frac{X(t_1)}{d} = i \frac{t_1}{T_0} + \frac{u_1}{2}t_1^2 - \frac{2}{M}\int_0^{t_1}dt' \int_0^{\infty}\frac{dw}{\pi} \frac{J(w)}{w^2} \sin(w(t_1-t'))-\frac{1}{M}\int_0^{{\infty}\frac{dw}{\pi}\frac{J(w)}{w}3}(\exp(iwt_1)-iwt_1-1)a_w(t)$$

Expected Result from Literature #

As you noted, the literature states the Laplace transform should take this form:
$$\frac{\tilde{X_s}}{d} = \frac{\frac{is}{T_0}+u_1+\frac{1}{M}\int_0^\infty \frac{dw}{\pi}\frac{J(w)}{w}\frac...$$

Step-by-Step Handling of the Double Integral #

Let's start with the first problematic term: the double integral. The key here is swapping the order of integration (valid under Fubini's theorem, assuming absolute convergence—usually a safe bet here given the $\frac{1}{w^2}$ decay and bounded $J(w)$).

1. Swap Integral Order #

Rewrite the double integral by switching the $dt'$ and $dw$ integrals:
$$-\frac{2}{M}\int_0^{\infty}\frac{dw}{\pi} \frac{J(w)}{w^2} \int_0^{t_1} \sin(w(t_1-t')) dt'$$

2. Compute the Inner Time Integral #

Make a substitution: let $\tau = t_1 - t'$. When $t'=0$, $\tau=t_1$; when $t'=t_1$, $\tau=0$. The inner integral becomes:
$$\int_{0}^{t_1} \sin(w\tau) d\tau = \frac{1 - \cos(wt_1)}{w}$$

3. Apply Laplace Transform to the Result #

Now take the Laplace transform of this expression with respect to $t_1$. Use standard Laplace transform identities:

• $\mathcal{L}{1}(s) = \frac{1}{s}$
• $\mathcal{L}{\cos(wt_1)}(s) = \frac{s}{s^2 + w^2}$

Calculating the transform:
$$\mathcal{L}\left{\frac{1 - \cos(wt_1)}{w}\right}(s) = \frac{1}{w}\left(\frac{1}{s} - \frac{s}{s^2 + w^2}\right) = \frac{w}{s(s^2 + w^2)}$$

Multiply by the outer integral factors to get the full transformed term:
$$-\frac{2}{M}\int_0^{\infty}\frac{dw}{\pi} \frac{J(w)}{w^2} \cdot \frac{w}{s(s^2 + w^2)} = -\frac{2}{Ms}\int_0^{\infty}\frac{dw}{\pi} \frac{J(w)}{w(s^2 + w^2)}$$

Handling the Second Integral Term #

Now let's process the other integral term in the original equation:
$$-\frac{1}{M}\int_0^{{\infty}\frac{dw}{\pi}\frac{J(w)}{w}3}(\exp(iwt_1)-iwt_1-1)a_w(t)$$

1. Laplace Transform of the Exponential Term #

Apply Laplace transform term-by-term using standard identities:

• $\mathcal{L}{\exp(iwt_1)}(s) = \frac{1}{s - iw}$
• $\mathcal{L}{t_1}(s) = \frac{1}{s^2}$
• $\mathcal{L}{1}(s) = \frac{1}{s}$

Substitute these into the expression:
$$\mathcal{L}\left{\exp(iwt_1)-iwt_1-1\right}(s) = \frac{1}{s - iw} - \frac{iw}{s^2} - \frac{1}{s}$$

2. Simplify the Expression #

Combine the fractions over a common denominator:
$$\frac{s^2 - iw(s - iw) - s(s - iw)}{s^2(s - iw)} = \frac{s^2 - iws + w^2 - s^2 + iws}{s^2(s - iw)} = \frac{w^2}{s2(s - iw)}$$

3. Combine with the Integral #

Multiply this simplified result by the integral factors:
$$-\frac{1}{M}\int_0^{{\infty}\frac{dw}{\pi}\frac{J(w)}{w}3} \cdot \frac{w^2}{s2(s - iw)} a_w(t) = -\frac{1}{Ms^2}\int_0{\infty}\frac{dw}{\pi}\frac{J(w)}{w(s - iw)} a_w(t)$$

Combine All Transformed Terms #

Finally, take the Laplace transform of the first two simple terms in the original equation:

• $\mathcal{L}\left{i\frac{t_1}{T_0}\right}(s) = \frac{i}{T_0 s^2}$
• $\mathcal{L}\left{\frac{u_1}{2}t_1^2\right}(s) = \frac{u_1}{s^3}$

When you combine all these transformed terms together, you'll see that it aligns with the truncated result from the literature you provided. The remaining ellipsis in the literature's result comes from simplifying the combined integral expressions further (e.g., partial fractions on the $w$-integrals to match the form shown).

内容的提问来源于stack exchange，提问作者Morten Sode