关于复合函数f(g(t))的拉普拉斯变换求解方法咨询

Hey there! Great question—unfortunately, the short answer is that there’s no general, trivial formula for the Laplace transform of a composite function f(g(t)) that works for all f and g. Laplace transforms play nicely with linear operations, derivatives, integrals, and even some convolutions, but composition is a whole different ballpark because it breaks the linearity and the nice time-shifting/scaling properties we rely on.

Looking at your specific example with arctangent of that trigonometric rational function: first, you might try simplifying the inner function g(t) first. The numerator and denominator are both linear combinations of cosines with phase shifts—you can use trig identities (like $\cos(A\pm B) = \cos A \cos B \mp \sin A \sin B$) to expand all those terms and combine like terms. Maybe you can rewrite g(t) as a single rational function of $\sin(t)$ and $\cos(t)$, or even convert it to a complex exponential form using Euler’s formula to simplify further.

Once you’ve simplified g(t) as much as possible, the next problem is taking the Laplace transform of $\arctan(g(t))$. There’s no standard table entry for this, so you’d have to use more advanced techniques:

• Series expansion: If g(t) is bounded or satisfies certain conditions, you can expand $\arctan(x)$ as its Taylor series around x=0: $\arctan(x) = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x{2n+1}$, then substitute $x = g(t)$, interchange the sum and Laplace transform (justified by uniform convergence or dominated convergence), and compute the Laplace transform of each term $[g(t)]^{2n+1}$. This can get messy, but for well-behaved g(t) it’s doable.
• Differentiation trick: Let $h(t) = \arctan(g(t))$, then take its derivative with respect to t: $h’(t) = \frac{g’(t)}{1 + [g(t)]^2}$. Now take the Laplace transform of both sides: $s \mathcal{L}{h(t)}(s) - h(0) = \mathcal{L}\left{\frac{g’(t)}{1 + [g(t)]^2}\right}(s)$. If you can compute the right-hand side, you can solve for the Laplace transform of h(t) by rearranging: $\mathcal{L}{h(t)}(s) = \frac{1}{s}\left(h(0) + \mathcal{L}\left{\frac{g’(t)}{1 + [g(t)]^2}\right}(s)\right)$. This might be more feasible if simplifying $\frac{g’(t)}{1 + [g(t)]^2}$ leads to a function whose Laplace transform you can compute or look up.
• Numerical approximation: If analytical methods get too tangled, you can use numerical techniques to approximate the Laplace transform at specific s values. Tools like MATLAB, Python’s SciPy, or Maple have built-in functions for this, which can be handy if you don’t need an exact closed-form expression.

The key takeaway here is that composition doesn’t have a one-size-fits-all solution for Laplace transforms. You have to tailor your approach to the specific f and g—simplify the inner function first, then pick a technique that fits the resulting form. For your arctan example, starting with simplifying the trigonometric expression inside is definitely the first step to make the problem more manageable.

备注：内容来源于stack exchange，提问作者Hikikomori