Consider a nonlinear system of the form
$$\dot{y}(t)=p(y(t)) + u(t)$$
where
$$p(q) = a_kq^k+a_{k-1}q^{k-1}+\ldots+a_1q$$ $$u(t) = -\left(\alpha_ky(t)^k+\ldots+\alpha_1y(t)\right)-y(t)$$
with $\dot{\alpha}_i=y(t)^{i+1}$ for $i=1,\ldots, k$. It can be shown that
$$\frac{\rm d}{{\rm d}t}\left[\frac12\left(y(t)^2+\sum_{i=1}^k(a_i-\alpha_i)^2\right)\right]=-y(t)^2\tag{1}$$
Using this information, show that $y(t)\to0$ as $t\to\infty$. [Hint: Use first principles.]
So I must show that solution $y(t)=0$ is asymptotically stable. Originally I tried Lyapunov theory but that didn't work (and is advised against). But how would I show this from first principles? I need to show that
$$\lim_{t\to\infty} y(t)=0$$
from the information given. How on earth would I do that?
These type of problems arise in adaptive control where the controller uses estimations (the $\alpha_i$ in your case) of the unknown system parameters ($a_i$ in the example). The convergence analysis follows from the so called Barbalat lemma which states the following:
If $f$ is uniformly continuous and the integral $\int_0^{\infty}{f(s)ds}$ exists and is finite then $\lim_{t\rightarrow\infty}f(t)=0$.
In the above example let us define $$S:=\frac{1}{2}y^2+\frac{1}{2}\sum_{i=1}^k{(a_i-\alpha_i)^2}$$ You have proved that $$\dot{S}\leq -y^2\leq 0\qquad\qquad (1)$$ i.e. $S$ is decreasing and lower bounded from zero and therefore it converges i.e. there exists some $S_{\infty}\geq 0$ such that $\lim_{t\rightarrow\infty}S(t)=S_{\infty}$. Integrating (1) over $[0,\infty)$ we obtain $$S_{\infty}-S(0)\leq -\int_0^{\infty}{y^2(t)dt}$$
Hence, for function $f(t)=y^2(t)$ the integral $\int_0^{\infty}{f(s)ds}$ exists and is finite. If we also prove that $f$ is uniformly continuous then from Barbalat's lemma the desired property $\lim_{t\rightarrow\infty}f(t)=0$ is directly obtained.
The boundedness of $S$ implies the boundedness of $y$ and $\alpha_i$ ($i=1,\ldots,k$). Thus, $u$ is bounded (see the definition of $u$) and therefore from the original equation describing the system dynamics, the derivative $\dot{y}$ is also bounded. Since every differentiable function with bounded derivative is uniformly continuous, $\dot{y}\in\mathcal{L}_{\infty}$ implies $y$ is uniformly continuous which combined with $y\in\mathcal{L}_{2}$ yields the desired conclusion.