Let $f:\mathbb R^n\to \mathbb R^n$ smooth. Let $\hat y\in\mathbb R^n$ be a stable equilibrium point for the ODE $y'(t)=f(y(t))$. Namely :
- $f(\hat y)=0$,
- for every $\epsilon>0$ there exists $\delta>0$ such that $|y(t)-\hat y|<\epsilon\ \ \forall\,t\geq0$ provided that $|y(0)-\hat y|<\delta$.
Now fix $\epsilon>0$, fix $y_0$ such that $|y_0-\hat y|<\delta$, and consider the Cauchy problem
$$ y'(t)=f(y(t)) \ ,\quad y(0)=y_0 \ .$$
Denote by $A:=D f(\hat y)$ the jacobian matrix of $f$ at the equilibrium point $\hat y$ and consider the linearized problem around $\hat y$: $$ z'(t) = A\, z(t)\ ,\quad z(0)=y_0-\hat y\ .$$
Suppose the matrix $A$ is negative definite. I would like to say that the linearized problem is a good approximation for the original one. A formal question could be something like: is there a uniform constant $C$ such that $$ |y(t)-\hat y-z(t)| \leq C\,\epsilon^2 \quad\forall t\geq0 \quad\forall y_0\in B(\hat y,\delta) \ ?$$
Tentative solution. Let $r>0$ and $\epsilon\in(0,r)$. By Taylor expansion of $f$ around $\hat y$, there exists a constant $C=C(r)$ such that $$ f(y) = A\,(y-\hat y) + \omega(y)\ ,\quad |\omega(y)|\leq C\, |y-\hat y|^2\ \forall y\in B(\hat y,r)\ .$$ Therefore $$ \frac{d}{d t}\big(y(t)-\hat y-z(t)\big) \,=\, A\,\big(y(t)-\hat y-z(t)\big) \,+\, \omega(y(t))\ .$$ Solving this as a linear equation in $y-\hat y-z$ with non-homegenous term $\omega(y(t))$ gives $$ y(t)-\hat y-z(t) \,=\, e^{tA}\,\big(y(0)-\hat y-z(0)\big) \,+\, \int_0^te^{(t-s) A}\,\omega(y(s))\,ds$$ The first term of the sum is zero. I'd like to bound the second term using the stability of equilibrium, indeed: $$ |\omega(y(s))| \leq C\, |y(s)-\hat y|^2 \leq C\,\epsilon^2 \quad\forall s\geq0 \,$$ and if could take this term out of the integral I would be left with $$ \Big\| \int_0^te^{(t-s) A} ds \Big\| \,=\, \| A^{-1}(e^{tA}-I) \| \,\leq\, \|A^{-1}\| $$ obtaining a uniform bound in $t\geq0$, of the type $$ |y(t)-\hat y-z(t)| \,\leq\, C\,\|A^{-1}\|\,\epsilon^2 \ .$$ The problem is I don't see how to split the remainder $\omega(y(s))$ from the rest of the integral.
Hint: Since $A$ is assumed to be negative definite you can start by proving something like: $$ \exists C, \lambda > 0, \forall \tau \geq 0, \forall w \in \mathbb{R}^n, \quad | e^{\tau A} w | \leq C e^{-\lambda \tau} |w|. $$ Then use this estimate to bound the integrand.