# Smooth dependence on the initial condition of the integral of an ODE

I am considering an ODE $\dot{x}=f(x)$, with $x\in\mathbb{R}^d$ and $d<\infty$. $f$ is a $C^k$ function and I denote by $\Phi_t x$ be the value of the solution at time $t$.

I assume that my ODE has a unique attractor that is exponentially stable: $\|\Phi_t x - x^*\| \le \beta e^{-\alpha t}$, with $\alpha>0$ and I define the function $G$ by: $$G x = \int_0^\infty (\Phi_t x - x^*) dt$$

I believe that the exponential stability of the ODE plus $f$ being $C^k$ suffices to show that $G$ is $C^k$ but I am not 100% sure and I am not able to find a reference.

Would anyone know a reference? (my main interest is the $C^2$ case).

I don't know a reference, but here's the rough proof that $G \in C^1$.

First, from the integral equality $$\Phi_t(x+h) - \Phi_t(x) = \int_0^1 D\Phi_t(x+\eta h) \cdot h \,\mathrm{d}\eta$$ follows the mean-value estimate $$\| \Phi_t(x+h) - \Phi_t(x) - D\Phi_t(x) \cdot h \| \le \sup_{\eta \in [0,1]} \| D\Phi_t(x+\eta h) - D\Phi_t(x) \|\,\|h\|.$$ This allows us to estimate $$\|G(x+h) - G(x) - \int_0^\infty D\Phi_t(x) \cdot h \,\mathrm{d}t\| \\ \le \int_0^\infty \| \sup_{\eta\in[0,1]} D\Phi_t(x+\eta h) - D\Phi_t(x) \|\,\mathrm{d}t\,\|h\|.$$ To show that this is $o(h)$, we split up the integral into intervals $[0,T]$ and $[T,\infty)$. For any finite $T$, the first part is small since $t \mapsto D\Phi_t(x)$ is uniformly continuous on $[0,T]$. The second part is exponentially small when $T \to \infty$ due to the exponential stability. Hence $G$ is differentiable and $$DG(x) = \int_0^\infty D\Phi_t(x) \,\mathrm{d}t.$$ Similar estimates can be repeated to prove that $G \in C^1$ and by induction that $G \in C^k$.

Extra details in reply to Willy Wong's comment:

Since $x^*$ is exponentially stable and $f \in C^1$, it follows that $x^*$ is linearly stable, hence $\|D\Phi_t(x^*)\| \le C e^{-\alpha t}$. Then for $x$ close to $x^*$, $D\Phi_t(x)$ is a small perturbation, hence the exponential estimates are approximately preserved. Since $\Phi_t(x) \to x^*$ and $D\Phi_t(x) = D\Phi_{t-T}(\Phi_T(x)) \cdot D\Phi_T(x)$, we see that when $T$ is large enough, essentially the same estimate holds for any initial $x$. Thus $\|D\Phi_t(x^*)\| \le \tilde{C} e^{-\tilde{\alpha} t}$ with $\tilde{\alpha} \approx \alpha$. This implies that the integral of $D\Phi_t(x^*)$ is well-defined.

Note that the fact that $f \in C^1$ at $x^*$ is also implicitly used to argue that $t \mapsto D\Phi_t(x)$ is uniformly (equi)continuous on any bounded interval.

• How do you justify $\int_0^\infty D\Phi_t(x) ~\mathrm{d}t$ exists? (Just being nitpicky here, since the proof should depend on the fact that $f$ is $C^k$ at $x^*$, otherwise there are trivial counterexamples.) – Willie Wong May 26 '16 at 14:06
• Thanks for this answer but I found it hard to generalize it to $C^2$. To me, the main difficulty is to prove that for $x$ close to $x^*$: $\| D^k \Phi_t(x) \| \le C^{-\alpha t}$. Once this is done, you proof seems easy to generalize to $C^k$. But even for the $C^1$ case, this is not obvious for me. Should it be? – N. Gast May 27 '16 at 12:16
• Once you know that $\|D\Phi_t(x)\| \le C e^{-\alpha t}$, then $\|D^k\Phi_t(x)\| \le C e^{-\alpha t}$ follows from a Gronwall-like argument. See Lemma C.1 in my book: jaapeldering.nl/includes/get.php?file=NHIM-noncompact-book.pdf – Jaap Eldering May 27 '16 at 19:27
• Thanks again for the pointer. It took me some time to take the time to understand your proof of Lemma C.1 of your book but now, I am convinced that it solves my question. – N. Gast Jul 21 '16 at 9:19