In this part we consider a map $\varphi$ from $\mathbf { R } ^ { n }$ to $\mathbf { R } ^ { n }$ of class $\mathcal { C } ^ { 1 }$ such that $\varphi ( 0 ) = 0$, and denoting $a = d \varphi ( 0 )$, such that all eigenvalues of $a$ have strictly negative real part. Let $b(x,y) = \int_0^{+\infty} \langle e^{ta}(x) \mid e^{ta}(y) \rangle\, dt$ and $q(x) = b(x,x)$. Let $x_0 \in \mathbf{R}^n$ and $f_{x_0}$ the solution of $y' = \varphi(y),\ y(0) = x_0$. We fix strictly positive real numbers $\alpha$ and $\beta$ such that for all $t \in \mathbf{R}_+$: $$q(f_{x_0}(t)) \leqslant \alpha \Rightarrow -\|f_{x_0}(t)\|^2 + 2b(f_{x_0}(t), \varepsilon(f_{x_0})(t)) \leqslant -\beta q(f_{x_0}(t))$$ Show then that: $$q \left( x _ { 0 } \right) < \alpha \quad \Rightarrow \quad \forall t \geqslant 0 , q \left( f _ { x _ { 0 } } \right) ( t ) \leqslant e ^ { - \beta t } q \left( x _ { 0 } \right) .$$
In this part we consider a map $\varphi$ from $\mathbf { R } ^ { n }$ to $\mathbf { R } ^ { n }$ of class $\mathcal { C } ^ { 1 }$ such that $\varphi ( 0 ) = 0$, and denoting $a = d \varphi ( 0 )$, such that all eigenvalues of $a$ have strictly negative real part. Let $b(x,y) = \int_0^{+\infty} \langle e^{ta}(x) \mid e^{ta}(y) \rangle\, dt$ and $q(x) = b(x,x)$. Let $x_0 \in \mathbf{R}^n$ and $f_{x_0}$ the solution of $y' = \varphi(y),\ y(0) = x_0$. We fix strictly positive real numbers $\alpha$ and $\beta$ such that for all $t \in \mathbf{R}_+$:
$$q(f_{x_0}(t)) \leqslant \alpha \Rightarrow -\|f_{x_0}(t)\|^2 + 2b(f_{x_0}(t), \varepsilon(f_{x_0})(t)) \leqslant -\beta q(f_{x_0}(t))$$
Show then that:
$$q \left( x _ { 0 } \right) < \alpha \quad \Rightarrow \quad \forall t \geqslant 0 , q \left( f _ { x _ { 0 } } \right) ( t ) \leqslant e ^ { - \beta t } q \left( x _ { 0 } \right) .$$