Let $M \in M_{n}(\mathbf{R})$. We denote by (S) the differential system: $$(\mathrm{S}) \quad X' = MX$$ where $X$ is a function from the variable $t$ in $\mathbf{R}$ to $\mathbf{R}^{n}$, differentiable on $\mathbf{R}$.
We assume $n = 2$, we then denote $X = (x; y)$ where $x$ and $y$ are two functions differentiable from $\mathbf{R}$ to $\mathbf{R}$ and we set $z = x + iy$.
Let $M \in M_{2}(\mathbf{R})$ be semi-simple. Give a necessary and sufficient condition, concerning the real and imaginary parts of the eigenvalues of $M$, for every solution of (S) to have each of its coordinates tend to 0 as $+\infty$.

In all that follows, $\mathbf{K = C}$. We set $E = \mathbf{C}^n$. We are given $A \in \mathcal{M}_n(\mathbf{C})$. We set $\alpha = \max_{\lambda \in \operatorname{Sp}(A)} \operatorname{Re}(\lambda)$. For all real $t$ and for $(i,j) \in \llbracket 1,n \rrbracket^2$, we denote by $v_{i,j}(t)$ the coefficient with indices $(i,j)$ of the matrix $e^{tA}$.
$\mathbf{22}$ ▷ Deduce from question 21) that there exists a natural integer $p$ such that, for all $(i,j) \in \llbracket 1,n \rrbracket^2$, we have $$v_{i,j}(t) = O\left(t^p e^{\alpha t}\right) \text{ as } t \rightarrow +\infty.$$

In all that follows, $\mathbf{K = C}$. We set $E = \mathbf{C}^n$. We are given $A \in \mathcal{M}_n(\mathbf{C})$. We set $\alpha = \max_{\lambda \in \operatorname{Sp}(A)} \operatorname{Re}(\lambda)$.
$\mathbf{23}$ ▷ Study the converse of question 19): that is, show that if $\alpha < 0$ then $\lim_{t \rightarrow +\infty} f_A(t) = 0_n$.

In all that follows, $\mathbf{K = C}$. We set $E = \mathbf{C}^n$. We are given $A \in \mathcal{M}_n(\mathbf{C})$. We introduce the following polynomials: $$\begin{aligned} & P_s(X) = \prod_{\substack{\lambda \in \operatorname{Sp}(A) \\ \operatorname{Re}(\lambda) < 0}} (X - \lambda)^{m_\lambda}, \\ & P_i(X) = \prod_{\substack{\lambda \in \operatorname{Sp}(A) \\ \operatorname{Re}(\lambda) > 0}} (X - \lambda)^{m_\lambda}, \\ & P_n(X) = \prod_{\substack{\lambda \in \operatorname{Sp}(A) \\ \operatorname{Re}(\lambda) = 0}} (X - \lambda)^{m_\lambda}, \end{aligned}$$ and the subspaces $E_s = \operatorname{Ker}(P_s(A))$, $E_i = \operatorname{Ker}(P_i(A))$ and $E_n = \operatorname{Ker}(P_n(A))$ of $E = \mathbf{C}^n$. We have $E = E_s \oplus E_i \oplus E_n$.
$\mathbf{26}$ ▷ Show that $$E_n = \left\{ X \in E \mid \exists C \in \mathbf{R}_+^* \quad \exists p \in \mathbf{N} \quad \forall t \in \mathbf{R} \quad \left\| e^{tA} X \right\|_E \leq C(1 + |t|)^p \right\}.$$

For $p \in \mathbb { R } ^ { * }$ we denote by $(E_p)$ the differential equation on $\mathbb { R } _ { + } ^ { * }$ $$\left( E _ { p } \right) : x \left( y ^ { \prime \prime } - y ^ { \prime } \right) + p y = 0 .$$ Let $p \in \mathbb { R } ^ { * }$ and $\left( a _ { n } \right) _ { n \in \mathbb { N } }$ be a sequence of real numbers. We assume that the power series $\sum _ { n \geqslant 0 } a _ { n } x ^ { n }$ has infinite radius of convergence. Show that the function $f : x \mapsto \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n }$ is a solution of $\left( E _ { p } \right)$ if and only if $$\left\{ \begin{array} { l } a _ { 0 } = 0 \\ n ( n + 1 ) a _ { n + 1 } = ( n - p ) a _ { n } , \quad \forall n \in \mathbb { N } ^ { * } \end{array} \right.$$

For $p \in \mathbb { R } ^ { * }$ we denote by $(E_p)$ the differential equation on $\mathbb { R } _ { + } ^ { * }$ $$\left( E _ { p } \right) : x \left( y ^ { \prime \prime } - y ^ { \prime } \right) + p y = 0 .$$ The coefficients of a power series solution satisfy $a_0 = 0$ and $n(n+1)a_{n+1} = (n-p)a_n$ for all $n \in \mathbb{N}^*$. Show that $(E_p)$ has non-identically zero polynomial solutions if and only if $p \in \mathbb { N } ^ { * }$. Show that then, the non-zero polynomial solutions of $(E_p)$ are of degree $p$ and belong to the vector space $E$.

For $p \in \mathbb { N } ^ { * }$, consider a polynomial $P \in \mathbb { R } [ X ]$ such that the polynomial function $x \mapsto P ( x )$ is a solution of the equation $\left( E _ { p } \right) : x \left( y ^ { \prime \prime } - y ^ { \prime } \right) + p y = 0$. For all $x \in \mathbb { R }$, we denote $h ( x ) = \mathrm { e } ^ { - x } P ( x )$. Show that the function $h$ is a solution of the differential equation $x \left( y ^ { \prime \prime } + y ^ { \prime } \right) + p y = 0$ on $\mathbb { R } _ { + } ^ { * }$.

For $p \in \mathbb { R } ^ { * }$ we denote by $(E_p)$ the differential equation on $\mathbb { R } _ { + } ^ { * }$ $$\left( E _ { p } \right) : x \left( y ^ { \prime \prime } - y ^ { \prime } \right) + p y = 0 .$$ We fix a non-zero real $p$ and assume that $p \notin \mathbb { N } ^ { * }$. The coefficients of a power series solution satisfy $a_0 = 0$ and $n(n+1)a_{n+1} = (n-p)a_n$ for all $n \in \mathbb{N}^*$. Justify the existence of sequences $\left( a _ { n } \right) _ { n \in \mathbb { N } } \in \mathbb { R } ^ { \mathbb { N } }$ not identically zero such that the power series $\sum _ { n \geqslant 0 } a _ { n } x ^ { n }$ has infinite radius of convergence and such that the function $x \mapsto \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n }$ is a solution of $(E_p)$.

In this part, $a$ denotes an endomorphism of $\mathbf { C } ^ { n }$. We use the decomposition $\mathbf { C } ^ { n } = \bigoplus _ { i = 1 } ^ { r } E _ { i }$ where $E _ { i } = \operatorname { Ker } \left( a - \lambda _ { i } id _ { \mathbf { C } ^ { n } } \right) ^ { m _ { i } }$, with $\|.\|_c$ the norm on $\mathcal{L}(\mathbf{C}^n)$.
Deduce the existence of a polynomial $P$ with real coefficients such that: $$\forall t \in \mathbf { R } , \quad \left\| e ^ { t a } \right\| _ { c } \leqslant P ( | t | ) \sum _ { i = 1 } ^ { r } e ^ { t \operatorname { Re } \left( \lambda _ { i } \right) }$$ where $\operatorname { Re } ( z )$ denotes the real part of a complex number $z$.

For any matrix $A \in \mathscr { M } _ { n } ( \mathbf { R } )$, we denote by $u _ { A }$ the endomorphism canonically associated with $A$ in $\mathbf { R } ^ { n }$ and $v _ { A }$ the endomorphism of $\mathbf { C } ^ { n }$ canonically associated with $A$, viewed as a matrix in $\mathscr { M } _ { n } ( \mathbf { C } )$. We keep the notation $\| . \| _ { c }$ for the norm introduced in part A on $\mathcal { L } \left( \mathbf { C } ^ { n } \right)$ and we use $\| . \| _ { r }$ on $\mathcal { L } \left( \mathbf { R } ^ { n } \right)$. Show that there exists $C > 0$ such that: $$\forall A \in \mathscr { M } _ { n } ( \mathbf { R } ) , \quad \forall t \in \mathbf { R } , \quad \left\| e ^ { t u _ { A } } \right\| _ { r } \leqslant C \left\| e ^ { t v _ { A } } \right\| _ { c }$$

We consider $u$ an endomorphism of $\mathbf { R } ^ { n }$, and $A \in \mathscr { M } _ { n } ( \mathbf { R } )$ its matrix in the canonical basis. We denote by $S p ( A )$ the complex spectrum of $A$. Let $g _ { x _ { 0 } }$ be the unique $\mathcal { C } ^ { 1 }$ solution on $\mathbf { R } _ { + }$ of: $$\left\{ \begin{aligned} y ^ { \prime } & = u ( y ) \\ y ( 0 ) & = x _ { 0 } \end{aligned} \right.$$
Show that: $$\forall x _ { 0 } \in \mathbf { R } ^ { n } , \quad \lim _ { t \rightarrow + \infty } \left\| g _ { x _ { 0 } } ( t ) \right\| = 0 \Longleftrightarrow S p ( A ) \subset \mathbf { R } _ { - } ^ { * } + i \mathbf { R } .$$

We consider $u$ an endomorphism of $\mathbf { R } ^ { n }$, and $A \in \mathscr { M } _ { n } ( \mathbf { R } )$ its matrix in the canonical basis. Let $g _ { x _ { 0 } }$ be the unique $\mathcal { C } ^ { 1 }$ solution on $\mathbf { R } _ { + }$ of: $$\left\{ \begin{aligned} y ^ { \prime } & = u ( y ) \\ y ( 0 ) & = x _ { 0 } \end{aligned} \right.$$
In this question we assume that all eigenvalues of $A$ have strictly negative real part. Show then that there exist two strictly positive constants $C _ { 2 }$ and $\alpha$ such that: $$\forall t \in \mathbf { R } _ { + } , \quad \left\| e ^ { t u } \right\| _ { r } \leqslant C _ { 2 } e ^ { - \alpha t }$$ and deduce a bound on $\left\| g _ { x _ { 0 } } ( t ) \right\|$ for $t \in \mathbf { R } _ { + }$.

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.
Show that the function $$b : \left\lvert \, \begin{array} { r l l } \mathbf { R } ^ { n } \times \mathbf { R } ^ { n } & \rightarrow & \mathbf { R } \\ ( x , y ) & \mapsto & \int _ { 0 } ^ { + \infty } \left\langle e ^ { t a } ( x ) \mid e ^ { t a } ( y ) \right\rangle d t \end{array} \right.$$ is well-defined and that it defines an inner product on $\mathbf { R } ^ { n }$.

We consider the Euclidean space $E = \mathscr{M}_{N,1}(\mathbf{R})$ equipped with the inner product $\langle X, Y \rangle = \sum_{i=1}^{N} X[i] Y[i] \pi[i]$, and the matrix $H_t$ defined by $\forall (i,j) \in \llbracket 1;N \rrbracket^2, H_t[i,j] = e^{-t} \sum_{n=0}^{+\infty} \frac{t^n K^n[i,j]}{n!}$. Let $X \in E$. We denote by $\psi_X$ the function defined from $\mathbf{R}$ to $E$ by $\psi_X : t \mapsto H_t X$ and $\varphi_X$ the function defined from $\mathbf{R}$ to $\mathbf{R}$ by $\varphi_X : t \mapsto \|H_t X\|^2$. Justify that $\psi_X$ is differentiable and that for all $t$ in $\mathbf{R}$, $$\psi_X'(t) = -(I_N - K) H_t X$$

Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation $$y'' + qy = 0 \tag{1}$$ Justify the existence of two solutions $y_1$ and $y_2$ in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$ to (1) such that: $$\left\{ \begin{array}{l} y_1(0) = 1 \\ y_1'(0) = 0 \end{array} \quad \text{and} \quad \left\{ \begin{array}{l} y_2(0) = 0 \\ y_2'(0) = 1 \end{array} \right. \right.$$ Justify that $\operatorname{Vect}(y_1, y_2)$ is the set of solutions of (1) in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$.

Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation $$y'' + qy = 0 \tag{1}$$ Let $y_1$ and $y_2$ be the solutions in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$ to (1) satisfying $$\left\{ \begin{array}{l} y_1(0) = 1 \\ y_1'(0) = 0 \end{array} \quad \text{and} \quad \left\{ \begin{array}{l} y_2(0) = 0 \\ y_2'(0) = 1 \end{array} \right. \right.$$ Show that: $$\forall t \in \mathbb{R}, \quad y_1(t) y_2'(t) - y_1'(t) y_2(t) = 1$$

Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation $$y'' + qy = 0. \tag{1}$$ Justify the existence of two solutions $y_1$ and $y_2$ in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$ to (1) such that: $$\left\{ \begin{array}{l} y_1(0) = 1 \\ y_1'(0) = 0 \end{array} \text{ and } \left\{ \begin{array}{l} y_2(0) = 0 \\ y_2'(0) = 1. \end{array} \right. \right.$$ Justify that $\operatorname{Vect}(y_1, y_2)$ is the set of solutions of (1) in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$.

Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation $$y'' + qy = 0. \tag{1}$$ Let $y_1$ and $y_2$ be the solutions of (1) with initial conditions $y_1(0)=1, y_1'(0)=0$ and $y_2(0)=0, y_2'(0)=1$. Show that: $$\forall t \in \mathbb{R}, \quad y_1(t) y_2'(t) - y_1'(t) y_2(t) = 1.$$

Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation $$y'' + qy = 0 \tag{1}$$ Let $y_1$ and $y_2$ be the solutions in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$ to (1) satisfying $$\left\{ \begin{array}{l} y_1(0) = 1 \\ y_1'(0) = 0 \end{array} \quad \text{and} \quad \left\{ \begin{array}{l} y_2(0) = 0 \\ y_2'(0) = 1 \end{array} \right. \right.$$ Show that if $y \in \mathscr{C}^2(\mathbb{R}, \mathbb{C})$ is a solution of (1), then the function $t \mapsto y(t+T)$ is also one. Deduce that for all $t \in \mathbb{R}$: $$y(t+T) = y(T) y_1(t) + y'(T) y_2(t)$$

Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation $$y'' + qy = 0. \tag{1}$$ Let $y_1$ and $y_2$ be the solutions of (1) with initial conditions $y_1(0)=1, y_1'(0)=0$ and $y_2(0)=0, y_2'(0)=1$. Show that if $y \in \mathscr{C}^2(\mathbb{R}, \mathbb{C})$ is a solution of (1), then the function $t \mapsto y(t+T)$ is also one. Deduce that for all $t \in \mathbb{R}$: $$y(t+T) = y(T) y_1(t) + y'(T) y_2(t).$$

Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation $$y'' + qy = 0 \tag{1}$$ Let $y_1$ and $y_2$ be the solutions in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$ to (1) satisfying $$\left\{ \begin{array}{l} y_1(0) = 1 \\ y_1'(0) = 0 \end{array} \quad \text{and} \quad \left\{ \begin{array}{l} y_2(0) = 0 \\ y_2'(0) = 1 \end{array} \right. \right.$$ Let $\mu \in \mathbb{C}^*$, and let $\lambda \in \mathbb{C}$ such that $\mu = e^{\lambda T}$. Show that the following three assertions are equivalent.
(a) Equation (1) has a non-zero solution $y \in \mathscr{C}^2(\mathbb{R}, \mathbb{C})$ that satisfies: $$\forall t \in \mathbb{R}, \quad y(t+T) = \mu y(t)$$
(b) The complex number $\mu$ is a solution of the equation with unknown $x$: $$x^2 - \left(y_1(T) + y_2'(T)\right) x + 1 = 0$$
(c) The differential equation (1) has a non-zero solution $y \in \mathscr{C}^2(\mathbb{R}, \mathbb{C})$ such that: $$\forall t \in \mathbb{R}, \quad y(t) = e^{\lambda t} u(t)$$ where $u \in \mathscr{C}^2(\mathbb{R}, \mathbb{C})$ is a $T$-periodic function.

Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation $$y'' + qy = 0. \tag{1}$$ Let $y_1$ and $y_2$ be the solutions of (1) with initial conditions $y_1(0)=1, y_1'(0)=0$ and $y_2(0)=0, y_2'(0)=1$. Let $\mu \in \mathbb{C}^*$, and let $\lambda \in \mathbb{C}$ such that $\mu = e^{\lambda T}$. Show that the following three assertions are equivalent.
(a) Equation (1) has a non-zero solution $y \in \mathscr{C}^2(\mathbb{R}, \mathbb{C})$ that satisfies: $$\forall t \in \mathbb{R}, \quad y(t+T) = \mu y(t).$$
(b) The complex number $\mu$ is a solution of the equation with unknown $x$: $$x^2 - \left(y_1(T) + y_2'(T)\right) x + 1 = 0.$$
(c) The differential equation (1) has a non-zero solution $y \in \mathscr{C}^2(\mathbb{R}, \mathbb{C})$ such that: $$\forall t \in \mathbb{R}, \quad y(t) = e^{\lambda t} u(t),$$ where $u \in \mathscr{C}^2(\mathbb{R}, \mathbb{C})$ is a $T$-periodic function.

Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation $$y'' + qy = 0 \tag{1}$$ Let $y_1$ and $y_2$ be the solutions in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$ to (1) satisfying $$\left\{ \begin{array}{l} y_1(0) = 1 \\ y_1'(0) = 0 \end{array} \quad \text{and} \quad \left\{ \begin{array}{l} y_2(0) = 0 \\ y_2'(0) = 1 \end{array} \right. \right.$$ Let $\mu_1, \mu_2$ be the complex roots of the equation with unknown $x$: $$x^2 - \left(y_1(T) + y_2'(T)\right) x + 1 = 0$$ Show that if $\mu_1 \neq \mu_2$ and if $\lambda$ is a complex number such that $\mu_1 = e^{\lambda T}$, then for any solution $y$ of (1), there exist two $T$-periodic functions $w_1$ and $w_2$, as well as two complex numbers $\alpha$ and $\beta$ such that $$\forall t \in \mathbb{R}, \quad y(t) = \alpha e^{\lambda t} w_1(t) + \beta e^{-\lambda t} w_2(t)$$

Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation $$y'' + qy = 0 \tag{1}$$ Let $y_1$ and $y_2$ be the solutions in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$ to (1) satisfying $$\left\{ \begin{array}{l} y_1(0) = 1 \\ y_1'(0) = 0 \end{array} \quad \text{and} \quad \left\{ \begin{array}{l} y_2(0) = 0 \\ y_2'(0) = 1 \end{array} \right. \right.$$ Let $\mu_1, \mu_2$ be the complex roots of the equation with unknown $x$: $$x^2 - \left(y_1(T) + y_2'(T)\right) x + 1 = 0$$ Suppose that $\mu_1 = \mu_2$. Show that $\mu_1 = \mu_2 = \pm 1$ and that equation (1) admits a periodic solution in $\mathscr{C}^2(\mathbb{R}, \mathbb{C})$.

Let $q : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous application, periodic with period $T > 0$. We consider the differential equation $$y'' + qy = 0. \tag{1}$$ Let $y_1$ and $y_2$ be the solutions of (1) with initial conditions $y_1(0)=1, y_1'(0)=0$ and $y_2(0)=0, y_2'(0)=1$. Let $\mu_1, \mu_2$ be the complex roots of the equation with unknown $x$: $$x^2 - \left(y_1(T) + y_2'(T)\right) x + 1 = 0.$$ Show that if $\mu_1 \neq \mu_2$ and if $\lambda$ is a complex number such that $\mu_1 = e^{\lambda T}$, then for any solution $y$ of (1), there exist two $T$-periodic functions $w_1$ and $w_2$, as well as two complex numbers $\alpha$ and $\beta$ such that $$\forall t \in \mathbb{R}, \quad y(t) = \alpha e^{\lambda t} w_1(t) + \beta e^{-\lambda t} w_2(t).$$