We denote by $A$ a real square matrix of size 2 and we set, for all $x$ in $\mathbb{R}^2$, $f(x) = Ax$. For $a$ in $\mathbb{R}^2$, we denote by $u_a(t)$ the solution on $\mathbb{R}$ of the Cauchy problem $$X' = AX, \quad X(0) = a$$ In other words, $u_a$ is the unique function of class $C^1$ from $\mathbb{R}$ to $\mathbb{R}^2$ such that $u_a(0) = a$ and, for every real $t$, $u_a'(t) = A u_a(t)$.
We assume $A$ is diagonal of the form $$A = \operatorname{diag}(\lambda_1, \lambda_2) = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$
Let $a$ and $b$ be two elements of $\mathbb{R}^2$ and let $t$ be a real number. Show that $$\operatorname{det}\left(u_a(t), u_b(t)\right) = \exp\left(t \operatorname{div}_f(a)\right) \operatorname{det}\left(u_a(0), u_b(0)\right)$$

We denote by $A$ a real square matrix of size 2 and we set, for all $x$ in $\mathbb{R}^2$, $f(x) = Ax$. For $a$ in $\mathbb{R}^2$, we denote by $u_a(t)$ the solution on $\mathbb{R}$ of the Cauchy problem $$X' = AX, \quad X(0) = a$$ In other words, $u_a$ is the unique function of class $C^1$ from $\mathbb{R}$ to $\mathbb{R}^2$ such that $u_a(0) = a$ and, for every real $t$, $u_a'(t) = A u_a(t)$.
We assume $A$ is diagonal of the form $$A = \operatorname{diag}(\lambda_1, \lambda_2) = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$
Use the result of II.B.2 to interpret the sign of $\operatorname{div}_f(a)$ in terms of the direction of variation of the area of a certain parallelogram as a function of $t$.

We denote by $A$ a real square matrix of size 2 and we set, for all $x$ in $\mathbb{R}^2$, $f(x) = Ax$. For $a$ in $\mathbb{R}^2$, we denote by $u_a(t)$ the solution on $\mathbb{R}$ of the Cauchy problem $$X' = AX, \quad X(0) = a$$ We still assume that $A = \operatorname{diag}(\lambda_1, \lambda_2)$.
We set $a = (a_1, a_2)$ and $u_a(t) = (x_1(t), x_2(t))$. We assume that $\lambda_1 \neq 0$ and $a_1 > 0$. Determine a function $\theta_a$ such that $x_2(t) = \theta_a(x_1(t))$ for every real $t$.

We denote by $A$ a real square matrix of size 2 and we set, for all $x$ in $\mathbb{R}^2$, $f(x) = Ax$. For $a$ in $\mathbb{R}^2$, we denote by $u_a(t)$ the solution on $\mathbb{R}$ of the Cauchy problem $$X' = AX, \quad X(0) = a$$ We still assume that $A = \operatorname{diag}(\lambda_1, \lambda_2)$.
In this question, $a = (2,1)$ and $b = (1,2)$.
For each of the following cases, illustrate on the same figure the graphs of the functions $\theta_a$, $\theta_b$ and $\theta_{a+b}$, as well as the parallelograms with vertices $(0,0)$, $u_a(t)$, $u_b(t)$ and $u_a(t) + u_b(t)$ for $t = 0$ and a strictly positive value of $t$.
a) $\lambda_1 = 1$ and $\lambda_2 = 2$.
b) $\lambda_1 = 1$ and $\lambda_2 = -2$.
c) $\lambda_1 = 1$ and $\lambda_2 = -1$.

We denote by $A$ a real square matrix of size 2 and we set, for all $x$ in $\mathbb{R}^2$, $f(x) = Ax$. For $a$ in $\mathbb{R}^2$, we denote by $u_a(t)$ the solution on $\mathbb{R}$ of the Cauchy problem $$X' = AX, \quad X(0) = a$$
Revisit questions II.B.1 and II.B.2 in the case where $A$ is triangular of the form $$A = \begin{pmatrix} \lambda & \mu \\ 0 & \lambda \end{pmatrix}$$

We denote by $A$ a real square matrix of size 2 and we set, for all $x$ in $\mathbb{R}^2$, $f(x) = Ax$. For $a$ in $\mathbb{R}^2$, we denote by $u_a(t)$ the solution on $\mathbb{R}$ of the Cauchy problem $$X' = AX, \quad X(0) = a$$
Show that the relation $$\operatorname{det}\left(u_a(t), u_b(t)\right) = \exp\left(t \operatorname{div}_f(a)\right) \operatorname{det}\left(u_a(0), u_b(0)\right)$$ holds when the matrix $A$ has a characteristic polynomial that splits over $\mathbb{R}$.

We denote by $A$ a real square matrix of size 2 and we set, for all $x$ in $\mathbb{R}^2$, $f(x) = Ax$. For $a$ in $\mathbb{R}^2$, we denote by $u_a(t)$ the solution on $\mathbb{R}$ of the Cauchy problem $$X' = AX, \quad X(0) = a$$
Extend the result $$\operatorname{det}\left(u_a(t), u_b(t)\right) = \exp\left(t \operatorname{div}_f(a)\right) \operatorname{det}\left(u_a(0), u_b(0)\right)$$ to the case of an arbitrary real $2 \times 2$ matrix.

In this question we consider the linear differential system $\mathcal{S} : X' = AX$ associated with the matrix $A = \begin{pmatrix} 1 & -4 & 0 \\ 1 & -2 & -1 \\ 1 & 1 & 0 \end{pmatrix}$.
We call trajectories of $\mathcal{S}$ the arcs of the space $\mathbb{R}^3$ parametrized by the solutions of $\mathcal{S}$. We want to determine the rectilinear trajectories and the planar trajectories of $\mathcal{S}$.
IV.F.1) Construct an invertible matrix $P$ and a matrix $T = \begin{pmatrix} \alpha & \beta & 0 \\ -\beta & \alpha & 0 \\ 0 & 0 & \gamma \end{pmatrix}$ with $(\alpha, \beta, \gamma)$ in $(\mathbb{R}^*)^3$ such that $P^{-1}AP = T$ and determine a plane $F$ and a line $G$ stable by the endomorphism of $\mathbb{R}^3$ canonically associated with $A$ and supplementary in $\mathbb{R}^3$.
IV.F.2) Determine the unique solution of the Cauchy problem $\mathcal{P}_U : \begin{cases} X' = AX \\ X(0) = U \end{cases}$ when $U$ belongs to $G$.
IV.F.3) For all $\sigma = (a, b)$ in $\mathbb{R}^2$, we consider the Cauchy problem $\mathcal{C}_{\sigma} : \begin{cases} x' = -x + 2y \\ y' = -2x - y \\ x(0) = a,\ y(0) = b \end{cases}$ and $\varphi = (x, y)$ in $\mathcal{C}^1(\mathbb{R}, \mathbb{R}^2)$ the unique solution of $\mathcal{C}_{\sigma}$.
Specify $x'(0)$ and $y'(0)$; show that $x$ and $y$ are solutions of the same linear homogeneous differential equation of second order with constant coefficients and thus deduce $\varphi$ as a function of $a$ and $b$.
IV.F.4) Determine the rectilinear trajectories and the planar trajectories of the differential system $X' = AX$.

Let $\alpha \in \mathbb{R}$. We say that $f$ is a solution of equation $\left(E_{\alpha}\right)$ if $f \in \mathcal{C}^{2}(\mathbb{R}, \mathbb{C})$ and $$\left(E_{\alpha}\right) \quad \forall x \in \mathbb{R}, f^{\prime\prime}(x) + \frac{ix}{2} f^{\prime}(x) + \frac{f(x)}{2}\left(\alpha|f(x)|^{2} + 1\right) = 0$$ In questions 1 to 7 of Part I, we assume that $\alpha > 0$. Moreover, we assume that there exists a solution $f$ of $(E_{\alpha})$ which satisfies $f(0) = 1$ and $f^{\prime}(0) = 0$.
Show that $\forall x \in \mathbb{R}, |f(x)| \leq 1$ and that $$\forall x \in \mathbb{R}, \left|f^{\prime}(x)\right| \leq \frac{1}{2\sqrt{\alpha}}(\alpha + 1)$$

Let $\alpha \in \mathbb{R}$. We say that $f$ is a solution of equation $\left(E_{\alpha}\right)$ if $f \in \mathcal{C}^{2}(\mathbb{R}, \mathbb{C})$ and $$\left(E_{\alpha}\right) \quad \forall x \in \mathbb{R}, f^{\prime\prime}(x) + \frac{ix}{2} f^{\prime}(x) + \frac{f(x)}{2}\left(\alpha|f(x)|^{2} + 1\right) = 0$$ In questions 1 to 7 of Part I, we assume that $\alpha > 0$. Moreover, we assume that there exists a solution $f$ of $(E_{\alpha})$ which satisfies $f(0) = 1$ and $f^{\prime}(0) = 0$.
The purpose of this question is to prove that there exist $(\ell, M_{0}) \in \mathbb{R}_{+}^{2}$ such that $$\forall x > 0, \left||f(x)|^{2} - 1\right| \leq \frac{M_{0}}{x}$$
(a) Show that $$\forall x \in \mathbb{R}, \Im\left(f^{\prime}(x)\overline{f(x)}\right) + \frac{x}{4}|f(x)|^{2} - \frac{1}{4}\int_{0}^{x}|f(t)|^{2}\,dt = 0$$
(b) Show that $$\forall x > 0, \frac{d}{dx}\left(\frac{1}{x}\int_{0}^{x}|f(t)|^{2}\,dt\right) = -\frac{4}{x^{2}}\Im\left(f^{\prime}(x)\overline{f(x)}\right)$$
(c) Deduce that there exists $\ell \in \mathbb{R}_{+}$ such that $$\lim_{x \rightarrow +\infty} \frac{1}{x}\int_{0}^{x}|f(t)|^{2}\,dt = \ell$$
(d) Show that there exists $M \in \mathbb{R}_{+}$ such that $$\forall x > 0, \left|\frac{1}{x}\int_{0}^{x}|f(t)|^{2}\,dt - \ell\right| \leq \frac{M}{x}$$
(e) Conclude.

Let $\alpha \in \mathbb{R}$. We say that $f$ is a solution of equation $\left(E_{\alpha}\right)$ if $f \in \mathcal{C}^{2}(\mathbb{R}, \mathbb{C})$ and $$\left(E_{\alpha}\right) \quad \forall x \in \mathbb{R}, f^{\prime\prime}(x) + \frac{ix}{2} f^{\prime}(x) + \frac{f(x)}{2}\left(\alpha|f(x)|^{2} + 1\right) = 0$$ In questions 1 to 7 of Part I, we assume that $\alpha > 0$. Moreover, we assume that there exists a solution $f$ of $(E_{\alpha})$ which satisfies $f(0) = 1$ and $f^{\prime}(0) = 0$.
(a) Suppose in this question that $\ell = 1$. Show that there exists $M_{1} \in \mathbb{R}_{+}$ such that $$\forall x > 0, \left||f(x)|^{2} - 1\right| \leq \frac{M_{1}}{x^{3/2}}$$
(b) Deduce that $\ell < 1$.

Let $\alpha \in \mathbb{R}$. We say that $f$ is a solution of equation $\left(E_{\alpha}\right)$ if $f \in \mathcal{C}^{2}(\mathbb{R}, \mathbb{C})$ and $$\left(E_{\alpha}\right) \quad \forall x \in \mathbb{R}, f^{\prime\prime}(x) + \frac{ix}{2} f^{\prime}(x) + \frac{f(x)}{2}\left(\alpha|f(x)|^{2} + 1\right) = 0$$ In questions 1 to 7 of Part I, we assume that $\alpha > 0$. Moreover, we assume that there exists a solution $f$ of $(E_{\alpha})$ which satisfies $f(0) = 1$ and $f^{\prime}(0) = 0$.
Show that $|f|$ is not periodic.

Let $\alpha \in \mathbb{R}$. We say that $f$ is a solution of equation $\left(E_{\alpha}\right)$ if $f \in \mathcal{C}^{2}(\mathbb{R}, \mathbb{C})$ and $$\left(E_{\alpha}\right) \quad \forall x \in \mathbb{R}, f^{\prime\prime}(x) + \frac{ix}{2} f^{\prime}(x) + \frac{f(x)}{2}\left(\alpha|f(x)|^{2} + 1\right) = 0$$ In questions 1 to 7 of Part I, we assume that $\alpha > 0$. Moreover, we assume that there exists a solution $f$ of $(E_{\alpha})$ which satisfies $f(0) = 1$ and $f^{\prime}(0) = 0$.
For $(\ell, x) \in \mathbb{R}_{+}^{*} \times \mathbb{R}$, we set: $$f_{\alpha}(x) = f(x)\exp\left(i\frac{x^{2}}{4}\right), \quad \Psi_{\alpha}(t, x) = \frac{1}{\sqrt{t}} f_{\alpha}\left(\frac{x}{\sqrt{t}}\right)$$
(a) Does there exist $t > 0$ such that $\Psi_{\alpha}(t, .)$ is periodic?
(b) Express $f_{\alpha}^{\prime}, f_{\alpha}^{\prime\prime}$ and $|f_{\alpha}|$ in terms of $f, f^{\prime}, f^{\prime\prime}$ and $|f|$.
(c) Justify that for all $(t, x) \in \mathbb{R}_{+}^{*} \times \mathbb{R}$, we have $\Psi_{\alpha}(., x) \in C^{1}(\mathbb{R}^{+*}, \mathbb{C})$ and $\Psi_{\alpha}(t, .) \in C^{2}(\mathbb{R}, \mathbb{C})$, then prove that $\Psi_{\alpha}$ satisfies equation $(F_{\alpha})$: $$\left(F_{\alpha}\right) \quad i\frac{\partial \Psi_{\alpha}}{\partial t}(t, x) + \frac{\partial^{2} \Psi_{\alpha}}{\partial x^{2}}(t, x) + \frac{1}{2}\Psi_{\alpha}(t, x)\left(\alpha\left|\Psi_{\alpha}(t, x)\right|^{2} + \frac{1}{t}\right) = 0$$

For $m \in \mathbb{R}$, we denote by $\mathcal{M}$ the matrix $$\mathcal{M} = \left(\begin{array}{ccc} 0 & -m & 0 \\ m & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$ We assume that $G \in C^{2}(\mathbb{R}, \mathbb{R}^{3})$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime}(x) = \frac{1}{2}\left((I_{3} + \mathcal{M})G(x)\right) \wedge G^{\prime}(x)$$ and that moreover $$\|G^{\prime}(0)\| = 1, \quad \left((I_{3} + \mathcal{M})G(0)\right) \cdot G^{\prime}(0) = 0$$ We set $\forall x \in \mathbb{R}, T(x) = G^{\prime}(x)$.
Show that for all $x \in \mathbb{R}$, $\|T(x)\| = 1$.

For $m \in \mathbb{R}$, we denote by $\mathcal{M}$ the matrix $$\mathcal{M} = \left(\begin{array}{ccc} 0 & -m & 0 \\ m & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$ We assume that $G \in C^{2}(\mathbb{R}, \mathbb{R}^{3})$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime}(x) = \frac{1}{2}\left((I_{3} + \mathcal{M})G(x)\right) \wedge G^{\prime}(x)$$ and that moreover $$\|G^{\prime}(0)\| = 1, \quad \left((I_{3} + \mathcal{M})G(0)\right) \cdot G^{\prime}(0) = 0$$ We set $\forall x \in \mathbb{R}, T(x) = G^{\prime}(x)$.
Show that for all $x \in \mathbb{R}$, $\left(I_{3} + \mathcal{M}\right)G(x) - x G^{\prime}(x) = 2 G^{\prime}(x) \wedge G^{\prime\prime}(x)$.

For $m \in \mathbb{R}$, we denote by $\mathcal{M}$ the matrix $$\mathcal{M} = \left(\begin{array}{ccc} 0 & -m & 0 \\ m & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$ We assume that $G \in C^{2}(\mathbb{R}, \mathbb{R}^{3})$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime}(x) = \frac{1}{2}\left((I_{3} + \mathcal{M})G(x)\right) \wedge G^{\prime}(x)$$ and that moreover $$\|G^{\prime}(0)\| = 1, \quad \left((I_{3} + \mathcal{M})G(0)\right) \cdot G^{\prime}(0) = 0$$ We set $\forall x \in \mathbb{R}, T(x) = G^{\prime}(x)$. $F(t)$ denotes the matrix exponential defined in question 9.
For $(t, x) \in \mathbb{R}_{+}^{*} \times \mathbb{R}$, we define $\tilde{G}(t, x) = \sqrt{t}\, F\!\left(\frac{\ln(t)}{2}\right) G\!\left(\frac{x}{\sqrt{t}}\right)$. Show that for all $(t, x) \in \mathbb{R}_{+}^{*} \times \mathbb{R}$ we have $\tilde{G}(., x) \in C^{1}(\mathbb{R}_{+}^{*}, \mathbb{R}^{3})$ and $\tilde{G}(t, .) \in C^{2}(\mathbb{R}, \mathbb{R}^{3})$, then establish that $$\forall (t, x) \in \mathbb{R}_{+}^{*} \times \mathbb{R}, \frac{\partial \tilde{G}}{\partial t}(x, t) = \frac{\partial \tilde{G}}{\partial x}(x, t) \wedge \frac{\partial^{2} \tilde{G}}{\partial x^{2}}(x, t)$$

We assume that $m = 0$, that is $$\mathcal{M} = \left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$ We assume that $G \in C^{2}(\mathbb{R}, \mathbb{R}^{3})$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime}(x) = \frac{1}{2}\left(G(x)\right) \wedge G^{\prime}(x)$$ and that moreover there exists $\lambda > 0$ such that $$G(0) = (0, 0, 2\lambda), \quad G^{\prime}(0) = (1, 0, 0)$$ We set $\forall x \in \mathbb{R}, T(x) = G^{\prime}(x)$.
Show that for all $x \in \mathbb{R}$, we have $|G_{1}(x)| \leq |x|$, where we denote by $G_{1} \in C^{2}(\mathbb{R}, \mathbb{R})$ the first coordinate of $G = (G_{1}, G_{2}, G_{3})$.

We assume that $m = 0$, that is $$\mathcal{M} = \left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$ We assume that $G \in C^{2}(\mathbb{R}, \mathbb{R}^{3})$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime}(x) = \frac{1}{2}\left(G(x)\right) \wedge G^{\prime}(x)$$ and that moreover there exists $\lambda > 0$ such that $$G(0) = (0, 0, 2\lambda), \quad G^{\prime}(0) = (1, 0, 0)$$
Show that $G \in C^{\infty}(\mathbb{R}, \mathbb{R}^{3})$.

We assume that $m = 0$, that is $$\mathcal{M} = \left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$ We assume that $G \in C^{\infty}(\mathbb{R}, \mathbb{R}^{3})$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime}(x) = \frac{1}{2}\left(G(x)\right) \wedge G^{\prime}(x)$$ and that moreover there exists $\lambda > 0$ such that $$G(0) = (0, 0, 2\lambda), \quad G^{\prime}(0) = (1, 0, 0)$$ We set $\forall x \in \mathbb{R}, T(x) = G^{\prime}(x)$.
Show that for all $x \in \mathbb{R}$, we have $\|T^{\prime}(x)\| = \lambda$.

We assume that $m = 0$, that is $$\mathcal{M} = \left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)$$ We assume that $G \in C^{\infty}(\mathbb{R}, \mathbb{R}^{3})$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime}(x) = \frac{1}{2}\left(G(x)\right) \wedge G^{\prime}(x)$$ and that moreover there exists $\lambda > 0$ such that $$G(0) = (0, 0, 2\lambda), \quad G^{\prime}(0) = (1, 0, 0)$$ We set $\forall x \in \mathbb{R}, T(x) = G^{\prime}(x)$. For $x \in \mathbb{R}$, we introduce the vectors $$n(x) = \frac{T^{\prime}(x)}{\lambda}, \quad b(x) = T(x) \wedge n(x)$$ so that $(T(x), n(x), b(x))$ forms a direct orthonormal basis.
(a) Using question 15, show that for all $x \in \mathbb{R}$, we have $2b^{\prime}(x) = -x\, n(x)$.
(b) Deduce that $n^{\prime}(x) = -\lambda T(x) + \frac{x}{2} b(x)$.
(c) Show that $G$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime\prime}(x) + \left(\lambda^{2} + \frac{x^{2}}{4}\right) G^{\prime}(x) - \frac{x}{4} G(x) = 0$$

We assume that $m = 0$ and there exists $\lambda > 0$ such that $G(0) = (0,0,2\lambda)$, $G^{\prime}(0) = (1,0,0)$, and $G$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime\prime}(x) + \left(\lambda^{2} + \frac{x^{2}}{4}\right) G^{\prime}(x) - \frac{x}{4} G(x) = 0$$
(a) Write the linear differential equation $Y^{\prime\prime\prime} + \left(\lambda^{2} + \frac{x^{2}}{4}\right) Y^{\prime} - \frac{x}{4} Y = 0$, where $Y \in C^{3}(\mathbb{R}, \mathbb{R}^{3})$, in the form of a differential system $X^{\prime} = AX$, where $X \in C^{1}(\mathbb{R}, \mathcal{M}_{n,1}(\mathbb{R}))$ and where $A \in C(\mathbb{R}, \mathcal{M}_{n}(\mathbb{R}))$, with $n \in \mathbb{N}^{*}$. We will specify $n$ and $A$.
(b) Show that the coordinates $G_{1}, G_{2}, G_{3}$ of $G$ satisfy $$\forall x \in \mathbb{R}, G_{1}(-x) = -G_{1}(x),\quad G_{2}(-x) = G_{2}(x),\quad G_{3}(-x) = G_{3}(x)$$
(c) Show that for all $x \in \mathbb{R}$, $\|G(x)\|^{2} = x^{2} + 4\lambda^{2}$.
(d) Establish that if $G_{1}$ does not vanish on $\mathbb{R}^{*}$, then $G$ is an injective application on $\mathbb{R}$.

We assume that $m = 0$ and there exists $\lambda > 0$ such that $G(0) = (0,0,2\lambda)$, $G^{\prime}(0) = (1,0,0)$, and $G$ satisfies $$\forall x \in \mathbb{R}, G^{\prime\prime\prime}(x) + \left(\lambda^{2} + \frac{x^{2}}{4}\right) G^{\prime}(x) - \frac{x}{4} G(x) = 0$$
(a) Show that $$\forall x \in \mathbb{R}, \left|G_{1}^{\prime}(x) - \cos(\lambda x)\right| \leq \frac{|x|^{3}}{6\lambda}$$ Hint: One may assume that for $r \in C(\mathbb{R}, \mathbb{R})$, if $y \in C^{2}(\mathbb{R}, \mathbb{R})$ satisfies $y^{\prime\prime}(x) + \lambda^{2} y(x) = r(x)$ for all $x \in \mathbb{R}$, then $$y(x) = \cos(\lambda x)\, y(0) + \frac{\sin(\lambda x)}{\lambda}\, y^{\prime}(0) + \frac{1}{\lambda}\int_{0}^{x} r(s)\sin(\lambda(x-s))\,ds$$
(b) Deduce that there exists $\lambda_{0} > 0$ such that if $\lambda > \lambda_{0}$ then there exists $x_{0} \neq 0$ such that $G_{1}(x_{0}) = 0$.

Let $\lambda \in \mathbb{C}$ such that $\operatorname{Re}(\lambda) > 0$. Let $u$ be a function with complex values of class $\mathcal{C}^{1}$ on $\mathbb{R}^{+}$.
Suppose that the function $v = u' + \lambda u$ is bounded on $\mathbb{R}^{+}$. Show that $u$ is bounded on $\mathbb{R}^{+}$.
One may consider the differential equation $y' + \lambda y = v$.

Let $T \in \mathcal{M}_{n}(\mathbb{C})$ be an upper triangular matrix with complex entries. Suppose that the diagonal entries of $T$ are complex numbers with strictly positive real part. Let $u_{1}, \ldots, u_{n}$ be functions with complex values, defined and of class $\mathcal{C}^{1}$ on $\mathbb{R}^{+}$ and let, for all $t \in \mathbb{R}^{+}$, $$U(t) = \begin{pmatrix} u_{1}(t) \\ \vdots \\ u_{n}(t) \end{pmatrix}$$
Suppose that, for all $t \in \mathbb{R}^{+}$, $U'(t) + TU(t) = 0$.
Show that the functions $u_{j}$, where $1 \leqslant j \leqslant n$, are bounded on $\mathbb{R}^{+}$.

We denote $Q = A _ { p - 1 } A _ { p - 2 } \cdots A _ { 0 }$. Show that if $| \operatorname { tr } Q | < 2$, then every solution of (II.2) is bounded.