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 $G \in C^{\infty}(\mathbb{R}, \mathbb{R}^{3})$ 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 $$\forall x \in \mathbb{R}, G^{\prime\prime\prime}(x) + \left(\lambda^{2} + \frac{x^{2}}{4}\right) G^{\prime}(x) - \frac{x}{4} G(x) = 0$$

A ball is thrown upward with an initial velocity $V _ { 0 }$ from the surface of the earth. The motion of the ball is affected by a drag force equal to $\mathrm { m } \gamma v ^ { 2 }$ (where m is mass of the ball, $v$ is its instantaneous velocity and $\gamma$ is a constant). Time taken by the ball to rise to its zenith is:
(1) $\frac { 1 } { \sqrt { \gamma g } } \ln \left( 1 + \sqrt { \frac { \gamma } { g } } \mathrm {~V} _ { 0 } \right)$
(2) $\frac { 1 } { \sqrt { \gamma g } } \tan ^ { - 1 } \left( \sqrt { \frac { \gamma } { g } } \mathrm {~V} _ { 0 } \right)$
(3) $\frac { 1 } { \sqrt { \gamma \mathrm {~g} } } \sin ^ { - 1 } \left( \sqrt { \frac { \gamma } { \mathrm {~g} } } \mathrm {~V} _ { 0 } \right)$
(4) $\frac { 1 } { \sqrt { 2 \gamma g } } \tan ^ { - 1 } \left( \sqrt { \frac { 2 \gamma } { g } } \mathrm {~V} _ { 0 } \right)$

A particle of mass $M$ originally at rest is subjected to a force whose direction is constant but magnitude varies with time according to the relation $F = F _ { 0 } \left[ 1 - \left( \frac { t - T } { T } \right) ^ { 2 } \right]$ where $F _ { 0 }$ and $T$ are constants. The force acts only for the time interval $2 T$. The velocity $v$ of the particle after time $2 T$ is:
(1) $\frac { 2 F _ { 0 } T } { M }$
(2) $\frac { F _ { 0 } T } { 2 M }$
(3) $\frac { 4 F _ { 0 } T } { 3 M }$
(4) $\frac { F _ { 0 } T } { 3 M }$