We define the function $\varphi : \mathbb { R } \rightarrow \mathbb { R }$ by $$\begin{cases} \varphi ( x ) = \exp \left( \frac { - x } { \sqrt { 1 - x } } \right) & \text { if } x < 1 \\ \varphi ( x ) = 0 & \text { if } x \geqslant 1 \end{cases}$$
Show that $\varphi$ is continuous on $\mathbb { R }$ and of class $C ^ { \infty }$ on $\mathbb { R } \backslash \{ 1 \}$.

We consider a general balanced urn. For all real $x, u$ and $v$, we set $$H(x, u, v) = \sum_{n=0}^{+\infty} P_{n}(u,v) \frac{x^{n}}{n!}$$ defined on $D_{\rho} = ]-\rho, \rho[ \times ]0,2[^{2}$ for $\rho$ sufficiently small.
Verify that $H(0, u, v) = u^{a_{0}} v^{b_{0}}$ and then that $H$ is a solution on $D_{\rho}$ of the partial differential equation $$\frac{\partial H}{\partial x}(x,u,v) = u^{a+1} v^{b} \frac{\partial H}{\partial u}(x,u,v) + u^{c} v^{d+1} \frac{\partial H}{\partial v}(x,u,v).$$

Construct a function $\rho \in \mathcal{C}^\infty(\mathbb{R})$, constant equal to 1 on $[-1,1]$ and constant equal to 0 on $\mathbb{R} \setminus [-2,2]$.

Let $r$ be the function from $\mathbb{R}$ to $\mathbb{C}$ such that, for all real $x$, $$r(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi} \rho(\xi) \,\mathrm{d}\xi$$ where $\rho \in \mathcal{C}^\infty(\mathbb{R})$ is constant equal to 1 on $[-1,1]$ and constant equal to 0 on $\mathbb{R} \setminus [-2,2]$.
Show that $r$ is differentiable on $\mathbb{R}$ and give an expression for its derivative function (possibly involving an integral).

Let $r$ be the function from $\mathbb{R}$ to $\mathbb{C}$ such that, for all real $x$, $$r(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi} \rho(\xi) \,\mathrm{d}\xi$$ where $\rho \in \mathcal{C}^\infty(\mathbb{R})$ is constant equal to 1 on $[-1,1]$ and constant equal to 0 on $\mathbb{R} \setminus [-2,2]$.
Show that $x \mapsto x^2 r(x)$ is bounded on $\mathbb{R}$ and deduce that $r$ is integrable and bounded on $\mathbb{R}$.

Let $\lambda > 0$ and let $f \in L^1(\mathbb{R}) \cap \mathcal{C}^1(\mathbb{R})$ such that $\hat{f} \in L^1(\mathbb{R})$ and such that $\hat{f}$ is zero outside the segment $[-\lambda, \lambda]$. We denote by $r_\lambda$ the function from $\mathbb{R}$ to $\mathbb{C}$ such that $r_\lambda(x) = r(\lambda x)$ for all real $x$, where $r(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi} \rho(\xi) \,\mathrm{d}\xi$.
We admit that $f * r_\lambda$ is integrable. Show that $f = \lambda f * r_\lambda$.

Construct a function $\rho \in \mathcal{C}^\infty(\mathbb{R})$, constant equal to 1 on $[-1,1]$ and constant equal to 0 on $\mathbb{R} \setminus [-2,2]$.

Let $\rho$ be the function constructed in question 30. Let $r$ be the function from $\mathbb{R}$ to $\mathbb{C}$ such that, for all real $x$, $$r(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi} \rho(\xi)\,\mathrm{d}\xi$$ Show that $r$ is differentiable on $\mathbb{R}$ and give an expression for its derivative function (possibly involving an integral).

Let $\rho$ be the function constructed in question 30. Let $r$ be the function from $\mathbb{R}$ to $\mathbb{C}$ such that, for all real $x$, $$r(x) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi} \rho(\xi)\,\mathrm{d}\xi$$ Show that $x \mapsto x^2 r(x)$ is bounded on $\mathbb{R}$ and deduce that $r$ is integrable and bounded on $\mathbb{R}$.

Let $\lambda > 0$ and let $f \in L^1(\mathbb{R}) \cap \mathcal{C}^1(\mathbb{R})$ such that $\hat{f} \in L^1(\mathbb{R})$ and such that $\hat{f}$ is zero outside the segment $[-\lambda, \lambda]$. We denote by $r_\lambda$ the function from $\mathbb{R}$ to $\mathbb{C}$ such that $r_\lambda(x) = r(\lambda x)$ for all real $x$, where $r(x) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} \mathrm{e}^{\mathrm{i}x\xi}\rho(\xi)\,\mathrm{d}\xi$ and $\rho$ is the function from question 30. We admit that $f * r_\lambda$ is integrable. Show that $f = \lambda f * r_\lambda$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ For any real number $x$, express $\Phi_n^{(n)}(x)$ and $\Phi_n^{(n+1)}(x)$ in terms of $L_n(x)$ and $L_n'(x)$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ Use the equality $\Phi_{n+1}^{(n+1)}(x) = \frac{\mathrm{d}^{n+1} x\Phi_n(x)}{\mathrm{d}x^{n+1}}$, which we will justify, to prove the equality $$L_{n+1}(x) = \left(1 - \frac{x}{n+1}\right) L_n(x) + \frac{x}{n+1} L_n'(x)$$ valid for any real number $x$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ Use the equality $\Phi_{n+1}^{(n+2)}(x) = \frac{\mathrm{d}^{n+1} \Phi_{n+1}^{(1)}}{\mathrm{d}x^{n+1}}(x)$ to prove the equality $$L_{n+1}'(x) = L_n'(x) - L_n(x)$$ valid for any real number $x$.

Let $n \in \mathbb{N}$ be a non-zero natural integer. We define, for any real number $x$, $$\Phi_n(x) = \mathrm{e}^{-x} x^n \quad \text{and} \quad L_n(x) = \frac{\mathrm{e}^x}{n!} \Phi_n^{(n)}(x).$$ The confluent hypergeometric function $M_{a,c}$ is the solution of $$x y''(x) + (c-x) y'(x) - a y(x) = 0$$ satisfying $M_{a,c}(0) = 1$. Show that $L_n$ is a confluent hypergeometric function.

We assume that $I = [a,b]$ with $a < b$, $\forall x \in I, w(x) = 1$ (general weight $w$ in the formula for $e(f)$), and that $f$ is of class $\mathcal{C}^{m+1}$ on $I$, where $m$ is the order of the quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$.
For every natural number $m$, consider the function $\varphi_m : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $$\forall (x,t) \in \mathbb{R}^2, \quad \varphi_m(x,t) = \begin{cases} (x-t)^m & \text{if } x \geqslant t, \\ 0 & \text{if } x < t. \end{cases}$$
Using the Taylor formula with integral remainder, show that $e(f) = e(R_m)$, where $R_m$ is defined by $$\forall x \in [a,b], \quad R_m(x) = \frac{1}{m!} \int_a^b \varphi_m(x,t) f^{(m+1)}(t)\,\mathrm{d}t.$$

We assume that $I = [a,b]$ with $a < b$, and that $f$ is of class $\mathcal{C}^{m+1}$ on $I$, where $m \geqslant 1$ is the order of the quadrature formula $I_n(f) = \sum_{j=0}^n \lambda_j f(x_j)$.
For every natural number $m$, consider the function $\varphi_m : \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $$\forall (x,t) \in \mathbb{R}^2, \quad \varphi_m(x,t) = \begin{cases} (x-t)^m & \text{if } x \geqslant t, \\ 0 & \text{if } x < t. \end{cases}$$
Deduce that, if $m \geqslant 1$, $$e(f) = \frac{1}{m!} \int_a^b K_m(t) f^{(m+1)}(t)\,\mathrm{d}t$$ where the function $K_m : [a,b] \rightarrow \mathbb{R}$ is defined by $$\forall t \in [a,b], \quad K_m(t) = e\left(x \mapsto \varphi_m(x,t)\right) = \int_a^b \varphi_m(x,t) w(x)\,\mathrm{d}x - \sum_{j=0}^n \lambda_j \varphi_m(x_j, t).$$ You may use the following admitted result: for every continuous function $g : [a,b]^2 \rightarrow \mathbb{R}$, we have $$\int_a^b \left(\int_a^b g(x,t)\,\mathrm{d}t\right)\mathrm{d}x = \int_a^b \left(\int_a^b g(x,t)\,\mathrm{d}x\right)\mathrm{d}t.$$

We assume that $I = [0,1]$, $\forall x \in I, w(x) = 1$, and we consider the quadrature formula $$I_1(g) = \frac{g(0) + g(1)}{2},$$ which is of order $m = 1$.
Calculate the associated Peano kernel $t \mapsto K_1(t)$ and show that, for every function $g$ of class $\mathcal{C}^2$ from $[0,1]$ to $\mathbb{R}$, we have the following bound on the associated quadrature error: $$|e(g)| \leqslant \frac{1}{12} \sup_{x \in [0,1]} |g''(x)|.$$

For all $n \in \mathbb{N}^{\star}$, justify that there exists a unique function $f_n \in \mathcal{C}^{2}(]0, +\infty[)$ satisfying $f_n(1) = 0$, $f_n(2) = 0$ and $f_n^{\prime\prime}(x) = (-1)^n 2^{-nx^2}$ for all $x > 0$.

For all $n \in \mathbb{N}^\star$, justify that there exists a unique function $f_n \in \mathcal{C}^2(]0,+\infty[)$ satisfying $f_n(1) = 0$, $f_n(2) = 0$ and $f_n^{\prime\prime}(x) = (-1)^n 2^{-nx^2}$ for all $x > 0$.

To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $$U ( f ) ( x ) = \int _ { 0 } ^ { x } \left( 1 - \mathrm { e } ^ { - t } \right) \frac { f ( t ) } { t } \mathrm {~d} t + \left( \mathrm { e } ^ { x } - 1 \right) \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$ Let $f \in E$. Show that $U ( f )$ is of class $\mathcal { C } ^ { 1 }$ on $\mathbb { R } _ { + } ^ { * }$ and satisfies, for all $x > 0$, $$( U ( f ) ) ^ { \prime } ( x ) = \mathrm { e } ^ { x } \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$

To each function $f \in E$, we associate the endomorphism $U$ of $E$ defined for all $x > 0$ by $$U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$ It has been shown that $U(f)$ satisfies $y'' - y' = -f(x)/x$ on $\mathbb{R}_+^*$. Let $\lambda \in \mathbb { R } ^ { * }$. We assume that $\lambda$ is an eigenvalue of $U$. Let $f$ be an eigenvector associated with it. Show that $f$ is a solution of the differential equation $(E_{1/\lambda}) : x(y'' - y') + \frac{1}{\lambda} y = 0$.

For $p \in \mathbb{N}^*$, let $P$ be a non-zero polynomial solution of $(E_p) : x(y'' - y') + py = 0$. It has been shown that $U(f)$ satisfies $y'' - y' = -f(x)/x$ and that $U$ is self-adjoint ($\langle f | U(g) \rangle = \langle U(f) | g \rangle$). Prove that the function $pU(P) - P$ satisfies on $\mathbb { R } _ { + } ^ { * }$ the differential equation $y ^ { \prime \prime } - y ^ { \prime } = 0$.

We consider $\mathcal{F} : \mathbb{R}^d \rightarrow \mathcal{P}_c(\mathbb{R}^d)$ taking values in the set $\mathcal{P}_c(\mathbb{R}^d)$ of compact subsets of $\mathbb{R}^d$, and the differential inclusion problem: $$\left\{\begin{array}{l} y'(t) \in \mathcal{F}(y(t)) \\ y(0) = y_{\text{init}} \end{array}\right.$$
Show that if for every compact $K \subset \mathbb{R}^d$, there exists $C_K > 0$ such that $\mathcal{F}$ satisfies: $$\forall x, y \in K, \forall v_x \in \mathcal{F}(x), \forall v_y \in \mathcal{F}(y), \quad \langle v_x - v_y, x - y \rangle \leqslant C_K \|x - y\|^2$$ then problem (2) admits at most one maximal solution. (Hint: You may look at $\|X(t) - Y(t)\|^2$ for $X$ and $Y$ two solutions.)

We consider the differential inclusion problem given by $d = 2$ and $\mathcal{F} : \mathbb{R}^2 \rightarrow \mathcal{P}_c(\mathbb{R}^2)$ defined for all $x = (x_1, x_2) \in \mathbb{R}^2$ by: $$\mathcal{F}(x) = \begin{cases} \{v^-\} & \text{if } x_1 < 0 \\ \{v^+\} & \text{if } x_1 > 0 \\ [v_1^+, v_1^-] \times [v_2^+, v_2^-] & \text{if } x_1 = 0 \end{cases}$$ where $v^- = (v_1^-, v_2^-) \in \mathbb{R}^2$ and $v^+ = (v_1^+, v_2^+) \in \mathbb{R}^2$ with $v_1^- \geqslant v_1^+$ and $v_2^- \geqslant v_2^+$.
We set $v^- = (1, 2)$ and $v^+ = (-1, 2)$.
(a) Show that $\mathcal{F}$ satisfies condition (3).
(b) We choose $y_{\text{init}} = (0, 0)$. Find all maximal solutions of problem (2).
(c) We choose $y_{\text{init}} = (1, 0)$. Find all maximal solutions of problem (2).

We consider the differential inclusion problem given by $d = 2$ and $\mathcal{F} : \mathbb{R}^2 \rightarrow \mathcal{P}_c(\mathbb{R}^2)$ defined for all $x = (x_1, x_2) \in \mathbb{R}^2$ by: $$\mathcal{F}(x) = \begin{cases} \{v^-\} & \text{if } x_1 < 0 \\ \{v^+\} & \text{if } x_1 > 0 \\ [v_1^+, v_1^-] \times [v_2^+, v_2^-] & \text{if } x_1 = 0 \end{cases}$$ where $v^- = (v_1^-, v_2^-) \in \mathbb{R}^2$ and $v^+ = (v_1^+, v_2^+) \in \mathbb{R}^2$ with $v_1^- \geqslant v_1^+$ and $v_2^- \geqslant v_2^+$.
We set $v^- = (0, 1)$ and $v^+ = (1, 1)$.
(a) Show that $\mathcal{F}$ does not satisfy condition (3).
(b) We choose $y_{\text{init}} = (1, 0)$. Find all maximal solutions of problem (2).
(c) We choose $y_{\text{init}} = (0, 0)$. Find all maximal solutions of problem (2).