Conclude that \(\max_n \vert y(t_n) - y_n\vert \to 0\) when \(h \to 0.\)
Exercise 2
Write a code to compute the solution of the Price-Demand system
by the Euler Method,
by the Runge-Kutta Method.
Exercise 3
One denote by \(\{e_1, e_2, e_3 \}\) the canonical basis of \(\mathbb{R}^3.\) For two vectors \(x, y\) of \(\mathbb{R}^3\) the exterior product \(x \wedge y\) is defined as \[x \wedge y =
\begin{vmatrix}
x_1 & y_1 & e_1 \\
x_2 & y_2 & e_2 \\
x_3 & y_3 & e_3
\end{vmatrix}
= \begin{pmatrix}
x_2 y_3 -y_2 x_3 \\
x_3 y_1 - y_3 x_1 \\
x_1 y_2 - y_1 x_2
\end{pmatrix}.\]
If \(x,y\) are differentiable show that \[\frac{d}{dt} x(t) \wedge y(t) = \dot{x} \wedge y + x \wedge \dot{y}.\] What generalisation of this formula can be provided ?
Hint: \(x \wedge y\) is bi-linear.
Exercise 4
Let \(u=(u_1, u_2, 0 )^t\) and \(v=(v_1, v_2, 0)^t\) two vectors of \(\mathbb{R}^3.\)
Triangle constructed with the vectors \(u\) and \(v.\)
Show that the area of the triangle defined by the origin \(O\) and the two vectors \(u\) and \(v\) (see Figure 1) is given by \[A = \frac{1}{2}\vert u_1 v_2 - v_1 u_2 \vert.\]
Hint: one can assume that the coordinates are chosen such that \(u\) lies on the \(x\)-axis of \(\mathbb{R}^2.\)
Consider \(x(t)\) a planar vector of \(\mathbb{R}^3\) depending of \(t.\)
Area swept between \(t\) and \(t+h\) by \(x(t).\)
From the previous point, show that the area swept between \(t\) and \(t+h\) by \(x(t)\) is given by \[A(t+h) - A(t) = \frac{1}{2}\left\vert x_1(t) \left(x_2(t+h) - x_2(t)\right) - \left( x_1(t+h) - x_1(t)\right)x_2(t)\right\vert.\] Derive that one has \[\frac{dA}{dt}(t) = \frac{1}{2} \left\vert x_1(t) x_2'(t)-x_1'(t)x_2(t)\right\vert.\]
Show that the area swept by \(x(t)\) between \(t_0\) and \(t\) is \[\displaystyle A(t) = \frac{1}{2} \int_{t_0}^t r^2(s)\theta'(s) ds\] where \(r\) and \(\theta\) are the polar coordinate of \(x(t)\) and \(s\) is the arc-length parameter.
Exercise 5
For \(a \geq b > 0\) we set \(c^2 = a^2 -b^2.\)
Polar coordinate of an ellipse.
In the plane \((x,y)\) one considers the curve defined by \(\displaystyle r(\theta) = \frac{b^2}{a-c \cos \theta},\) see Figure 5.
Show that \(\displaystyle\frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1,\) i.e. the point \(P\) is on an ellipse.
Conversely, one considers the ellipse \(\displaystyle\frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1.\) One sets \[x = r \cos \theta - c \quad , \quad y = r\sin \theta.\]
Show that if \((x,y)\) is located on the ellipse one has \(\displaystyle r = r(\theta) = \frac{b^2}{a-c \cos \theta}.\)
Show that the curve \[r(\theta) = \frac{h^2}{m} \frac{1}{1 - \varepsilon \cos \theta}, \qquad \text{with} \;\varepsilon<1\] is an ellipse.