A particle moves in two-dimensional space. Its position is given by coordinates \((x, y)\) which satisfy
\[\frac{\mathrm{d}x}{\mathrm{d}t} = -x + 3y + u\]
\[\frac{\mathrm{d}y}{\mathrm{d}t} = x + y + u\]
where \(t\) is the time and \(u\) is a function of time. At time \(t = 0\), the particle has position \((x_0,\, y_0)\).
By considering \(\dfrac{\mathrm{d}x}{\mathrm{d}t} - \dfrac{\mathrm{d}y}{\mathrm{d}t}\), show that if the particle is at the origin \((0,0)\) at some time \(t > 0\), then it is necessary that \(x_0 = y_0\).
Given that \(x_0 = y_0\), find a constant value of \(u\) that ensures that the particle is at the origin at a time \(t = T\), where \(T > 0\).
A particle whose position in three-dimensional space is given by co-ordinates \((x, y, z)\) moves with time \(t\) such that
\[\frac{\mathrm{d}x}{\mathrm{d}t} = 4y - 5z + u\]
\[\frac{\mathrm{d}y}{\mathrm{d}t} = x - 2z + u\]
\[\frac{\mathrm{d}z}{\mathrm{d}t} = x - 2y + u\]
where \(u\) is a function of time. At time \(t = 0\), the particle has position \((x_0,\, y_0,\, z_0)\).
Show that, if the particle is at the origin \((0,0,0)\) at some time \(t > 0\), it is necessary that \(y_0\) is the mean of \(x_0\) and \(z_0\).
Show further that, if the particle is at the origin \((0,0,0)\) at some time \(t > 0\), it is necessary that \(x_0 = y_0 = z_0\).
Given that \(x_0 = y_0 = z_0\), find a constant value of \(u\) that ensures that the particle is at the origin at a time \(t = T\), where \(T > 0\).
Solution:
\(\,\) \begin{align*}
&& \frac{\d x}{\d t} - \frac{\d y}{\d t} &= -2x + 2y \\
&&&= -2(x-y) \\
\Rightarrow && \dot{z} &= -2z \tag{\(z = x-y\)} \\
\Rightarrow && z &= Ae^{-2t} \\
z = 0, t : && A &= 0 \\
\Rightarrow && z &= 0 \quad \forall t \\
\Rightarrow && x &= y \quad \forall t \\
\Rightarrow && x_0 &= y_0
\end{align*}
Since \(x = y\) for all \(t\) our equation can we written as \(\frac{\d x}{\d t} = 2x + u\). This has solution \(x = Ae^{2t} - \frac{u}{2}\) we also have \begin{align*}
t = 0: && x_0 &= A - \frac{u}{2} \\
t = T: && 0 &= Ae^{2T} - \frac{u}{2} \\
\Rightarrow && x_0 &= \frac{u}{2} - e^{-2T}\frac{u}{2} \\
\Rightarrow && u &= \frac{2x_0}{1-e^{-2T}}
\end{align*}
Let \(w = y - \frac12(x+z)\) then \begin{align*}
&& \dot{w} &= \dot{y} - \tfrac12(\dot{x}+\dot{z}) \\
&&&= (x-2z+u) - \tfrac12(4y-5z+u+x-2y+u) \\
&&&= -y + \tfrac12(x+z) \\
&&&= -w \\
\Rightarrow && w &= Ae^{-t} \\
\text{at origin}: && w &= 0 \\
\Rightarrow && y &= \tfrac12(x+z)
\end{align*}
We now have \(\dot{x} = 2x+2z-5z+u = 2x-3z+u\) and \(\dot{z} = x - x- z +u = - z+u\) so in particular \((x - z)' = 2(x-z)\) or \(x-z = Ae^{2t}\) and since we hit the origin \(x = z\) for all \(t\) and so \(y = x = z\) for all \(t\).
Notice we now have \(\dot{x} = -x + u\) or \(x = Ae^{-t} + u\)
\begin{align*}
t = 0: && x_0 &= A + u \\
t = T: && 0 &= Ae^{-T} + u \\
\Rightarrow && x_0 &= u-e^{T} u \\
\Rightarrow && u &= \frac{x_0}{1-e^{T}}
\end{align*}