- #1
Benny
- 584
- 0
Q. Suppose two populations are governed by the equations
[tex]
\mathop x\limits^ \bullet = x - \frac{1}{4}x^2 - \frac{1}{4}y^2
[/tex]
[tex]
\mathop y\limits^ \bullet = xy - y - \frac{1}{2}y^2
[/tex]
(i) Show that the relevant non-zero critical points are (4,0) and (2,2).
(ii) Find the equation of the particular trajectory that starts at (x,y) = (3.9,0.1)
(iii) Find and sketch the trajectories in the neighbourhood of (2,2).
The first one is just solving the simultaneous equations [tex]\mathop x\limits^ \bullet = \mathop y\limits^ \bullet = 0[/tex].
I can't find a way to do (ii) properly though because the answer I obtain doesn't appear to allow me to find a relationship between x and y which is independent of t (ie. the required trajectory).
My attempt at (ii) is as follows. I linearize the system near the critical point (x,y) = (4,0) by setting x = 4 + X and y = Y. Plugging this back into the original system (top of this message) I obtain
[tex]
\begin{array}{l}
\mathop X\limits^ \bullet = - X - Y \\
\mathop Y\limits^ \bullet = 3Y \\
\end{array}
[/tex]
So the linear system associated with the original system in a neighbourhood of the critical point (x,y) = (4,0) is
[tex]
\left[ {\begin{array}{*{20}c}
{\mathop X\limits^ \bullet } \\
{\mathop Y\limits^ \bullet } \\
\end{array}} \right] = \left[ {\begin{array}{*{20}c}
{ - 1} & { - 1} \\
0 & 3 \\
\end{array}} \right]\left[ {\begin{array}{*{20}c}
X \\
Y \\
\end{array}} \right]
[/tex]
Let C denote the coefficient matrix of the linear system. I found the eigenvalues of C to be [tex]\lambda _1 = - 1[/tex] and [tex]\lambda _2 = 3[/tex] with corresponding eigenvectors
[tex]
\mathop {v_1 }\limits^ \to = \left[ {\begin{array}{*{20}c}
1 \\
0 \\
\end{array}} \right]
[/tex]
and
[tex]
\mathop {v_2 }\limits^ \to = \left[ {\begin{array}{*{20}c}
1 \\
{ - 4} \\
\end{array}} \right]
[/tex]
respectively.
So the general solution is:
[tex]
\left[ {\begin{array}{*{20}c}
X \\
Y \\
\end{array}} \right] = A\left[ {\begin{array}{*{20}c}
1 \\
0 \\
\end{array}} \right]e^{ - t} + B\left[ {\begin{array}{*{20}c}
1 \\
{ - 4} \\
\end{array}} \right]e^{3t}
[/tex]
I need the particular trajectory that starts at (x,y) = (3.9,0.1)...in other words at (X,Y) = (-0.1,0.1) since x = 4 + X and y = Y.
Using (X,Y) = (-0.1,0.1) at t = 0 I get
[tex]
\left[ {\begin{array}{*{20}c}
X \\
Y \\
\end{array}} \right] = 0.075\left[ {\begin{array}{*{20}c}
{ - 1} \\
0 \\
\end{array}} \right]e^{ - t} - 0.025\left[ {\begin{array}{*{20}c}
1 \\
{ - 4} \\
\end{array}} \right]e^{3t}
[/tex]
There certainly doesn't look like a relationship between X and Y which is independent of t which means that I can't find the trajectory. So I must have gone wrong somewhere. I've gone over my working many times but I don't know what I'm missing. Can someone please help me out with part (ii)?
[tex]
\mathop x\limits^ \bullet = x - \frac{1}{4}x^2 - \frac{1}{4}y^2
[/tex]
[tex]
\mathop y\limits^ \bullet = xy - y - \frac{1}{2}y^2
[/tex]
(i) Show that the relevant non-zero critical points are (4,0) and (2,2).
(ii) Find the equation of the particular trajectory that starts at (x,y) = (3.9,0.1)
(iii) Find and sketch the trajectories in the neighbourhood of (2,2).
The first one is just solving the simultaneous equations [tex]\mathop x\limits^ \bullet = \mathop y\limits^ \bullet = 0[/tex].
I can't find a way to do (ii) properly though because the answer I obtain doesn't appear to allow me to find a relationship between x and y which is independent of t (ie. the required trajectory).
My attempt at (ii) is as follows. I linearize the system near the critical point (x,y) = (4,0) by setting x = 4 + X and y = Y. Plugging this back into the original system (top of this message) I obtain
[tex]
\begin{array}{l}
\mathop X\limits^ \bullet = - X - Y \\
\mathop Y\limits^ \bullet = 3Y \\
\end{array}
[/tex]
So the linear system associated with the original system in a neighbourhood of the critical point (x,y) = (4,0) is
[tex]
\left[ {\begin{array}{*{20}c}
{\mathop X\limits^ \bullet } \\
{\mathop Y\limits^ \bullet } \\
\end{array}} \right] = \left[ {\begin{array}{*{20}c}
{ - 1} & { - 1} \\
0 & 3 \\
\end{array}} \right]\left[ {\begin{array}{*{20}c}
X \\
Y \\
\end{array}} \right]
[/tex]
Let C denote the coefficient matrix of the linear system. I found the eigenvalues of C to be [tex]\lambda _1 = - 1[/tex] and [tex]\lambda _2 = 3[/tex] with corresponding eigenvectors
[tex]
\mathop {v_1 }\limits^ \to = \left[ {\begin{array}{*{20}c}
1 \\
0 \\
\end{array}} \right]
[/tex]
and
[tex]
\mathop {v_2 }\limits^ \to = \left[ {\begin{array}{*{20}c}
1 \\
{ - 4} \\
\end{array}} \right]
[/tex]
respectively.
So the general solution is:
[tex]
\left[ {\begin{array}{*{20}c}
X \\
Y \\
\end{array}} \right] = A\left[ {\begin{array}{*{20}c}
1 \\
0 \\
\end{array}} \right]e^{ - t} + B\left[ {\begin{array}{*{20}c}
1 \\
{ - 4} \\
\end{array}} \right]e^{3t}
[/tex]
I need the particular trajectory that starts at (x,y) = (3.9,0.1)...in other words at (X,Y) = (-0.1,0.1) since x = 4 + X and y = Y.
Using (X,Y) = (-0.1,0.1) at t = 0 I get
[tex]
\left[ {\begin{array}{*{20}c}
X \\
Y \\
\end{array}} \right] = 0.075\left[ {\begin{array}{*{20}c}
{ - 1} \\
0 \\
\end{array}} \right]e^{ - t} - 0.025\left[ {\begin{array}{*{20}c}
1 \\
{ - 4} \\
\end{array}} \right]e^{3t}
[/tex]
There certainly doesn't look like a relationship between X and Y which is independent of t which means that I can't find the trajectory. So I must have gone wrong somewhere. I've gone over my working many times but I don't know what I'm missing. Can someone please help me out with part (ii)?
Last edited: