- #1

- 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: