0
$\begingroup$

Consider the system of differential equations:

$$x'=-x-y+4$$ $$y'=3-xy$$

a. Find the fixed points.

$x'=-x-y+4$

$x+y=4$

$x+3/x=4$

x=3,x=1

$y'=3-xy$

$y=3/x$

fixed points: (1,3), (3,1)

b. Determine the type of the linearized system at each fixed point.

calculating the Jacobian:

\begin{array}{cc} -1 & -1 \\ -y & -x \\ \end{array}

for the fixed point (1,3): \begin{array}{cc} -1 & -1 \\ -3 & -1 \\ \end{array}

calculating eigenvalues:

$λ_1=-1-\sqrt 3$ (could be positive or negative) $λ_2=\sqrt 3-1$ (negative)

So it is unstable (I think, because if we use the negative root of 3, then the first eigenvalue is positive, is this correct?

for the fixed point (3,1): \begin{array}{cc} -1 & -1 \\ -1 & -3 \\ \end{array}

calculating eigenvalues:

$λ_1=-2-\sqrt 2$ (negative) $λ_2=\sqrt 2-2$ (negative)

So it is stable

C. Determine the nullclines and the signs of $x'$ and $y'$ on the nullclines and in the various regions determined by them. (I'm not sure I am calculating the signs of $x'$ and $y'$ correctly)

y-nullcine: $y'=3/x$ x-nullcine: $y=4-x$

R1

$x'<00$ $y'>0$

R2

$x'<0$ $y'>0$

R3

$x'>0$ $y'<0$

R4

$x'<00$ $y'<0$

R5

$x'<0$ $y'<0$

R6

$x'0$ $y'<0$

enter image description here

d. Draw the phase plane portraitenter image description here

$\endgroup$
  • $\begingroup$ So is the general method to look at the phase portrait to determine the positive and negative roots, or the opposite, do I need to determine the roots to draw the phase portrait? $\endgroup$ – Math Major Nov 14 '14 at 16:27
  • $\begingroup$ I get everything about the critical points and nullclines, but I guess the problem is I am still a little confused about how to draw the direction fields or how I can determine the direction lines if I don't have the phase portrait to work backwards from $\endgroup$ – Math Major Nov 14 '14 at 16:33
1
$\begingroup$

Phase portrait:

$\qquad\qquad\qquad\qquad\qquad$enter image description here

Fixed points:

  • At $(3,1)$, Jacobian matrix $\begin{pmatrix}-1&-1\\-1&-3\end{pmatrix}$, trace $-4$ (negative), determinant $+2$ (positive), discriminant $(-4)^2-4\cdot(+2)=8$ (positive), hence two real negative eigenvalues: the point $(3,1)$ is a stable node
  • At $(1,3)$, Jacobian matrix $\begin{pmatrix}-1&-1\\-3&+1\end{pmatrix}$, trace $0$, determinant $-4$ (negative), hence two real eigenvalues of opposite signs: the point $(1,3)$ is a saddle point

...As explained there:

$\qquad$enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.