MAT244-2018S > Quiz-7

Q-T5101

(1/1)

**Victor Ivrii**:

Problem

a. Determine all critical points of the given system of equations.

b. Find the corresponding linear system near each critical point.

c. Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?

d. Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the

nonlinear system.

$$

\left\{\begin{aligned}

&\frac{dx}{dt} = (2 + x)( y - x)\\

&\frac{dy}{dt} = (4 - x)( y + x)

\end{aligned}\right.$$

**Darren Zhang**:

(a)The critical points consist of the solution set of the equations. $$(2+x)(y-x) = 0$$ $$(4-x)(y+x) = 0$$, the only critical points are at (0,0), (4,4)and (-2,2).

(b,c) First note that $F(x) = (2+x)(y-x) $ ,$G(x,y) = (4-x)(y+x)$ . The Jacobian matrix of the vector field is $$J = \begin{pmatrix} -2-2x+y & 2+x \\ 4-2y-2x & 4-x \end{pmatrix}$$

At the origin, the coefficient matrix of the linearized system is $$J(0,0) = \begin{pmatrix} -2 & 2 \\ 4 & 4 \end{pmatrix}$$

with eigenvalues $r_1 = 1-\sqrt{17}$, $r_2 = 1+\sqrt{17}$ . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point (-2,2), the coefficient matrix of the linearized system is $$J(-2,2) = \begin{pmatrix} 4 & 0 \\ 6 & 6 \end{pmatrix}$$

with eigenvalues $r_1 = 4$ and $r_2 = 6$ . The eigenvalues are real, unequal and positive, hence the critical point is an unstable node. At the point (4,4), the coefficient matrix of the linearized system is $$J(4,4) = \begin{pmatrix} -6 & 6 \\ -8 & 0 \end{pmatrix}$$

with complex conjugate eigenvalues $r_1 = -3 + \sqrt{39}i$, $r_2 = -3 - \sqrt{39}i$, The critical point is a stable spiral, which is asymptotically stable.

Attached is the part (d)

**Syed Hasnain**:

i think that in the first J ,

Gx should be 4-y-2x instead of 4-2y-2x

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