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Messages - Junjie Zhang

Pages: [1] 2
1
Quiz-7 / Re: Q7-T0801
« on: March 30, 2018, 01:26:18 PM »
(a)The critical points are given by the solution set of the equations $$-2x-x-y-x(x^2+y^2) = 0$$ $$x-y+y(x^2+y^2) = 0$$ It is clear that the origin is a critical point. Solving the first equation for C , we find that $$y = \frac{-1 \pm \sqrt{1-8x^2-4x^4}}{2x}$$
Substitution of these relations into the second equation results in two equations of the form $f_{1}(x) = 0$ and $f_{2}(x) = 0$ . Plotting these functions, we note that only$f_{1}(x) = 0$ has real roots. ('It follows that the additional critical points are at (-0.33076,1.0924) and (0.33076,-1.0924)
(b,c) Given that $$F(x,y) = -2x-x-y-x(x^2+y^2)$$ $$G(x,y) = x-y+y(x^2+y^2)$$
the Jacobian matrix of the vector field is $$\begin{pmatrix} -2-3x^2-y^2 & -1-2xy \\ 1+2xy & -1+x^2+3y^2 \end{pmatrix}$$
with complex conjugate eigenvalues $r_{1,2} =((-3 \pm i \sqrt{3})/2) $ . Hence the critical point is a stable spiral, which is asymptotically stable. At the point (-0.33076,1.0924) , the coefficient matrix of the linearized system is $$J(-0.33076,1.0924) = \begin{pmatrix} -3.5216 & -0.27735 \\ 0.27735 & 2.6895 \end{pmatrix} $$
With eigenvalues $r_1 = -3.5092$ and $r_2 = 2.6771$ The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable. Identical results hold for the point at (0.33076,-1.0924).

Attached is the part(d)

2
Quiz-7 / Re: Q-T5101
« on: March 30, 2018, 01:24:57 PM »
(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)

3
Quiz-7 / Re: Q7-T0501
« on: March 30, 2018, 01:13:14 PM »
(a) The critical points are solutions of the equations,$1-y = 0$, $(x-y)(x+y) = 0$
The first equation requires that y= 1 . Based on the second equation, x =1/-1. Hence the critical points are (-1,1) and (1,1).
(b,c) $F(x,y) = 1-y$ and $G(x,y) = x^2-y^2 $. The Jacobian matrix of the vector field is $$J = \begin{pmatrix} 0 & -1 \\ 2x & 2y \end{pmatrix}$$
At the critical point $(-1,1)$, the coefficient matrix of the linearized system is $$J(-1,1) = \begin{pmatrix} 0 & -1 \\ -2 & -2 \end{pmatrix}$$
with eigenvalues $r_1 = -1-\sqrt{3}$ and $r_2 = -1+\sqrt{3} $ . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point (1,1), the coefficient matrix of the linearized system is  $$J = \begin{pmatrix} 0 & -1 \\ 2 & -2 \end{pmatrix}$$ with complex conjugate eigenvalues $r_1 = -1+i$, $r_2 = -1-i$. The critical point is stable spiral, which is asymptotically stable.
Attached is the part(d)

4
Quiz-7 / Re: Q7-T0401
« on: March 30, 2018, 01:01:22 PM »
(a)The critical points are given by the solution set of the equations.$$1-xy = 0$$ $$x-y^3=0$$
After multiplying the second equation by y , it follows that $y=1/-1$ . Hence the critical points of the system are at (-1,1) and (-1,-1).

(b,c) Note that $F(x,y) = 1-xy$ and $G(x,y) = x-y^3$ . The Jacobian matrix of the vector field is $$ J = \begin{pmatrix} -y & -x \\ 1 & -3y^2 \end{pmatrix} $$
At the critical point (1,1), the coefficient matrix of the linearized system is$$ J(1,1) = \begin{pmatrix} -1 & -1 \\ 1 & -3 \end{pmatrix}$$

with eigenvalues  $r_1 = r_2 = -2$ . The eigenvalues are real and equal. It is easy to show that there is only one linearly independent eigenvector. Hence the critical point is a stable improper node. 

 At the point (-1,-1), the coefficient matrix of the linearized system is $$ J(-1,-1) = \begin{pmatrix} 1 & 1 \\ 1 & -3 \end{pmatrix}$$

with eigenvalues $r_1 = -1+\sqrt{5}$, $r_2 = -1- \sqrt{5}$. The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable.

Attached is the part(d).

5
Quiz-7 / Re: Q7-T0301
« on: March 30, 2018, 12:45:51 PM »
(a)The critical points are solns of these questions.
$$y+x(1-x^2-y^2) = 0$$ $$-x+y(1-x^2-y^2)=0$$
Multiply the first equation by y and the second equation by x . The difference of the two equations gives $x^2+y^2=0$. Hence the only critical point is at the origin.
(b,c) With $F(x,y) = y+x(1-x^2-y^2)$ and $G(x,y) = -x+y(1-x^2-y^2)$, the Jacobian matrix of the vector field is
$$ J = \begin{pmatrix} 1-3x^2-y^2 & 1-2xy \\ -1-2xy & 1-x^2-3y^2 \end{pmatrix} $$
At the origin, the coefficient matrix of the linearized system is
$$ J(0,0) = \begin{pmatrix} 1 & 1  \\ -1 & 1\end{pmatrix} $$
with complex conjugate eigenvalues $r_{1} = 1+i$, $r_{2} = 1-i$,  . Hence the origin is an unstable spiral.

Attached is the part (d)

6
Web Bonus Problems / Re: Phaseportrait
« on: March 19, 2018, 11:48:27 AM »
Attached is the phase portrait.
I use Matlab
The type of this is centre, as is shown in handout 9, 3b http://www.math.toronto.edu/courses/mat244h1/20181/LN/Ch7-LN9.pdf.
It's purely imaginary numbers, and it's stable.

7
Quiz-6 / Re: Q6--T0801
« on: March 17, 2018, 12:32:00 PM »
Setting $x = \xi t^r$ results in the algebraic equations
$$\begin{pmatrix}1-r & -1\\\ 5 & -3-r\end{pmatrix} \begin{pmatrix}\xi_1 \\ \xi_2 \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}$$
The characteristic equation is $r^2+2r+2=0$, with root $r = -1+i, r = -1-i$. Substituting reduces $r = -1-i$ the system of equations to $(2+i)\xi_{1}-\xi_{2}=0$,The eigenvectors are $\xi^{(1)} = (1,2+i)^T$, $\xi^{(2)} = (1,2-i)^T$.Hence one of the complex-valued solutions is given by $$x^{(1)} = \begin{pmatrix}1 \\\ 2+i \end{pmatrix}e^{-(1+i)t} =\begin{pmatrix}1 \\\ 2+i \end{pmatrix}e^{-t}(\cos t - i \sin t) $$
Therefore the general solution is $$x = c_1 e^{-t}\begin{pmatrix}\cos t \\\ 2\cos t + \sin t \end{pmatrix} + c_2 e^{-t}\begin{pmatrix}\sin t \\\ -\cos t + 2\sin t \end{pmatrix}$$.
Attached is the graph.

8
Quiz-5 / Re: Q5--T0501, T5101
« on: March 09, 2018, 06:02:12 PM »
Also, we can Sketch the graph.
(c) Attached is the graph
Edit: Glad to see Junya add the graph afterwards.

9
Quiz-5 / Re: Q5--T0401, T0901
« on: March 09, 2018, 05:57:49 PM »
Solving the first equation for B , we have $x_{2} = x_{1}'/2$. Substitution into the second equation results in $$x_{1}''/2 = -2x_{1}$$
The resulting equation is $x_1'+4x_1 = 0$, with general soln. $$x_{1}(t) = c_{1}cos2t+c_{2}sin2t$$
With $x_2$ given in terms of x_{1}, it follows that $$x_2(t) = -c_{1}sin2t+c_{2}cos2t$$.
Imposing the specific conditions, we have $c_1=3,c_2 = 4$.
Therefore, $x_1(t) = 3cos2t+4sin2t$ and $x_2(t)=-3sin2t+4cos2t$
Attached is the graph.

10
Quiz-5 / Re: Q5--T0101, T0801
« on: March 09, 2018, 05:53:39 PM »
Solving the first equation for B , we obtain $x_2 = x_1'/2+x_1/4$. Substitution into the second equation results in $$x_1''/2+x_1'/4 = -2x_1-(x_1'/2+x_1'/4)/2$$.
Rearranging the terms, the single differential equation for $x_1$ is $$x_1''+x_1'+\frac{17}{4}x_1=0$$
The general soln is $$x_1(t) = e^{-t/2}(c_1cos2t+c_2sin2t)$$. With $x_2$ given in terms of $x_1$, it has
$$x_2(t) = e^{-t/2}(-c_1cos2t+c_2sin2t)$$.
Imposing the specified initial conditions, we obtain $c_1 = -2, c_2 = 2$. Hence,
$$x_1(t) = e^{-t/2}(-2cos2t+2sin2t)$$.
$$x_2(t) = e^{-t/2}(2cos2t+2sin2t)$$.
Attached is the graph.

11
Quiz-3 / Re: Q3-T0801
« on: February 10, 2018, 06:24:35 PM »
The characteristic equation is $$2r^2+r-4r=0$$ with roots $$r = -\frac{1}{4}+\frac{\sqrt{33}}{4}, -\frac{1}{4}-\frac{\sqrt{33}}{4}$$
Therefore, the general equation is $$y = c_{1}exp(-1-\sqrt{33})t/4+c_{2}exp(-1+\sqrt{33})t/4$$
$$y = (-\frac{1}{4}+\frac{\sqrt{33}}{4})c_{1}exp(-1-\sqrt{33})t/4+ (-\frac{1}{4}-\frac{\sqrt{33}}{4})c_{2}exp(-1+\sqrt{33})t/4$$
In order to satisfy the initial conditions, we require that $$c_{1}+c_{2}=0$$
$$(-\frac{1}{4}+\frac{\sqrt{33}}{4})c_{1}+(-\frac{1}{4}-\frac{\sqrt{33}}{4})c_{2}=1$$.
Therefore, we can get $$c_{1}=\frac{-2}{\sqrt{33}},  c_{2}=\frac{2}{\sqrt{33}}$$.
Therefore, the final solution is $$y = \frac{-2}{\sqrt{33}}exp(-1-\sqrt{33})t/4+\frac{2}{\sqrt{33}}exp(-1+\sqrt{33})t/4$$,
y->\infty as t -> \infty

12
Quiz-3 / Re: Q3-T0401
« on: February 10, 2018, 06:08:19 PM »
Substitution of the assumed solution $$y = e^{rt}$$ results in the characteristic equation $r^2+4r+3=0$, The roots of the equation are  r = -1/-3. Hence the general solution is $$y = c_{1}e^{-t}+c_{2}e^{-3t}$$. Its derivative is $$y' = -c_{1}e^{-t}+-3*c_{2}e^{-3t}$$
Based on the first equation y(0)=1, we can know that $$c_{1}+c_{2} = 2$$.
Based on the second equation y'(0)=-1, we can know that $$-c_{1}-3c_{2}=-1$$, therefore, $$c_{1}= \frac{5}{2}, c_{2} = -\frac{1}{2}$$
Therefore we can get the equation that  $$y = \frac{5}{2}e^{-t}-\frac{1}{2}e^{-3t}$$.
y ->0 when t -> \infty

13
Quiz-3 / Re: Q3-T5101
« on: February 10, 2018, 06:04:13 PM »
The characteristic equation is $r^2-2r-1=0$, with root of $r = 1 + \sqrt{3}, 1- \sqrt{3}$.
Hence the general solution is  $$y = c_{1}exp(1-\sqrt{3})t+c_{2}exp(1+\sqrt{3})t$$

14
Quiz-3 / Re: Q3-T0501
« on: February 10, 2018, 05:57:47 PM »
Substitution of the assumed solution $y=e^{rt}$ results in the characteristic equation $$2r^2-3r+1=0$$
 The roots of the equation are $r = \frac{1}{2}, 1$. Hence the general solution is $y = c_{1}e^{\frac{t}{2}}+c_{2}e^{\frac{3t}{2}}$

15
Quiz-3 / Re: Q3-T0701
« on: February 10, 2018, 05:54:40 PM »
The characteristic equation is $r^2-2r-2=0$, with root of $r = 1+\sqrt{3}, 1-\sqrt{3}$.
Hence, the general solution should be $y = c_{1}exp(1-\sqrt{3})t+c_{2}(1+\sqrt{3})t$

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