Author Topic: Q7-0901  (Read 3089 times)

Victor Ivrii

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Q7-0901
« on: March 30, 2018, 12:24:45 PM »
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} = (1 + x) \sin (y)\\ &\frac{dy}{dt} = 1 - x - \cos (y) \end{aligned}\right.
« Last Edit: March 31, 2018, 08:20:52 AM by Victor Ivrii »

Jared Jubas-Malz

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Re: Q7-0901
« Reply #1 on: March 30, 2018, 01:21:33 PM »
The critical points would be when $x'=0$ and $y'=0$:
\begin{align}0=(1+x)\sin(y)\end{align}
\begin{align}0=1-x-\cos(y)\end{align}
From $(1)$, $\sin(y)$ would be $0$ when $y=n\pi$ where $n=0, 1, 2, 3,...$. Since there is a $\cos(y)$ in $(2)$, the previous equation can be split into two cases: $y=2n\pi$ where $n=0, 1, 2, 3,....$ and $y=n\pi$ where $n=1, 3, 5,....$. Then in case 1 $\cos(y)=1$ and in case 2 $cos(y)=-1$. Plugging both of these into $(2)$ gives the critical points:
\begin{align}(0, 2n\pi) \quad where \quad n=0, 1, 2, 3,...\end{align}
\begin{align}(2, n\pi) \quad where \quad n=1, 3, 5,...\end{align}
The Jacobian matrix would be:
\begin{align}J=\begin{pmatrix}F_{x}&F_{y}\\G_{x}&G_{y}\end{pmatrix}=\begin{pmatrix}\sin(y)&-\cos(y)(1+x)\\-1&\sin(y)\end{pmatrix}\end{align}
At $(0, 2n\pi)$:
\begin{align}J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\end{align}
This gives eigenvalues $r_{1}=i$ and $r_{2}=-i$. Since $r_{1}, r_{2}=\lambda\pm i\mu$ where $\lambda=0$ the nonlinear system would be an indeterminate center or spiral point.
At $(2, n\pi)$:
\begin{align}J=\begin{pmatrix}0&-3\\-1&0\end{pmatrix}\end{align}
The eigenvalues would be $r_{1}=\sqrt{3}$ and $r_{2}=-\sqrt{3}$. Since $r_{2}<0<r_{1}$ the nonlinear system would be an unstable saddle point.
« Last Edit: March 30, 2018, 01:24:02 PM by Jared Jubas-Malz »

Meng Wu

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Re: Q7-0901
« Reply #2 on: March 30, 2018, 02:25:18 PM »
Also, the critical points could be $(0,-2n\pi)$, where $0,1,2,\dots$ and $(2, -n\pi)$ where $n=1,3,5,\dots$.

Victor Ivrii

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Re: Q7-0901
« Reply #3 on: March 31, 2018, 08:20:30 AM »
Actually $n=0,\pm1,\pm2,\ldots$.

Jared, on the plot you need emphasize the saddle point (just making lines very close to it).

From our system we can exclude $dt$:
$$- \bigl(1-x-\cos(y)\bigr)\,dx +(1 + x) \sin (y)\,dy =0.$$
This is not exact.

Bonus problem
Find integrating factor $\mu=\mu(x)$ and integrate to $H(x,y)=C$. This will prove that indeed $(0,2\pi n)$ are centers, not the focal points
« Last Edit: March 31, 2018, 08:26:34 AM by Victor Ivrii »

Jared Jubas-Malz

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Re: Q7-0901
« Reply #4 on: March 31, 2018, 10:52:58 PM »
I attached another phase portrait at the end of this post with the saddle point emphasized.
For the bonus, this is what I did. I'm not entirely sure if it's correct.
Begin by setting $M=x+\cos(y)-1$ and $N=(1+x)\sin(y)$. Since it's not exact, we look for an integrating factor $\mu=\mu(x)$:
\begin{align}\frac{d\mu}{dx}=\frac{M_{y}-N_{x}}{N}\mu\end{align}
\begin{align}\frac{d\mu}{dx}=\frac{M_{y}-N_{x}}{N}=\frac{-\sin(y)-\sin(y)}{(1+x)\sin(y)}=\frac{-2}{1+x}\end{align}
Therefore $(8 )$ becomes:
\begin{align}\frac{d\mu}{dx}=\frac{-2}{1+x}\mu\end{align}
Separating and integrating gives:
\begin{align}\mu=\frac{1}{(1+x)^2}\end{align}
Plugging the integrating factor back into the original equation:
\begin{align}\frac{x+\cos(y)-1}{(1+x)^2}dx+\frac{\sin(y)}{1+x}dy\end{align}
which is exact. Then,
\begin{align}H_{x}=\frac{x+\cos(y)-1}{(1+x)^2}\end{align}
\begin{align}H_{y}=\frac{\sin(y)}{1+x}\end{align}
Integrating $(13)$ with respect to x:
\begin{align}H(x,y)=\ln(x+1)-\frac{\cos(y)-2}{x+1}+h(y)\end{align}
Plugging into $(14)$:
\begin{align}H_{y}=\frac{\partial}{\partial(y)}(\ln(x+1)-\frac{\cos(y)-2}{x+1}+h(y))=\frac{\sin(y)(x+1)}{(x+1)^2}+h'(y)=\frac{\sin(y)}{1+x}\end{align}
so $h'(y)=0\implies h(y)=c$.
Therefore this implies that
\begin{align}H(x,y)=\ln(x+1)-\frac{\cos(y)-2}{x+1}=C\end{align}
« Last Edit: March 31, 2018, 10:56:18 PM by Jared Jubas-Malz »