- #1
zenterix
- 702
- 84
- Homework Statement
- Consider the system of nonlinear differential equations
- Relevant Equations
- ##x'=-x+xy##
##y'=-2y+xy##
The critical points are ##(0,0)## and ##(2,1)##.
The linearization of these equations is
$$\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}-1+y_0&x_0\\y_0&x_0-2\end{bmatrix}\begin{bmatrix}x-x_0\\y-y_0\end{bmatrix}$$
At ##(0,0)## we have
$$\begin{bmatrix}x'\\ y'\end{bmatrix}=\begin{bmatrix}-1&0\\0&-2\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}$$
The eigenvalues are ##-1## and ##-2## and associated eigenvectors are ##\hat{i}## and ##\hat{j}##.
The general solution is
$$\vec{x}=c_1e^{-t}\hat{i}+c_2e^{-2t}\hat{j}$$
and here is a phase portrait
At ##(2,1)## we have
$$\begin{bmatrix}x'\\ y'\end{bmatrix}=\begin{bmatrix}0&2\\1&0\end{bmatrix}\begin{bmatrix}x-2\\ y-1\end{bmatrix}$$
Eigenvalues are ##\pm\sqrt{2}## and associated eigenvectors are ##\langle 1,\pm\sqrt{2}\rangle##.
The general solution is
$$\begin{bmatrix}x-2\\ y-1\end{bmatrix}=c_1e^{\sqrt{2}t}\begin{bmatrix}1\\\sqrt{2}\end{bmatrix}+c_2e^{-\sqrt{2}t}\begin{bmatrix} 1\\ -\sqrt{2}\end{bmatrix}$$
Here is a phase portrait
My question is how to draw the phase portrait of the original system by combining the two phase portraits I drew.
Note that the phase portraits represent linearized behavior of the system about the critical points. As far as I know, though I drew the phase portraits showing how linear solutions behave for all $t$, we are only using these linear solutions to get an idea for the behavior near the critical points.
There is an accompanying example in some notes I am following and they draw the phase portrait as
It is not clear to me how some of these solutions were obtained.
For example, the one in the bottom right.
So far I have the following
I guess my doubt is about what happens, for example, around region A above.
But I realize now that the behavior far away from the critical points is not important, or at least not confidently characterizable by these linear approximations, which are maybe accurate only near the critical points.
The linearization of these equations is
$$\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}-1+y_0&x_0\\y_0&x_0-2\end{bmatrix}\begin{bmatrix}x-x_0\\y-y_0\end{bmatrix}$$
At ##(0,0)## we have
$$\begin{bmatrix}x'\\ y'\end{bmatrix}=\begin{bmatrix}-1&0\\0&-2\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}$$
The eigenvalues are ##-1## and ##-2## and associated eigenvectors are ##\hat{i}## and ##\hat{j}##.
The general solution is
$$\vec{x}=c_1e^{-t}\hat{i}+c_2e^{-2t}\hat{j}$$
and here is a phase portrait
At ##(2,1)## we have
$$\begin{bmatrix}x'\\ y'\end{bmatrix}=\begin{bmatrix}0&2\\1&0\end{bmatrix}\begin{bmatrix}x-2\\ y-1\end{bmatrix}$$
Eigenvalues are ##\pm\sqrt{2}## and associated eigenvectors are ##\langle 1,\pm\sqrt{2}\rangle##.
The general solution is
$$\begin{bmatrix}x-2\\ y-1\end{bmatrix}=c_1e^{\sqrt{2}t}\begin{bmatrix}1\\\sqrt{2}\end{bmatrix}+c_2e^{-\sqrt{2}t}\begin{bmatrix} 1\\ -\sqrt{2}\end{bmatrix}$$
Here is a phase portrait
My question is how to draw the phase portrait of the original system by combining the two phase portraits I drew.
Note that the phase portraits represent linearized behavior of the system about the critical points. As far as I know, though I drew the phase portraits showing how linear solutions behave for all $t$, we are only using these linear solutions to get an idea for the behavior near the critical points.
There is an accompanying example in some notes I am following and they draw the phase portrait as
It is not clear to me how some of these solutions were obtained.
For example, the one in the bottom right.
So far I have the following
I guess my doubt is about what happens, for example, around region A above.
But I realize now that the behavior far away from the critical points is not important, or at least not confidently characterizable by these linear approximations, which are maybe accurate only near the critical points.
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