
#1
Apr1312, 06:09 AM

P: 110

So I want to be able to draw the phase portrait for linear systems such as:
x'=x2y y'=3x4y I am completely confused, but this is what I have come up with so far: Step 1: Write out the system in the form of a matrix. Step 2: Find the eigenvalues and eigenvectors for the matrix. Step 3: Using the eigenvectors draw the eigenlines. Step 4: Using the the eigenvalues label the direction of the eigenlines[(+) = away, ()= towards] Step 5: Using the eigenvalues determine the type of the system. Eg: node, star, spiral etc. Step 6: Fill in a few trajectories. My issues appear at steps 5 and 6, I can't figure out how to draw the trajectories. Which eigenline should they be based around ? Also with complex eigenvalues or when there is 1 eigenvector what do I have to do ? Any help or link to a webpage would be greatly appreciated. 



#2
Apr1312, 07:58 AM

P: 367

since you have a linear system, as with all linear systems, you want to put it in a "cannonical form", and draw your phase portrait as a linear shift from the cannonical form phase portrait ( with the change of basis operator as your shift )
the cannonical forms are exactly the situations where you have 2 different eigenvalues, 1 eigenvector... et c if you haven't learnt about these, I will elaborate 



#3
Apr1312, 08:08 AM

P: 110





#4
Apr1312, 09:22 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,900

Procedure to draw a phase portrait by hand ?[tex]\begin{bmatrix}x' \\ y'\end{bmatrix}= \begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}[/tex] [quote]Step 2: Find the eigenvalues and eigenvectors for the matrix.[quote] The eigenvalue equation for this matrix is [tex]\left\begin{array}{cc}1\lambda & 2 \\ 3 & 4\lambda\end{array}\right= \lambda^2+ 3\lambda+ 2= (\lambda+ 1)(\lambda+ 2)= 0[/tex] so the eigenvalues are 1 and 2. If <x y> is an eigenvector corresponding to eigenvalue 1, then [tex]\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x 2y \\ 3x 4y\end{bmatrix}= \begin{bmatrix}x \\ y\end{bmatrix}[/tex] which gives the two equations x 2y= x and 3x 4y= y both of which reduce to y=x. An eigenvector corresponding to eigenvalue 1 is <1, 1>. If <x y> is an eigenvector corresponding to eigenvalue 2, then [tex]\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x 2y \\ 3x 4y\end{bmatrix}= \begin{bmatrix}2x \\ 2y\end{bmatrix}[/tex] which gives the two equations x 2y= 2x and 3x 4y= 2y both of which reduce to 2y=3x. An eigenvector corresponding to eigenvalue 2 is <2, 3>. The "interesting" case is when one eigenvalue is positive and the other negative. In that case, trajectories close to the line with negative eigenvalue must be toward the origin and trajectories close to the line with positive eigenvalue must be away form the origin. Also, they cannot pass through the origin yet must be continuous lines the only way that can happen is if the curve close to the origin so that a line toward the origin turns into a line away from the origin. Again, trajectories cannot cross except at the origin. 



#5
Feb2413, 05:57 AM

P: 77

@HallsofIvy: Thank you so much! That was such an elaborate reply! helped me a lot!



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