Changing parameters in a 3x3 Matrix trying to find the general solution.

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Solution 1

If you look at that matrix right, its determinant is obviously $(b-\lambda)(\lambda^2-a^2)$, which should make it a breeze to find eigenvalues and eigenvectors.

Solution 2

$X^{'}= \begin{pmatrix} 0&0&a \\ 0&b&0\\ a&0&0 \end{pmatrix}X$ , $X=\begin{pmatrix} x\\y\\z\end{pmatrix}$$\begin{pmatrix} x'\\y'\\z'\end{pmatrix} =\begin{pmatrix} 0&0&a \\ 0&b&0\\ a&0&0 \end{pmatrix}\begin{pmatrix} x\\y\\z\end{pmatrix}$$then we have $$\begin{cases} x'=az & \text{} \\ y'=by& \text{} \\ z'=ax & \text{} \\ \end{cases}$$ w e have$y=e^{bt+d_1}$$x'=az\to x''=az'\to x''=a^2x$$ easily find x then by substitution find z

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Faust

Updated on March 05, 2020

• Faust over 3 years

Consider the system $$X^{'}= \begin{pmatrix} 0&0&a \\ 0&b&0\\ a&0&0 \end{pmatrix}X$$

depending on the two parameters a and b.

1) find the general solution of this system. \begin{pmatrix} -\lambda&0&a \\ 0&(b-\lambda)&0\\ a&0&-\lambda \end{pmatrix} $-\lambda(b-\lambda) (-\lambda)+ a(b-\lambda) (-\lambda) = -\lambda^{3} +(b+a) \lambda^{2}-ab\lambda$

$(\lambda^{2} -(b+a) \lambda+ab) (-\lambda)$

$(\lambda -[(b+a)/2])^{2} = -ab + [(a+b)^{2}/4]$

$\lambda_{1} = (-ab + [(a+b)/4])^{1/2}$ $\lambda_{2} = -(-ab + [(a+b)/4])^{1/2}$ $\lambda_{3} = 0$

im not sure how to force the eigenvalues out of this and subbing this mess into that equation for each eigenvalue is there an easier way to compute this?

EDIT as pointed out below this is a mess by expanding down the second column of the matrix we have $\lambda_{1} = b$ $\lambda_{2} = -a$ $\lambda_{3} = a$

Where $V_{1}=<0,1,0>$ $V_{2}=<1,0,(-1)>$ $V_{3}=<1,0,(1)>$

Why isnt this the general solution?

$C_{1} e^{bt} V_{1}$ + $C_{2} e^{-at} V_{2}$ + $C_{3} e^{at} V_{3}$

2) sketch the region in the ab plane where this system has different types of phase portraits.

• Faust over 10 years
sorry i don't really understand how u got all those things. or in general what that even says, perhaps its time for me to go to sleep...
• Faust over 10 years
im not sure how you got that but, that would mean $\lambda$ = b,a,-a Edit nvm im a moron
• Gerry Myerson over 10 years
Evaluate the determinant by expansion along the second row.
• Gerry Myerson over 10 years
@julien, $z'=ax$.
• Julien over 10 years
@GerryMyerson Yes, sure, I added a typo to a typo...Thanks.
• Julien over 10 years
Nice observation, Maisam. So doing this, we find necessary conditions on $x,y,z$. And we need to check that they work afterwards. +1.
• Faust over 10 years
i follow the train of logic to get to $x^{''}=a^{2}x$
• M.H over 10 years
@Faust7:use characteristics polynomial to solve it $(D^2-a^2)x=0$
• Faust over 10 years
Sorry i have never seen anything like this before i am not sure why your trying to find x?
• Faust over 10 years
are you talking about the vectors?
• M.H over 10 years
@Faust7:i offer look on this book differential equation and dynamical system an introduction to chaos of Morris sorry i dont have time to explain i must go university
• Faust over 10 years
haha, you have been more then helpful sir thank you very much for all your help!