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The question is as follow's consider the Matrix A

$$A = \begin{pmatrix} \lambda & 0 &0\\ 1 & \lambda&0 \\ 0&1& \lambda \end{pmatrix},$$

Compute $e^{tA}$, and use it to solver the initial value problem

$X^{'}$ = AX,

$X(0)$=$X_{0}$=$(x_{0},y_{0},z_{0})$

Question (ii)

Under what conditions on $\lambda$ and $X_{0}$ does the solution $X(t)$ of the initial value problem converges to the origin as t tends to $\infty$ Justify.

For part 1 i know what it says, it wants me to write $e^{tA}$ where A is actually my matrix i was sick for class and the notes i have are barely legible and as per usual my textbook doesn't have any worked out example's only a proof.

So if someone could show me how to set this up and/or point me in the direction of some dummy level examples with a nice explanation i would appreciate that.

For part 2 i think id like to construct a stable star node but any value where all the eigenvalues $\lambda <0$ should converge to (0,0,0) in infinite time ?

also why does the IVP matter in this case everything should move to (0,0,0) in eithier finite or infinite time if all the eigenvalues are negative?

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To compute the powers of $A$, it may help to write $A=\lambda I_3 +N$ where $N$ is nilpotent and commutes, of course, with $\lambda I_3$. Once you have that, you'll get $\exp (tA)$. –  1015 Mar 4 '13 at 1:03
    
Whats nilpotent mean? –  Faust7 Mar 4 '13 at 1:12
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$M$ is nilpotent $\Leftrightarrow \exists k\in\mathbb{N}, M^k=0_n$ –  xavierm02 Mar 4 '13 at 1:15
1  
It makes it really easy to calculate the exponential because you have a finite number of terms to add since $\sum\limits_{k=0}^{+\infty}\cfrac{M^k}{k!}=\sum\limits_{k=0}^{n-1}\cfrac{M^k}{k‌​!}+\sum\limits_{k=n}^{+\infty}\cfrac{M^k}{k!}=\sum\limits_{k=0}^{n-1}\cfrac{M^k}{‌​k!}+\sum\limits_{k=n}^{+\infty}\cfrac{0_n}{k!}$$=\sum\limits_{k=0}^{n-1}\cfrac{M^‌​k}{k!}$ –  xavierm02 Mar 4 '13 at 1:17
    
ah, ic and $0_{n}$ is the 0 matrix where n is the number of times we multiply it by itself to get the 0 matrix. –  Faust7 Mar 4 '13 at 1:19

1 Answer 1

up vote 3 down vote accepted

$A = \begin{pmatrix} \lambda & 0 &0\\ 1 & \lambda&0 \\ 0&1& \lambda \end{pmatrix} = \lambda\begin{pmatrix} 1 & 0 &0\\ 0 & 1&0 \\ 0&0& 1 \end{pmatrix}+ \begin{pmatrix} 0& 0 &0\\ 1 & 0&0 \\ 0&1& 0 \end{pmatrix} = \lambda I_3 + N$

$e^A = e^{\lambda I_3 + N}$

Since the two matrices commute, ie $(\lambda I_3) N = N(\lambda I_3)$

$e^A = e^{\lambda I_3}e^{N}$

$e^{\lambda I_3}=\sum\limits_{n=0}^{+\infty}\cfrac{\left(\lambda I_3\right)^n}{n!}=\sum\limits_{n=0}^{+\infty}\cfrac{\lambda^n I_3^n}{n!}=\sum\limits_{n=0}^{+\infty}\cfrac{\lambda^n I_3}{n!}=\left(\sum\limits_{n=0}^{+\infty}\cfrac{\lambda^n }{n!}\right)I_3 = e^\lambda I_3$

$N^2=\begin{pmatrix} 0& 0 &0\\ 1 & 0&0 \\ 0&1& 0 \end{pmatrix}\begin{pmatrix} 0& 0 &0\\ 1 & 0&0 \\ 0&1& 0 \end{pmatrix} = \begin{pmatrix} 0& 0 &0\\ 0 & 0&0 \\ 1&0& 0 \end{pmatrix}$

$N^3=N N^2 =\begin{pmatrix} 0& 0 &0\\ 1 & 0&0 \\ 0&1& 0 \end{pmatrix}\begin{pmatrix} 0& 0 &0\\ 0 & 0&0 \\ 1&0& 0 \end{pmatrix} = \begin{pmatrix} 0& 0 &0\\ 0 & 0&0 \\ 0&0& 0 \end{pmatrix} = 0_3$

$e^N = \sum\limits_{n=0}^{+\infty}\cfrac{\left(\lambda N\right)^n}{n!}=\sum\limits_{n=0}^{2}\cfrac{\left(\lambda N\right)^n}{n!}+\sum\limits_{n=3}^{+\infty}\cfrac{\left(\lambda N\right)^n}{n!}=\sum\limits_{n=0}^{2}\cfrac{\left(\lambda N\right)^n}{n!} = \cfrac{I_3}{0!}+\cfrac{N}{1!}+\cfrac{N^2}{2!}=\begin{pmatrix} 1 & 0 &0\\ 0 & 1&0 \\ 0&0& 1 \end{pmatrix}+\begin{pmatrix} 0& 0 &0\\ 1 & 0&0 \\ 0&1& 0 \end{pmatrix}+\cfrac{1}{2}\begin{pmatrix} 0& 0 &0\\ 0 & 0&0 \\ 1&0& 0 \end{pmatrix}=\cfrac{1}{2}\begin{pmatrix} 2 & 0 &0\\ 2 & 2&0 \\ 1&2& 2 \end{pmatrix}$

$e^A = e^{\lambda I_3}e^{N}=\cfrac{e^{\lambda}}{2}\begin{pmatrix} 2 & 0 &0\\ 2 & 2&0 \\ 1&2& 2 \end{pmatrix}$


$e^{tA}=e^{t\lambda I_3 + tN}=e^{t\lambda I_3}e^{tN}=\cfrac{e^{\lambda t}}{2}\begin{pmatrix} 2 & 0 &0\\ 2t & 2&0 \\ t^2&2t& 2 \end{pmatrix}$

$X(0)=X_0=\begin{pmatrix} x_0\\ y_0 \\ z_0 \end{pmatrix}$

$X'=AX\tag{E}$

The unique solution is $X(t)=e^{tA}X_0=\cfrac{e^{\lambda t}}{2}\begin{pmatrix} 2 & 0 &0\\ 2t & 2&0 \\ t^2&2t& 2 \end{pmatrix}\begin{pmatrix} x_0\\ y_0 \\ z_0 \end{pmatrix}=\cfrac{e^{\lambda t}}{2}\begin{pmatrix} 2x_0\\ 2tx_0+2y_0 \\ t^2 x_0 + 2t y_0 + 2z_0 \end{pmatrix}=\begin{pmatrix} e^{\lambda t}x_0\\ e^{\lambda t}\left(tx_0+y_0\right) \\ e^{\lambda t}\left(\cfrac{t^2}{2} x_0 + t y_0 + z_0\right) \end{pmatrix}$


We want $\begin{pmatrix} e^{\lambda t}x_0\\ e^{\lambda t}\left(tx_0+y_0\right) \\ e^{\lambda t}\left(\cfrac{t^2}{2} x_0 + t y_0 + z_0\right) \end{pmatrix}\underset{t\to +\infty}{\longrightarrow} \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$

Since the space is finite dimensional ($3 < +\infty$), convergence is the same thing as convergence coordinate wise.

So $\begin{pmatrix} e^{\lambda t}x_0\\ e^{\lambda t}\left(tx_0+y_0\right) \\ e^{\lambda t}\left(\cfrac{t^2}{2} x_0 + t y_0 + z_0\right) \end{pmatrix}\underset{t\to +\infty}{\longrightarrow} \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}\Leftrightarrow \left\{ \begin{array}{ll} e^{\lambda t}x_0 \underset{t\to +\infty}{\longrightarrow}0 \\ e^{\lambda t}\left(tx_0+y_0\right)\underset{t\to +\infty}{\longrightarrow}0\\ e^{\lambda t}\left(\cfrac{t^2}{2} x_0 + t y_0 + z_0\right)\underset{t\to +\infty}{\longrightarrow}0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{ll} \lambda < 0 \mbox{ or } x_0 = 0\\ \lambda < 0 \mbox{ or } x_0=y_0 = 0\\ \lambda < 0 \mbox{ or } x_0=y_0=z_0 = 0 \end{array} \right.\Leftrightarrow \lambda < 0 \mbox{ or } x_0=y_0=z_0 = 0$

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Wow that's extremely straight forward... –  Faust7 Mar 4 '13 at 1:12
1  
Well usually the tricky part is to decompose the matrix into a diagonal matrix (because you know that $e^{diag(a_1,\dots,a_n)}=diag(e^{a_1},\dots,e^{a_n})$) and a nilpotent matrix (because then you only have a finite number of terms to add). And in your case, the matrix was made to make this decomposition easy to find. –  xavierm02 Mar 4 '13 at 1:21
    
    
That doesn't surprise me the textbook talks about this idea for about, 3 pages. any hints for the IVP n why i can't just pick anything? –  Faust7 Mar 4 '13 at 1:26
    
@Faust7: I edited. –  xavierm02 Mar 4 '13 at 1:55

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