How do i find $e^{tA}$?

Question

I've been trying to solve the following system of linear differential equations

$$\begin{aligned} \dot x_1(t) &= 10 x_1(t) + 5 x_2(t) − 5 x_3(t)\\ \dot x_2(t) &= −5 x_1(t) − x_2(t) + 6 x_3(t)\\ \dot x_3(t) &= −5 x_1(t) − 6 x_2(t) + 11 x_3(t)\\ \end{aligned}$$

with initial state $x (0) = (1,0,1)$, and I found out that the solution should be $x(t) = e^{tA}x_0$, where

$$A = \begin{pmatrix} 10 & 5 & -5 \\\ -5 & -1 & 6 \\\ -5 & -6 & 11 \end{pmatrix}$$

How do I calculate the matrix exponential $e^{tA}$?

I found that $f(A) = C f(J) C^{-1}$ but that still leaves me with the question of how to calculate $f(J)$?

This answer should be helpful. – Git Gud Apr 08 '18 at 22:17 — Git Gud, Apr 08 '18 at 22:17
the initial condition is missing for equation 2 – user577215664 Apr 08 '18 at 22:20 — user577215664, Apr 08 '18 at 22:20
Thanks, I fixed it. – Alexander Kraynov Apr 08 '18 at 22:26 — Alexander Kraynov, Apr 08 '18 at 22:26

score 5 · Accepted Answer · answered Apr 08 '18 at 22:22

Let $v_1=(-1,1,1)$, let $v_2=(0,1,1)$, and let $v_3=(1,-1,0)$. Then $A.v_1=10v_1$, $A.v_2=5v_2$, and $A.v_3=v_2+5v_3$, Let$$M=\begin{pmatrix}-1&0&1\\1&1&-1\\1&1&0\end{pmatrix};$$the columns of $M$ are $v_1$, $v_2$ and $v_3$. Therefore$$M^{-1}.A.M=\begin{pmatrix}10&0&0\\0&5&1\\0&0&5\end{pmatrix},$$from which it follows that\begin{align}e^{tA}&=M.\exp\left(t\begin{pmatrix}10&0&0\\0&5&1\\0&0&5\end{pmatrix}\right).M^{-1}\\&=M.\begin{pmatrix}e^{10 t} & 0 & 0 \\ 0 & e^{5 t} & te^{5 t} \\ 0 & 0 & e^{5 t}\end{pmatrix}.M^{-1}\\&=\begin{pmatrix}e^{10 t} & e^{5 t} \left(e^{5 t}-1\right) & e^{5 t}-e^{10 t} \\ e^{5 t}-e^{10 t} & -e^{5 t} \left(t+e^{5 t}-2\right) & e^{5 t} \left(t+e^{5 t}-1\right) \\ e^{5 t}-e^{10 t} & -e^{5 t} \left(t+e^{5 t}-1\right) & e^{5 t} \left(t+e^{5 t}\right)\end{pmatrix}.\end{align}

How do i find $e^{tA}$?

1 Answers1