Compute $e^{tA}$

Question

When I do my homework (stability theory), I must use the knowledge to the matrix. But I don't remember it :(.

Here's my problem:

For the system of equations: $$\begin{cases} & \text{ } \dot{x}(t)=\dfrac{dx}{dt}= 4x-y\\ & \text{ } \dot{y}(t)=\dfrac{dy}{dt}= 3x+y-z\\ & \text{ } \dot{z}(t)=\dfrac{dz}{dt}= x+z \end{cases} \tag{I}$$ Compute $e^{tA}$.

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Here's my solution:

We have (I) iff : $$\dot{v}=Av$$, where $v=\begin{bmatrix} x\\ y\\ z\\ \end{bmatrix}$ and $A=\begin{pmatrix} & 4 &-1&0 \\ & 3 &1&-1 \\ & 1 & 0&1 \end{pmatrix}$

• I compute the eigenvalues ​​of matrix $A$:

Since $$\left | A-\lambda I \right |=0 \implies \lambda_{1,2,3}=2$$

• We suppose that $A=PJP^{-1}$, where $J: diag(J_1(\lambda_1),J_1(\lambda_2),J_1(\lambda_3))$,

whence $e^{tA}=P\cdot e^{tJ}\cdot P^{-1}$.

• I compute $$e^{tJ}(2)=e^{tJ^1}\cdot \exp(2t)=\exp(2t)\cdot \begin{pmatrix} & 1 &t&\dfrac{t^2}{2} \\ & 0 &1&t \\ & 0 &0&1 \end{pmatrix}$$

• But I have stuck when I find matrix $P$.

Anyone can help me!

Any help will be appreciated. Thanks!

Answer 1

Hint:

$P$ is made up of the columns from your eigenvectors. Here, you have two generalized eigenvectors.

We have (recall, these are not unique):

• $v_1 = (1,2,1)$
• $v_2 = (1,1,0)$
• $v_3 = (0,-1,0)$

You just find the inverse for the $P^{-1}$ for the other matrix.

Update

You should arrive at:

$$e^{tA} = \dfrac{e^{2 t}}{2}\begin{bmatrix} t^2+4 t+2 & -t^2-2 t & t^2 \\ 2 (t^2+3 t) & -2 (t^2+t-1) & 2 (t^2-t) \\ t^2+2 t & -t^2 & t^2-2 t+2 \end{bmatrix}$$