Systems of first order equations with constant coefficient

Question

Find the fundamental matrix for the system $$\dot x_1 = x_1 + x_2$$ $$\dot x_2 = x_2 + x_3$$ $$\dot x_3 = x_3$$

I attempted this question using the steps outlined below,

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

The eigenvalues are determined as, $$det(A-I\lambda) = \begin{vmatrix}1-\lambda & 1 & 0\\\ 0 & 1-\lambda & 1\\\ 0 & 0 & 1-\lambda\end{vmatrix} = 0$$

From the above, I get repeated eigenvalues of 1 (i.e. $\lambda$ = 1).

I am stuck with how to determine the solutions and eigenvectors to this problem since the eigenvalues are repeated. Any help on how to do this will be appreciated.

Answer 1

This is a triangular system. The last ODE is independent from the others. Once you solve $\dot x_3 = x_3$ as $\hat x_3(t)$ you can substitute in the previous ODE $\dot x_2 = x_2 + \hat x_3$ that will depend only on $\dot x_2$ determining then $\hat x_2(t)$ and finally, the first one as $\dot x_1 = x_1 + \hat x_2$

Answer 2

$\hat{D}=d/dt$

$I: (\hat{D}-1)x_3=0$

$II:(\hat{D}-1)x_2=x_3\implies (\hat{D}-1)^2x_2=0$

$III:(\hat{D}-1)x_1=x_2\implies (\hat{D}-1)^3x_1=0$

$I$ tells us $x_3=C_1e^t$.

I forget the theorem, but there's one that goes if differential operator $\hat{O}$ has repeated eigenvalues, the corresponding eigenvectors are a power of the independent variable times the base case. Info and proof here: Repeated roots

Combined with $II$ that tells us $x_2=C_1'e^t+C_2'te^t$

We also apply the principle to $III$ to get $x_1=C_3'e^t+C_4'te^t +C_5't^2e^t$

With this being a system of equations, one can find the relationship between the constants and match against initial values to solve for them.