System of 3 variable differential equations with 3 repeating eigen values

Question

The system I am trying to solve is of this form

$x' =ax+5y$

$y' =ay+2z$

$z' =az$

so finding eigen values we solve

$ \begin{bmatrix} a-\lambda &5 &0 \\ 0 &a-\lambda &2 \\ 0 &0 &a-\lambda \end{bmatrix} $

and we will get $\lambda_{1,2,3}=a$

How do I continue on from here? Will the general solution be of the form

$Y = c_1K_1e^{at}+tc_2K_2e^{at}+t^2c_1K_3e^{at}$, where $K_1=K_2=K_3$ are the same eigen vectors since same eigen values

Note then $K_1=K_2=K_3 = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ , which isn't what the answer in the book is :(

Answer 1

Here's a guide on how to deal with repeated eigenvalues. The idea is to use generalized eigenvectors such that

$$ \begin{aligned} (\textbf{A}-\lambda\textbf{I})\vec{v}_1 &= \vec{0} \\ (\textbf{A}-\lambda\textbf{I})\vec{v}_2 &= \vec{v}_1 \\ (\textbf{A}-\lambda\textbf{I})\vec{v}_3 &= \vec{v}_2 \end{aligned} \tag{1} $$

EDIT: I made a mistake in my initial answer. The third linearly independent solution isn't a straightforward combination of the powers of $t^ne^{\lambda t}$. However, we can still apply the same technique by generalizing the constants

$$ \vec{r}_3 = \big(\vec{v}_1t^2 + \vec{v}_2 Bt + \vec{v}_3 C\big) e^{\lambda t} $$

Plugging this in, we obtain

\begin{align} \vec{r}_3' &= \big(2\vec{v}_1t + \lambda\vec{v}_1 t^2 + B\vec{v}_2 + \lambda B \vec{v}_2t + \lambda C \vec{v}_3\big)e^{\lambda t} \\ &= \big(\lambda \vec{v_1}\big) t^2 e^{\lambda t} + \big(2 \vec{v}_1 + \lambda B \vec{v}_2 \big)t e^{\lambda t} + \big(B\vec{v}_2 + \lambda C\vec{v}_3 \big) e^{\lambda t} \end{align}

$$ \implies \left\{ \begin{aligned} \textbf{A}\vec{v}_1 &= \lambda \vec{v}_1 \\ B\textbf{A}\vec{v}_2 &= 2 \vec{v}_1 + \lambda B \vec{v}_2 \\ C\textbf{A}\vec{v}_3 &= B\vec{v}_2 + \lambda C\vec{v}_3 \end{aligned} \right. \implies \left\{ \begin{aligned} \big(\textbf{A} - \lambda \textbf{I}\big)\vec{v}_1 &= \vec{0} \\ \big(\textbf{A} - \lambda \textbf{I}\big)\vec{v}_2 &= \frac{2}{B} \vec{v}_1 \\ \big(\textbf{A} - \lambda \textbf{I}\big)\vec{v}_3 &= \frac{B}{C}\vec{v}_2 \end{aligned} \right. $$

To get to the same form as $(1)$, we set $B=C=2$, arriving at the general solution in $(2)$.

$$ \begin{bmatrix}x \\ y \\ z \end{bmatrix} = c_1\vec{v}_1e^{\lambda t} + c_2\big(\vec{v}_1te^{\lambda t} + \vec{v}_2e^{\lambda t}\big) + c_3 \big(\vec{v}_1t^2e^{\lambda t} + \vec{v}_2 2te^{\lambda t} + \vec{v}_3 2 e^{\lambda t}\big) \tag{2} $$

---

Let's go on to find these vectors. Note the product first $$ (\textbf{A}-\lambda\textbf{I})\vec{v} = \begin{bmatrix} 0 & 5 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5y \\ 2z \\ 0 \end{bmatrix} $$

The first eigenvector satisfies

$$ (\textbf{A}-\lambda\textbf{I})\vec{v}_1 = \begin{bmatrix} 5y_1 \\ 2z_1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

which gives $y_1 = z_1 = 0$. There is no restriction on $x_1$, so we can pick any value we want. Note that we need $x_1 \ne 0$ to avoid a trivial solution, so let's say $x_1 = 1$

Then, the first generalized eigenvector must satisfy

$$ (\textbf{A}-\lambda\textbf{I})\vec{v}_2 = \begin{bmatrix} 5y_2 \\ 2z_2 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $$

This time, $y_2=1/5$, $z_2=0$, and again any value for $x_2$ will work. We pick $x_2=0$ since there's no possibility of getting a zero vector.

You can do the last case. The 3 vectors are

$$ \vec{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} 0 \\ 1/5 \\ 0 \end{bmatrix}, \quad \vec{v}_3 = \begin{bmatrix} 0 \\ 0 \\ 1/10 \end{bmatrix} $$

Plugging in $(2)$, we get the full solution $$ \begin{bmatrix}x \\ y \\ z \end{bmatrix} = \begin{bmatrix} c_1 e^{at} + c_2 te^{at} + c_3t^2e^{at} \\ \frac{1}{5}\big(c_2e^{at} + 2c_3te^{at}\big) \\ \frac{1}{5}c_3e^{at} \end{bmatrix} $$

Answer 2

Another way to solve this system is to solve $$ z'=az$$ and plug the result in $$ y' =ay+2z$$

Then solve $ y' =ay+2z$ for $y$ and plug the results in $$x' =ax+5y$$ and solve for $x$

Answer 3

Dylan’s answer takes you through the general method of dealing with eigenvalues for which the geometric multiplicity is less than the algebraic multiplicity, but in this case there’s a much more direct way to find a solution, one that doesn’t require computing any eigenvectors whatsoever.

Note first that since the coefficient matrix $A$ is triangular, its eigenvalues are just its diagonal elements, so they can be read directly from $A$ without going through the usual process of finding the roots of the characteristic polynomial. The only eigenvalue is $a$, so you can decompose $A$ into the sum of the diagonal matrix $aI$ and $N=A-aI$. These two matrices commute, which means that $$e^{tA} = e^{t(aI+N)} = e^{taI}e^{tN}.$$ Now, $N^2\ne0$ and $N^3=0$, so the power series for $e^{tN}$ will have only three terms: $$e^{tN} = I+tN+\frac12 t^2 N^2.$$ On the other hand, $e^{taI}=e^{at}I$, therefore the general solution to the system is $$e^{at}\left(I+tN+\frac12 t^2 N^2\right)\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix} = e^{at} \begin{bmatrix}1 & 5t & 5t^2 \\ 0 & 1 & 2t \\ 0&0&1\end{bmatrix}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix},$$ with the constants determined by the initial conditions, as usual. This likely doesn’t look exactly like the book answer, but keep in mind that the three constants are arbitrary, so you can rename combinations of them as needed.