Let $A$ be a single $p\times p$ Jordan block. Find general solution to $\dfrac{dx}{dt} = Ax$

Question

Let $A$ be a single $p\times p$ Jordan block. Find the general solution to $\,\dfrac{dx}{dt} = Ax$.

What should I approach first? Please help!

Answer 1

If $$ A=\left(\begin{matrix} \lambda & 1 & 0 & \cdots & 0&0 \\ 0 & \lambda & 1 & \cdots &0& 0 \\ \vdots &\vdots&\vdots&&\vdots&\vdots \\ 0 & 0 &0 &\cdots &1&0 \\ 0 & 0 &0 &\cdots &\lambda &1 \\ 0 & 0 &0 &\cdots &0 &\lambda \end{matrix} \right)=\lambda I+J\in\mathbb R^{n\times n}. $$ Then $$\mathrm{e}^{tA}=\mathrm{e}^{\lambda t I+tJ} =\mathrm{e}^{\lambda t}\mathrm{e}^{tJ}, $$ as $I$ and $J$ commute and $\mathrm{e}^{aI}=\mathrm{e}^aI$. Also, $$ J=\left(\begin{matrix} 0 & 1 & 0 & \cdots & 0&0 \\ 0 & 0 & 1 & \cdots &0& 0 \\ \vdots &\vdots&\vdots&&\vdots&\vdots \\ 0 & 0 &0 &\cdots &1&0 \\ 0 & 0 &0 &\cdots &0 &1 \\ 0 & 0 &0 &\cdots &0 &0 \end{matrix} \right),\quad J^2=\left(\begin{matrix} 0 & 0 & 1 & \cdots & 0&0 \\ 0 & 0 & 0 & \cdots &0& 0 \\ \vdots &\vdots&\vdots&&\vdots&\vdots \\ 0 & 0 &0 &\cdots &0&1 \\ 0 & 0 &0 &\cdots &0 &0 \\ 0 & 0 &0 &\cdots &0 &0 \end{matrix} \right) $$ and $$ J^{n-1}=\left( \begin{matrix} 0 & 0 & 0 & \cdots & 0&1 \\ 0 & 0 & 0 & \cdots &0& 0 \\ \vdots &\vdots&\vdots&&\vdots&\vdots \\ 0 & 0 &0 &\cdots &0&0 \\ 0 & 0 &0 &\cdots &0 &0 \\ 0 & 0 &0 &\cdots &0 &0 \end{matrix} \right) $$ and $J^n=J^{n+1}=\cdots=0$. Thus $$ \mathrm{e}^{tJ}=I+tJ+\frac{t^2 J}{2}+\cdots+\frac{t^{n-1}J^{n-1}}{(n-1)!}= \left( \begin{matrix} 1 & t & \frac{t^2}{2!}& \cdots & \frac{t^{n-2}}{(n-2)!}&\frac{t^{n-1}}{(n-1)!} \\ 0 & 1 & t & \cdots &\frac{t^{n-3}}{(n-3)!}&\frac{t^{n-2}}{(n-2)!} \\ \vdots &\vdots&\vdots&&\vdots&\vdots \\ 0 & 0 &0 &\cdots &t&\frac{t^2}{2!} \\ 0 & 0 &0 &\cdots &1 &t \\ 0 & 0 &0 &\cdots &0 &1 \end{matrix} \right) $$

Answer 2

Here's how I do it (with thanks to Yiorgos S. Smyrlis):

Note first that the equation

$\dfrac{dx}{dt} = Ax \tag{1}$

always has the solution

$x(t) = e^{A(t - t_0)} x(t_0) \tag{2}$

for $x(t)$ with initial condition $x(t_0)$ at $t = t_0$. Next observe that a $p \times p$ Jordan block $A$ is a matrix of the form

$A = \lambda I + N, \tag{3}$

where $N = [n_{ij}]$ is the $p \times p$ matrix with

$n_{ij} = 1 \; \; \text{if} \; \; j = i + 1, \tag{3}$

$n_{ij} = 0 \; \; \text{otherwise}. \tag{4}$

Since $\lambda I$ and $N$ commute, that is $(\lambda I)N = N(\lambda I)$, we have that

$e^{A(t - t_0)} = e^{(\lambda I + N)(t - t_0)} = e^{\lambda I(t - t_0)}e^{N(t - t_0)}; \tag{5}$

that this holds may be seen in my answer to this question. Now observe that

$e^{\lambda I(t - t_0)} = e^{\lambda(t - t_0)}I \tag{6}$

and

$e^{N(t - t_0)} = \sum_0^\infty \dfrac{N^n(t - t_0)^n}{n!} = \sum_0^{p - 1} \dfrac{N^n(t - t_0)^n}{n!}, \tag{7}$

since, as is easily seen (and worked out in detail by Yiorgos S. Smyrlis in his answser),

$N^p = 0. \tag{8}$

It follows that $e^{N(t - t_0)}$ is upper triangular and the entries of $e^{N(t - t_0)}$ are monomials in $t - t_0$; we thank Yiorgos for taking the trouble to write them out in full. In any event, we have

$e^{A(t - t_0)} = e^{\lambda(t - t_0)}\sum_0^{p - 1} \dfrac{N^n(t - t_0)^n}{n!}, \tag{9}$

and from here $e^{A(t - t_0)}x(t_0)$ is easy to write down.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!