How can we prove this matrix exponential formula

Question

I am working on the reduction of second ordre differential equation to a first order system of equations, and I encounter this formula which we are trying to generalise.

It is known that for $\alpha \in \mathbb{C}$ then,

$$\exp \left(t \begin{bmatrix} 0 & 1 \\ \alpha & 0 \\ \end{bmatrix}\right) = \begin{bmatrix} cosh(t\alpha^\frac{1}{2}) & \int_0^t{cosh(s\alpha^\frac{1}{2})ds} \\ \frac{d}{dt}cosh(t\alpha^\frac{1}{2}) & cosh(t\alpha^\frac{1}{2}) \\ \end{bmatrix} \ \ (t \in \mathbb{R})\ \ \ (1.1)$$

where

$$cosh(t\alpha^\frac{1}{2}) = \sum_{n=0}^\infty\frac{t^{2n}}{(2n)!}\alpha^n$$

I proved it when $\alpha=1$ and $\alpha=-1$, but how to do it in the general case?

You can find this formula in this paper.

Edit:

I found this result in Klaus Nagel book, but is there any other nice method to prove the formula?

2.7 More Examples. (iii) Take an arbitrary 2 x 2 matrix $A = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$, define $\delta := ad - bc$, $\tau := a + d$, and take $\gamma \in \mathbb{C}$ such that $\gamma^2 = \frac{1}{4}(\gamma^2 - 4\delta)$. Then the (semi) group generated by A is given by the matrices (2.5)

$$e^{tA} = \begin{cases} e^{\frac{t\tau}{2}}(\frac{1}{\tau}sinh(t\gamma)A + (cosh(t\gamma) - \frac{2\tau}{\gamma}sinh(t\gamma))I) & if\ \gamma \ne 0, \\ e^{\frac{t\tau}{2}}(tA + (1 - \frac{t\tau}{2})I) & if \gamma = 0 \end{cases}$$

Answer 1

$$ A = \left(\begin{array}{cc} 0 & 1 \\ a & 0\end{array}\right) \;\;\; A^2 = \left(\begin{array}{cc} a & 0 \\ 0 & a\end{array}\right) \;\;\; A^3 = \left(\begin{array}{cc} 0 & a \\ a^2 & 0\end{array}\right) \;\;\; A^4 = \left(\begin{array}{cc} a^2 & 0 \\ 0 & a^2\end{array}\right) $$ $$ A^{2k} = \left(\begin{array}{cc} a^k & 0 \\ 0 & a^k\end{array}\right) \;\;\; A^{2k+1} = \left(\begin{array}{cc} 0 & a^k \\ a^{k+1} & 0\end{array}\right) $$ $$ e^{tA} = \sum_{n=0}^\infty \frac{t^n A^n}{n!} = \sum_{k=0}^\infty \frac{t^{2k} A^{2k}}{(2k)!} + \sum_{k=0}^\infty \frac{t^{2k+1} A^{2k+1}}{(2k+1)!} $$ $$ \left(\begin{array}{cc} \sum_{k=0}^\infty \frac{t^{2k}a^k}{(2k)!} & \sum_{k=0}^\infty \frac{t^{2k+1}a^k}{(2k+1)!} \\ \sum_{k=0}^\infty \frac{t^{2k+1}a^{k+1}}{(2k+1)!} & \sum_{k=0}^\infty \frac{t^{2k}a^k}{(2k)!} \end{array}\right) $$ Assuming that integration and differentiation commute with these inifinite sums we can write: $$ \frac{d}{dt} \sum_{k=0}^\infty \frac{t^{2k}a^k}{(2k)!} = \sum_{k=1}^\infty \frac{2k t^{2k-1}a^k}{(2k)!} = \sum_{k=1}^\infty \frac{ t^{2k-1}a^k}{(2k-1)!} = \sum_{k=0}^\infty \frac{ t^{2k+1}a^k}{(2k+1)!} = $$ $$ \int_0^t \sum_{k=0}^\infty \frac{s^{2k}a^k}{(2k)!} = \sum_{k=0}^\infty \frac{s^{2k+1}a^k}{(2k+1)(2k)!}\arrowvert_0^t = \sum_{k=0}^\infty \frac{t^{2k+1}a^k}{(2k+1)!} = $$

Answer 2

Generally, the ODE

$$ \frac{\partial}{\partial t} f(t) = kf(t) $$

has a solution which looks like $f(t) = e^{tk}f(0) = Ce^{tk}$ for arbitrary constant $C$. This solution corresponds to the initial condition $f(0)=C$ and $\dot f(0) = Ck$.

In our case, the function $f(t) = C e^{t k}$ for $k=\begin{bmatrix}0 & 1 \\ \alpha & 0\end{bmatrix}$ and $C=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$ is the solution to

$$ \frac{\partial}{\partial t} f(t) = \begin{bmatrix}0 & 1 \\ \alpha & 0\end{bmatrix} f(t) $$

with the initial condition $f(0) = C$ and $\dot f(0) = Ck = k$.

The solution to the equation is $f(t)=\begin{bmatrix}p(t) & q(t) \\ r(t) & s(t)\end{bmatrix}$, where

$$ \begin{cases} \dot p(t) = r(t), & p(0) = 1, & \dot p(0) = 0, \\ \dot q(t) = s(t), & q(0) = 0, & \dot q(0) = 1, \\ \dot r(t) = \alpha p(t), & r(0) = 0, & \dot r(0) = \alpha, \\ \dot s(t) = \alpha q(t), & s(0) = 1, & \dot s(0) = 0. \end{cases} $$

From which you quickly get $\ddot p(t) = \alpha p(t)$ and $\ddot s(t) = \alpha s(t)$, hence $p(t) = s(t) = \cosh (t\sqrt\alpha)$, and everything else follows from it, so we obtain the result

$$ f(t) = \begin{bmatrix}\cosh(t \sqrt \alpha) & \frac{1}{\sqrt\alpha}\sinh(t\sqrt \alpha) \\ \sqrt\alpha\sinh(t \sqrt \alpha) & \cosh(t \sqrt \alpha) \end{bmatrix}. $$

Taking into consideration $\frac{\partial}{\partial x} \cosh x = \sinh x$ and $\frac{\partial}{\partial x} \sinh x = \cosh x$ it leads to the same answer, as in your initial question.