Some linear differential equations generated by matrices

1.   Introduction

There is much research of linear differential equations involving or generated by matrices, see Derevenskii ^[2] and Elishevich ^[3]. But these and other papers are limited to the first few orders. We are not aware of papers giving solutions for linear differential equations of general order generated by matrices. In the paper, we provide solutions for such a class of linear differential equations.

Let $R$ and $C$ denote the real and complex numbers, respectively. For any function $f(t)$ of $t\in R$, set $f_{\centerdot n}(t) = (d/dt)^nf(t)$. Let $Y(t) = Y = \left(Y_1, \ldots, Y_r \right)\in C^{s\times r}$ be a function of $t\in R$, and a constant $N = \left(N_{j, k} \right)\in C^{r\times r}$, such that

Choose $q\geq 1$. Suppose that for $0\leq n\leq q$, and $T_n(t)\in C^{s\times s}$, there is a constant $S_n = \left(S_{n, j, k} \right) \in C^{r\times r}$ such that

If $r = s$, as in the abstract, we can take

If $r > s$, we can take $Y(t)$ as the first $s$ rows of $e^{tN}$. We wish to solve the differential equation $Lf = 0_s$ for $f = f(t):R\rightarrow C^s$, where

that is,

Theorem 2.1 gives solutions in terms of the roots of $d = d(\nu)$, where

We call $D(\nu)$ the characteristic matrix of the operator $L$. If $\nu$ is a multiple root, then other solutions are given by Theorem 2.2.

Section 3 chooses $r = s = 2$ and $N$ of (3.1) below. $T_n(t)$ of (1.3) are then linear in $\cos 2t$ and $\sin 2t$ and so have period $\pi$. Example 3.2 gives for the first time the full solution of a well known example. Example 3.3 solves a differential equation that arises in the theory of planetary perturbations, and was the incentive for this paper. Section 4 shows how $D (\nu)$ and $T_n(t)$ are given by the Jordan form of $N$ when $r = s$.

Section 5 solves $Lf = g$, any given function in $C^s$, when $r = s$ and $T_q = I_s$, by converting it to the standard form $X_{\centerdot 1} = AX+F$. Its solution is then given by the variation of constants formula in terms of the solution of $Lf = 0_s$. When $A$ is periodic, Floquet's theorem only gives the form of the solution of $X_{\centerdot 1} = AX$. We give the actual solution. Examples 3.1-3.5 all have periodic $A$. Set $i = \sqrt{-1}$. Set $\delta_{j, k} = 1$ or $0$ for $j = k$ or $j\neq k$.

2.   Main results

Theorem 2.1. Suppose that $Y(t)$, $N$ satisfy (1.1), and that$T_n(t)$, $S_n$ satisfy (1.2) for $0\leq n\leq q$.Let $\nu$ be any of the $qr$ roots of $d(\nu) = 0$for $d(\nu)$of (1.5).Choose $a(\nu) = a = \left(a_1, \ldots, a_r\right)'\in C^r$ such that $E(\nu) = 0_r$, where

Then for $f = f(t)\in C^s$ and$L$ of (1.4), a solution of $Lf = 0_s$ is

Proof: Since $Y_{\centerdot n} = YN^n$, by Leibniz' rule, the $n$th derivative of $e^{\nu t}Y$ is

So,

and $L\ f_0 = e^{\nu t}Y\ D(\nu)a(\nu) = 0_r$.

When the roots of $d(\nu)$ are distinct, this gives $qr$ independent solutions of (2.1).

Theorem 2.2. Take $\nu$, $D = D(\nu)$, $a = a(\nu)$ and $E = E(\nu)$ of Theorem 2.1.Suppose that for some $k\geq 1$,

for $0 \leq m < k$, where $E_{\centerdot m}(\nu) = \partial_\nu^m E(\nu)$, and $\partial_\nu = d/d \nu$.Then for $0\leq n < k$, $Lf = 0_s$ has solutions

where $z_{n} = z_{n}(t, \nu) = \left(t+\partial_\nu\right)^n\ a(\nu)$.

Proof: By (2.2) and Leibniz' rule,

where

Transform from $m$ to $c = m+r$. Then ${n\choose m}{n-m\choose r} = {n\choose c} {c\choose r}$. So,

where

by (2.3).

We call $f_n$ a characteristic solution of $Lf = 0_s$. $D(\nu)$ and $d(\nu)$ expand as

where

So,

Corollary 2.1. Take $Lf$ of (1.4).Consider the exceptional case when $D(\nu)\equiv 0_{s\times s}$.Suppose that

for $0_q\leq m\leq n$.Then for any $a\in C^s$, a characteristic solution of $Lf = 0_s$ is

We now transfer the condition (2.3) from $E(\nu)$ to $d(\nu)$. Let $M$ be the adjoint of $D$: $(-1)^{j+k}M_{k, j}$ is the determinant of $D$ with its $j$th row and $k$th column deleted. If $d\neq 0$, then $M = d D^{-1}$.

Corollary 2.2. Let $e_{1, s}, \ldots, e_{2, s}$ be any basis for $R^s$.For $D$ and $d$ of (1.5), set $a_{(j)} = Me_{j, s}$ and $E_{(j)} = Da_{(j)} = d e_{j, s}$.So, for $m\in Z^q$, $E_{(j)\centerdot m} = d_{\centerdot m} e_{j, s}$.Choose $\nu$ so that $d = 0$.Given $1\leq j\leq s$ and $a = a_{(j)}$, $f_{n, j}(t) = f_n(t, \nu)$ of (2.4) is a characteristic solution of $Lf = 0_s$ if

for $0 \leq m \leq n$.

By (2.5), (2.6) does not extend to $n = rq$ if $d_{r, q}\neq 0$. We now take $e_{j, s}$ as the $j$th unit vector in $R^s$. So, for $1\leq j\leq s$, $a_{(j)}$ is the $j$th column of $M$. For $1\leq k\leq s$, its $k$th element is $a_{(j)k} = M_{k, j} = a_{j, k}$ say. Corollary 2.2 breaks the solution $f_n(x, \nu)$ into $s$ basis solutions $f_{n, j}(x)$, $1\leq j\leq s$, of $Lf = 0_s$. For example, if $s = 2$ then

$d = 0$ implies

$d = d_{\centerdot 1} = 0$ implies

$d = d_{\centerdot 1} = d_{\centerdot 2} = 0$ implies

$d_{\centerdot m} = 0$ for $0\leq m\leq 3$ implies $f_{3, j} = e^{\nu t} \left(z_{3, 1} Y_1+z_{3, 2} Y_2\right)$, where, for $j = 1$,

and, for $j = 2$,

Generally, each $f_{n, j}$ is a linear combination of $\left(f_{m, 1}, \ 0\leq m\leq n\right)$. We shall give details in a later paper. For example, $M_{1, 1}\neq 0$ implies, for $2\leq j\leq s$, $f_{0, j}(t) = f_{0, 1}(t)M_{j, 1}/M_{1, 1}$. $s = 2$ and $D_{2, 2}\neq 0$ imply $f_{0, 2}(t) = -f_{0, 1}(t)D_{2, 1}/D_{2, 2}$.

For $L(t) = L$ of (1.4) and $\tau \neq 0$, set

where $T_{\tau, n}(t) = \tau^{q-n}T_n(\tau t)$. So, $T_q = I_s$ implies $T_{\tau, q} = I_s$. For example, $q = s = 2$, $T_2 = I_2$ imply $L_\tau (t) = (d/dt)^2+\tau T_1(\tau t)+\tau^2T_0(\tau t)$.

Corollary 2.3. Take $\nu$, $D = D(\nu)$, $a = a(\nu)$ of Theorems 2.1-2.2, and$f(t) = f_0(t, \nu)$ of (2.1), or $f(t) = f_n(t, \nu)$ of Theorem 2.2.Then a solution of $L_\tau (t)X(t) = 0_s$ is $X(t) = f_n(\tau t, \nu)$.

We have not assumed that $r = s$ or $T_q = I_s$. However, if $r = s$ and $T_q = I_s$, then $Lf = 0_s$ can be written in the standard form $X_{\centerdot 1} = AX$, where

(to be read as $A = -T_0$ if $q = 1$), and each $0$ is $s\times s$. So, (2.15) with $f$ of Theorems 2.1, 2.2 give all $qs$ linearly independent solutions of $X_{\centerdot 1} = AX$.

3.   An application to the unit circle

Set $c_t = \cos t$, $s_t = \sin t$. Here, we take $r = s = 2$ and

So,

Set

$N$, $R(t)$, $Q(t)$, $\Lambda$ and $J$ all have determinant $\pm 1$. Some properties are:

(1.3) holds for

So, (1.3) also holds for any linear combination of these, say

for any constants $b_{n, j} \in C$. Any $2\times 2$ matrix $S_n$ can be put in this form: set

So,

where

Corollary 3.1. For $T_n(t)$ of (3.3), a solution of $Lf = 0_2$ is (2.1) with$\nu$ any of the $2q$ roots of $d = 0$,

where

where

where the real and imaginary parts are taken as if $\nu$ were real.

Proof: Since $N^2 = -I_2$,

for $g_n$ and $h_n$ of (3.7), (3.8). So, for $S_n$ of (3.2),

So, by (1.5), (2.5) holds. $g_n$, $h_n$, $c_{n, j}$, $c_j$, $D$, $d$ are polynomials in $\nu$ of degree $n$, $n-1$, $n$, $q$, $q$, $2q$. So, $d = 0$ has $2q$ roots $\nu$.

By (3.7)-(3.8), when $q = 1$, the $c_j$ needed for (3.5) are

By (3.7)-(3.8), when $q = 2$, the $c_j$ needed for (3.5) are

When $q = 3$, we add $c_{3, j}$ of (3.6) to $c_j$ for $j = 1, 2, 3, 4$.

$Lf = 0_2$ can only be reduced to the form $X_{\centerdot 1} = AX$ if $e_q\neq 0$, where $e_n = \det\ T_n(t) = \det\ S_n = b_{n, 1}^2 +b_{n, 2}^2 -b_{n, 3}^2 -b_{n, 4}^2$, since then we can reduce $T_q(t)$ to $I_2$ by multiplying by $T_q(t)^{-1}$. Set

where $\overline{T_n}(t) = T_q(t)^{-1}T_n(t)$ is a linear combination of $1$, $s_{2, t}$, $c_{2, t}$, $s_{4, t}$, $c_{4, t}$.

Example 3.1. Take $q = 1$, $S_1 = I_2$, $S_0 = \lambda\ \Lambda$. So, (1.2) holds with $T_1 = I_2$ and $T_0(t) = \lambda R(2t)$. Further,

and $d = \nu^2-\lambda^2+1$ with roots $\nu = \pm \left(\lambda^2-1\right)^{1/2} = \nu_1$, $\nu_2$, say. By (2.7), $a = a_{(1)} = {\nu-\lambda\choose -1}$ implies $a_{\centerdot 1} = {1\choose 0}$, $z_0 = {\nu-\lambda\choose -1}$ and $z_1 = {t(\nu-\lambda)+1\choose -t}$; $a = a_{(2)} = {1\choose \nu+\lambda}$ implies $a_{\centerdot 1} = {0\choose 1}$, $z_0 = {1\choose \nu+\lambda}$ and $z_1 = {t\choose t(\nu+\lambda)+1}$; $d = 0$ implies $\nu = \pm \left(1-\lambda^2\right)^{1/2}$. So, by (2.8), solutions are

If $d = 0$ and $\lambda = \pm 1$, then $d_{\centerdot 1} = \nu = 0$ so by (2.9) and (2.10) solutions are

An extension is

Example 3.2. Given scalars $b_0$, $b_1$, $b_2$, we solve $f_{\centerdot 1}(t) = A(t)f(t)$ for

So, $Lf = f_{\centerdot 1} -A(t)f$. Take $q = 1$, $S_1 = I_2$ and

Then (1.4) holds with $T_1(t) = I_2$, $T_0(t) = -A(t)$. So,

and $d(\nu) = \left(\nu-b_0\right)^2 -b_1^2+\left(b_2-1\right)^2$ has roots

for $\delta = b_1^2-\left(b_2-1\right)^2$. By (2.8), if $\nu = \nu_1$ or $\nu_2$, then $d = 0$ and solutions are

Consider the case $\delta = 0$. So, $b_2 = 1+\lambda b_1$, where $\lambda = \pm 1$,

Let us rescale this example by transforming to $T = t/\tau$, $x(T) = f(\tau T)$ for $\tau\neq 0$, then replacing $T$ by $t$.

Example 3.3. Set $A_\tau (t) = \tau A(\tau t)$. For $A(t)$ of (3.9), ${x}_{\centerdot 1}(t) = A_\tau (t)x(t)$ has solutions $f_{0, j}(\tau t)$ for $f_{0, j}(t)$ of (3.11) with $\nu$ of (3.10). If $\delta = 0$, other solutions are $f_{1, j}(\tau t)$ for $f_{1, j}(t)$ of (2.9) and (2.10). We consider two cases. The first case is that $\tau = 1/2$. In this case,

The second case is that $\tau = -1$, $b_0 = 1/4$, $b_2 = 1$ and $b_1 = -3/4$. Then $\nu_1 = -1/2$, $\nu_2 = 1$ and independent solutions are

The other two $f_{0, j}$ are $0_2$. In this case, $A_\tau(t) = -B-b_1 R(-2t)$ can be written

with period $T = \pi$. Markus and Yamabe ^[5] used this form of $A_\tau(t)$ but only gave the first solution $f_{0, 1}(-t)$. This example is quoted by Chicone ^[1], but again the second solution $f_{0, 2}(-t)$ is not given.

Example 3.4. Take $q = 2$, $S_2 = I_2$, $S_1 = 0_{2\times 2}$ and $S_0 = {\rm diag}\ \left(b_0+b_1, b_0-b_1\right)$ for scalars $b_0$, $b_1$. Then $T_2 = I_2$, $T_1 = 0_{2\times 2}$ and

Then, $L = I_2 (d/dt)^2+T_0$. (1.2) holds with

where $b = b_0+1$, $c = \left(b_0-1\right)^2-b_1^2$. $d$ has four roots, $\nu = \pm \nu_1$, $\pm \nu_2$, where

So, $\nu^2-1+b_0 = -2\pm \delta^{1/2}$ and solutions are given by $f_{0, 1}$, $f_{0, 2}$ of (2.8) with $D_{2, 2} = \nu^2-1+b_0-b_1$, $D_{2, 1} = 2\nu$, $D_{1, 2} = -2\nu$ and $D_{1, 1} = \nu^2-1+b_0+b_1$. If $\delta$ of (3.12) is 0 and $b\neq 0$, then there are two roots of multiplicity two, $\nu = \lambda \nu_0$, where $\lambda = \pm 1$, $\nu_0 = (-b)^{1/2}$; so other solutions are $f_{1, 1}$ of (2.9) with $\nu = \lambda \nu_0$,

and $f_{1, 2}$ of (2.10) with $\nu = \lambda \nu_0$,

Now suppose that $b_0 = -1$, $b_1 = 2\lambda$, where $\lambda = \pm 1$. Then $\nu = 0$ has multiplicity 4. So, other solutions are $f_{2, 1}$ of (2.11), (3.13) with $D_{2, 2\centerdot 2} = 2$, $D_{2, 1\centerdot 2} = 0$; $f_{2, 2}$ of (2.12), (3.14) with $D_{1, 2\centerdot 2} = 0$, $D_{1, 1\centerdot 2} = 2$; $f_{3, 1}$ of (2.13) with

and $f_{3, 2}$ of (2.14) with

To solve $Lf = g$, a given function in $C^2$, Section 5 will need its derivative, $\partial_t\ {f}_{0}(t, \nu) = e^{\nu t}v(t, \nu)$, where

By (2.15), $Lf = 0_2$ can be written $X_{\centerdot 1} = AX$, where

Now suppose that $b_0$, $b_1$ are real. If $\delta < 0$, all four roots are complex. Suppose that $\delta\geq 0$. Then $\pm \nu_1$ are real if $b\leq \delta^{1/2}$, and $\pm \nu_2$ are real if $\delta^{1/2}\leq -b$. So, if $b\leq \delta^{1/2}\leq -b$ then all four roots are real.

The differential equation, $Lf = 0_2$, arises in the theory of planetary perturbations with $b_0 = -\gamma/2$, $b_1 = -3\gamma/2$, so that $b = 1-\gamma/2$, $c = -2\gamma^2$, $\delta = 9\gamma(\gamma-8/9)/4$, and $\gamma > 0$ real. So, $\delta\geq 0$ if and only if $\gamma\geq 8/9$. Also $\gamma\geq 1$ if and only if $\nu_1^2\geq 0$. For $\gamma\geq 8/9$, $\nu_2^2 < 0$. So, $\gamma < 8/9$ implies four complex roots; $8/9\leq \gamma < 1$ implies four imaginary roots; $1\leq \gamma$ implies $\nu_1^2\geq 0 > \nu_2^2$ so that two roots are real and two are imaginary. For large $\gamma$, $\nu_1$ and $\nu_2$ of (3.12) satisfy

By Corollary 2.3, we have

Example 3.5. For $L$ of Example 3.4, $L_\tau (t) = (d/dt)^2+\tau^2T_0(\tau t)$, and $L_\tau (t)X(t) = 0_2$ has solution $X(t) = f_{n, j}(\tau t)$ when $f_{n, j}(t)$ is a solution to Example 3.4.

4.   Using the Jordan form when $r=s$

Suppose that $r = s$. Then $D$ and $T_n(t)$ can be easily written using the Jordan form of $N$, in terms of block matrices, or scalars when the Jordan form is diagonal. Suppose that $N$ has Jordan form $N = PJP^{-1}$, where $J = {\rm diag}\ \left(J_1, \ldots, J_p\right)$, $J_j = J_{m_j} \left(\lambda_j\right)$, $J_{m}(\lambda) = \lambda I_m +U_m$, and $U_m$ is the $m\times m$ matrix of zeros except for ones on the first super-diagonal, that is, $\left(U_m\right)_{j, k} = \delta_{k, j+1}$ for $1\leq j < m$. So, $s = m_1 + \cdots + m_p$ and for $0\leq n < m$, $U_m^n$ is the $m\times m$ matrix of zeros except for ones on the $n$th super-diagonal, that is, $\left(U_m^n\right)_{j, k} = \delta_{k, j+n}$ for $1\leq j < m-n$. Also for $n\geq m$, $U_m^n = 0_{m\times m}$, and

where

So, $V_m(t)$ has zeros on its subdiagonals, and the elements of its $j$th superdiagonal are all $t^j/j$. That is, $V_m(t)_{j, j+n} = t^j/j!$ for $1\leq j < m-n$. For example,

Partition $P$, $P^{-1}$, $S_n$ and $T_n = T_n(t)$ of (1.3) as $m_j\times m_k$ blocks, $P = \left(P_{j, k}\right)$, $P^{-1} = \left(P^{j, k}\right)$, $S_n = \left(S_{n, j, k}\right)$, $T_n(t) = \left(T_n(t)_{j, k}\right)$ for $1\leq j$, $k\leq p$, and do similarly for $D$, $N^n$ and $Y = Y(t) = e^{tN}$. Then

where

and $e^{tJ_c}$ is given by (4.2) with $m = m_c$, $\lambda = \lambda_c$. So, $T_n(t)$ is a mixture of polynomials in $t$ and factors $e^{\left(\lambda_c-\lambda_d\right)t}$. For example, if $P = I_s$, then

Consider the two extremes: first the one Jordan block case $J = J_s(\lambda)$. Then $p = 1$, $m_1 = s$, $P$ is scalar, say $1$,

Second the diagonal Jordan form with $J = {\rm diag}\ \left(\lambda_1, \ldots, \lambda_s\right)$. Then $p = s$, $m_j\equiv 1$, $J_c = \lambda_c$,

with all components scalar, and

For example, if $P = I_s$, then

Example 4.1. Take $J = {\rm diag}\ \left(\lambda_1, \lambda_2\right)$. So, $s = 2$. Set $\delta = \lambda_1-\lambda_2$. Then

and

where

and $q_n = P^{-1}S_nP$. For $N$ of Section 3, we can take

In Section 3, we avoided having to use $P$ and these eigenvalues by using $N^2 = -I_2$.

5.   Solving $Lf=g$ when $T_q=I_s$ and $r=s$

We now give the solution of $Lf = g$ when $T_q = I_s$ and $r = s$. This can be written $X_{\centerdot 1} = AX+F$ for $X$, $A$ of (2.15) and $F = \left(0_s', 0_s', \ldots, g'\right)'$ as noted on page 90 of Hale ^[4] for the case $s = 1$.

For $j = 1, \ldots, qr$, set $f_j = f\left(t, \nu_j\right)$ of (2.1) if $\nu_j$ are distinct. Otherwise choose $f_j$ using Theorem 2.2. Set

where

So, $U_j\in C^{qs}$. For example, if $s = 2$ and $\left\{\nu_j\right\}$ are distinct, we can take $a = a_{(1)}$ or $a_{(2)}$ of (2.7). Now suppose that $r = s$. Then $U(t) = U = \left(U_1, \ldots, U_{qs}\right)\in C^{qs\times qs}$ is a fundamental matrix solution of $X_{\centerdot 1} = AX$. That is, $\det\ U(t)\neq 0$. Its $(j, k)$ element is

So, $\widetilde{U}(t) = U(t)U(0)^{-1}$ is the principal matrix solution at 0 as it satisfies $U(0) = I_{qs}$: see, for example, page 80 of Hale ^[4]. So, by the variation of constants formula, see, for example, his page 81, $X_{\centerdot 1} = AX+F$ has solution

where $\widetilde{X}(0) = U(0)^{-1}X(0)$. Write $U^{-1}$ as a $2\times 2$ block matrix with $(j, k)$ element $U^{j, k}\in C^{s\times s}$. Then $U^{-1}F = {U^{1, 2}\choose U^{2, 2}}f$, so that $Lf = g$ has solution

the first of the two block rows of

where $\left(\widetilde{U}_{1}, \widetilde{U}_{2} \right) = U^{-1}$. So, the solutions (5.2) and (5.3) have $qs$ unknowns $X(0)$ for $X$ of (2.15), that is, the initial values of $f$ and its first $q-1$ derivatives.

We now illustrate a use of Floquet's theorem. Set $r = qs$. Suppose that $r = s, T_q = I_s$, and $\left\{T_n\right\}$ have period $T$. (This holds for Example 1.1 and Section 3 with $T = \pi$ or $\pi/ \tau$ for Example 3.4.) (2.15) puts $Lf = 0_s$ in the standard form $X_{\centerdot 1} = AX$, where $A$ is periodic. According to Floquet's theorem (see, for example, page 118 of Hale ^[4] or page 164 of Chicone ^[1]), since $U(t)$ is a fundamental matrix solution of $X_{\centerdot 1} = AX\in C^{r\times r}$ for $(X, A)$ of (2.15), and $A$ has period $T$, there exists a constant $B\in C^{r\times r}$ and $P = P(t)$ with period $T$ such that $U(t) = P(t)e^{Bt}$. However, Floquet's theorem does not give $P(t)$, $B$ while our method does, as we now show. $\left\{\nu_j\right\}$ are the eigenvalues of $B$. Since these eigenvalues are distinct, $B$ has diagonal Jordan form, say $Q\Lambda Q^{-1}$, where $\Lambda = {\rm diag}\ \left(\nu_1, \ldots, \nu_r\right)$. So,

where $R(t) = PQ$. The $(j, k)$ element of $U(t)Q$ is

So, $R_{j, k}(t)$ is the coefficient of $e^{\nu_k t}$ in $U(t)Q$. This gives $R(t)$. Set $\left(Q^{j, k} \right) = Q^{-1}$. By (5.1), the coefficient of $e^{\nu_b t}$ in

is $\delta_{b, j} x_{j, k}(t) = R_{j, b}(t)Q^{b, k}$. So,

$P(t) = x(t) = \left(y_k \left(t, \nu_j \right) \right)$ is of (5.1). Also, $U(t) = P(t)e^{Bt}$, so that $Qe^{\Lambda t}Q^{-1} = e^{Bt} = UP(t)^{-1} = Q(t)$ say. So, $Qe^{\Lambda t} = Q(t)Q$. Write $Q$ as $\left(q_1, \ldots, q_r\right)$. So, for all $t$, $q_j$ is an eigenvector of $Q(t)$ with eigenvalue $e^{\nu_j t}$. So, we can take $q_j$ as the eigenvector of $Q(T)$ with eigenvalue $e^{\nu_j T}$. So, now we have $\Lambda$, $Q$ and $B$.

Conflict of interest

The authors declare no conflicts of interest in this paper.