# The vectorial ODE $D\left( X \right) = AX$, with $A$ a constant coefficient matrix

Let $X(t)$ a vector function in $R^n$, and let A be an $n \times n$ matrix with constant coefficients. Let us use $D(X)$ to denote the derivative with respect to $t$ of the function $X$ (this will be the derivative at each coordinate). Does anyone know any material where the general case of the equation $D(X) = AX$ is solved? I could not find good material for this, and I really need it.

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As Didier Piau said, the solution is $X(t)=e^{tA}X(0)$ for every $t\in\mathbb R$.

Moreover, $e^{tA}$ can be computed as follows, for any $n$ by $n$ matrix $A$ with complex coefficients.

Let $X$ be an indeterminate; let $$p(X)=\prod_{j=1}^r\ (X-a_j)^{m_j}\ \in\mathbb C[X]$$ be the minimal polynomial of $A$, where the $a_j$ are the distinct eigenvalues of $A$; and let $t$ be a real number. For each $j$ put $$e_j(X):=e^{a_jt}\ \sum_{k=0}^{m_j-1}\ \frac{t^k}{k!}\ (X-a_j)^k,$$ and for each rational fraction fraction $f(X)$ defined at $a_j$ let $T_j(f(X))$ be the degree less than $m_j$ Taylor approximation of $f(X)$ at $X=a_j$.

Then we have $e^{tA}=e(A)$ where $e(X)\in\mathbb C[X]$ is given by $$e(X)=\sum_j\ T_j\left(e_j(X)\frac{(X-a_j)^{m_j}}{p(X)}\right)\ \frac{p(X)}{(X-a_j)^{m_j}}\quad.$$ Moreover $e(X)$ is the only polynomial of degree less than $\deg p(X)$ which satisfies $e^{tA}=e(A)$.

EDIT. It's easy to see that $e(X)$ is the unique degree less than $\deg p(X)$ solution to the congruences $e(X)\equiv e_j(X)$ mod $(X-a_j)^{m_j}$.

The above formula for $e(X)$ is essentially contained in Article 36 of Gauss's Disquisitiones Arithmeticae. A French translation is available at Internet Archive and at Google:

Article 36 can be paraphrased as follows.

Let $A$ be a principal ideal domain; let $p_1$, ..., $p_r$ be distinct irreducible elements of $A$; let $m_1$, ... , $m_r$ be positive integers; and let $x_1$, ... , $x_r$ be elements of $A$. Then the congruences $$x\equiv x_i\bmod p_j^{m_j}$$ for $j=1,\dots,r$, can be solved by putting $a:=p_1^{m_1}\cdots p_r^{m_r}$ and $$x:=\sum_{j=1}^r\ T_j\left(x_j\ \frac{p_j^{m_j}}{a}\right)\cdot\frac{a}{p_j^{m_j}}\quad,$$ where $T_j(x_j\,p_j^{m_j}/a)$ is any solution $y$ to the congruence $$\frac{a}{p_j^{m_j}}\ y\equiv x_j\bmod p_j^{m_j}.$$ Such a solution exists because $p_j$ doesn't divide $a/p_j^{m_j}$.

This applies in particular to the case $A=K[Z]$, where $K$ is a field and $Z$ an indeterminate. Then our congruences $$x(Z)\equiv x_j(Z)\bmod p_j(Z)^{m_j}\tag{1}$$ have a unique solution $x(Z)$ satisfying $$\deg x(Z) < \deg a(Z),\tag{2}$$ where $a(Z)$ is the product of the $p_j(Z)^{m_j}$.

Extending $K$ if necessary, we can assume $p_j(Z)=Z-a_j$, where the $a_j$ are distinct elements of $K$.

If we define $T_j(f(Z))$ as the degree less than $m_j$ Taylor approximation of $f(Z)$ at $Z=a_j$, where $f(Z)\in K(Z)$ is any rational fraction $f(Z)\in K(Z)$ defined at $a_j$, then $$\sum_{j=1}^r\ T_j\left(x_j(Z)\ \frac{p_j(Z)^{m_j}}{a(Z)}\right)\cdot\frac{a(Z)}{p_j(Z)^{m_j}}$$ will be the solution satisfying $(2)$.

Many elementary problems boil down to solving congruences of the form $(1)$.

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I think you will like this ;) – user13838 Oct 4 '11 at 17:56
Dear @percusse: Thank you very much! – Pierre-Yves Gaillard Oct 4 '11 at 18:17

The solution is $X(t)=\mathrm e^{tA}X(0)$ for every $t\geqslant0$, very much like in the one dimensional case. The matrix $\mathrm e^{tA}$ may be defined as $\mathrm e^{tA}=\sum\limits_{n=0}^{+\infty}\dfrac{t^n}{n!}A^n$. See here and especially this paragraph.

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but if have this form of series, can I represent it on a matrix, in a easy way? – August Oct 2 '11 at 13:29
Did you read the WP page? This paragraph answers your question. – Did Oct 2 '11 at 13:49
Dear @Didier: Your link doesn't work for me. – Pierre-Yves Gaillard Oct 2 '11 at 14:04
Sorry, this is the link. – Did Oct 2 '11 at 14:12
@Didier: Thanks. Is it on purpose that you wrote $t\ge0$ and not $t\in\mathbb R$? – Pierre-Yves Gaillard Oct 2 '11 at 14:16