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I am wondering what can be said in general about the fundamental matrix of a system of linear differential equations. For simplicity, let $A(t)\in\mathbb{C}^{n\times n}$ be a time dependent matrix, smooth on some interval $I$, say $[0,t_I]$.

Is is true that there is no closed form formula for the fundamental matrix $M(t)$ of the equation $A(t)v(t)=\dot{v}(t)$, with $v(t)$ a time dependent vector ? What is the best I can hope for when solving such systems in general (suppose $A(t)$ does not commute with itself at different times) ?

In other words, if I have a given $A(t)$, which happens not to commute with iself at different time, should I hope to obtain an analytical expression for $M(t)$? Which approach would you try if say $A(t)$ was at least sparse?

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    $\begingroup$ No hope. Say, any second order linear equation can be converted to a system like yours. But even simplest equations don't have "closed form" solutions. $\endgroup$
    – Alex Degtyarev
    Commented Feb 20, 2014 at 21:15
  • $\begingroup$ You have a series expansion for $M$, and in general I'm afraid this is "the best one can hope for". $\endgroup$
    – Pietro Majer
    Commented Feb 21, 2014 at 7:13
  • $\begingroup$ Is this series expansion explicitly known at all orders ? And I don't mean something like the Dyson series, I mean something that actually allows you to obtain an explicit analytical form for order $n$. $\endgroup$
    – PLG
    Commented Feb 21, 2014 at 10:24
  • $\begingroup$ Relevant:mathoverflow.net/questions/140849/solution-of-linear-ode/… $\endgroup$
    – Piyush Grover
    Commented Feb 21, 2014 at 13:31
  • $\begingroup$ Thanks for the link! However it seems to me that Picard and Vessiot's theory answers the question negatively when it comes to elementary means but, as pointed by Bryant below, a solution is exactly expressible if one allows for integration procedures. I just realised that indeed, the solution to any system $A(t)v(t)=\dot{v}(t)$ is exactly expressible in terms of the solutions to a $finite$ number of linear Volterra equations of the 2nd kind. $\endgroup$
    – PLG
    Commented Feb 22, 2014 at 15:10

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Assuming that $A(t)$ is analytic at $t=0$, getting a power series solution $M(t)$ at $t=0$ of $M'(t) = A(t)M(t)$ is trivial, of course. You just write $A(t) = A_0 + A_1 t + \cdots$ and $M(t) = I_n + M_1 t + \cdots $ and then recursively solve, for $k>0$ $$ k\ M_k = A_{k-1} + A_{k-2}M_1 + \cdots + A_0 M_{k-1}\ , $$ so $M_1 = A_0$, $M_2 = \tfrac12(A_1+A_0^2)$, $M_3 = \tfrac16(2A_2 + 2A_1A_0 + A_0A_1+A_0^3)$, etc. If $A(t)$ is sparse, then computing the various products of the $A_i$ might not be hard; it depends on the matrix.

One case in which one can integrate the fundamental equation by elementary means is when $A$ takes values in a solvable Lie subalgebra $\frak{s}$ of $\frak{gl}(n,\mathbb{R})$. In this case, by a theorem of Lie and Engel, $M(t)$ can be computed from $A(t)$ by a sequence of quadratures. (The abelian case is, of course, the simplest case of this.)

Meanwhile, it is a classical theorem that, when $n\ge 2$, there is no sequence of algebraic operations and quadratures that will produce $M(t)$ starting from the general $A(t)$.

More generally, if you add to your allowable operations, an 'integration procedure' $IP_{\frak{g}}$ for each simple Lie algebra $\frak{g}$ that will solve the equation $M'(t) = M(t)A(t)$ for any curve $A:\mathbb{R}\to\frak{g}$ for the corresponding $M:\mathbb{R}\to G$, then the equation $M'(t) = M(t)A(t)$ can be solved for any $A$ taking values in any Lie algebra $\frak{a}$ by a sequence of applications of the $IP_{\frak{g}}$ for the simple $\frak{g}$ that appear in the Levi decomposition of $\frak{a}$, followed by a sequence of quadratures.

Look in any good book on differential algebra for a discussion of this, as well as the independence of the operations $IP_{\frak{g}}$ for different simple Lie algebras. The case ${\frak{g}} = {\frak{sl}}(2,\mathbb{R})$ is, of course, the theory of the Riccati equation.

By the way, this result about solvability was one of the primary motivations for wanting to classify the simple Lie algebras. In fact, all of the classical techniques (such as variation of parameters, getting the general solution of the Riccati equation by quadrature from a single solution, etc.), are special cases of general techniques for reducing or solving equations by the use of group actions, and this was the original motivation for studying Lie group actions and homogeneous spaces of Lie groups.

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  • $\begingroup$ Alright so, if I understand you well, allowing for integration, it is possible to find $M(t)$ from $A(t)$ analytically and in a \emph{finite} number of steps (say 1 integral = 1 step), for all $A(t)$ (finite and smooth)? I am asking because it seems to me that what you said gives rise to the Magnus series for $M(t)$, which is still not (in general) computable exactly in a finite number of steps. In any case thanks for your enlightning answer! I am going straight away to read more on this. $\endgroup$
    – PLG
    Commented Feb 21, 2014 at 13:37
  • $\begingroup$ Well, that depends on whether you are willing to count each application of $IP_{\frak{g}}$ as some 'finite number of steps' and whether you regard this as 'finding the solution analytically in closed form'. Even for the first simple Lie algebra ${\frak{sl}}(2,\mathbb{R})$, which is the theory of the Riccati equation, many people (including some of the other commenters) do not regard $IP_{\frak{sl}(2,\mathbb{R})}$ as giving a 'closed-form' solution to the Riccati equation, since the resulting functions aren't 'elementary'. The point is that it's a fundamentally new operation beyond quadrature. $\endgroup$
    – Robert Bryant
    Commented Feb 21, 2014 at 13:58
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Forget about finding a closed analytical expression for the fundamental matrix. Think about the simple case of a second order scalar equation such as the Airy equation $\ddot x=tx.$

There is nevertheless a very remarkable explicit expression for the determinant of the fundamental matrix $M(t)$ (i.e. the solution of $\dot M=AM, M(0)=Id$ ). In fact, we have $$ \det M=\exp\int_0^t\text{trace } A(s) ds. $$ Very remarkable indeed since the matrix $M$ is not known explicitly but its determinant has a simple closed expression: to prove the above formula is not difficult, since you just have to verify $$ \frac d{dt} \det M=(\text{trace } A) \det M, $$ which is a direct consequence of multilinearity of $$\det M= C_1\wedge \dots\wedge C_n, $$ where the $C_j$ are the columns of the matrix $M$. This (non-commutative) exterior product behaves simply with respect to differentiation and the derivative of the column $C_j$ can be found using the ODE satisfied by $M$.

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  • $\begingroup$ But the Airy equation is typically a case that I would say has a closed form expression. Indeed the power series expansion of the Airy function solution to this equation is known analytically at all order. Similarly if the solution turns out to be hypergeometric etc. then I would say it is closed form. Maybe my use of "closed form" is not correct but I really mean that the solution would at least be known at all orders of its power series expansion. If instead, there is no general way of finding at least all orders for an artbirary $M(t)$, then I would say there is no closed form. $\endgroup$
    – PLG
    Commented Feb 21, 2014 at 10:22
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Try by hand a piecewise constant matrix, even with say 4 pieces on [0,1]. You can find a closed form expression, using maple or mathematica, or your pen, by what good will it do? The expression, while closed, will be completely useless. Of course you can take a $C^\infty$ approximation of it, and it will be as close as you wish to it.

Why do like closed forms? It is

  • Because you can compute a solution with arbitrary precision ? Numerical methods for ODEs do that for you.
  • Because you want to understand how it behaves ? Special functions (or combination of special functions) often have intricate behaviors, you will find out more either numerically or by using the ODE directly to prove things.
  • Because you will know that the solution exists / is unique? That's what Peano/ Cauchy do for you...
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  • $\begingroup$ Good point. I think I like closed form expressions for what they are, not for what you can do with them. I don't quite agree with your second point however, because I always found it easier to understand how the solution behaves from a closed form expression than from a numerical evaluation. $\endgroup$
    – PLG
    Commented Feb 22, 2014 at 15:05
  • $\begingroup$ @Pierre-LouisGiscard on the example I suggest, how does that work for you? $\endgroup$
    – username
    Commented Feb 22, 2014 at 17:23
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    $\begingroup$ Rather well: since an analytical closed form expression for the fundamental matrix is available, I can for example easily see the solution dependency on, say, certain parameters. If the problem comes up in a particular application, it is most likely that the values that $A(t)$ take have a meaning (the parameters). Then, the consequences of changing these parameters are more easily obtained: a good example is 1D quantum random walks. Having a Bessel function solution enables one to demonstrate balistic spread. This is much harder to justify from a numerical fit... $\endgroup$
    – PLG
    Commented Feb 24, 2014 at 9:40

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