Systems of Linear Differential Equations: Finding the General Solution from Particular Solutions

线性微分方程组：已知特解求通解 苏剑林 · 2014-06-18 · Mathematics integral differential equation symmetry

Given a first-order system of ordinary differential equations consisting of $n$ first-order ODEs
$$\dot{\boldsymbol{x}}=\boldsymbol{A}\boldsymbol{x}$$
where $\boldsymbol{x}=(x_1(t),\dots,x_n(t))^{T}$ is the unknown function (vector) to be solved for, and $\boldsymbol{A}=(a_{ij}(t))_{n\times n}$ is a known matrix of functions. Suppose we already know $n-1$ linearly independent particular solutions $\boldsymbol{x}_1,\boldsymbol{x}_2,\dots,\boldsymbol{x}_{n-1}$ (column vectors) of this system. Find the general solution.

This is a problem a classmate of mine asked me on June 5th. At the time, I glanced at it and felt it could probably be tackled quite easily via the method of Lie symmetry. That same evening I did a quick analysis and found that, based on the idea of Lie symmetry, the information given above is indeed in principle sufficient to construct the general solution. However, I spent several days trying various approaches — geometric, algebraic — without managing to construct the appropriate canonical variables, and so I never wrote down an explicit solution, and the problem was set aside. Today, while revisiting the analysis, I stumbled — quite by accident — upon a solution that I'm fairly satisfied with. More below.

Feasibility analysis

(Readers unfamiliar with Lie symmetry can skip this section.)

Let's analyze the symmetries possessed by this system of equations. First, since this is a linear system, one of the most basic infinitesimal generators is

$$X=\boldsymbol{x}\frac{\partial}{\partial \boldsymbol{x}}$$

Since we already know $n-1$ particular solutions, by the superposition principle the system also has the symmetries

$$X_i=\boldsymbol{x}_i\frac{\partial}{\partial \boldsymbol{x}}$$

At this point we already have $n$ symmetries of this ODE system, and upon inspection one finds that these $n$ infinitesimal generators form precisely a $n$-dimensional solvable Lie algebra. Hence, starting from these conditions, solving the general solution of this system is in principle feasible. However, finding a suitable transformation is not straightforward.

The solution process

What kind of transformation is suitable? Today I found that the following procedure helps simplify the original problem. Let $\boldsymbol{\xi}$ be any column vector linearly independent of $\boldsymbol{x}_1,\dots,\boldsymbol{x}_{n-1}$ (it may be a constant vector, or a function of $t$). Consider the transformation

$$\boldsymbol{x}=[\boldsymbol{x}_1,\dots,\boldsymbol{x}_{n-1},\boldsymbol{\xi}]\boldsymbol{u}$$

Substituting into the original equation

$$\begin{aligned} \dot{\boldsymbol{x}}& =[\dot{\boldsymbol{x}}_1,\dots,\dot{\boldsymbol{x}}_{n-1},\dot{\boldsymbol{\xi}}]\boldsymbol{u}+[\boldsymbol{x}_1,\dots,\boldsymbol{x}_{n-1},\boldsymbol{\xi}]\dot{\boldsymbol{u}} \\ &=\boldsymbol{A}[\boldsymbol{x}_1,\dots,\boldsymbol{x}_{n-1},\boldsymbol{\xi}]\boldsymbol{u}\\ &=[\boldsymbol{A}\boldsymbol{x}_1,\dots,\boldsymbol{A}\boldsymbol{x}_{n-1},\boldsymbol{A}\boldsymbol{\xi}]\boldsymbol{u}\\ &=[\dot{\boldsymbol{x}}_1,\dots,\dot{\boldsymbol{x}}_{n-1},\boldsymbol{A}\boldsymbol{\xi}]\boldsymbol{u}\end{aligned}$$

we get

$$\begin{aligned}\dot{\boldsymbol{u}} &=[\boldsymbol{x}_1,\dots,\boldsymbol{x}_{n-1},\boldsymbol{\xi}]^{-1}[\boldsymbol{0},\dots,\boldsymbol{0},\boldsymbol{A}\boldsymbol{\xi}-\dot{\boldsymbol{\xi}}]\boldsymbol{u}\\ &=[\boldsymbol{x}_1,\dots,\boldsymbol{x}_{n-1},\boldsymbol{\xi}]^{-1}\left(\boldsymbol{A}\boldsymbol{\xi}-\dot{\boldsymbol{\xi}}\right)u_n \end{aligned}$$

Note that the right-hand side involves only $u_n$ and $t$. If we denote

$$[\boldsymbol{x}_1,\dots,\boldsymbol{x}_{n-1},\boldsymbol{\xi}]^{-1}\left(\boldsymbol{A}\boldsymbol{\xi}-\dot{\boldsymbol{\xi}}\right)=\boldsymbol{\eta}=(\eta_1,\eta_2,\dots,\eta_n)^T$$

then

$$\begin{aligned} \dot{u}_n&=\eta_n u_n\\ \dot{u}_{n-1}&=\eta_{n-1} u_n\\ \dot{u}_{n-2}&=\eta_{n-2} u_n\\ &\dots\\ \dot{u}_1&=\eta_1 u_n \end{aligned}$$

Integrating this yields the general solution of the original equation. This splits into two cases. In the first case, it's easy to see that $u_n\equiv 0$ is a solution of the above system, from which we solve for $u_i=c_i,\quad i=1,2,\dots,n-1$, recovering the original $n-1$ particular solutions. In the second case, direct integration gives

$$u_n=\exp\left(\int \eta_n(t)dt\right)$$

so that we can integrate step by step

$$u_i=\int \eta_i(t)\exp\left(\int \eta_n(t)dt\right)dt,\quad i=1,2,\dots,n-1$$

This produces new solutions linearly independent of the previous $n-1$ particular solutions. (Why must they be linearly independent? Here's one way to see it: each time I compute a $u_i$, I must integrate once, and each integration introduces an independent constant. So altogether the various $u_i$ carry $n$ independent constants! Hence what I obtain in this way is indeed the general solution!! And the general solution naturally contains both the particular solutions and the new linearly independent solutions.)

Analysis of the result

Since the choice of the vector $\boldsymbol{\xi}$ is arbitrary, we can choose $\boldsymbol{\xi}$ so as to make the computation as simple as possible, given the known particular solutions. In most cases, in fact, we can choose

$$\boldsymbol{\xi}=(\underbrace {0, \cdots ,0}_{n - 1\text{count}},1)^T$$

In this case we find that

$$\boldsymbol{A}\boldsymbol{\xi}=(a_{1n},a_{2n},\dots,a_{nn})^T$$

that is, only the last column survives. From $\dot{\boldsymbol{u}} =[\boldsymbol{x}_1,\dots,\boldsymbol{x}_{n-1},\boldsymbol{\xi}]^{-1}[\boldsymbol{0},\dots,\boldsymbol{0},\boldsymbol{A}\boldsymbol{\xi}-\dot{\boldsymbol{\xi}}]\boldsymbol{u}$ we can see that, given $n-1$ known particular solutions, most of the information in the matrix $\boldsymbol{A}$ turns out to be redundant. This also shows that each column of $\boldsymbol{A}$ carries equal amounts of information.

A simple example

Suppose we know a nonzero particular solution $x=\phi(t)$ of the differential equation

$$\ddot{x}+a(t)\dot{x}+b(t)x=0$$

and we wish to find its general solution. We rewrite the above equation as a first-order system

$$\frac{d}{{dt}}\left( {\begin{array}{\cdot {20}{c}} x_1\\ x_2 \end{array}} \right) = \left[ {\begin{array}{\cdot {20}{c}} 0&1\\ { - b}&{ - a} \end{array}} \right]\left( {\begin{array}{\cdot {20}{c}} x_1\\ x_2 \end{array}} \right)$$

The particular solution is $[x_1,x_2]=[\phi(t),\phi'(t)]$. Applying the transformation

$$\left( {\begin{array}{\cdot {20}{c}} x_1\\ x_2 \end{array}} \right) = \left[ {\begin{array}{\cdot {20}{c}} \phi &0\\ \phi' &1 \end{array}} \right]\left( {\begin{array}{\cdot {20}{c}} u_1\\ u_2 \end{array}} \right)$$

since $\dot{\boldsymbol{u}} =[\boldsymbol{x}_1,\dots,\boldsymbol{x}_{n-1},\boldsymbol{\xi}]^{-1}[\boldsymbol{0},\dots,\boldsymbol{0},\boldsymbol{A}\boldsymbol{\xi}-\dot{\boldsymbol{\xi}}]\boldsymbol{u}$, we have

$$\frac{d}{dt}\left( {\begin{array}{\cdot {20}{c}} u_1\\ u_2 \end{array}} \right)=\left[ {\begin{array}{\cdot {20}{c}} \phi &0\\ \phi' &1 \end{array}} \right]^{-1}\left[ {\begin{array}{\cdot {20}{c}} 0 &1\\ 0 &-a \end{array}} \right]\left( {\begin{array}{\cdot {20}{c}} u_1\\ u_2 \end{array}} \right)$$

that is,

$$\frac{d}{dt}\left( {\begin{array}{\cdot {20}{c}} u_1\\ u_2 \end{array}} \right)=\left[ {\begin{array}{\cdot {20}{c}} 0 &1/\phi\\ 0 &-a-\phi'/\phi \end{array}} \right]\left( {\begin{array}{\cdot {20}{c}} u_1\\ u_2 \end{array}} \right)$$

which gives

$$\begin{aligned} u_2&=\frac{\exp\left(-\int a dt\right)}{\phi}\\ u_1&=\int\frac{u_2}{\phi} dt=\int\frac{\exp\left(-\int a dt\right)}{\phi^2}dt\\ x=x_1&=\phi u_1=\phi\int\frac{\exp\left(-\int a dt\right)}{\phi^2}dt \end{aligned}$$