Consider a system of n first-order ordinary differential equations: \dot{\boldsymbol{x}}=\boldsymbol{A}\boldsymbol{x} where \boldsymbol{x}=(x_1(t),\dots,x_n(t))^{T} is the vector of unknown functions, and \boldsymbol{A}=(a_{ij}(t))_{n\times n} is a known matrix of functions. Suppose n-1 linearly independent particular solutions \boldsymbol{x}_1, \boldsymbol{x}_2, \dots, \boldsymbol{x}_{n-1} (column vectors of solutions) are known. Find the general solution of the system.
This is a problem a classmate asked me on June 5th. At the time, I felt that the solution could be easily constructed using Lie symmetry methods. That evening, I performed a brief analysis and found that based on the ideas of Lie symmetry, the provided information is indeed sufficient to construct the general solution. However, I spent several days trying various geometric and algebraic approaches without successfully constructing the appropriate canonical variables, and thus failed to write down an explicit solution, so I set it aside. Today, while re-analyzing this problem, I unexpectedly constructed a solution that I find quite satisfying.
Feasibility Analysis
(Readers who are not familiar with Lie symmetry can skip this section directly.)
Let us analyze the symmetries possessed by this system of equations. First, since this is a system of linear equations, a fundamental infinitesimal generator is: X=\boldsymbol{x}\frac{\partial}{\partial \boldsymbol{x}} Since n-1 particular solutions are already known, according to the principle of superposition, the system also possesses the symmetries: X_i=\boldsymbol{x}_i\frac{\partial}{\partial \boldsymbol{x}} Thus, we already possess n symmetries for this system of ordinary differential equations. Verification shows that these n infinitesimal generators form an n-dimensional solvable Lie algebra. Therefore, starting from these conditions, solving for the general solution of this system is theoretically feasible. However, finding the appropriate transformation is not always straightforward.
Solution Process
What kind of transformation is suitable? I found today that the following process helps simplify the original problem. Let \boldsymbol{\xi} be any column vector that is linearly independent of \boldsymbol{x}_1, \dots, \boldsymbol{x}_{n-1} (it can 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 this 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} Therefore: \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} (\boldsymbol{A}\boldsymbol{\xi} - \dot{\boldsymbol{\xi}}) u_n \end{aligned} Notice that the right-hand side only contains u_n and t. If we denote: [\boldsymbol{x}_1, \dots, \boldsymbol{x}_{n-1}, \boldsymbol{\xi}]^{-1} (\boldsymbol{A}\boldsymbol{\xi} - \dot{\boldsymbol{\xi}}) = \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 these yields the general solution of the original equation. This results in two cases. The first is the trivial case where u_n \equiv 0 is a solution to the above system, which leads to u_i = c_i for i=1, 2, \dots, n-1; this recovers the original n-1 particular solutions. The second case is direct integration: u_n = \exp\left(\int \eta_n(t)dt\right) From which 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 leads to a new solution that is linearly independent of the first n-1 particular solutions. (Why must it be linearly independent? One way to understand this is that for every u_i calculated, an integration is required, and each integration introduces an independent constant. Thus, the set of u_i contains n independent constants in total! Therefore, what we have found is the general solution. The general solution naturally includes the particular solutions as well as the new linearly independent solution.)
Analysis of Results
Since the choice of the vector \boldsymbol{\xi} is arbitrary, we can choose \boldsymbol{\xi} to make the calculation as simple as possible based on the known particular solutions. In most cases, we can choose: \boldsymbol{\xi}=(\underbrace{0, \dots, 0}_{n-1}, 1)^T In this case, we find: \boldsymbol{A}\boldsymbol{\xi}=(a_{1n}, a_{2n}, \dots, a_{nn})^T That is, only the last column of \boldsymbol{A} is preserved. 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 particular solutions, most of the information in matrix \boldsymbol{A} is redundant. At the same time, this indicates that the information contained in each column of \boldsymbol{A} is equivalent.
A Simple Example
Suppose we are given a differential equation: \ddot{x}+a(t)\dot{x}+b(t)x=0 and a known non-zero particular solution x=\phi(t). Find the general solution. We rewrite the equation as a system of first-order equations: \frac{d}{dt} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -b & -a \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} The particular solution is [x_1, x_2]^T = [\phi(t), \phi'(t)]^T. Apply the transformation: \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} \phi & 0 \\ \phi' & 1 \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} 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} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \begin{pmatrix} \phi & 0 \\ \phi' & 1 \end{pmatrix}^{-1} \begin{pmatrix} 0 & 1 \\ 0 & -a \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} Which simplifies to: \frac{d}{dt} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} = \begin{pmatrix} 0 & 1/\phi \\ 0 & -a - \phi'/\phi \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} Solving this 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}
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