How to Use Variation of Parameters for a 3×3 System

Title: How to Use Variation of Parameters for a 3×3 System Category: mathematics

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A 3×3 matrix by itself does not determine a differential-equation solution: you also need an equation showing how the unknown vector changes, and possibly a forcing term or initial condition. The usual “changing parameters” method for a linear system is called variation of parameters (or variation of constants).

For a nonhomogeneous linear system

$$x'(t)=A(t)x(t)+f(t),$$

where $x(t)$ and $f(t)$ are 3-component column vectors and $A(t)$ is a 3×3 matrix, assume that $A(t)$ and $f(t)$ are continuous on an interval of interest. The method gives the general solution once you know a fundamental matrix for the homogeneous system.

The fundamental-matrix formula

First solve the associated homogeneous system:

$$x'(t)=A(t)x(t).$$

A fundamental matrix $\Phi(t)$ is a 3×3 matrix whose columns are three linearly independent solutions of this homogeneous system. It is invertible on the interval of interest. To vary the constants in the homogeneous solution, write

$$x(t)=\Phi(t)u(t),$$

where $u(t)$ is now an unknown vector of functions. Differentiating and substituting into the system gives

$$\Phi'(t)u(t)+\Phi(t)u'(t)=A(t)\Phi(t)u(t)+f(t).$$

Since $\Phi'=A\Phi$, those terms cancel, leaving

$$u'(t)=\Phi(t)^{-1}f(t).$$

Integrating gives the general solution

$$x(t)=\Phi(t)\left(c+\int_{t_0}^{t}\Phi(s)^{-1}f(s)\,ds\right),$$

where $c$ is a constant 3-vector and $t_0$ is a fixed reference time. Changing $t_0$ changes the constants’ representation, not the family of solutions. If you are given $x(t_0)=x_0$, then $c=\Phi(t_0)^{-1}x_0$.

Worked 3×3 example

Consider the system

$$x'(t)= \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&-2 \end{pmatrix}x(t) + \begin{pmatrix} e^t\\ 1\\ t \end{pmatrix}.$$

The homogeneous equations are $x_1'=x_1$, $x_2'=0$, and $x_3'=-2x_3$. A fundamental matrix is therefore

$$\Phi(t)= \begin{pmatrix} e^t&0&0\\ 0&1&0\\ 0&0&e^{-2t} \end{pmatrix}.$$

Its inverse is diagonal with entries $e^{-t},1,e^{2t}$. Multiplying the forcing vector by this inverse gives

$$\Phi(t)^{-1}f(t)= \begin{pmatrix} 1\\ 1\\ te^{2t} \end{pmatrix}.$$

Integrate each component from $0$ to $t$:

$$\int_0^t\Phi(s)^{-1}f(s)\,ds= \begin{pmatrix} t\\ t\\ e^{2t}(t/2-1/4)+1/4 \end{pmatrix}.$$

Multiplying by $\Phi(t)$ and absorbing constant terms into three arbitrary constants gives

$$x(t)= \begin{pmatrix} e^t(C_1+t)\\ C_2+t\\ C_3e^{-2t}+t/2-1/4 \end{pmatrix}.$$

You can check this by differentiating each component and substituting it into its equation. For example, the third component satisfies $x_3'+2x_3=t$.

Constant coefficient matrices

When $A$ is constant, one common choice is the matrix exponential $\Phi(t)=e^{At}$. The formula becomes

$$x(t)=e^{At}\left(c+\int_{t_0}^{t}e^{-As}f(s)\,ds\right).$$

This remains valid whether or not $A$ is invertible. If $f(t)=0$, the integral vanishes and the general solution is just $x(t)=e^{At}c$. For a specified initial value at $t_0$, the equivalent initial-value formula is

$$x(t)=e^{A(t-t_0)}x(t_0)+\int_{t_0}^{t}e^{A(t-s)}f(s)\,ds.$$

Common mistakes

• Using a matrix without specifying the system. You need an equation such as $x'=Ax+f(t)$; a matrix alone does not tell you the unknown function or forcing.
• Putting the forcing term in the wrong place. The variation equation is $u'=\Phi^{-1}f$, not $u'=f$ in general.
• Forgetting the integration constants. A general solution of a 3-component first-order system has three arbitrary constants. Initial conditions fix their values.
• Assuming every system is linear. The formula applies to linear systems. A nonlinear system generally needs different methods, though linearization can be useful near a particular solution.
• Confusing the matrix exponential with entrywise exponentiation. $e^{At}$ is defined by the matrix power series, not by exponentiating each entry of $A$ separately.