Construction of the General Solution of a System of Differential Equations Using the Jordan Form

We consider again a system of $$n$$ linear homogeneous differential equations with constant coefficients:

$\mathbf{X}’\left( t \right) = A\mathbf{X}\left( t \right),$

where

${\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}} {{x_1}\left( t \right)}\\ {{x_2}\left( t \right)}\\ \vdots \\ {{x_n}\left( t \right)} \end{array}} \right],\;\;}\kern-0.3pt {A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}}\\ {{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}}\\ \cdots & \cdots & \cdots & \cdots \\ {{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{nn}}} \end{array}} \right].}$

A fundamental set of solutions of the system must include $$n$$ linearly independent functions. When constructing a solution using the eigenvalues and eigenvectors, it often appears that the number of eigenvectors is less than n, i.e. for such systems, there is no basis consisting only of eigenvectors. In this case, the solution can be sought, for example, by the method of undetermined coefficients. However, there is a more general and more elegant way of constructing the solution. It is based on the fact that any square matrix can be reduced to the so-called Jordan canonical form (strictly speaking, this is true over the complex numbers). Knowing the Jordan form of a matrix and the Jordan basis, you can get the general solution of the system.

Consider this solving technique in more detail. First we introduce some basic definitions.

Jordan Form of a Matrix

Jordan form can be viewed as a generalization of the square diagonal matrix. The so-called Jordan blocks corresponding to the eigenvalues $${\lambda _i}$$ of the original matrix are placed on its diagonal. The eigenvalues $${\lambda _i}$$ can be equal in different blocks. Jordan matrix structure might look like this:

${J \text{ = }}\kern0pt{\left( {\begin{array}{*{20}{c}} \color{blue}{{\lambda _1}}& \color{blue}{1}&{0}&{0}&{0}&{0}\\ \color{blue}{0}&\color{blue}{{\lambda _1}}&{0}&{0}& 0 &{0}\\ {0}&{0}&\color{red}{{\lambda _2}}&{0}&{0}&{0}\\ {0}&{0}&{0}&\color{Green}{{\lambda _3}}&\color{Green}1&\color{Green}0\\ {0}& 0 &{0}&\color{Green}0&\color{Green}{{\lambda _3}}&\color{Green}1\\ {0}&{0}&{0}&\color{Green}0&\color{Green}0&\color{Green}{{\lambda _3}} \end{array}} \right),}$

The eigenvalues $${\lambda _i}$$ themselves are on the main diagonal. Each eigenvalue $${\lambda _i}$$ occurs as many times as its algebraic multiplicity $${k_i}.$$ In each block of size more than $$1,$$ there is a parallel diagonal above the main one, consisting of units. All other elements of the Jordan matrix are zero. The order of the Jordan blocks in the matrix is not unique.

Generalized Eigenvectors and Jordan Chains

Consider a Jordan block of size $$k$$ associated with an eigenvalue $${\lambda}.$$ Such block has $$k$$ basis vectors $${\mathbf{V}_1},$$ $${\mathbf{V}_2}, \ldots ,$$ $${\mathbf{V}_k}.$$ Vector $${\mathbf{V}_1}$$ $$\left( {{\mathbf{V}_1} \ne 0} \right)$$ is the eigenvector and satisfies the equation

${A{\mathbf{V}_1} = \lambda {\mathbf{V}_1},\;\; }\Rightarrow {\left( {A – \lambda I} \right){\mathbf{V}_1} = \mathbf{0}.}$

Vector $${\mathbf{V}_2}$$ $$\left( {{\mathbf{V}_2} \ne 0} \right)$$ is determined from the equation

$\left( {A – \lambda I} \right){\mathbf{V}_2} = {\mathbf{V}_1}$

and is called the generalized eigenvector of first order. Similarly, one can find the other generalized eigenvectors of higher order:

$\left( {A – \lambda I} \right){\mathbf{V}_3} = {\mathbf{V}_2},$

$\ldots\ldots\ldots\ldots\ldots\ldots$

$\left( {A – \lambda I} \right){\mathbf{V}_k} = {\mathbf{V}_{k-1}}.$

Note that from the relationships

${\left( {A – \lambda I} \right){\mathbf{V}_1} = \mathbf{0}\;\;\text{and}\;\;}\kern-0.3pt{\left( {A – \lambda I} \right){\mathbf{V}_2} = {\mathbf{V}_1}}$

it follows that

${\left( {A – \lambda I} \right)^2}{\mathbf{V}_2} = \mathbf{0}.$

For a generalized eigenvector $${\mathbf{V}_k}$$ of order $$k$$ the following equality holds

${\left( {A – \lambda I} \right)^k}{\mathbf{V}_k} = \mathbf{0}.$

The set of vectors $${\mathbf{V}_1},$$ $${\mathbf{V}_2}, \ldots ,$$ $${\mathbf{V}_k}$$ consisting of eigenvector $${\mathbf{V}_1}$$ and generalized eigenvectors $${\mathbf{V}_2}, \ldots ,$$ $${\mathbf{V}_k}$$ is linearly independent and is called a Jordan chain.

Each Jordan chain of length $$k$$ corresponds to $$k$$ linearly independent solutions of the homogeneous system in the form

${\mathbf{X}_1} = {e^{\lambda t}}{\mathbf{V}_1},$

${{\mathbf{X}_2} }={ {e^{\lambda t}}\left( {\frac{t}{{1!}}{\mathbf{V}_1} + {\mathbf{V}_2}} \right),}$

${{\mathbf{X}_3} }={ {e^{\lambda t}}\left( {\frac{{{t^2}}}{{2!}}{\mathbf{V}_1} + \frac{t}{{1!}}{\mathbf{V}_2} + {\mathbf{V}_3}} \right),}$

$\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots$

${{\mathbf{X}_k} }={ {e^{\lambda t}}\left( {\frac{{{t^{k – 1}}}}{{\left( {k – 1} \right)!}}{\mathbf{V}_1} + \cdots }\right.}+{\left.{ \frac{t}{{1!}}{\mathbf{V}_{k – 1}} + {\mathbf{V}_k}} \right).}$

The total number of solutions is equal to the sum of the lengths of the Jordan chains for all blocks, that is equal to the size of the matrix $$n.$$ The set of such linearly independent vector functions is a fundamental system of solutions.

The General Solution for $$2 \times 2$$ and $$3 \times 3$$ Matrices

In practice, the most common are systems of differential equations of the 2nd and 3rd order. We consider all cases of Jordan form, which can be encountered in such systems and the corresponding formulas for the general solution. In total there are eight different cases ($$3$$ for the $$2 \times 2$$ matrix and $$5$$ for the $$3 \times 3$$ matrix). It is convenient to illustrate this classification by the following table:

We now discuss how to calculate the eigenvectors and generalized eigenvectors in these cases and construct the general solution.

Case $$1.$$ Matrix $$2 \times 2.$$ Two Distinct Eigenvalues $${\lambda _1},{\lambda _2}$$

In this case, the Jordan normal form is diagonal. Each eigenvalue $${\lambda _i}$$ has one eigenvector $${\mathbf{V}_i},$$ which can be found from the matrix equation

$\left( {A – {\lambda _i}I} \right){\mathbf{V}_i} = \mathbf{0}.$

The general solution is given by

${\mathbf{X}\left( t \right) }={ {C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} + {C_2}{e^{{\lambda _2}t}}{\mathbf{V}_2}.}$

Case $$2.$$ Matrix $$2 \times 2.$$ One Eigenvalue $${\lambda _1}$$ $$\left( {{k_1} = 2,{s_1} = 2} \right)$$

This matrix has a single eigenvalue of multiplicity $$2.$$ The matrix rank for this value of $${\lambda _1}$$ is $$0.$$ Therefore, the geometric multiplicity is equal to

${{s_1} }={ n – \text{rank}\left( {A – {\lambda _1}I} \right) }={ 2 – 0 = 2,}$

i.e. when solving the equation

$\left( {A – {\lambda _1}I} \right){\mathbf{V}} = \mathbf{0}$

we obtain two linearly independent eigenvectors $${\mathbf{V}_1}$$ and $${\mathbf{V}_2}.$$ The general solution has almost the same form as in Case $$1:$$

${\mathbf{X}\left( t \right) }={ {C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} + {C_2}{e^{{\lambda _1}t}}{\mathbf{V}_2}.}$

Case $$3.$$ Matrix $$2 \times 2.$$ One Eigenvalue $${\lambda _1}$$ $$\left( {{k_1} = 2,{s_1} = 1} \right)$$

Here the matrix rank is $$1.$$ Hence the geometric multiplicity of the eigenvalue $${\lambda _1}$$ and the number of eigenvectors is equal

${{s_1} }={ n – \text{rank}\left( {A – {\lambda _1}I} \right) }={ 2 – 1 = 1.}$

This eigenvector $${\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}$$ can be found from the equation

$\left( {A – {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0}.$

To construct a fundamental system of solutions we need one more linearly independent vector. As such, we take the generalized eigenvector $${\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T}$$ satisfying the equation

$\left( {A – {\lambda _1}I} \right){\mathbf{V}_2} = {\mathbf{V}_1}.$

We can compose the matrix $$H$$ based on the regular and generalized eigenvectors:

$H = \left[ {\begin{array}{*{20}{c}} {{V_{11}}}&{{V_{12}}}\\ {{V_{21}}}&{{V_{22}}} \end{array}} \right],$

Then the Jordan form $$J$$ can be found using the relationship

${H^{ – 1}}AH = J,$

where $${H^{ – 1}}$$ is the inverse matrix of $$H.$$ This property can be used for validation of regular and generalized eigenvectors.

The general solution is represented as

${\mathbf{X}\left( t \right) }={ {C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} }+{ {C_2}{e^{{\lambda _1}t}}\left( {t{\mathbf{V}_1} + {\mathbf{V}_2}} \right).}$

Case $$4.$$ Matrix $$3 \times 3.$$ Three Distinct Eigenvalues $${\lambda _1},{\lambda _2}, {\lambda _3}$$

Here the Jordan form is diagonal. Each eigenvalue $${\lambda _i}$$ has its eigenvector $${\mathbf{V}_i},$$ which is determined from the equation

$\left( {A – {\lambda _i}I} \right){\mathbf{V}_i} = \mathbf{0}.$

The general solution of the system of 3 differential equations can be written as

${\mathbf{X}\left( t \right) }={ {C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} }+{ {C_2}{e^{{\lambda _2}t}}{\mathbf{V}_2} } + {{C_3}{e^{{\lambda _3}t}}{\mathbf{V}_3}.}$

Case $$5.$$ Matrix $$3 \times 3.$$ Two Eigenvalues $${\lambda _1}\left( {{k_1} = 2,{s_1} = 2} \right),$$ $${\lambda _2}\left( {{k_2} = 1,{s_2} = 1} \right)$$

In this case, the characteristic equation has two roots, one of which has multiplicity $${{k_1} = 2}.$$ When substituting this multiple root $${\lambda _1},$$ the matrix $$A – {\lambda _1}I$$ has rank $$1.$$ As a result, the geometric multiplicity of $${\lambda _1}$$ and the number of associated eigenvectors is equal to

${{s_1} }={ n – \text{rank}\left( {A – {\lambda _1}I} \right) }={ 3 – 1 = 2.}$

Both the linearly independent eigenvectors $${\mathbf{V}_1}$$ and $${\mathbf{V}_2}$$ (they correspond to two Jordan blocks) are determined from the equation

$\left( {A – {\lambda _1}I} \right){\mathbf{V}} = \mathbf{0}.$

The third block in the Jordan form consists of a simple eigenvalue $${\lambda _2}\left( {{k_2} = 1,{s_2} = 1} \right).$$ The eigenvector $${\mathbf{V}_3}$$ for this number can be found from the equation

$\left( {A – {\lambda _2}I} \right){\mathbf{V}_3} = \mathbf{0}.$

The general solution is given by

${\mathbf{X}\left( t \right) }={ {C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} }+{ {C_2}{e^{{\lambda _1}t}}{\mathbf{V}_2} } + {{C_3}{e^{{\lambda _2}t}}{\mathbf{V}_3}.}$

Case $$6.$$ Matrix $$3 \times 3.$$ Two Eigenvalues $${\lambda _1}\left( {{k_1} = 2,{s_1} = 1} \right),$$ $${\lambda _2}\left( {{k_2} = 1,{s_2} = 1} \right)$$

This case differs from the previous one in that the first eigenvalue $${\lambda _1}$$ has only one eigenvector $${\mathbf{V}_1},$$ which satisfies the equation

$\left( {A – {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0}.$

The matrix rank for the number $${\lambda _1}$$ is $$2:$$

${\text{rank}\left( {A – {\lambda _1}I} \right) = 2,\;\;}\Rightarrow {{{s_1} }= {n – \text{rank}\left( {A – {\lambda _1}I} \right) }={ 3 – 2 = 1.}}$

The missing linearly independent vector can be found as a generalized eigenvector $${\mathbf{V}_2},$$ connected to $${\mathbf{V}_1}:$$

$\left( {A – {\lambda _1}I} \right){\mathbf{V}_2} = {\mathbf{V}_1}.$

The other eigenvalue $${\lambda _2}$$ (corresponding to the second Jordan block) provides one more eigenvector $${\mathbf{V}_3}.$$ The general solution has the form:

$\require{AMSmath.js} {\mathbf{X}\left( t \right) \text{ = }}\kern0pt{ \underbrace {{C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} + {C_2}{e^{{\lambda _1}t}}\left( {t{\mathbf{V}_1} + {\mathbf{V}_2}} \right)}_{\substack{ \text{1st Jordan block}}} } + {\underbrace {{C_3}{e^{{\lambda _2}t}}{\mathbf{V}_3}}_{\substack{ \text{2nd Jordan block}}}.}$

Case $$7.$$ Matrix $$3 \times 3.$$ One Eigenvalue $${\lambda _1}$$ $$\left( {{k_1} = 3,{s_1} = 2} \right)$$

Here the Jordan form consists of two blocks with the same eigenvalue $${\lambda _1}.$$ The first block has one regular eigenvector $${\mathbf{V}_1}$$ and one generalized eigenvector $${\mathbf{V}_2}.$$ They can be found from the relationships

${\left( {A – {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0},\;\;}\kern-0.3pt {\left( {A – {\lambda _1}I} \right){\mathbf{V}_2} = {\mathbf{V}_1}.}$

The first equation has two solutions in the form of two eigenvectors (since $$\text{rank}\left( {A – {\lambda _1}I} \right) = 1$$). The second regular eigenvector (denoted as $${\mathbf{V}_3}$$) is associated with the second Jordan block.

The general solution is described by

${\mathbf{X}\left( t \right) \text{ = }}\kern0pt{ \underbrace {{C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} + {C_2}{e^{{\lambda _1}t}}\left( {t{\mathbf{V}_1} + {\mathbf{V}_2}} \right)}_{\substack{ \text{1st Jordan block}}} } + {\underbrace {{C_3}{e^{{\lambda _1}t}}{\mathbf{V}_3}}_{\substack{ \text{2nd Jordan block}}}.}$

Case $$8.$$ Matrix $$3 \times 3.$$ One Eigenvalue $${\lambda _1}$$ $$\left( {{k_1} = 3,{s_1} = 1} \right)$$

In this case, the linear operator $$A$$ has one eigenvalue $${\lambda _1}$$ of multiplicity $${k_1} = 3.$$ The rank of the matrix $$\left( {A – {\lambda _1}I} \right)$$ is $$2.$$ This leads to the fact that the equation

$\left( {A – {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0}$

has a solution in the form of a single eigenvector $${\mathbf{V}_1}.$$ The missing two linearly independent vectors are determined as generalized eigenvectors from the chain of relations

${\left( {A – {\lambda _1}I} \right){\mathbf{V}_2} = {\mathbf{V}_1},\;\;}\kern-0.3pt {\left( {A – {\lambda _1}I} \right){\mathbf{V}_3} = {\mathbf{V}_2}.}$

The general solution is

${\mathbf{X}\left( t \right) }={ {C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} }+{ {C_2}{e^{{\lambda _1}t}}\left( {t{\mathbf{V}_1} + {\mathbf{V}_2}} \right) } + {{C_3}{e^{{\lambda _1}t}}\left( {\frac{{{t^2}}}{{2!}}{\mathbf{V}_1} + t{\mathbf{V}_2} + {\mathbf{V}_3}} \right).}$

Below we consider examples of systems of equations corresponding to the Cases $$1 – 8.$$ Cases $$1,2,4,5$$ with a “sufficient” number of eigenvectors are also presented on the web page Method of Eigenvalues and Eigenvectors.

Solved Problems

Click or tap a problem to see the solution.

Example 1

Solve the system of equations
${\frac{{dx}}{{dt}} = 2x – 3y,\;\;}\kern-0.3pt{\frac{{dy}}{{dt}} = – x + 4y.}$

Example 2

Find the general solution of the system
${\frac{{dx}}{{dt}} = – x,\;\;}\kern-0.3pt{\frac{{dy}}{{dt}} = – y.}$

Example 3

Find the general solution of the system
${\frac{{dx}}{{dt}} = 2x – y,\;\;}\kern-0.3pt{\frac{{dy}}{{dt}} = x + 4y.}$

Example 4

Solve the system of equations
${\frac{{dx}}{{dt}} = – 4x – 6y – 6z,\;\;}\kern-0.3pt {\frac{{dy}}{{dt}} = x + 3y + z,\;\;}\kern-0.3pt {\frac{{dz}}{{dt}} = 2x + 4z.}$

Example 5

Find the general solution of the system of differential equations
${\frac{{dx}}{{dt}} = x – y – z,\;\;}\kern-0.3pt {\frac{{dy}}{{dt}} = – x + y – z,\;\;}\kern-0.3pt {\frac{{dz}}{{dt}} = – x – y + z.}$

Example 6

Find the general solution of the system
${\frac{{dx}}{{dt}} = – 3x – 6y + 6z,\;\;}\kern-0.3pt {\frac{{dy}}{{dt}} = x + 6z,\;\;}\kern-0.3pt {\frac{{dz}}{{dt}} = – y + 4z.}$

Example 7

Find the general solution of the system of linear differential equations
${\frac{{dx}}{{dt}} = 4x + 6y – 15z,\;\;}\kern-0.3pt {\frac{{dy}}{{dt}} = x + 3y – 5z,\;\;}\kern-0.3pt {\frac{{dz}}{{dt}} = x + 2y – 4z.}$

Example 8

Solve the system of linear homogeneous equations
${\frac{{dx}}{{dt}} = – 7x – 5y – 3z,\;\;}\kern-0.3pt {\frac{{dy}}{{dt}} = 2x – 2y – 3z,\;\;}\kern-0.3pt {\frac{{dz}}{{dt}} = y.}$
Page 1
Concept
Page 2
Problems 1-8