Differential Equations

Systems of Equations

Method of Matrix Exponential

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Problems 1-3

Definition and Properties of the Matrix Exponential

Consider a square matrix \(A\) of size \(n \times n,\) elements of which may be either real or complex numbers. Since the matrix \(A\) is square, the operation of raising to a power is defined, i.e. we can calculate the matrices
{{A^0} = I,\;\;{A^1} = A,\;\;}\kern-0.3pt
{{A^2} = A \cdot A,\;\;}\kern-0.3pt
{{A^3} = {A^2} \cdot A,\; \ldots ,}\kern-0.3pt
{{A^k} = \underbrace {A \cdot A \cdots A}_\text{k times},}
\] where \(I\) denotes a unit matrix of order \(n.\)

We form the infinite matrix power series
\[{I + \frac{t}{{1!}}A + \frac{{{t^2}}}{{2!}}{A^2} }+{ \frac{{{t^3}}}{{3!}}{A^3} + \cdots }+{ \frac{{{t^k}}}{{k!}}{A^k} + \cdots }\] The sum of the infinite series is called the matrix exponential and denoted as \({e^{tA}}:\)
\[{e^{tA}} = \sum\limits_{k = 0}^\infty {\frac{{{t^k}}}{{k!}}{A^k}} .\] This series is absolutely convergent.

In the limiting case, when the matrix consists of a single number \(a,\) i.e. has a size of \(1 \times 1,\) this formula is converted into a known formula for expanding the exponential function \({e^{at}}\) in a Maclaurin series:
{{e^{at}} = 1 + at + \frac{{{a^2}{t^2}}}{{2!}} + \frac{{{a^3}{t^3}}}{{3!}} + \cdots }
= {\sum\limits_{k = 0}^\infty {\frac{{{a^k}{t^k}}}{{k!}}} .}
\] The matrix exponential has the following main properties:

  • If \(A\) is a zero matrix, then \({e^{tA}} = {e^0} = I;\) (\(I\) is the identity matrix);
  • If \(A = I,\) then \({e^{tI}} = {e^t}I;\)
  • If \(A\) has an inverse matrix \({A^{ – 1}},\) then \({e^A}{e^{ – A}} = I;\)
  • \({e^{mA}}{e^{nA}} = {e^{\left( {m + n} \right)A}},\) where \(m, n\) are arbitrary real or complex numbers;
  • The derivative of the matrix exponential is given by the formula
    \[\frac{d}{{dt}}\left( {{e^{tA}}} \right) = A{e^{tA}}.\]
  • Let \(H\) be a nonsingular linear transformation. If \(A = HM{H^{ – 1}},\) then \({e^{tA}} = H{e^{tM}}{H^{ – 1}}.\)

The Use of the Matrix Exponential for Solving Homogeneous Linear Systems with Constant Coefficients

The matrix exponential can be successfully used for solving systems of differential equations. Consider a system of linear homogeneous equations, which in matrix form can be written as follows:
\[\mathbf{X}’\left( t \right) = A\mathbf{X}\left( t \right).\] The general solution of this system is represented in terms of the matrix exponential as
\[\mathbf{X}\left( t \right) = {e^{tA}}\mathbf{C},\] where \(\mathbf{C} =\) \( {\left( {{C_1},{C_2}, \ldots ,{C_n}} \right)^T}\) is an arbitrary \(n\)-dimensional vector. The symbol \(^T\) denotes transposition. In this formula, we cannot write the vector \(\mathbf{C}\) in front of the matrix exponential as the matrix product \(\mathop {\mathbf{C}}\limits_{\left[ {n \times 1} \right]} \mathop {{e^{tA}}}\limits_{\left[ {n \times n} \right]} \) is not defined.

For an initial value problem (Cauchy problem), the components of \(\mathbf{C}\) are expressed in terms of the initial conditions. In this case, the solution of the homogeneous system can be written as
\[{\mathbf{X}\left( t \right) = {e^{tA}}{\mathbf{X}_0},\;\;}\kern-0.3pt{\text{where}\;\;}\kern-0.3pt{{\mathbf{X}_0} = \mathbf{X}\left( {t = {t_0}} \right).}\] Thus, the solution of the homogeneous system becomes known, if we calculate the corresponding matrix exponential. To calculate it, we can use the infinite series, which is contained in the definition of the matrix exponential. Often, however, this allows us to find the matrix exponential only approximately. To solve the problem, one can also use an algebraic method based on the latest property listed above. Consider this method and the general pattern of solution in more detail.

Algorithm for Solving the System of Equations Using the Matrix Exponential

  1. We first find the eigenvalues \({\lambda _i}\)of the matrix (linear operator) \(A;\)
  2. Calculate the eigenvectors and (in the case of multiple eigenvalues) generalized eigenvectors;
  3. Construct the nonsingular linear transformation matrix \(H\) using the found regular and generalized eigenvectors. Compute the corresponding inverse matrix \({H^{ – 1}}\);
  4. Find the Jordan normal form J for the given matrix \(A,\) using the formula
    \[J = {H^{ – 1}}AH.\] Note: In the process of finding the regular and generalized eigenvectors, the structure of each Jordan block often becomes clear. This allows to write the Jordan form without calculation by the above formula.
  5. Knowing the Jordan form \(J,\) we compose the matrix \({e^{tJ}}.\) The corresponding formulas for this conversion are derived from the definition of the matrix exponential. The matrices \({e^{tJ}}\) for some simple Jordan forms are shown in the following table:
The matrices exp(tJ) for some Jordan forms

Figure 1.

  1. Compute the matrix exponential \({e^{tA}}\) by the formula
    \[{e^{tA}} = H{e^{tJ}}{H^{ – 1}}.\]
  2. Write the general solution of the system:
    \[\mathbf{X}\left( t \right) = {e^{tA}}\mathbf{C}.\] For a second order system, the general solution is given by
    \[{\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
    \end{array}} \right] }={ {e^{tA}}\left[ {\begin{array}{*{20}{c}}
    \end{array}} \right],}\] where \({C_1},{C_2}\) are arbitrary constants.

Solved Problems

Click on problem description to see solution.

 Example 1

Find the general solution of the system, using the matrix exponential:
\[{\frac{{dx}}{{dt}} = 2x + 3y,\;\;}\kern-0.3pt{\frac{{dy}}{{dt}} = 3x + 2y.}\]

 Example 2

Solve the system of equations by the method of matrix exponential:
\[{\frac{{dx}}{{dt}} = 4x,\;\;}\kern-0.3pt{\frac{{dy}}{{dt}} = x + 4y.}\]

 Example 3

Solve the system of equations using the matrix exponential:
\[{\frac{{dx}}{{dt}} = x + y,\;\;}\kern-0.3pt{\frac{{dy}}{{dt}} = – x + y.}\]

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Problems 1-3