Editing Matrix exponential (section)

=== Linear differential equations ===

The matrix exponential has applications to systems of [[linear differential equation]]s.  (See also [[matrix differential equation]].)  Recall from earlier in this article that a ''homogeneous'' differential equation of the form
<math display="block"> \mathbf{y}' = A\mathbf{y} </math>
has solution {{math|''e''<sup>''At''</sup> '''y'''(0)}}.

If we consider the vector
<math display="block"> \mathbf{y}(t) = \begin{bmatrix} y_1(t) \\ \vdots \\y_n(t) \end{bmatrix}   ~,</math>
we can express a system of ''inhomogeneous'' coupled linear differential equations as
<math display="block"> \mathbf{y}'(t) = A\mathbf{y}(t)+\mathbf{b}(t).</math>
Making an [[ansatz]] to use an integrating factor of {{math|''e''<sup>−''At''</sup>}} and multiplying throughout, yields
<math display="block">\begin{align}
  &            &       e^{-At}\mathbf{y}'-e^{-At}A\mathbf{y} &= e^{-At}\mathbf{b} \\
  &\Rightarrow &       e^{-At}\mathbf{y}'-Ae^{-At}\mathbf{y} &= e^{-At}\mathbf{b} \\
  &\Rightarrow & \frac{d}{dt} \left(e^{-At}\mathbf{y}\right) &= e^{-At}\mathbf{b}~.
\end{align}</math>

The second step is possible due to the fact that, if {{math|1=''AB'' = ''BA''}}, then {{math|1=''e''<sup>''At''</sup>''B'' = ''Be''<sup>''At''</sup>}}. So, calculating {{math|''e''<sup>''At''</sup>}} leads to the solution to the system, by simply integrating the third step with respect to {{mvar|t}}.

A solution to this can be obtained by integrating and multiplying by <math>e^{\textbf{A}t}</math> to eliminate the exponent in the LHS. Notice that while <math>e^{\textbf{A}t}</math> is a matrix, given that it is a matrix exponential, we can say that <math>e^{\textbf{A}t} e^{-\textbf{A}t} = I</math>. In other words, <math>\exp{\textbf{A}t} = \exp{{(-\textbf{A}t)}^{-1}}</math>.

==== Example (homogeneous) ====
Consider the system
<math display="block">\begin{matrix}
  x' &=& 2x & -y &  +z \\
  y' &=&    & 3y & -1z \\
  z' &=& 2x & +y & +3z
\end{matrix}~.</math>

The associated [[defective matrix]] is
<math display="block">A = \begin{bmatrix}
  2 & -1 &  1 \\
  0 &  3 & -1 \\
  2 &  1 &  3
\end{bmatrix}~.</math>

The matrix exponential is
<math display="block">e^{tA} = \frac{1}{2}\begin{bmatrix}
    e^{2t}\left( 1 + e^{2t} - 2t\right) & -2te^{2t}       &  e^{2t}\left(-1 + e^{2t}\right) \\
   -e^{2t}\left(-1 + e^{2t} - 2t\right) &  2(t + 1)e^{2t} & -e^{2t}\left(-1 + e^{2t}\right) \\
    e^{2t}\left(-1 + e^{2t} + 2t\right) &  2te^{2t}       &  e^{2t}\left( 1 + e^{2t}\right)
\end{bmatrix}~,</math>

so that the general solution of the homogeneous system is
<math display="block">\begin{bmatrix}x \\y \\ z\end{bmatrix} =
    \frac{x(0)}{2}\begin{bmatrix}e^{2t}\left(1 + e^{2t} - 2t\right) \\ -e^{2t}\left(-1 + e^{2t} - 2t\right) \\ e^{2t}\left(-1 + e^{2t} + 2t\right)\end{bmatrix}
  + \frac{y(0)}{2}\begin{bmatrix}-2te^{2t} \\ 2(t + 1)e^{2t} \\ 2te^{2t}\end{bmatrix}
  + \frac{z(0)}{2}\begin{bmatrix}e^{2t}\left(-1 + e^{2t}\right) \\ -e^{2t}\left(-1 + e^{2t}\right) \\ e^{2t}\left(1 + e^{2t}\right)\end{bmatrix} ~,
</math>

amounting to
<math display="block">\begin{align}
  2x &= x(0)e^{2t}\left(1 + e^{2t} - 2t\right) + y(0)\left(-2te^{2t}\right) + z(0)e^{2t}\left(-1 + e^{2t}\right) \\[2pt]
  2y &= x(0)\left(-e^{2t}\right)\left(-1 + e^{2t} - 2t\right) + y(0)2(t + 1)e^{2t} + z(0)\left(-e^{2t}\right)\left(-1 + e^{2t}\right) \\[2pt]
  2z &= x(0)e^{2t}\left(-1 + e^{2t} + 2t\right) + y(0)2te^{2t} + z(0)e^{2t}\left(1 + e^{2t}\right) ~.
\end{align}</math>

==== Example (inhomogeneous) ====
Consider now the inhomogeneous system
<math display="block">\begin{matrix}
  x' &=& 2x & - & y & + & z & + & e^{2t} \\
  y' &=&    &   & 3y& - & z & \\
  z' &=& 2x & + & y & + & 3z & + & e^{2t}
\end{matrix} ~.</math>

We again have
<math display="block">A = \left[\begin{array}{rrr}
  2 & -1 &  1 \\
  0 &  3 & -1 \\
  2 &  1 &  3
\end{array}\right] ~,</math>

and
<math display="block">\mathbf{b} = e^{2t}\begin{bmatrix}1 \\0\\1\end{bmatrix}.</math>

From before, we already have the general solution to the homogeneous equation. Since the sum of the homogeneous and particular solutions give the general solution to the inhomogeneous problem, we now only need find the particular solution.

We have, by above,
<math display="block">\begin{align}
  \mathbf{y}_p
  &= e^{tA}\int_0^t e^{(-u)A}\begin{bmatrix}e^{2u} \\0\\e^{2u}\end{bmatrix}\,du+e^{tA}\mathbf{c} \\[6pt]
  &= e^{tA}\int_0^t \begin{bmatrix}
         2e^u - 2ue^{2u} & -2ue^{2u}    & 0 \\
    -2e^u + 2(u+1)e^{2u} & 2(u+1)e^{2u} & 0 \\
                2ue^{2u} & 2ue^{2u}     & 2e^u
  \end{bmatrix}\begin{bmatrix}e^{2u} \\0 \\e^{2u}\end{bmatrix}\,du + e^{tA}\mathbf{c} \\[6pt]
  &= e^{tA}\int_0^t \begin{bmatrix}
     e^{2u}\left( 2e^u - 2ue^{2u}\right) \\
     e^{2u}\left(-2e^u + 2(1 + u)e^{2u}\right) \\
    2e^{3u} + 2ue^{4u}
  \end{bmatrix}\,du + e^{tA}\mathbf{c} \\[6pt]
  &= e^{tA}\begin{bmatrix}
    -{1 \over 24}e^{3t}\left(3e^t(4t - 1) - 16\right) \\
     {1 \over 24}e^{3t}\left(3e^t(4t + 4) - 16\right) \\
     {1 \over 24}e^{3t}\left(3e^t(4t - 1) - 16\right)
  \end{bmatrix} + \begin{bmatrix}
           2e^t - 2te^{2t} &      -2te^{2t} & 0 \\
    -2e^t + 2(t + 1)e^{2t} & 2(t + 1)e^{2t} & 0 \\
                  2te^{2t} &       2te^{2t} & 2e^t
  \end{bmatrix}\begin{bmatrix}c_1 \\c_2 \\c_3\end{bmatrix} ~,
\end{align}</math>
which could be further simplified to get the requisite particular solution determined through variation of parameters.
Note '''c''' = '''y'''<sub>''p''</sub>(0). For more rigor, see the following generalization.