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8.4 Matrix Exponential: Class Notes

8.4a

Recall that ${\displaystyle e^x=\sum _{k=0}^{\infty }\frac{x^k}{k!}=1+x+\frac{1}{2}{x}^2+\frac{1}{6}{x}^3+\cdots }$ This is the power series representation for the exponential function. In the same way, we can define a matrix exponential.

Definition

For any $n\times n$ matrix $𝐀$,

$${\displaystyle e^{𝐀t}=𝐈+𝐀t+{𝐀}^2\frac{t^2}{2!}+\cdots +{𝐀}^k\frac{t^k}{k!}+\cdots =\sum _{k=0}^{\infty }{𝐀}^k\frac{t^k}{k!}}$$

Example

Compute $e^{𝐀t}$ and $e^{-𝐀t}$.

$$𝐀=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$$

Solution

Compute the powers of $𝐀$:

$${𝐀}^2=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=𝐈$$

$${𝐀}^3={𝐀}^2𝐀=𝐈𝐀=𝐀,{𝐀}^4=𝐈,{𝐀}^5=𝐀,\dots$$

The even powers are $𝐈$ and the odd powers are $𝐀$. In the series, the even powers put their terms on the main diagonal and the odd powers put theirs off the diagonal:

$${\displaystyle e^{𝐀t}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+\left(\begin{array}{cc}0& t\\ t& 0\end{array}\right)+\left(\begin{array}{cc}{t}^2/2& 0\\ 0& t^2/2\end{array}\right)+\left(\begin{array}{cc}0& \frac{1}{6}{t}^3\\ \frac{1}{6}{t}^3& 0\end{array}\right)+\cdots }$$

$${\displaystyle e^{𝐀t}=\left(\begin{array}{cc}1+\frac{1}{2}{t}^2+\frac{1}{4!}{t}^4+\cdots & t+\frac{1}{3!}{t}^3+\frac{1}{5!}{t}^5+\cdots \\ t+\frac{1}{3!}{t}^3+\frac{1}{5!}{t}^5+\cdots & 1+\frac{1}{2}{t}^2+\frac{1}{4!}{t}^4+\cdots \end{array}\right)}$$

The even series is the power series for $\cosh t$ and the odd series is the power series for $\sinh t$:

Replacing $t$ with $-t$ (cosh is even, sinh is odd):

8.4b

Consider $𝐗\prime =𝐀𝐗$. If $𝐗=e^{𝐀t}𝐂$, then ${\displaystyle 𝐗\prime =\frac{d}{dt}\left(e^{𝐀t}𝐂\right)}$. To find ${\displaystyle \frac{d}{dt}\left(e^{𝐀t}\right)}$, we use the definition:

$${\displaystyle \frac{d}{dt}\left(e^{𝐀t}\right)=\frac{d}{dt}(𝐈+𝐀t+{𝐀}^2\frac{t^2}{2!}+{𝐀}^3\frac{t^3}{3!}+\cdots )}$$

Differentiate term by term ($𝐈$ is constant), then factor $𝐀$ out of every term:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle \frac{d}{dt}\left(e^{𝐀t}\right)}& {\displaystyle =𝐀+{𝐀}^{2}t+\frac{1}{2}{𝐀}^3{t}^2+\frac{1}{3!}{𝐀}^4{t}^3+\cdots }\hfill \\ \hfill {\displaystyle }& {\displaystyle =𝐀(𝐈+𝐀t+\frac{1}{2}{𝐀}^2{t}^2+\frac{1}{3!}{𝐀}^3{t}^3+\cdots )}\hfill \\ \hfill {\displaystyle }& {\displaystyle =𝐀e^{𝐀t}}\hfill \end{array}}$$

$𝐗\prime =𝐀e^{𝐀t}𝐂$, which is $𝐗\prime =𝐀𝐗$.

So for $𝐗\prime =𝐀𝐗$, with $𝐀$ containing constant entries, $𝐗=e^{𝐀t}𝐂$ is a solution. (Note: $𝐂$ is a column matrix of arbitrary coefficients.)

Example

Use the matrix exponential to find the general solution of the given system.

$$𝐗\prime =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)𝐗$$

Solution

This is the matrix from Example 1:

$$𝐀=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)⟹e^{𝐀t}=\left(\begin{array}{cc}\cosh t& \sinh t\\ \sinh t& \cosh t\end{array}\right)$$

Initial-value problems

$e^{𝐀t}=𝚽$ is the fundamental matrix for the system, so variation of parameters yields, the general solution

$${\displaystyle 𝐗=e^{𝐀t}𝐂+e^{𝐀t}{\int }_{t_0}^t{e}^{-𝐀s}𝐅(s)ds}$$

Here $e^{-𝐀s}={\left(e^{𝐀s}\right)}^{-1}$, playing the role of ${𝚽}^{-1}$.

Note: $e^{-𝐀s}$ is $e^{𝐀t}$ with $t$ replaced by $-s$. In our work, we will take $t_0=0$.

Example

Find the general solution of the given system.

$$𝐗\prime =\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)𝐗+\left(\begin{array}{c}3\\ -1\end{array}\right)$$

Solution

For a diagonal matrix, each power just raises the diagonal entries to that power:

$$𝐀=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)⟹{𝐀}^2=\left(\begin{array}{cc}1& 0\\ 0& 2^2\end{array}\right)⟹{𝐀}^3=\left(\begin{array}{cc}1& 0\\ 0& 2^3\end{array}\right)$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle e^{𝐀t}}& {\displaystyle =𝐈+𝐀t+\frac{1}{2}{𝐀}^2{t}^2+\cdots }\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+\left(\begin{array}{cc}t& 0\\ 0& 2t\end{array}\right)+\left(\begin{array}{cc}\frac{1}{2}{t}^2& 0\\ 0& \frac{2^2{t}^2}{2}\end{array}\right)+\left(\begin{array}{cc}\frac{1}{3!}{t}^3& 0\\ 0& \frac{2^3{t}^3}{3!}\end{array}\right)+\cdots }\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{cc}1+t+\frac{1}{2}{t}^2+\frac{1}{3!}{t}^3+\cdots & 0\\ 0& 1+2t+\frac{2^2{t}^2}{2}+\frac{2^3{t}^3}{3!}+\cdots \end{array}\right)}\hfill \end{array}}$$

The diagonal entries are the series for $e^t$ and $e^{2t}$:

$$e^{𝐀t}=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)$$

$${\displaystyle {𝐗}_c=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)𝐂;{𝐗}_p=e^{𝐀t}{\int }_{t_0}^t{e}^{-𝐀s}𝐅(s)ds}$$

Replace $t$ with $-s$ to get $e^{-𝐀s}$:

$$e^{-𝐀s}𝐅=\left(\begin{array}{cc}{e}^{-s}& 0\\ 0& e^{-2s}\end{array}\right)\left(\begin{array}{c}3\\ -1\end{array}\right)=\left(\begin{array}{c}3e^{-s}\\ -e^{-2s}\end{array}\right)$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle e^{𝐀t}{\int }_0^t\left(\begin{array}{c}3e^{-s}\\ -e^{-2s}\end{array}\right)ds}& {\displaystyle =e^{𝐀t}{\left(\begin{array}{c}-3e^{-s}\\ \frac{1}{2}{e}^{-2s}\end{array}\right)|}_0^t}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)\left(\begin{array}{c}-3e^{-t}+3\\ \frac{1}{2}{e}^{-2t}-\frac{1}{2}\end{array}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{c}-3+3e^t\\ \frac{1}{2}-\frac{1}{2}{e}^{2t}\end{array}\right)=\left(\begin{array}{c}-3\\ 1/2\end{array}\right)+\left(\begin{array}{c}3e^t\\ -\frac{1}{2}{e}^{2t}\end{array}\right)}\hfill \end{array}}$$

Add ${𝐗}_c$ and ${𝐗}_p$:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle 𝐗}& {\displaystyle =\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{2t}\end{array}\right)\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)+\left(\begin{array}{c}-3\\ 1/2\end{array}\right)+\left(\begin{array}{c}3e^t\\ -\frac{1}{2}{e}^{2t}\end{array}\right)}\hfill \\ \hfill {\displaystyle }& {\displaystyle =\left(\begin{array}{c}{c}_1{e}^t\\ c_2{e}^{2t}\end{array}\right)+\left(\begin{array}{c}-3\\ 1/2\end{array}\right)+\left(\begin{array}{c}3e^t\\ -\frac{1}{2}{e}^{2t}\end{array}\right)}\hfill \end{array}}$$

The $e^t$ terms combine, and so do the $e^{2t}$ terms. Since $c_1$ and $c_2$ are arbitrary, rename $c_3=c_1+3$ and ${\displaystyle c_4=c_2-\frac{1}{2}}$: