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\centerline{{\sc Theory of first-order systems.}}

\b


We will be looking at ODEs which can be written in the form \[\mathbf{x'}(t) = A\mathbf{x}(t) + \mathbf{g}(t),\] where $A$ is an $n \times n$ matrix and $\mathbf{g}(t)$ is an $n \times 1$ column vector.	If $\mathbf{g}(t)$ is the zero vector, then the system is \textit{homogeneous}.  In other words, the system
 \[\mathbf{x'}(t) = A\mathbf{x}(t)\] is homogeneous.

\b

\textbf{Theorem.} (Superposition Theorem)  If $\mathbf{f_1}$ and  $\mathbf{f_2}$ are solutions of the homogeneous system $\mathbf{x'}(t) = A\mathbf{x}(t)$, then for any constants $C_1$ and $C_2$, $C_1\mathbf{f_1} + C_2\mathbf{f_2}$ is also a solution of the homogeneous system.

\b

\textbf{Example.} The system \begin{eqnarray*} \d{x}{t} & = & -x -2y\\
\d{y}{t} & = & 3x + 4y
\end{eqnarray*}
can be written as $\mathbf{x'} = A\mathbf{x}$ where $A = \pmatrix{-1 & -2 \cr 3 & 4}$.	One solution is given by $\mathbf{f_1} = \pmatrix{x \cr y} = \pmatrix{3e^t-2e^{2t} \cr -3e^t + 3e^{2t}}$.  Another solution is given by $\mathbf{f_2} = \pmatrix{x \cr y} = \pmatrix{2e^t-2e^{2t} \cr -2e^t + 3e^{2t}}$.	It follows from the theorem that any function of the form $\mathbf{x} =  \pmatrix{C_1(3e^t-2e^{2t})+C_2(-2e^t-2e^{2t})\cr C_1(-3e^t + 3e^{2t})+C_2(2e^t + 3e^{2t})}$ is also a solution.

\b


The \textit{Wronskian} $W(\mathbf{f_1} , \cdots , \mathbf{f_n} (t)$ of the vector functions $\mathbf{f_1} , \cdots ,\mathbf{f_n}$ is the determinant of the matrix whose columns are the vectors $\mathbf{f_1}(t) , \cdots ,\mathbf{f_n}(t)$.

\b

\textbf{Example.} If $\mathbf{f_1} =  \pmatrix{3e^t-2e^{2t} \cr -3e^t + 3e^{2t}}$ and $\mathbf{f_2} = \pmatrix{2e^t-2e^{2t} \cr -2e^t + 3e^{2t}}$ then $W(\mathbf{f_1},\mathbf{f_t})(1) = det \pmatrix{3e-2e^2 & 2e -2e^2 \cr -3e+3e^2 & -2e+3e^2} = e^3$.

\b

\textbf{Theorem.} If $W(\mathbf{f_1} , \cdots , \mathbf{f_n} (t) \ne 0$ for some $t$ in an interval $I = (a,b)$, then the functions $\mathbf{f_1} , \cdots , \mathbf{f_n} (t)$ are linearly independent on $I$.  Furthermore, if each of the functions $\mathbf{f_1} , \cdots , \mathbf{f_n} (t)$ is a solution of an ODE
$\mathbf{x'}(t) = A\mathbf{x}(t)$ on $I$, then either $W(\mathbf{f_1} , \cdots , \mathbf{f_n} (t) \ne 0$ for all $t$ in $I$ or $W(\mathbf{f_1} , \cdots , \mathbf{f_n} (t) = 0$ for all $t$ in $I$.


\b

\textbf{Theorem.} Suppose that $\mathbf{x'}(t) = A\mathbf{x}(t)$ is a system of $n$ ODE's in $n$ unknowns $x_1, \cdots ,x_n$.  If $\mathbf{f_1}, \cdots , \mathbf{f_n}$ are linearly independent solutions of the ODE, then every solution has the form $C_1\mathbf{f_1} + \cdots + C_n\mathbf{f_n}$.


\b

We say that	 $C_1\mathbf{f_1} + \cdots + C_n\mathbf{f_n}$ is the \textit{general solution} of the ODE, and we say that $\{\mathbf{f_1}, \cdots , \mathbf{f_n}\}$ is a \textit{fundamental set of solutions}.

\b

\textbf{Example.} In the last example, we saw that $\pmatrix{3e^t-2e^{2t} \cr -3e^t + 3e^{2t}}$ and $\pmatrix{x \cr y} = \pmatrix{2e^t-2e^{2t} \cr -2e^t + 3e^{2t}}$ are solutions of the system \begin{eqnarray*} \d{x}{t} & = & -x -2y.\\
\d{y}{t} & = & 3x + 4y.
\end{eqnarray*}

These solutions are linearly independent, because we saw that their Wronskian is not $0$.	 Therefore, every solution of the system has the form	$\mathbf{x} =  \pmatrix{C_1(3e^t-2e^{2t})+C_2(2e^t-2e^{2t})\cr C_1(-3e^t + 3e^{2t})+C_2(-2e^t + 3e^{2t})}$.

\b


\textbf{Theorem.} Every solution of the nonhomogeneous ODE $\mathbf{x'}(t) = A\mathbf{x}(t) + \mathbf{g}(t)$ has the form $\mathbf{x} = C_1\mathbf{f_1} + \cdots + C_n\mathbf{f_n} + \mathbf{h}$, where $C_1\mathbf{f_1} + \cdots + C_n\mathbf{f_n}$ is the general solution of the homogeneous ODE $\mathbf{x'}(t) = A\mathbf{x}(t)$ and $\mathbf{h}$ is any solution of the nonhomogeneous ODE $\mathbf{x'}(t) = A\mathbf{x}(t) + \mathbf{g}(t)$.


\b

$ C_1f_1 + \cdots + C_nf_n + h_n$ is called the \textit{general solution} of the nonhomogeneous ODE.


\b

\{ Page 389 \hspace{1in} 6(a-c), 7(a-c) \\


\pagebreak

\centerline{{\sc Finding solutions with distinct real eigenvalues.}}

\b

Recall that the general solution of the ODE $\d{x}{t} = ax$, where $a$	is a constant,  is $x = Ce^{ta}$.  Therefore, a solution of the decoupled system \begin{eqnarray*} \d{x}{t} & =	& ax\\
                  \d{y}{t} & =	& by
\end{eqnarray*}
is given by $x = C_1e^{ta}$, $y = C_2e^{tb}$.

We can write this in matrix form as:
\b

The solution of the system $\pmatrix{x' \cr y'} = \pmatrix{a & 0 \cr 0 & b}\pmatrix{x \cr y}$ is $\pmatrix{x \cr y} = \pmatrix{e^{at} & 0 \cr 0 & e^{bt}}\pmatrix{C_1 \cr C_2}$.	Notice that $\pmatrix{e^{at} & 0 \cr 0 & e^{bt}} = \exp\pmatrix{at & 0 \cr 0 & bt}$.

\b

This generalizes to the case where the system is not decoupled.

\b

\textbf{Theorem.} Suppose that the $n \times n$ matrix $A$ has $n$ distinct eigenvalues.  Then the columns of the matrix $exp(tA)$ form a fundamental set of solutions of the ODE $\mathbf{x'} = A\mathbf{x}$.  Furthermore, the general solution of the ODE $\mathbf{x'} = A\mathbf{x}$ is given by $\mathbf{x} = exp(tA)\mathbf{C}$, where $\mathbf{C} = \pmatrix{C_1 \cr \vdots \cr C_n}$ is a constant column vector.

\b

\textbf{Examples.}

\begin{enumerate}


\item Solve the system \begin{eqnarray*} \d{x}{t} & = & -4x - 10y\\
                                         \d{y}{t} & = & 3 x + 7y
\end{eqnarray*}

\b

This system can be written as $\mathbf{x'} = A\mathbf{x}$ where $A$ is the matrix $A = \pmatrix{-4 & -10 \cr 3 & 7}$. The characteristic polynomial of $A$ is $p(\lambda) = det \pmatrix{-4 - \lambda & -10 \cr 3 & 7 - \lambda} = \lambda^2 -3\lambda + 2 = (\lambda - 1)(\lambda - 2)$.	Therefore, the eigenvalues are $\lambda = 1$ and $\lambda = 2$.  In particular, there are two distinct eigenvalues.

Therefore, the general solution is given by $\mathbf{x} = exp(tA)\pmatrix{C_1 \cr C_2}$.	We must compute $exp(tA)$.	An eigenvector corresponding to $\lambda = 1$ is $\pmatrix{2 \cr -1}$.	 An eigenvector corresponding to $\lambda = 2$ is $\pmatrix{-5 \cr 3}$.	 Therefore, $P^{-1}AP = \pmatrix{1 & 0 \cr 0 & 2}$, where $P = \pmatrix{2 & -5 \cr -1 & 3}$ (so $P^{-1} = \pmatrix{3 & 5 \cr 1 & 2}$.)	 Therefore, $A = P\pmatrix{1 & 0 \cr 0 & 2}P^{-1}$, so $tA = P\pmatrix{t & 0 \cr 0 & 2t}P^{-1}$.	 This gives $\exp(tA) = P \exp\pmatrix{t & 0 \cr 0 & 2t}P^{-1} = \pmatrix{2 & -5 \cr -1 & 3}\pmatrix{e^t & 0 \cr 0 & e^{2t}} \pmatrix{3 & 5 \cr 1 & 2} = \pmatrix{6e^t - 5e^{2t} & 10e^t -10e^2t \cr -3e^t + 3e^{2t} & -5e^t + 6e^{2t}}$.	Therefore, the general solution is given by \[\pmatrix{x \cr y} = \pmatrix{6e^t - 5e^{2t} & 10e^t -10e^2t \cr -3e^t + 3e^{2t} & -5e^t + 6e^{2t}}\pmatrix{C_1 \cr C_2},\] or \[ x = C_1(6e^t - 5e^{2t}) + C_2(10e^t -10e^2t),\] \[y = C_1(-3e^t + 3e^{2t}) + C_2(-5e^t + 6e^{2t}).\]


\item Solve the initial value problem \begin{eqnarray*} \d{x}{t} & = & -4x - 10y,\\
                                         \d{y}{t} & = & 3 x + 7y,
\end{eqnarray*} $x(0) = 1$, $y(0) = -1$.

\b

We know from the last example that the general solution of the ODE is given by	\[ x = C_1(6e^t - 5e^{2t}) + C_2(10e^t -10e^2t),\] \[y = C_1(-3e^t + 3e^{2t}) + C_2(-5e^t + 6e^{2t}).\]	From the initial conditions we get \begin{eqnarray*}
C_1 & = & 1 .\\
C_2 & = & -1.
\end{eqnarray*}

Therefore, the solution is given by \[ x = (6e^t - 5e^{2t}) -(10e^t -10e^{2t}) = -4e^t + 5e^{2t},\] \[y = (-3e^t + 3e^{2t}) -(-5e^t + 6e^{2t}) = 2e^t-3e^{2t}.\]

\b

We can check that this works.




\end{enumerate}



\b

When we first discussed systems, we looked at how to get a system of $2$ first order equations from a second order equation.  Let's compare two methods of solving.

\b

\textbf{Example.} Solve $\dd{x}{t} -5\d{x}{t} + 6x = 0$.

\b

Old method:	 The characteristic polynomial is $p(\lambda) = \lambda^2 - 5\lambda + 6 = (\lambda - 2)(\lambda - 3)$, so the general solution is $x = C_1e^{2t} + C_2e^{3t}$.

\b

Using systems: This equation corresponds to the system \begin{eqnarray*} \d{x}{t} & =  & y. \\
\d{y}{t}& = & -6x + 5y.
\end{eqnarray*}

 This system can be written as $\mathbf{x'} = A \mathbf{x}$, where $A = \pmatrix{ 0 & 1 \cr -6 & 5 }$.	The characteristic polynomial of $A$ is $p(\lambda) = \lambda^2 - 5\lambda + 6 = (\lambda - 2)(\lambda - 3)$ (which is the same as in the other method).  Eigenvectors corresponding to $2$ and $3$ respectively are $\pmatrix{1 \cr 2}$ and $\pmatrix{1 & 3}$ respectively.  Therefore, $\pmatrix{2 & 0 \cr 0 & 3} = P^{-1}AP$ where $P = \pmatrix{1 & 1 \cr 2 & 3}$, so $P^{-1} = \pmatrix{3 & -1 \cr -2 & 1}$, so $tA = P\pmatrix{2t & 0 \cr 0 & 3t}P^{-1}$.	 The general solution of the ODE is $\pmatrix{x \cr y} = exp(tA)\pmatrix{C_1 \cr C_2} = P \left(exp\pmatrix{2t & 0 \cr 0 & 3t}\right)P^{-1}\pmatrix{C_1 \cr C_2} = \pmatrix{1 & 1 \cr 2 & 3}\pmatrix{e^{2t} & 0 \cr 0 & e^{3t}}\pmatrix{3 & -1 \cr -2 & 1}\pmatrix{C_1 \cr C_2} =	\pmatrix{C_1(3e^{2t}-2e^{3t}) -C_2(e^{2t}-e^{3t}) \cr C_1(6e^{2t}-6e^{3t})-C_2(2e^{2t}-3e^{3t})}.$


\b

Notice that the solution we get by this method, in addition to being more work, comes up with an answer which looks different (but in fact is not) from the other method.

\b

\{ Page 398 \hspace{1in} 1-4, 15-17	 \\


\pagebreak

\centerline{{\sc Finding solutions with distinct complex eigenvalues.}}


\b

If all eigenvalues of the matrix $A$ are distinct, even if some of them are complex, we can again use the exponential function to find the general solution of $\mathbf{x'} = A \mathbf{x}$.  Again, we use Euler's formula: $e^{i\theta} = \cos \theta + i\sin \theta$.

\b

\textbf{Example.}




 Solve the system \begin{eqnarray*}
\d{x}{t} & = & x - 2y \\
\d{y}{t} & = & x - y
\end{eqnarray*}


\b

This system can be written as $\mathbf{x'} = A \mathbf{x}$ where $A = \pmatrix{1 & -2 \cr 1 & -1}$.  The characteristic polynomial of $A$ is $p(\lambda) = \lambda^2 + 1$, which has roots $\pm i$.	Eigenvectors associated to $-i$ and $i$ are $\pmatrix{2 \cr 1+i}$ and $\pmatrix{2 \cr 1-i}$ respectively.	 If $P = \pmatrix{2 & 2 \cr 1+i & 1-i}$, then $P^{-1} = \pmatrix{\frac{1+i}{4} & \frac{-i}{2} \cr \frac{1-i}{4} & \frac{i}{2}}$. Then $P^{-1}AP = D$, where $D = \pmatrix{-i & 0 \cr 0 & i}$, so $A = PDP^{-1}$.	Therefore, $\exp(tA) = P \exp(tD)P^{-1} = P\pmatrix{e^{-it} & 0 \cr 0 & e^{it}}P^{-1} = $\\ $\pmatrix{(\frac{1+i}{2})e^{-it}+(\frac{1-i}{2})e^{it} & (e^{it}-e^{-it})i \cr \frac{(e^{-it}-e^{it})i}{2} & (\frac{1-i}{2})e^{-it}+(\frac{1+i}{2})e^{it}}.$

This expression can be simplified.	For example, $(\frac{1+i}{2})e^{-it}+(\frac{1-i}{2})e^{it}$ can be written as $2 \cos t + 2 \sin t$.  With such simplification, we get \[\exp(tA) = \pmatrix{\cos t + \sin t & 2 \sin t \cr - \sin t & \cos t - \sin t}.\] Therefore, the general solution is \[\pmatrix{x \cr y} = \pmatrix{\cos t + \sin t & 2 \sin t \cr - \sin t & \cos t - \sin t}\pmatrix{C_1 \cr C_2} = \pmatrix{C_1(\cos t + \sin t) + 2C_2 \sin t \cr C_1(- \sin t) +C_2( \cos t - \sin t)},\] or \[\pmatrix{x \cr y} = \pmatrix{(C_1 + 2C_2)\cos t + C_1 \sin t \cr C_2 \cos t - (C_1 + C_2) \sin t}.\]

\b

\{p. 409 \hspace{1in} 1(a), 2(a), 3(a), 9

















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