Initial Value Problems Revisited

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To summarize the ideas developed in this chapter, we review the method that we have developed to solve the system of differential equations

$\begin{equation} \label{E:2dode} \begin{array}{rcl} \dot{x} & = & ax+by \\ \dot{y} & = & cx+dy \end{array} \end{equation}$ satisfying the initial conditions $\begin{equation} \label{E:2dic} \begin{array}{rcl} x(0) & = & x_0 \\ y(0) & = & y_0. \end{array} \end{equation}$

Begin by rewriting (??) in matrix form

$\begin{equation} \label{E:2dodeM} \dot{X} = CX \end{equation}$ where $C = \mattwo {a}{b}{c}{d} \AND X(t)=\vectwo {x(t)}{y(t)}.$ Rewrite the initial conditions (??) in vector form $\begin{equation} \label{E:2dicV} X(0) = X_0 \end{equation}$ where $X_0=\vectwo {x_0}{y_0}.$

When the eigenvalues of $C$ are real and distinct we now know how to solve the initial value problem (??) and (??). This solution is found in four steps.

Find the eigenvalues $\lambda _1$ and $\lambda _2$ of $C$ .

These eigenvalues are the roots of the characteristic polynomial as given by (??): $p_C(\lambda )=\lambda ^2-\trace (C)\lambda +\det (C).$ These roots may be found either by factoring $p_C$ or by using the quadratic formula. The roots are real and distinct when the discriminant $D=\trace (C)^2-4\det (C)>0.$ Recall (??) and Theorem ??.

Find eigenvectors $v_1$ and $v_2$ of $C$ associated with the eigenvalues $\lambda _1$ and $\lambda _2$ .

For $j=1$ and $j=2$ , the eigenvector $v_j$ is found by solving the homogeneous system of linear equations

$\begin{equation} \label{E:eeqn} (C-\lambda_j I_2)v = 0 \end{equation}$ for one nonzero solution. Lemma ?? tells us that there is always a nonzero solution to (??) since $\lambda _j$ is an eigenvalue of $C$ .

Step 3: Using superposition, write the general solution to the system of ODEs (??) as

$\begin{equation} \label{E:gensoln} X(t) = \alpha_1e^{\lambda_1 t}v_1 + \alpha_2 e^{\lambda_2 t}v_2, \end{equation}$ where $\alpha _1,\alpha _2\in \R$ .

Theorem ?? tells us that for $j=1,2$ $X_j(t) = e^{\lambda _j t}v_j$ is a solution to (??). The principle of superposition (see Section ??) allows us to conclude that $X(t) = \alpha _1X_1(t) + \alpha _2X_2(t)$ is also a solution to (??) for any scalars $\alpha _1,\alpha _2\in \R$ . Thus, (??) is valid.

Note that the initial condition corresponding to the general solution (??) is

$\begin{equation} \label{E:geninit} X(0) = \alpha_1v_1 + \alpha_2v_2, \end{equation}$ since $e^0=1$ .

Step 4: Solve the initial value problem by solving the system of linear equations

$\begin{equation} \label{E:geninit2} X_0 = \alpha_1v_1 + \alpha_2v_2 \end{equation}$ for $\alpha _1$ and $\alpha _2$ (see (??)).

Let $A$ be the $2\times 2$ matrix whose columns are $v_1$ and $v_2$ . That is,

$\begin{equation} \label{E:Av1v2} A = (v_1|v_2). \end{equation}$ Then we may rewrite (??) in the form $\begin{equation} \label{E:solveinit} A\vectwo{\alpha_1}{\alpha_2} = X_0. \end{equation}$

We claim that the matrix $A=(v_1|v_2)$ (defined in (??)) is always invertible. Recall Lemma ?? which states that if $w$ is a nonzero multiple of $v_2$ , then $w$ is also an eigenvector of $A$ associated to the eigenvalue $\lambda _2$ . Since the eigenvalues $\lambda _1$ and $\lambda _2$ are distinct, it follows that the eigenvector $v_1$ is not a scalar multiple of the eigenvector $v_2$ (see Lemma ??). Therefore, the area of the parallelogram spanned by $v_1$ and $v_2$ is nonzero and the determinant of $A$ is nonzero by Theorem ?? of Chapter ??. Corollary ?? of Chapter ?? now implies that $A$ is invertible. Thus, the unique solution to (??) is $\vectwo {\alpha _1}{\alpha _2} = A\inv X_0.$ This equation is easily solved since we have an explicit formula for $A\inv$ when $A$ is a $2\times 2$ matrix (see (??) in Section ??). Indeed, $A\inv = \frac {1}{\det (A)}\mattwo {d}{-b}{-c}{a}.$

An Initial Value Problem Solved by Hand

Solve the linear system of differential equations

$\begin{equation} \label{E:ivpbh} \begin{array}{rcl} \dot{x} & = & 3x-y \\ \dot{y} & = & 4x-2y \end{array} \end{equation}$ with initial conditions $\begin{equation} \label{E:ivpbhic} \begin{array}{rcc} x(0) & = & 2 \\ y(0) & = & -3. \end{array} \end{equation}$

Rewrite the system (??) in matrix form as $\dot {X} = CX$ where $C = \mattwo {3}{-1}{4}{-2}.$ Rewrite the initial conditions (??) in vector form $X(0) = X_0 = \vectwo {2}{-3}.$

Now proceed through the four steps outlined previously.

Find the eigenvalues of $C$ .

The characteristic polynomial of $C$ is $p_C(\lambda ) = \lambda ^2 -\trace (C)\lambda +\det (C) = \lambda ^2-\lambda -2 = (\lambda -2)(\lambda +1).$ Therefore, the eigenvalues of $C$ are $\lambda _1 = 2 \AND \lambda _2=-1.$

Find the eigenvectors of $C$ .

Find an eigenvector associated with the eigenvalue $\lambda _1 = 2$ by solving the system of equations

$\begin{align*} (C-\lambda_1I_2)v &= \left(\mattwo{3}{-1}{4}{-2}-\mattwo{2}{0}{0}{2}\right)v \\ &= \mattwo{1}{-1}{4}{-4}v = 0. \end{align*}$

One particular solution to this system is $v_1 = \vectwo {1}{1}.$

Similarly, find an eigenvector associated with the eigenvalue $\lambda _2 = -1$ by solving the system of equations

$\begin{align*} (C-\lambda_2I_2)v &= \left(\mattwo{3}{-1}{4}{-2}-\mattwo{-1}{0}{0}{-1}\right)v \\ &= \mattwo{4}{-1}{4}{-1}v = 0. \end{align*}$

One particular solution to this system is $v_2 = \vectwo {1}{4}.$

Step 3: Write the general solution to the system of differential equations.

Using superposition the general solution to the system (??) is: $X(t) = \alpha _1e^{2t}v_1 + \alpha _2e^{-t}v_2 = \alpha _1e^{2t}\vectwo {1}{1} + \alpha _2e^{-t}\vectwo {1}{4},$ where $\alpha _1,\alpha _2\in \R$ . Note that the initial state of this solution is: $X(0) = \alpha _1\vectwo {1}{1}+\alpha _2\vectwo {1}{4} = \vectwoc {\alpha _1 + \alpha _2}{\alpha _1+4\alpha _2}.$

Step 4: Solve the initial value problem.

Let $A=(v_1|v_2) = \mattwo {1}{1}{1}{4}.$ The equation for the initial condition is $A\vectwo {\alpha _1}{\alpha _2} = X_0.$ See (??).

We can write the inverse of $A$ by formula as $A\inv = \frac {1}{3}\mattwo {4}{-1}{-1}{1}.$ It follows that we solve for the coefficients $\alpha _j$ as $\vectwo {\alpha _1}{\alpha _2} = A\inv X_0 = \frac {1}{3}\mattwo {4}{-1}{-1}{1} \vectwo {2}{-3} = \frac {1}{3} \vectwo {11}{-5}.$ In coordinates $\alpha _1 = \frac {11}{3} \AND \alpha _2=-\frac {5}{3}.$

The solution to the initial value problem (??) and (??) is:

$\begin{align*} X(t) &= \frac{1}{3} \left(11e^{2t}v_1 - 5e^{-t}v_2\right) \\ &= \frac{1}{3} \left(11e^{2t}\vectwo{1}{1} - 5e^{-t}\vectwo{1}{4} \right). \end{align*}$

Expressing the solution in coordinates, we obtain:

$\begin{eqnarray*} x(t) & = & \frac{1}{3} \left(11e^{2t} - 5e^{-t} \right)\\ y(t) & = & \frac{1}{3} \left(11e^{2t} -20e^{-t}\right). \end{eqnarray*}$

An Initial Value Problem Solved using MATLAB

Next, solve the system of ODEs

$\begin{eqnarray*} \dot{x} & = & 1.7x+3.5y \\ \dot{y} & = & 1.3x-4.6y \end{eqnarray*}$

with initial conditions

$\begin{eqnarray*} x(0) & = & 2.7 \\ y(0) & = & 1.1\;. \end{eqnarray*}$

Rewrite this system in matrix form as $\dot {X} = CX$ where $C = \mattwo {1.7}{3.5}{1.3}{-4.6}.$ Rewrite the initial conditions in vector form $X_0 = \vectwo {2.7}{1.1}.$

Now proceed through the four steps outlined previously. In MATLAB begin by typing

C  = [1.7 3.5; 1.3 -4.6]
 
X0 = [2.7; 1.1]

Find the eigenvalues of $C$ by typing

lambda = eig(C)

and obtaining

lambda =
2.3543
-5.2543

So the eigenvalues of $C$ are real and distinct.

To find the eigenvectors of $C$ we need to solve two homogeneous systems of linear equations. The matrix associated with the first system is obtained by typing

C1 = C - lambda(1)*eye(2)

which yields

C1 =
-0.6543 3.5000
1.3000 -6.9543

We can solve the homogeneous system $({\tt C1})x=0$ by row reduction — but MATLAB has this process preprogrammed in the command null. So type

v1 = null(C1)

and obtain

Similarly, to find an eigenvector associated to the eigenvalue $\lambda _2$ type

C2 = C - lambda(2)*eye(2);
 
v2 = null(C2)

and obtain

v2 =
-0.4496
0.8932

The general solution to this system of differential equations is: $X(t) = \alpha _1 e^{2.3543t}\vectwo {-0.9830}{-0.1838} + \alpha _2e^{-5.2543t}\vectwo {-0.4496}{0.8932}.$

Solve the initial value problem by finding the scalars $\alpha _1$ and $\alpha _2$ . Form the matrix $A$ by typing

A = [v1 v2]

Then solve for the $\alpha$ ’s by typing

alpha = inv(A)*X0

obtaining

alpha =
-3.0253
0.6091

Therefore, the closed form solution to the initial value problem is:

$\begin{align*} X(t) = & 3.0253 e^{2.3543t}\vectwo{0.9830}{0.1838} \\ + 0.6091 e^{-5.2543t}\vectwo{-0.4496}{0.8932}. \end{align*}$

Exercises

In Exercises ?? – ?? find the solution to the system of differential equations $\dot {X} = CX$ satisfying $X(0)=X_0$ .

$C = \mattwo {1}{1}{0}{2}$ and $X_0 = \vectwo {1}{4}$ .

$C = \mattwo {2}{-3}{0}{-1}$ and $X_0 = \vectwo {1}{-2}$ .

$C = \mattwo {-3}{2}{-2}{2}$ and $X_0 = \vectwo {-1}{3}$ .

$C = \mattwo {2}{1}{1}{2}$ and $X_0 = \vectwo {1}{2}$ .

Solve the initial value problem $\dot {X}=CX$ where $X_0=e_1$ given that

$X(t) = e^{-t}\vectwo {1}{2}$ is a solution,
$\trace (C)=3$ , and
$C$ is a symmetric matrix.

In Exercises ?? – ??, with MATLAB assistance, find the solution to the system of differential equations $\dot {X} = CX$ satisfying $X(0)=X_0$ .

$C = \mattwo {1.76}{4.65}{0.23}{1.11}$ and $X_0 = \vectwo {0.34}{-0.50}$ .

$C = \mattwo {1.23}{2\pi }{\pi /2}{1.45}$ and $X_0 = \vectwo {1.2}{1.6}$ .

In Exercises ?? – ??, find the solution to $\dot {X} = CX$ satisfying $X(0)=X_0$ in two different ways, as follows.

Use pplane5 to find $X(0.5)$ . Hint: Use the Specify a computation interval option in the PPLANE5 Keyboard input window to compute the solution to $t=0.5$ . Then use the zoom in square feature to determine an answer to three decimal places.
Next use MATLAB to find the eigenvalues and eigenvectors of $C$ and to find a closed form solution $X(t)$ . Use this formula to evaluate $X(0.5)$ to three decimal places.
Do the two answers agree?

$C = \mattwo {2.65}{-2.34}{-1.5}{-1.2} \AND X_0=\vectwo {0.5}{0.1}$ .

$C = \mattwo {1.2}{2.4}{0.6}{-3.5} \AND X_0=\vectwo {0.5}{0.7}$ .