Variation of Parameters for Systems

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In this section we consider solving inhomogeneous systems of linear differential equations of the form

$\begin{equation} \label{eq:linihsys} \frac{dX}{dt} = A(t)X + G(t) \end{equation}$ where $A(t)=(a_{ij}(t))$ is an $n\times n$ matrix and $G(t)=(g_1(t),\ldots ,g_n(t))^t$ is a vector of continuous functions. The system is homogeneous when $G(t)=0$ .

We divide this discussion into two parts: finding the general solution of the homogeneous equation and finding a particular solution to the inhomogeneous equation using variation of parameters.

Homogeneous Nonconstant Coefficient Systems

From the theory of constant coefficient systems of linear differential equations, we know abstractly how to write a basis of solutions to the homogeneous system $\dot {X}=AX$ where $A$ is an $n\times n$ constant matrix. That basis is $X_j(t) = e^{tA}v_j,$ where $\{v_1,\ldots ,v_n\}$ is a basis of $\R ^n$ . For the homogeneous system of $n$ nonconstant coefficient linear differential equations

$\begin{equation} \label{E:NCCH} \dot{X} = A(t) X \end{equation}$ a similar statement is true. It should be noted, however, that only in special cases can (??) be solved in closed form.

Let $v_1,\ldots ,v_n$ be a basis for $\R ^n$ . Then there exist solutions $X_1(t),\ldots ,X_n(t)$ of (??) satisfying $X_j(0)=v_j$ and these solutions form a basis of solutions to (??).

Proof: The theory of differential equations implies that there exists a unique solution to (??) satisfying the initial condition $X(0)=X_0$ for any $X_0\in \R ^n$ . Therefore, there exist solutions $X_j(t)$ such that $X_1(0)=v_j$ . We claim that these solutions form a basis for the vector space of all solutions to (??).
Let $X(t)$ be any solution to (??) with initial condition $X(0)=X_0$ . Since $\{X_1(0),\ldots ,X_n(0)\}$ is a basis for $\R ^n$ , we can find scalars $\alpha _j$ such that $X_0 = \alpha _1X_1(0) + \cdots + \alpha _nX_n(0).$ Linearity of (??) implies that $\alpha _1X_1(t) + \cdots + \alpha _nX_n(t)$ is a solution. Therefore, uniqueness of solutions to (??) implies that $X(t) = \alpha _1X_1(t) + \cdots + \alpha _nX_n(t)$ for all time $t$ , which proves the proposition. $\blacksquare$

Existence of solutions to the initial value problem $X(t_0)=X_0$ also guarantees that the vectors $X_1(t_0),\ldots ,X_n(t_0)$ must be linearly independent. Indeed, since $X_1(t),\ldots ,X_n(t)$ is a basis of solutions to (??), we must be able to write $X_0 = \alpha _1X_1(t_0) + \cdots + \alpha _nX_n(t_0)$ for scalars $\alpha _1,\ldots ,\alpha _n$ in order for a solution to this initial value problem to exist. That is, the vectors $X_1(t_0),\ldots ,X_n(t_0)$ must span $\R ^n$ . Hence they are linearly independent. We have proved the following:

Let $X_1,\ldots ,X_n$ be a linearly independent set of solutions to (??). Then, for every $t$ , the matrix $\begin{equation} \label{E:Y(t)} Y(t) = \left(X_1(t)|\cdots |X_n(t)\right) \end{equation}$ is invertible.

Proof: Proposition ?? implies that the columns of $Y(t)$ are linearly independent. Hence $Y(t)$ is invertible. $\blacksquare$

Rather than relying on the general existence and uniqueness theory for systems of differential equations, there is a computationally direct way of proving Lemma ??. It is possible to show that the determinant of the matrix $Y(t)$ in (??) is nonzero and hence $Y(t)$ is invertible. This approach is carried out in Section ??

The Theory of Variation of Parameters

Let $X_1(t),\ldots ,X_n(t)$ be a basis of solutions to the homogeneous system $\dot {X}=A(t)X$ . In the method of variation of parameters we look for solutions to the inhomogeneous system (??) of the form $X(t) = c_1(t)X_1(t) + \cdots + c_n(t)X_n(t),$ where $c_1(t),\ldots ,c_n(t)$ are differentiable functions. In order to find a method for determining the functions $c_j$ , we assume that $X$ is a solution to (??).

Use the product rule to compute

$\begin{eqnarray*} \dps\frac{dX}{dt} & = &\sum_{j=1}^n\left(c_j\frac{dX_j}{dt} +\frac{dc_j}{dt} X_j\right)\\ & = & \sum_{j=1}^n\left( c_jAX_j+\frac{dc_j}{dt}X_j\right)\\ & = & AX+\sum_{j=1}^n \frac{dc_j}{dt}X_j. \end{eqnarray*}$

It follows that $X$ is a solution of (??) if and only if $\begin{equation} \label{eq:dcjXj} G(t) = \frac{dc_1}{dt}(t) X_1(t) + \cdots + \frac{dc_n}{dt}(t) X_n(t). \end{equation}$ So variation of parameters works if we can find functions $c_j$ that satisfy (??).

We claim that it is always possible to find functions $d_j(t)$ so that

$\begin{equation} \label{E:djXj} G(t) = d_1(t)X_1(t) + \cdots + d_n(t)X_n(t). \end{equation}$ To verify (??), note that (??) can be rewritten in matrix form as $Y(t) D(t) = G(t)$ where $Y(t)$ is defined in (??) and $D(t)=(d_1(t),\ldots ,d_n(t))^t.$ Using the invertibility of $Y(t)$ , as proved in Lemma ??, we find that $D(t) = Y(t)\inv G(t).$ It now follows that (??) is satisfied by integrating the differential equations $\dot {c}_j=d_j$ , where the functions $d_j(t)$ are defined in (??), to find the functions $c_j$ .

We summarize this discussion as follows.

Variation of Parameters: Systems Consider $\begin{equation} \label{eq:VPS} \begin{array}{rcl} \dps \frac{dX}{dt} & = & A(t)X+G(t)\\ X(t_0) & = & X_0. \end{array} \end{equation}$

Let $X_1(t),\ldots ,X_n(t)$ be a basis of solutions to the homogeneous differential equation $\dot {X}=A(t)X$ .
Let $Y(t)$ be defined as in (??) and define $D(t)=(d_1(t),\ldots ,d_n(t))^t$ by $\begin{equation} \label{e:D(t)} D(t) = Y(t)\inv G(t). \end{equation}$
Let the functions $c_1(t),\ldots ,c_n(t)$ satisfy the initial value problem $\begin{equation} \label{eq:cj0} \frac{dc_j}{dt}=d_j \AND c_1(t_0)X_1(t_0) + \cdots +c_n(t_0)X_n(t_0) = X_0. \end{equation}$

Then $X(t) = c_1(t)X_1(t) + \cdots + c_n(t)X_n(t)$ is the unique solution of (??).

Note that when $n\ge 3$ , the hand computation of $Y(t)\inv$ can be painful.

Examples with Constant Coefficient Matrix $A$ when $n=2$

We consider two examples of variation of parameters applied to forced systems of differential equations of the form $\dot {X} = AX + G(t).$

An Example with Two Real Simple Eigenvalues

Consider the initial value problem (??) where

$\begin{equation} \label{eq:exinhom2} A=\mattwo{-1}{-8}{-16}{7}, \quad G(t)= \vectwo{1-t}{-2-t}, \quad X_0 = \vectwo{-1}{-1}. \end{equation}$ Variation of parameters proceeds in four steps:

The eigenvalues and eigenvectors of $A$ can be found either by using MATLAB or by hand. The eigenvalues are $\lambda _1=-9$ and $\lambda _2=15$ , and the associated eigenvectors are $v_1=(1,1)^t$ and $v_2=(1,-2)^t$ . Therefore, solutions of the homogeneous equation are: $X_1(t) = e^{-9t}\vectwo {1}{1} \AND X_2(t) = e^{15t}\vectwo {1}{-2}.$

We compute the inverse $Y(t)\inv = \mattwo {e^{-9t}}{e^{15t}}{e^{-9t}}{-2e^{15t}}\inv = -\frac {1}{3e^{6t}}\mattwo {-2e^{15t}}{-e^{15t}}{-e^{-9t}}{e^{-9t}} = \frac {1}{3}\mattwo {2e^{9t}}{e^{9t}}{e^{-15t}}{-e^{-15t}}.$ The vector $D(t)$ is: $D(t) = Y(t)\inv G(t) = \frac {1}{3}\mattwo {2e^{9t}}{e^{9t}}{e^{-15t}}{-e^{-15t}}\vectwo {1-t}{-2-t} = \vectwo {-te^{9t}}{e^{-15t}}$

By assumption $X_0=c_1(0)X_1(0)+c_2(0)X_2(0).$ Therefore $\vectwo {-1}{-1} = c_1(0)\vectwo {1}{1} + c_2(0)\vectwo {1}{-2},$ from which it follows that $c_1(0)=-1$ and $c_2(0)=0$ . Since $\dot {c}_1 = d_1 = -te^{9t} \AND \dot {c}_2 = d_2 = e^{-15t},$ it follows that $\begin {array}{rclcl} c_1(t) & = & \int _0^t (-\tau ) e^{9\tau }d\tau - 1 & = & -\frac {1}{9}\left (t-\frac {1}{9}\right )e^{9t}-\frac {82}{81} \\ c_2(t) & = & \int _0^t e^{-15t}d\tau & = & \frac {1}{15}\left (1-e^{-15t}\right ). \end {array}$

The solution $X(t)$ of the initial value problem (??) is now given by

$\begin{eqnarray*} X(t) & = & c_1(t)e^{-9t}\vectwo{1}{1} + c_2(t)e^{15t}\vectwo{1}{-2} \\ & = & \left(\begin{array}{c} -\frac{1}{9}\left(t-\frac{1}{9}\right)- \frac{82}{81}e^{-9t}+\frac{1}{15}\left( e^{15t}-1\right)\\ -\frac{1}{9}\left(t-\frac{1}{9}\right)- \frac{82}{81}e^{-9t}-\frac{2}{15}\left( e^{15t}-1\right) \end{array}\right). \end{eqnarray*}$

An Example with a Nontrivial Jordan Block

Solve the system

$\begin{equation} \label{e:2djb} \begin{array}{rcl} \dot{X} & = & A X + (t,1)^t \\ X(0) & = & (1,-1)^t, \end{array} \end{equation}$ where $A = \mattwo {2}{1}{0}{2}.$

A basis for solutions to the homogeneous equation $\dot {X}=AX$ is $X_1(t) = e^{2t}\vectwo {1}{0} \AND X_2(t) = e^{2t}\vectwo {t}{1}.$

We compute $D(t) = e^{-2t}\mattwo {1}{t}{0}{1}\inv G(t) = e^{-2t}\mattwo {1}{-t}{0}{1}\vectwo {t}{1} = e^{-2t}\vectwo {0}{1}.$

Note that $X(0) = \vectwo {1}{-1} = c_1(0)\vectwo {1}{0} + c_2(0)\vectwo {0}{1}.$ So $c_1(0)=1$ and $c_2(0)=-1$ . Now solve the differential equations $\begin {array}{rclcl} \dot {c}_1 & = & d_1(t) & = & 0 \\ \dot {c}_2 & = & d_2(t) & = & e^{-2t}, \end {array}$ obtaining

$\begin{eqnarray*} c_1(t) & = & 1 \\ c_2(t) & = & -\frac{1}{2}(e^{-2t}+1). \end{eqnarray*}$

It follows from Theorem ?? that the solution to (??) is $X(t) = c_1(t)X_1(t) + c_2(t)X_2(t) = -\frac {1}{2}\left (\begin {array}{c} t(1+e^{2t})-2e^{2t} \\ e^{2t}+1 \end {array}\right ).$

Exercises

In Exercises ?? – ?? solve the given initial value problems by variation of parameters.

$\frac {dX}{dt}=\frac {1}{2}\mattwo {1}{1}{1}{1} X + \vectwo {e^t}{0},\quad X(0) = \vectwo {1}{-1}.$

$\frac {dX}{dt}=\mattwo {1}{3}{3}{1} X + \vectwo {2t}{t},\quad X(0) = \vectwo {0}{1}.$

Find the general solution to the system of differential equations $\frac {dX}{dt}= \mattwo {-\frac {1}{t}}{\frac {1}{t^2}}{1}{0}X+\vectwo {1}{3t}.$ Hint: Verify that $X_1(t) = \vectwoc {1}{t} \AND X_2(t) = \vectwo {\frac {1}{t^2}}{-\frac {1}{t}}$ are solutions to the homogeneous equation and then use variation of parameters.

Find the general solution to the system of differential equations $\frac {dX}{dt}= \mattwoc {0}{1}{\frac {3}{t^2}}{\frac {1}{t}}X+\vectwoc {t}{2t}.$ Hint: Verify that $X_1(t) = \vectwoc {t^3}{3t^2}\AND X_2(t) = \vectwoc {\frac {1}{t}}{-\frac {1}{t^2}}$ are solutions to the homogeneous equation and then use variation of parameters.