7.5 Inhomogeneous Systems

1Reading¶

Material related to this page, as well as additional exercises, can be found in ALA 10.4 and ALA 10.6.

2Learning Objectives¶

By the end of this page, you should know:

the equation of an inhomogeneous system
the solution of a scalar inhomogeneous system
the solution of a vector inhomogeneous system

3Definition¶

Many systems of interest can be modeled by first-order inhomogeneous linear systems of ordinary differential equations of the form:

\dot{\vv x} = A\vv x + \vv f, \quad (\text{SYS})

(1)

where $A \in \mathbb{R}^{n \times n}$ , $\vv x(t) \in \mathbb{R}^n$ is the solution, and $\vv f(t) \in \mathbb{R}^n$ is a known vector-valued function of $t$ acting as an external force on the system.

4Solution for the Scalar¶

We rely on our work on general linear systems to know that the solution to (SYS) will have the general form given by the superposition:

\vv x(t) = \vv x^*(t) + \vv z(t),

(2)

where $\vv x^*(t)$ is a particular solution representing a response to the forcing, and $\vv z(t)$ is a solution to the corresponding homogeneous system $\dot{\vv z} = A\vv z$ .

Since we already know that $\vv z(t) = e^{At}\vv z(0)$ , all that’s left to work out is the particular solution $\vv x^*(t)$ in response to the forcing $\vv f(t)$ .

We’ll start, as before, by recalling the scalar case. To solve $\dot{ x} = a x + f$ , we set $x(t) = e^{at}v(t)$ , where $v(t)$ is a function to be determined. Differentiating, we compute

\dot{x} = ae^{at}v(t) + e^{at}\dot{v} = ax + e^{at}\dot{v},

(3)

from which we conclude $x(t)$ satisfies $\dot{x} = ax + f$ if and only if $\dot{v} = e^{-ta}f$ , i.e., if and only if $v(t) = \int_{t_0}^t e^{-\tau a}f(\tau)d\tau + v(t_0)$ , where $v(0) = e^{-t_0a}x(t_0)$ .

5Solution for the Vector¶

Perhaps unsurprisingly, we can extend this method to the vector setting via the matrix exponential.

We replace our initial guess with $\vv x(t) = e^{At}\vv v(t)$ , where $\vv v(t)$ is a vector-valued function to be determined. We compute the derivative $\dot{\vv x}$ as:

\dot{\vv x} = \frac{d}{dt}(e^{At}\vv v) = \frac{d}{dt}(e^{At})\vv v + e^{At}\frac{d\vv v}{dt} = Ae^{At}\vv v + e^{At}\dot{\vv v} = A\vv x + e^{At}\dot{\vv v}

(4)

from which we conclude that $\dot{\vv v} = e^{-At}\vv f$ . We can integrate both sides to obtain, via the Fundamental Theorem of Calculus, that:

\vv v(t) = \vv v(t_0) + \int_{t_0}^t e^{-A \tau}f(\tau)d\tau, \quad \text{where } \vv v(t_0) = e^{-t_0A}x(t_0)

(5)

This is the last piece we needed to write the general solution to the inhomogeneous initial value problem.

Theorem 1 (The solution to an inhomogeneous linear ODE)

The solution to the initial value problem $\dot{\vv x} = A\vv x + \vv f$ with initial conditions $\vv x(t_0) = \vv b$ is

\vv x(t) = e^{(t-t_0)A}\vv b + \int_{t_0}^t e^{(t-\tau)A}f(\tau)d\tau.

(6)

Here, $e^{(t - t_0) A}\vv b$ is the solution to the homogeneous system $\dot{\vv x} = A \vv x$ with the initial conditions $\vv x(t_0) = \vv b$ ; and $\int_{t_0}^t e^{(t-\tau)A}f(\tau)d\tau$ is the solution to the inhomogeneous system $\dot{\vv x} = A\vv x + \vv f$ with initial conditions $\vv x(t_0) = \vv 0$ (we proved this above). Adding the two will yield a solution satisfying both the initial conditions $\vv x(t_0) = \vv b$ as well as the equality $\dot{\vv x} = A \vv x + \vv f$ .

Example 1 (A

2\times 2

inhomogeneous system (ALA Example 10.36))

In this example, we’ll solve a $2\times 2$ inhomogeneous system using the theorem we just proved:

\begin{align*} \dot x_1 &= 2x_1 - x_2,\\ \dot x_2 &= 4x_1 - 3x_2 + e^t \end{align*}

(7)

with initial conditions $x_1(0) = 1, x_2(0) = 0$ . We can rewrite this sytem as

\dot{\vv x} = \bm 2 & -1\\4 & -3 \em \vv x + \bm 0 \\ e^t \em := A\vv x + f(t)

(8)

Recall that solving this system proceeds in two steps: first, we solve the homogeneous system $\dot{\vv x} = A\vv x$ with initial conditions $\vv {x}(0) = \bm 1\\0\em$ ; second, we find a particular solution to the inhomogeneous system $\dot{\vv x} = A\vv x + \vv f$ with initial conditions $\vv x(0) = \vv 0$ .

First, recall that the solution to the homogenous system with initial conditions $\bm 1\\0\em$ is found using the matrix exponential,

\vv z(t) = e^{At}\bm 1\\0 \em

(9)

Solving for the eigenvalues and eigenvectors of $A$ , we find that $A$ has

\lambda_1 = -2, \quad \vv{v_1} = \bm 1\\4\em,\\ \lambda_2 = 1, \quad \vv{v_2} = \bm 1\\1\em.

(10)

Hence $A$ is diagonalizable as $\bm 1&1\\4&1 \em \bm -2 &0\\0&1\em \bm 1&1\\4&1\em^{-1}$ . Recall from our cases for the matrix exponential that this implies that $e^{At} = \bm 1&1\\4&1 \em \bm e^{-2t} &0\\0&e^t\em \bm 1&1\\4&1\em^{-1}$ , and hence

\vv z(t) = \bm 1&1\\4&1 \em \bm e^{-2t} &0\\0&e^t\em \bm 1&1\\4&1\em^{-1}\bm 1\\0 \em = \bm \frac 4 3 e^t - \frac 1 3 e^{-2t} \\ \frac 4 3 e^t - \frac 4 3 e^{-2t} \em.

(11)

Second, the solution to the inhomogeneous solution with initial conditions $\vv 0$ is given by

\begin{align*} \vv{x^*}(t) &= \int_{0}^{t}{e^{A(t - \tau)}f(\tau)\ d\tau}\\ &= \int_{0}^{t}{\bm 1&1\\4&1 \em \bm e^{-2(t - \tau)} &0\\0&e^{t - \tau}\em \bm 1&1\\4&1\em^{-1}\bm 0\\e^\tau \em \ d\tau}\\ &= \int_{0}^{t}{\bm -\frac 1 3 e^t + \frac 1 3 e^{-2t + 3\tau} \\ -\frac 1 3 e^t + \frac 4 3 e^{-2t + 3\tau} \em \ d\tau}\\ &= \bm -\frac 1 3 te^{t} + \frac 1 9 (e^{t} - e^{-2t}) \\ -\frac 1 3 te^t + \frac 4 9 (e^t - e^{-2t}) \em. \end{align*}

(12)

By our theorem for the solution to an inhomogeneous linear ODE, the solution to $\dot{x} = A\vv x + \vv f$ with initial conditions $\bm 1\\0\em$ is the sum $\vv{x^*}(t) + \vv z(t)$ ,

\vv x(t) = \bm -\frac 1 3 te^t + \frac {13}{9} e^t - \frac 4 9 e^{-2t} \\ -\frac 1 3 te^t + \frac {16}{9} e^t - \frac{16}{9}e^{-2t}\em.

(13)