Systems and Conversion from Higher-Order Equations, differential equations lesson

A mass on a spring passes through its rest position. Where will it be a moment later? Its position alone cannot say, because the mass may be moving left or right, slowly or quickly, and only position and velocity together decide the motion. Collecting such quantities into one vector rewrites an equation of higher order as a system of first-order equations, here always with $t$ as the independent variable and a prime meaning $d/dt$ . This lesson shows how to make that conversion, how to carry the initial data across in the right order, how to write a linear system with a matrix, and how to tell autonomous systems from time-dependent ones.

Key terms

State vector

Definition Let $y$ be an unknown function of $t$ that solves a scalar equation of order $n$ . Its state vector is the column

X(t) = \begin{bmatrix} y(t) \\ y'(t) \\ \vdots \\ y^{(n-1)}(t) \end{bmatrix},

which lists the unknown and its first $n - 1$ derivatives in this fixed order. When the equation can be written as $y^{(n)} = g(t, y, y', \ldots, y^{(n-1)})$ with $g$ and its partial derivatives in the last $n$ arguments continuous near $(t_0, X(t_0))$ , the value $X(t_0)$ at one time determines the solution on an interval around $t_0$ , and that is what makes $X$ a state. More generally, a state vector is the ordered column $X = (x_1, \ldots, x_n)^{T}$ of the unknowns of a first-order system. Its entries $x_k$ are scalar functions and $X$ itself is a vector function, so the state $X$ is kept apart from the scalar unknown $y$ .

In plain words The state is the list of numbers you need at one instant to predict what happens next. For a second-order equation that is the value and the slope, and for a third-order equation the second derivative joins them.

Example The functions $\cos 2t$ and $\cos 2t + \sin 2t$ both solve $y'' + 4y = 0$ , and both equal $1$ at $t = 0$ . At $t = \pi/4$ the first equals $0$ and the second equals $1$ , so the value of $y$ alone at one instant does not fix the future. Their state vectors at $t = 0$ are $(1, 0)^{T}$ and $(1, 2)^{T}$ , and they differ in the velocity entry. For a mass on a spring the state is the column of position and velocity.

First-order system

Definition A first-order system of $n$ equations for unknown functions $x_1, \ldots, x_n$ of $t$ is

\begin{gathered} x_k' = f_k(t, x_1, \ldots, x_n), \\[0.4em] k = 1, \ldots, n, \end{gathered}

written compactly as $X' = F(t, X)$ with $X = (x_1, \ldots, x_n)^{T}$ and $F = (f_1, \ldots, f_n)^{T}$ , where $F$ is defined on an open set $D$ of points $(t, X)$ with $X$ in $\mathbb{R}^n$ . A vector function is differentiated entry by entry. A solution on an open interval $I$ is a differentiable vector function $X$ such that $(t, X(t))$ lies in $D$ and $X'(t) = F(t, X(t))$ for every $t$ in $I$ . Each equation contains one first derivative, but its right side may involve every unknown, so the equations are usually coupled.

In plain words Each unknown gets its own rate equation, and the rate of one unknown may depend on all the others. Solving the system means finding all the unknowns at once.

Example The system $x_1' = x_2$ , $x_2' = -x_1$ is coupled, since each rate involves the other unknown. The vector function $X(t) = (\sin t, \cos t)^{T}$ solves it on $\mathbb{R}$ , because $(\sin t)' = \cos t = x_2$ and $(\cos t)' = -\sin t = -x_1$ . Right sides need not be linear, as in the predator and prey model $x_1' = x_1(1 - x_2)$ , $x_2' = x_2(x_1 - 1)$ .

Conversion of a higher-order equation to a system

Definition Suppose an equation of order $n$ has been solved for its highest derivative,

y^{(n)} = g\big(t, y, y', \ldots, y^{(n-1)}\big).

Set $x_1 = y$ , $x_2 = y'$ and so on up to $x_n = y^{(n-1)}$ . The definitions give the first $n - 1$ equations and the original equation gives the last,

\begin{aligned} x_k' &= x_{k+1} \quad (k < n), \\[0.4em] x_n' &= g(t, x_1, \ldots, x_n). \end{aligned}

The two problems are equivalent on every interval $I$ . If $y$ solves the equation on $I$ , then $X = (y, y', \ldots, y^{(n-1)})^{T}$ solves the system on $I$ . Conversely, if $X$ solves the system on $I$ , the first $n - 1$ rows force $x_2 = x_1'$ , $x_3 = x_1''$ and so on up to $x_n = x_1^{(n-1)}$ , so $y = x_1$ is $n$ times differentiable and the last row is the original equation. An equation written as $a_n(t)\,y^{(n)} + \cdots = \ldots$ must first be divided by $a_n(t)$ , which is possible only where $a_n(t) \ne 0$ . The conversion is then made separately on each interval where the leading coefficient does not vanish.

In plain words Give names to the unknown and to each derivative below the order. Most rows of the system only say that one name is the derivative of the previous one, and the last row carries the equation itself.

Example Take $y''' + 2y'' - y' - 2y = 0$ . Solving for the highest derivative gives

y''' = 2y + y' - 2y''.

With $x_1 = y$ , $x_2 = y'$ and $x_3 = y''$ , the system is

\begin{aligned} x_1' &= x_2, \\ x_2' &= x_3, \\ x_3' &= 2x_1 + x_2 - 2x_3. \end{aligned}

Since the leading coefficient is $1$ , the weights in the last row are the negatives of the other coefficients of the original equation, because those terms moved across the equals sign. The nonlinear pendulum equation $\theta'' + \sin\theta = 0$ converts the same way. With $x_1 = \theta$ and $x_2 = \theta'$ it becomes $x_1' = x_2$ , $x_2' = -\sin x_1$ .

Vector initial data

Definition An initial condition for $X' = F(t, X)$ prescribes the whole state at one time, $X(t_0) = X_0$ with $X_0$ in $\mathbb{R}^n$ . For a converted equation of order $n$ , the scalar conditions $y^{(j)}(t_0) = k_j$ for $j = 0, 1, \ldots, n - 1$ become

X(t_0) = (k_0, k_1, \ldots, k_{n-1})^{T},

in the same order as the entries of $X$ . The existence and uniqueness theorem carries over to systems. If $F$ and all the partial derivatives $\partial f_i/\partial x_j$ are continuous on an open set containing $(t_0, X_0)$ , the problem has a solution on some open interval containing $t_0$ , and two solutions agree on every interval containing $t_0$ where both are defined. That interval may be smaller than the set of times where $F$ is defined.

In plain words One vector replaces the list of starting values, and its entries have to line up with the entries of the state. An equation of order $n$ needs exactly $n$ starting numbers, the value and the first $n - 1$ derivatives at one time.

Example For the third-order equation above, take $y(0) = 2$ , $y'(0) = 1$ and $y''(0) = -1$ , so that $X(0) = (2, 1, -1)^{T}$ . The characteristic polynomial $r^3 + 2r^2 - r - 2 = (r - 1)(r + 1)(r + 2)$ suggests the candidate

y = e^t + 2e^{-t} - e^{-2t}.

Its derivatives are

\begin{aligned} y' &= e^t - 2e^{-t} + 2e^{-2t}, \\[0.4em] y'' &= e^t + 2e^{-t} - 4e^{-2t}, \\[0.4em] y''' &= e^t - 2e^{-t} + 8e^{-2t}. \end{aligned}

At $t = 0$ the entries of $(y, y', y'')^{T}$ are $1 + 2 - 1 = 2$ , then $1 - 2 + 2 = 1$ and $1 + 2 - 4 = -1$ , which is the initial vector. In $y''' + 2y'' - y' - 2y$ the coefficients of $e^t$ add up to $1 + 2 - 1 - 2 = 0$ , those of $e^{-t}$ to $-2 + 4 + 2 - 4 = 0$ and those of $e^{-2t}$ to $8 - 8 - 2 + 2 = 0$ . So the scalar equation holds for every $t$ , which is the last row of the system, and the first two rows hold because each entry of $X$ is the derivative of the one before. Listing the data as $(2, -1, 1)^{T}$ instead would pose the problem with $y'(0) = -1$ and $y''(0) = 1$ , whose solution is a different function.

Matrix form of a linear system

Definition A first-order system is linear when each right side is a linear combination of the unknowns plus a term that depends on $t$ alone. It can then be written

X' = A(t)\,X + G(t),

where the coefficient matrix $A(t)$ is an $n \times n$ matrix of functions of $t$ and the forcing $G(t)$ is a column of functions. Entry $i$ of the product $A(t)X$ is $a_{i1}(t)x_1 + \cdots + a_{in}(t)x_n$ , so row $i$ of the matrix holds the weights of equation $i$ . The system is homogeneous when $G(t) = 0$ for every $t$ in the interval. If $A$ and $G$ are continuous on an open interval $I$ , then for every $t_0$ in $I$ and every $X_0$ the problem with $X(t_0) = X_0$ has exactly one solution, and that solution is defined on all of $I$ . Converting the normalized linear equation $y^{(n)} + p_{n-1}(t)\,y^{(n-1)} + \cdots + p_0(t)\,y = g(t)$ gives a matrix with ones just above the diagonal, the last row $(-p_0, -p_1, \ldots, -p_{n-1})$ and zeros elsewhere, together with the forcing $G = (0, \ldots, 0, g)^{T}$ .

In plain words The matrix is a compact table of the weights in the equations. Multiplying it into the state column rebuilds every equation, one row at a time.

Example The third-order example has $G = 0$ and

A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 2 & 1 & -2 \end{bmatrix}.

Multiplying out gives $AX = (x_2, x_3, 2x_1 + x_2 - 2x_3)^{T}$ , the three right sides found before. The coefficients are constant, so every solution exists for all real $t$ . For the forced equation $y'' + 3y' + 2y = \cos t$ , the state $X = (y, y')^{T}$ gives

\begin{aligned} X' &= \begin{bmatrix} 0 & 1 \\ -2 & -3 \end{bmatrix} X \\[0.4em] &\quad + \begin{bmatrix} 0 \\ \cos t \end{bmatrix}. \end{aligned}

Autonomous and time-dependent systems

Definition A system $X' = F(t, X)$ is autonomous if its right side does not depend on $t$ explicitly, so that it has the form $X' = F(X)$ . Otherwise it is time-dependent, also called nonautonomous. A linear system $X' = A(t)X + G(t)$ is autonomous exactly when $A$ and $G$ are constant. For an autonomous system the arrow $F(X)$ attached to a state is the same whenever that state is reached, and if $X(t)$ is a solution, then so is the shifted function $X(t - c)$ for every constant $c$ , on the shifted interval.

In plain words In an autonomous system the rule for change looks only at the current state, never at the clock. In a time-dependent system the same state can move differently at different times, because a coefficient or an outside input has changed.

Example The constant-matrix system above and the pendulum system $x_1' = x_2$ , $x_2' = -\sin x_1$ are autonomous. The equation $y'' + (1 + t)\,y' + y = 0$ , whose damping coefficient grows with time, gives

A(t) = \begin{bmatrix} 0 & 1 \\ -1 & -(1 + t) \end{bmatrix}.

At the state $(0, 1)^{T}$ the arrow $A(t)X$ is $(1, -1)^{T}$ at $t = 0$ but $(1, -2)^{T}$ at $t = 1$ , so the system is time-dependent. The forced equation $y'' + 3y' + 2y = \cos t$ is time-dependent for another reason. Its matrix is constant, but its forcing changes with $t$ .

Check yourself

Write $y'' + 5y' + 6y = 0$ with $y(0) = 1$ and $y'(0) = -4$ as a system $X' = AX$ with an initial vector, and check that $y = 2e^{-3t} - e^{-2t}$ gives a solution of the system.

Show the answer

Solving for the highest derivative gives $y'' = -6y - 5y'$ . With $X = (y, y')^{T}$ ,

\begin{gathered} X' = \begin{bmatrix} 0 & 1 \\ -6 & -5 \end{bmatrix} X, \\[0.4em] X(0) = \begin{bmatrix} 1 \\ -4 \end{bmatrix}. \end{gathered}

The candidate has $y' = -6e^{-3t} + 2e^{-2t}$ and $y'' = 18e^{-3t} - 4e^{-2t}$ . The first row says $x_1' = x_2$ , which holds because the second entry of $X$ is the derivative of the first. The second row needs $x_2' = y''$ to equal

\begin{aligned} -6y - 5y' &= (-12 + 30)\,e^{-3t} \\[0.4em] &\quad + (6 - 10)\,e^{-2t}, \end{aligned}

which is $18e^{-3t} - 4e^{-2t}$ , so it holds too. At $t = 0$ the state is $(2 - 1, -6 + 2)^{T} = (1, -4)^{T}$ , the required initial vector.

Write $(t - 1)\,y'' - t\,y' + y = 0$ with $y(0) = 1$ and $y'(0) = 2$ as a first-order system. Is the system autonomous, and on which interval does the existence theorem for linear systems guarantee exactly one solution?

Show the answer

The leading coefficient $t - 1$ vanishes at $t = 1$ , so the equation can be solved for $y''$ only where $t \ne 1$ , giving $y'' = (t\,y' - y)/(t - 1)$ . With $X = (y, y')^{T}$ ,

X' = \begin{bmatrix} 0 & 1 \\[0.2em] -\dfrac{1}{t - 1} & \dfrac{t}{t - 1} \end{bmatrix} X,

with $X(0) = (1, 2)^{T}$ . The matrix depends on $t$ , so the system is time-dependent. Its entries are continuous on $(-\infty, 1)$ and on $(1, \infty)$ , and the initial time $0$ lies in the first interval, so the theorem guarantees exactly one solution on $(-\infty, 1)$ . As a check, $y = e^t + t$ has $y(0) = 1$ , $y'(0) = 2$ and $(t - 1)e^t - t(e^t + 1) + e^t + t = 0$ , so it is that solution.

The equation $y''' - y' = 0$ with $y(0) = 0$ , $y'(0) = 1$ and $y''(0) = 0$ is converted with the state $X = (y, y', y'')^{T}$ . A classmate writes the initial vector as $(0, 0, 1)^{T}$ . Solve the correct problem and the classmate’s problem, and compare the solutions.

Show the answer

The system is $x_1' = x_2$ , $x_2' = x_3$ , $x_3' = x_2$ , and the correct initial vector is $(0, 1, 0)^{T}$ . The characteristic polynomial $r^3 - r = r(r - 1)(r + 1)$ gives

y = C_1 + C_2e^t + C_3e^{-t},

with $y' = C_2e^t - C_3e^{-t}$ and $y'' = C_2e^t + C_3e^{-t}$ . The correct data give $C_1 + C_2 + C_3 = 0$ , $C_2 - C_3 = 1$ and $C_2 + C_3 = 0$ , so $C_2 = \tfrac{1}{2}$ , $C_3 = -\tfrac{1}{2}$ , $C_1 = 0$ and $y = \sinh t$ . The classmate’s vector means $y'(0) = 0$ and $y''(0) = 1$ , which give $C_2 = C_3 = \tfrac{1}{2}$ and $C_1 = -1$ , so $y = \cosh t - 1$ . The two functions agree only at $t = 0$ , since $\sinh t = \cosh t - 1$ reduces to $e^{-t} = 1$ . Swapping two entries of the initial vector poses a different problem.

Systems and Conversion from Higher-Order Equations

Key terms

State vector

First-order system

Conversion of a higher-order equation to a system

Vector initial data

Matrix form of a linear system

Autonomous and time-dependent systems

Check yourself

How this lesson connects

Builds on

Used later in