Nonhomogeneous and Time-Dependent Linear Systems, differential equations lesson

With $t$ as the independent variable and a prime meaning $d/dt$ , a homogeneous system $X' = A(t)X$ describes how states evolve on their own, and a fundamental matrix carries every initial state forward. Most models also have a prescribed input, such as a steady inflow or an applied force, which adds a term $b(t)$ to the right side. This lesson lets the constants in $X = \Phi(t)c$ vary with time and derives the variation of constants formula for systems. It then solves a forced system whose coefficient matrix depends on $t$ and whose answer contains an integral with no elementary antiderivative.

Key terms

Nonhomogeneous linear system

Definition A linear system of first-order equations has the form

X' = A(t)X + b(t),

where the state $X(t)$ has $n$ components, $A(t)$ is an $n \times n$ coefficient matrix and $b(t)$ is a column with $n$ entries, called the forcing or input. All entries are assumed continuous on an open interval $I$ . The system is nonhomogeneous when $b$ is not identically zero, and $X' = A(t)X$ is the associated homogeneous system. For every $t_0$ in $I$ and every initial state $X_0$ there is exactly one solution with $X(t_0) = X_0$ , and it exists on all of $I$ . If $X_p$ is one solution and $\Phi$ is a fundamental matrix of the homogeneous system, then every solution on $I$ is $X = \Phi(t)c + X_p$ for exactly one constant vector $c$ .

In plain words The matrix term says how the components act on one another, and the forcing is an input that does not depend on the state. A linear system cannot blow up while its coefficients and forcing stay continuous, so its solutions last as long as the interval does. Two solutions differ by a solution of the homogeneous system. The sum of two solutions is not a solution, because it satisfies the system with the forcing $2b(t)$ in place of $b(t)$ .

Example For the state $X = (x_1, x_2)^{T}$ , the equations $x_1' = tx_1 + 1$ and $x_2' = -x_2 + 2$ form the system with $b = (1, 2)^{T}$ and

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

Both are continuous on $\mathbb{R}$ , so every initial state at any time $t_0$ gives a solution on all of $\mathbb{R}$ . The constant $x_2 = 2$ and the function $x_2 = 2 - e^{-t}$ both solve the second equation, and their difference $e^{-t}$ solves the homogeneous equation $x_2' = -x_2$ .

Time-dependent coefficient matrix

Definition A coefficient matrix $A(t)$ is time dependent when its entries change with $t$ . If $A$ is continuous on an open interval $I$ , the homogeneous system $X' = A(t)X$ still has fundamental matrices on $I$ and a principal fundamental matrix $U(t, t_0)$ , but there is no general formula for them. The matrix $e^{tA(t)}$ is not a solution in general. The exponential of $\int_{t_0}^{t} A(\tau)\,d\tau$ is the principal fundamental matrix when $A(t)A(s) = A(s)A(t)$ for all $t$ and $s$ in $I$ , but not in general. The eigenvalues of $A(t)$ at each fixed $t$ do not determine how the solutions behave. In practice a fundamental matrix is found from special structure, such as a diagonal or triangular matrix, or it is computed numerically.

In plain words With a changing matrix, the shortcuts for constant matrices stop working. A fundamental matrix still exists and still carries every initial state forward, but finding it takes the structure of the particular system or a numerical method.

Example For $A(t) = \operatorname{diag}(t, -1)$ the equations separate. The solution of $x_1' = tx_1$ with $x_1(0) = 1$ is $e^{t^2/2}$ , and the solution of $x_2' = -x_2$ with $x_2(0) = 1$ is $e^{-t}$ , so

\Phi(t) = \begin{pmatrix} e^{t^2/2} & 0 \\ 0 & e^{-t} \end{pmatrix}

is the principal fundamental matrix at $0$ . Diagonal matrices commute, and $\Phi(t)$ is indeed the exponential of the integral $\operatorname{diag}(t^2/2, -t)$ . A second example shows that eigenvalues at frozen times do not decide whether solutions decay. For

A(t) = \begin{pmatrix} -1 & e^{2t} \\ 0 & -1 \end{pmatrix},

each matrix $A(t)$ has the single eigenvalue $-1$ . Yet $X(t) = (\tfrac12e^t, e^{-t})^{T}$ is a solution, because the second component satisfies $x_2' = -x_2$ and the derivative of the first component is $\tfrac12e^t = -\tfrac12e^t + e^{2t}e^{-t}$ . Its first component grows without bound.

Inverse fundamental matrix

Definition Let $\Phi$ be a fundamental matrix of $X' = A(t)X$ on $I$ . By Liouville’s formula $\det\Phi(t) \ne 0$ at every point of $I$ , so the inverse fundamental matrix $\Phi(t)^{-1}$ exists for every $t$ in $I$ . It is differentiable, and differentiating $\Phi\Phi^{-1} = I$ gives $(\Phi^{-1})' = -\Phi^{-1}A(t)$ . For each $t$ it converts a state into coordinates with respect to the columns of $\Phi(t)$ , by $c = \Phi(t)^{-1}X$ . In particular $c_0 = \Phi(t_0)^{-1}X_0$ gives the constants of the homogeneous solution with $X(t_0) = X_0$ . For a constant matrix and $\Phi = e^{tA}$ the inverse is $e^{-tA}$ , but for a time-dependent matrix $\Phi(t)^{-1}$ is generally not $\Phi(-t)$ . For a $2 \times 2$ matrix $M$ with rows $(\alpha, \beta)$ and $(\gamma, \delta)$ and nonzero determinant $\det M = \alpha\delta - \beta\gamma$ , the inverse $M^{-1}$ is the matrix with rows $(\delta, -\beta)$ and $(-\gamma, \alpha)$ divided by $\det M$ .

In plain words The inverse fundamental matrix reads off how much of each basis solution a state contains at a given time. It undoes $\Phi(t)$ at the same time $t$ , which is not the same as running time backward to $-t$ .

Example For $\Phi(t) = \operatorname{diag}(e^{t^2/2}, e^{-t})$ the inverse is $\operatorname{diag}(e^{-t^2/2}, e^t)$ , while $\Phi(-t) = \operatorname{diag}(e^{t^2/2}, e^t)$ has the wrong first entry for every $t \ne 0$ . For the oscillator matrix $A$ with rows $(0, 1)$ and $(-4, 0)$ , the principal fundamental matrix $e^{tA}$ has rows $(\cos 2t, \tfrac12\sin 2t)$ and $(-2\sin 2t, \cos 2t)$ and determinant $1$ . The formula above keeps the equal diagonal entries and changes the signs of the other two, so $e^{-tA}$ has rows $(\cos 2t, -\tfrac12\sin 2t)$ and $(2\sin 2t, \cos 2t)$ . This is $e^{tA}$ with $t$ replaced by $-t$ , as it must be for a constant matrix.

Variation of constants for systems

Definition Let $A(t)$ and $b(t)$ be continuous on an open interval $I$ , let $\Phi$ be a fundamental matrix of $X' = A(t)X$ on $I$ , and let $t_0$ be a point of $I$ . The unique solution of $X' = A(t)X + b(t)$ with $X(t_0) = X_0$ is $X(t) = \Phi(t)c(t)$ for every $t$ in $I$ , where

\begin{aligned} c(t) &= \Phi(t_0)^{-1}X_0 \\[0.4em] &\quad + \int_{t_0}^{t} \Phi(\tau)^{-1}b(\tau)\,d\tau. \end{aligned}

To derive it, substitute $X = \Phi(t)c(t)$ . The product rule gives $X' = \Phi'c + \Phi c' = A(t)\Phi c + \Phi c'$ , and the term $A(t)\Phi c$ matches the homogeneous part of the equation, which leaves $\Phi(t)c'(t) = b(t)$ . Hence $c' = \Phi^{-1}b$ , and integrating from $t_0$ with $c(t_0) = \Phi(t_0)^{-1}X_0$ gives the formula for $c(t)$ . For a constant matrix with $\Phi = e^{tA}$ it reads

\begin{aligned} X(t) &= e^{(t - t_0)A}X_0 \\[0.4em] &\quad + \int_{t_0}^{t} e^{(t - \tau)A}b(\tau)\,d\tau. \end{aligned}

The order of the factors matters. The inverse $\Phi(\tau)^{-1}$ acts on the input $b(\tau)$ at the time $\tau$ when it arrives, and $\Phi(t)$ stands to the left of the integral. Variation of parameters for $y'' + p(t)y' + q(t)y = g(t)$ is the special case of the first-order form with state $(y, y')^{T}$ and forcing $(0, g)^{T}$ .

In plain words The homogeneous solutions form a moving basis, and the forcing is absorbed by letting the coordinates in that basis drift. The rate of drift is the input expressed in the basis at the moment it arrives, and the final state is rebuilt from the accumulated coordinates with the basis at the time of observation.

Example Solve $x_1' = tx_1 + 1$ and $x_2' = -x_2 + 2$ with $X(0) = (1, 1)^{T}$ , using $\Phi(t) = \operatorname{diag}(e^{t^2/2}, e^{-t})$ , whose value at $0$ is $I$ . The integrand is

\Phi(\tau)^{-1}b = \begin{pmatrix} e^{-\tau^2/2} \\ 2e^{\tau} \end{pmatrix}.

The second entry integrates to $2e^t - 2$ , so $x_2 = e^{-t}(1 + 2e^t - 2) = 2 - e^{-t}$ . The first entry has no elementary antiderivative and stays as an integral,

\begin{aligned} x_1(t) &= e^{t^2/2}\big(1 + G(t)\big), \\[0.4em] G(t) &= \int_0^t e^{-\tau^2/2}\,d\tau. \end{aligned}

The formula holds for every real $t$ , since $A$ and $b$ are continuous on $\mathbb{R}$ . As $t$ grows, $x_2$ approaches $2$ while $x_1$ grows without bound.

Non-elementary integrals

Definition An elementary function is built from constants, powers, roots, exponentials, logarithms, trigonometric functions and their inverses by finitely many arithmetic operations and compositions. An integral is non-elementary when its integrand has no elementary antiderivative, as for $e^{-\tau^2/2}$ , $e^{\tau^2}$ and $\sin\tau/\tau$ . For $f$ continuous on an interval $I$ and $t_0$ in $I$ , the function

F(t) = \int_{t_0}^{t} f(\tau)\,d\tau

is still defined and differentiable on $I$ , with $F' = f$ by the fundamental theorem of calculus. A solution formula that contains such an integral is exact. Its values can be computed to any accuracy by numerical quadrature or through a named function such as the error function

\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-s^2}\,ds.

In plain words An integral that cannot be written with familiar functions is still a perfectly good function. It can be differentiated, checked and evaluated, and a name such as erf only abbreviates the integral.

Example The first component $x_1 = e^{t^2/2}(1 + G)$ of the example above satisfies its equation, since $G'(t) = e^{-t^2/2}$ and the product rule gives

\begin{aligned} x_1' &= t\,x_1 + e^{t^2/2}e^{-t^2/2} \\[0.4em] &= t\,x_1 + 1. \end{aligned}

Also $x_1(0) = 1$ , because $G(0) = 0$ . The substitution $\tau = \sqrt{2}\,s$ gives

G(t) = \sqrt{\pi/2}\,\operatorname{erf}\!\left(t/\sqrt{2}\right),

which is about $0.8556$ at $t = 1$ , so $x_1(1) \approx 3.0594$ . The improper integral of $e^{-\tau^2/2}$ over $[0, \infty)$ has the closed-form value $\sqrt{\pi/2}$ , so $e^{-t^2/2}x_1(t) \to 1 + \sqrt{\pi/2} \approx 2.2533$ as $t \to \infty$ . The exact formula shows how fast $x_1$ grows, even though $x_1$ itself is not an elementary function.

Check yourself

Let $A$ have rows $(0, 1)$ and $(0, 0)$ and let $b(t) = (0, 6t)^{T}$ . Solve $X' = AX + b(t)$ with $X(0) = (0, 0)^{T}$ , and compare the result with the equation $y'' = 6t$ .

Show the answer

Since $A^2 = 0$ , the exponential is $e^{tA} = I + tA$ , and its inverse at time $\tau$ is $e^{-\tau A} = I - \tau A$ . The integrand $e^{-\tau A}b(\tau) = (-6\tau^2, 6\tau)^{T}$ integrates from $0$ to $t$ to $(-2t^3, 3t^2)^{T}$ . Multiplying by $e^{tA}$ gives

\begin{aligned} X(t) &= \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}\begin{pmatrix} -2t^3 \\ 3t^2 \end{pmatrix} \\[0.4em] &= \begin{pmatrix} t^3 \\ 3t^2 \end{pmatrix}. \end{aligned}

The system is the first-order form of $y'' = 6t$ with state $(y, y')^{T}$ and $y(0) = y'(0) = 0$ . Its solution $y = t^3$ has $y' = 3t^2$ , which matches both components.

Let $B$ have rows $(0, 1)$ and $(1, 0)$ . The matrix $\Phi(t)$ with rows $(t, 1/t)$ and $(t, -1/t)$ is a fundamental matrix of $tX' = BX$ on $(0, \infty)$ . Solve $tX' = BX + (2t^2, 0)^{T}$ with $X(1) = (0, 0)^{T}$ , and explain what goes wrong if the forcing $(2t^2, 0)^{T}$ is used in the formula without first dividing by $t$ .

Show the answer

The formula needs the form $X' = A(t)X + b(t)$ . Dividing by $t$ gives $A(t) = B/t$ and $b(t) = (2t, 0)^{T}$ , which are continuous on $(0, \infty)$ , while $A(t)$ is not defined at $t = 0$ . So the solution lives on $(0, \infty)$ . The determinant of $\Phi(t)$ is $-1 - 1 = -2$ , and the $2 \times 2$ inverse formula gives

\Phi(t)^{-1} = \begin{pmatrix} \tfrac{1}{2t} & \tfrac{1}{2t} \\[0.2em] \tfrac{t}{2} & -\tfrac{t}{2} \end{pmatrix},

so $\Phi(\tau)^{-1}b(\tau) = (1, \tau^2)^{T}$ . Since $X(1) = 0$ , only the integral term remains. The integral from $1$ to $t$ is $(t - 1, (t^3 - 1)/3)^{T}$ , and multiplying by $\Phi(t)$ gives

X(t) = \begin{pmatrix} \tfrac43t^2 - t - \tfrac{1}{3t} \\[0.2em] \tfrac23t^2 - t + \tfrac{1}{3t} \end{pmatrix}.

To check, $t$ times the derivative of the first component is $\tfrac83t^2 - t + \tfrac{1}{3t}$ , which equals the second component plus $2t^2$ . Also $t$ times the derivative of the second component is $\tfrac43t^2 - t - \tfrac{1}{3t}$ , which is the first component, and both components vanish at $t = 1$ . With the forcing $(2t^2, 0)^{T}$ left undivided, the formula solves $X' = (B/t)X + (2t^2, 0)^{T}$ instead. That is the equation $tX' = BX + (2t^3, 0)^{T}$ , not the one we were given.

Solve $X' = A(t)X + b$ with $A(t) = \operatorname{diag}(-2t, -1)$ , $b = (1, 1)^{T}$ and $X(0) = (0, 0)^{T}$ . Explain why the answer is exact although one of its integrals is non-elementary, and find the limit of each component as $t \to \infty$ .

Show the answer

The principal fundamental matrix at $0$ is $\Phi(t) = \operatorname{diag}(e^{-t^2}, e^{-t})$ , and $\Phi(\tau)^{-1}b = (e^{\tau^2}, e^{\tau})^{T}$ . With $X(0) = 0$ the formula gives $x_2(t) = e^{-t}(e^t - 1)$ and

\begin{gathered} x_1(t) = e^{-t^2}\int_0^t e^{\tau^2}\,d\tau, \\[0.4em] x_2(t) = 1 - e^{-t}. \end{gathered}

The integrand $e^{\tau^2}$ has no elementary antiderivative, but it is continuous, so the integral is a differentiable function of $t$ with derivative $e^{t^2}$ . The product rule then confirms $x_1' = -2tx_1 + 1$ , and $x_1(0) = 0$ . The second component tends to $1$ . For the first, the numerator $\int_0^t e^{\tau^2}\,d\tau$ and the denominator $e^{t^2}$ of the quotient both tend to infinity, and L’Hôpital’s rule gives

\lim_{t \to \infty} x_1(t) = \lim_{t \to \infty} \frac{e^{t^2}}{2te^{t^2}} = 0.

In fact $x_1(t)$ is close to $1/(2t)$ for large $t$ , and $x_1(10) \approx 0.0503$ .

Nonhomogeneous and Time-Dependent Linear Systems

Key terms

Nonhomogeneous linear system

Time-dependent coefficient matrix

Inverse fundamental matrix

Variation of constants for systems

Non-elementary integrals

Check yourself

How this lesson connects

Builds on

Used later in