Linear ODEs, Evolution Operators, and Wronskians

A linear ordinary differential equation is organized most cleanly by its state at one time and the linear map that transports that state to another. A fundamental matrix packages a basis of homogeneous solutions, its Wronskian detects whether that basis remains independent, and variation of constants adds forcing. Boundary and matching conditions then become finite linear systems built from the same evolution map.

The basic results below assume continuous coefficients on a real interval. Piecewise-continuous coefficients are handled interval by interval, with matching conditions derived from the differential equation. Singular points, vanishing leading coefficients, and distributional sources require separate care.

From a scalar equation to a first-order state

Consider a normalized $n$th-order linear equation on an interval $I$,

$$y^{(n)} +a_{n-1}(t)y^{(n-1)} +\cdots +a_1(t)y' +a_0(t)y =r(t),$$

with continuous coefficients and forcing. Introduce the state

$$X(t) = \begin{pmatrix} y(t)\\ y'(t)\\ \vdots\\ y^{(n-1)}(t) \end{pmatrix}.$$

Then the equation is equivalent to

$$\dot X(t)=A(t)X(t)+J(t),$$

where $A$ is the companion matrix and $J=(0,\ldots,0,r)^{\mathsf T}$. For each $t_0\in I$ and initial state $X(t_0)=X_0$, continuity of $A$ and $J$ gives one solution on all of $I$. For fixed forcing, the map from $X_0$ to $X(t)$ is affine, with linear part given by the homogeneous evolution operator. Equivalently, the difference of two solutions with the same forcing depends linearly on the difference of their initial states.

If the original leading coefficient can vanish, dividing by it is illegal there. Such a point is a singular point of the normalized equation, and the initial-value theorem just quoted does not cross it automatically.

Evolution operators and composition

For the homogeneous system

$$\dot X(t)=A(t)X(t),$$

define the evolution operator $\mathsf U(t,t_0)$ by

$$\frac{\partial}{\partial t}\mathsf U(t,t_0) = A(t)\mathsf U(t,t_0), \qquad \mathsf U(t_0,t_0)=\mathbb 1.$$

The solution with initial state $X_0$ is

$$X(t)=\mathsf U(t,t_0)X_0.$$

Uniqueness gives the two structural identities

$$\mathsf U(t,s)\mathsf U(s,t_0) = \mathsf U(t,t_0), \qquad \mathsf U(t,t_0)^{-1} = \mathsf U(t_0,t).$$

Thus evolution on a regular interval is always invertible, even when some solutions grow or decay strongly.

For constant $A$,

$$\mathsf U(t,t_0)=e^{(t-t_0)A}.$$

For time-dependent matrices, one may write an ordinary exponential only if $[A(t),A(s)]=0$ for all relevant $s,t$. In general, for $t\geq t_0$ the Peano–Baker series is

$$\begin{aligned} \mathsf U(t,t_0) ={}& \mathbb 1 +\int_{t_0}^{t}A(s_1)\,\mathrm ds_1\\ &+ \int_{t_0\leq s_2\leq s_1\leq t} A(s_1)A(s_2)\, \mathrm ds_1\,\mathrm ds_2 +\cdots . \end{aligned}$$

The ordering of the matrices follows the ordering of their times. This series is often abbreviated as a time-ordered exponential,

$$\mathsf U(t,t_0) = \mathcal T \exp\!\left( \int_{t_0}^{t}A(s)\,\mathrm ds \right),$$

but the series, not the symbol, defines what $\mathcal T$ means. Teschl, Teschl 2012, Chapter 3, §3.4, PDF, develops the principal matrix solution, composition law, and the noncommutativity qualification.

Fundamental matrices and Wronskians

Let $X_1,\ldots,X_n$ be solutions of the homogeneous first-order system and put them into columns:

$$F(t) = \begin{pmatrix} |&&|\\ X_1(t)&\cdots&X_n(t)\\ |&&| \end{pmatrix}.$$

Then

$$\dot F(t)=A(t)F(t).$$

The matrix $F$ is a fundamental matrix when its columns are linearly independent. Its Wronskian is

$$W(t)=\det F(t).$$

If $F(t_0)$ is invertible, the evolution operator is recovered as

$$\mathsf U(t,t_0) = F(t)F(t_0)^{-1}.$$

The determinant obeys Abel’s identity, also called Liouville’s formula:

$$W(t) = W(t_0) \exp\!\left( \int_{t_0}^{t} \operatorname{tr}A(s)\,\mathrm ds \right).$$

One proof uses Jacobi’s determinant formula wherever $F$ is invertible:

$$\begin{aligned} \frac{\mathrm d}{\mathrm dt}\det F &= \det F\, \operatorname{tr}(F^{-1}\dot F)\\ &= \det F\, \operatorname{tr}(F^{-1}AF)\\ &= \operatorname{tr}A\, \det F. \end{aligned}$$

Solving this scalar equation gives the result. Consequently, a Wronskian that is nonzero at one time is nonzero throughout the regular interval. A candidate solution basis therefore needs to be checked at only one convenient point. These statements and the proof are given in Teschl, §3.4, Eqs. (3.88)–(3.91).

For the second-order homogeneous equation

$$y''+p(t)y'+q(t)y=0,$$

take $X=(y,y')^{\mathsf T}$. Then

$$A(t) = \begin{pmatrix} 0&1\\ -q(t)&-p(t) \end{pmatrix},$$

and two scalar solutions have Wronskian

$$W[y_1,y_2] = y_1y_2'-y_1'y_2.$$

Since $\operatorname{tr}A=-p$,

$$W(t) = W(t_0) \exp\!\left( -\int_{t_0}^{t}p(s)\,\mathrm ds \right).$$

In particular, $W$ is constant when the first-derivative term is absent. The statement “zero Wronskian implies dependence” is reliable here because $y_1,y_2$ solve the same regular linear equation. It is false for arbitrary differentiable functions without that shared-equation hypothesis.

Forcing and variation of constants

For

$$\dot X=A(t)X+J(t), \qquad X(t_0)=X_0,$$

the solution is

$$X(t) = \mathsf U(t,t_0)X_0 + \int_{t_0}^{t} \mathsf U(t,s)J(s)\,\mathrm ds.$$

The first term transports the initial state. The second transports each infinitesimal source contribution from its insertion time $s$ to the observation time $t$. Differentiation verifies the formula: the moving upper limit supplies $J(t)$, while $\partial_t\mathsf U(t,s)=A(t)\mathsf U(t,s)$ supplies the homogeneous part. This is the variation-of-constants formula in Teschl, §3.4, Eqs. (3.92)–(3.97).

The formula already has the structure later used for retarded Green operators, but it does not select a retarded, advanced, Feynman, or other physical boundary condition by itself. That selection belongs to the Fundamental Solutions and Green Operators and Hyperbolic Equations and Causal Propagators.

Boundary data as a finite linear system

Initial data specify the full state at one time. A boundary-value problem instead imposes linear conditions at two or more points. On $[a,b]$, suppose

$$B_aX(a)+B_bX(b)=c.$$

Variation of constants gives

$$X(b) = \mathsf U(b,a)X(a)+G_b, \qquad G_b = \int_a^b\mathsf U(b,s)J(s)\,\mathrm ds.$$

Therefore the unknown initial state must solve

$$\left[ B_a+B_b\mathsf U(b,a) \right]X(a) = c-B_bG_b.$$

When the total number of independent boundary conditions equals the state dimension, this is a square finite-dimensional system. The boundary-value problem has a unique solution exactly when its matrix is invertible. If the matrix is singular, there may be no solution or a family of solutions, depending on compatibility with the right-hand side. This is why uniqueness of every initial-value problem does not imply uniqueness of every boundary-value problem.

When such a boundary system comes from a regular weighted second-order eigenvalue problem, continue to Sturm–Liouville Problems and Eigenfunction Expansions for the self-adjoint domain, orthogonality, completeness, and spectral Green kernel.

Smallest example: the harmonic oscillator

For

$$y''+\omega^2y=0, \qquad \omega>0,$$

the state $X=(y,\dot y)^{\mathsf T}$ evolves over $\Delta=t-t_0$ by

$$\mathsf U_\omega(\Delta) = \begin{pmatrix} \cos(\omega\Delta)& \dfrac{\sin(\omega\Delta)}{\omega}\\[6pt] -\omega\sin(\omega\Delta)& \cos(\omega\Delta) \end{pmatrix}.$$

Direct multiplication verifies

$$\mathsf U_\omega(\Delta_1) \mathsf U_\omega(\Delta_2) = \mathsf U_\omega(\Delta_1+\Delta_2),$$

and

$$\det\mathsf U_\omega(\Delta)=1.$$

The determinant is the Wronskian statement $\operatorname{tr}A=0$. In the limit $\omega\to0$,

$$\mathsf U_0(\Delta) = \begin{pmatrix} 1&\Delta\\ 0&1 \end{pmatrix},$$

which correctly transports the solutions of $y''=0$.

QFT-facing example: mode matching across a frequency step

With the site’s $(+---)$ metric, the free Klein–Gordon equation is

$$\left( \partial_t^2-\nabla^2+m^2 \right)\phi=0.$$

Each spatial Fourier mode obeys

$$\ddot\phi_{\mathbf k}(t) +\omega_{\mathbf k}^2\phi_{\mathbf k}(t)=0, \qquad \omega_{\mathbf k} = \sqrt{\mathbf k^2+m^2}.$$

Thus every momentum label carries a two-dimensional ODE state. A standard complex mode

$$u_{\mathbf k}(t) = \frac{e^{-i\omega_{\mathbf k}t}} {\sqrt{2\omega_{\mathbf k}}}$$

has the constant Wronskian

$$W[u_{\mathbf k},u_{\mathbf k}^*]=i.$$

The value $i$ is a chosen normalization, not a consequence of linear independence alone.

For a controlled matching example, let the squared frequency jump at $t=0$:

$$\omega(t) = \begin{cases} \omega_-,&t<0,\\ \omega_+,&t>0, \end{cases} \qquad \omega_\pm>0.$$

There is no delta source, so integrating the equation through $t=0$ requires both $u$ and $\dot u$ to be continuous. Start with

$$u(t) = \frac{e^{-i\omega_-t}}{\sqrt{2\omega_-}}, \qquad t<0,$$

and write the solution for $t>0$ as

$$u(t) = \frac{ \alpha e^{-i\omega_+t} +\beta e^{+i\omega_+t}} {\sqrt{2\omega_+}}.$$

The two matching equations give

$$\begin{aligned} \alpha &= \frac12 \left( \sqrt{\frac{\omega_+}{\omega_-}} + \sqrt{\frac{\omega_-}{\omega_+}} \right),\\ \beta &= \frac12 \left( \sqrt{\frac{\omega_+}{\omega_-}} - \sqrt{\frac{\omega_-}{\omega_+}} \right). \end{aligned}$$

Wronskian preservation supplies an independent check:

$$|\alpha|^2-|\beta|^2=1.$$

This page uses the step only to demonstrate evolution and matching. The physical interpretation of mode normalization and positive frequency belongs to The Klein–Gordon Field and Its Modes. Schwartz 2014, §§2.3 and 3.1 supplies the free-field decomposition into harmonic-oscillator modes and the Klein–Gordon equation.

Matching conditions and singular sources

A coefficient jump does not by itself license a jump in every state component. Derive matching conditions by integrating the equation across the interface. For example,

$$y''+\omega(t)^2y=0$$

with bounded piecewise-continuous $\omega^2$ requires $y$ and $y'$ to be continuous. If instead

$$y''+\omega(t)^2y = J\,\delta(t-t_*),$$

then, under the assumption that $y$ itself remains continuous,

$$y'(t_*^+)-y'(t_*^-)=J.$$

For a divergence-form equation, the naturally continuous flux may be $p(t)y'$ rather than $y'$. The differential equation, including its leading coefficient and singular sources, decides which quantities match.

Failure modes

A time-dependent matrix is not an ordinary exponential. The formula $\exp(\int A)$ generally fails when matrices at different times do not commute. Use the ordered series or solve the matrix initial-value problem.

A Wronskian test needs a shared equation. For solutions of one regular linear system, a nonzero Wronskian at one time proves independence everywhere. For arbitrary differentiable functions, an identically zero Wronskian need not prove linear dependence.

Initial and boundary data are not interchangeable. A complete initial state has a unique evolution. Separated boundary conditions can be incompatible or can leave a nullspace; inspect the boundary matrix.

A singular point breaks the regular theorem. If the leading coefficient vanishes or another coefficient diverges, fundamental-matrix transport across that point needs a new local analysis.

Matching the wrong variables changes the problem. Continuity of $y'$ is correct for a unit leading coefficient without a delta source. In divergence form or with singular sources, integrate first and match the resulting flux or jump.

Wronskian conservation does not choose a physical basis. It checks normalization and independence. Positive-frequency, vacuum, incoming, or outgoing conditions require additional physical input.

Exercises

1. Verify directly that $\mathsf U_\omega(\Delta)$ solves the oscillator matrix equation, has determinant $1$, and composes by addition of time intervals.

   Check

   Differentiate the matrix and compare with

   $$A = \begin{pmatrix} 0&1\\ -\omega^2&0 \end{pmatrix}.$$

   The determinant is $\cos^2(\omega\Delta)+\sin^2(\omega\Delta)=1$. The composition law follows from the sine and cosine addition formulas.

2. For $y''+p(t)y'+q(t)y=0$, differentiate $W=y_1y_2'-y_1'y_2$ and recover Abel’s identity.

   Check

   Substitution of $y_j''=-p\,y_j'-q\,y_j$ gives

   $$W' = y_1y_2''-y_1''y_2 = -pW.$$

   Hence $W(t)=W(t_0)\exp[-\int_{t_0}^t p(s)\,\mathrm ds]$.

3. Derive the coefficients $\alpha,\beta$ for the frequency step and verify $|\alpha|^2-|\beta|^2=1$.

   Check

   Continuity at zero gives

   $$\alpha+\beta = \sqrt{\frac{\omega_+}{\omega_-}}, \qquad \alpha-\beta = \sqrt{\frac{\omega_-}{\omega_+}}.$$

   Adding and subtracting yields the displayed coefficients. Writing $r=\sqrt{\omega_+/\omega_-}$ gives $\alpha=(r+r^{-1})/2$ and $\beta=(r-r^{-1})/2$, so $\alpha^2-\beta^2=1$.

References

• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, §§2.3 and 3.1, Cambridge University Press, 2014. Book record. This is the QFT source for free fields as families of oscillator modes and for the Klein–Gordon mode equation.
• Gerald Teschl, Ordinary Differential Equations and Dynamical Systems, Chapter 3, American Mathematical Society, 2012. Open author edition, PDF; AMS book record. This is the structural and teaching source for linear systems, principal evolution matrices, Abel’s identity, and variation of constants.