The Wronskian and independence

Theorem

The Wronskian and independence

T-065Home MU-205Threads change

Statement

Let \(I\subseteq\mathbb{R}\) be an interval and let \(p_0,p_1,\dots,p_{n-1}:I\to\mathbb{R}\) be continuous. Consider the \(n\)-th order linear homogeneous ODE \(L[y]:=y^{(n)}+p_{n-1}(x)y^{(n-1)}+\cdots+p_1(x)y'+p_0(x)y=0\) on \(I\), and let \(y_1,\dots,y_n:I\to\mathbb{R}\) be solutions of \(L[y]=0\). Define the Wronskian \(W(x)=W[y_1,\dots,y_n](x):=\det\big(y_j^{(i)}(x)\big)_{0\le i\le n-1,\,1\le j\le n}\). Then (a) [Abel's identity] \(W\) satisfies \(W'(x)=-p_{n-1}(x)W(x)\) on \(I\), so \(W(x)=W(x_0)\exp\!\left(-\int_{x_0}^{x}p_{n-1}(t)\,dt\right)\) for all \(x,x_0\in I\); in particular \(W\) is either identically zero on \(I\) or vanishes nowhere on \(I\). (b) \(y_1,\dots,y_n\) are linearly independent on \(I\) if and only if \(W(x_0)\neq0\) for some \(x_0\in I\) (equivalently, for every \(x_0\in I\)).

Why it matters

Linear independence of functions is, by definition, a statement about every point of \(I\) simultaneously — a genuinely infinite-dimensional condition. The Wronskian collapses this to a single determinant evaluated at one point, turning an existence-of-nontrivial-relation question into finite-dimensional linear algebra. This is what makes it possible to certify, in one line of algebra, that \(n\) solutions of an \(n\)-th order linear ODE form a fundamental set — a basis for the solution space — which is the entire basis for the theory of general solutions \(y=c_1y_1+\cdots+c_ny_n\).

The theorem also explains a phenomenon that looks miraculous the first time you see it: for solutions of a common linear ODE you never need to check the Wronskian on the whole interval, only at one convenient point (often \(x=0\), or wherever initial data is imposed), because Abel's identity forces it to be nonvanishing everywhere or nowhere at once.

Hypotheses

Continuity of \(p_0,\dots,p_{n-1}\) on \(I\).Continuity guarantees solutions exist on all of \(I\) and are \(n\) times continuously differentiable (so the Wronskian matrix is well defined), and it makes \(t\mapsto p_{n-1}(t)\) integrable so Abel's formula \(W(x)=W(x_0)e^{-\int p_{n-1}}\) makes sense. If \(p_{n-1}\) has, say, a non-integrable singularity at some \(x^\ast\in I\), the integrating factor argument in Step 6 breaks down and \(W\) can vanish at isolated points while being nonzero elsewhere. \(y_1,\dots,y_n\) are solutions of the same linear ODE \(L[y]=0\).This is the hypothesis that makes the converse direction of (b) true. Without it, "linearly independent" no longer forces \(W\) to avoid zero: \(f(x)=x^2\) and \(g(x)=x|x|\) on \(\mathbb{R}\) are linearly independent (neither is a scalar multiple of the other), yet \(W[f,g](x)=f g'-f'g\equiv0\) for all \(x\), because they are not both solutions of one fixed second-order linear ODE with continuous coefficients on all of \(\mathbb{R}\) (any ODE with \(x^2\) as a solution forces \(x|x|\) to fail to be a solution near \(0\), or vice versa). Exactly \(n\) functions for an \(n\)-th order equation.The Wronskian is only defined as a determinant when the number of functions equals the order of the ODE (so the matrix is square). With fewer than \(n\) solutions the construction does not apply; with more than \(n\) solutions of an \(n\)-th order equation they are automatically dependent (the solution space has dimension \(n\)), so the theorem is stated for the borderline, informative case of exactly \(n\). \(I\) is an interval (connected).Abel's formula integrates \(p_{n-1}\) along a path in \(I\); if the domain were a union of disjoint intervals, \(W\) could vanish identically on one component and be nonvanishing on another, and "some \(x_0\)" would no longer imply "every \(x_0\)" globally. Restricting to a connected interval is what lets the dichotomy in part (a) be a genuine dichotomy on the whole domain.

Proof

W(x) = \det\begin{pmatrix} y_1(x) & \cdots & y_n(x)\\ y_1'(x) & \cdots & y_n'(x)\\ \vdots & & \vdots\\ y_1^{(n-1)}(x) & \cdots & y_n^{(n-1)}(x)\end{pmatrix}

Definition of the Wronskian matrix; each \(y_j\in C^{n}(I)\) since it solves an \(n\)-th order ODE with continuous coefficients, so every entry is well defined and continuous on \(I\). A

\text{Suppose, for contradiction, } y_1,\dots,y_n \text{ are linearly dependent on } I.

We prove "\(W(x_0)\neq0\) for some \(x_0\) \(\Rightarrow\) independence" by its contrapositive: assume dependence and derive \(W(x_0)=0\) for every \(x_0\). B

\exists\, c=(c_1,\dots,c_n)\neq 0 \text{ with } \sum_{j=1}^{n} c_j y_j(x) = 0 \ \ \forall x\in I.

Definition of linear dependence of a finite set of functions on \(I\). A

\sum_{j=1}^{n} c_j y_j^{(k)}(x) = 0 \ \ \forall x\in I,\quad k=0,1,\dots,n-1.

Differentiate the identity in Step 3 termwise \(k\) times; legal because each \(y_j\in C^{n}(I)\) and differentiation is linear, so an identically-zero finite linear combination of functions has an identically-zero derivative. A

M(x_0)\,c = 0, \qquad M(x_0):=\big(y_j^{(i)}(x_0)\big)_{0\le i\le n-1,\,1\le j\le n},

Evaluate the \(n\) identities of Step 4 at any fixed \(x_0\in I\); this is exactly the matrix–vector equation \(M(x_0)c=0\) with \(M(x_0)\) the Wronskian matrix at \(x_0\) and \(c\) the fixed nonzero vector from Step 3. B

c\neq 0 \text{ solves } M(x_0)c=0 \ \Longrightarrow\ \det M(x_0)=0, \text{ i.e. } W(x_0)=0.

Standard linear algebra: a square matrix with a nontrivial kernel is singular (invertibility criterion — if \(\det M(x_0)\neq0\) then \(M(x_0)\) is invertible and \(c=M(x_0)^{-1}0=0\), contradicting \(c\neq0\)). Since \(x_0\in I\) was arbitrary, \(W\equiv0\) on \(I\). B

\text{Hence: } W(x_0)\neq 0 \text{ for some } x_0\in I \ \Longrightarrow\ y_1,\dots,y_n \text{ linearly independent on } I.

Steps 2–6 show dependence forces \(W\equiv0\); contrapositive gives the claim. This half needs no ODE structure at all — it holds for any \((n-1)\)-times differentiable functions. B

W'(x) = \det\begin{pmatrix} y_1 & \cdots & y_n\\ \vdots & & \vdots\\ y_1^{(n-2)} & \cdots & y_n^{(n-2)}\\ y_1^{(n)} & \cdots & y_n^{(n)}\end{pmatrix}

The derivative of a determinant is the sum, over each row, of the determinant with only that row differentiated (multilinearity of \(\det\) in its rows). For rows \(0,\dots,n-2\), differentiating row \(i\) produces row \(i+1\), which already appears elsewhere in the matrix, so that summand is a determinant with a repeated row and vanishes. Only differentiating the last row survives. C

y_j^{(n)} = -p_{n-1}y_j^{(n-1)} - p_{n-2}y_j^{(n-2)} - \cdots - p_0 y_j, \quad j=1,\dots,n.

Each \(y_j\) solves \(L[y_j]=0\); rearrange to express the top derivative. A

W'(x) = -p_{n-1}(x)\,W(x).

Substitute Step 9 into the last row of Step 8 and expand by multilinearity in that row: every term \(-p_i y_j^{(i)}\) with \(i\lt n-1\) reproduces a determinant with two equal rows (row \(i\) already present) and vanishes; only the \(-p_{n-1}y_j^{(n-1)}\) term survives, giving \(-p_{n-1}\) times the original Wronskian determinant. C

W(x) = W(x_0)\exp\!\left(-\int_{x_0}^{x} p_{n-1}(t)\,dt\right), \qquad x,x_0\in I.

Step 10 is a first-order linear ODE for the scalar function \(W\); \(p_{n-1}\) continuous on the interval \(I\) makes it integrable, and solving by the standard integrating-factor method gives this closed form. Since \(\exp\) is never zero, \(W(x)=0 \iff W(x_0)=0\), for any pair \(x,x_0\in I\); this proves part (a): \(W\equiv0\) on \(I\) or \(W\) vanishes nowhere on \(I\). C

\text{If } y_1,\dots,y_n \text{ are linearly dependent, then } W\equiv0 \text{ on } I.

This is exactly Steps 2–6 run with \(x_0\) left arbitrary: the dependence relation of Step 3 holds identically on \(I\), so Step 6's conclusion \(W(x_0)=0\) holds for every \(x_0\in I\) at once. A

\therefore\quad y_1,\dots,y_n \text{ independent on } I \iff W(x_0)\neq0 \text{ for some (every) } x_0\in I.

Combine: Step 7 gives (\(W(x_0)\neq0\) for some \(x_0\)) \(\Rightarrow\) independence. Step 12 gives its contrapositive (dependence \(\Rightarrow W\equiv0\)), so independence \(\Rightarrow W\) is not identically \(0\), which by Step 11's dichotomy means \(W(x_0)\neq0\) for every \(x_0\). This closes the biconditional and proves part (b). B

Result

y_1,\dots,y_n \text{ solve } L[y]=0 \text{ on } I \ \Longrightarrow\ \Big(\text{independent on } I \iff W(x_0)\neq0 \text{ for some } x_0\in I\Big)

Reading. For \(n\) solutions of a single \(n\)-th order linear homogeneous ODE with continuous coefficients, you may test linear independence by evaluating one determinant at one convenient point. Abel's identity guarantees the answer would be the same at every other point too, so there is no risk of an unlucky choice of \(x_0\).

Scope. The "\(W(x_0)\neq0\Rightarrow\) independent" half holds for any \((n-1)\)-times differentiable functions, no ODE required. The converse ("independent \(\Rightarrow W\) never zero") genuinely needs the functions to be solutions of one common linear ODE with continuous coefficients on a connected interval \(I\) — it is false for arbitrary independent functions.

Corollaries & converses

Fundamental set test. \(n\) solutions of an \(n\)-th order linear homogeneous ODE with \(W(x_0)\neq0\) at one point form a fundamental set (a basis for the \(n\)-dimensional solution space); every solution is uniquely \(c_1y_1+\cdots+c_ny_n\).
Trichotomy collapses to a dichotomy. For solutions of a common linear ODE, "\(W\) vanishes somewhere but not everywhere" is impossible — Abel's identity rules it out.
Contrapositive as a dependence test. If \(W(x_0)=0\) at even one point, the solutions are automatically dependent (equivalently \(W\equiv0\)); no further checking is needed.
Converse for general functions fails. Independence of arbitrary smooth functions does not imply nonvanishing Wronskian — see Fails without below. The hypothesis "solutions of a common linear ODE" cannot be dropped.
Order-reduction use. Abel's identity (part a) is itself the engine behind reduction of order: knowing one solution \(y_1\) and the coefficient \(p_{n-1}\) pins down \(W\) up to a constant, which yields a first-order equation for a second solution.

Fails without

Dropping "solutions of a common linear ODE": \(f(x)=x^2\), \(g(x)=x|x|\) on \(I=\mathbb{R}\) are linearly independent (no constant multiple relates them, since \(g\) is odd-like in sign while \(f\) is not), yet \(W[f,g](x)=f(x)g'(x)-f'(x)g(x)\equiv0\) for all \(x\) (direct computation: \(g'(x)=2|x|\), so \(fg'-f'g = x^2\cdot2|x| - 2x\cdot x|x| = 2x^2|x|-2x^2|x|=0\)). The certificate "\(W\) nonzero \(\Rightarrow\) independent" still works vacuously (its converse simply cannot be invoked here), but the biconditional of part (b) fails outright.
Dropping continuity of the coefficients: take the Cauchy–Euler-type equation \(x^2y''-2xy'+2y=0\) but now imagine forcing a coefficient with a non-continuous jump discontinuity at some \(x^\ast\) inside \(I\) (rather than the classical singular endpoint); Abel's integrating-factor argument (Step 11) requires \(p_{n-1}\) integrable on the path from \(x_0\) to \(x\), and a genuine discontinuity can make \(\int p_{n-1}\) fail to exist or the solution space itself fail to be \(n\)-dimensional across \(x^\ast\), so the "vanishes nowhere or vanishes everywhere" dichotomy need not hold across the discontinuity.
Using fewer solutions than the order: for \(y''+y=0\) (order \(2\)), a single solution such as \(y_1=\cos x\) has no well-defined \(2\times2\) Wronskian test for "independence" of a one-element set in the relevant sense — the theorem's determinant construction simply does not apply until you have exactly \(n=2\) functions.

Common errors

Applying the "\(W\equiv0\Rightarrow\) dependent" direction to functions that are not known solutions of a common linear ODE (e.g. concluding \(x^2\), \(x|x|\) are dependent because their Wronskian is zero).
Checking \(W\) at only one point and treating that as sufficient evidence even when the functions are not stipulated to solve a linear ODE — the single-point shortcut is only valid under the ODE hypothesis.
Building the Wronskian matrix with rows and columns transposed inconsistently between problems, then mis-expanding the determinant; the value \(\det\) is unaffected by transposition, but sign errors creep in from careless cofactor expansion.
Forgetting that nonvanishing Wronskian is a sufficient, not necessary, condition for independence when the ODE hypothesis is absent, and wrongly asserting "independent implies \(W\) never zero" for arbitrary function families.
Misidentifying \(p_{n-1}\) in Abel's formula — using the coefficient of \(y\) or \(y'\) instead of \(y^{(n-1)}\) (i.e. picking the wrong coefficient out of \(L[y]=y^{(n)}+p_{n-1}y^{(n-1)}+\cdots\)).

Discussion

The name honours Józef Hoene-Wroński, who introduced the determinant in the 1810s, though its role as an independence certificate for solutions of linear ODEs was clarified only later, notably once Abel's identity (1827) made explicit why the determinant cannot vanish at an isolated point without vanishing everywhere. Giuseppe Peano in the 1880s sharpened the picture further by exhibiting exactly the kind of counterexample used above (\(x^2\) and \(x|x|\)-type functions) to show that \(W\equiv0\) does not imply dependence for functions in general — correcting a folk belief that had crept into some 19th-century textbooks.

Structurally, the theorem is best understood through the companion (first-order) system \(\mathbf{y}'=A(x)\mathbf{y}\) obtained by setting \(\mathbf{y}=(y,y',\dots,y^{(n-1)})^{\top}\); the Wronskian matrix is then precisely a fundamental matrix of this system evaluated at \(x\), and Abel's identity \(W'=-p_{n-1}W\) is the scalar case of Liouville's formula \(\det\Phi(x)=\det\Phi(x_0)\exp\!\big(\int_{x_0}^x \operatorname{tr}A(t)\,dt\big)\) for fundamental matrices of linear systems, since \(\operatorname{tr}A=-p_{n-1}\) for the companion matrix. Seeing it this way removes the mystery of "why does one point suffice" — it is the same phenomenon as a linear flow map either always preserving orientation-with-nonzero-volume or always collapsing volume to zero, never switching between the two along a continuous path of invertible-or-singular matrices without discontinuity in the coefficients.

The theorem is also the backbone of the method of variation of parameters: solving for the particular solution of an inhomogeneous linear ODE requires inverting the Wronskian matrix, and the guarantee that this matrix is invertible at every point (given independence of the homogeneous solutions) is exactly this theorem.

A subtlety worth internalizing: the theorem does not claim the Wronskian is a "measure" of how independent the functions are, nor that a small \(|W(x_0)|\) signals near-dependence in any uniform sense — \(W\) can be rescaled by reparametrizing the \(y_i\) (replacing \(y_1\) by \(2y_1\) doubles \(W\)) without changing independence at all. Its only load-bearing content is the sign of the boolean "zero or not," not its magnitude.

Common misconception. Students often invert the logical direction and believe "linearly independent solutions must have nonzero Wronskian" is obvious/definitional. It is a theorem, not a definition, and its proof genuinely uses that the functions solve a common linear ODE — see the Fails-without example, which is independent functions with identically zero Wronskian outside that hypothesis.

Worked examples

y'' + y = 0 \text{ on } I=\mathbb{R}, \quad y_1=\cos x,\ y_2=\sin x.

Both solve the constant-coefficient ODE \(L[y]=y''+0\cdot y'+1\cdot y=0\), coefficients continuous on all of \(\mathbb{R}\); hypotheses of the theorem hold with \(n=2\), \(p_1\equiv0\). A

W(x)=\det\begin{pmatrix}\cos x & \sin x\\ -\sin x & \cos x\end{pmatrix} = \cos^2x+\sin^2x = 1.

Direct computation of the \(2\times2\) determinant from the definition in Step 1 of the proof. A

W(0) = 1 \neq 0.

Evaluate at the convenient point \(x_0=0\); by the theorem this single evaluation certifies independence on all of \(\mathbb{R}\), and by Abel's identity \(W\equiv1\) everywhere (consistent with \(p_1\equiv0\) forcing \(W'=0\)). B

\cos x,\ \sin x \text{ are linearly independent on } \mathbb{R}, \text{ hence a fundamental set for } y''+y=0.

Reading. The general solution of \(y''+y=0\) is exactly \(y=c_1\cos x+c_2\sin x\), \(c_1,c_2\in\mathbb{R}\), because these two solutions span the full 2-dimensional solution space.

x^2y''-2xy'+2y=0 \text{ on } I=(0,\infty), \quad y_1=x,\ y_2=x^2.

Rewrite in the theorem's normal form: \(y''-\tfrac{2}{x}y'+\tfrac{2}{x^2}y=0\), so \(p_1(x)=-2/x\) is continuous on \(I=(0,\infty)\); check \(y_1=x\): \(x^2\cdot0-2x\cdot1+2x=0\)\(\checkmark\); \(y_2=x^2\): \(x^2\cdot2-2x\cdot2x+2x^2=2x^2-4x^2+2x^2=0\)\(\checkmark\). A

W(x)=\det\begin{pmatrix}x & x^2\\ 1 & 2x\end{pmatrix}=2x^2-x^2=x^2.

Direct computation; note \(W(x)=x^2\neq0\) for every \(x\in(0,\infty)\), consistent with the dichotomy of part (a) — it never touches zero on this interval. A

W(x) \overset{?}{=} W(1)\exp\!\left(-\int_1^x \left(-\tfrac{2}{t}\right)dt\right) = 1\cdot\exp(2\ln x) = x^2.

Cross-check against Abel's identity (Step 11 of the proof) with \(p_1(t)=-2/t\), \(x_0=1\): the closed form matches the direct computation exactly, confirming the formula and the fact that \(W\) is nonvanishing throughout \((0,\infty)\). C

x,\ x^2 \text{ are linearly independent on } (0,\infty), \text{ hence a fundamental set there.}

Reading. Every solution of this Cauchy–Euler equation on \((0,\infty)\) has the form \(y=c_1x+c_2x^2\); the Wronskian test avoided having to verify independence "by hand" (checking no constant multiple relation holds for all \(x\)) and instead reduced to one determinant evaluation.

Problems

Compute \(W[e^{x},e^{2x}](x)\) and use it to show \(e^x,e^{2x}\) are independent solutions of \(y''-3y'+2y=0\) on \(\mathbb{R}\).
Solution
\(W=\det\begin{pmatrix}e^x & e^{2x}\\ e^x & 2e^{2x}\end{pmatrix}=2e^{3x}-e^{3x}=e^{3x}\). Since \(e^{3x}\neq0\) for every \(x\) (in particular at \(x_0=0\), \(W(0)=1\neq0\)), and both functions solve \(y''-3y'+2y=0\) (characteristic roots \(1,2\)) with continuous (constant) coefficients on \(\mathbb{R}=I\), the theorem gives independence on \(\mathbb{R}\).
Let \(y_1,y_2,y_3\) solve a third-order linear ODE with continuous coefficients on \(\mathbb{R}\), and suppose \(W(0)=0\). What can you conclude about \(W(5)\), and why?
Solution
\(W(5)=0\) as well. By part (a) (Abel's identity), \(W\) is either identically zero on the whole interval or never zero; since \(W(0)=0\), the "never zero" branch is excluded, so \(W\equiv0\) on \(\mathbb{R}\), in particular \(W(5)=0\). (Equivalently: \(W(0)=0\) forces dependence by the contrapositive of the theorem, and dependent solutions have \(W\equiv0\) everywhere by Step 12 of the proof.)
Explain precisely why \(f(x)=x^3\) and \(g(x)=x^2|x|\) can be linearly independent on \(\mathbb{R}\) while \(W[f,g]\equiv0\), without contradicting this theorem.
Solution
Compute for \(x\geq0\): \(g(x)=x^3\), so \(f,g\) agree and \(W=fg'-f'g=0\) trivially there. For \(x\lt0\): \(g(x)=-x^3\), \(g'(x)=-3x^2\); \(W=x^3(-3x^2)-3x^2(-x^3)=-3x^5+3x^5=0\). So \(W\equiv0\) on \(\mathbb{R}\) yet \(f,g\) are independent (no constant \(k\) has \(g=kf\) on all of \(\mathbb{R}\), since the sign relationship flips at \(x=0\)). This does not contradict the theorem because the theorem's converse ("independent \(\Rightarrow W\) never zero") requires \(f,g\) to be solutions of one common linear ODE with continuous coefficients on \(\mathbb{R}\); \(f\) and \(g\) here are not both solutions of any single such ODE on all of \(\mathbb{R}\) (any linear ODE with \(x^3\) as a solution near \(0\) forces its solution space to be smooth in a way \(x^2|x|\) is not, since \(x^2|x|\) is only \(C^2\), not \(C^3\), at \(x=0\)).
For \(y'' + p_1(x)y' + p_0(x)y=0\) with \(p_1\) continuous on \(I\), suppose \(y_1\) is a known nonvanishing solution on \(I\). Use Abel's identity to derive a formula for a second, independent solution \(y_2\) (reduction of order), and confirm \(W[y_1,y_2]\) matches Abel's prediction.
Solution
Abel gives \(W(x)=y_1y_2'-y_1'y_2=C\exp\!\left(-\int p_1\right)\) for a constant \(C\); take \(C=1\) (any nonzero \(C\) just rescales \(y_2\)). Dividing by \(y_1^2\): \(\left(\dfrac{y_2}{y_1}\right)'=\dfrac{y_1y_2'-y_1'y_2}{y_1^2}=\dfrac{e^{-\int p_1}}{y_1^2}\), so \(y_2=y_1\displaystyle\int \frac{e^{-\int p_1(x)dx}}{y_1(x)^2}\,dx\). By construction \(W[y_1,y_2]=e^{-\int p_1}\neq0\) on \(I\) (since the exponential is never zero and \(y_1\) is nonvanishing), matching Abel's identity exactly, so by the theorem \(y_1,y_2\) are independent — a genuine second solution, not a multiple of \(y_1\).
True or false, with justification: "If \(y_1,\dots,y_n\) are any \(n\) solutions of an \(n\)-th order linear homogeneous ODE with continuous coefficients on \(I\), and they are linearly independent, then their Wronskian is nonzero at every point of \(I\) except possibly finitely many."
Solution
False as stated (the "except possibly finitely many" hedge is not needed, but more importantly is not what the theorem gives — it gives something stronger and cleaner). By part (a), for solutions of a common linear ODE with continuous coefficients on an interval, \(W\) is either identically zero or nowhere zero — there is no intermediate case with isolated zeros. So the correct strengthening is: if the solutions are independent, \(W\) is nonzero at every single point of \(I\), with zero exceptions, not "all but finitely many." The proposed statement understates what the theorem actually delivers.