= Wronskian
{c}
{title2=$W(f,g)=fg'-f'g$}
{wiki}
For two <differentiable functions> $f,g$ on an <interval> $I$, their <Wronskian> is the <function>
$$
W(f,g)(x)=\det\begin{pmatrix}f(x)&g(x)\\f'(x)&g'(x)\end{pmatrix}=f(x)g'(x)-f'(x)g(x).
$$
The order matters: exchanging $f$ and $g$ reverses the sign. For $n$ sufficiently <differentiable functions>, the <Wronskian> is the <determinant> $\det(f_j^{(i-1)}(x))_{i,j=1}^n$.<NIST Digital Library of Mathematical Functions/1 Algebraic and Analytic Methods>[NIST Digital Library of Mathematical Functions; §1.13(i), equation 1.13.4 and the following n-argument definition]{ref}
The <Wronskian> measures whether the <vectors> $(f(x),f'(x))$ and $(g(x),g'(x))$ supply independent value-and-<derivative> data at that point. For real-valued <functions>, its absolute value is the <area> of the <parallelogram> they span. If a constant <linear combination> $af+bg$ is zero throughout $I$, then $af'+bg'=0$ too. A nonzero <Wronskian> at even one point makes this pair of <linear equations> invertible and forces $a=b=0$. Thus a nonzero <Wronskian> somewhere always proves <linear independence> of the <functions> on $I$.
For an explicit example, the <exponential functions>
$$
y_1(x)=e^{-x},\qquad y_2(x)=e^{2x}
$$
solve the same <homogeneous linear differential equation> $y''-y'-2y=0$ on the whole real line: substitution gives $1+1-2=0$ and $4-2-2=0$, respectively. Their <Wronskian> is
$$
W(y_1,y_2)=e^{-x}(2e^{2x})-(-e^{-x})e^{2x}=3e^x>0.
$$
For the <initial value problem> $y(0)=2$, $y'(0)=-1$, the <linear combination> $y=c_1y_1+c_2y_2$ requires $c_1+c_2=2$ and $-c_1+2c_2=-1$. Hence $c_1=5/3$, $c_2=1/3$, and $y=\frac53e^{-x}+\frac13e^{2x}$. The nonzero <Wronskian> makes the two <initial conditions> independently adjustable.
\Image[/wronskian-exponential-solutions-and-area.png]
{title=Independent <exponential functions> supply independent initial data, with <Wronskian> equal to the oriented <parallelogram> area}
The stronger converse requires a common equation. Suppose $y_1,y_2$ solve $y''+p(x)y'+q(x)y=0$ with <continuous functions> $p,q$ on one <interval> $I$. Then the <functions> are <linearly independent> if and only if their <Wronskian> is nonzero at one point, equivalently at every point. If $W(x_0)=0$, choose constants $a,b$, not both zero, making $z=ay_1+by_2$ satisfy $z(x_0)=z'(x_0)=0$. The <superposition principle> makes $z$ a <solution of a differential equation> for the same equation, and the <uniqueness theorem for ordinary differential equations> forces $z$ to be identically zero. This proves <linear dependence>.<18.03 Supplementary Notes by Haynes Miller/The Wronskian>[18.03 Supplementary Notes by Haynes Miller; Chapter 15, The Wronskian, printed pp. 72–73: initial-data system (3) and the zero-or-nowhere-zero theorem]{ref}
For arbitrary <differentiable functions>, an identically zero <Wronskian> need not imply <linear dependence>. Consider, on $I=(-2,2)$,
$$
f(x)=x^4,\qquad g(x)=x|x|^3=\begin{cases}-x^4,&x<0,\\x^4,&x\geq0.\end{cases}
$$
Both <functions> have three continuous <derivatives>. On each side of zero, $g$ is a constant multiple of $f$, so $W(f,g)=0$; at zero their values and first <derivatives> are all zero, so the <Wronskian> is zero there too. Nevertheless $af+bg=0$ evaluated at $x=-1$ gives $a-b=0$, while evaluation at $x=1$ gives $a+b=0$. Hence $a=b=0$: these <functions> are <linearly independent> on the whole <interval>. There is no single constant multiplier valid on both sides of the common zero.
\Image[/wronskian-zero-with-independent-functions.png]
{title=An identically zero <Wronskian> can coexist with <linear independence> when the common regular equation hypothesis is absent}
These <functions> cannot both solve any equation $y''+py'+qy=0$ with <continuous functions> $p,q$ throughout $(-2,2)$: even $f$ alone would contradict uniqueness because $f(0)=f'(0)=0$ but $f$ is not identically zero. They do solve the singular equation $x^2y''-3xy'=0$, whose leading coefficient vanishes at zero. Dividing by $x^2$ gives $p=-3/x$ only away from zero; <Abel's identity> cannot be applied across that point. The example pinpoints why checking the equation and its <interval> is essential before using a zero <Wronskian> to infer <linear dependence>.
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