Author: Eiko

Tags: Differential Algebra, Differential Equations, Wronskian, Determinant, Linearly Dependent

Wronskian

The Wronskian Matrix

For a differential field $K$ and some elements $y_1,\dots ,y_n\in K$, their associated Wronskian matrix is defined as

$$W(y_1,\dots ,y_n)=\left(\begin{array}{cccc}{y}_1& y_2& \cdots & y_n\\ y_1^{\prime }& y_2^{\prime }& \cdots & y_n^{\prime }\\ ⋮& ⋮& \ddots & ⋮\\ y_1^{(n-1)}& y_2^{(n-1)}& \cdots & y_n^{(n-1)}\end{array}\right)$$

where $y_i^{(j)}$ denotes the $j$-th derivative of $y_i$. It is used to determine if the elements $y_i$ are linearly dependent over the constant field $C_K=K^{\partial =0}$.

Curious Thought: Why do we need these derivatives?

One might wonder what is the point of all these derivatives. In linear algebra to tell whether a set of vector is linear dependent we only have to verify some determinants on its coordinates. But here the space of vectors are ‘functions’, for actual differential equations such spaces are infinite dimensional. So functions being linearly dependent is a pretty strong condition.

After you realize the above, you will be curious again, functions being linearly dependent is stronger than saying $(y_i,y_i^{\prime },\dots ,y_i^{(n-1)}),i=1,\dots ,n$ are linearly dependent as vectors (on certain points). But for the Wronskian to vanish as an element in $K$, we are requiring them being dependent over all points. The theory of differential algebra tells you that you don’t need to consider functions, you just need to consider them as abstract elements in the differential field for the story to work.

The Wronskian (Determinant)

Denote by

$$|W|=detW(y_1,\dots ,y_n).$$

It enjoys the following properties:

• If there is a linear transform $z=Cy$ where $c$ is a matrix consists of $C_K$, then

  $$|W(z)|=det(C)|W(y)|.$$

• If $y_1,\dots ,y_n$ are linearly dependent over $C_K$, i.e. there exists not all zero coefficients $c_i$ such that $\sum c_i{y}_i=0$, then $|W(y_1,\dots ,y_n)|=0$.

• If $|W(y_I)|=0$ and there exists a non-vanishing Wronskian on a subset $|W(y_{I^{\prime }})|\ne 0$ where $I^{\prime }\subset I$ is a subset of $n-1$ elements, then $y_i$ are linearly dependent over $C_K$.

• If $|W(y_I)|=0$, then they are linearly dependent over $C_K$.

Proof

Only the last two properties need proof.

• If $|W|=0$ in $K$, we should have a dependence relation in $K$

  $$\sum c_i{y}_i^{(k)}=0,k=0,\dots ,n-1.$$

  The problem is $c_i$ are not necessarily constants. If we differentiate the equation, we shall get

  $$\sum c_i^{\prime }{y}_i^{(k)}+\sum c_i{y}_i^{(k+1)}=0,k=0,\dots n-1.$$

  But the second term is zero for $k=0,\dots ,n-2$, this gives us

  $$\sum c_i^{\prime }{y}_i^{(k)}=0,k=0,\dots ,n-2,$$

  which is impossible unless $c_i^{\prime }$ are all $0$, since our assumption implies that the matrix $(y_i^{(j)}{)}_{0\le j\le n-2}$ is full rank (which is $n-1$).

• For the last property, consider the largest subset $I^{\prime }$ where $W(y_{I^{\prime }})\ne 0$and a subset $I^{″}\supset I^{\prime }$ strictly and slightly larger than $I^{\prime }$, then $W(y_{I^{″}})=0$. By the previous property, $y_{I^{″}}$ are linearly dependent over $C_K$, so is $y_I$.

Relation with Differential Equations

Let $K$ be a differential field, consider a differential equation in $K$,

$$PY=Y^{(n)}+a_{n-1}{Y}^{(n-1)}+\cdots +a_{0}Y=0,$$

where $P$ can be thought as an element of $K[\partial ]$.

• If $PY=0$ has solutions $y_1,\dots ,y_{n+1}$ in any possibly extension field $L\supset K$, then $|W|=0$.

  This follows from the fact that the differential equation expresses the last row as a linear combination of the previous rows.

• In any $L\supset K$, there can be atmost $n$ $C_L$-linearly independent solutions to $PY=0$.

  This is a corollary of the previous property and the last property of the Wronskian.