Wronskian: Definition from Answers.com

In mathematics, the Wronskian is a function named after the Polish mathematician Józef Hoene-Wroński. It is especially important in the study of differential equations, where it can be used to determine whether a set of solutions is linearly independent.

1 Definition
2 The Wronskian and linear independence
3 Examples
4 Proof: Wronskian and linear independence
5 Abstract definition
6 External links

Definition

For n real- or complex-valued functions f₁, ..., f_n, which are n − 1 times differentiable on an interval I, the Wronskian W(f₁, ..., f_n) as a function on I is defined by

$W(f_1, \ldots, f_n) (x)= \begin{vmatrix} f_1(x) & f_2(x) & \cdots & f_n(x) \\ f_1'(x) & f_2'(x) & \cdots & f_n' (x)\\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(n-1)}(x)& f_2^{(n-1)}(x) & \cdots & f_n^{(n-1)}(x) \end{vmatrix},\qquad x\in I.$

That is, it is the determinant of the matrix constructed by placing the functions in the first row, the first derivative of each function in the second row, and so on through the (n - 1)st derivative, thus forming a square matrix sometimes called a fundamental matrix.

In linear differential equations, the Wronskian can be computed more easily by Abel's identity.

The Wronskian and linear independence

The Wronskian can be used to determine whether a set of differentiable functions is linearly independent on a given interval:

If the Wronskian is non-zero at some point in an interval, then the associated functions are linearly independent on the interval.

This is useful in many situations. For example, if we wish to verify that two solutions of a second-order differential equation are independent, we may use the Wronskian. Note that if the Wronskian is zero everywhere in the interval, the functions may or may not be linearly independent. A common misconception is that W = 0 everywhere implies linear dependence; the third example below shows that this is not true.

Examples

Consider the functions f₁(x) = 1, f₂(x) = x and f₃(x) = x³, defined for every real number x. Calculate the Wronskian:

$W(f_1,f_2,f_3)(x) = \begin{vmatrix} 1 & x & x^3\\ 0 & 1 & 3x^2 \\ 0 & 0 & 6x \end{vmatrix} = 6x, \qquad x\in\mathbb R.$

We see that there is at least one real x where the Wronskian is not zero (indeed it's non-zero for all x ≠ 0), so these three functions must be linearly independent.

Consider the functions f₁(x) = 1, f₂(x) = x² and f₃(x) = 3 + 2x², defined for every real number x. These functions are clearly dependent, since f₃(x) = 3f₁(x) + 2f₂(x) everywhere on the real line. Thus, the Wronskian must be zero everywhere, which is indeed that case as the following quick calculation shows:

$W(f_1,f_2,f_3)(x) = \begin{vmatrix} 1 & x^2 & 3+2x^2 \\ 0 & 2x & 4x \\ 0 & 2 & 4 \end{vmatrix} = 8x-8x = 0,\qquad x\in\mathbb R.$

The third column vector is three times the first plus two times the second one, hence the column vectors are linearly dependent and the determinant has to be zero. This linear relation between the column vectors has the same coefficients as the linear relation between the functions. The proof below generalizes this observation.

As mentioned above, if the Wronskian is zero everywhere, it does not mean in general that the functions involved are linearly dependent. As an example, consider the functions f₁(x) = x² and

$f_2(x) = \begin{cases} -f_1(x) & \mathrm{for} \; x < 0, \\ f_1(x) & \mathrm{for} \; x \ge 0. \end{cases}$

Note that f₂ is differentiable at zero. Since f₂ is not a constant multiple of f₁ (same constant for all real x), these two functions are linear independent over the set of real numbers. However, their Wronskian is zero everywhere:

$W(f_1,f_2)(x) = \begin{cases} \begin{vmatrix} x^2 & -x^2 \\ 2x & -2x \end{vmatrix} = 0 & \mathrm{for} \; x < 0, \\[15pt] \begin{vmatrix} x^2 & x^2 \\ 2x & 2x \end{vmatrix} = 0 & \mathrm{for} \; x \ge 0 \end{cases}$

(the determinants are zero because the column vectors are linearly dependent in both cases).

Proof: Wronskian and linear independence

Suppose that the functions are linearly dependent over the interval I, hence there are real (or complex) numbers λ₁, ..., λ_n (not all of them zero) such that

$\lambda_1f_1(x)+\cdots+\lambda_nf_n(x)=0,\qquad x\in I.$

Since differentiation is a linear operation, this equation also holds for all the n − 1 derivatives. Hence this linear combination of the columns of the associated Wronskian matrix is also zero for every x in I, which means that the columns are always linearly dependent. Consequently, the Wronskian determinant is zero at all points of the interval. Therefore, if the Wronskian determinant is nonzero at some point, the functions have to be linearly independent.

Abstract definition

There is a sense in which the Wronskian of an n-th order linear differential equation is its n-th exterior power. For that idea to be implemented one must be working with some formulation in which differential equations are sufficiently like vector spaces: for example in the language of vector bundles carrying a connection.

External links

This article incorporates material from Wronskian on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.
More on the Wronskian, Paul's Online Math Notes
Wronskian defined on Mathworld.wolfram.com.

This entry is from Wikipedia, the leading user-contributed encyclopedia. It may not have been reviewed by professional editors (see full disclaimer)

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