Vronsky determinant

With the help of the Wronski determinant , which was named after the Polish mathematician Josef Hoëné-Wroński (1776-1853), scalar functions can be tested for linear independence if they are sufficiently often differentiable . This can be particularly useful when solving an ordinary differential equation .

definition

For real or complex valued functions on an interval the Wronsky determinant is defined by ${\displaystyle n}$ ${\displaystyle f_{1},\dots ,f_{n}}$ ${\displaystyle I}$

${\displaystyle W(f_{1},\dots ,f_{n})(t)={\begin{vmatrix}f_{1}(t)&f_{2}(t)&\dots &f_{n}(t)\\f_{1}^{(1)}(t)&f_{2}^{(1)}(t)&\dots &f_{n}^{(1)}(t)\\\vdots &\vdots &\vdots &\vdots \\f_{1}^{(n-1)}(t)&f_{2}^{(n-1)}(t)&\dots &f_{n}^{(n-1)}(t)\end{vmatrix}},\qquad t\in I,}$

The functions are in the first line and the superscript numbers in brackets denote the first to th derivative in the following lines . ${\displaystyle (n-1)}$

properties

The calculation of the Wronsky determinant of linear, ordinary differential equations of the second order can be simplified by the application of the Abelian identity if the coefficient is in the representation in the fundamental system . ${\displaystyle y''(t)-a_{1}y'(t)-a_{0}y(t)=0}$ ${\displaystyle a_{1}\neq 0}$

Criterion for linear independence

Applies for a , then the functions are on the interval linearly independent . On the other hand, the linear dependence of the functions does not follow for all of them . That is, the equality does not result in a linear dependence on the interval . Because it applies that the functions can be locally linearly independent (see counterexample). ${\displaystyle W(f_{1},\ldots ,f_{n})(t_{0})\neq 0}$ ${\displaystyle t_{0}\in I}$ ${\displaystyle f_{1},\ldots ,f_{n}}$ ${\displaystyle I}$ ${\displaystyle W(f_{1},\ldots ,f_{n})(t)=0}$ ${\displaystyle t\in I}$ ${\displaystyle f_{1},\ldots ,f_{n}}$ ${\displaystyle I}$

example

Starting from the Sturm-Liouville problem , the differential equation becomes second order

${\displaystyle -\psi ''=\lambda \psi }$

considered with the boundary conditions . Anyone is chosen as the solution for and . Because of the boundary conditions is and so and therefore for . The solution is therefore ${\displaystyle \psi (0)=\psi (\pi )=0}$ ${\displaystyle \psi (t)=\alpha \sin({\sqrt {\lambda }}t)+\beta \cos({\sqrt {\lambda }}t)}$ ${\displaystyle \lambda >0}$ ${\displaystyle \alpha ,\beta \in \mathbb {R} }$ ${\displaystyle \psi (0)=\psi (\pi )=0}$ ${\displaystyle \alpha \neq 0,\beta =0}$ ${\displaystyle \sin({\sqrt {\lambda }}\pi )=0,}$ ${\displaystyle {\sqrt {\lambda }}\pi =n\pi }$ ${\displaystyle \lambda =n^{2}}$ ${\displaystyle n\in \mathbb {N} }$

${\displaystyle \psi (t)=\alpha \sin(nt)}$

elected. Since another solution of this differential equation is given by with (see Sturm-Liuoville problem), the second solution ${\displaystyle \phi =p\psi '=\psi '}$ ${\displaystyle p=1}$

${\displaystyle \phi (t)=\alpha n\cos(nt)}$

assumed and checked for linear independence using the Wronsky determinant. It follows

${\displaystyle W(\phi ,\psi )(t)={\begin{vmatrix}\phi (t)&\psi (t)\\\phi '(t)&\psi '(t)\end{vmatrix}}={\begin{vmatrix}\alpha n\cos(nt)&\alpha \sin(nt)\\-\alpha n^{2}\sin(nt)&\alpha n\cos(nt)\end{vmatrix}}=(\alpha n)^{2}\cos ^{2}(nt)+(\alpha n)^{2}\sin ^{2}(nt)=(\alpha n)^{2}>0}$.

So is for (more precisely for all ) and the linear independence of the functions is given. ${\displaystyle W(\phi ,\psi )(t)=\left|{\begin{smallmatrix}\phi (t)&\psi (t)\\\phi '(t)&\psi '(t)\end{smallmatrix}}\right|>0}$ ${\displaystyle t\geq 0}$ ${\displaystyle t\in \mathbb {R} }$ ${\displaystyle \phi (t),\psi (t)}$

Counterexample

The functions defined on the real numbers serve as a counterexample

${\displaystyle f_{1}(t)={\begin{cases}0\ ,&{\mbox{falls }}t\leq 0,\\t^{2}\ ,&{\mbox{falls }}t>0,\end{cases}}\qquad {\mbox{und}}\qquad f_{2}(t)={\begin{cases}t^{2}\ ,&{\mbox{falls }}t\leq 0,\\0\ ,&{\mbox{falls }}t>0.\end{cases}}}$

For all true ${\displaystyle t\in \mathbb {R} }$

${\displaystyle W(f_{1},f_{2})(t)={\begin{vmatrix}f_{1}(t)&f_{2}(t)\\f'_{1}(t)&f'_{2}(t)\end{vmatrix}}=0.}$

But leads for to and for to what the linear independence on or for the two functions implies. For true and what linear dependence means. ${\displaystyle \lambda \ f_{1}(t)+\mu \ f_{2}(t)=0}$ ${\displaystyle t=1}$ ${\displaystyle \lambda =0}$ ${\displaystyle t=-1}$ ${\displaystyle \mu =0}$ ${\displaystyle t=1}$ ${\displaystyle t=-1}$ ${\displaystyle t=0}$ ${\displaystyle f_{1}(0)=0}$ ${\displaystyle f_{2}(0)=0}$ ${\displaystyle t=0}$

literature

• H. Heuser: Ordinary differential equations . Teubner, 1995, ISBN 3-519-22227-2 , p. 250.

Web links

• Eric W. Weisstein : Wronskian . In: MathWorld (English).