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| Title of object: |
Wronskian determinant |
| Canonical Name: |
WronskianDeterminant |
| Type: |
Definition |
| Created on: |
2002-02-19 01:41:25 |
| Modified on: |
2003-11-07 09:56:11 |
| Classification: |
msc:34-00 |
| Synonyms: |
Wronskian determinant=Wronskian |
Revision comment (for changes between this and next version):
| Changes for correction #10908 ('Typographic corrections'). |
Preamble:
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Content:
If we have some functions $f_1, f_2, \dots f_n$ then the \emph{Wronskian determinant} (or simply the Wronskian) $W(f_1,f_2,f_3 \dots f_n)$ is the determinant of the square matrix
$$W(f_1,f_2,f_3 \dots f_n) = \left |
\begin{array}{ccccc}
f_1 & f_2 & f_3 & \dots & f_n \\
f_1' & f_2' & f_3' & \dots & f_n' \\
f_1'' & f_2'' & f_3'' & \dots & f_n'' \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
f_1^{(n-1)} & f_2^{(n-1)} & f_3^{(n-1)} & \dots & f_n^{(n-1)} \\
\end{array} \right |
$$
where $f^{(k)}$ indicates the $k$th derivative of $f$ (not exponentiation).
The Wronskian of a set of functions $F$ is another function, which is zero over any interval where $F$ is linearly dependent. Just as a set of vectors is said to be linearly dependent whenever one vector may by expressed as a linear combination of a finite subset of the others, a set of functions $\{ f_1, f_2, f_3 \dots f_n\}$ is said to be dependent over an interval $I$ if one of the functions can be expressed as a linear combination of a finite subset of the others, i.e,
$$a_1f_1(t) + a_2f_2(t) + \cdots a_nf_n(t) = 0$$
for some $a_1, a_2, \dots a_n$, not all zero, at any $t \in I$.
Therefore the Wronskian can be used to determine if functions are independent. This is useful in many situations. For example, if we wish to determine if two solutions of a second-order differential equation are independent, we may use the Wronskian.
\paragraph{Examples}
Consider the functions $x^2$, $x$, and $1$. Take the Wronskian:
$$W = \left |
\begin{array}{ccc}
x^2 & x & 1 \\
2x & 1 & 0 \\
2 & 0 & 0 \\
\end{array} \right | = -2$$
Note that $W$ is always non-zero, so these functions are independent everywhere. Consider, however, $x^2$ and $x$:
$$W = \left |
\begin{array}{cc}
x^2 & x \\
2x & 1 \\
\end{array} \right | = x^2-2x^2 = -x^2$$
Here $W = 0$ only when $x = 0$. Therefore $x^2$ and $x$ are independent except at $x=0$.
Consider $2x^2+3$, $x^2$, and $1$:
$$W = \left |
\begin{array}{ccc}
2x^2 + 3 & x^2 & 1 \\
4x & 2x & 0 \\
4 & 2 & 0 \\
\end{array} \right | = 8x-8x = 0$$
Here $W$ is always zero, so these functions are always dependent. This is intuitively obvious, of course, since
$$2x^2 + 3 = 2(x^2) + 3(1)$$ |
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