## 8.12. Nonsingularity Of The Wronskian Matrix Of N Independent Solutions Of A Homogeneous Linear Equation

Consider the Wronskian matrix W of n independent solutions u 1 ,, u n  of the homogeneous equation

L( y )= y ( n ) + k=1 n P k ( x ) y ( nk ) =0                      (8.23a)

on an interval J.  Let w( x )=detW( x ) .  Then we have:

### Theorem 8.12.

w + P 1 ( x )w=0                        (8.23)

on J.  Hence,   cJ , we have

w( x )=w( c )exp[ c x dt    P 1 ( t ) ]         [ Abel’s formula ]          (8.24)

Furthermore, w( x )0     xJ  and W( x )  is nonsingular on J.

#### Proof

Let u be the row vector u=( u 1 ,, u n ) .  Hence, we can write

w( x )=detW =det( u, u ,, u ( n2 ) , u ( n1 ) )

According to Ex.7, §5.22, we have

w =det( u, u ,, u ( n2 ) , u ( n ) )

On the other hand,

P 1 ( x )w=det( u, u ,, u ( n2 ) , P 1 ( x ) u ( n1 ) )

Hence,

w + P 1 ( x )w=det( u, u ,, u ( n2 ) , u ( n ) + P 1 ( x ) u ( n1 ) )

Since uk are solutions of (8.23a), the rows of the determinant are dependent.  This proves (8.23).  The Abel’s formula follows by integrating (8.23) over [ c,x ] .

Since e x 0  for all real x, we see that w( x )0  for all x if there exists a cJ  such that w( c )0 .  The existence of c can be shown by a contradiction argument.  Thus, assume that w( t )=0     tJ .  We then choose a fixed point t= t 0  and consider the system of linear equations

W( t 0 )X=O                    (a)

where X is a column vector.  Since w( t 0 )=detW( t 0 )=0 , there exists a nonzero solution, say, X= ( c 1 ,, c n ) t O .  Consider then the linear combination

f( t )= k=1 n c k u k ( t )

Obviously, L( f )=0  since L( u k )=0  for all k.  Also, putting X into eq(a) gives

k=1 n [ W( t 0 ) ] ik c k = k=1 n u k ( i1 ) ( t 0 ) c k   = f ( i1 ) ( t 0 )=0              for i=1,,n

which means the initial-value vector of f at t= t 0  is O.  By the uniqueness theorem, f( t )=O  for all t, which means c k =0  for all k, contrary to our original assumption.  QED.