The Jacobian Determinant   

The determinant of the Jacobian matrix of a transformation is given by
  det
( J) = ê
ê
ê 		 
xu
xv
yu
yv
 ê
ê
ê 		   =    x
u
    y
v
   -    x
v
    y
u
  
However, we often use a notation for det( J) that is more suggestive of how the determinant is calculated.
   ( x,y)
( u,v)
  =    x
u
    y
v
   -    x
v
    y
u
  
The remainder of this section explores the Jacobian determinant and some of its more important properties.        

 

EXAMPLE 3    Calculate the Jacobian Determinant of

T( u,v) = á u2-v,u2+v ñ
Solution:  If we identify x = u2-v and y = u2+v, then
   ( x,y)
( u,v)
 
=
   x
u
    y
v
 -  x
v
    y
u
  
 
=
( 2u) ( 1) -( -1) ( 2u)
 
=
4u

   

Before we consider applications of the Jacobian determinant, let's develop some of the its properties.  To begin with, if x(u, v) and y(u, v) are differentiable functions, then
   ( y,x)
( u,v)
 
=
   y
u
    x
v
 -  y
v
    x
u
  
 
=
- æ
è 		   x
u
    y
v
 -  x
v
    y
u
 ö
ø 		 
 
=
-  ( x,y)
( u,v)
 
from which it follows immediately that
   ( x,x)
( u,v)
=  ( y,y)
( u,v)
= 0
Similarly, if f( u,v) , g( u,v) , and h(u,v) are differentiable, then
   ( f+g,h)
( u,v)
 
=
   ( f+g)
u
    h
v
 -  ( f+g)
v
    h
u
  
 
=
   f
u
    h
v
 +  g
u
    h
v
 -  f
v
    h
u
 -  g
v
    h
u
  
 
=
  æ
è 		   f
u
    h
v
 -  f
v
    h
u
 ö
ø 		+ æ
è 		   g
u
    h
v
 -  g
v
    h
u
 ö
ø 		 
 
=
   ( f,h)
( u,v)
+  ( g,h)
( u,v)
 
The remaining properties in the next theorem can be obtained in similar fashion.       

Theorem 3.2: If f( u,v) , g( u,v) , and h( u,v) are differentiable functions and k is a number, then
 
   ( g,f)
( u,v)
= -  ( f,g)
( u,v)
 
 
   ( f+g,h)
( u,v)
=  ( f,h)
( u,v)
+  (g,h)
( u,v)
 
   ( f,f)
( u,v)
= 0
 
   ( f-g,h)
( u,v)
=  ( f,h)
( u,v)
+  ( g,h)
( u,v)
 
   ( kf,g)
( u,v)
= k  ( f,g)
( u,v)
 
 
   ( fg,h)
( u,v)
=  ( f,h)
( u,v)
 g+f   (g,h)
( u,v)
 
  

These and additional properties will be explored in the exercises.    

 

EXAMPLE 4    Verify the property

   ( fg,h)
( u,v)
=  ( f,h)
( u,v)
 g  +  f   ( g,h)
( u,v)
 
Solution:  Direct calculation leads to
   ( fg,h)
( u,v)
 
=
   ( fg)
u
    h
v
 -  ( fg)
v
    h
u
  
 
=
  æ
è 		   f
u
  g+f   g
u
 ö
ø 		   h
v
 - æ
è 		   f
v
  g+f   g
v
 ö
ø 		   h
u
  
 
=
  æ
è 		   f
u
    h
v
 -  f
v
    h
u
 ö
ø 		g+f æ
è 		   g
u
    h
v
 -  g
v
    h
u
 ö
ø 		 
 
=
   ( f,h)
( u,v)
 g+f   ( g,h)
( u,v)
 

        Check your Reading: If k is constant and f(u,v) is differentiable, then what is
   ( k,f)
( u,v)
?