Differential Forms

Stoke's theorem and Green's theorem can be both unified and extended. However, before doing so, it is advantageous to define a new notation for derivatives and Jacobian determinants.

To begin with, if every possible partial derivative of a function exists and is continuous, then we say that the function is C^¥.  If  x(u,v),  y(u,v), and z(u,v) are C^¥ functions of u and v, then let us define the wedge products or exterior products by

dx^dy =  ¶( x,y) ¶(u,v) dudv,        dy^dz =  ¶(y,z) ¶( u,v) dudv,        dz^dx =  ¶( z,x) ¶( u,v) dudv

The properties of the wedge product follow from the Jacobian determinant. For example,

dy^dx =  ¶( y,x) ¶(u,v) dudv =  -¶( x,y) ¶(u,v) dudv = -dx^dy

As a result, dx^dx = -dx^dx, so that dx^dx = 0. Similarly, the exterior product has all of the following properties:

dy^dx = -dx^dy,        dz^dy = -dy^dz,        dx^dz = -dz^dx

dx^dx = dy^dy = dz^dz = 0

We also define the wedge-product to distribute over addition, so that for example

dx^( dy+dz) = dx^dy + dx^dz

Secondly, we use differentials and the exterior product to define differential forms. To begin with, 0-forms are simply functions U( x,y,z) , and one-forms are are defined

F = M( x,y,z) dx + N( x,y,z) dy + P( x,y,z) dz

(1-forms are closely related to vector fields). In addition, two-forms are of the form

w = A( x,y,z) dy^dz  +  B( x,y,z) dz^dx  +  C( x,y,z) dx^dy

Finally, let us define a differential operator on differential forms. To begin with, if U( x,y,z) is a 0-form, then its differential dU is the 1-form given by

dU = Uxdx + Uydy + Uzdz

Likewise, if F = Mdx + Ndy + Pdz is a 1-form, then its differential dF is the 2-form given by

dF = dM^dx + dN^dy + dP^dz

where dM, dN, and dP are differentials of the 0-forms M, N, and P.      

EXAMPLE 1    Evaluate dU when U( x,y,z) = x²y+z².       

Solution: Since U is a 0-form, its differential is

dU =  ¶ ¶x ( x2y+z2) dx  +    ¶ ¶y ( x2y+z2) dy  +    ¶ ¶z ( x2y+z2) dz = 2xydx+x2dy+2zdz

Notice that dU is a 1-form.       

EXAMPLE 2    Evaluate dw when w = x²dx+xy dy        

Solution: Since w is a 1-form, its differential is

dw = d( x2) ^dx+d( xy) ^dy

However, d( x²) = 2xdx and d( xy) = ¶_x( xy) dx + ¶_y( xy) dy = ydx + xdy. Thus,

dw = ( 2xdx) ^dx+[ ydx+xdy] ^dy = 2x  dx^dx+y  dx^dy+x  dy^dy = 2x·0+y  dx^dy+x·0

That is, dw = y  dx^dy, which is a 2-form.