The Jacobian of a Transformation

A coordinate transformation T( u,v) = áx( u,v) ,y( u,v) ñ is said to be smooth if x( u,v) and y( u,v) have continuous partial derivatives. The matrix

J( u,v) = é ê ë xu xv yu yv ù ú û

is called the Jacobian Matrix of a smooth coordinate transformation T( u,v) . The Jacobian J( u,v) is in some sense the "derivative" of T( u,v) , and the sense in which it is the derivative will be explored in this section.       

EXAMPLE 1    What is the Jacobian matrix for the polar coordinate transformation?    

Solution: Since x = rcos( q) and y = rsin( q) , the Jacobian matrix is

J( r,q) = é ê ë xr xq yr yq ù ú û = é ê ë cos( q) -rsin( q) sin( q) rcos( q) ù ú û

A transformation T( u,v) = á x(u,v) ,y( u,v) ñ can be written in matrix form as

T( u,v) = é ê ë x( u,v) y( u,v) ù ú û

If u( t) = á u( t) ,v(t) ñ is a curve in the uv-plane, then x( t) = T( u( t) ,v( t) ) is the image of u( t) in the xy-plane. Moreover,

dx  dt  =  é ê ê ê ê ë xu  du dt +xv  dv dt yu  du dt +yv  dv dt ù ú ú ú ú û = é ê ë xu xv yu yv ù ú û é ê ê ê ê ë  du dt  dv dt ù ú ú ú ú û

The last vector is du/dt. Thus, we have shown that if r( t) = T( u( t) ) , then

dx  dt  = J(u)  du  dt

That is, the Jacobian maps tangent vectors to curves in the uv-plane to tangent vectors to curves in the xy-plane.
It is in this sense that J( u,v) is the derivative of T(u,v) .        

EXAMPLE 2    Let T( u,v) = áu²-v²,2uv ñ

    a) Find the velocity of u( t) = át,t² ñ when t = 1.

    b) Find the Jacobian and apply it to the vector in a)

    c) Find x( t) = T( u( t)) in the xy-plane and then find its velocity vector at t = 1. Compare to the result in (b).        

Solution: a) Since u' ( t) = á 1,2t ñ , the velocity at t = 1 is u' ( 1) = á 1,2 ñ .

b) Since x( u,v) = u²-v² and y( u,v) = 2uv, the Jacobian of T( u,v) is

J( u,v) = é ê ë xu xv yu yv ù ú û = é ê ë 2u -2v 2v 2u ù ú û

which at the point ( 1,1) is given by

J( 1,1) = é ê ë 2 -2 2 2 ù ú û

Identifying u' ( 1) = á1,2 ñ with [ 1,2] ^t leads to

J( 1,1) u' ( 1) = é ê ë 2 -2 2 2 ù ú û é ê ë 1 2 ù ú û = é ê ë -2 6 ù ú û

c) Substituting u = t, v = t² into T( u,v) = áu²-v²,2uv ñ results in

x( t) = ( t2-t4,2t3)

which has a velocity of x' ( t) = á2t-4t³,6t² ñ . Moreover, x' (1) = á -2,6 ñ