Prove the Second Derivative of a Multivariate Function Using the Chain Rule

[Yagoda]

Homework Statement

Let [itex] h(u,v) = f(a(u,v), b(u,v))[/itex], where [itex] a_u = b_v[/itex] and [itex]a_v = -b_u[/itex].
Show that [itex]h_{uu} + h_{vv} = (f_{xx} + f_{yy}) (a^2_u + a^2_v)[/itex].

Homework Equations

The Attempt at a Solution

I suppose my first question is where the x's and y's come from. (I thought at first it was a typo in the problem, but this type of setup appears in several other exercises in the book).
To try to make it easier to understand I tried letting the a's and x's and b's be y's so that we get [itex]h(u,v) = f(x(u,v), y(u,v)) [/itex], but then I realized that to prove the result we need apparently both a's, b's, x's and y's.

To compute [itex]h_{uu}[/itex] we would begin by getting [itex]h_u[/itex], but I'm having trouble figuring this out since I think all the letters are tripping me up.

 

[jedishrfu]

x=a(u,v) and y=b(u,v)

First revisit the simple chain rule:

y = g(f(x)) which we can write as y = g(u) with u=f(x)

dy/dx = dg/du * du/dx

and so in your case they want you to compute the second derivative of h

ph/pu = pf/px * px/pu + pf/py * py/pu (think p as the partial derivative operator)