Re: difficult question

In article <1167566101.056056.262260@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
tamiry <tamir.yehuda@xxxxxxxxx> wrote:
Hi,
Can any one please help with this:

Prove that the F(x,y) = C can be expressed as h(x,y) = C', where h is
harmonic function,
iff (@^2F/@x^2 + @^2F/@y^2)/(((@F/@x)^2)+((@F/@x)^2)) is a function of
F

F, h are real functions.
C, C' - constants.
@ - partial derivative.
@^2F/@x^2 - second order partial derivative of F by respect to x.


The basic fact you need to use is that for any function f of one variable,
we have:

Delta(f(F)) = f'(F) Delta F + f''(F) ||grad F||^2

Take the first part of the problem: if for all C we have F(x,y)=C iff
h(x,y)=C' for some harmonic function h (note that h will depend on C),
let f be the function that maps C to C' in this arrangement. We then have
f(F(x,y))=h(x,y)=C'. Although h depends on C (or equivalently C'), we
will always have that f(F) is harmonic and thus Delta(f(F))=0. The
identity above then gives the result.

For the converse, suppose that Delta(F) = f(F) ||grad F||^2. Multiplying
this by the integrating factor exp(-integral f)(F) gives:

g'(F) Delta F + g''(F) ||grad F||^2 = 0

where g(F) = integral( exp(-integral f)). Thus Delta(g(F))=0, and if
F(x,y)=C, then g(F(x,y))=g(C) with g(F(x,y)) being harmonic.

--
Daniel Mayost

.