Question

\[{∂^2 \over ∂x^2} \ln(x^2 + y^2)\] \[{∂^2 \over ∂x^2} \ln(x^2 + y^2)\] @Mathematics

Answer 1

fxx

Answer 2

derivate with respect X, treating y or anything else as a constant

Answer 3

you start with fx, fy, then you have fxx, fxy,fyx,fyy , it turns out that
fxy = fyx which is called fubini's theorem

Answer 4

\[F_x={∂ \over ∂x} ({2x\over x^2 +y^2 })\]

Answer 5

right so far?

Answer 6

yes

Answer 7

cool

Answer 8

\[= {2 \over x^2 +y^2}+ {-4x^2 \over (x^2+y^2)^2}\]

Answer 9

\[= {2 (x^2 + y^2) -4x^2 \over (x^2 +y^2)^2}\]

Answer 10

is this correct ?

Answer 11

...

Answer 12

Yes and simplify.

Answer 13

\[=-{2(x^2-y^2) \over (x^2 + y^2)^2}\]

Answer 14

Sure, or bring the - inside the numerator.

Answer 15

yeah i could do that but i am trying to prove that The Laplacian of (ln(x^2 + y^2) =0

Answer 16

Which should now be obvious as the above result is anti-symmetric in x and y.

Answer 17

yeah i got it now, Thanks a million JamesJ

Answer 18

The other way to do it is to use the Laplacian in polar coordinates, as your function is ln(r^2) = 2 ln(r).

Answer 19

For the record, in polars,
\[ \nabla^2 f(r,\theta) = f_{rr} + \frac{1}{r}f_r + \frac{1}{r^2}f_{\theta \theta} \]

Answer 20

With you function f = 2 ln r, hence \( f_r = 2/r , f_{rr} = -2/r^2 \) and therefore
\[ \nabla^2 f = -2/r^2 + (1/r) . (2/r) + 0 = 0 \]

Answer 21

However, if you don't already have the Laplacian in polars, the Cartesian calculation you're doing is much easier than proving the polar Laplacian! :-)

Answer 22

i diden't even think of polar coordinates, but that looks like a much more elegant way so solve the question. thanks