# Leibniz' rule

• Nov 4th 2009, 11:00 AM
pepsi
Leibniz' rule
hello

i've been trying for ages and this one question i can't answer, show that:

u(x,y)=(y/pi) times integral , from - to + infinity, of {f(t)dt/[(x-t)^2 + y^2]}

satisfies

'second partial of u w.r.t x' + 'second partial of u w.r.t y' = 0

i've tried

du =
{-(y/pi)times integral, from -to+ inf, of {2(x-t)f(t)dt/[(x-t)^2 + y^2]^2}}dx
+
{(1/pi)times integral , from - to + infinity, of {f(t)dt/[(x-t)^2 + y^2]}
(y/pi)times integral, from -to+ infinity, of {2yf(t)dt/[(x-t)^2 + y^2]^2}}dy

and then take the second derivatives of the partials and add them but it don't work.

NOW I AM SURE i am going about this the wrong way, BUT HOPEFULLY some brighter soul out there will see what i am trying to do and any way tell me what is the right way
• Nov 4th 2009, 12:08 PM
shawsend
I'll do the first one (assuming everything sufficient to take the partials):

$\frac{\partial}{\partial x} u=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{\partial}{\partial x}\left(\frac{f(t)}{(x-t)^2+y^2}\right)dx=-\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{2(x-t)f(t)}{\left[(x-t)^2+y^2\right]^2}dt$

$\frac{\partial^2 u}{\partial x^2}=\frac{y}{\pi}\int_{-\infty}^{\infty} \frac{8(x-t)^2f(t)}{((x-t)^2+y^2)^3}-\frac{2f(t)}{((x-t)^2+y^2)^2}dt=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{[(x-t)^2-2y^2]f(t)}{[(x-t)^2+y^2)^3}dt$

I did check the results in Mathematica and the Laplace expression does equate to zero.
• Nov 4th 2009, 12:20 PM
pepsi
Quote:

Originally Posted by shawsend
I'll do the first one (assuming everything sufficient to take the partials):

$\frac{\partial}{\partial x} u=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{\partial}{\partial x}\left(\frac{f(t)}{(x-t)^2+y^2}\right)dx=-\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{2(x-t)f(t)}{\left[(x-t)^2+y^2\right]^2}dt$

$\frac{\partial^2 u}{\partial x^2}=\frac{y}{\pi}\int_{-\infty}^{\infty} \frac{8(x-t)^2f(t)}{((x-t)^2+y^2)^3}-\frac{2f(t)}{((x-t)^2+y^2)^2}dt=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{[(x-t)^2-2y^2]f(t)}{[(x-t)^2+y^2)^3}dt$

I did check the results in Mathematica and the Laplace expression does equate to zero.

yes thanks, i do appreciate this . however i did get that far my self so my issue is with the second one.

p.s. don't think i am pushing to get the full answer, it simply the case and that's how it is. anyway, i do appreciate your help
• Nov 5th 2009, 06:47 AM
shawsend
Quote:

Originally Posted by shawsend
$\frac{\partial^2 u}{\partial x^2}=\frac{y}{\pi}\int_{-\infty}^{\infty} \frac{8(x-t)^2f(t)}{((x-t)^2+y^2)^3}-\frac{2f(t)}{((x-t)^2+y^2)^2}dt=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{[(x-t)^2-2y^2]f(t)}{[(x-t)^2+y^2)^3}dt$

That should be:

$\frac{\partial^2 u}{\partial x^2}=\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{f(t)\left[6y(x-t)^2-2y^3\right]}{((x-t)^2+y^2)^3}dt$

and I think you got the other one by now. It's messy and I had problems with it too but here it is:

$\frac{\partial u}{\partial y}=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{-2y f(t)}{((x-t)^2+y^2)^2}+\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f(t)}{(x-t)^2+y^2}$

$\frac{\partial^2 u}{\partial y^2}=\frac{y}{\pi}\left[\int_{-\infty}^{\infty} \frac{f(t)\left[(x-t)^2+y^2\right]^2(-2)+8y^2\left[(x-t)^2+y^2\right]}{((x-t)^2+y^2)^4}\right]$ $-\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{4yf(t)}{(x-t)^2+y^2)^2}$

$=\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{f(t)\left[-2y^3-2y(x-t)^2+8y^3-\left[4y((x-t)^2+y^2)\right]\right]}{((x-t)^2+y^2)^3}$

$=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f(t)\left[2y^3-6y(x-t)^2\right]}{((x-t)^2+y^2)^3}$

which cancels the other one.