Differential equation in a distribution space

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• Apr 4th 2010, 11:22 AM
mambutu
Differential equation in a distribution space
Hi everybody!

I need your help to solve the following exercise:

find the general solution of the equation
x^2*u'(x) = pv(1/x)
where u'(x) is the derivative of u with respect to x and pv(1/x) stands for 'principal value of 1/x'.

The answer should be u(x) = c1 + c2*theta(x) + c3*delta(x) - 1/2*fp(1/x^2), where theta(x) is the Heaviside's function and fp(1/x^2) is the 'finite part of 1/x^2'.

For the sake of clarity, I solved similar equations in distributions spaces making use of Green's functions, Fourier transforms and convolutions, etc., but I can't figure out how to approach this one.

Thanks in advance for your suggestions!
• Apr 13th 2010, 01:36 PM
mambutu
I hope someone can help me...
• Apr 13th 2010, 02:24 PM
lvleph
Most people here seem to do Pure Mathematics, so you will be hard pressed to find help. Unfortunately, we don't cover Green's Functions until tomorrow, so I can't help yet.
• Jul 25th 2010, 07:49 AM
skyscreper
I think it's too late for givin a usefull answer to mambutu, however i write for next viewer:
First observe that: if we indicate with $\displaystyle f^{(m)}$ the $\displaystyle m$-order derivative of $\displaystyle f$
if $\displaystyle m<n$
then $\displaystyle x^n \delta^{(m)}=0$
In fact $\displaystyle (x^n \delta^{(m)}, \phi)=(\delta^{(m)},x^n \phi)=\pm(\delta,(x^n\phi)^{(m)}) = \pm(\delta, Ax^n+...+Bx^{n-m})=0$
So if u want to solve
$\displaystyle x^2 u^{(1)}=0$
we remember that $\displaystyle \theta^{(1)}=\delta$
then
$\displaystyle u=c_1 + c_2\theta(x) + c_3\delta(x)$
this is the general solution.
The particular solution is $\displaystyle - 1/2 fp(\frac 1 {x^2})$
The full solution is the sum of both....

i hope it's clear.