# Diracs delta equation - general interpretation

1. Apr 30, 2012

### finitefemmet

Im really just searching for a general explanation!

If you are solving a pretty standard left hand side differential equation, but a diracs delta function on the right hand side. I am abit confused about how to interpret this.

If this is the case for the right hand side:

r(t) = Diracs (t) ,for 0≤ t<T with the period T=2∏

Think of this as an periodic outside force on a spring system, now I dont know how to interpret this. Does this mean that r(t) repeats itself, at t=0, t=2∏ and so on. Or that the diracs delta equation only excists between 0 and 2∏?

Since its a diracs delta equation, it cannot work over a longer time interval? Since its an instant impuls over an extremely small time space.

If anyone could shed some light over this, I would be most gratefull.
I am not looking for a solution, just general information on how to interpret this information with the diracs delta function

Thank you, and excuse my poor english!

Last edited: Apr 30, 2012
2. Apr 30, 2012

### Office_Shredder

Staff Emeritus
They mean that the function repeats itself

3. Apr 30, 2012

### HallsofIvy

Staff Emeritus
For every positive integer k, let fk(t)=
0 for 2(n-1)pi+ 1/k to 2npi- 1/k,
k/2 for 2npi- 1/k to 2npi+ 1/k

for n any positive integer. The periodic "delta function" is the limit of fk(x) as k goes to infinity. Essentially, that gives a "delta function" at every multiple of 2pi.

4. May 1, 2012

### finitefemmet

Thank you both;)