I am considering the following forcing function (part of a differential equation) $$s(x)=\delta(x)-\delta(x-\frac{1}{2})$$ I was just wondering if someone could provide me with some intuition of what this 'function' looks like. I am thinking that it is zero everywhere except at $x=0,x=\frac{1}{2}$, where it is infinite.
I need to expand this as a Fourier series in order to solve the differential equation, and I think this intuition may help me.
The differential equation is: $$y''(x)+\omega^2=s(x)$$ If you're interested.
Any help is appreciated, Thanks :)
Ok we have: $y''(x)+\omega^2=\delta(x)-\delta(x- 1/2).$ The integral of $\delta(x)$ is $H(x)$ which is the heavyside function: $H(x)=1, \forall x\ge 0$ and $H(x)=0 \; \forall x<0$. Now we integrate twice and we get: $$y(x)=-\omega^2 \frac{x^2}{2}+c_2+c_1x+xH(x)-\left(x-\frac{1}{2}\right)H(x-\frac{1}{2}).$$ The integrate of $H(x)$ should be easy to compute, just draw $H(x)$ if you are not getting it.