1. complex PDE

hi,
i just started learning about PDE in complex variables and im very confused by it.

why isit that for f(x) = integrate from 2pi to 0 ( H(x-t)) dt

that if t<x, H(x-t) = 1 and for t>x, H(x-t) = 0?

what does that mean?

also, is there any books on notes that anyone can recommend which explains this concept becos i cant seem to find any materials on this.

thanks

2. Your H function behaves exactly like the Heaviside Step Function.

in my notes, it was stated that

f(x) = [ integrate from x to 0 {-i exp(ia(x-t))g(t) dt }] = [ integrate from 2pi to 0 { -i exp(ia(x-t) H(x-t) g(t) dt}]

may i know why H(x-t) is brought into this equation?

4. Hmm. Gotta see real integrals here. So you're asking why it is that

$\displaystyle{f(x)=-i\int_{x}^{0}e^{ia(x-t)}g(t)\,dt=-i\int_{2\pi}^{0}e^{ia(x-t)}H(x-t)\,g(t)\,dt.}$ Is that right? If it is, could you please provide some more context? Statement of the original problem, perhaps? Thanks!

5. yes that is what im saying=)

the full thing would be

let D = id/dx + a.

let h(x) = exp(iax). solve for df=g where f(2pi) =f(0).

below is a picture of the steps..which i do not understand.

thanks!

6. Ok, the real integral being claimed is

$\displaystyle{\int_{0}^{x}e^{ia(x-t)}g(t)\,dt=\int_{0}^{2\pi}e^{ia(x-t)}H(x-t)g(t)\,dt.}$

The reason you can do this is that the Heaviside step function crunches everything down to zero as soon as $t$ reaches $x$. Therefore, integrating beyond $x$ shouldn't pick up any more area. Think of it this way:

$\displaystyle{\int_{0}^{2\pi}e^{ia(x-t)}H(x-t)g(t)\,dt=\int_{0}^{x}e^{ia(x-t)}H(x-t)g(t)\,dt+\int_{x}^{2\pi}e^{ia(x-t)}H(x-t)g(t)\,dt}$

$\displaystyle{=\int_{0}^{x}e^{ia(x-t)}1\cdot g(t)\,dt+\int_{x}^{2\pi}e^{ia(x-t)}0\cdot g(t)\,dt=\int_{0}^{x}e^{ia(x-t)}g(t)\,dt.}$

Make sense now?

7. does that mean that i can always change the limits of my integration and assume that when x>t when integrating from x to 0 dt that it will be 0 using H(x-t)?

or is there some special cases that this works?

thank you!

8. As far as I know, you can always increase the upper limit of an integral if you ensure that beyond the original upper limit, the integrand is zero. That's essentially what's happening here.

9. sorry i think im having a mental block. how do you know that in this integration, that if t>x, the answer of this integration will be 0 and hence H(x-t) can be used? i thought it was based on assumption that it will be 0

10. how do you know that in this integration, that if t>x, the answer of this integration will be 0
Because the upper limit of the original integral is $x$. If the variable of integration is $t$, and the lower limit is $0$ and the upper limit is $x$, then it follows that for that integral, $0\le t\le x.$

I wouldn't say that "the answer of this integration will be 0". Because I'm not saying that, in your case, the integral from 0 to x of your integrand is zero. I'm saying that if you extend the upper limit beyond the original upper limit, you'd better have an integrand of zero, or you'll change the value of the integral.