Recent content by bdforbes

My current thinking on this is that we could start with f=\frac{1}{2}xsgn(x) + x + 1 but as soon as we take the derivatives, we lose the last two terms, and the Fourier transform is then effectively operating on the C=D=0 solution. So it's not really surprising that we only get one solution...

If I solve a simple 2nd order ODE using a Fourier transform, I only get one solution. E.g.: \frac{d^2f}{dx^2}=\delta (2\pi ik)^2\tilde{f}=1 \tilde{f}=\frac{1}{(2\pi ik)^2} f = \frac{1}{2}xsgn(x) However, the general solution is f = \frac{1}{2}xsgn(x) + Cx + D Why do I...

Is it possible that this Fourier transform is defined purely through the Helmholtz equation? I.e. no definition of a Fourier transform can be directly applied to the function itself?

Thanks for your reply JasonRF. I am actually specifically trying to forward transform the Green's function, as opposed to inverting the transform, which I am relatively comfortable with. Does your method still apply? I will try to get a hold of that text.

By taking the Fourier transform of the fundamental Helmholtz equation (\nabla^2+k^2)G(\vec{x})=-\delta(\vec{x}), one finds that G(\vec{x})=\frac{e^{ikr}}{r} and \tilde{G}(\vec{\xi})=\frac{1}{k^2-\xi^2}. However, I can't figure out how to directly confirm that this Fourier...

I have come to the conclusion that the Cauchy Principal Value is the appropriate type of improper integral to use. This is because Fourier transforms originate in the consideration of periodic functions; we should consider some approximation to the sinc function which is periodic, i.e. cut the...

Thanks Ray, I'll be interested to see the results. I'm quite busy at the moment, so I'm happy to wait until you've had the chance to write it up.

I had considered that earlier. We arrive at the result which we know is correct for all values of x, yet the derivation only applies to certain values of x. It would be great if we could make an argument at this stage to extend to all values of x.

Does this mean this method is only applicable to expressions of the form (1+\alpha x)^{-\gamma} with \gamma>0? So I would effectively end up with this? M\{H(1-x\alpha)(1-\alpha x)^{-\gamma}\}(p)=\int\limits_0^{1/\alpha}(1-\alpha x)^{-\gamma}x^{p-1}dx How would I proceed from here?

I can easily find the Fourier transform of rect(x) to be 2sinc(2\pi k) using particular conventions (irrelevant here). But when I attempt to inverse Fourier transform the sinc function, I find I have to resort to contour integration and Cauchy principal values. This is troubling to me. It...

I know how to derive the binomial approximation for (1+\alpha x)^{\gamma} using a Mellin transform, but for (1-\alpha x)^{\gamma} the method appears to fail because I can't take x to infinity. Here is the basics of the method. Take the Mellin transform of (1+\alpha x)^{\gamma}: M(p) =...

I've noticed another big problem with the Fourier transform approach. The Fourier transform of G(r)=\frac{e^{ikr}}{r} does not actually exist, it diverges. I assume there is a sense in which we are able to work with the object \tilde{G}(\vec{\xi}) and still achieve reasonable results...

When I substitute G(r)=e^{ikr}/r into the first integral, I get \frac{2}{\xi\epsilon^3}\left(ik\epsilon e^{ik\epsilon}-e^{ik\epsilon}\right)sin(2\pi\xi\epsilon) This diverges as \epsilon\to 0, so I must have made an error early on. Should I not have ignored the surface at infinity when I...

Upon further expansion we get: \begin{align*} 0&=\left[k^2-4\pi^2\xi^2\right]\tilde{G}(\vec{\xi})+\lim_{\epsilon\to 0}\left\{\int_{\partial B^c_\epsilon}e^{-2\pi i\vec{\xi}\cdot\vec{r}}\nabla G(\vec{r})\cdot d\vec{A}+2\pi i\vec{\xi}\cdot\int_{\partial B^c_\epsilon}G(\vec{r})e^{-2\pi...

If we set that whole mess to some constant A independent of \xi, then proceed with the contour integration, we will arrive at the correct result. We can then substitute the Green's function back in and confirm that the limit does exist. Is it a problem to assume there is no \xi dependence?