4

(This question inspired by question A specific 1st order PDE which looks almost like a linear PDE.)

Solve integral equation $$ g(x)=\int\limits_{-\infty}^\infty \rho(\omega)\left[e^{i\omega x} - \frac{1}{1 + k i\omega e^{-i\omega}}\right]d\omega, $$ where $g(x)\colon\mathbb R\to\mathbb R$ is given and $k$ is a real constant; $\rho$ is unknown function.

I tried expand $1/(1 + k\ldots)$ into series or differentiate w.r.t. $k$, but it's useless. Since first part (with $e^{i\omega x}$) is a inverse Fourier transform, I tried to apply Fourier transform, but what we should do with integral $$ \int\limits_{-\infty}^\infty e^{-i\eta x}d\eta\int\limits_{-\infty}^\infty \frac{\rho(\omega)d\omega}{1 + ki\omega e^{-i\omega}}? $$

Part with $k$ is just a number, but we can't denote $g(x)=\mathcal F^{-1}[\rho] + \mathrm{const}$. I have tried, and it is not a true constant anyway (depends on $\rho$).

Community
  • 1
Michael Galuza
  • 4,730

3 Answers3

3

This is an attempt to answer to the tricky question of the determining of the constant :

enter image description here

This draw me to think that :

  • If $\int_{-\infty}^{\infty}\frac{\varphi (\omega)}{1+k i\omega e^{-i\omega}}d\omega =0 $, then the problem has a solution, which is the inverse Fourier transform of $g(x)$.

  • If not, there is no function $\rho(\omega)$ satisfying the given equation.

Since I am not quite sure of the validity of the argument, I think that the question remains open.

JJacquelin
  • 66,221
  • 3
  • 37
  • 87
  • $$\int\limits_{-\infty}^\infty \left(1-\frac{1}{1+ki\omega e^{-i\omega}}\right)\delta(\omega),d\omega = 0,$$ isn't it? – Michael Galuza Jul 27 '15 at 02:12
  • @ Michael Galuza : Ho yes ! I missed it. I will change the end of my first answer. So, the question remains open. – JJacquelin Jul 27 '15 at 06:14
2

This is not a solution.

It would have been desirable to have some understanding of the nature of the operation of multiplying by $\frac{1}{1 + k i\omega e^{-i\omega}}$ in frequency domain, which I don't. I tried to inverse transform it, it is probably going to be a distribution/generalized function and I am unable to find what it would be.

Numerically, it looks something like this

enter image description here

An example where I do (Using non unitary angular frequency transform and using $g(w)\rightarrow G(x),\rho (w) \rightarrow R(x)$):

$$G(x)=\int\limits_{-\infty}^\infty \rho(w)\left[e^{i w x}+ \frac{1}{k + i w}\right]dw=2\pi R(x)+\int ^0 _{-\infty}R(x)e^{kx}dx$$

For the second term, I used the convolution theorem and the fact that the integral in one domain is equivalent to its transform evaluated at 0.

Lets pick $k=1$ and $G(x)=e^{-x^2/2}$

$$G(x)=2\pi R(x)+\int ^0 _{-\infty}R(x)e^{x}dx$$

G and R will differ by a simple constant. Say $$R(x)=\frac{e^{-x^2/2}-A}{2 \pi}$$

Now it is a trivial matter to find A, after that you would inverse transform $R$ to get $\rho$

$$e^{-x^2/2}=\int_{-\infty }^{\infty } \left(e^{i w x}+\frac{1}{1+i w}\right) \left(\frac{e^{-\frac{w^2}{2}}}{\sqrt{2 \pi }}-\sqrt{\frac{e \pi }{2}} \text{erfc}\left(\frac{1}{\sqrt{2}}\right) \delta (w)\right) \, dw$$

Gappy Hilmore
  • 3,937
1

Okay, I mentioned that I was unable to inverse transform by hand (and I doubt that it is possible at all) however this process can be done numerically.

Using the same convention:

using non-unitary angular frequency transform and using $g(w)\rightarrow G(x),\rho (w) \rightarrow R(x)$

$$h(w)=-\frac{1}{1 + ik w e^{-iw}}$$

$H(x)$ is the (numerical) inverse frequency transform of this function. This is a known function. For $k=1/2$ it looks like:

enter image description here

Using convolution theorem and the fact that the integral in one domain is equivalent to its transform evaluated at 0.

$$G(x)=2\pi R(x)+\int^\infty _{-\infty}R(\xi)H(x-\xi)d\xi \bigg|_{x=0}$$

$$R(x)=\frac{G(x)-A}{2\pi}$$

$$A=\frac{1}{2\pi}\int^\infty _{-\infty}G(\xi)H(-\xi)d\xi -\frac{A}{2\pi}\int^\infty _{-\infty}H(-\xi)d\xi$$

All the integrals are known numbers, that can easily be found numerically. From those, you can solve for A. And then, you can easily find $\rho$

You can continue solving using these if you'd like

n1=18;
start1=2^7;
x=linspace(-start1,start1,2^n1+1);
x=x(1:2^n1);
G=sech(x);

w=AngularFFTIndex(x);
k=1/2;
h=-1./(1+ 1i* k *w.*exp(- 1i *w));
[~,H ]=EasyIFFT(w,h);

-

function [ x,f ] = EasyIFFT(  k, F )

x=AngularFFTIndex(k);
f=ifftshift(ifft(F));

Enf=sum(abs(f).^2)*(x(2)-x(1));
EnF=sum(abs(F).^2)/2/pi*(k(2)-k(1));

f=f*sqrt(EnF/Enf).*exp(-1i*k*x(1));


end

-

function [ k,F ] = EasyFFT(  x, f )%just make sure number of points is a power of 2

k=AngularFFTIndex(x);
F=fftshift(fft(f));

Enf=sum(abs(f).^2)*(x(2)-x(1));
EnF=sum(abs(F).^2)/2/pi*(k(2)-k(1));

F=F*sqrt(Enf/EnF).*exp(1i*k*x(1));


end
Gappy Hilmore
  • 3,937